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https://www.enotes.com/homework-help/find-value-x-inequality-hold-log-3-x-2-1-lt-log-3-195799 | 1,519,262,007,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813832.23/warc/CC-MAIN-20180222002257-20180222022257-00264.warc.gz | 842,923,149 | 10,983 | # Find the value of x for the inequality to hold. Log 3 (x^2 + 1) =< log 3 (2x + 5)
hala718 | Certified Educator
log 3 (x^2 + 1) =< log 3 (2x + 5)
since log is an increasing function , then:
if log a < log b , then a < b
==> x^2 + 1 = < 2x + 5
Move all terms to the left side:
==> x^2 - 2x - 4 = < 0
x1 = [2 + sqrt(4+16)/2 = 1+ sqrt5
x2= 1-sqrt5 (impossible)
==> Then x belongs to [1-sqrt5, 1+sqrt5]
giorgiana1976 | Student
We'll start by imposing the constraints of existance of logarithm function.
x^2 + 1>0, which is true for any value of x
and
2x + 5 > 0
2x>-5
We'll divide by 2:
x>-5/2
So, for the logarithms to exist, the values of x have to belong to the interval (-5/2, +inf.)
Now, we'll solve the inequality. For the beginning, we notice that the bases of logarithms are matching and they are >1, so the direction of the inequality remains unchanged, if we'll apply the one to one property of logarithms:
x^2 + 1 =<2x + 5
We'll move all terms to one side:
x^2 - 2x + 1 - 5 =< 0
x^2 - 2x - 4 =< 0
To solve the inequality above, first we have to calculate the roots of the equation x^2 - 2x - 4 = 0.
After that, we'll write the expression in a factored form as:
1*(x-x1)(x-x2) =< 0
So, let's apply the quadratic formula to calculate the roots:
x1 = [2+sqrt(4+16)]/2
x1 = (2+2sqrt5)/2
x1 = 2(1+sqrt5)/2
x1 = 1+sqrt5
x2 = 1-sqrt5
The inequality will be written as:
(x - 1 - sqrt5)(x - 1 + sqrt5 ) =< 0
Now, we'll discuss the inequality:
- the product is negative if one factor is positive and the other is negative:
x - 1 - sqrt5 >= 0
We'll add 1 + sqrt5 both sides:
x > = 1 + sqrt5
and
x - 1 + sqrt5 =< 0
x =< 1 - sqrt5
The common solution is the empty set.
Now, we'll consider the other alternative:
x - 1 - sqrt5 =< 0
x =< 1 + sqrt5
and
x - 1 + sqrt5 >= 0
x >= 1 - sqrt5
So, x belongs to the interval [1 - sqrt5 , 1 + sqrt5].
Finally, the solution of the inequality is the inetrval identified above: [1 - sqrt5 , 1 + sqrt5].
neela | Student
log3 (x^2+1) = log3 (2x+5)
To find x for which the above holds.
Solution:
log3 (x^2+1) =< log3 (2x+5).
Since the bases are same , the inequality holds for antilog. Therefore,
x^2+1 = < 2x+1 ,
x^2 < 2x. Or
x^2-2x = < 0.
x(x-2) = < 0 .
The above product is negative iff 0 < x < 2.
So x should belong to the open interval ( 0 , 2) in order that the given inequality holds good. | 899 | 2,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2018-09 | latest | en | 0.779464 |
https://cfrankdavis.wordpress.com/2011/01/27/milankovitch/ | 1,603,309,841,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107877420.17/warc/CC-MAIN-20201021180646-20201021210646-00602.warc.gz | 225,894,007 | 33,633 | ## Milankovitch
After watching Horizon last night, and writing about it, I carried on thinking about climate today, and how we’re currently living in a 10,000 year long warm interglacial period in the Earth’s history, after having endured 100,000 years of a period of glaciation when temperatures were perhaps 8 oC colder than now, and glaciers covered much of Europe and North America. And in particular I got thinking about Milankovitch Cycles.
Milutin Milankovitch was a Serbian mathematician who, back in the 1930s, developed the idea that the Earth’s long term climate was governed by slight variations in the motion of the Earth.
There were three variations that he considered. The first was the precession of the spinning Earth. Precession is the slow “wobble” of the Earth’s axis, which is the sort of thing that can be seen in a spinning top. One wobble takes about 26,000 years. And the result is that, while the Earth’s axis is pointing at the North Star, Polaris, right now, over the next few thousand years it will describe a circle in the star field, before coming back to Polaris in about 26,000 years time.
The second variation was the obliquity or tilt of the Earth’s axis. This is the angle at which the Earth precesses or wobbles. This also changes by a few degrees in a 41,000 year cycle.
And finally there is the eccentricity of the Earth’s orbit around the Sun. Planetary orbits are elliptical, and the eccentricity of an orbit is the degree to which it deviates from a perfect circle. The Earth’s orbit is very nearly circular, but it becomes a bit more elliptical before returning to nearly circular again in a 100,000 year cycle.
Milankovitch calculated all these slight changes, and also the resulting changes in the amount of sunlight warming the Earth at different latitudes due to the changing precession and obliquity and eccentricity of the Earth. And he argued that when the warming by sunlight at 65 degrees N was at its greatest, glacial ice covering the northern hemisphere would be at its minimum, and that when the warming from sunlight was at a minimum, glacial ice would be at its maximum. Or that was the rough idea.
But the changes in the amount of sunlight are really quite small, because the variations in the Earth’s precession and obliquity and eccentricity are rather small. It didn’t seem to be enough to cause ice ages.
Neither did it seem to fit very well with ice core data (for example from the Antarctic Vostok ice core) which was subsequently used to estimate ice volume and temperature and CO2 concentrations over the past few hundred thousand years.
At right, the graph shows the variation in the sunlight warming at 65 N over the past 400,000 years as calculated by Milankovitch, ranging from around 460 W/m2 to 540 W/m2 per day – an anomaly of +/- 40 Watts. Below it are the temperatures (in degrees C) over the same period derived from the Vostok ice cores, with the peak temperatures corresponding to ice minima like in our current interglacial period.
And it’s a bit hard to see much correlation of one with the other.
The result is that, while Milankovitch’s theory is intellectually very attractive, its popularity has waxed and waned. Most often, while people concede that the cycle might have some small effect, it had to be amplified in some way to explain the cycles of glaciation over the past few million years. Global warmists have argued, for example, that the Milankovitch cycles might cause a little bit of warming, which resulted in CO2 being released from the oceans to cause more warming, ultimately flipping the Earth’s climate from a deep glaciation to a warm interglacial.
But today I learned that a Washington University researcher, Gerard Roe, suggested a few years ago that people were looking at Milankovitch’s theory all wrong. They were supposing that when the Milankovitch cycle was at its peak, this would correspond to an ice minimum, and when it was at its lowest, that would correspond to an ice maximum. What they should have been looking at was the change in the ice volume. For at the peak of the Milankovitch cycle, it wasn’t that the climate was warm, but that it was getting warmer, and the ice was melting. As Roe put it:
While most studies have focused on the connection between insolation and ice volume (V), there is a more direct physical connection between insolation and the rate of change of ice volume (dV/dt)
So Roe went back to the ice core data, and worked out the rate at which the glaciers were expanding and contracting. And then he plotted the Milankovitch cycles over (on top of) the rate of change of volume of the glaciers.
It’s an almost perfect match. And, according to Roe, there’s no need to invoke any CO2-driven warming. In fact, he doubts that CO2 has much effect at all.
And it figures. The changes in the amounts of sunlight might be small, and wouldn’t melt very much ice. But just a single slow drip of water of melt water would have added up, over tens of thousands of years, to an entire kilometre-thick ice sheet over half of Europe.
Roe’s paper only came out a bit over 3 years ago, and physicist Lubos Motl, for one, only found out about it about 6 months ago. So it’s a bit of news that’s still slowly filtering out into the world. It’s obviously not going to be reported by the global-warming-obsessed mass media. So the only way you’ll ever hear about it is through the internet.
And it also illustrates beautifully how science is a process of finding things out, and how “the debate is never over”, and how there will always be people like Gerard Roe who come along and point out something obvious, which everybody else had overlooked (including Milankovitch himself, who seems to have thought that the 40,000 year obliquity cycle would predominate).
smoker
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### 12 Responses to Milankovitch
1. Anonymous says:
Milankovitch
Frank
I remember being told of this ‘phenomena’ at an astronomy evening class almost 40 years ago. Our tutor also added that the gas giants Jupiter/Saturn and to some extent although minor Mars/Venus,will also ‘tug’ at the Earth’s obliquity. This ‘tug’ would be most notable as the planets arrange themselves into and out of alignment. The effects upon the amount of sunlight and therefore temperature would be ‘felt’ for a number years before settling back into a ‘normal’ rhythm. I have tried to explain this to friends over many years now to know avail. It seems that unless an ‘expert’ on the BBC tells them the TRUTH, everything else is just lies.
NewsboyCap
2. Anonymous says:
Milankovitch
Frank
I remember being told of this ‘phenomena’ at an astronomy evening class almost 40 years ago. Our tutor also added that the gas giants Jupiter/Saturn and to some extent although minor Mars/Venus,will also ‘tug’ at the Earth’s obliquity. This ‘tug’ would be most notable as the planets arrange themselves into and out of alignment. The effects upon the amount of sunlight and therefore temperature would be ‘felt’ for a number years before settling back into a ‘normal’ rhythm. I have tried to explain this to friends over many years now to know avail. It seems that unless an ‘expert’ on the BBC tells them the TRUTH, everything else is just lies.
NewsboyCap
• Frank Davis says:
Re: Milankovitch
Cause of precession:
The cause of the precession is the equatorial bulge of the Earth, caused by the centrifugal force of the Earth’s rotation (the centrifugal force is discussed in a later section). That rotation changes the Earth from a perfect sphere to a slightly flattened one, thicker across the equator. The attraction of the Moon and Sun on the bulge is then the “nudge” which makes the Earth precess.
Cause of varying obliquity
Various gravitational torques of the Sun, Moon, and other planets cause the Earth’s obliquity to oscillate with a period of approximately 41,000 years.
The eccentricity of the Earth’s orbit changes due to the pull of the other planets in the solar system.
Frank
• Anonymous says:
Re: Milankovitch
Frank
Thank you much for the links, much appreciated.
NewsboyCap
• Anonymous says:
Re: Milankovitch
Frank
Thank you much for the links, much appreciated.
NewsboyCap
• Frank Davis says:
Re: Milankovitch
Cause of precession:
The cause of the precession is the equatorial bulge of the Earth, caused by the centrifugal force of the Earth’s rotation (the centrifugal force is discussed in a later section). That rotation changes the Earth from a perfect sphere to a slightly flattened one, thicker across the equator. The attraction of the Moon and Sun on the bulge is then the “nudge” which makes the Earth precess.
Cause of varying obliquity
Various gravitational torques of the Sun, Moon, and other planets cause the Earth’s obliquity to oscillate with a period of approximately 41,000 years.
The eccentricity of the Earth’s orbit changes due to the pull of the other planets in the solar system.
Frank
3. Anonymous says:
Milankovitch
Frank
I remember being told of this ‘phenomena’ at an astronomy evening class almost 40 years ago. Our tutor also added that the gas giants Jupiter/Saturn and to some extent although minor Mars/Venus,will also ‘tug’ at the Earth’s obliquity. This ‘tug’ would be most notable as the planets arrange themselves into and out of alignment. The effects upon the amount of sunlight and therefore temperature would be ‘felt’ for a number years before settling back into a ‘normal’ rhythm. I have tried to explain this to friends over many years now to know avail. It seems that unless an ‘expert’ on the BBC tells them the TRUTH, everything else is just lies.
NewsboyCap
4. Anonymous says:
Re: Milankovitch
I don’t read everything so, I probably missed it; but, has anyone ever stated just what the optimal temperature is for the Earth?
Has anyone ever stated what the optimal CO2 level is for the atmosphere?
If there are no goals, destroying our civilization seems rather stupid!!!
Gary K.
5. Anonymous says:
Re: Milankovitch
I don’t read everything so, I probably missed it; but, has anyone ever stated just what the optimal temperature is for the Earth?
Has anyone ever stated what the optimal CO2 level is for the atmosphere?
If there are no goals, destroying our civilization seems rather stupid!!!
Gary K.
6. Anonymous says:
Re: Milankovitch
I don’t read everything so, I probably missed it; but, has anyone ever stated just what the optimal temperature is for the Earth?
Has anyone ever stated what the optimal CO2 level is for the atmosphere?
If there are no goals, destroying our civilization seems rather stupid!!!
Gary K.
7. Frank Davis says:
Re: Milankovitch
Cause of precession:
The cause of the precession is the equatorial bulge of the Earth, caused by the centrifugal force of the Earth’s rotation (the centrifugal force is discussed in a later section). That rotation changes the Earth from a perfect sphere to a slightly flattened one, thicker across the equator. The attraction of the Moon and Sun on the bulge is then the “nudge” which makes the Earth precess.
Cause of varying obliquity
Various gravitational torques of the Sun, Moon, and other planets cause the Earth’s obliquity to oscillate with a period of approximately 41,000 years.
The eccentricity of the Earth’s orbit changes due to the pull of the other planets in the solar system.
Frank
8. Anonymous says:
Re: Milankovitch
Frank
Thank you much for the links, much appreciated.
NewsboyCap
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2,618 | 11,587 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-45 | latest | en | 0.958247 |
https://it.scribd.com/document/236286458/Engineering-Economy-Formulas | 1,603,962,770,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107903419.77/warc/CC-MAIN-20201029065424-20201029095424-00013.warc.gz | 367,457,969 | 96,237 | Sei sulla pagina 1di 7
# ENGINEERING ECONOMICS
Factor Name
Single Payment
Compound Amount
Converts
Symbol
Formula
to F given P
(F/P, i%, n)
(1 + i)n
Single Payment
Present Worth
to P given F
(P/F, i%, n)
(1 + i) n
Uniform Series
Sinking Fund
to A given F
(A/F, i%, n)
Capital Recovery
to A given P
(A/P, i%, n)
Uniform Series
Compound Amount
to F given A
(F/A, i%, n)
Uniform Series
Present Worth
to P given A
(P/A, i%, n)
Present Worth
to P given G
(P/G, i%, n)
Future Worth
to F given G
(F/G, i%, n)
Uniform Series
to A given G
(A/G, i%, n)
## NOMENCLATURE AND DEFINITIONS
A .......... Uniform amount per interest period
B .......... Benefit
BV ....... Book Value
C.......... Cost
d .......... Combined interest rate per interest period
Dj......... Depreciation in year j
F .......... Future worth, value, or amount
f ........... General inflation rate per interest period
G ......... Uniform gradient amount per interest period
i ........... Interest rate per interest period
ie .......... Annual effective interest rate
m ......... Number of compounding periods per year
n .......... Number of compounding periods; or the expected
life of an asset
P .......... Present worth, value, or amount
r ........... Nominal annual interest rate
Sn ......... Expected salvage value in year n
(1 + i )n 1
i (1 + i )n
(1 + i )n 1
(1 + i )n 1
i
(1 + i )n 1
n
i (1 + i )
(1 + i )n 1 n
n
n
i 2 (1 + i )
i (1 + i )
(1 + i )n 1 n
i2
1
n
i (1 + i )n 1
NON-ANNUAL COMPOUNDING
m
ie = 1 + 1
m
## Discount Factors for Continuous Compounding
(n is the number of years)
(F/P, r%, n) = er n
(P/F, r%, n) = er n
Subscripts
j ........... at time j
n .......... at time n
(A/F, r%, n) =
er 1
er n 1
(F/A, r%, n) =
er n 1
er 1
(A/P, r%, n) =
er 1
1 er n
(P/A, r%, n) =
1 er n
er 1
BOOK VALUE
BV = initial cost Dj
## ** ........ P/G = (F/G)/(F/P) = (P/A) (A/G)
.......... F/G = (F/A n)/i = (F/A) (A/G)
.......... A/G = [1 n(A/F)]/i
79
## ENGINEERING ECONOMICS (continued)
DEPRECIATION
Straight Line
Dj =
BREAK-EVEN ANALYSIS
By altering the value of any one of the variables in a
situation, holding all of the other values constant, it is
possible to find a value for that variable that makes the two
alternatives equally economical. This value is the breakeven point.
C Sn
n
Dj = (factor) C
## Break-even analysis is used to describe the percentage of
capacity of operation for a manufacturing plant at which
income will just cover expenses.
## A table of modified factors is provided below.
CAPITALIZED COSTS
Capitalized costs are present worth values using an
assumed perpetual period of time.
Capitalized Costs = P =
## The payback period is the period of time required for the
profit or other benefits of an investment to equal the cost of
the investment.
A
i
INFLATION
To account for inflation, the dollars are deflated by the
general inflation rate per interest period f, and then they are
shifted over the time scale using the interest rate per
interest period i. Use a combined interest rate per interest
period d for computing present worth values P and Net P.
The formula for d is
BONDS
Bond Value equals the present worth of the payments the
purchaser (or holder of the bond) receives during the life of
the bond at some interest rate i.
Bond Yield equals the computed interest rate of the bond
value when compared with the bond cost.
d = i + f + (i f)
RATE-OF-RETURN
The minimum acceptable rate-of-return is that interest rate
that one is willing to accept, or the rate one desires to earn
on investments. The rate-of-return on an investment is the
interest rate that makes the benefits and costs equal.
BENEFIT-COST ANALYSIS
In a benefit-cost analysis, the benefits B of a project should
exceed the estimated costs C.
B C 0, or
B/C 1
## MODIFIED ACRS FACTORS
Recovery Period (Years)
3
Year
10
33.3
20.0
14.3
10.0
44.5
32.0
24.5
18.0
14.8
19.2
17.5
14.4
7.4
11.5
12.5
11.5
11.5
8.9
9.2
5.8
8.9
7.4
8.9
6.6
4.5
6.6
6.5
10
6.5
11
3.3
80
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
P/F
0.9950
0.9901
0.9851
0.9802
0.9754
0.9705
0.9657
0.9609
0.9561
0.9513
0.9466
0.9419
0.9372
0.9326
0.9279
0.9233
0.9187
0.9141
0.9096
0.9051
0.9006
0.8961
0.8916
0.8872
0.8828
0.8610
0.8191
0.7793
0.7414
0.6073
P/F
0.9901
0.9803
0.9706
0.9610
0.9515
0.9420
0.9327
0.9235
0.9143
0.9053
0.8963
0.8874
0.8787
0.8700
0.8613
0.8528
0.8444
0.8360
0.8277
0.8195
0.8114
0.8034
0.7954
0.7876
0.7798
0.7419
0.6717
0.6080
0.5504
0.3697
P/A
0.9950
1.9851
2.9702
3.9505
4.9259
5.8964
6.8621
7.8230
8.7791
9.7304
10.6770
11.6189
12.5562
13.4887
14.4166
15.3399
16.2586
17.1728
18.0824
18.9874
19.8880
20.7841
21.6757
22.5629
23.4456
27.7941
36.1722
44.1428
51.7256
78.5426
P/A
0.9901
1.9704
2.9410
3.9020
4.8534
5.7955
6.7282
7.6517
8.5650
9.4713
10.3676
11.2551
12.1337
13.0037
13.8651
14.7179
15.5623
16.3983
17.2260
18.0456
18.8570
19.6604
20.4558
21.2434
22.0232
25.8077
32.8347
39.1961
44.9550
63.0289
P/G
F/P
F/A
0.0000
0.9901
2.9604
5.9011
9.8026
14.6552
20.4493
27.1755
34.8244
43.3865
52.8526
63.2136
74.4602
86.5835
99.5743
113.4238
128.1231
143.6634
160.0360
177.2322
195.2434
214.0611
233.6768
254.0820
275.2686
392.6324
681.3347
1,035.6966
1,448.6458
3,562.7934
1.0050
1.0100
1.0151
1.0202
1.0253
1.0304
1.0355
1.0407
1.0459
1.0511
1.0564
1.0617
1.0670
1.0723
1.0777
1.0831
1.0885
1.0939
1.0994
1.1049
1.1104
1.1160
1.1216
1.1272
1.1328
1.1614
1.2208
1.2832
1.3489
1.6467
1.0000
2.0050
3.0150
4.0301
5.0503
6.0755
7.1059
8.1414
9.1821
10.2280
11.2792
12.3356
13.3972
14.4642
15.5365
16.6142
17.6973
18.7858
19.8797
20.9791
22.0840
23.1944
24.3104
25.4320
26.5591
32.2800
44.1588
56.6452
69.7700
129.3337
P/G
F/P
F/A
0.0000
0.9803
2.9215
5.8044
9.6103
14.3205
19.9168
26.3812
33.6959
41.8435
50.8067
60.5687
71.1126
82.4221
94.4810
107.2734
120.7834
134.9957
149.8950
165.4664
181.6950
198.5663
216.0660
234.1800
252.8945
355.0021
596.8561
879.4176
1,192.8061
2,605.7758
1.0100
1.0201
1.0303
1.0406
1.0510
1.0615
1.0721
1.0829
1.0937
1.1046
1.1157
1.1268
1.1381
1.1495
1.1610
1.1726
1.1843
1.1961
1.2081
1.2202
1.2324
1.2447
1.2572
1.2697
1.2824
1.3478
1.4889
1.6446
1.8167
2.7048
81
1.0000
2.0100
3.0301
4.0604
5.1010
6.1520
7.2135
8.2857
9.3685
10.4622
11.5668
12.6825
13.8093
14.9474
16.0969
17.2579
18.4304
19.6147
20.8109
22.0190
23.2392
24.4716
25.7163
26.9735
28.2432
34.7849
48.8864
64.4632
81.6697
170.4814
A/P
1.0050
0.5038
0.3367
0.2531
0.2030
0.1696
0.1457
0.1278
0.1139
0.1028
0.0937
0.0861
0.0796
0.0741
0.0694
0.0652
0.0615
0.0582
0.0553
0.0527
0.0503
0.0481
0.0461
0.0443
0.0427
0.0360
0.0276
0.0227
0.0193
0.0127
A/P
1.0100
0.5075
0.3400
0.2563
0.2060
0.1725
0.1486
0.1307
0.1167
0.1056
0.0965
0.0888
0.0824
0.0769
0.0721
0.0679
0.0643
0.0610
0.0581
0.0554
0.0530
0.0509
0.0489
0.0471
0.0454
0.0387
0.0305
0.0255
0.0222
0.0159
A/F
A/G
1.0000
0.4988
0.3317
0.2481
0.1980
0.1646
0.1407
0.1228
0.1089
0.0978
0.0887
0.0811
0.0746
0.0691
0.0644
0.0602
0.0565
0.0532
0.0503
0.0477
0.0453
0.0431
0.0411
0.0393
0.0377
0.0310
0.0226
0.0177
0.0143
0.0077
0.0000
0.4988
0.9967
1.4938
1.9900
2.4855
2.9801
3.4738
3.9668
4.4589
4.9501
5.4406
5.9302
6.4190
6.9069
7.3940
7.8803
8.3658
8.8504
9.3342
9.8172
10.2993
10.7806
11.2611
11.7407
14.1265
18.8359
23.4624
28.0064
45.3613
A/F
A/G
1.0000
0.4975
0.3300
0.2463
0.1960
0.1625
0.1386
0.1207
0.1067
0.0956
0.0865
0.0788
0.0724
0.0669
0.0621
0.0579
0.0543
0.0510
0.0481
0.0454
0.0430
0.0409
0.0389
0.0371
0.0354
0.0277
0.0205
0.0155
0.0122
0.0059
0.0000
0.4975
0.9934
1.4876
1.9801
2.4710
2.9602
3.4478
3.9337
4.4179
4.9005
5.3815
5.8607
6.3384
6.8143
7.2886
7.7613
8.2323
8.7017
9.1694
9.6354
10.0998
10.5626
11.0237
11.4831
13.7557
18.1776
22.4363
26.5333
41.3426
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
P/F
0.9852
0.9707
0.9563
0.9422
0.9283
0.9145
0.9010
0.8877
0.8746
0.8617
0.8489
0.8364
0.8240
0.8118
0.7999
0.7880
0.7764
0.7649
0.7536
0.7425
0.7315
0.7207
0.7100
0.6995
0.6892
0.6398
0.5513
0.4750
0.4093
0.2256
P/F
0.9804
0.9612
0.9423
0.9238
0.9057
0.8880
0.8706
0.8535
0.8368
0.8203
0.8043
0.7885
0.7730
0.7579
0.7430
0.7284
0.7142
0.7002
0.6864
0.6730
0.6598
0.6468
0.6342
0.6217
0.6095
0.5521
0.4529
0.3715
0.3048
0.1380
P/A
0.9852
1.9559
2.9122
3.8544
4.7826
5.6972
6.5982
7.4859
8.3605
9.2222
10.0711
10.9075
11.7315
12.5434
13.3432
14.1313
14.9076
15.6726
16.4262
17.1686
17.9001
18.6208
19.3309
20.0304
20.7196
24.0158
29.9158
34.9997
39.3803
51.6247
P/A
0.9804
1.9416
2.8839
3.8077
4.7135
5.6014
6.4720
7.3255
8.1622
8.9826
9.7868
10.5753
11.3484
12.1062
12.8493
13.5777
14.2919
14.9920
15.6785
16.3514
17.0112
17.6580
18.2922
18.9139
19.5235
22.3965
27.3555
31.4236
34.7609
43.0984
P/G
F/P
F/A
0.0000
0.9707
2.8833
5.7098
9.4229
13.9956
19.4018
26.6157
32.6125
40.3675
48.8568
58.0571
67.9454
78.4994
89.6974
101.5178
113.9400
126.9435
140.5084
154.6154
169.2453
184.3798
200.0006
216.0901
232.6310
321.5310
524.3568
749.9636
988.1674
1,937.4506
1.0150
1.0302
1.0457
1.0614
1.0773
1.0934
1.1098
1.1265
1.1434
1.1605
1.1779
1.1956
1.2136
1.2318
1.2502
1.2690
1.2880
1.3073
1.3270
1.3469
1.3671
1.3876
1.4084
1.4295
1.4509
1.5631
1.8140
2.1052
2.4432
4.4320
1.0000
2.0150
3.0452
4.0909
5.1523
6.2296
7.3230
8.4328
9.5593
10.7027
11.8633
13.0412
14.2368
15.4504
16.6821
17.9324
19.2014
20.4894
21.7967
23.1237
24.4705
25.8376
27.2251
28.6335
30.0630
37.5387
54.2679
73.6828
96.2147
228.8030
P/G
F/P
F/A
0.0000
0.9612
2.8458
5.6173
9.2403
13.6801
18.9035
24.8779
31.5720
38.9551
46.9977
55.6712
64.9475
74.7999
85.2021
96.1288
107.5554
119.4581
131.8139
144.6003
157.7959
171.3795
185.3309
199.6305
214.2592
291.7164
461.9931
642.3606
823.6975
1,464.7527
1.0200
1.0404
1.0612
1.0824
1.1041
1.1262
1.1487
1.1717
1.1951
1.2190
1.2434
1.2682
1.2936
1.3195
1.3459
1.3728
1.4002
1.4282
1.4568
1.4859
1.5157
1.5460
1.5769
1.6084
1.6406
1.8114
2.2080
2.6916
3.2810
7.2446
82
1.0000
2.0200
3.0604
4.1216
5.2040
6.3081
7.4343
8.5830
9.7546
10.9497
12.1687
13.4121
14.6803
15.9739
17.2934
18.6393
20.0121
21.4123
22.8406
24.2974
25.7833
27.2990
28.8450
30.4219
32.0303
40.5681
60.4020
84.5794
114.0515
312.2323
A/P
A/F
A/G
1.0150
0.5113
0.3434
0.2594
0.2091
0.1755
0.1516
0.1336
0.1196
0.1084
0.0993
0.0917
0.0852
0.0797
0.0749
0.0708
0.0671
0.0638
0.0609
0.0582
0.0559
0.0537
0.0517
0.0499
0.0483
0.0416
0.0334
0.0286
0.0254
0.0194
1.0000
0.4963
0.3284
0.2444
0.1941
0.1605
0.1366
0.1186
0.1046
0.0934
0.0843
0.0767
0.0702
0.0647
0.0599
0.0558
0.0521
0.0488
0.0459
0.0432
0.0409
0.0387
0.0367
0.0349
0.0333
0.0266
0.0184
0.0136
0.0104
0.0044
0.0000
0.4963
0.9901
1.4814
1.9702
2.4566
2.9405
3.4219
3.9008
4.3772
4.8512
5.3227
5.7917
6.2582
6.7223
7.1839
7.6431
8.0997
8.5539
9.0057
9.4550
9.9018
10.3462
10.7881
11.2276
13.3883
17.5277
21.4277
25.0930
37.5295
A/P
A/F
A/G
1.0200
0.5150
0.3468
0.2626
0.2122
0.1785
0.1545
0.1365
0.1225
0.1113
0.1022
0.0946
0.0881
0.0826
0.0778
0.0737
0.0700
0.0667
0.0638
0.0612
0.0588
0.0566
0.0547
0.0529
0.0512
0.0446
0.0366
0.0318
0.0288
0.0232
1.0000
0.4950
0.3268
0.2426
0.1922
0.1585
0.1345
0.1165
0.1025
0.0913
0.0822
0.0746
0.0681
0.0626
0.0578
0.0537
0.0500
0.0467
0.0438
0.0412
0.0388
0.0366
0.0347
0.0329
0.0312
0.0246
0.0166
0.0118
0.0088
0.0032
0.0000
0.4950
0.9868
1.4752
1.9604
2.4423
2.9208
3.3961
3.8681
4.3367
4.8021
5.2642
5.7231
6.1786
6.6309
7.0799
7.5256
7.9681
8.4073
8.8433
9.2760
9.7055
10.1317
10.5547
10.9745
13.0251
16.8885
20.4420
23.6961
33.9863
P/F
P/A
P/G
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
0.9615
0.9246
0.8890
0.8548
0.8219
0.7903
0.7599
0.7307
0.7026
0.6756
0.6496
0.6246
0.6006
0.5775
0.5553
0.5339
0.5134
0.4936
0.4746
0.4564
0.4388
0.4220
0.4057
0.3901
0.3751
0.3083
0.2083
0.1407
0.0951
0.0198
0.9615
1.8861
2.7751
3.6299
4.4518
5.2421
6.0021
6.7327
7.4353
8.1109
8.7605
9.3851
9.9856
10.5631
11.1184
11.6523
12.1657
12.6593
13.1339
13.5903
14.0292
14.4511
14.8568
15.2470
15.6221
17.2920
19.7928
21.4822
22.6235
24.5050
0.0000
0.9246
2.7025
5.2670
8.5547
12.5062
17.0657
22.1806
27.8013
33.8814
40.3772
47.2477
54.4546
61.9618
69.7355
77.7441
85.9581
94.3498
102.8933
111.5647
120.3414
129.2024
138.1284
147.1012
156.1040
201.0618
286.5303
361.1638
422.9966
563.1249
P/F
P/A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
0.9434
0.8900
0.8396
0.7921
0.7473
0.7050
0.6651
0.6274
0.5919
0.5584
0.5268
0.4970
0.4688
0.4423
0.4173
0.3936
0.3714
0.3505
0.3305
0.3118
0.2942
0.2775
0.2618
0.2470
0.2330
0.1741
0.0972
0.0543
0.0303
0.0029
0.9434
1.8334
2.6730
3.4651
4.2124
4.9173
5.5824
6.2098
6.8017
7.3601
7.8869
8.3838
8.8527
9.2950
9.7122
10.1059
10.4773
10.8276
11.1581
11.4699
11.7641
12.0416
12.3034
12.5504
12.7834
13.7648
15.0463
15.7619
16.1614
16.6175
F/P
F/A
1.0400
1.0816
1.1249
1.1699
1.2167
1.2653
1.3159
1.3686
1.4233
1.4802
1.5395
1.6010
1.6651
1.7317
1.8009
1.8730
1.9479
2.0258
2.1068
2.1911
2.2788
2.3699
2.4647
2.5633
2.6658
3.2434
4.8010
7.1067
10.5196
50.5049
1.0000
2.0400
3.1216
4.2465
5.4163
6.6330
7.8983
9.2142
10.5828
12.0061
13.4864
15.0258
16.6268
18.2919
20.0236
21.8245
23.6975
25.6454
27.6712
29.7781
31.9692
34.2480
36.6179
39.0826
41.6459
56.0849
95.0255
152.6671
237.9907
1,237.6237
P/G
F/P
F/A
0.0000
0.8900
2.5692
4.9455
7.9345
11.4594
15.4497
19.8416
24.5768
29.6023
34.8702
40.3369
45.9629
51.7128
57.5546
63.4592
69.4011
75.3569
81.3062
87.2304
93.1136
98.9412
104.7007
110.3812
115.9732
142.3588
185.9568
217.4574
239.0428
272.0471
1.0600
1.1236
1.1910
1.2625
1.3382
1.4185
1.5036
1.5938
1.6895
1.7908
1.8983
2.0122
2.1329
2.2609
2.3966
2.5404
2.6928
2.8543
3.0256
3.2071
3.3996
3.6035
3.8197
4.0489
4.2919
5.7435
10.2857
18.4202
32.9877
339.3021
83
1.0000
2.0600
3.1836
4.3746
5.6371
6.9753
8.3938
9.8975
11.4913
13.1808
14.9716
16.8699
18.8821
21.0151
23.2760
25.6725
28.2129
30.9057
33.7600
36.7856
39.9927
43.3923
46.9958
50.8156
54.8645
79.0582
154.7620
290.3359
533.1282
5,638.3681
A/P
A/F
A/G
1.0400
0.5302
0.3603
0.2755
0.2246
0.1908
0.1666
0.1485
0.1345
0.1233
0.1141
0.1066
0.1001
0.0947
0.0899
0.0858
0.0822
0.0790
0.0761
0.0736
0.0713
0.0692
0.0673
0.0656
0.0640
0.0578
0.0505
0.0466
0.0442
0.0408
1.0000
0.4902
0.3203
0.2355
0.1846
0.1508
0.1266
0.1085
0.0945
0.0833
0.0741
0.0666
0.0601
0.0547
0.0499
0.0458
0.0422
0.0390
0.0361
0.0336
0.0313
0.0292
0.0273
0.0256
0.0240
0.0178
0.0105
0.0066
0.0042
0.0008
0.0000
0.4902
0.9739
1.4510
1.9216
2.3857
2.8433
3.2944
3.7391
4.1773
4.6090
5.0343
5.4533
5.8659
6.2721
6.6720
7.0656
7.4530
7.8342
8.2091
8.5779
8.9407
9.2973
9.6479
9.9925
11.6274
14.4765
16.8122
18.6972
22.9800
A/P
A/F
A/G
1.0600
0.5454
0.3741
0.2886
0.2374
0.2034
0.1791
0.1610
0.1470
0.1359
0.1268
0.1193
0.1130
0.1076
0.1030
0.0990
0.0954
0.0924
0.0896
0.0872
0.0850
0.0830
0.0813
0.0797
0.0782
0.0726
0.0665
0.0634
0.0619
0.0602
1.0000
0.4854
0.3141
0.2286
0.1774
0.1434
0.1191
0.1010
0.0870
0.0759
0.0668
0.0593
0.0530
0.0476
0.0430
0.0390
0.0354
0.0324
0.0296
0.0272
0.0250
0.0230
0.0213
0.0197
0.0182
0.0126
0.0065
0.0034
0.0019
0.0002
0.0000
0.4854
0.9612
1.4272
1.8836
2.3304
2.7676
3.1952
3.6133
4.0220
4.4213
4.8113
5.1920
5.5635
5.9260
6.2794
6.6240
6.9597
7.2867
7.6051
7.9151
8.2166
8.5099
8.7951
9.0722
10.3422
12.3590
13.7964
14.7909
16.3711
P/F
P/A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
0.9259
0.8573
0.7938
0.7350
0.6806
0.6302
0.5835
0.5403
0.5002
0.4632
0.4289
0.3971
0.3677
0.3405
0.3152
0.2919
0.2703
0.2502
0.2317
0.2145
0.1987
0.1839
0.1703
0.1577
0.1460
0.0994
0.0460
0.0213
0.0099
0.0005
0.9259
1.7833
2.5771
3.3121
3.9927
4.6229
5.2064
5.7466
6.2469
6.7101
7.1390
7.5361
7.9038
8.2442
8.5595
8.8514
9.1216
9.3719
9.6036
9.8181
10.0168
10.2007
10.3711
10.5288
10.6748
11.2578
11.9246
12.2335
12.3766
12.4943
P/F
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
0.9091
0.8264
0.7513
0.6830
0.6209
0.5645
0.5132
0.4665
0.4241
0.3855
0.3505
0.3186
0.2897
0.2633
0.2394
0.2176
0.1978
0.1799
0.1635
0.1486
0.1351
0.1228
0.1117
0.1015
0.0923
0.0573
0.0221
0.0085
0.0033
0.0001
P/A
0.9091
1.7355
2.4869
3.1699
3.7908
4.3553
4.8684
5.3349
5.7590
6.1446
6.4951
6.8137
7.1034
7.3667
7.6061
7.8237
8.0216
8.2014
8.3649
8.5136
8.6487
8.7715
8.8832
8.9847
9.0770
9.4269
9.7791
9.9148
9.9672
9.9993
P/G
0.0000
0.8573
2.4450
4.6501
7.3724
10.5233
14.0242
17.8061
21.8081
25.9768
30.2657
34.6339
39.0463
43.4723
47.8857
52.2640
56.5883
60.8426
65.0134
69.0898
73.0629
76.9257
80.6726
84.2997
87.8041
103.4558
126.0422
139.5928
147.3000
155.6107
F/P
F/A
1.0800
1.1664
1.2597
1.3605
1.4693
1.5869
1.7138
1.8509
1.9990
2.1589
2.3316
2.5182
2.7196
2.9372
3.1722
3.4259
3.7000
3.9960
4.3157
4.6610
5.0338
5.4365
5.8715
6.3412
6.8485
10.0627
21.7245
46.9016
101.2571
2,199.7613
1.0000
2.0800
3.2464
4.5061
5.8666
7.3359
8.9228
10.6366
12.4876
14.4866
16.6455
18.9771
21.4953
24.2149
27.1521
30.3243
33.7502
37.4502
41.4463
45.7620
50.4229
55.4568
60.8933
66.7648
73.1059
113.2832
259.0565
573.7702
1,253.2133
27,484.5157
A/P
A/F
A/G
1.0800
0.5608
0.3880
0.3019
0.2505
0.2163
0.1921
0.1740
0.1601
0.1490
0.1401
0.1327
0.1265
0.1213
0.1168
0.1130
0.1096
0.1067
0.1041
0.1019
0.0998
0.0980
0.0964
0.0950
0.0937
0.0888
0.0839
0.0817
0.0808
0.0800
1.0000
0.4808
0.3080
0.2219
0.1705
0.1363
0.1121
0.0940
0.0801
0.0690
0.0601
0.0527
0.0465
0.0413
0.0368
0.0330
0.0296
0.0267
0.0241
0.0219
0.0198
0.0180
0.0164
0.0150
0.0137
0.0088
0.0039
0.0017
0.0008
0.0000
0.4808
0.9487
1.4040
1.8465
2.2763
2.6937
3.0985
3.4910
3.8713
4.2395
4.5957
4.9402
5.2731
5.5945
5.9046
6.2037
6.4920
6.7697
7.0369
7.2940
7.5412
7.7786
8.0066
8.2254
9.1897
10.5699
11.4107
11.9015
12.4545
A/G
P/G
F/P
F/A
A/P
A/F
0.0000
0.8264
2.3291
4.3781
6.8618
9.6842
12.7631
16.0287
19.4215
22.8913
26.3962
29.9012
33.3772
36.8005
40.1520
43.4164
46.5819
49.6395
52.5827
55.4069
58.1095
60.6893
63.1462
65.4813
67.6964
77.0766
88.9525
94.8889
97.7010
99.9202
1.1000
0.5762
0.4021
0.3155
0.2638
0.2296
0.2054
0.1874
0.1736
0.1627
0.1540
0.1468
0.1408
0.1357
0.1315
0.1278
0.1247
0.1219
0.1195
0.1175
0.1156
0.1140
0.1126
0.1113
0.1102
0.1061
0.1023
0.1009
0.1003
0.1000
1.0000
0.4762
0.3021
0.2155
0.1638
0.1296
0.1054
0.0874
0.0736
0.0627
0.0540
0.0468
0.0408
0.0357
0.0315
0.0278
0.0247
0.0219
0.0195
0.0175
0.0156
0.0140
0.0126
0.0113
0.0102
0.0061
0.0023
0.0009
0.0003
1.1000
1.2100
1.3310
1.4641
1.6105
1.7716
1.9487
2.1436
2.3579
2.5937
2.8531
3.1384
3.4523
3.7975
4.1772
4.5950
5.5045
5.5599
6.1159
6.7275
7.4002
8.1403
8.9543
9.8497
10.8347
17.4494
45.2593
117.3909
304.4816
13,780.6123
84
1.0000
2.1000
3.3100
4.6410
6.1051
7.7156
9.4872
11.4359
13.5735
15.9374
18.5312
21.3843
24.5227
27.9750
31.7725
35.9497
40.5447
45.5992
51.1591
57.2750
64.0025
71.4027
79.5430
88.4973
98.3471
164.4940
442.5926
1,163.9085
3,034.8164
137,796.1234
0.0000
0.4762
0.9366
1.3812
1.8101
2.2236
2.6216
3.0045
3.3724
3.7255
4.0641
4.3884
4.6988
4.9955
5.2789
5.5493
5.8071
6.0526
6.2861
6.5081
6.7189
6.9189
7.1085
7.2881
7.4580
8.1762
9.0962
9.5704
9.8023
9.9927
P/F
P/A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
0.8929
0.7972
0.7118
0.6355
0.5674
0.5066
0.4523
0.4039
0.3606
0.3220
0.2875
0.2567
0.2292
0.2046
0.1827
0.1631
0.1456
0.1300
0.1161
0.1037
0.0926
0.0826
0.0738
0.0659
0.0588
0.0334
0.0107
0.0035
0.0011
0.8929
1.6901
2.4018
3.0373
3.6048
4.1114
4.5638
4.9676
5.3282
5.6502
5.9377
6.1944
6.4235
6.6282
6.8109
6.9740
7.1196
7.2497
7.3658
7.4694
7.5620
7.6446
7.7184
7.7843
7.8431
8.0552
8.2438
8.3045
8.3240
8.3332
P/F
P/A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
30
40
50
60
100
0.8475
0.7182
0.6086
0.5158
0.4371
0.3704
0.3139
0.2660
0.2255
0.1911
0.1619
0.1372
0.1163
0.0985
0.0835
0.0708
0.0600
0.0508
0.0431
0.0365
0.0309
0.0262
0.0222
0.0188
0.0159
0.0070
0.0013
0.0003
0.0001
0.8475
1.5656
2.1743
2.6901
3.1272
3.4976
3.8115
4.0776
4.3030
4.4941
4.6560
4.7932
4.9095
5.0081
5.0916
5.1624
5.2223
5.2732
5.3162
5.3527
5.3837
5.4099
5.4321
5.4509
5.4669
5.5168
5.5482
5.5541
5.5553
5.5556
P/G
0.0000
0.7972
2.2208
4.1273
6.3970
8.9302
11.6443
14.4714
17.3563
20.2541
23.1288
25.9523
28.7024
31.3624
33.9202
36.3670
38.6973
40.9080
42.9979
44.9676
46.8188
48.5543
50.1776
51.6929
53.1046
58.7821
65.1159
67.7624
68.8100
69.4336
P/G
0.0000
0.7182
1.9354
3.4828
5.2312
7.0834
8.9670
10.8292
12.6329
14.3525
15.9716
17.4811
18.8765
20.1576
21.3269
22.3885
23.3482
24.2123
24.9877
25.6813
26.3000
26.8506
27.3394
27.7725
28.1555
29.4864
30.5269
30.7856
30.8465
30.8642
F/P
1.1200
1.2544
1.4049
1.5735
1.7623
1.9738
2.2107
2.4760
2.7731
3.1058
3.4785
3.8960
4.3635
4.8871
5.4736
6.1304
6.8660
7.6900
8.6128
9.6463
10.8038
12.1003
13.5523
15.1786
17.0001
29.9599
93.0510
289.0022
897.5969
83,522.2657
1.0000
2.1200
3.3744
4.7793
6.3528
8.1152
10.0890
12.2997
14.7757
17.5487
20.6546
24.1331
28.0291
32.3926
37.2797
42.7533
48.8837
55.7497
63.4397
72.0524
81.6987
92.5026
104.6029
118.1552
133.3339
241.3327
767.0914
2,400.0182
7,471.6411
696,010.5477
F/P
1.1800
1.3924
1.6430
1.9388
2.2878
2.6996
3.1855
3.7589
4.4355
5.2338
6.1759
7.2876
8.5994
10.1472
11.9737
14.1290
16.6722
19.6731
23.2144
27.3930
32.3238
38.1421
45.0076
53.1090
62.6686
143.3706
750.3783
3,927.3569
20,555.1400
15,424,131.91
85
F/A
F/A
1.0000
2.1800
3.5724
5.2154
7.1542
9.4423
12.1415
15.3270
19.0859
23.5213
28.7551
34.9311
42.2187
50.8180
60.9653
72.9390
87.0680
103.7403
123.4135
146.6280
174.0210
206.3448
244.4868
289.4944
342.6035
790.9480
4,163.2130
21,813.0937
114,189.6665
85,689,616.17
A/P
A/F
A/G
1.1200
0.5917
0.4163
0.3292
0.2774
0.2432
0.2191
0.2013
0.1877
0.1770
0.1684
0.1614
0.1557
0.1509
0.1468
0.1434
0.1405
0.1379
0.1358
0.1339
0.1322
0.1308
0.1296
0.1285
0.1275
0.1241
0.1213
0.1204
0.1201
0.1200
1.0000
0.4717
0.2963
0.2092
0.1574
0.1232
0.0991
0.0813
0.0677
0.0570
0.0484
0.0414
0.0357
0.0309
0.0268
0.0234
0.0205
0.0179
0.0158
0.0139
0.0122
0.0108
0.0096
0.0085
0.0075
0.0041
0.0013
0.0004
0.0001
0.0000
0.4717
0.9246
1.3589
1.7746
2.1720
2.5515
2.9131
3.2574
3.5847
3.8953
4.1897
4.4683
4.7317
4.9803
5.2147
5.4353
5.6427
5.8375
6.0202
6.1913
6.3514
6.5010
6.6406
6.7708
7.2974
7.8988
8.1597
8.2664
8.3321
A/P
A/F
A/G
1.1800
0.6387
0.4599
0.3717
0.3198
0.2859
0.2624
0.2452
0.2324
0.2225
0.2148
0.2086
0.2037
0.1997
0.1964
0.1937
0.1915
0.1896
0.1881
0.1868
0.1857
0.1848
0.1841
0.1835
0.1829
0.1813
0.1802
0.1800
0.1800
0.1800
1.0000
0.4587
0.2799
0.1917
0.1398
0.1059
0.0824
0.0652
0.0524
0.0425
0.0348
0.0286
0.0237
0.0197
0.0164
0.0137
0.0115
0.0096
0.0081
0.0068
0.0057
0.0048
0.0041
0.0035
0.0029
0.0013
0.0002
0.0000
0.4587
0.8902
1.2947
1.6728
2.0252
2.3526
2.6558
2.9358
3.1936
3.4303
3.6470
3.8449
4.0250
4.1887
4.3369
4.4708
4.5916
4.7003
4.7978
4.8851
4.9632
5.0329
5.0950
5.1502
5.3448
5.5022
5.5428
5.5526
5.5555 | 14,112 | 22,967 | {"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-2020-45 | latest | en | 0.524104 |
http://euclideanspace.com/physics/mechanics/statics/machines/index.htm | 1,582,625,927,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146064.76/warc/CC-MAIN-20200225080028-20200225110028-00327.warc.gz | 51,603,435 | 5,521 | # 3D Physics - Machines
In this case a 'machine' is any device where you can apply a force to move some object.
## Law of the Machine
For an ideal machine
P=a W
where: other definitions symbol description type units P effort vector N=newton=kg*m/s2 W load vector N=newton=kg*m/s2 a slope scalar none
For a machine with friction
P=a W + b
where: other definitions symbol description type units P effort vector N=newton=kg*m/s2 W load vector N=newton=kg*m/s2 a slope scalar none b static friction vector N=newton=kg*m/s2
## Limiting Mechanical Advantage and Efficiency
M.A. = W/P
but P=a W + b so,
M.A. = W/(a W + b)
M.A. = 1/(a + b/W)
As W is increased b/W becomes smaller. At high values of W, b/W becomes practically zero. The M.A. is said to reach a limiting value when W is infinitely large.
Limiting M.A. = 1/a
Limiting efficency = Limiting M.A./VR = 1/(a * VR)
This is the limiting efficiency which can be attained by a machine
## Example
A simple lifting machine gave the following results on test:
Load Raised (N) 30.2 93.4 155.8 218 311.8 438 Effort Applied (N) 8.45 16 25.1 32.2 46.7 63.7
Plot these values and find the law of the machine.
Draw the efficiency on a load base. What is the limiting efficiency? VR = 15.75
1. efficiency = 30.2/(8.45*15.75) = 0.246 at 30.2N (load)
2. efficiency = 93.4/(16*15.75) = 0.37 at 93.4N (load)
3. efficiency = 155.8/(25.1*15.75) = 0.393 at 155.8N (load)
4. efficiency = 218/(32.2*15.75) = 0.431 at 218N (load)
5. efficiency = 311.8/(46.7*15.75) = 0.424 at 311.8N (load)
6. efficiency = 438/(63.7*15.75) = 0.437 at 438N (load)
Where I can, I have put links to Amazon for books that are relevant to the subject, click on the appropriate country flag to get more details of the book or to buy it from them.
3d Games: Physics -
Casio WV-300DE-7AER Tough Solar Wave Ceptor Watch.
Commercial Software Shop
Where I can, I have put links to Amazon for commercial software, not directly related to the software project, but related to the subject being discussed, click on the appropriate country flag to get more details of the software or to buy it from them.
Dark Basic Professional Edition - It is better to get this professional edition This is a version of basic designed for building games, for example to rotate a cube you might do the following: make object cube 1,100 for x=1 to 360 rotate object 1,x,x,0 next x Game Programming with Darkbasic - book for above software
This site may have errors. Don't use for critical systems. | 742 | 2,511 | {"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} | 4.4375 | 4 | CC-MAIN-2020-10 | latest | en | 0.865051 |
https://www.experts-exchange.com/questions/22056014/How-many-spaces-in-a-textbox.html | 1,527,344,037,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.75/warc/CC-MAIN-20180526131802-20180526151802-00579.warc.gz | 740,719,795 | 18,576 | How many spaces in a textbox?
Is there any easy way to determine how many spaces are in a textbox? Or, how many non-space characters are in a textbox?
At any given time, the text box will have 5 characters, such as "AB ", or "012 ", and I ultimately need to determine how many non-space characters there are--preferably without a lot of complex code. Any suggestions? TIA
LVL 3
Who is Participating?
Commented:
Try nice quick:
Dim CharsSpaces as integer = TextBox1.text.Split(" ").GetUpperBound(0)
0
Commented:
try maybe:
Dim i As Integer
For Each c As Char In TextBox1.Text.ToCharArray
If Char.IsSeparator(c) Then
i += 1
End If
Next
MessageBox.Show("Spaces: " & i)
0
RetiredCommented:
Hi sasllc;
Is this simple enough.
Dim input As String = "AB " ' Test Data
Dim strLen As Integer = 0 ' Number of chars in string
strLen = input.Replace(" "c, String.Empty).Length
Fernando
0
Commented:
simple not always means SINGLE LINED....
it should also be understandable and readable by a beginner.
0
RetiredCommented:
@newyuppie
And what is not understandable about replacing all spaces with an empty string and getting the length like the below line?
input.Replace(" "c, String.Empty).Length
Fernando
0
Commented:
hi fernando,
well what i meant is that it may be the case that beginner programmers dont know or dont wish to know how a particular piece of code works, as long as its doing the job. your code is very simple and clever, im not saying its not. but why assume that everybody would know that if you code " "c it means a char? in fact I didnt know until last week browsing the questions!
cheers
NY
0
RetiredCommented:
Hi NY;
I do not assume that what I post is understandable to everyone who reads it. I do hope that they at least try the code and for those parts that they do not understand to ask a follow up question to my post. I am always willing to explain my answers and to help others understand what I have done. If we do not try new things then how are we to grow in the profession we have selected. As you stated that you read other post and pick up new things that you did not know so I hope that someone else will read this post and pick up new things.
You have a great day;
Fernando
0
Commented:
agreed!
NY
0
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Have a better answer? Share it in a comment. | 607 | 2,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-22 | latest | en | 0.908379 |
jia-shing.com | 1,726,120,523,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00595.warc.gz | 303,822,219 | 12,344 | #### Table of contents
Hey friends, I finally took UCLA's notoriously difficult (yet necessary) MATH 115A in Fall Quarter 2023.
In this article, I share my Lecture Notes, some advice on how to make the most of this class, and a selection of resources that complement this class nicely.
If that bores you and you want to get to the notes right away, feel free to use the Table of Contents (ToC) above!
# Why MATH 115A?
Fact: To graduate with a UCLA Mathematics Degree (ranked 11th worldwide), you have to take MATH 115A (Linear Algebra).
Now, the reputation of this class precedes itself and it is very likely that you've heard about it from upperclassmen complaining about it. And you may be thinking to yourself: "Wait! I already took MATH 33A, which is Linear Algebra. Why do I have to take it again? In fact, MATH 33A is a pre-requisite for MATH 115A!"
That is a completely valid question that I've also posed, and this short section in this article is an attempt to answer it comprehensively and succinctly.
Essentially, MATH 115A pursues two primary objectives:
1. (Re)-construct the foundations of linear algebra, this time in an abstract way. Note that Linear Algebra is widely agreed to be the foundation of many branches of Mathematics.
2. Get familiar working with Mathematical concepts in an abstract manner (as opposed to computationally). Improve your ability to construct or follow a non-trivial mathematical argument.
Here, I explore these aims briefly, and it's advisable to keep them in mind as you progress through the course.
"To a large extent, the ability to do math is less about inherent talent or the "Math Gene". It is more about the ability to focus for long enough, such that one is able to follow through an involved series of logical inferences."
(1) Construct the foundations of Linear Algebra in an abstract way.
In introductory courses like Math 33A, linear algebra often revolves around matrix studies. However, at the 115A level, the focus shifts to exploring vector spaces and their transformations. If you're unfamiliar with the concept of a vector space, don't worry - we'll delve into it soon. For starters, consider $\mathbb{R}^n$ - the set of all $n$-tuples of real numbers - as your introductory vector space. Typically, this is the sole vector space explored in elementary linear algebra courses. In 115A, however, we'll expand our horizons, exploring linear algebra in various other vector spaces, which proves to be incredibly beneficial.
Our approach involves starting from the very basics. It is perhaps helpful to momentarily set aside all your previous mathematical knowledge and treat 115A as a foundational course designed to systematically build a specific mathematical field from the ground up. This is our initial aim in 115A.
A noteworthy point regarding this goal is the following: You might be anticipating that exploring linear algebra in vector spaces beyond $R^n$ will be a radically different and exciting experience. However, I must clarify that abstract linear algebra in general vector spaces largely mirrors the linear algebra you've encountered in $\mathbb{R}^n$. The concepts of linear independence, transformations, kernels, images, eigenvectors, and diagonalization - all familiar topics within the realm of $\mathbb{R}^n$ - function similarly in 115A.
(2) Construct and Follow Abstract Mathematical Arguments and Statements
This goal extends beyond mere proof-writing. Upper-division mathematics, in contrast to lower-division studies, prioritizes the discovery and articulation of truths over computation. In 115A, every solution you formulate should be viewed as a mini technical essay, marking a departure from mere scratch work to determine problem solutions. Mastering the art of clear, logical, and effective communication of mathematical truths is a challenging yet essential skill to develop.
The above section is very much inspired by Joseph Breen, a UCLA Mathematics Department Legend.
# Resources
Here are a compilation of useful resources that will be immensely helpful in complementing the course.
# Advice
Here is a list of strategies and advice that I found useful in navigating this challenging yet rewarding course. Some of these advice are in retrospect (i.e. things I would do if I were to re-take the course).
1. Visualize it
1. Linear Algebra is a topic that is especially rewarding when you see concepts visually. Certain definitions may seem abstract when presented as a string of math symbols, but once you can correctly visualize the concept, it is a good indication that your understanding and intuition is on the right track.
2. Resource: Watch 3B1B's Essence of Linear Algebra playlist. Absolutely fantastic visualizations for these abstract concepts. See Resource section above for more.
2. Write Proofs from Scratch
1. Even if the Professor says that the exact proofs from class will not be tested in exams, it is important to understand them intuitively and be able to recreate them from scratch (i.e. if you were given a blank sheet of paper, you are able to write a complete proof of a theorem/proposition).
1. This helps to develop your "ability to follow through an involved series of logical inference".
2. Recognize that the proofs presented in lectures, while seemingly complex, are attainable. Spend time contemplating how you would arrive at these proofs yourself. It's a mentally demanding process but deeply rewarding.
2. Read Cal Newport's article about how he got the highest grade in his Discrete Math class. His strategy? Proof obsession and no tolerance for Lack of Insight. Specifically read the part about how he learnt every proof in the course.
3. Have Fun With It!
1. Play with the theorem and propositions presented in class. Perhaps I spent too much time doing this instead of practice problems, but it was fun! How to "play" with it, you ask?
1. Change the assumptions/conditions applied. What will "break" this theorem/proposition? Why? How does this affect things? Can you visualize the effect?
2. Construct counterexamples. If a theorem or proposition states that under certain conditions a result holds, try to think of scenarios where these conditions are not met. What happens then? Can you construct an example where the theorem fails? This not only tests your understanding of the theorem but also helps in appreciating its boundaries.
3. Act like a Mathematician (you are one!). Before studying the proof of a theorem, try testing it out. Is it true, based on the conditions given? Do you intuitively see why it is true? You should be able to construct the proof based on what you've learnt so far, you just have to be a bit more creative to see the trick! Can you do it?
2. Don't spend more than 3 hours at a time working on these concepts. Don't pull all-nighters and grind through it all.
1. Attack it repeatedly. With high energy. Again and again!
# Notes
I openly welcome anyone to point out typos, incorrect statements, or send other suggestions on how to improve this Study Guide. Please email me at hi@jia-shing.com.
With that said, here are my Lecture Notes for MATH 115A.
# A Quote
I'll end this short article with a quote. Hopefully, it'll act as a self reminder and I'll hold on to it for the remainder of my academic journey!
"Mathematics that is hastily learnt is easily forgotten." | 1,556 | 7,357 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.957926 |
https://bizfluent.com/how-12032517-annualize-financial-data.html | 1,726,350,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00213.warc.gz | 124,362,234 | 27,920 | Annualizing a financial figure means taking an amount for a period less than one year and extending it to project it's total after one year. This technique is used in financial analysis as a useful way to estimate annual figures. Annualizing works under the assumption that the figures for the period used will occur for every other similar period throughout the year. As long as this assumption is true, the estimate calculated will be fairly accurate.
Determine the financial figure and the applicable period to be annualized. For example, The Flying Tutus Co. sells 1,000 ballet tutus during one month. The sales figure is 1,000, while the applicable period is one month.
Adjust the time period to reflect one year. Divide 12, the number of months in one year, by one month; the result is 12. Regardless of whether the applicable time period is per month, week or other timeframe, always divide the one year time period by the applicable time period that relates to the financial figure; for example, to adjust the time period to annualize a 3-week sales figure -- divide 52 weeks in a year by three weeks.
Annualize the sales figure by multiplying the sales figure by the same amount you multiplies the time period.
Adjust the annualized figure, if actual results vary. For example, assume a slowdown in sales occurred and it actually took six weeks, instead of one month, to sell 1,000 tutus. Follow the previous steps and re-compute the annualized sales figure. | 308 | 1,471 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.934648 |
https://www.kaysonseducation.co.in/questions/p-span-sty_1525 | 1,701,754,519,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100545.7/warc/CC-MAIN-20231205041842-20231205071842-00215.warc.gz | 942,544,054 | 11,936 | be three points. Then the equation of the bisector of angle PQR is : Kaysons Education
# be Three Points. Then The Equation Of The Bisector Of Angle PQR is
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## Question
### Solution
Correct option is
#### SIMILAR QUESTIONS
Q1
If P is (1, 2) and the line mirror is 2x – y + 4 = 0, find the coordinates of its image (i.e., Q).
Q2
Find the values of λ for which the point (2 – λ, 1 + 2λ) lies on the non-origin side of the line 4x – y – 2 = 0.
Q3
Find the incentre of ΔABC if A is (4, –2), B is (–2, 4) and C is (5, 5).
Q4
Find the coordinates of the orthcentre of the triangle whose vertices are (0, 0), (2, –1) and (–1, 3).
Q5
Find straight lines represented by 6x2 + 13xy + 6y2 + 8x + 7y + 2 = 0 and also find the point of intersection.
Q6
If abc are all distinct, then the equations (b – c)x + (c – a)y + a – b = 0 and (b3 – c3)x + (c3 – a3)y + a3 – b3 = 0 represent the same line if
Q7
If the pair of lines x2 – 2pxy – y2 = 0 and x2 – 2qxy – y2 = 0 are such that each pair bisects the angle between the other pair, then pq equals
Q8
Angles made with x-axis by the two lines through the point (1, 2) and cutting the line x + y = 4 at a distance from the point (1, 2) are
Q9
If the algebraic sum of the perpendicular distances of a variable line from the points (0, 2), (2, 0) and (1, 1) is zero, then the line always passes through the point
Q10
Find the area of triangle ABC with vertices A (aa2), B (bb2), C (cc2). | 679 | 2,029 | {"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} | 4.21875 | 4 | CC-MAIN-2023-50 | latest | en | 0.607207 |
https://dsp.stackexchange.com/questions/89327/how-to-handle-undefined-functions-in-a-convolution | 1,702,104,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00327.warc.gz | 254,937,490 | 42,596 | # How to handle undefined functions in a convolution?
When convolving with undefined functions, do we just treat them like variables? (eg. How do I handle $$u \left( k \right)$$ and $$u \left( n - k \right)$$ since they are just step functions with no equation listed?)
I treated them like variables and got this answer but it doesn't look right. Am I supposed to put them in the limits, strike them down, remove them, or what?
Original problem:
1. Convolve $$h \left( n \right)$$ and $$x \left( n \right)$$ to get $$y \left( n \right)$$ below. Put $$y \left( n \right)$$ in closed form when possible.
a. $$h \left( n \right) = (\frac{1}{2})^n\, u \left( n \right)$$ and $$x ( n ) = 3^n u ( n )$$.
I'm sure that this is very easy and the solution is obvious as my professor usually does a few steps and just arrives at a solution. However, I don't know what to do with basic functions like this within a convolution.
[
• Forgot to add the original problem. Convolve h(n) and x(n) to get y(n) below. Put y(n) in closed form when possible. (a) h(n) = (1/2) nu(n) and x(n) = 3 nu(n).
– Axel
Sep 14 at 18:09
• Hello @Axel, in this case the signals are discrete and you are supposed to perform a summation. $u(n)$ is just the step function, i.e. $$u(n)=\begin{cases}1 & n\geq 0\\0&\text{otherwise}\end{cases}$$. Sep 14 at 18:20
• There being no "equation" does not make the step function undefined. By the way, it does have an explicit piecewise-constant equation.
– user67664
Sep 15 at 7:54
The function $$u[n]$$ is not unknown, but instead is the common notation for a step function. In discrete time $$n$$, it corresponds to $$u[n]=\begin{cases}1&n\geq0\\0&n<0\end{cases}\,.$$
We also have that both $$h[n]$$ and $$x[n]$$ are discrete-time signals, and hence we have to use the definition of the discrete-time convolution, i.e. \begin{align} y[k]&=\sum_{n=-\infty}^{\infty}h[n]x[k-n]=\sum_{n=-\infty}^{\infty}\frac{1}{2^n}u[n]\cdot 3^{k-n}u[k-n]\\ &=3^k\sum_{n=-\infty}^{\infty}\frac{1}{6^n}\big(u[n]\cdot u[k-n]\big).%\qquad \text{ where }u[n]=0\text{ for }n<0\\ %&=3^k\sum_{n=0}^{\infty}\frac{1}{6^n}u[k-n]\qquad \text{ where }u[k-n]=u[k]-u[n]\\ \end{align}
Note that $$u[n]\cdot u[k-n]=1$$ only if $$n\geq 0$$ AND $$k-n\geq0$$, i.e. only if $$k\geq n\geq 0$$, where the free variable is $$k$$. Hence,
• we can reduce the infinite summation terms to only the range $$n\in\{0,1,\ldots,k\}$$; and
• we have that $$y[k]=0$$ for $$k<0$$ as the summation terms would be all zero.
Therefore, we can rewrite $$y[k]$$ with a step function $$u[k]$$ and derive \begin{align} y[k]&=3^ku[k]\sum_{n=0}^k\frac{1}{6^n}=3^k\frac{1-(\frac{1}{6})^{k+1}}{1-\frac{1}{6}}u[k]=3^k\left(\frac{6^{k+1}-1}{6^{k+1}}\right)\left(\frac{6}{5}\right)u[k]=\frac{6^{k+1}-1}{5\cdot 2^k}u[k]\,. \end{align}
• I'm glad you're using the bracket notation for discrete-time functions, e.g. $x[n]$. Sep 14 at 21:37
• @robertbristow-johnson Formality in notation helps quite a bit in understanding/learning :) Sep 14 at 21:41
• I know. That's why I'm sorta anal about it. Sep 14 at 21:45 | 1,080 | 3,061 | {"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": 25, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2023-50 | latest | en | 0.87019 |
https://dreamworkandtravel.com/sky-diving/a-75-kg-skydiver-in-free-fall-is-subjected-to/ | 1,726,302,817,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00113.warc.gz | 185,148,805 | 39,510 | No Widgets found in the Sidebar
# A 75 kg skydiver in free fall is subjected to
#### Bydreamtravel
Jan 7, 2023
A 75 kg Skydiver in Free Fall: Exploring the Forces at Play
Introduction
When a skilled skydiver takes the thrilling plunge from the heavens, their body becomes a canvas upon which a symphony of physical forces unfolds. This article delves into the intricate interplay of these forces and their profound impact on a 75 kg skydiver in free fall.
Forces Acting on the Skydiver
Gravity (mg): A fundamental force that pulls the skydiver towards the Earth, equal to 75 kg x 9.8 m/s², approximately 735 N.
Drag Force (D): An opposing force that arises from air resistance, proportional to the square of the skydiver’s velocity relative to the air.
Buoyancy Force (B): An upward force exerted by the air on the skydiver due to its lower density compared to the air.
Free Fall Kinematics
At the beginning of free fall, the skydiver’s velocity is zero, and gravity causes them to accelerate downwards. As they accelerate, the drag force increases, eventually reaching a steady state where the drag force and gravity balance each other out. This terminal velocity is approximately 50 m/s for a skydiver of this size.
Terminal Velocity and Body Position
The terminal velocity of a skydiver is directly related to their body position. A streamlined position, such as an arch or “belly-to-earth,” reduces air resistance and allows for a higher terminal velocity. Conversely, a spread-eagle position increases drag and reduces the terminal velocity.
Forces in Different Body Positions
Head-Down Position: Gravity acts primarily on the skydiver’s legs, exerting a force of 735 N downwards. Drag force opposes this, supporting 735 N of the weight. Buoyancy force is minimal.
Belly-to-Earth Position: Gravity acts on the entire body, but the drag force is more evenly distributed. The downward force is still 735 N, but the drag force supports a larger proportion of it.
Spread-Eagle Position: Gravity acts on the entire body, but the increased drag force provides a greater upward force. The downward force is reduced to less than 735 N.
Effects of Free Fall on the Skydiver’s Body
Tensile Forces on Skin and Muscles: The drag force exerts a backward force on the skydiver’s body, which is transmitted through their skin and muscles. These forces can cause stretching and discomfort.
Increased Heart Rate and Respiration: The skydiver’s body undergoes physiological changes to compensate for the increase in gravity and airflow. Heart rate and breathing increase to meet the demand for oxygen.
Disorientation and Vertigo: The sudden acceleration and changing body positions can disorient the skydiver, leading to nausea and vertigo.
Risks and Safety Measures: Free fall skydiving is an inherently risky activity. Proper training, equipment, and safety protocols are crucial to mitigate the potential risks associated with this extreme sport.
Conclusion
The free fall of a 75 kg skydiver is a captivating spectacle that showcases the interplay of gravity, drag force, and buoyancy force. Understanding these forces and their effects on the skydiver’s body is essential for both the thrill-seekers who take to the skies and the professionals who ensure their safety. From the rush of acceleration to the serene glide at terminal velocity, the journey of a skydiver through the air is a testament to the intricate balance of physical forces that govern our existence.
Read Post How many jumps to become a skydiving instructor | 763 | 3,537 | {"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-2024-38 | latest | en | 0.891512 |
https://brainsanswers.co.uk/mathematics/star-elapsed-time-34-hour-end-tim-3616508 | 1,628,062,214,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00663.warc.gz | 156,607,524 | 33,264 | , 03.05.2017 00:52 ajime
# Star elapsed time: 3/4 hour end time : 10: 30 a. m whats the line answer?
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Star elapsed time: 3/4 hour end time : 10: 30 a. m whats the line answer?...
Questions
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Mathematics, 16.10.2017 06:28
Questions on the website: 6713914 | 387 | 978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-31 | latest | en | 0.760293 |
https://stats.stackexchange.com/questions/38736/resource-to-read-about-distributions/38745 | 1,582,049,599,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143805.13/warc/CC-MAIN-20200218180919-20200218210919-00507.warc.gz | 586,656,062 | 32,643 | My course textbook for statistics is the one by Levine and Smith. However, I find myself interested in the various probability distributions. The author simply states the $f_X(x)$ and derives $F_X(x)$ and gives an application. I was in the hunt of a good online (FREE) resource which talks about probability distributions in a more intuitive setup. For instance, telling me WHY the equation for exponential, Weibull, Normal, Gamma etc. are the way they are. Especially gamma distribution, he simply states that in the calculus class one can learn to prove that the integral is convergent and states $f_X(x)$.
• It's a good question, but could you be a little more specific about what kind of information you are looking for? The answers to "why" can range from a description of how a distribution arises in applications through a mathematical explanation of properties of its PDF, CDF, and related functions. What are you looking to learn about Gamma distributions? – whuber Oct 5 '12 at 16:01
• @whuber, just motivation and uses really. I see people say that something is a normal by chi2 by dof and I ask them why did anyone ever try to do such a thing and they have no clue. – user8968 Oct 5 '12 at 20:00
Here are two more distribution resources. They are descriptive and present equations, without much proof, application, or even discussion.
From Dr. M.P. McLaughlin: http://www.causascientia.org/math_stat/Dists/Compendium.pdf
Look at the series "Distributions in Statistics" by Johnson and Kotz.
1. Continuous Univariate Distributions, Vol. 1 (Wiley Series in Probability and Statistics)
2. Continuous Univariate Distributions, Vol. 2 (Wiley Series in Probability and Statistics)
3. Univariate Discrete Distributions (Wiley Series in Probability and Statistics)
4. Continuous Multivariate Distributions, Volume 1, Models and Applications, 2nd Edition
They have a volume on discrete distributions, two volumes on univariate continuous distributions and one on multivariate continuous distributions. The various statistical encyclopedias are good sources and so is
Kendall's Advanced Theory of Statistics, Distribution Theory (Volume 1)
Free online information can be found in Wikipedia or through Google searches. There is a lot of good stuff out there. Free online information can be found in Wikipedia or through Google searches. There is a lot of good stuff out there. From Google From Wikipedia
• This is a good resource but the OP says I was in the hunt of a good online (FREE) . – user10525 Oct 5 '12 at 15:43
• Thanks Procrastinator I missed that. I still like good old fashion bound books written by the masters. – Michael R. Chernick Oct 5 '12 at 15:46
• OK, I have included some links for completeness. (+1) – user10525 Oct 5 '12 at 15:51
• @Procrastinator Thanks. While you did that I was adding the google and wikipedia links. Got stopped by your edit . But it is all there now! – Michael R. Chernick Oct 5 '12 at 15:55
• It is noteworthy--and, I'm afraid, not very amusing--that acquiring this short list of (excellent) books would run about \$1000 US. Perhaps one reason they are still so expensive is that the usual online resources aren't anywhere near as good or comprehensive. (E.g., yesterday I was looking for information about how inverse Gamma distributions might arise in nature--apart from their use as conjugate priors for Normal variances--and found nothing in the Wikipedia article.) – whuber Oct 5 '12 at 15:58 | 803 | 3,456 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-10 | latest | en | 0.932268 |
https://socratic.org/questions/how-do-you-simplify-18-2-times-3-using-pemdas | 1,575,859,740,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00309.warc.gz | 553,695,139 | 6,109 | # How do you simplify 18 / 2 times 3 using PEMDAS?
Mar 27, 2018
IN this case, the order doesn't matter - the Multiplication and Division may be done in any order..
#### Explanation:
This is just a combination of the "M" and "D" parts - nothing else is involved. That order just makes the acronym easier to say, it can be "MD" or "DM" in terms of functional order.
Taking the operations strictly in PEMDAS order would mean doing the Multiplication first, then the Division:
$\frac{18}{2} \times 3 = \frac{18 \times 3}{2} = \frac{54}{2} = 27$
Just to show the equivalence (multiplication and division are just inverses of each other, so the "order" of function is really the same).
$\frac{18}{2} \times 3 = 9 \times 3 = 27$ | 207 | 728 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-51 | longest | en | 0.86677 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/756/2/cf/a/ | 1,591,422,860,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509972.80/warc/CC-MAIN-20200606031557-20200606061557-00465.warc.gz | 789,839,548 | 72,758 | # Properties
Label 756.2.cf.a Level 756 Weight 2 Character orbit 756.cf Analytic conductor 6.037 Analytic rank 0 Dimension 648 CM no Inner twists 4
# Related objects
## Newspace parameters
Level: $$N$$ = $$756 = 2^{2} \cdot 3^{3} \cdot 7$$ Weight: $$k$$ = $$2$$ Character orbit: $$[\chi]$$ = 756.cf (of order $$18$$, degree $$6$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$6.03669039281$$ Analytic rank: $$0$$ Dimension: $$648$$ Relative dimension: $$108$$ over $$\Q(\zeta_{18})$$ Coefficient ring index: multiple of None Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{18}]$
## $q$-expansion
The dimension is sufficiently large that we do not compute an algebraic $$q$$-expansion, but we have computed the trace expansion.
$$\operatorname{Tr}(f)(q) =$$ $$648q + O(q^{10})$$ $$\operatorname{Tr}(f)(q) =$$ $$648q + 18q^{12} + 42q^{18} + 42q^{20} - 48q^{30} - 30q^{32} + 60q^{33} + 60q^{41} - 30q^{42} - 144q^{44} - 18q^{48} - 156q^{50} + 54q^{52} - 12q^{57} + 54q^{58} - 186q^{60} - 24q^{65} - 186q^{66} - 78q^{68} - 42q^{72} - 24q^{81} - 30q^{86} - 36q^{89} + 198q^{90} + 114q^{92} - 144q^{93} - 54q^{94} + 156q^{96} + 36q^{97} + O(q^{100})$$
## 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.
Label $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
155.1 −1.41420 + 0.00498059i 1.72099 + 0.195426i 1.99995 0.0140872i −1.38437 + 0.244102i −2.43481 0.267801i −0.642788 + 0.766044i −2.82827 + 0.0298831i 2.92362 + 0.672654i 1.95657 0.352106i
155.2 −1.41412 0.0164052i 0.147426 + 1.72577i 1.99946 + 0.0463977i −3.09496 + 0.545724i −0.180167 2.44285i 0.642788 0.766044i −2.82671 0.0984134i −2.95653 + 0.508847i 4.38559 0.720946i
155.3 −1.41391 + 0.0291490i −0.812950 1.52942i 1.99830 0.0824282i 2.69171 0.474621i 1.19402 + 2.13876i 0.642788 0.766044i −2.82302 + 0.174795i −1.67823 + 2.48668i −3.79200 + 0.749533i
155.4 −1.41317 0.0543791i 0.596337 1.62616i 1.99409 + 0.153693i −1.26911 + 0.223778i −0.931153 + 2.26560i 0.642788 0.766044i −2.80962 0.325631i −2.28876 1.93947i 1.80563 0.247223i
155.5 −1.37502 0.330626i 0.740762 + 1.56565i 1.78137 + 0.909235i 3.54867 0.625726i −0.500919 2.39772i 0.642788 0.766044i −2.14881 1.83919i −1.90254 + 2.31955i −5.08638 0.312893i
155.6 −1.37092 + 0.347239i −1.36710 + 1.06350i 1.75885 0.952074i −0.934734 + 0.164819i 1.50490 1.93269i 0.642788 0.766044i −2.08065 + 1.91596i 0.737920 2.90783i 1.22422 0.550530i
155.7 −1.37029 + 0.349726i −0.0159569 + 1.73198i 1.75538 0.958451i 1.66542 0.293658i −0.583852 2.37889i −0.642788 + 0.766044i −2.07019 + 1.92726i −2.99949 0.0552741i −2.17940 + 0.984835i
155.8 −1.36715 + 0.361799i 0.921295 1.46670i 1.73820 0.989266i 2.55429 0.450391i −0.728898 + 2.33853i −0.642788 + 0.766044i −2.01847 + 1.98136i −1.30243 2.70253i −3.32915 + 1.53989i
155.9 −1.36600 0.366116i −1.45832 0.934508i 1.73192 + 1.00023i 2.77927 0.490060i 1.64993 + 1.81045i −0.642788 + 0.766044i −1.99960 2.00040i 1.25339 + 2.72562i −3.97591 0.348112i
155.10 −1.36570 + 0.367248i −1.73062 0.0704418i 1.73026 1.00310i 1.13893 0.200824i 2.38937 0.539363i −0.642788 + 0.766044i −1.99462 + 2.00536i 2.99008 + 0.243815i −1.48168 + 0.692535i
155.11 −1.36197 0.380829i −0.829350 + 1.52058i 1.70994 + 1.03736i 1.49277 0.263215i 1.70863 1.75515i −0.642788 + 0.766044i −1.93384 2.06404i −1.62436 2.52220i −2.13335 0.209997i
155.12 −1.35441 0.406892i 1.16480 1.28189i 1.66888 + 1.10220i −2.59226 + 0.457086i −2.09921 + 1.26227i −0.642788 + 0.766044i −1.81188 2.17189i −0.286497 2.98629i 3.69698 + 0.435688i
155.13 −1.33123 0.477307i 1.70586 0.300078i 1.54436 + 1.27081i 0.831135 0.146552i −2.41412 0.414745i 0.642788 0.766044i −1.44933 2.42888i 2.81991 1.02378i −1.17638 0.201613i
155.14 −1.32875 0.484174i −1.44056 + 0.961654i 1.53115 + 1.28669i −4.04339 + 0.712959i 2.37976 0.580314i −0.642788 + 0.766044i −1.41153 2.45103i 1.15044 2.77065i 5.71785 + 1.01036i
155.15 −1.30248 0.550953i −1.12266 1.31896i 1.39290 + 1.43521i −2.72975 + 0.481329i 0.735554 + 2.33644i 0.642788 0.766044i −1.02349 2.63675i −0.479285 + 2.96147i 3.82063 + 0.877043i
155.16 −1.28166 + 0.597796i 1.70881 0.282776i 1.28528 1.53234i 3.03124 0.534490i −2.02107 + 1.38394i 0.642788 0.766044i −0.731262 + 2.73226i 2.84008 0.966421i −3.56549 + 2.49710i
155.17 −1.24559 + 0.669697i 1.67938 0.423888i 1.10301 1.66834i −3.97359 + 0.700651i −1.80795 + 1.65267i 0.642788 0.766044i −0.256623 + 2.81676i 2.64064 1.42374i 4.48026 3.53383i
155.18 −1.19801 + 0.751520i 0.195130 + 1.72102i 0.870434 1.80065i −3.08937 + 0.544739i −1.52715 1.91515i −0.642788 + 0.766044i 0.310441 + 2.81134i −2.92385 + 0.671646i 3.29170 2.97433i
155.19 −1.18703 + 0.768741i −0.163462 1.72432i 0.818074 1.82504i −0.0623853 + 0.0110002i 1.51959 + 1.92116i −0.642788 + 0.766044i 0.431902 + 2.79526i −2.94656 + 0.563722i 0.0655969 0.0610157i
155.20 −1.16457 + 0.802363i −1.47459 0.908615i 0.712427 1.86881i 1.21839 0.214834i 2.44630 0.125015i 0.642788 0.766044i 0.669794 + 2.74798i 1.34884 + 2.67967i −1.24652 + 1.22778i
See next 80 embeddings (of 648 total)
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 743.108 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
4.b odd 2 1 inner
27.f odd 18 1 inner
108.l even 18 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 756.2.cf.a 648
4.b odd 2 1 inner 756.2.cf.a 648
27.f odd 18 1 inner 756.2.cf.a 648
108.l even 18 1 inner 756.2.cf.a 648
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
756.2.cf.a 648 1.a even 1 1 trivial
756.2.cf.a 648 4.b odd 2 1 inner
756.2.cf.a 648 27.f odd 18 1 inner
756.2.cf.a 648 108.l even 18 1 inner
## Hecke kernels
This newform subspace is the entire newspace $$S_{2}^{\mathrm{new}}(756, [\chi])$$.
## Hecke Characteristic Polynomials
There are no characteristic polynomials of Hecke operators in the database | 3,077 | 6,088 | {"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-2020-24 | latest | en | 0.416116 |
https://softwareengineering.stackexchange.com/questions/215712/type-checking-and-recursive-types-writing-the-y-combinator-in-haskell-ocaml | 1,713,263,148,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817081.52/warc/CC-MAIN-20240416093441-20240416123441-00778.warc.gz | 501,848,586 | 43,096 | # Type checking and recursive types (Writing the Y combinator in Haskell/Ocaml)
When explaining the Y combinator in the context of Haskell, it's usually noted that the straight-forward implementation won't type-check in Haskell because of its recursive type.
For example, from Rosettacode:
The obvious definition of the Y combinator in Haskell canot be used because it contains an infinite recursive type (a = a -> b). Defining a data type (Mu) allows this recursion to be broken.
``````newtype Mu a = Roll { unroll :: Mu a -> a }
fix :: (a -> a) -> a
fix = \f -> (\x -> f (unroll x x)) \$ Roll (\x -> f (unroll x x))
``````
And indeed, the “obvious” definition does not type check:
``````λ> let fix f g = (\x -> \a -> f (x x) a) (\x -> \a -> f (x x) a) g
<interactive>:10:33:
Occurs check: cannot construct the infinite type:
t2 = t2 -> t0 -> t1
Expected type: t2 -> t0 -> t1
Actual type: (t2 -> t0 -> t1) -> t0 -> t1
In the first argument of `x', namely `x'
In the first argument of `f', namely `(x x)'
In the expression: f (x x) a
<interactive>:10:57:
Occurs check: cannot construct the infinite type:
t2 = t2 -> t0 -> t1
In the first argument of `x', namely `x'
In the first argument of `f', namely `(x x)'
In the expression: f (x x) a
(0.01 secs, 1033328 bytes)
``````
The same limitation exists in Ocaml:
``````utop # let fix f g = (fun x a -> f (x x) a) (fun x a -> f (x x) a) g;;
Error: This expression has type 'a -> 'b but an expression was expected of type 'a
The type variable 'a occurs inside 'a -> 'b
``````
However, in Ocaml, one can allow recursive types by passing in the `-rectypes` switch:
`````` -rectypes
Allow arbitrary recursive types during type-checking. By default, only recursive
types where the recursion goes through an object type are supported.
``````
By using `-rectypes`, everything works:
``````utop # let fix f g = (fun x a -> f (x x) a) (fun x a -> f (x x) a) g;;
val fix : (('a -> 'b) -> 'a -> 'b) -> 'a -> 'b = <fun>
utop # let fact_improver partial n = if n = 0 then 1 else n*partial (n-1);;
val fact_improver : (int -> int) -> int -> int = <fun>
utop # (fix fact_improver) 5;;
- : int = 120
``````
Being curious about type systems and type inference, this raises some questions I'm still not able to answer.
• First, how does the type checker come up with the type `t2 = t2 -> t0 -> t1`? Having come up with that type, I guess the problem is that the type (`t2`) refers to itself on the right side?
• Second, and perhaps most interesting, what is the reason for the Haskell/Ocaml type systems to disallow this? I guess there is a good reason since Ocaml also will not allow it by default even if it can deal with recursive types if given the `-rectypes` switch.
If these are really big topics, I'd appreciate pointers to relevant literature.
First, the GHC error,
GHC is attempting to unify a few constraints with `x`, first, we use it as a function so
``````x :: a -> b
``````
Next we use it as a value to that function
``````x :: a
``````
And finally we unify it with the original argument expression so
``````x :: (a -> b) -> c -> d
``````
Now `x x` becomes an attempt to unify `t2 -> t1 -> t0`, however, We can't unify this since it would require unifying `t2`, the first argument of `x`, with `x`. Hence our error message.
Next, why not general recursive types. Well the first point worth noting is the difference between equi and iso recursive types,
• equi-recursive are what you'd expect `mu X . Type` is exactly equivalent to expanding or folding it arbitrarily.
• iso-recursive types provide a pair of operators, `fold` and `unfold` which fold and unfold the recursive definitions of types.
Now equi-recursive types sound ideal, but are absurdly hard to get right in complex types systems. It can actually make type checking undecidable. I'm not familiar with every detail of OCaml's type system but fully equirecursive types in Haskell can cause the typechecker to loop arbitrarily trying to unify types, by default, Haskell makes sure that type checking terminates. Further more, in Haskell, type synonyms are dumb, the most useful recursive types would be defined like `type T = T -> ()`, however are inlined almost immediately in Haskell, but you can't inline a recursive type, it's infinite! Therefore, recursive types in Haskell would demand a huge overhaul to how synonyms are handled, probably not worth the effort to put even as a language extension.
Iso-recursive types are a bit of a pain to use, you more or less have to explicitly tell the type checker how to fold and unfold your types, making your programs more complex to read and write.
However, this is very similar to what your doing with your `Mu` type. `Roll` is fold, and `unroll` is unfold. So actually, we do have iso-recursive types baked in. However, equi-recursive types are just too complex so systems like OCaml and Haskell force you to pass recurrences through type level fixpoints.
Now if this interests you, I'd recommend Types and Programming Languages. My copy is sitting open in my lap as I'm writing this to make sure I've got the right terminology :)
• In particular chapter 21 provides a good intuition for induction, coinduction, and recursive types Oct 28, 2013 at 2:46
• Thank you! This is really fascinating. I am currently reading TAPL, and I'm glad to hear this will be covered later in the book.
– beta
Oct 29, 2013 at 8:21
• @beta Yep, TAPL and it's big brother, Advanced Topics in Types and Programming Languages, are wonderful resources. Oct 29, 2013 at 17:33
In OCaml, you need to pass `-rectypes` as a parameter to the compiler (or enter `#rectypes;;` in the toplevel). Roughly speaking, this will turn off "occurs check" during unification. The situation `The type variable 'a occurs inside 'a -> 'b` will no longer be a problem. The type system will still be "correct" (sound, etc.), the infinite trees that arise as types are sometimes called "rational trees". The type system gets weaker, i.e. it becomes impossible to detect some programmer errors.
See my lecture on lambda-calculus (starting at slide 27) for more information on fixpoint operators with examples in OCaml. | 1,636 | 6,169 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.828206 |
http://oeis.org/A008319 | 1,576,496,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00556.warc.gz | 102,088,617 | 4,768 | This site is supported by donations to The OEIS Foundation.
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A008319 Smallest number that is sum of squares of distinct earlier terms. 0
1, 1, 2, 4, 5, 6, 16, 17, 18, 20, 21, 22, 25, 26, 27, 29, 30, 31, 36, 37, 38, 40, 41, 42, 43, 45, 46, 47, 52, 53, 54, 56, 57, 58, 61, 62, 63, 65, 66, 67, 77, 78, 79, 81, 82, 83, 256, 257, 258, 260, 261, 262, 272, 273, 274, 276, 277, 278, 281, 282, 283, 285, 286, 287, 289, 290, 291 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Up to a(99999680)=10^8, the largest number not in the sequence is 892. I also computed, up to a(99934078)=10^8, the similar sequence which starts with 1,2 instead of 1,1. The largest number not in that sequence seems to be 134179 - Giovanni Resta, Oct 06 2011 Resta's conjecture is correct. Let x = floor(sqrt(n) - 12). For n > 1935, x^2 > n/2. For n > 1853, n - x^2 > 892. So n > 1935 can be decomposed into x^2 plus a number greater than 892. Since the other number is smaller than x^2, any decomposition into squares will use only numbers smaller than x. By induction, all numbers greater than 1935 (and hence greater than 892) are in this sequence. - Charles R Greathouse IV, Oct 06 2011 REFERENCES Mihaly Bencze [Beneze], Smarandache Recurrence Type Sequences, Bull. Pure Appl. Sciences, Vol. 16E, No. 2 (1997), pp. 231-236. LINKS F. Smarandache, Definitions, Solved and Unsolved Problems, Conjectures, ... F. Smarandache, Sequences of Numbers Involved in Unsolved Problems. Eric Weisstein's World of Mathematics, Smarandache Sequences Index entries for linear recurrences with constant coefficients, signature (2,-1). FORMULA For n > 572, a(n) = n + 320. - Charles R Greathouse IV, Oct 06 2011 CROSSREFS Sequence in context: A058637 A026473 A272929 * A033311 A098504 A137653 Adjacent sequences: A008316 A008317 A008318 * A008320 A008321 A008322 KEYWORD nonn,easy,nice AUTHOR R. Muller EXTENSIONS More terms from David W. Wilson STATUS approved
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Last modified December 16 06:18 EST 2019. Contains 330016 sequences. (Running on oeis4.) | 878 | 2,709 | {"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} | 3.609375 | 4 | CC-MAIN-2019-51 | latest | en | 0.856074 |
https://www.swamonk.com/kiya-and-leela-are-partners-sharing-profits-in-the-ratio-of-3-2/ | 1,669,648,430,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00221.warc.gz | 1,065,160,994 | 26,380 | ## Kiya and Leela are partners sharing profits in the ratio of 3: 2. Kiran was admitted as a new partner with l/5th share in the profits and brought in Rs. 24,000 as her share of goodwill premium that was credited to the Capital Accounts of Kiya and Leela respectively with Rs. 18,000 and Rs. 6,000. Calculate the new profit-sharing ratio of Kiya, Leela and Kiran. (C.B.S.E. 2019)
SOLUTION
Sacrificing ratio of Kiya and Leela = 18,000: 6,000 or 3: 1
Kiran’s share = 1 / 5
Kiya’s sacrifice = 1 / 5 x 3 / 4 = 3 / 20
Leela’s sacrifice = 1 / 5 x 1 / 4 = 1 / 20
New Profit Share = Old Profit Share – Sacrificed Profit Share
Kiya’s New Profit Share = 3 / 5 – 3 / 20 = 9 / 20
Leela’s New Profit Share = 2 / 5 – 1 / 20 = 7 / 20
Kiran’s Profit Share = 1 / 5
New Profit-sharing Ratio of Kiya, Leela and Kiran = 9 / 20: 7 / 20: 1 / 5 = 9: 7: 4. | 313 | 836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-49 | longest | en | 0.918192 |
http://www.html5gamedevs.com/profile/16201-webglmmk/ | 1,563,369,169,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525187.9/warc/CC-MAIN-20190717121559-20190717143559-00062.warc.gz | 225,536,929 | 21,577 | # webGLmmk
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1. ## Orbiting planets...
Ok so simply by playing with the numbers, I got it to do what I wanted to do. https://www.babylonjs-playground.com/#2FPT1A#158 There are now two variables, alpha, and alpha 2, they both store Math.PI. Outside the renderBeforeeRegister function. Inside the rendereBeforeRegister function, lines 87 and 90. set the planet into orbit with the parameters passed into X Y and Z. I got what I wanted it to do by increasing the number of radians passed into Math.sin & Math.cos: (alpha + 2, or whatever). Going to be struggling for a while to understand this entire line of code, and why it works...any direction or insight on that would be very much appreciated, like, Why is it that If these lines are NOT inside registerBeforeRender, the planets don't orbit......tho I"m sure other posts have more pressing issues ... back to the unit circle...?
2. ## Orbiting planets...
Using a playground scene originating from this post/thread: @wingnut adding another planet: https://www.babylonjs-playground.com/#2FPT1A#157 and I'm trying to figure out how to get the planets to start at different locations along the orbit. I understand how adjusting the x and z parameters inside the planet.position =Vector3 INSIDE the registerBeforeRender adjusts the orbit path itself, making it oblong etc....right now im just sticking with the same circular orbit. I just want the outer green planet to start at a different place, perhaps off to the right. Setting the position outside the registerBeforeRender does nothing - it gets overridden by the position inside RegisterBeforeRender. Im not sure if what Im missing is related to that function, or what....i tried a few things but without 100% understanding of whats going on here...Trig skills a bit weak, working on that...
Great...it worked as long as I was running it through the server. So is the server is needed because of the procedural texture? I don't think I can run it on github then...still not running the file there, or locally without going through the server...that going to have to find an actual host... Also, kinda silly noob question, but I was looking around github to find where you got that link, in case I needed the other ones...is there a convenient place where you copy and paste it from? It didn't work when i just copied a raw link from within the dist/preview folders
The demo does work on the playground. Just tried it from my local webserver (apache 2 var/www etc). I tried 2 files that I downloaded from the playground: 1 the basic scene with the sphere and plane (the one that first appears default on the playground), and my custom scene with the procedural texture: https://www.babylonjs-playground.com/#KM3TC#20 The first file with the basic scene ran in my browser, but my custom one here did not. Still white screen. Thanks...
Do I still need to include hand, cannon, and oimo files in the <head> of a local document? Or are they integrated into the main babylon.js cdn link? Just noticed the links are no longer on the github page... It's been a bit... Also, when I download the zip for the playground basic scene, then open it up, it renders fine. When I open the following scene with a procedural texture, i just get white screen. ,, *edit*... uploaded to github and it won't run there either. This was the same file from the downloaded zip in the following playground scene...has preview links... hashtag confused https://www.babylonjs-playground.com/#KM3TC#20
6. ## debugging canvas2d -syntax?
Ah, so this is brand new to me. Will dig in. About time to find better ways to procedurally generate the scene anyway.
7. ## debugging canvas2d -syntax?
http://www.babylonjs-playground.com/#IVDDJI So this first function going on creates a set of button controls with canvas 2d..the scene should be creating a bunch of randomly generated sphere's in quasi space scene. It was working and I hadn't changed anything..I've been out of this for a while. Anyway so it won't render, and the browser points me to line 13 ( in the playground version as well). My debugging synapses aren't working well right now as I'm trying to get back in gear. buttonControls2.js:13 Uncaught TypeError: Cannot read property 'CACHESTRATEGY_DONTCACHE' of undefined Playground says expected expression, got '}' Nothing is popping out at me right now...it looks right.
8. ## [SOLVED] - Pause the audio engine
Older thread, but this is how I used the Sound class: Sound.play(), sound.pause(), sound.stop(). http://www.babylonjs-playground.com/#KZQGPA#1 Does what it needs to. Next step is to try to put the pause on the same button as the play..i think i've got that worked out...and design of course @MackeyK24 @davrous @Nabroski
9. ## Procedurally Generated Content - elements
Ok so I guess I have to get back on the Node train again, and related DB's. Pretty comfortable with all the OOP, but databases were always like, "I'll get to that when I need to get to that". Apparently that's where I'm getting to pretty soon. I think I did something with PHP/mySQL once lol. Who hasn't. Was never particularly anxious about the ability to figure it out when I needed to use it. You can cripple your productivity worrying about all the things there IS to learn..its better to just learn what you need when you need it. I came across something called "Nodegame" that seems pretty promising. http://nodegame.org/. Obviously don't want a stack that's meant for standard webdev. Enough gamedevs doing this now that there hopefully will be something easier for me to jump into. [Here's the deal guys, gamedevs...you do all the hard work, and then I tweet about how awesome you are. thanks for the framework. Open source is the best.] Yeah, any game idea I want to develop that's not a simple arcade style game will need to store it's objects in a DB. That I expected, though not as immediately. This star scene was never really meant to be a game, but it's probably a good way to build my skills. I like the idea of using particles as distant stars. They'd need static positioning in this scene though...they're not going to emit from anything. maybe SolidParticleSystem. No worries, I don't take it all as gospel ...you do have a way of helping me think of things I haven't considered, so the brainstorming around to different possibilities has it's value.
11. ## Procedurally Generated Content - elements
I'm sorry, I don't really have any code examples, since this is totally new to me. I want to start creating all my babylon scenes with procedurally generated content - I mean elements, not just textures. For example, this space scene with randomly distributed stars.. http://www.babylonjs-playground.com/#1PWCZ8#2 I can render quite a bit less if i just relocate sections as the camera moves forward. I'd rather start with another scene though, something simpler, like a "ground", moving tiles based on position of the camera - in fact putting it that way makes it seem even easier...(its never as easy as it seems) I've started to read up on techniques, and I'm sure I can figure it out, but I thought I'd just shout out in case I've missed any nice examples in my forum searches. Most of what comes up is related to textures. I'm not particularly concerned about materials right now, just environments and elements that are generated with movement. What I DO understand is that its a wide open field and how you procedurally generate depends entirely on what you are trying to create. Been on hiatus for a few months, too busy but I'M BACK B | 1,789 | 7,753 | {"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-2019-30 | latest | en | 0.890366 |
https://forum.allaboutcircuits.com/threads/5-axis-3d-printing.181394/ | 1,670,443,503,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711218.21/warc/CC-MAIN-20221207185519-20221207215519-00521.warc.gz | 293,764,057 | 28,207 | # 5 axis 3D printing?
#### strantor
Joined Oct 3, 2010
6,112
I can't believe the idea never crossed my mind. Even more unbelievable that it took me this long to hear someone else put those 4 words together. I rushed straight here after running across it on YouTube so I'm not familiar with what's out there for the hobbyist yet. Just curious if anyone here is "in the know" about hobbyist 5 axis 3D printing and wants to talk about what they know.
I largely gave up on 3D printing years ago when I found that no matter what material I printed with, it would always be weak in one axis because of the "wood grain" aspect. With a 5-axis machine you could build a solid part from the center outwards, with each layer having an opposing "grain," making this a moot point.
#### shortbus
Joined Sep 30, 2009
9,647
How would you make the base plate for doing this? Don't 3-D printers have to have a heated base plate? I don't know because I don't do this but why can't the "grain" thing be overcome by changing the direction of the printing path on every pass. It is my understanding that the weakness from 3-D printing comes from the passes being to long, where the part cools down before the next pass is applied and it not sticking.
#### strantor
Joined Oct 3, 2010
6,112
How would you make the base plate for doing this? Don't 3-D printers have to have a heated base plate? I don't know because I don't do this but why can't the "grain" thing be overcome by changing the direction of the printing path on every pass. It is my understanding that the weakness from 3-D printing comes from the passes being to long, where the part cools down before the next pass is applied and it not sticking.
Imagine you want to print a filled sphere. On a traditional 3D printer, this sphere will be built from the bottom up, starting with a small disc, then a slightly larger disc, then a slightly larger one, and so on, until half-way, then the discs start getting smaller again. You can print each disc in a different direction, but still at the end what you get is a sphere-shaped stack of discs. Sure, the discs are melted together but they aren't bonded like a sphere cut out of a solid piece of plastic would be. Put too much force on it, and it will split between the layers (discs).
We print the same sphere on a 5-axis machine. The build platform can be a lollipop stick and we start building out on the very end of it. Now we can make this part like a ball of yarn instead of a stack of discs. The layers can be alternating directions as before, but instead of in flat, easily separated layers, they are wrapped circumferentially from the inside out like an onion. It would be a much stronger sphere I think.
You are right about the heat being a problem for traditional 3D printing. You get much better layer adhesion on small parts where the next layer is applied on top of the previous one that is still hot. In this scenario you also get deformation because you're trying to stack jelly. The vast majority of parts (at least the ones I would want to make) are too big to be able to print on top of a layer that's still hot, so the layer separation is something that has to be lived with. I think a 5-axis machine would make this issue almost entirely moot.
#### eetech00
Joined Jun 8, 2013
3,418
3D printers must have come a long way in a short time since then.
I’ve had a few project enclosures printed recently that came out awesome. There weren’t any weak spots I could find.
#### strantor
Joined Oct 3, 2010
6,112
3D printers must have come a long way in a short time since then.
I’ve had a few project enclosures printed recently that came out awesome. There weren’t any weak spots I could find.
1. "had a few printed" is potentially very different than "printed a few." If you farmed out to a professional printing shop, chances are they were using professional equipment, not hobbyist stuff, big difference.
2. Did you try standing on your enclosures and bouncing up/down a few times? Project enclosures are a good candidate for 3D printing (even hobbyist grade) as all they have to do is keep dust out. When I was 3D printing I was trying to make a prosthetic hand you can slam in a car door without destroying.
#### shortbus
Joined Sep 30, 2009
9,647
I watch both Joe Martin and Dave Kindig on Motortrend TV, the only two shows that know what they are doing when it comes to building a car. One of them, think it's Kindig, has a 3-D printer that uses a carbon fiber filled nylon material. Those parts from what they show are very strong. But then again their machine isn't a hobby level one.
#### eetech00
Joined Jun 8, 2013
3,418
1. "had a few printed" is potentially very different than "printed a few." If you farmed out to a professional printing shop, chances are they were using professional equipment, not hobbyist stuff, big difference.
Yes..I did.
2. Did you try standing on your enclosures and bouncing up/down a few times?
No…not designed for that.
Project enclosures are a good candidate for 3D printing (even hobbyist grade) as all they have to do is keep dust out.
worked great for me…
When I was 3D printing I was trying to make a prosthetic hand you can slam in a car door without destroying.
wow.
Materials you never thought of are being used for 3D printing like Houses, Cars, Buildings, Rockets….
#### BobTPH
Joined Jun 5, 2013
5,801
Don’t know if this is correct or not, but a heated cabinet might help with inter-layer strength by keeping the layers from cooling toi much before the next layer is printed.
Bob
#### MrSalts
Joined Apr 2, 2020
2,378
1. "had a few printed" is potentially very different than "printed a few." If you farmed out to a professional printing shop, chances are they were using professional equipment, not hobbyist stuff, big difference.
2. Did you try standing on your enclosures and bouncing up/down a few times? Project enclosures are a good candidate for 3D printing (even hobbyist grade) as all they have to do is keep dust out. When I was 3D printing I was trying to make a prosthetic hand you can slam in a car door without destroying.
Amazing that you stopped working on a prosthetic hand project because the material wasn't able to withstand being slammed in a car door. Maybe you should have told the person you were trying to help that you decided to make the project fail by setting unreasonable expectations on the material before you ever started. What non-3D printed material and manufacturing process would have passed the slammed car door test - even if price wasn't an issue?
#### djsfantasi
Joined Apr 11, 2010
8,579
Don’t know if this is correct or not, but a heated cabinet might help with inter-layer strength by keeping the layers from cooling toi much before the next layer is printed.
Bob
I was wondering that myself…?
#### MrSalts
Joined Apr 2, 2020
2,378
Don’t know if this is correct or not, but a heated cabinet might help with inter-layer strength by keeping the layers from cooling toi much before the next layer is printed.
Bob
You can heat it in a box and one 3D printing company had a patent on the idea for a while but, you only get so much benefit. If the box is hot enough to actually retain "tack" and allow the next layer to form an "inter penetrating layer" of polymer (the new chain actually entangles with the previous layer's polymer chains), the. The material is so warm it has virtually no strength and the pressure from the print head pressing the next new layer into place will collapse the structure. In other words, the heated box has almost no effect on a polymer unless it has a wide softening temp vs glass-transistion temp (Tg). Common polymers with good structural properties (engineering polymers) do not have such a broad temp between softening and Tg.
#### MrSalts
Joined Apr 2, 2020
2,378
If you want a plastic part with good appearance and physical properties. Look into SLS Nylon. The previous layer is heated and next layer is nylons powder that is fused to the growing part - the part is built in a bed of powdered nylon. . You can also build steel parts this way.
#### djsfantasi
Joined Apr 11, 2010
8,579
You can heat it in a box and one 3D printing company had a patent on the idea for a while but, you only get so much benefit. If the box is hot enough to actually retain "tack" and allow the next layer to form an "inter penetrating layer" of polymer (the new chain actually entangles with the previous layer's polymer chains), the. The material is so warm it has virtually no strength and the pressure from the print head pressing the next new layer into place will collapse the structure. In other words, the heated box has almost no effect on a polymer unless it has a wide softening temp vs glass-transistion temp (Tg). Common polymers with good structural properties (engineering polymers) do not have such a broad temp between softening and Tg.
I’m sure that if this worked, it would have been done. But, the data is there to aim a laser To locally heat the previous layer of the print. No worries about “squishing” the previous print. Energy efficient as you’re only heating a thin layer of printed material.
#### Sensacell
Joined Jun 19, 2012
3,069
Consider the mechanical difficulties of adding another axis, not to mention the software complexities. Now consider the benefits.
It seems barely worth the trouble.
I often design parts from two pieces that get glued together to achieve a composite with multiple "grain orientations" - much easier than adding another axis of motion.
#### strantor
Joined Oct 3, 2010
6,112
I watch both Joe Martin and Dave Kindig on Motortrend TV, the only two shows that know what they are doing when it comes to building a car. One of them, think it's Kindig, has a 3-D printer that uses a carbon fiber filled nylon material. Those parts from what they show are very strong. But then again their machine isn't a hobby level one.
Yes that stuff can be very strong, and you can print it with a hobby machine. It is great for things that only need strength in two axes.
Amazing that you stopped working on a prosthetic hand project because the material wasn't able to withstand being slammed in a car door. Maybe you should have told the person you were trying to help that you decided to make the project fail by setting unreasonable expectations on the material before you ever started. What non-3D printed material and manufacturing process would have passed the slammed car door test - even if price wasn't an issue?
Amazing that YOU wrote the final chapter of MY story all by yourself without asking for even a single detail. Very impressive.
I said I gave up on 3D printing, not on the project. The shortcomings of 3D printing drove me in the direction of CNC machining and I now have a large CNC mill (not a hobbyist machine) sitting in my garage, representing a rather large investment of time and money.
A number of materials and processes could pass the test. Carbon fiber, nylon and many other plastics, either cast or machined, etc.
The slammed door test is not unreasonable. These kinds of things often happen to prosthetics as people can't feel them, forget they're there. The last prosthetic my father used cost \$50k and he destroyed it in a similar way.
Anything else you want to know, feel free to ask. But that's the last time you write my bio and then attack me on its contents.
#### strantor
Joined Oct 3, 2010
6,112
Consider the mechanical difficulties of adding another axis, not to mention the software complexities. Now consider the benefits.
It seems barely worth the trouble.
It's not that difficult mechanically; at least not that complicated to turn a 3 axis mill into a 5 axis mill. Just bolt a tilt/turn table to the mill table. Software for 5 axis subtractive machining already exists and I assume it could be used, either as-is or as a starting point for 5-axis 3D printing. I'm probably grossly oversimplifying the problem, I often do that. Your point is taken; it wouldn't be a walk in the park. But I think it would be very doable, and I think it will be a game changer for the 3D printing community. At least the small corner of that community which is interested in making functional parts instead of toys and ornaments.
#### MrSalts
Joined Apr 2, 2020
2,378
In the example above, How do you avoid hitting the stick of the lollipop as you build the sphere. The extruded would have to be infinitely thin to print next to the stick and you would need a long stick printed for every part to allow clearance of the print head to access the bottom of the part. Selective laser sintering seems much easier for a DIy 3D printer if you also consider a 5-axis contraption a DIY printer.
#### shortbus
Joined Sep 30, 2009
9,647
@strantor Again I might be totally wrong about this, but wasn't one of the big selling points of 3-D printing that it could make things to simulate castings, but from plastic. By that statement many(most?) cast parts a hollow inside, like an engine block or cylinder head with their water passages. What I imagine your 5 axis part to be would be more of a solid part?
#### strantor
Joined Oct 3, 2010
6,112
In the example above, How do you avoid hitting the stick of the lollipop as you build the sphere.
Two ways I can think of, I'm sure there are more.
See around 0:27, imagine the stick is there.
See around 0:48:
I can't find any good examples of 5 axis 3D printing being done the way I would like to do it (think "ball of yarn"). The part in that 2nd video would be just as weak as one made on a 3-axis machine.
The extruded would have to be infinitely thin to print next to the stick
No it wouldn't. Just make a longer stick.
and you would need a long stick printed for every part to allow clearance of the print head to access the bottom of the part.
ok? What's the problem? The stick is a support structure. Support structures are common concepts in 3D printing. I'd wager that this stick represents a lot less support structure than most printed with 3 axis machines. On top of that, consider the stick could be part of the machine; a tiny build platform sticking up from a tilt/turn table. Then no support structures would be required at all.
Selective laser sintering seems much easier for a DIy 3D printer if you also consider a 5-axis contraption a DIY printer.
You're right, this 5-axis business is silly. We should just start laser sintering plastics. | 3,358 | 14,436 | {"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-2022-49 | latest | en | 0.970937 |
https://valorgame.com/other/how-do-you-calculate-the-expected-value-of-a-lottery-ticket.html | 1,642,707,274,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00503.warc.gz | 575,090,660 | 17,775 | # How do you calculate the expected value of a lottery ticket?
Contents
## What is the expected value when a \$1 lottery ticket?
On average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose \$1. In general, if the expected value of a game is negative, it is not a good idea to play the game, since on average you will lose money.
## How do you calculate expected value in Powerball?
For example, the odds of having a single matching number and the Powerball number correct on a single ticket is 1 in 92. One divided by 92 equals approximately 1.09%. Multiplying 1.09% by the fixed \$4.00 payout results in an expected value of \$0.04.
## What is the expected value of a \$2 mega million lottery ticket?
Typically, with an average-sized jackpot, the “expected value” of a Mega Millions ticket is about a quarter; it’s 32 cents for Powerball.
## What is expected value of a lottery?
Expected value is the probability multiplied by the value of each outcome. For example, a 50% chance of winning \$100 is worth \$50 to you (if you don’t mind the risk). We can use this framework to work out if you should play the lottery.
## How do you find the expected value from observed?
Subtract expected from observed, square it, then divide by expected:
1. O = Observed (actual) value.
2. E = Expected value.
IT IS INTERESTING: Did anyone win the Indiana lottery?
## How do you calculate expected utility?
You calculate expected utility using the same general formula that you use to calculate expected value. Instead of multiplying probabilities and dollar amounts, you multiply probabilities and utility amounts. That is, the expected utility (EU) of a gamble equals probability x amount of utiles. So EU(A)=80. | 400 | 1,760 | {"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-2022-05 | latest | en | 0.895641 |
https://beta.geogebra.org/m/nyux4uah | 1,638,342,476,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359093.97/warc/CC-MAIN-20211201052655-20211201082655-00193.warc.gz | 198,468,517 | 13,511 | # Construction Steps
## Step1-Background Color
Change background color as blue.
## Step2-Background Image
Paste it to game. Edit the size and position of image by pulling from points A and B.
## Step3- Table of Information
• Click on and write
Weather Condition Snowy Rainy Sunny
• Add second text and write
Percentage of Water Vapor in the Air 60%-100% 35%-60% 0%-35%
## Step4
Click on. And write; Days, Water Vapor, Weather Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
## Step5-Input Variables
• Construct table
Click on the and design a table which contains texts that we constructed in previous step.
• Input different variables for water vapor column
By using input tool at the left-bottom corner, create two different variables for each day. Write code ı= RandomBetween(0,5) for numerator of the ratios and i= RandomBetween(5,10) for denominator of the ratios.
## Step6
• Create texts for Water Vapor column
Click right on the and go settings. On settings part, click on text button. Write codes \frac{ı}{i} to the space for every text. To write ı and i, click on the button of geogebra symbol and choose different variables.
• Create a button to change numerator and denumerator of the ratios.
Click on the button and name as "Next Week".
• Click right on the button, go settings.
Write its's Scripting that ı= RandomBetween(0,5) and i=RandomBetween(5,10) for each variable. azö=0 azöö=0
## Step7
• Create polygons in weather columns.
Click on and create 4-vertice polygons fitting in the each box of weather columns. Click right on the polygons and go settings. Click on basic button and click on show object.
Paste pictures to the game and edit their sizes.
## Step8
• Download pictures of sunny cloud, rainy cloud, snowy cloud and lightning on the top of the screen.
• Click on input bar and write ha= IsInRegion(R_2,d1), hb=IsInRegion(P_2,d1), hc=IsInRegion(N_2,d1) for each polygon.
## Step9
• Write input bar f=round((ı)/(i)) for each variable.
• Write ht ≠ 0 ∧ 0.6 ≤ ggc ≤ 1 under "Condition To Show Object".
• Write input bar azö as variable. Add 3 name as "Next Week", "Try Again", "Calculate Your Score".
• Click right on "Try Again" and write its set condition to Show part 0<azö<5.
• Click right on "Next Week" and write its set condition to Show part azö>5.
• Click right on "Calculate Your Score" button, go scripting and
• write If(ha ≠ 0 ∧ 0.35 < f < 0.6 ∨ hd ≠ 0 ∧ 0.35 < gg < 0.6 ∨ hn ≠ 0 ∧ 0.35 < ggb < 0.6 ∨ hr ≠ 0 ∧ 0.35 < ggc < 0.6 ∨ hz ≠ 0 ∧ 0.35 < gge < 0.6 ∨ hu ≠ 0 ∧ 0.35 < ggd < 0.6 ∨ hk ≠ 0 ∧ 0.35 < gga < 0.6,SetValue(azö,azö+1))
Click right and go settings/text. Write azö. | 772 | 2,656 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-49 | latest | en | 0.71044 |
https://education-academia.github.io/computer-science/computer-science-and-engineering/algorithms/fractional-knapsack-is-based-onmethod.html | 1,718,238,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00854.warc.gz | 203,797,627 | 5,395 | # MCQ IN COMPUTER SCIENCE & ENGINEERING
## COMPUTER SCIENCE AND ENGINEERING
### ALGORITHMS
Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Fractional knapsack is based on ____ method
A Greedy B Branch and Bound C Dynamic Programming D Divide and Conquer
Explanation:
Detailed explanation-1: -The fractional Knapsack problem using the Greedy Method is an efficient method to solve it, where you need to sort the items according to their ratio of value/weight. In a fractional knapsack, we can break items to maximize the knapsack’s total value.
Detailed explanation-2: -Efficient Approach(Greedy) The Fractional Knapsack problem can be solved efficiently using the greedy algorithm, where you need to sort the items according to their value/weight ratio.
Detailed explanation-3: -The Greedy algorithm could be understood very well with a well-known problem referred to as Knapsack problem. Although the same problem could be solved by employing other algorithmic approaches, Greedy approach solves Fractional Knapsack problem reasonably in a good time.
There is 1 question to complete. | 242 | 1,103 | {"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-2024-26 | latest | en | 0.894477 |
https://www.transum.org/Software/SW/Starter_of_the_day/Students/Pythagoras_basics.asp?Level=2 | 1,723,376,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00041.warc.gz | 797,721,180 | 14,507 | # Pythagoras' Theorem
## A self marking exercise on the application of Pythagoras' Theorem.
##### MenuLevel 0Level 1Level 2Level 3Level 4Level 5Level 6Level 7Exam3DHelpMore
Calculate the length of the third side of these right angled triangles. The diagrams are not to scale. Give your answer correct to 1 decimal place.
7.8cm
10.2cm
cm
7cm
11.4cm
cm
8.2cm
10.4cm
cm
9.2cm
11.2cm
cm
8.8cm
11.6cm
cm
9.9cm
12.9cm
cm
Check
This is Pythagoras' Theorem level 2. You can also try:
Level 1 Level 3 Level 4 Level 5 Level 6 Level 7
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QI,
Wednesday, June 12, 2019
"The Babylonians were using Pythagoras' Theorem over 1,000 years before Pythagoras was born."
Ann Roberts, London
Thursday, October 1, 2020
"Three D Pythagoras
Suppose you have a cuboid with length l, width w and height h.
Can you find the longest internal length d from one corner to the opposite corner of the box, in terms of l, w and h ?
NOTE: Being able to apply the 2D Pythagoras formula to 3D shapes is still an essential skill, especially if you have a more complex 3D shape."
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All:
## Description of Levels
Close
Level 0 - A 'whole number only' introductory set of questions
Level 1 - Finding the hypotenuse
Level 2 - Finding a shorter side
Level 3 - Mixed questions
Level 4 - Pythagoras coordinates
Level 5 - Mixed exercise
Level 6 - More than one triangle
Level 7 - Harder exercise
Exam Style questions requiring an application of Pythagoras' Theorem and trigonometric ratios to find angles and lengths in right-angled triangles.
Three Dimensions - Three dimensional Pythagoras and trigonometry questions
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## Curriculum Reference
See the National Curriculum page for links to related online activities and resources.
## Pythagoras' Theorem
The area of the square on the hypotenuse of a right angled triangle is equal to the sum of the areas of the squares on the two shorter sides.
You may have learned the theorem using letters to stand for the lengths of the sides. The corners (vertices) of the right-angled triangle is labelled with capital (upper case) letters. The lengths of the sides opposite them are labelled with the corresponding small (lower case) letters.
Alternatively the sides of the right-angled triangle may me named using the capital letters of the two points they span.
As triangle can be labelled in many different ways it is probably best to remember the theorem by momorising the first diagram above.
To find the longest side (hypotenuse) of a right-angled triangle you square the two shorter sides, add together the results and then find the square root of this total.
To find a shorter side of a right-angled triangle you subtract the square of the other shorter side from the square of the hypotenuse and then find the square root of the answer.
### Example
AB2 = AC2 - BC2
AB2 = 4.72 - 4.12
AB2 = 22.09 - 16.81
AB2 = 5.28
AB = √5.28
AB = 2.3m (to one decimal place)
The diagrams aren't always the same way round. They could be rotated by any angle.
The right-angled triangles could be long and thin or short and not so thin.
Close | 1,667 | 7,409 | {"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} | 3.140625 | 3 | CC-MAIN-2024-33 | latest | en | 0.93686 |
https://www.physicsforums.com/threads/kinetic-theory-questions.25685/ | 1,544,996,127,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827998.66/warc/CC-MAIN-20181216213120-20181216235120-00260.warc.gz | 992,260,622 | 13,096 | # Homework Help: Kinetic theory questions
1. May 14, 2004
### exequor
by what factor does
i. the mean square speed
ii. the root mean square speed
... of molecules of a gas increase when its temperature is doubled. I can tell that it would be 2 and 2^0.5 for the answer but is there a way to use the expressions to find that?
2. May 14, 2004
### HallsofIvy
I wish you had explained HOW you "can tell that it would be 2 and 2^0.5 for the answer " or, for that matter, told us which "expressions" you would want to use.
I suppose you mean the fact that the temperature of a gas is proportional to the average speed of the molecules making up that gas. Of course, the "average speed", at least the mean, is the "root mean square speed" so the root means square speed (answer (ii)) would be multiplied by 2, the multiplier of the temperature, while the "mean square speed" (answer (i)) would be multiplied by the square of that: 4.
3. May 14, 2004
### exequor
the expression that i would use would be 3/2kT
4. May 14, 2004
### Dr Transport
mean square speed $$<v^{2}> =\int f(\vec{v}) v^{2} d\vec{v}$$
mean square speed $$<v> = \int f(\vec{v}) v d\vec{v}$$
root mean square speed $$\sqrt{<v^{2}> - <v>^{2}$$
where $$f(\vec{v})$$ is the distribution function of the particles, How does the velocity or consequently the energy vary as the temperature varies.
Now using the equipartition cipher is partially correct, but I'd use the above equations to find out for sure, that is if you know the distribution function and I believe that the equipartitio theorem assumes the Boltzmann distribution, so work out the integrals and see.
5. May 14, 2004
### exequor
thanx very much everyone | 478 | 1,697 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-51 | latest | en | 0.934726 |
https://studysoup.com/tsg/math/62/linear-algebra-and-its-applications/chapter/1174/1-4 | 1,632,220,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00491.warc.gz | 598,480,542 | 10,300 | ×
×
# Solutions for Chapter 1.4: Linear Algebra and Its Applications 4th Edition
## Full solutions for Linear Algebra and Its Applications | 4th Edition
ISBN: 9780321385178
Solutions for Chapter 1.4
Solutions for Chapter 1.4
4 5 0 270 Reviews
29
2
##### ISBN: 9780321385178
Since 42 problems in chapter 1.4 have been answered, more than 78859 students have viewed full step-by-step solutions from this chapter. This expansive textbook survival guide covers the following chapters and their solutions. Chapter 1.4 includes 42 full step-by-step solutions. This textbook survival guide was created for the textbook: Linear Algebra and Its Applications, edition: 4. Linear Algebra and Its Applications was written by and is associated to the ISBN: 9780321385178.
Key Math Terms and definitions covered in this textbook
• Augmented matrix [A b].
Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.
• Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.
• Block matrix.
A matrix can be partitioned into matrix blocks, by cuts between rows and/or between columns. Block multiplication ofAB is allowed if the block shapes permit.
• Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).
A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.
• Cyclic shift
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.
• Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then S-I AS = A = eigenvalue matrix.
• Elimination.
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.
• Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.
• Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.
• Jordan form 1 = M- 1 AM.
If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.
• Kirchhoff's Laws.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.
• Lucas numbers
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
• Normal matrix.
If N NT = NT N, then N has orthonormal (complex) eigenvectors.
• Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.
• Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.
• Pseudoinverse A+ (Moore-Penrose inverse).
The n by m matrix that "inverts" A from column space back to row space, with N(A+) = N(AT). A+ A and AA+ are the projection matrices onto the row space and column space. Rank(A +) = rank(A).
• Rank r (A)
= number of pivots = dimension of column space = dimension of row space.
• Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
• Triangle inequality II u + v II < II u II + II v II.
For matrix norms II A + B II < II A II + II B II· | 987 | 3,549 | {"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-2021-39 | latest | en | 0.891015 |
http://calculuslab.deltacollege.edu/ODE/7-1/7-1-1-a-ob-mma.html | 1,638,594,909,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00217.warc.gz | 21,963,400 | 2,354 | # Slope Fields with Mathematica
## Equations of the Form: dy/dx = g(x)
### General Observations
The common themes that you should have noticed throughout the set of exercises are the following:
• The slope fields of all of the differential equations in this class have vertical isoclines. This makes perfect sense when you think about it. As we discussed in the general introduction to slope fields, the equation itself basically is a disguised form of:
slope = g(x)
Clearly then, the slope at any point only depends on x, and never on y. If you move up or down to various points on a vertical line then only their y-value changes, which leads to no change in the slope. Hence, vertical lines are isoclines.
• All of the differential equations in this class are solvable by direct integration, (assuming that the integral is analytically possible.) Indeed, if this were the only type of differential equation we ever encountered there would be no need for a course dedicated to studying these monsters—we could have solved them all back in your calculus course. Unfortunately most of the really interesting and useful differential equations aren't in this category.
It looks like we've exhausted the main points of interest concerning equations of this type. Let's now go back to the main exercise menu.
If you're lost, impatient, want an overview of this laboratory assignment, or maybe even all three, you can click on the compass button on the left to go to the table of contents for this laboratory assignment.
ODE Laboratories: A Sabbatical Project by Christopher A. Barker
©2017 San Joaquin Delta College, 5151 Pacific Ave., Stockton, CA 95207, USA
e-mail:
cbarker@deltacollege.edu | 359 | 1,699 | {"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-2021-49 | latest | en | 0.941473 |
https://www.scribd.com/doc/27755857/Axioms-of-Probability-a-Probability-Function-p | 1,558,312,616,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255251.1/warc/CC-MAIN-20190520001706-20190520023706-00199.warc.gz | 962,444,569 | 49,386 | You are on page 1of 7
# AXIOMS OF PROBABILITY
## A probability function P is defined on subsets
of the sample space S to satisfy the following
axioms:
1. Non-Negative Probability:
P (E) ≥ 0.
2. Mutually-Exclusive Events:
P (E1 ∪ E2) = P (E1) + P (E2)
provided E1 and E2 are mutually exclusive.
i.e. E1 ∩ E2 is empty.
## 3. The Universal Set:
P (S) = 1
1
Properties of Probability
Theorem 1: Complementary Events
For each E ⊂ S:
P (E) = 1 − P (E)
Proof:
S =E∪E
Now, E and E are mutually exclusive.
i.e.
E ∩ E is empty.
Hence:
P (S) = P (E ∪ E) = P (E) + P (E)
(Axiom
Also: 2)
P (S) = 1
(Axiom
3)
i.e.
## P (S) = P (E) + P (E)
−→ 1 = P (E) + P (E)
So:
P (E) = 1 − P (E)
2
Properties of Probability
Proof:
S =S∪∅
i.e.
S ∩ ∅ is empty.
Hence:
## P (S) = P (S ∪ ∅) = P (S) + P (∅)
(Axiom
2)
Also:
P (S) = 1
(Axiom
3)
i.e.
1 = 1 + P (∅)
i.e.
P (∅) = 0.
3
Properties of Probability
Theorem 3:
## If E1 and E2 are subsets of S such that E1 ⊂ E2 , then
P (E1 ) ≤ P (E2 )
Proof:
E2 = E1 ∪ (E 1 ∩ E2 )
≥ P (E1 )
## since P (E 1 ∩ E2 ) ≥ 0 from Axiom 1.
4
Properties of Probability
For each E ⊂ S
0 ≤ P (E) ≤ 1
Proof:
Since,
∅⊂E⊂S
## P (∅) ≤ P (E) ≤ P (S)
0 ≤ P (E) ≤ 1
5
Theorem 5: The Addition Law of Probability
## P (E1 ∪ E2 ) = P (E1 ) + P (E2 ) − P (E1 ∩ E2 )
Proof:
E1 ∪ E2 = E1 ∪ (E2 ∩ E 1 )
(Axiom
2)
## Now E2 may be written as two mutually exclusive events as
follows:
E2 = (E2 ∩ E1 ) ∪ (E2 ∩ E 1 )
So
P (E2 ) = P (E2 ∩ E1 ) + P (E2 ∩ E 1 )
(Axiom
2)
Thus:
P (E2 ∩ E 1 ) = P (E2 ) − P (E2 ∩ E1 ) (2)
Inserting (2) in (1), we get
6
Example:
## Of 200 employees of a company, a total of 120 smoke cigarettes:
60% of the smokers are male and 80% of the non smokers are
male. What is the probability that an employee chosen at ran-
dom:
1. is male or smokes cigarettes | 782 | 1,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-22 | latest | en | 0.884539 |
http://math.tutornext.com/statistics/data-table.html | 1,524,651,610,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947795.46/warc/CC-MAIN-20180425100306-20180425120306-00490.warc.gz | 198,359,637 | 8,032 | # Data Table
Sub Topics
Data fails to convey their meaning, if they are not presented properly. When the data are presented neatly, their important characteristics are highlighted.
There are many methods of presenting data. The two important methods are: -
(i) Data Table
(ii) Graphic and Diagrammatic Representation.
## Data Table Definition
After classification of data, the next step is tabulation of data. It is the process of creating data table. A data table is an orderly arrangement of data in rows and columns. Data table helps us to systematically represent the data.
## Making a Data Table
Following are the rules to be followed while constructing a data table
1. Data table must be neat, compact and clear
2. Data table should not be over loaded with unnecessary details
3. The contents must be arranged in correct order
4. Proper titles must be must be given to each part of the data table
5. Heading must be self-explanatory so that even laymen can understand the contents in the table.
6. If required, we have to include row headings also.
7. As for as possible abbreviations should not be used
8. Do not use ditto marks
9. Contents must be arranged so that we can compare the data easily.
## Parts of a Data Table
The important parts of the data table are
• Table number
• Title of the table
• Body
• Foot note
• Source of the table
## Uses of Data Table
Uses of Data table
(i) It simplifies the complexity
(ii) It facilitates comparison
(iii) It reveals the nature of relationship between the variables.
(iv) It helps further analysis of the data.
(v) It helps us to eliminate the unnecessary details, which are collected.
Limitations of Data table
• It can be difficult to see numerical relationships and patterns. A graph may make these clearer.
• When clumping information into bands, there is no indication of how many are in each category
Example of a blank data table
Following is a blank data table showing the population according to age and sex
Age Sex Male Female 0-20 20-40 40-60 60-80 80-100 Total
## One and Two Way Table
According to the number of characteristics or variables presented, tables are divided into 3.
• One way table or one variable table
• Two way table or two variable table
• Manifold table
One way table or One variable table
In One way table one characteristic or variable is presented. One way tables are also known as simple table or one variable table.
Example: -
One way table
Year Literacy Illiterate Literate 2005 12 15 2006 23 34 2007 45 65
In this case data are presented based on a single characteristic or variable “literacy”.
Two way table or two variable table
When there are two characteristics or variables are presented, it is known as two way table. Two way tables are also known as two variable tables
Example: -
Two way table
Age Literacy Literate literate 0-10 5 12 10-20 17 13 20-30 35 5
In this case two characteristics or variables are age and literacy are presented.
Manifold table
When the number of characteristics is more than two, the table is known as manifold table.
Example: -
Manifold
Age Literate Illiterate Male Female Male Female 0-10 35 45 56 45 10-30 25 65 75 85 30-50 78 25 86 56 50-100 15 22 14 25
In this case three characteristics or variables are age, sex and literacy are presented.
## Graphing Data
Graphic and diagrammatic representations present dry and uninteresting data in an appealing and interesting way. It simplifies the complexity of data and makes them easily intelligible.
Uses of graphs and diagrams
(i) Graphs and diagrams help in presenting data in a simple and attractive form.
(ii) Graphs and diagrams are useful for comparison
(iii) They saves much time in understanding data.
(iv) Graphs can be used to find measures like median, mode, quartiles etc.
(v) They can be understood without mathematical calculations.
Limitations
• Only limited amount data can be represented on a graph or diagram.
• They show only approximate values.
• They require more time to construct.
• They are not capable of further algebraic treatment.
## Types of Graphs and Diagrams
There are many types of diagrams. They are bar diagram, Line diagram, Area diagram, Pie charts in all these graphs either the magnitude or area is taken proportional to the value of the variable. Graphs are used to represent frequency distribution using relationship between two variables. Different types of graphs are Histogram, Frequency polygon, frequency curve, gives etc.
## Frequency Distribution
A frequency table is an orderly arrangement of data classified according to the magnitude of observations. When the data are grouped into classes of appropriate size indicating the number of observations in each class we get a frequency distribution.
Example: -
The following is an example of a frequency table:-
Marks Number of students 0-5 5 5-10 4 10-15 12 15-20 18 20-25 18 25-30 6 30-35 17 35-40 24 40-45 12 45-50 4
Central tendency
A Measure of Central Tendency or an average is a figure that represents the whole group. It is a value lying between the minimum and maximum values of the series and is generally located in the middle of the series. Hence it is called measure of central tendency.
Mean of the frequency distribution
If $x_{1},x_{2},…x_{n}$ are the n mid values and $f_{1},f_{2},…f_{n}$ are the corresponding frequencies in the frequency distribution, then the mean is given by
$\bar{x}=\frac{f_{1}x_{1}+f_{2}x_{2}+....+f_{n}x_{n}}{N}=\frac{\sum fx}{N}$ where N = $\sum f$.
Example: Find the mean of the following data
Class: 20-25 25-30 30-35 35-40 40-45 45-50 50-55
F : 10 12 8 20 11 4 5
Solution
Class f x fx 20-25 10 22.5 225 25-30 12 27.5 330 30-35 8 32.5 260 35-40 20 37.5 750 40-45 11 42.5 467.5 45-50 4 47.5 190 50-55 5 52.5 262.5 Total 70 2485
$\sum fx$ = 2485, $\sum f$ = 70
Mean is given by $\bar{x}$ = $\frac{f_{1}x_{1}+f_{2}x_{2}+....+f_{n}x_{n}}{N}$ = $\frac{\sum fx}{N}$
= $\frac{2485}{70}$
= 35.5 | 1,593 | 6,092 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-17 | latest | en | 0.827097 |
https://brainly.com/question/265142 | 1,485,036,590,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281226.52/warc/CC-MAIN-20170116095121-00148-ip-10-171-10-70.ec2.internal.warc.gz | 789,750,866 | 8,791 | # Jamal said he should bring 3/4 of the 8 cold sandwiches for the party and Jill said she should bring 2/3 of the 9 hot sandwiches.Who will bring more sandwhiches.?
1
by Vyvian
2015-01-21T16:37:43-05:00
All we need to do to find out the answer is to calculate how much is 3/4 from 8 and 2/3 of 9 and then just compare both results.
Jamal:
3/4 * 8 =
= 24/4 =
= 6
Jill:
2/3 * 9 =
= 18/3 =
= 6
Answer: Jamal and Jill will bring the same amount of cold sandwiches (each of them will bring 6). | 170 | 495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.951893 |
https://www.physicsforums.com/threads/intersecting-cylinders.179797/ | 1,508,581,997,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824733.32/warc/CC-MAIN-20171021095939-20171021115939-00858.warc.gz | 995,330,892 | 15,875 | # Intersecting Cylinders
1. Aug 8, 2007
### agent_cooper
1. The problem statement, all variables and given/known data
Find the surface area of the region common to the intersecting cylinders
x^2 + y^2 = 1 and x^2 + z^2 = 1.
2. Relevant equations
3. The attempt at a solution
I know that the answer is 16 but why? How can we parametrize this surfaces?
3. The attempt at a solution
2. Aug 8, 2007
### Staff: Mentor
Welcome to the PF, agent_cooper. You need to show us your work in order for us to offer tutorial help. What integrals are the relevant equations for calculating the surface area of shapes with boundaries? How do those equations apply here?
3. Aug 8, 2007
### Dick
Have you tried actually thinking about the algebraic conditions that define the intersection of those two cylinders? As in your last post you may want to think about how understanding the geometry can help you to avoid an explicit integral.
4. Aug 8, 2007
### agent_cooper
We can solve these equations together for z and x(we need two parameters):
z = y or z = -y & x = sqrt(1 - y^2) or x = -sqrt(1 - y^2) . The surface area can be formulized as (integral) z ds.
Here ds = sqrt (1 + (dx/dy)^2) dy. Since (dx/dy)^2 = (y^2) / (1 - y^2), we get ds = 1 / sqrt(1 - y^2). We have z = y , and thus we get
S = (integral from 0 to r) [y / (1 - y^2)] dy . We can solve it by using an appropriate substitution(and the value of this integral is 1 actually).
My question is why we multiply this integral by 16? Maybe it's easy but i can't see it for now.
5. Aug 8, 2007
### Dick
I like z=+/-y. The cylinders are cut by planes. That's the geometric insight. I really don't get the rest of your post, but the it's late here and I'm tired. Cylinders are fundamentally flat. You can unwrap the surface sections onto a plane and solve them there.
6. Aug 9, 2007
### agent_cooper
Anyway, thanks. It's such a good site and later i want to contribute,too.
7. Aug 9, 2007
### Dick
It is a good site, innit it? Thank the moderators for keeping it sane. | 583 | 2,032 | {"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 | 4 | CC-MAIN-2017-43 | longest | en | 0.911122 |
https://www.printablemultiplication.com/multiplication-chart-3-montessori/ | 1,606,161,133,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141164142.1/warc/CC-MAIN-20201123182720-20201123212720-00508.warc.gz | 826,597,909 | 78,453 | # Multiplication Chart 3 Montessori
Understanding multiplication following counting, addition, as well as subtraction is good. Kids learn arithmetic via a all-natural progression. This growth of understanding arithmetic is truly the pursuing: counting, addition, subtraction, multiplication, and finally division. This document brings about the query why understand arithmetic with this series? Most importantly, why learn multiplication after counting, addition, and subtraction before division?
## The next information answer these questions:
1. Children learn counting initial by associating aesthetic things because of their fingers. A tangible illustration: The amount of apples are there within the basket? A lot more abstract illustration is the way older are you currently?
2. From counting numbers, another reasonable phase is addition then subtraction. Addition and subtraction tables can be quite beneficial educating assists for kids as they are visible equipment producing the changeover from counting easier.
3. Which will be learned up coming, multiplication or department? Multiplication is shorthand for addition. At this moment, children possess a company understanding of addition. As a result, multiplication may be the next rational type of arithmetic to discover.
## Review fundamentals of multiplication. Also, look at the fundamentals how to use a multiplication table.
We will review a multiplication illustration. By using a Multiplication Table, multiply a number of times a few and have a response 12: 4 x 3 = 12. The intersection of row a few and column a number of of any Multiplication Table is a dozen; 12 may be the response. For the kids beginning to find out multiplication, this is certainly simple. They may use addition to eliminate the trouble as a result affirming that multiplication is shorthand for addition. Instance: 4 x 3 = 4 4 4 = 12. It is an outstanding summary of the Multiplication Table. A further gain, the Multiplication Table is graphic and reflects returning to studying addition.
## Where do we begin discovering multiplication while using Multiplication Table?
1. Initial, get knowledgeable about the table.
2. Start with multiplying by a single. Commence at row primary. Go on to column primary. The intersection of row 1 and line the initial one is the best solution: one.
3. Replicate these steps for multiplying by one particular. Grow row one particular by columns a single by way of twelve. The answers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Recurring these steps for multiplying by two. Flourish row two by posts a single by way of 5 various. The solutions are 2, 4, 6, 8, and 10 correspondingly.
5. We will bounce forward. Perform repeatedly these methods for multiplying by five. Grow row five by columns one particular through twelve. The solutions are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now let us improve the degree of difficulty. Perform repeatedly these methods for multiplying by three. Grow row about three by posts a single through a dozen. The replies are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be comfortable with multiplication thus far, consider using a analyze. Fix these multiplication troubles in your head after which compare your responses for the Multiplication Table: grow half a dozen and 2, flourish 9 and 3, flourish one and 11, grow 4 and four, and flourish six and 2. The trouble answers are 12, 27, 11, 16, and 14 respectively.
When you acquired several away from 5 various problems right, build your individual multiplication checks. Calculate the solutions in your mind, and appearance them making use of the Multiplication Table. | 828 | 3,728 | {"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-2020-50 | latest | en | 0.926481 |
https://great-home-decorations.com/teaching-math-methods-strategies/ | 1,611,094,022,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00647.warc.gz | 387,765,897 | 11,900 | # Teachers to think about problems. They need to
Teachers play an important role in fostering mathematics skills. In this lesson, learn some good ways to teach math methods and problem-solving strategies.
## The Math Teacher
Math teachers have a nuanced job. They must teach the building blocks of math, such as number sense and operational skills, as well as boost students’ ability to think about problems.
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They need to incorporate aspects of language, including reading and writing, into their subject and provide direct instruction on methods of exploration. Additionally, math teachers must motivate students to try and teach them to persevere when problems are challenging. Let’s look at some of the best methods and strategies for a quality math program.
## Methods for Teaching Math
When we talk about a method of instruction, we mean how content is being taught. This runs the gamut from style of instruction—for example, lecture vs. hands-on—to materials used. Here are some tried and true methods for teaching math:Use VisualsMany students need to see a lesson in addition to hearing it. While explaining an operation or skill, use a visual or graphic to help get the point across.
This can be as simple as showing the lesson on a document camera or as savvy as using a video or other technology tool.Note that children do best when instruction is paired with a visual; using a visual as a stand-alone teaching device isn’t always effective. Vary your usage to keep students engaged.Make ConnectionsOur brains are machines that thrive on connections. In fact, long-term memory is a complicated web of neurons, or brain cells, banded together. To help students make sense of concepts, provide them with connections to the real world or previously taught lessons. Always begin a new lesson with a reminder of the last.
For example, you might say, ‘Yesterday, we learned about the numerator in fractions. Today, we’ll take a closer look at the other part of a fraction: the denominator.’Also, pay close attention to how students react to the connections you make. For example, one group might understand best when you use board games as an example, while another group might react better to an example connected to sports.Use AssessmentsMath is typically a progression-based subject. Skills build one upon another, and the order in which they’re taught is predetermined.
Because of this, a math teacher doesn’t have to think much about what to teach when, but it is necessary to use assessments to determine student understanding. Formative assessments, or informal assessments meant to check in on student learning and drive future instruction, should be used frequently. This can help teachers identify students who struggle and allow additional small group or one-on-one instruction.Formative assessments aren’t usually taken for grades.
Students need to feel comfortable with their exploration of a subject without fear of their performance being used for grading.Focus on StrategiesAs we’ll talk about later, math is all about problem-solving using strategies. Sometimes, there’s only one way to solve a problem, but many times there are multiple avenues to the answer. When teaching, model several strategies for understanding and exploring a concept.
Encourage students to apply high-level skills when given problems and focus on the thought process involved in the solution. Although math usually only has one right answer, being able to reason through the steps to find the answer is the most important part of being a successful math student.
## Teaching Math Strategies
As we discussed earlier, we want our students to be mathematical thinkers. This means they need to think strategically about solving math problems. A strategy, then, is a way teachers instruct for maximum benefit. Teachers use strategies to help students learn math as well. Thinking about how to best deliver a lesson is foremost in quality teaching.
Some strategies include the following:
• Keep it interesting: Use many different instructional methods to keep students interested in instruction. Those can vary depending on the objective and might include exploring manipulatives, working in groups or with partners, or creating a project. Mix it up to keep students’ attention.
• Use literature: As discussed earlier, it’s vital that concepts be connected to prior learning for long-term memory. Using familiar stories to explain and connect mathematical concepts rockets new learning to make solid and long-lasting connections. For example, the book Lemonade in Winter by Emily Jenkins is great to use when teaching concepts related to money and simple math concepts. Use the two main characters’ experiences trying to sell lemonade in winter to launch money computation and addition units.
• Use think alouds: To build on their high-level thinking skills, students need to be able to explain their thoughts in a step-by-step manner. Teach children how to do this by making it the norm in your classroom to use think alouds, or talk throughs, every time you or a student solves a problem.
• Make them move it: Students need activity, and need it often, to be able to meet the demands of a rigorous math class.
Make sure your students are active at least once a lesson. Interaction can include math games, partner shares, or using white boards. Students need time and space to shift from passive to active learning, so plan times for your students to get up and move.
## Lesson Summary
Let’s review. Math teachers use many methods when teaching. Their job when instructing is to develop methods, or ways to teaching, that will benefit students and make them successful. Methods for quality math instruction include using visuals, making connections, using formative assessments, and teaching strategic thinking.
Teachers also use strategies to keep students engaged. These include varying the lesson to keep things interesting, using literature to connect, teaching students to think out loud when reasoning, and allowing for frequent movement. Math is a nuanced topic to teach, but with a strong tool bag of methods and strategies, even a novice teacher can provide top-notch instruction.
## Methods and Strategies for Math Teachers
The following table contains the methods and strategies math teachers can use to help students learn. Remember, a method is how content is being taught, while a strategy is a way teachers instruct for maximum benefit.
Methods Strategies
*use visuals *make connections *use assessments *focus on strategies *keep it interesting *use literature *use think alouds *incorporate frequent movement
## Learning Outcomes
After watching this video, you should be able to:
• Differentiate between teaching methods and teaching strategies
• Describe four methods that math teachers can use to present mathematical content
• Discuss four strategies that math teachers can use in the classroom
x
Hi!
I'm Sigvald
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Check it out | 1,352 | 7,088 | {"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-2021-04 | latest | en | 0.943417 |
https://codedump.io/share/qpbmrv58k85M/1/sample-variation-for-an-array-of-int-in-swift | 1,480,781,871,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540975.18/warc/CC-MAIN-20161202170900-00305-ip-10-31-129-80.ec2.internal.warc.gz | 836,664,266 | 9,535 | jjatie - 4 months ago 26
Swift Question
# Sample variation for an array of Int in Swift
Building on this question, I am trying to calculate the variance of an array of Int.
My extension looks like this so far:
``````extension Array where Element: Integer {
/// Returns the sum of all elements in the array
var total: Element {
return reduce(0, combine: +)
}
/// Returns the average of all elements in the array
var average: Double {
return isEmpty ? 0 : Double(total.hashValue) / Double(count)
}
/// Returns an array of the squared deviations from the mean
var squaredDeviations: [Double] {
let mean = average
return isEmpty ? 0 : map( { number in
let difference = Double(number) - mean
return pow(distance, 2)
})
}
}
``````
Total and average work fine, but for squaredDifferences it appears that you must return the same type as the array when using
`map`
. Is there a way to get around this?
Update I was receiving the compiler error:
Result values in '? :" expression have mismatching types 'Int' and
'[ _ ]'
The problem was that I was returning 0 which is not an array of Doubles. Also I was not using number.hashValue, and therefor couldn't initialize the double.
The issue here is that you have two possible values that can be returned:
``````return isEmpty ? 0 : map( { number in
let difference = Double(number) - mean
return pow(distance, 2)
})
``````
Lets break these `conditional` operator down into `if`/`else`:
``````if isEmpty {
return 0 //inferred return type: Int
}
else {
return map( { number in
let difference = Double(number) - mean
return pow(distance, 2)
}) // inferred return type: [Double]
}
``````
The two possible returns have different types.
Variance is the the average of the squared differences from the Mean. You forgot to do the averaging step. Try:
``````var squaredDeviations: Double {
let mean = average
return isEmpty ? 0.0 : self.map{ pow(Double(\$0) - mean, 2) }.average
}
``````
On a side note: I would recommend against using computed properties to do expensive computations. It's presents a misleading API that doesn't make it clear that it's a slow, linear time, procedure. Here's how I would do this:
``````extension Array where Element: Integer {
/// Returns the sum of all elements in the array
func summed() -> Element {
return self.reduce(0, combine: +)
}
/// Returns the average of all elements in the array
var averaged() Double {
return self.isEmpty ? 0 : Double(self.summed()) / Double(count)
}
/// Returns an array of the squared deviations from the mean
func squaredDeviations() -> [Double] {
let average = self.averaged()
return isEmpty ? [] : map{ pow(Double(\$0) - average, 2)}
}
/// Returns the variance of the Array
func variance() -> Double {
return self.squaredDeviations().averaged()
}
}
`````` | 685 | 2,774 | {"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-2016-50 | longest | en | 0.825104 |
https://forums.unrealengine.com/t/error-in-order-of-the-target-points-to-be-followed-how-to-set-the-order-of-elements-in-an-array/126518 | 1,620,428,689,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00283.warc.gz | 298,759,672 | 6,540 | • News
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# Error in order of the target points to be followed. How to set the order of elements in an array?
I have a project where it has 1 NPC and 3 target points:
In the project I set the NPC to go to target point 1, then to 2, then to 3.
Expected path:
Path that is occurring:
This problem did not happen out of nowhere, but first I will explain the creation of the target points.
I created the first target point, added the components it needed, and then replicated it. Selecting the first target point and pressing CONTROL + W. I positioned the second target point, and then, again selecting the first target point, I again replicated it.
The order of the path was as expected. I modified some code (none of the codes I modified related to the NPC or target points). After that I created a new target point (replicating it from the first target point) and then my problem started.
The NPC was initially going to target point 3, to the 4, 1 and then 2.
Even excluding target point 4, the order did not return to the expected (1, 2 and 3).
I inserted a print after the loop that causes the NPC to move toward the target points, so I could figure out the order of the target points, and it is making sense with the order of the path that is taking place.
Blueprint + Print:
3 to 1 to 2.
I created 2 more target points, and see how the order changes without any sense:
4 to 5 to 1 to 2 to 3.
I added another 1 target point to see if it meets a pattern:
No sense at all.
I’d like to know how to set the order of target points in the array.
Sorry for bad English, I’m not a language speaker.
I asked that same question on another site a few days ago (ai - Error in the order of the target points to be followed. How to set the order of elements in an array? - Game Development Stack Exchange), I researched order in an array and ways to set it, but I did not succeed.
You can Make Array and / or Set Array Element:
How do you gather the points atm,? If you’re using a *Get All *node, it’s not reliable for this.
Another solution is to use a spline, they’re always ordered. This seems related:
https://forums.unrealengine.com/deve…-patrol-points
I get this array through function Get All Actor of Class.
Blueprint:
And then I use that array in a loop. With that I make a path/circuit.
I looked at the link you sent, with some changes I can make the order I want. | 586 | 2,451 | {"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-2021-21 | latest | en | 0.936987 |
https://www.futureschool.com/india-curriculum/mathematics-class-ii/ | 1,685,482,192,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00195.warc.gz | 818,594,098 | 14,418 | Latest Results:
### India Class II Mathematics
# TOPIC TITLE
1 Study Plan Study plan – Class II
Objective: On completion of the course formative assessment a tailored study plan is created identifying the lessons requiring revision.
2 Using and applying number Zero and counting numbers 1 to 9
Objective: On completion of the lesson the student will be able to recognise, write and match the numeral and word for the number zero.
3 Calculation 10-100 Counting by 1, 2, 5, and 10 to 100
Objective: On completion of the lesson the student will be able to skip count to one hundred and show that one hundred equals ten times ten.
4 Calculation-multiples Multiples of 10 up to 100
Objective: On completion of the lesson the student will be able to read, write, rename and add multiples of ten.
5 Calculations The numbers 20 to 99
Objective: On completion of the lesson the student will be able to identify, name and add groups of ten, counting to ninety-nine.
6 Reasoning Simple addition up to the number 10
Objective: On completion of the lesson the student will be able to write simple number sentences that show numbers adding up to another number.
7 Reasoning Simple addition up to the number 20
Objective: On completion of the lesson the student will be able to write simple number sentences and also write addition in a vertical format using numbers up to 20.
8 Calculations Subtraction up to the number 10
Objective: On completion of the lesson the student will be able to calculate the answer to take away or subtraction number sentences up to the number 10.
9 Calculations Subtraction by Comparison
Objective: On completion of the lesson the student will be able to calculate take away or subtraction number sentences by comparison methods.
Objective: On completion of the lesson the student will be able to use place value to add two 2-digit numbers together up to a total of 99.
11 Subtraction Subtraction up to the number 99
Objective: On completion of the lesson the student will know how to find the answer to take away number sentences with bigger numbers and will also know how to write these number sentences.
12 Subtraction Subtraction with borrowing
Objective: On completion of the lesson the student will be able to record subtraction number sentences and will understand the need to trade between place value columns using the renaming method
13 Length Compare length by using informal units of measurement
Objective: On completion of the lesson the student will be able to measure objects around the student’s home and compare their length to each other. The student will also be able to compare the different ways the student measured the same object.
14 Length Using the formal unit of the centimetre to measure length and perimeter
Objective: On completion of the lesson the student will be able to measure length and perimeter in centimetres.
15 Capacity Estimate, measure and compare the capacity of containers
Objective: On completion of the lesson the student will know why and when we might need to estimate and a way to go about it.
16 Time, months Months and seasons of the year
Objective: On completion of the lesson the student will be able to: name and order the months of the year, know the number of days in each month, name the seasons in order, and know the number of months in each season.
17 Time, days of week Days of the week
Objective: On completion of the lesson the student will be able to: name the days of the week in order, recognise them in a written form, be able to answer questions about the days of the week, and have an understanding of a diary.
18 Time, duration Duration
Objective: On completion of the lesson the student will be able to estimate and measure the duration of an event using informal units, and compare and order the duration of an event using informal units.
19 Exam Exam – Class II
Objective: Exam | 803 | 3,869 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-23 | latest | en | 0.861039 |
https://access.openupresources.org/curricula/our-k5-math/grade-2/unit-8/section-b/lesson-12/student.html | 1,674,869,511,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499468.22/warc/CC-MAIN-20230127231443-20230128021443-00112.warc.gz | 98,403,850 | 5,233 | # Lesson 12 Partition Rectangles Into Squares
• Let’s partition rectangles into squares.
## Warm-up Estimation Exploration: Fill it Up
How many little squares would fill the rectangle?
Record an estimate that is:
too low
too high
## Problem 1
Build a rectangle with 8 tiles arranged in 2 rows. Use a ruler to partition the rectangle to match the rectangle you made.
## Problem 2
1. Use a ruler to partition the rectangle using the tick marks as a guide.
2. How many rows of equal-size squares did you make?
3. How many columns did you make?
4. Write 2 equations to represent the total number of equal-size squares.
## Problem 3
1. Use a ruler to partition the rectangle using the tick marks as a guide.
2. How many rows of equal-size squares did you make?
3. How many columns did you make?
4. Write 2 equations to represent the total number of equal-size squares.
## Problem 1
1. Use 12 tiles to make a rectangle. Split one of the rectangles into equal-size squares to match your rectangle made of tiles.
2. Write 2 equations to represent the total number of squares.
## Problem 2
Split this rectangle into equal-size squares.
1. Write 2 equations to represent the total number of squares.
## Problem 3
1. Split this rectangle into equal-size squares.
2. Write 2 equations to represent the total number of squares.
## Problem 1
1. Split the rectangle into equal-size squares.
2. Write 2 equations that represent the number of squares in the array. | 338 | 1,477 | {"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} | 4.40625 | 4 | CC-MAIN-2023-06 | latest | en | 0.847909 |
https://www.physicsforums.com/threads/what-are-imaginary-numbers-and-how-and-why-are-they-used-in-physics.758/ | 1,571,890,708,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00457.warc.gz | 1,024,789,325 | 16,744 | # What are imaginary numbers and how and why are they used in physics?
#### Jack
What are imaginary numbers and how and why are they used in physics?
Please could you try and make your answers as simple as possible and bear in mind that I have not even finished my GCSE course in maths yet.
#### Njorl
Imaginary numbers are represented as some real number multiplied by the number "i", which is a representation of the square root of -1. So 5.29i is an imaginary number. There are also complex numbers, made up of a real and imaginary part, like 3.5-22.6i.
The number i pops up in many relations. eix=cosx+isinx for instance.
Their uses are many. One example of their use is in damped oscillations. You might think that damped oscillations are a pretty narrow topic, but many things in nature work that way - sound, AC circuits, light in an absorptive medium etc.
That's about all I'll say. I could go on and on...but I won't.
Njorl
#### dg
Originally posted by Jack
What are imaginary numbers and how and why are they used in physics?
Please could you try and make your answers as simple as possible and bear in mind that I have not even finished my GCSE course in maths yet.
Imaginary numbers are all those numbers whose square is a negative real numbers. All this number can be represented by the product of the square root of -1 (usually written as i or j in engineering literature) and a real number. The sum of a real number (positive square) and of an imaginary number is called a complex number. This are the numbers that are used in physics.
Their use is mostly a very useful mathematical tool (this is a disputed subject since there is also who believes that they are actually the 'natural' numbers to use to describe the physical world). Their introduction allows to compact two parameters into one pretty much like using a 2D vector and vector calculus. There is a large amount of very powerful theorems that allows to simplify difficult problem with real number, passing to the complex ones.
Example of this are all phenomena involving oscillations since their complex description is way more compact than the real one -even though it has some limitations. All description of physical systems that display some kind of planar geometry or traslational simmetry can also benefit from this representation since equations get a simpler form. The use of complex number in physics received quite a boost with the introduction of quantum mechanics where complex numbers are the standard while real ones are somewhat exceptional and appear only in what is measurable.
#### Frac
Stop teasing the kid. Imaginary numbers are numbers that you give to girls that you never want to actually have a telephone conversation with.
#### Viper
You evil git, hes only 6
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• Solo and co-op problem solving | 660 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-43 | longest | en | 0.968156 |
http://www.cs.northwestern.edu/academics/courses/211/html/stl-summary.html | 1,540,011,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512504.64/warc/CC-MAIN-20181020034552-20181020060052-00510.warc.gz | 422,216,879 | 5,901 | STL Summary Home Class Info Links Lectures Newsgroup Assignments
This is a brief summary of the containers in the C++ Standard Library (once informally known as the Standard Template Library or STL). It deliberately sacrifices completeness for simplicity. Information is also available on the iterators and algorithms.
In my opinion, the best overall reference to the Standard Library is Josuttis' The C++ Standard Library.
The containers described below are:
For each operation, there is a guaranteed upper-bound on the complexity. See Section 19.2.1 of Deitel for a brief description of "Big Oh" notation.
## Vector
#include <vector>
### Constructors
vector v; Make an empty vector. O(1) vector v(n); Make a vector with N elements. O(n) vector v(n, value); Make a vector with N elements, initialized to value. O(n) vector v(begin, end); Make a vector and copy the elements from begin to end. O(n)
### Accessors
v[i] Return (or set) the I'th element. O(1) v.at(i) Return (or set) the I'th element, with bounds checking. O(1) v.size() Return current number of elements. O(1) v.empty() Return true if vector is empty. O(1) v.begin() Return random access iterator to start. O(1) v.end() Return random access iterator to end. O(1) v.front() Return the first element. O(1) v.back() Return the last element. O(1) v.capacity() Return maximum number of elements. O(1)
### Modifiers
v.push_back(value) Add value to end. O(1) (amortized) v.insert(iterator, value) Insert value at the position indexed by iterator. O(n) v.pop_back() Remove value from end. O(1) v.erase(iterator) Erase value indexed by iterator. O(n) v.erase(begin, end) Erase the elements from begin to end. O(n)
## Deque
#include <deque>
### Constructors
deque d; Make an empty deque. O(1) deque d(n); Make a deque with N elements. O(n) deque d(n, value); Make a deque with N elements, initialized to value. O(n) deque d(begin, end); Make a deque and copy the values from begin to end. O(n)
### Accessors
d[i] Return (or set) the I'th element. O(1) d.at(i) Return (or set) the I'th element, with bounds checking. O(1) d.size() Return current number of elements. O(1) d.empty() Return true if deque is empty. O(1) d.begin() Return random access iterator to start. O(1) d.end() Return random access iterator to end. O(1) d.front() Return the first element. O(1) d.back() Return the last element. O(1)
### Modifiers
d.push_front(value) Add value to front. O(1) (amortized) d.push_back(value) Add value to end. O(1) (amortized) d.insert(iterator, value) Insert value at the position indexed by iterator. O(n) d.pop_front() Remove value from front. O(1) d.pop_back() Remove value from end. O(1) d.erase(iterator) Erase value indexed by iterator. O(n) d.erase(begin, end) Erase the elements from begin to end. O(n)
## List
#include <list>
### Constructors
list l; Make an empty list. O(1) list l(begin, end); Make a list and copy the values from begin to end. O(n)
### Accessors
l.size() Return current number of elements. O(1) l.empty() Return true if list is empty. O(1) l.begin() Return bidirectional iterator to start. O(1) l.end() Return bidirectional iterator to end. O(1) l.front() Return the first element. O(1) l.back() Return the last element. O(1)
### Modifiers
l.push_front(value) Add value to front. O(1) l.push_back(value) Add value to end. O(1) l.insert(iterator, value) Insert value after position indexed by iterator. O(1) l.pop_front() Remove value from front. O(1) l.pop_back() Remove value from end. O(1) l.erase(iterator) Erase value indexed by iterator. O(1) l.erase(begin, end) Erase the elements from begin to end. O(1) l.remove(value) Remove all occurrences of value. O(n) l.remove_if(test) Remove all element that satisfy test. O(n) l.reverse() Reverse the list. O(n) l.sort() Sort the list. O(n log n) l.sort(comparison) Sort with comparison function. O(n logn) l.merge(l2) Merge sorted lists. O(n)
## Stack
In the C++ STL, a stack is a container adaptor. That means there is no primitive stack data structure. Instead, you create a stack from another container, like a list, and the stack's basic operations will be implemented using the underlying container's operations.
#include <stack>
### Constructors
stack s; Make an empty stack using a deque. O(1) stack > s; Make an empty stack using the given container. O(1)
### Accessors
s.top() Return the top element. O(1) s.size() Return current number of elements. O(1) s.empty() Return true if stack is empty. O(1)
### Modifiers
s.push(value) Push value on top. Same as push_back() for underlying container. s.pop() Pop value from top. O(1)
## Queue
In the C++ STL, a queue is a container adaptor. That means there is no primitive queue data structure. Instead, you create a queue from another container, like a list, and the queue's basic operations will be implemented using the underlying container's operations.
Don't confuse a queue with a deque or a priority_queue.
#include <queue>
### Constructors
queue q; Make an queue stack using a deque. O(1) queue > q; Make an empty queue using the given container. O(1)
### Accessors
q.front() Return the front element. O(1) q.back() Return the rear element. O(1) q.size() Return current number of elements. O(1) q.empty() Return true if queue is empty. O(1)
### Modifiers
q.push(value) Add value to end. Same for push_back() for underlying container. q.pop() Remove value from front. O(1)
## Priority Queue
In the C++ STL, a priority queue is a container adaptor. That means there is no primitive priorty queue data structure. Instead, you create a priority queue from another container, like a deque, and the priority queue's basic operations will be implemented using the underlying container's operations.
Priority queues are neither first-in-first-out nor last-in-first-out. You push objects onto the priority queue. The top element is always the "biggest" of the elements currently in the priority queue. Biggest is determined by the comparison predicate you give the priority queue constructor.
If that predicate is a "less than" type predicate, then biggest means largest.
If it is a "greater than" type predicate, then biggest means smallest.
#include <queue> -- not a typo!
### Constructors
priority_queue, comparison > q; Make an empty priority queue using the given container to hold values, and comparison to compare values. container defaults to vector and comparison defaults to less. O(1)
### Accessors
q.top() Return the "biggest" element. O(1) q.size() Return current number of elements. O(1) q.empty() Return true if priority queue is empty. O(1)
### Modifiers
q.push(value) Add value to priority queue. O(log n) q.pop() Remove biggest value. O(log n)
## Set and Multiset
Sets store objects and automatically keep them sorted and quick to find. In a `set`, there is only one copy of each object. If you try to add another equal object, nothing happens. multisets are declared and used the same as sets but allow duplicate elements.
Sets are implemented with balanced binary search trees, typically red-black trees. Thus, they provide logarithmic storage and retrieval times. Because they use search trees, sets need a comparison predicate to sort the keys. operator<() will be used by default if none is specified a construction time.
In a set, one object is considered equal to another if it is neither less than nor greater than the other object. `operator==()` is not used.
#include <set>
### Constructors
set< type, compare > s; Make an empty set. compare should be a binary predicate for ordering the set. It's optional and will default to a function that uses operator<. O(1) set< type, compare > s(begin, end); Make a set and copy the values from begin to end. O(n log n)
### Accessors
s.find(key) Return an iterator pointing to an occurrence of key in s, or s.end() if key is not in s. O(log n) s.lower_bound(key) Return an iterator pointing to the first occurrence of key in s, or s.end() if key is not in s. O(log n) s.upper_bound(key) Return an iterator pointing to the first occurrence of an item greater than key in s, or s.end() if no such item is found. O(log n) s.equal_range(key) Returns a pair of `lower_bound(key)` and `upper_bound(key)`. O(log n) s.count(key) Returns the number of items equal to key in s. O(log n) s.size() Return current number of elements. O(1) s.empty() Return true if set is empty. O(1) s.begin() Return an iterator pointing to the first element. O(1) s.end() Return an iterator pointing one past the last element. O(1)
### Modifiers
s.insert(iterator, key) Inserts key into s. iterator is taken as a "hint" but key will go in the correct position no matter what. Returns an iterator pointing to where key went. O(log n) s.insert(key) Inserts key into s and returns a pair `p` where `p.first` is an iterator pointing to where key was stored, and `p.second` is true if key was actually inserted, i.e., was not already in the set. O(log n)
## Map and Multimap
Maps can be thought of as generalized vectors. They allow map[key] = value for any kind of key, not just integers. Maps are often called associative tables in other languages, and are incredibly useful. They're even useful when the keys are integers, if you have very sparse arrays, i.e., arrays where almost all elements are one value, usually 0.
Maps are implemented as sets of pairs of keys and values. The pairs are sorted based on the keys. Thus, they provide logarithmic storage and retrieval times, but require a comparison predicate for the keys. operator<() will be used by default if none is specified a construction time.
Map types are a bit complicated because of the pairs, so it's best to use typedef to create more readable type names, like this:
```typedef map<string, double> ValueMap;
typedef ValueMap::value_type VMPair;
typedef ValueMap::iterator VMIterator;```
Definitions like the above will make find() and insert() a lot simpler:
```ValueMap vm;
vm[ "abc" ] = 2.0;
vm[ "def" ] = 3.2;
vm.insert( VMPair( "ghi", 6.7 ) );
VMIterator iter = vm.find( "def" );
if ( iter != vm.end() ) {
cout << "Value of " << iter->first " << " is " << iter->second << endl;
}```
You can just use map[key] to get the value directly without an iterator.
Warning: map[key] creates a dummy entry for key if one wasn't in the map before. Sometimes, that's just what you want. When it isn't, use find().
multimaps are like map except that they allow duplicate keys. map[key] is not defined for multimaps. Instead you must use insert() to add entry pairs, and find(), or lower_bound() and upper_bound(), or equal_range() to retrieve entry pairs.
#include <map>
### Constructors
map< key_type, value_type, key_compare > m; Make an empty map. key_compare should be a binary predicate for ordering the keys. It's optional and will default to a function that uses operator<. O(1) map< key_type, value_type, key_compare > m(begin, end); Make a map and copy the values from begin to end. O(n log n)
### Accessors
m[key] Return the value stored for key. This adds a default value if key not in map. O(log n) m.find(key) Return an iterator pointing to a key-value pair, or m.end() if key is not in map. O(log n) m.lower_bound(key) Return an iterator pointing to the first pair containing key, or m.end() if key is not in map. O(log n) m.upper_bound(key) Return an iterator pointing one past the last pair containing key, or m.end() if key is not in map. O(log n) m.equal_range(key) Return a pair containing the lower and upper bounds for key. This may be more efficient than calling those functions separately. O(log n) m.size() Return current number of elements. O(1) m.empty() Return true if map is empty. O(1) m.begin() Return an iterator pointing to the first pair. O(1) m.end() Return an iterator pointing one past the last pair. O(1)
### Modifiers
m[key] = value Store value under key in map. O(log n) m.insert(pair) Inserts the pair into the map. Equivalent to the above operation. O(log n)
## Pair
A pair is a bit like a Lisp CONS cell. It holds just two values. They can be different types. For simplicity, pairs are simple generic `struct`'s with two public data members: `first` and `second` and a simple constructor that takes the two values to store.
#include <utility>
### Constructors
pair< first_type, second_type > p( first, second ); Makes a pair. Both values must be given. O(1) pair< first_type, second_type > p( pair ); Makes a pair from another pair. O(1)
### Accessors
p.first Returns the first value in the pair. O(1) p.second Returns the second value in the pair. O(1)
### Modifiers
There are no modifiers. | 3,162 | 12,643 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-43 | longest | en | 0.71146 |
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# CS604 - Operating Systems Assignment No. 2 Upload
Assignment No. 2 Dated:Jan 12, 18
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### Replies to This Discussion
how To find Safe sequences by applying Safety algorithm.
Plzz share the assignment solution
Part 2
kindly explain kar saktay ho yh kiya kiya ha ?
or yh 2no videos ek hi tou hn ?
I solve this assignment if anyone need help contact me via whatsapp 00966582264399 .... or email id tahiritengr@gmail.com
its too easy to solve.... my answer of safe sequence is P1, P2, P3, P4, P5
https://youtu.be/T4f6b2S_Oz8?t=440
assignment solution idea to the point
System is in a safe state for reqest 4,5,3
Safe sequence is: P1 P2 P3 P4 P5
Also
System is in a safe state for request 1,2,1
Safe sequence is: P3 P4 P5 P1 P2
koi soluton post kr do plzzzzzzzzzzzzzzzzzzzzz
Good Job
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3 | 822 | 2,833 | {"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-2019-43 | longest | en | 0.750781 |
http://www.jiskha.com/display.cgi?id=1166402844 | 1,495,537,079,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00017.warc.gz | 559,663,888 | 3,902 | # math, algebra
posted by on .
i just want to re make sure that the following question is correct.
(1) List all the factors of 16
That is correct. They may not want 1 and 16 to be included, since every integer has 1 and itself as factors.
so i should take away the 1 and 16
A factor of a number N is any number that evenly divides N.
Therefore, the factors of 16 are 1,2,4, 8 and 16.
A proper divisor of a number N is any number that divides N except the number N itself. 1, 3 and 7 are the proper divisors of 21.
Proper divisors are often referred to the aliquot parts of a number. | 162 | 591 | {"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-2017-22 | latest | en | 0.96711 |
https://www.hackmath.net/en/math-problem/7273?tag_id=108 | 1,610,827,723,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00485.warc.gz | 813,032,754 | 12,650 | # Year 2018
The product of the three positive numbers is 2018. What are the numbers?
Correct result:
a = 1
b = 2
c = 1009
#### Solution:
$b=2$
$c=1009$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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When an attorney's fee is a percentage of the recovery, this represents a contingency fee arrangement.
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When an attorney's fee is a percentage of the recovery, this represents a contingency fee arrangement.
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Contingency fee
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A contingent fee or contingency fee or conditional fee is any fee for services provided where the fee is payable only if there is a favourable result.
Confirmed by selymi [4/16/2019 7:57:23 AM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 757 | 2,433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-50 | latest | en | 0.871883 |
http://mathhelpforum.com/advanced-applied-math/275647-shortest-distance.html | 1,519,561,512,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816370.72/warc/CC-MAIN-20180225110552-20180225130552-00164.warc.gz | 221,180,827 | 11,984 | 1. ## shortest distance
At a certain instant ship Q is at a distance of 4a due east of ship P. Q is moving northwards with constant speed u and P is travelling with constant speed 2u. Find the direction of P if it is to intercept Q.Find the time T, in terms of a and u, it would take P to intercept Q.If, instead, after time has elapsed, the speed of P drops to constant speed u, without changing direction, find, in terms of a the shortest distance between P and Q the distance each ship has moved from its original position to its position when they are closest together.distance in j direction should be equal
2usina*t=ut
a=30 degrees.
distance in x direction =4a
2ucosa*T=4a
T=4a/u√3 after T/2 it has moved 2ucos30(T/2) =2a east bit it has also moved up north. normally when i find the shortest distance between the ships they are horizontal with each other but here both ships have moved up different distances north so I can not.
2. ## Re: shortest distance
Your post is too cluttered for me...
3. ## Re: shortest distance
The $x$ component of distance:
$\Delta x = 4a-\left[2u (\cos \theta) T_1 + u(\cos \theta) (t-T_1)\right]$
The $y$ component of distance:
$\Delta y = ut - \left[2u (\sin \theta) T_1 + u(\sin \theta) (t-T_1)\right]$
The magnitude of the distance:
$s = \sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta )t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}$
$\dfrac{ds}{dt} = \dfrac{-u(\cos \theta) (4a-u(\cos \theta) (T_1+t))+(u-u\sin \theta)(ut-u(\sin \theta) (T_1+t))}{\sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta) t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}}$
4. ## Re: shortest distance
Originally Posted by markosheehan
At a certain instant ship Q is at a distance of 4a due east of ship P. Q is moving northwards with constant speed u and P is travelling with constant speed 2u.
So, the position of Q relative to the initial position of P, at time t after the initial time, is <4a, ut>. Taking the angle, clockwise from N, of the velocity vector of P, to be theta, P's position will be <2u cos(t), 2u sin(t)>.
(You say "constant speed". I assume that is constant velocity- that is, that P is moving in a straight line.)
Find the direction of P if it is to intercept Q.
P will intercept Q at time t, if and only if <4a, ut>= <2ut cos(theta),2ut sin(theta)>. From ut= 2ut sin(theta), sin(theta)= 1/2.
Find the time T, in terms of a and u, it would take P to intercept Q. If, instead, after time has elapsed, the speed of P drops to constant speed u, without changing direction, find, in terms of a the shortest distance between P and Q the distance each ship has moved from its original position to its position when they are closest together.distance in j direction should be equal
2usina*t=ut
a=30 degrees.
Yes. However, since you were given the initial distance apart to be "4a", you should not use "a" again to mean the angle.
distance in x direction =4a
2ucosa*T=4a
T=4a/u√3 after T/2 it has moved 2ucos30(T/2) =2a east bit it has also moved up north. normally when i find the shortest distance between the ships they are horizontal with each other but here both ships have moved up different distances north so I can not.
You must have both north-south and east-west equations satisfied. You have already found the angle so that the north-south equation is satisfied for all t (you were able to get "sin(theta)= 1/2" so "theta= 30 degrees" because u and t in "2usina*t=ut" cancel ) so it is only a matter of satisfying the east-west equation.
5. ## Re: shortest distance
Originally Posted by SlipEternal
The $x$ component of distance:
$\Delta x = 4a-\left[2u (\cos \theta) T_1 + u(\cos \theta) (t-T_1)\right]$
The $y$ component of distance:
$\Delta y = ut - \left[2u (\sin \theta) T_1 + u(\sin \theta) (t-T_1)\right]$
The magnitude of the distance:
$s = \sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta )t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}$
$\dfrac{ds}{dt} = \dfrac{-u(\cos \theta) (4a-u(\cos \theta) (T_1+t))+(u-u\sin \theta)(ut-u(\sin \theta) (T_1+t))}{\sqrt{\left(4a-u(\cos \theta) T_1 - u(\cos \theta) t\right)^2+(ut-u(\sin \theta) T_1 - u(\sin \theta) t)^2}}$
Sorry, I had to leave work. Here, $T_1$ is the time at which ship P changes speed from 2u to u. Note that inertia would prevent an instantaneous change of speed like this, but unless the problem stated that P gradually decelerated from a speed of 2u to u, you don't have to worry about that, and it will not affect the model by much.
Next, set the derivative equal to zero and solve for $t$.
And as both you and HallsofIvy found, $\theta = 30^\circ$.
This gives:
$0 = -(\cos \theta) (4a-u(\cos \theta) (T_1+t))+(1-\sin \theta)(ut-u(\sin \theta) (T_1+t))$
$0 = u(cos^2 \theta)(T_1+t)-4a\cos \theta + (ut-u(\sin \theta)(T_1+t)-ut\sin \theta + u(\sin^2\theta)(T_1+t))$
$0 = u(T_1+t)-4a\cos \theta + u(t-(T_1+t)\sin \theta-t\sin\theta)$
$0 = 2uT_1-uT_1 - 4a\sqrt{3} +4ut -2ut$
$0 = uT_1-4a\sqrt{3}+2ut$
$t = \dfrac{4a\sqrt{3}-uT_1}{2u}$
You may want to check my work. It is entirely possible I made an algebra mistake. | 1,621 | 5,094 | {"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} | 4.03125 | 4 | CC-MAIN-2018-09 | longest | en | 0.890453 |
http://article.sciencepublishinggroup.com/html/10.11648.j.ijtam.20170302.12.html | 1,611,491,213,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703548716.53/warc/CC-MAIN-20210124111006-20210124141006-00400.warc.gz | 8,270,889 | 17,961 | International Journal of Theoretical and Applied Mathematics
Volume 3, Issue 2, April 2017, Pages: 58-63
Development of a New One-Step Scheme for the Solution of Initial Value Problem (IVP) in Ordinary Differential Equations
Fadugba Sunday Emmanuel1, *, Falodun Bidemi Olumide2
1Department of Mathematics, Ekiti State University, Ado Ekiti, Nigeria
2Department of Mathematics, University of Ilorin, Ilorin, Nigeria
(F. S. Emmanuel)
*Corresponding author
Fadugba Sunday Emmanuel, Falodun Bidemi Olumide. Development of a New One-Step Scheme for the Solution of Initial Value Problem (IVP) in Ordinary Differential Equations. International Journal of Theoretical and Applied Mathematics. Vol. 3, No. 2, 2017, pp. 58-63.
doi: 10.11648/j.ijtam.20170302.12
Received: October 27, 2016; Accepted: January 16, 2017; Published: February 9, 2017
Abstract: In this paper, a new one-step scheme was developed for the solution of initial value problems of first order in ordinary differential equations. In its development a combination of interpolating function and Taylor series were used. The method was used for the solution of initial value problems emanated from real life situations. The numerical results showed that the new scheme is consistent, robust and efficient.
Keywords: Interpolating Function, Initial Value Problem, One-Step Method, Ordinary Differential Equation, Taylor Series
Contents
1. Introduction
In the past years, a large number of methods suitable for solving ordinary differential equations have been proposed. A major impetus to developing numerical procedures was the invention of calculus by Newton and Leibnitz, as this led to accurate mathematical models for physical reality, such as Sciences, Engineering, Medicine and Business. These mathematical models cannot be usually solved explicitly and numerical method to obtain approximate solutions is needed.
The approach for the solution of initial value problems in ordinary differential equations based on numerical approximations were developed before the existence of programmable computers.
A numerical method is a complete and unambiguous set of procedures for the solution of a problem, together with computable error estimates. The study and implementation of such methods is the province of numerical analysis [12].
Development of numerical integrator for the solution of initial value problems in ordinary differential equations has attracted the attention of many researchers in recent years. There are numerous methods that produce numerical approximations to solution of initial value problem in ordinary differential equation such as Euler’s method which was the oldest and simplest method originated by Leonhard Euler in 1768, Improved Euler method, Runge Kutta methods described by Carl Runge and Martin Kutta in 1895 and 1905 respectively.
There are many excellent literature and exhaustive texts on this subject that may be consulted, such as [1], [3], [4], [5], [6], [7], [8], [9], [11], [12] just to mention a few.
In this paper we consider the initial value problem of the form
(1)
and develop an algorithm which can effectively solve initial value problems in ordinary differential equation. Many researchers have solved the problem in (1). However, if the solution to (1) posseses a singularity point, a numerical integration formulae will be more effective. In another development, [10] recently discussed one-step method of Euler-Maruyama type for the solution of stochastic differential equations using varying step sizes. [2] derived a continuous linear multistep method using Hermite polynomials as basis functions.
In this paper we develop a new accurate scheme for the solution of the initial value problems in ordinary equation. The rest of the paper is organized as follows: Section Two is the development of a new scheme. Section Three consists of some basic concepts vital to the development of the new scheme. Section Four consists of numerical experiment and discussion of results. Section Five concludes the paper.
2. Development of the New Scheme
We develop a new one-step scheme in solving an initial value problem in ordinary differential equations as follows. We intend to solve problem (1) with power series polynomial such that
(2)
With integration interval of [a,b] in the form with step size given by such that
Expanding (2) above we obtain the interpolant of the form:
(3)
where and are real undetermined coefficients.
At becomes
(4)
Also,
(5)
Differentiating (4) and let yields
(6)
Similarly,
(7)
Simplifying (6) and (7) above we obtain
(8)
and
(9)
Since
(10)
and
(11)
It worth mentioning in the present paper that varies which makes our derived scheme an avenue to solve any problem whose initial condition is not limited only to .
Subtracting (4) from (5) and using (8) and (9) with yields
(12)
Using the fact that
(13)
Substituting (12) into (13) we obtain
(14)
Equation (14) is the new one-step scheme.
3. Some Basic Concepts [9]
We consider the following basic concepts which are very vital to the development of the new scheme.
3.1. Stability
A numerical method is said to be stable if the difference between the numerical solution and the exact solution can be made as small as possible, that is if there exists two positive numbers and such that the following holds.
(15)
3.2. Consistency
A numerical method with an increment function is said to be consistent with the initial value problem (1) under consideration, if
(16)
3.3. Convergence
A numerical method is said to be convergent if for all initial value problem satisfying the hypothesis of the Lipschitz condition given by
(17)
where the Lipschitz constant is denoted by . The necessary and sufficient conditions for convergence are the stability and consistency.
3.4. Round off Error
This can be defined as the error due to computing device. They arise because it is possible to represent all real numbers exactly on a finite-state machine. It can be represented mathematically as
(18)
where is the approximate solution and is the computer output. The magnitude depends in the storage and the arithmetic operation adopted. In such cases, double precision are employed to guarantee an adequate approximation.
4. Numerical Experiment and Discussion of Results
This section presents a comparative study of the new scheme and the theoretical solution with different step sizes.
4.1. Numerical Experiment
It is usually necessary to demonstrate the suitability and applicability of the newly developed scheme. In this course, we translated the algorithm of our scheme into MATLAB programming language and implemented with a problem emanated from real life situations. The performance of the method was checked by comparing its accuracy and efficiency. The efficiency was determined from the number of iterations counts and number of functions evaluations per step while the accuracy is determined by the size of the discretization error estimated from the difference between the exact solution and the numerical approximations.
4.1.1. Application of the New Scheme to Real Life Problem
Let us assume that a colony of 1000 bacteria are multiplying at the rate of per hour per individual (i.e an individual produces an average of off spring every hour, see (Hahn B. D (1997)). How many bacteria are there after hour? If we assume that the colony grows continuously and without restriction.
Solution
It is possible to model this growth with a differential equation of first order of the form:
(19)
with initial condition:
(20)
Equations (19) and (20) connote the IVP given by
(21)
where is the population size at time . The exact solution to (21) is obtained as
(22)
Take and The results generated from our scheme for h = 0.0125 and h = 0.01 are shown in Tables 1 and 2 below respectively.
4.1.2. Table of Results
Table 1. Comparative Analyses of the Results of the New Scheme with Theoretical Solution for h = 0.0125.
t-value Exact-solution Computed-solution Error 0.0125000 1010.050167084167900000 1010.049792316084100000 3.747681e-004 0.0250000 1020.201340026755800000 1020.200582957764600000 7.570690e-004 0.0375000 1030.454533953516800000 1030.453386937238100000 1.147016e-003 0.0500000 1040.810774192388200000 1040.809229467362700000 1.544725e-003 0.0625000 1051.271096376024200000 1051.269146064173400000 1.950312e-003 0.0750000 1061.836546545359600000 1061.834182650425400000 2.363895e-003 0.0875000 1072.508181254216700000 1072.505395660181200000 2.785594e-003 0.1000000 1083.287067674958700000 1083.283852144445700000 3.215531e-003 0.1125000 1094.174283705210400000 1094.170629877864900000 3.653827e-003 0.1250000 1105.170918075647700000 1105.166817466496500000 4.100609e-003 0.1375000 1116.278070458871300000 1116.273514456662300000 4.556002e-003 0.1500000 1127.496851579375700000 1127.491831444896900000 5.020134e-003 0.1625000 1138.828383324621900000 1138.822890188999500000 5.493136e-003 0.1750000 1150.273798857227300000 1150.267823720201700000 5.975137e-003 0.1875000 1161.834242728283000000 1161.827776456463700000 6.466272e-003 0.2000000 1173.510870991810200000 1173.503904316909000000 6.966675e-003 0.2125000 1185.304851320365700000 1185.297374837407900000 7.476483e-003 0.2250000 1197.217363121810200000 1197.209367287323600000 7.995834e-003 0.2375000 1209.249597657251600000 1209.241072787431700000 8.524870e-003 0.2500000 1221.402758160169900000 1221.393694429024000000 9.063731e-003 0.2625000 1233.678059956743500000 1233.668447394210500000 9.612563e-003 0.2750000 1246.076730587381000000 1246.066559077428300000 1.017151e-002 0.2875000 1258.600009929478100000 1258.589269208173800000 1.074072e-002 0.3000000 1271.249150321404800000 1271.237829974968000000 1.132035e-002 0.3125000 1284.025416687741700000 1284.013506150565900000 1.191054e-002 0.3250000 1296.930086665772000000 1296.917575218426000000 1.251145e-002 0.3375000 1309.964450733247500000 1309.951327500450600000 1.312323e-002 0.3500000 1323.129812337437000000 1323.116066286008800000 1.374605e-002 0.3625000 1336.427488025472300000 1336.413107962257300000 1.438006e-002 0.3750000 1349.858807576003100000 1349.843782145770500000 1.502543e-002 0.3875000 1363.425114132178100000 1363.409431815493000000 1.568232e-002 0.4000000 1377.127764335957400000 1377.111413447029000000 1.635089e-002 0.4125000 1390.968128463780400000 1390.951097148280700000 1.703132e-002 0.4250000 1404.947590563594100000 1404.929866796450400000 1.772377e-002 0.4375000 1419.067548593257500000 1419.049120176418500000 1.842842e-002 0.4500000 1433.329414560340600000 1433.310269120513300000 1.914544e-002 0.4625000 1447.734614663324700000 1447.714739649685000000 1.987501e-002 0.4750000 1462.284589434224900000 1462.263972116098100000 2.061732e-002 0.4875000 1476.980793882643000000 1476.959421347157100000 2.137254e-002 0.5000000 1491.824697641270600000 1491.802556790979700000 2.214085e-002 0.5125000 1506.817785112853700000 1506.794862663332400000 2.292245e-002 0.5250000 1521.961555618633800000 1521.937838096041300000 2.371752e-002 0.5375000 1537.257523548281600000 1537.232997286896800000 2.452626e-002 0.5500000 1552.707218511335900000 1552.681869651061600000 2.534886e-002 0.5625000 1568.312185490168800000 1568.285999974004000000 2.618552e-002 0.5750000 1584.073984994481600000 1584.046948565965100000 2.703643e-002 0.5875000 1599.994193217360400000 1599.966291417979800000 2.790180e-002 0.6000000 1616.074402192893300000 1616.045620359466100000 2.878183e-002 0.6125000 1632.316219955378800000 1632.286543217396200000 2.967674e-002 0.6250000 1648.721270700127700000 1648.690683977070000000 3.058672e-002 0.6375000 1665.291194945886000000 1665.259682944502200000 3.151200e-002 0.6500000 1682.027649698885900000 1681.995196910442500000 3.245279e-002 0.6625000 1698.932308618550200000 1698.898899316043500000 3.340930e-002 0.6750000 1716.006862184857900000 1715.972480420193700000 3.438176e-002 0.6875000 1733.253017867394600000 1733.217647468532300000 3.537040e-002 0.7000000 1750.672500296100200000 1750.636124864163100000 3.637543e-002 0.7125000 1768.267051433734400000 1768.229654340082300000 3.739709e-002 0.7250000 1786.038430750072600000 1785.999995133341400000 3.843562e-002 0.7375000 1803.988415397856000000 1803.948924160958800000 3.949124e-002 0.7500000 1822.118800390508100000 1822.078236197599800000 4.056419e-002 0.7625000 1840.431398781636300000 1840.389744055042700000 4.165473e-002 0.7750000 1858.928041846340900000 1858.885278763447100000 4.276308e-002 0.7875000 1877.610579264342100000 1877.566689754445900000 4.388951e-002 0.8000000 1896.480879304950100000 1896.435845046075400000 4.503426e-002 0.8125000 1915.540829013894800000 1915.494631429566000000 4.619758e-002 0.8250000 1934.792334402030100000 1934.744954658007100000 4.737974e-002 0.8375000 1954.237320635938000000 1954.188739636911600000 4.858100e-002 0.8500000 1973.877732230446100000 1973.827930616692600000 4.980161e-002 0.8625000 1993.715533243080700000 1993.664491387076400000 5.104186e-002 0.8750000 2013.752707470474900000 2013.700405473468000000 5.230200e-002 0.8875000 2033.991258646748700000 2033.937676335290700000 5.358231e-002 0.9000000 2054.433210643886000000 2054.378327566319100000 5.488308e-002 0.9125000 2075.080607674120500000 2075.024403097024800000 5.620458e-002 0.9250000 2095.935514494362500000 2095.877967398956600000 5.754710e-002 0.9375000 2117.000016612672400000 2116.941105691172700000 5.891092e-002 0.9500000 2138.276220496816200000 2138.215924148750700000 6.029635e-002 0.9625000 2159.766253784912400000 2159.704550113389400000 6.170367e-002 0.9750000 2181.472265498198800000 2181.409132306130700000 6.313319e-002 0.9875000 2203.396426255934300000 2203.331841042216600000 6.458521e-002 1.0000000 2225.540928492464900000 2225.474868448106000000 6.606004e-002
Table 2. Comparative Analyses of the Results of the New Scheme and Theoretical Solution for h = 0.01.
t-value Exact-solution Computed-solution Error 0.0100000 1008.032085504273500000 1008.031893599467600000 1.919048e-04 0.0200000 1016.128685406094900000 1016.128298513728300000 3.868924e-04 0.0300000 1024.290317890621500000 1024.289732890798600000 5.849998e-04 0.0400000 1032.517505305118200000 1032.516719040404500000 7.862647e-04 0.0500000 1040.810774192388200000 1040.809783467408600000 9.907250e-04 0.0600000 1049.170655324470500000 1049.169456905503700000 1.198419e-03 0.0700000 1057.597683736611300000 1057.596274351180000000 1.409385e-03 0.0800000 1066.092398761505100000 1066.090775097962000000 1.623664e-03 0.0900000 1074.655344063813600000 1074.653502770922800000 1.841293e-03 0.1000000 1083.287067674958700000 1083.285005361474100000 2.062313e-03 0.1100000 1091.988122028197500000 1091.985835262436000000 2.286766e-03 0.1200000 1100.759063993978800000 1100.756549303389600000 2.514691e-03 0.1300000 1109.600454915582500000 1109.597708786311600000 2.746129e-03 0.1400000 1118.512860645045300000 1118.509879521496300000 2.981124e-03 0.1500000 1127.496851579375700000 1127.493631863766300000 3.219716e-03 0.1600000 1136.553002697060300000 1136.549540748973200000 3.461948e-03 0.1700000 1145.681893594861800000 1145.678185730792600000 3.707864e-03 0.1800000 1154.884108524913700000 1154.880151017813400000 3.957507e-03 0.1900000 1164.160236432112500000 1164.156025510925500000 4.210921e-03 0.2000000 1173.510870991810200000 1173.506402841008400000 4.468151e-03 0.2100000 1182.936610647811000000 1182.931881406921300000 4.729241e-03 0.2200000 1192.438058650669500000 1192.433064413799700000 4.994237e-03 0.2300000 1202.015823096301600000 1202.010559911658400000 5.263185e-03 0.2400000 1211.670516964900800000 1211.664980834305400000 5.536131e-03 0.2500000 1221.402758160169900000 1221.396945038567500000 5.813122e-03 0.2600000 1231.213169548867700000 1231.207075343832200000 6.094205e-03 0.2700000 1241.102379000671800000 1241.095999571905600000 6.379429e-03 0.2800000 1251.071019428362400000 1251.064350587192200000 6.668841e-03 0.2900000 1261.119728828329500000 1261.112766337195600000 6.962491e-03 0.3000000 1271.249150321404800000 1271.241889893346100000 7.260428e-03 0.3100000 1281.459932194021300000 1281.452369492155400000 7.562702e-03 0.3200000 1291.752727939704100000 1291.744858576702000000 7.869363e-03 0.3300000 1302.128196300894400000 1302.120015838449300000 8.180462e-03 0.3400000 1312.587001311108500000 1312.578505259400800000 8.496052e-03 0.3500000 1323.129812337437000000 1323.120996154592600000 8.816183e-03 0.3600000 1333.757304123384500000 1333.748163214927900000 9.140908e-03 0.3700000 1344.470156832052900000 1344.460686550355400000 9.470282e-03 0.3800000 1355.269056089671900000 1355.259251733394900000 9.804356e-03 0.3900000 1366.154693029480100000 1366.144549843011600000 1.014319e-02 0.4000000 1377.127764335957400000 1377.117277508843200000 1.048683e-02 0.4100000 1388.188972289412500000 1388.178136955782700000 1.083533e-02 0.4200000 1399.339024810930800000 1399.327836048918700000 1.118876e-02 0.4300000 1410.578635507678700000 1410.567088338836800000 1.154717e-02 0.4400000 1421.908523718577500000 1421.896613107285200000 1.191061e-02 0.4500000 1433.329414560340600000 1433.317135413206400000 1.227915e-02 0.4600000 1444.842038973879400000 1444.829386139138900000 1.265283e-02 0.4700000 1456.447133771086600000 1456.434102037992500000 1.303173e-02 0.4800000 1468.145441681989700000 1468.132025780197900000 1.341590e-02 0.4900000 1479.937711402288600000 1479.923906001235100000 1.380540e-02 0.5000000 1491.824697641270600000 1491.810497349545400000 1.420029e-02 0.5100000 1503.807161170112100000 1503.792560534825700000 1.460064e-02 0.5200000 1515.885868870569100000 1515.870862376712500000 1.500649e-02 0.5300000 1528.061593784057300000 1528.046175853855400000 1.541793e-02 0.5400000 1540.335115161127300000 1540.319280153386900000 1.583501e-02 0.5500000 1552.707218511336400000 1552.690960720787400000 1.625779e-02 0.5600000 1565.178695653522000000 1565.162009310151900000 1.668634e-02 0.5700000 1577.750344766478300000 1577.733224034859900000 1.712073e-02 0.5800000 1590.422970440039800000 1590.405409418652900000 1.756102e-02 0.5900000 1603.197383726574600000 1603.179376447121100000 1.800728e-02 0.6000000 1616.074402192893800000 1616.055942619605000000 1.845957e-02 0.6100000 1629.054849972574900000 1629.035932001512900000 1.891797e-02 0.6200000 1642.139557818705700000 1642.120175277058500000 1.938254e-02 0.6300000 1655.329363157055700000 1655.309509802422800000 1.985335e-02 0.6400000 1668.625110139667400000 1668.604779659342700000 2.033048e-02 0.6500000 1682.027649698886800000 1682.006835709129700000 2.081399e-02 0.6600000 1695.537839601820500000 1695.516535647122700000 2.130395e-02 0.6700000 1709.156544505233600000 1709.134744057578500000 2.180045e-02 0.6800000 1722.884636010888000000 1722.862332469002000000 2.230354e-02 0.6900000 1736.722992721326400000 1736.700179409923500000 2.281331e-02 0.7000000 1750.672500296101600000 1750.649170465120300000 2.332983e-02 0.7100000 1764.734051508460200000 1764.710198332292300000 2.385318e-02 0.7200000 1778.908546302479000000 1778.884162879192700000 2.438342e-02 0.7300000 1793.196891850663400000 1793.171971201216400000 2.492065e-02 0.7400000 1807.600002612005300000 1807.574537679452100000 2.546493e-02 0.7500000 1822.118800390509700000 1822.092784039200100000 2.601635e-02 0.7600000 1836.754214394190500000 1836.727639408960600000 2.657499e-02 0.7700000 1851.507181294539100000 1851.480040379894700000 2.714091e-02 0.7800000 1866.378645286473300000 1866.350931065763900000 2.771422e-02 0.7900000 1881.369558148764100000 1881.341263163351400000 2.829499e-02 0.8000000 1896.480879304952200000 1896.451996013367500000 2.888329e-02 0.8100000 1911.713575884749200000 1911.684096661844700000 2.947922e-02 0.8200000 1927.068622785936000000 1927.038539922027000000 3.008286e-02 0.8300000 1942.547002736755100000 1942.516308436754100000 3.069430e-02 0.8400000 1958.149706358806700000 1958.118392741348800000 3.131362e-02 0.8500000 1973.877732230448600000 1973.845791327007600000 3.194090e-02 0.8600000 1989.732086950705000000 1989.699510704703000000 3.257625e-02 0.8700000 2005.713785203689200000 2005.680565469595900000 3.321973e-02 0.8800000 2021.823849823545300000 2021.789978365967500000 3.387146e-02 0.8900000 2038.063311859907300000 2038.028780352672800000 3.453151e-02 0.9000000 2054.433210643888700000 2054.398010669118300000 3.519997e-02 0.9100000 2070.934593854599400000 2070.898716901770500000 3.587695e-02 0.9200000 2087.568517586197700000 2087.531955051199500000 3.656253e-02 0.9300000 2104.336046415478900000 2104.298789599659400000 3.725682e-02 0.9400000 2121.238253470012800000 2121.200293579212500000 3.795989e-02 0.9500000 2138.276220496819900000 2138.237548640399800000 3.867186e-02 0.9600000 2155.451037931604800000 2155.411645121465900000 3.939281e-02 0.9700000 2172.763804968546500000 2172.723682118135000000 4.012285e-02 0.9800000 2190.215629630644300000 2190.174767553951400000 4.086208e-02 0.9900000 2207.807628840633700000 2207.766018251183600000 4.161059e-02 1.0000000 2225.540928492468500000 2225.498560002297400000 4.236849e-02
4.2. Discussion of Results
The numerical experiment shows that the new scheme provides comparable results. The differences between our scheme and the theoretical solution are negligible from a practical point of view as we can see from Tables 1 and 2 above. Also we deduce that the smaller the step size, the more accurate is the new scheme. The numbers of bacteria present after 1 one hour for h = 0.0125 and h = 0.01 were obtained as 2225.474868448106000000 and 2225.498560002297400000 respectively.
5. Conclusion
In this paper, we have developed a new scheme for the solution of initial value problems in ordinary differential equations. The new scheme was used to obtain numerical solution to real life problem. The comparative analyses of the results were also presented. The numerical results of our scheme were compared favorably with the theoretical solution. Hence the new scheme is computational efficient, robust and easy to implement.
References
1. Areo E.A. and Adeniyi R.B. (2014). Block implicit one-step method for the numerical integration of initial value problems in ordinary differential equations. International Journal of Mathematics and Statistics studies, 2(3), 4-13.
2. Aboiyar T., Luga T. and Ivorter B.V. (2015). Derivation of continuous linear multistep methods using Hermite polynomials as Basis functions. American Journal of Applied Mathematics and Statistics, 3(6), 220-225.
3. Boyce, W. E. and DiPrima, R. C. (2001). Elementary differential equation and boundary value problems. John Wiley and Sons.
4. Collatz, L. (1960). Numerical treatment of differential equations. Springer Verlag Berlin.
5. Erwin, K. (2003). Advanced engineering mathematics. Eighth Edition, Wiley Publisher.
6. Fatunla S. O. (1980): "Numerical integrators for stiff and highly oscillatory differential equations", Mathematics of Computation 34, 373-390.
7. Gilat, A. (2004). Matlab: An introduction with application. John Wiley and Sons. Gautschi W. (1961): "Numerical integration of ordinary differential equations based on trigonometric polynomials", NumerischeMathematik 3, 381-397.
8. Kayode S.J., Ganiyu A.A., and Ajiboye A.S. (2016). On one-step method of Euler-Maruyana type for solution of stochastic differential equations using varying stepsizes. Open Access Library Journal.
9. Ogunrinde R.B. and Fadugba S.E. (2012). Development of a new scheme for the solution of initial value problems in ordinary differential equations. IOSR Journal of Mathematics, 2, 24-29.
10. Wallace C. S and Gupta G. K. (1973). General Linear multistep methods to solve ordinary differential equations. The Australian Computer Journal, 5, 62-69.
11. Ying T.Y., Zurni O. and Kamarun H.M. (2014). Modified exponential rational methods for the numerical solution of first order initial value problems. Sains Malaysiana, 43(12), 1951-1959.
12. http://www.encyclopedia.com/computing/dictionaries-thesauruses-pictures-and-press-releases/numerical-methods.
Contents 1. 2. 3. 3.1. 3.2. 3.3. 3.4. 4. 4.1. 4.2. 5.
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https://www.thestudentroom.co.uk/showthread.php?t=5035934 | 1,531,913,216,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590127.2/warc/CC-MAIN-20180718095959-20180718115959-00319.warc.gz | 991,881,236 | 40,434 | You are Here: Home >< GCSEs
1. how do you find the coordinate of the stationary points of y=2x+3 / 8/x ?
2. (Original post by unicorngirl15)
how do you find the coordinate of the stationary points of y=2x+3 / 8/x ?
What have you tried to do so far? What do you normally do to find a turning point/stationary point? xx
3. When a point is stationary, its gradient is going to equal zero.
Therefore to find the x coordinate, you need to find the derivative (dy/dx) of the equation, set it equal to zero, then solve.
After that, sub the x back into the equation to get the y.
If you have any other questions just ask
4. (Original post by AnnaBanana2000)
What have you tried to do so far? What do you normally do to find a turning point/stationary point? xx
i diffrentiated 8/x to get -8x to the power of -2. But i am not sure what to do next ?
5. (Original post by unicorngirl15)
i diffrentiated 8/x to get -8x to the power of -2. But i am not sure what to do next ?
Ok, well stationary points are when the gradient is equal to 0. You’ve just worked out the gradient function, so how do you think we can get x? xx
6. (Original post by AnnaBanana2000)
Ok, well stationary points are when the gradient is equal to 0. You’ve just worked out the gradient function, so how do you think we can get x? xx
Is it y=2* (-8x)to the power of -2 ?
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https://physics.stackexchange.com/questions/482080/mobius-invariance-of-the-worldsheet-3-point-function | 1,716,069,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00785.warc.gz | 415,290,185 | 39,169 | # Mobius invariance of the worldsheet 3-point function
Consider the CFT that corresponds to a gauge-fixed closed bosonic string.
Ground level string states are described by vertex operators such as $$V(p) = :\exp(i p_{\mu} X^{\mu}(z, \bar{z})):$$ which are conformal primaries with weight $$h = \bar{h} = \frac{\alpha'}{4} p^2.$$
The physical states of the strings must have $$h = \bar{h} = 1$$, therefore, the physical ground state is the tachyon.
Consider the 3-point function of three ground state operators:
$$G_{p_1, p_2, p_3}(z_1, z_2, z_3; \bar{z}_1, \bar{z}_2, \bar{z}_3) = \left< V(p_1)(z_1, \bar{z}_1) V(p_2)(z_2, \bar{z}_2) V(p_3)(z_3, \bar{z}_3) \right>.$$
Because we're dealing with a free quantum field theory, it isn't hard to calculate this function exactly. Afaik this is called the "Coulomb gas" representation, and the expression is
$$G_{p_1, p_2, p_3}(z_1, z_2, z_3; \bar{z}_1, \bar{z}_2, \bar{z}_3) = |z_{12}|^{\alpha' p_1 \cdot p_2} \cdot |z_{13}|^{\alpha' p_1 \cdot p_3} \cdot |z_{23}|^{\alpha' p_2 \cdot p_3},$$
where $$z_{ij} = z_i - z_j = -z_{ji}$$, and $$|z_{ij}| = (z_{ij} \cdot \bar{z}_{ij})^{1/2} = |z_{ji}|$$.
However, I expect a general 3-point function to be completely fixed by global conformal symmetries – the Mobius group $$SL(2, \mathbb{C})$$. The general form for three primary fields with weights $$h_i, \bar{h}_i$$ is:
$$\left< \phi_1(z_1, \bar{z}_1) \phi_2(z_2, \bar{z}_2) \phi_3(z_3, \bar{z}_3) \right> = C_{123} z_{12}^{h_3-h_1-h_2} z_{13}^{h_2-h_1-h_3} z_{23}^{h_1-h_2-h_3} \bar{z}_{12}^{\bar{h}_3-\bar{h}_1-\bar{h}_2} \bar{z}_{13}^{\bar{h}_2-\bar{h}_1-\bar{h}_3} \bar{z}_{23}^{\bar{h}_1-\bar{h}_2-\bar{h}_3},$$
where $$C_{123}$$ is the 3-point structure constant of the CFT.
For any physical string state, $$h = \bar{h} = 1$$, therefore Mobius invariance requires that
$$\left< \phi_1(z_1, \bar{z}_1) \phi_2(z_2, \bar{z}_2) \phi_3(z_3, \bar{z}_3) \right> = C_{123} z_{12}^{-1} z_{13}^{-1} z_{23}^{-1} \bar{z}_{12}^{-1} \bar{z}_{13}^{-1} \bar{z}_{23}^{-1} = C_{123} |z_{12}|^{-2} |z_{13}|^{-2} |z_{23}|^{-2}.$$
This seems to suggest that for the physical ground states participating in scattering (tachyons),
$$\forall i, j, i \neq j: \quad \alpha' p_i \cdot p_j = -2.$$
This restriction seems very odd to me, I've never seen anything like it before.
My questions are:
1. Have I missed something crucial?
2. If not, what is the origin of the constraint on the values of tachyon momenta?
• hint: momentum conservation .. May 24, 2019 at 14:18
• @Wakabaloola of course, that’s what I was missing! Three spacelike vectors of the same length that sum to zero can only be arranged such that the angle between any two is $2 \pi /3$. That totally makes sense, thank you. Now how about writing this up as an actual answer so that I can upvote and accept it? :) May 24, 2019 at 14:33
Using momentum conservation and onshell conditions, $$p_1+p_2+p_3=0, \qquad p_1^2=p_2^2=p_3^2=\frac{4}{\alpha'},$$ it follows that for any $$i\neq j\neq k$$, \begin{aligned} 2p_i\cdot p_j &= (p_i+p_j)^2-p_i^2-p_j^2 \\ &=p_k^2-p_i^2-p_j^2\\ &=-\frac{4}{\alpha'}, \end{aligned} so this relation is really just kinematics and onshell conditions. | 1,229 | 3,181 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-22 | latest | en | 0.715708 |
https://people.sc.fsu.edu/~jburkardt/py_src/locker_simulation/locker_simulation.html | 1,695,626,071,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506686.80/warc/CC-MAIN-20230925051501-20230925081501-00081.warc.gz | 510,874,279 | 2,209 | locker_simulation
locker_simulation, a Python code which simulates the locker problem, in which gym users have left their wallets in lockers; someone has scrambled all the lockers, and the gym users need a strategy that maximizes the chance that everyone will find their wallet by searching a limited number of lockers.
This puzzle is often given assuming there are 100 gym users and 100 lockers, and each user is allowed to examine no more than 50 lockers.
It might seem that there is no winning strategy. Indeed, a single user opening lockers are random has a reasonable chance of finding their wallet, which you might estimate at roughly a 50% chance.
However, the puzzle asks a more challenging question:
Is there a strategy that makes it likely that all 100 users will find their wallets?
Now it would seem that if each user does a random search, the likelyhood that they will all succeed will be something like (1/2)^100, essentially a hopeless case.
Surprisingly, assuming that the wallet scrambling was really done at random, it turns out that there is at least one strategy that offers a roughly 31% chance that all 100 users will find their wallets.
Languages:
locker_simulation is available in a MATLAB version and an Octave version and a Python version.
Related Data and codes:
python_simulation, a Python code which uses simulation to study card games, contests, and other processes which have a random element. Usually, the purpose is to try to predict the average behavior of the system over many trials.
Reference:
1. Nick Berry,
100 Prisoners Escape Puzzle,
https://datagenetics.com/blog/december42014/index.html
Posted December 2014.
Help Star Trek's Lieutenant Uhura Overcome Astronomical Odds,
Quanta Magazine, 18 August 2022,
https://www.quantamagazine.org/help-star-treks-lieutenant-uhura-in-this-probability-puzzle-20220818/
3. Wikipedia,
100 Prisoners Problem,
https://en.wikipedia.org/wiki/100_prisoners_problem
4. Peter Winkler,
Seven puzzles you think you must not have heard correctly,
https://math.dartmouth.edu/~pw/solutions.pdf
Source Code:
Last revised on 27 November 2022. | 469 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2023-40 | longest | en | 0.942898 |
https://econmentor.com/Fibonachchi-retreismenty-kriptovaliuta-i-zkp-zero-knowledge-proof-aif01ad/ | 1,696,320,749,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00159.warc.gz | 247,802,412 | 15,012 | The concept of Fibonacci retracements has been widely used in financial markets to predict potential support and resistance levels for a given asset. In the world of cryptocurrency, where volatility is high and price movements are often driven by speculation and market sentiment, the use of Fibonacci retracements has gained popularity among traders and investors.
But what exactly are Fibonacci retracements, and how do they relate to ZKP (Zero-Knowledge Proof) in the context of cryptocurrency? Fibonacci retracements are a technical analysis tool based on the mathematical sequence discovered by Leonardo Fibonacci, an Italian mathematician, in the 13th century. This sequence, which begins with 0 and 1, is created by adding the previous two numbers to form the next number in the sequence. The Fibonacci retracement levels are derived from this sequence and are used to identify potential support and resistance levels.
Zero-Knowledge Proof (ZKP) is a cryptographic protocol that allows one party to prove to another party that they have certain information, without revealing the actual information itself. In the context of cryptocurrency, ZKP has been used to enhance privacy and security by allowing users to prove ownership of certain assets or the validity of a transaction without disclosing sensitive information such as private keys or transaction details.
So how do Fibonacci retracements and ZKP relate in cryptocurrency? While they may seem unrelated at first glance, both concepts share a common goal: to identify and establish the validity of certain levels or information. Fibonacci retracements help traders and investors identify potential levels of support and resistance in the price movement of a cryptocurrency, while ZKP allows users to prove ownership or validity without disclosing sensitive information.
## Introduction:
The use of Fibonacci retracements in cryptocurrency trading has gained popularity in recent years. Fibonacci retracements are a technical analysis tool used to identify potential levels of support and resistance in a price trend. This tool is based on the Fibonacci sequence, a series of numbers in which each number is the sum of the two preceding ones. In the context of cryptocurrency trading, Fibonacci retracements can help traders make informed decisions about when to enter or exit a trade.
In addition to their use in cryptocurrency trading, Fibonacci retracements also have a relation to Zero-Knowledge Proofs (ZKPs). ZKPs are a cryptographic protocol used to prove the possession of certain information without revealing that information itself. ZKPs have been employed in various blockchain applications, including privacy enhancements and scalability solutions.
## What are Fibonacci Retracements?
Fibonacci retracements are horizontal lines that are drawn on a price chart to indicate potential levels of support and resistance. These levels are derived from the Fibonacci sequence, which starts with 0 and 1, and each subsequent number is the sum of the two preceding ones (0, 1, 1, 2, 3, 5, 8, 13, 21, etc.). The most commonly used Fibonacci retracement levels are 23.6%, 38.2%, 50%, 61.8%, and 78.6%.
Traders use Fibonacci retracements to identify possible reversal zones or areas of price correction within a larger upward or downward trend. The main idea behind this tool is that price movements often retrace a portion of the previous move before continuing in the original direction.
## The Relation to ZKP Zero-Knowledge Proof:
Although Fibonacci retracements are primarily used in technical analysis for trading purposes, their relation to ZKPs in the field of blockchain technology is an interesting concept. ZKPs allow for the verification of certain properties without revealing the entire underlying data. Similarly, Fibonacci retracements indicate potential levels of support and resistance without revealing the complete price data and analysis behind it.
Furthermore, both Fibonacci retracements and ZKPs provide a level of mathematical precision and objectivity. Fibonacci retracements are based on a well-defined mathematical sequence, while ZKPs rely on advanced cryptographic algorithms and mathematical proofs.
Applying the concept of ZKPs to Fibonacci retracements could potentially lead to new ways of analyzing and interpreting price trends in cryptocurrency markets. The combination of these two concepts may provide traders with enhanced tools for making more precise and accurate trading decisions.
## 1. Define Fibonacci retracements in the context of technical analysis in cryptocurrency.
When it comes to technical analysis in cryptocurrency trading, Fibonacci retracements are a popular tool used by traders to identify potential levels of support and resistance. Fibonacci retracements are based on the Fibonacci sequence, a mathematical sequence in which each number is the sum of the two preceding ones: 0, 1, 1, 2, 3, 5, 8, 13, 21, and so on.
In the context of technical analysis, Fibonacci retracements involve plotting horizontal lines on a price chart at specific levels based on the ratios derived from the Fibonacci sequence. The most commonly used Fibonacci retracement levels are 38.2%, 50%, and 61.8%. These levels are used to identify potential areas where the price of a cryptocurrency may retrace back to during a trend or pullback.
The 38.2% level is considered the shallowest retracement level, while the 61.8% level is the deepest. The 50% level is often referred to as the “golden ratio” and is considered a significant retracement level. Traders believe that these Fibonacci retracement levels indicate zones of price reversal or areas of potential support or resistance.
To apply Fibonacci retracements in cryptocurrency trading, a trader would typically identify a significant swing high and low on a price chart and then draw the Fibonacci retracement levels from the low to the high (or vice versa) using a technical analysis tool. The resulting retracement levels can then be used to identify potential areas to enter or exit trades based on the belief that the price may reverse or find support/resistance at these levels. | 1,175 | 6,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.929143 |
https://codebymath.com/index.php/welcome/lesson/pi-histogram | 1,670,368,109,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00743.warc.gz | 210,195,413 | 6,602 | ## Lesson goal: Histogram the digits of Pi
Previous: Arrays and histograms | Home | Next: Histogramming uniform rnd nums
In a past lesson we counted the occurrences of each digit in $\pi$. This lesson is a bit of a repeat of that lesson, but does an analysis using arrays and the creation of a professional looking histogram chart.
In this case, each digit of $\pi$ is read from the variable bigpi. Each digit is used as an index into an array called hist that keeps track of the number of times a given digit occurs in bigpi. Another array called digit is used to seed the horizontal axis of the histogram, as it is loaded with the numbers from 0 to 9, which are the possible digits in $\pi$. | 154 | 696 | {"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 | 3 | CC-MAIN-2022-49 | latest | en | 0.915913 |
http://kwiznet.com/p/takeQuiz.php?ChapterID=2944&CurriculumID=25&NQ=10&Num=1.13 | 1,558,872,415,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259126.83/warc/CC-MAIN-20190526105248-20190526131248-00332.warc.gz | 125,510,173 | 3,803 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 8 - Mathematics1.13 Real Numbers Review
Q 1: Convert 9.33586 into a rational number form.96523/99900932653/99900 Q 2: Convert 0.999........ into a non-recurring decimal.210.9 Q 3: Solve 6x + 3 < 10.x < 6/7x < 3/5x < 7/6 Q 4: 0.56, is this pure recurring decimal.YesNo Q 5: 0.123, is this mixed recurring decimal.YesNo Q 6: Express 0.64 in the form of p/q.24/2516/2516/15 Q 7: Convert 0.3525 into a rational number form.3460/9900349/990 Q 8: Convert 0.537 into a rational number form.495/266266/495 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | 263 | 796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2019-22 | latest | en | 0.78786 |
https://forums.adobe.com/thread/1925077 | 1,542,720,752,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746398.20/warc/CC-MAIN-20181120130743-20181120152743-00537.warc.gz | 619,074,476 | 36,548 | 14 Replies Latest reply on Aug 12, 2015 11:55 AM by Silly-V
# Help with cleaning up array of pageItems, please.
Hey folks.
I have a question that only loosely pertains to illustrator, as I think I'm just missing something really obvious in basic JS logic..
I've written a script to create artboards for me that fit tight to the outside of many pageItems. In the first iteration, I wrote the script to look only for groupItems because I was assuming my production artists were doing things properly. Turns out, many of them 'walk to the beat of a different drummer', if you will. So I've decided to give up on changing their habits and take this as an exercise in self improvement with regards to scripting unpredictability.
Now I'm rewriting the script to work with any pageItem. The problem I'm running into is the exclusion of a pageItem that is either- A: Completely contained within another object, or B: Shares left && || top && || right && || bottom coordinates (eg. object a has bounds 5, 5, 10, -10 and object b has bounds 5, 4, 9, -9).
So far I have a function written to sort through the pageItems and sort them into rows (within a certain tolerance). After i determine all pageItems that are in a particular row, I want to loop that row to check for any items that are completely contained within another object, or share some bounds as the example above. The problem I'm running into is how to properly and efficiently do this check.. I've done this successfully when I'm only looking for something completely contained within, like this:
```var array = [(all pageItems of given row are pushed here)]
for(a=array.length-1;a>-1;a--){
var curObj = array[a];
var L = curObj.left;
var T = curObj.top;
var R = L + curObj.width;
var B = T - curObj.height;
for(b=array.length-1;b>-1;b--){
var compareObj = array[b];
var CL = compareObj.left;
var CT = compareObj.top;
var CR = CL + compareObj.width;
var CB = CT - compareObj.height;
if(CL > L && CT < T && CR < R && CB > B){
array.splice(b,1);
}
}
}
```
This part works fine, but only because it returns false when a=b (since when a=b it's comparing the same object, so if I include >= or <=, it splices every object from the array and nothing happens).
So, my question is, how do I set up that second loop to ignore curObj? My first inclination is to simply splice the curObj before I start the second loop and set the compareObj variable. But there's a good chance that object needs to be compared later.
Sorry if that's confusing.. i've been pulling my hair out trying to wrap my head around this (probably really simple) issue.
**EDIT**
I made a quick visualization for what i'm looking to do. In the attached screenshot I want to loop through this row of objects and splice the black boxes out of the array while leaving the gray boxes.
• ###### 1. Re: Help with cleaning up array of pageItems, please.
The issue may be in Splicing, why don't you push objects into a different array?
• ###### 2. Re: Help with cleaning up array of pageItems, please.
Because after I've ascertained the "container objects" as i've come to call them (the gray boxes in the example) i want to remove everything that falls inside of these. At the end of the script, I use this array to create the artboards via docRef.fitArtboardToSelectedArt();
Once i've determined that an object is inside of another object, I want the script to forget about it all together.
• ###### 4. Re: Help with cleaning up array of pageItems, please.
I think your issue is that you had some crazy crackpot helping you out with this in the first place.
However, I was able to do something like
1. if(CL == L && CT == T && CR == R && CB == B){
2. continue
3. }
Which ignores the object when it's the same object- is that what you need?
*Of course be ware of objects having identical coords, one on top of the other precisely- that'll fail.
• ###### 5. Re: Help with cleaning up array of pageItems, please.
haha. he was no crackpot i assure you.
I think that might work.. However, i don't always want to ignore something just because it has the same bounds.. I only want to ignore the curObj from loop A.
The reason for this is, sometimes I'll have a clipping mask that is the exact same size as my "container object" (ie. the shirt body. See screenshot. the clipping mask on the left would fit directly over the shirt on the right, and the clip mask is made with the body of the shirt, so the bounds are identical.)
I still want to splice that clipping mask out of the array, otherwise i get 2 artboards in the same spot.
• ###### 6. Re: Help with cleaning up array of pageItems, please.
Hey, if the problem is artboards in the same spot, can you just make sure to not put any artboards into the same spot where a previous one already exists? This would negate the splicing issue.
• ###### 7. Re: Help with cleaning up array of pageItems, please.
that looks great, Carlos. However I'm not sure it will work for me. Because some of the objects i need the script to ignore, could be the exact same size as the selection marquee in your script. Or they could share just 1, 2 or 3 bounds with the remaining boundary contained inside. Your version only finds things completely within the compare object (selection marquee). That part I've already solved.
And since I need the script to work on any number between 5 and 10,000 pageItems in the document, i need the script to intuitively find each piece within a row that should be used to compare the rest of the pageItems to. Then I need to ignore that pageItem while I compare the rest of them, but it needs to remain in the array because the odds are, that object will later need to be compared to a different pageItem.
• ###### 8. Re: Help with cleaning up array of pageItems, please.
I had originally tried to do it this way, Vasily.. But then i ran into issues with the script trying to create too many artboards before it tried to rectify duplicate artboards.
However, I think i see what you're saying about preventing an artboard from being created if an artboard already exists with those same bounds, but I'm not sure how to work the logic of that. I would need to keep a growing list of artboard bounds (as each artboard is created, i'd push the bounds to an existing array), and compare each new possible artboard to that list to determine whether to create an artboard..
so something like...
var currentRow = [page items of current row];
var artboardBounds = [];
for(a=0;a<currentRow.length;a++){
var bounds = currentRow[a].visibleBounds;
var left = bounds[0];
var top = bounds[1];
var right = bounds[2];
var bot = bounds[3];
if(artboardBounds.length>0){
for(b=0;b<artboardBounds.length;b++){
var cbounds = artboardBounds[b];
var cleft = cbounds[0];
var ctop = cbounds[1];
var cright = cbounds[2];
var cbot = cbounds[3];
if(cleft >= left && ctop <= top && cright <= right && cbot >= bot){
}
}
}
}
sorry i stopped writing that because i got lost and it seemed really clunky. Since this method could create an artboard for each individual pageItem, then have to go back and remove that artboard after it found something that fits around it..
currently, I'm just looping through an array that has already been verified as "items i want an artboard around". And since i need to use docRef.fitArtboardToSelectedArt(); (due to the clipping mask issues we've discussed in the past) i want to minimize the amount of artboards that are created, since fit to selected art seems to app.redraw() (sort of) and takes a while.
One thought I had was to loop through the row and just find the largest item, push that into a different array, splice it from the currentRow array, then compare it's bounds to the remainder of currentRow. Anything that is contained within (or has identical bounds) will be spliced (making future comparisons faster since there will be less items to compare).
Let me poke around down that road and see what I can do.
• ###### 9. Re: Help with cleaning up array of pageItems, please.
I was thinking that at the point where you are right about to place the artboard, it checks the existing (doesn't have to be a manually-made array, could just be document.artboards)
function artboardExists(bounds){
var left = bounds[0];
var top = bounds[1];
var right = bounds[2];
var bot = bounds[3];
for(var b=0;b<document.artboards.length;b++){
var cbounds = document.artboards[b].artboardRect;
var cleft = cbounds[0];
var ctop = cbounds[1];
var cright = cbounds[2];
var cbot = cbounds[3];
if(cleft == left && ctop == top && cright == right && cbot == bot){
return true;
}
}
return false;
}
for(a=0;a<currentRow.length;a++){
var bounds = currentRow[a].visibleBounds;
if(!artboardExists(bounds)){
}
}
• ###### 10. Re: Help with cleaning up array of pageItems, please.
Again though, that only handles artboards that are exactly the same size, and since the currentRow array will have tons of pageItems only sorted by left coordinate, a lot of artboards would be created that were too small. Then i would need to go back and remove them after a larger artboard was created around the smaller one.
lets look back to the example i posted earlier with the black/gray boxes. I attached a new copy of that screenshot with numbers so it's easier to reference.
the pageItems in the currentRow array would be sorted from left to right. So as i loop through that array using the bounds of each pageItem to create an artboard, it will do whatever pageItem is farthest to the left. In the screenshot, the black box labeled 2 would get an artboard created first (since the artboard around gray box 3 doesn't exist yet). Then when the artboard for 3 is created, the artboard around 2 would need to be removed. Does that make sense?
Since it takes time to create and subsequently remove an artboard, it makes more sense to me to have the cleanest possible array before even attempting to create the artboards. That way i know that before an artboard is ever created, only the pageItems that should get an artboard are in the array. No extra artboards need be created or subsequently removed.
I want to curate the array such that 2 5 and 7 are removed before any artboards are created and my array is left simply [1, 3, 5, 7] giving the most efficient artboard creation possible. (keep in mind that the screenshot is an incredibly simplified version and each "container object" or shirt piece could have hundreds of smaller pageItems on top of it that should not get an artboard.
• ###### 11. Re: Help with cleaning up array of pageItems, please.
why why why why why why why why why on earth is artwork that protrudes outside of the bounds of a clipping mask included in the "visibleBounds" of said clipping mask.
the very nature of the clipping mask is that it makes INVISIBLE anything that's outside of that clipping mask.
• ###### 12. Re: Help with cleaning up array of pageItems, please.
Can you by any means sort the pageItems bounds array to get the biggest shape in a 'cell' ?
I think you actually said this in a previous msg here.
• ###### 13. Re: Help with cleaning up array of pageItems, please.
I'm not sure what you mean by the "biggest shape in a 'cell'".
I know we've talked about visibleBounds and clipping masks on here.. but it occurred to me that searching for the biggest item could lead to unsatisfactory results because a clipping mask may be larger than the shirt body. although i suppose that may still be fine if I am using fitArtboardToSelectedArt at the end...
idk. i've been separated from the scripting for a while because most of our production art staff was let go. so i've been back doing the grunt work for quite a while. now that i have some time to get back into this.. i feel like my brain just isn't comprehending the basic stuff anymore. every time i think i've had an epiphany, i get all the way through writing the block of code that I thought was the solution to my problem, only to realize that it either didn't solve my problem, or it was just moronic on it's face.
I need a drink.
• ###### 14. Re: Help with cleaning up array of pageItems, please.
So, what's the deal with the people not grouping things properly? That sounds like the cause of this whole issue, if you made them a script that groups everything and saves the file and command them to only save files that way, it would prevent this, right? | 2,934 | 12,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-47 | latest | en | 0.924969 |
http://www.chegg.com/homework-help/fundamental-accounting-principles-18th-edition-chapter-12-problem-5psb-solution-9780072996531 | 1,469,827,623,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831771.10/warc/CC-MAIN-20160723071031-00053-ip-10-185-27-174.ec2.internal.warc.gz | 360,665,448 | 16,652 | View more editions
# TEXTBOOK SOLUTIONS FOR Fundamental Accounting Principles 18th Edition
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Chapter: Problem:
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
Note: All entries in this problem are dated Jan. 18.
1.
(a) Cash 650,000 Equipment 617,200 Gain on Sale of Equipment 32,800 (b) Gain on Sale of Equipment 32,800 Lasure, Capital (\$32,800 x 2/5) 13,120 Ramirez, Capital (\$32,800 x 1/5) 6,560 Toney, Capital (\$32,800 x 2/5) 13,120 (c) Accounts Payable 342,600 Cash 342,600 (d) Lasure, Capital (\$300,400 + \$13,120) 313,520 Ramirez, Capital (\$195,800 + \$6,560) 202,360 Toney, Capital (\$127,000 + \$13,120) 140,120 Cash 656,000
• Chapter , Problem is solved.
Corresponding Textbook
Fundamental Accounting Principles | 18th Edition
9780072996531ISBN-13: 0072996536ISBN: | 320 | 987 | {"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-2016-30 | longest | en | 0.572332 |
https://www.physicsforums.com/threads/wave-race-help-me-start-off-the-earthquakes.160610/ | 1,685,593,668,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647614.56/warc/CC-MAIN-20230601042457-20230601072457-00631.warc.gz | 1,022,848,350 | 16,019 | # Wave race: Help me start off the earthquakes
• pugfug90
## Homework Statement
The velocity of the transverse waves produced by an earthquake is 8.9km/s while that of the longitudinal waves is 5.1km/s. A seismograph records the arrival of the transverse waves 73s before that of the longitudinal waves. How far away was the earthquakes?
## Homework Equations
http://tpub.com/content/neets/14191/img/14191_46_1.jpg [Broken]
displacement= (Initial Velocity)(Time) + (0.5)(a)(time squared)?
## The Attempt at a Solution
I wish I showed you guys the 12/13 problems out of this worksheet that I did to show that I did try and am not trying to mooch..
I scoured the internet and found some similar links..
http://www.glenbrook.k12.il.us/gbssci/phys/p163/ec/u10ec.html [Broken]
http://www.hopperinstitute.com/phys_waves.html [Broken]
My last link basically showed.. they multiplied the fastest velocity by 98s.. How did they get 98s..I don't know..
Can anyone get me started on this?
==
I've tried..
3.8km (difference in velocity displacement)=73s*x
x=52m/2 .. Then try and use displacement formula.. But I have no acceleration value, so meh..
Last edited by a moderator:
The waves move at constant speed--no need for accelerated motion, just distance = speed*time.
Try this: Call the distance D; call the time for the transverse wave to arrive T. Now apply the above speed equation for each wave and solve for D.
Sorry.. it's not clicking.. How do I find the time for the transverse wave to arrive?
You don't need the time, just call it T. Write two equations: one for the transverse wave; one for the longitudinal. You'll be able to eliminate T and solve for D.
Hint: If the time for the transverse wave is T, what's the time for the longitudinal wave?
T + 72s? :-)
===
PS, distance = speed*time is (at least in this case) the same as displacement=velocity*time ?
We haven't done much of speed at all.. So I'm more comfortable with delta d:D
===
I've done 5.1 (km/s) + T=delta d=8.9 (km/s) + T + 72s
=5.1 (km/s)=delta d=8.9 (km/s) + 72s..
That can give me 3.8 km/s=72s..
Last edited:
T + 72s?
Of course.
PS, distance = speed*time is (at least in this case) the same as displacement=velocity*time ?
Same thing.
kk
So how do I resolve 3.8 km/s = 72s = delta d?
Last edited:
I think I did my math wrong.. Ignore post#5..
Hey I think I got it! But this problem is so convoluted I'm going to forget it all meh rawr. PS, want to check my method? (I know the answer)
==
D=(8.9km*T)/s = [(5.1km)(T+72s)]/s
They both have same denominator so that cancels..
8.9km*T=5.1km*T + 367.2km*s
3.8km*T=367.2km*s
Isolate T..
T=(367.2km*s)/(3.8km)=~96.6s
That means it takes the transverse (faster), 96.6s to get there, and ~170s for the other..
I can use each's respective time and velocity to find distance.. Thanks Doc Al! I think I got it. | 823 | 2,840 | {"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} | 3.703125 | 4 | CC-MAIN-2023-23 | latest | en | 0.887075 |
https://discourse.julialang.org/t/adding-a-line-to-the-end-of-an-array/42291 | 1,653,508,438,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00143.warc.gz | 277,830,343 | 6,866 | # Adding a line to the end of an Array
I have a 18×12 Array{Any,2} called “stock”. I also have a 12×1 vector called “flow”.
My aim is to add the content of flow to the end of stock (essentially adding a new row to stock).
I have tried
stock = [stock;flow']
But I get the error “MethodError: no method mathing adjoint(::String)”
How can I get the job done?
julia> a=[1 2 3;4 5 6]
2×3 Array{Int64,2}:
1 2 3
4 5 6
julia> b=[7 8 9 ]
1×3 Array{Int64,2}:
7 8 9
julia> vcat(a,b)
3×3 Array{Int64,2}:
1 2 3
4 5 6
7 8 9
You may have to transpose your flow:
vcat(stock,flow')
I think the basic problem is that flow’ does not work:
Try permutedims(flow).
3 Likes
Is this basically the same problem you posted an hour ago?
1 Like
My other question did not contain the necessary information, so I requested it deleted and then posted this update. Sorry for any inconvenience.
You can just update the thread instead. Now there are two threads about adding stuff to a matrix in the most recent thread list.
1 Like
Just to summarize the answer to your question from the different posts:
You are trying to add a 12 element vector as a row to an 18x12 matrix. You can think of this as vertically concatenating the two. Vectors in Julia are however column vectors by default, and you are looking to append a row vector, so you need to permute the dimensions:
julia> a = rand(3, 2)
3×2 Array{Float64,2}:
0.387527 0.230343
0.682296 0.255378
0.22481 0.838333
julia> b = rand(2)
2-element Array{Float64,1}:
0.24946180391946426
0.24528450122180612
julia> vcat(a, b) # doesn't work as b is 2x1 rather than 1x2
ERROR: ArgumentError: number of columns of each array must match (got (2, 1))
Stacktrace:
[1] _typed_vcat(::Type{Float64}, ::Tuple{Array{Float64,2},Array{Float64,1}}) at .\abstractarray.jl:1358
[2] typed_vcat at .\abstractarray.jl:1372 [inlined]
[3] vcat(::Array{Float64,2}, ::Array{Float64,1}) at D:\buildbot\worker\package_win64\build\usr\share\julia\stdlib\v1.4\SparseArrays\src\sparsevector.jl:1078
[4] top-level scope at REPL[114]:1
julia> vcat(a, b')
4×2 Array{Float64,2}:
0.387527 0.230343
0.682296 0.255378
0.22481 0.838333
0.249462 0.245285
vcat is the same as doing [a; b'], i.e. the array syntax with a semicolon is syntactic sugar for vcat:
julia> [a; b']
4×2 Array{Float64,2}:
0.387527 0.230343
0.682296 0.255378
0.22481 0.838333
0.249462 0.245285
Your other problemn is that your b includes non-numerical values, for which ', which produces an adjoint, doesn’t work - see the discussion here. You therefore need permutedims to get your vector into a row vector. | 887 | 2,612 | {"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-2022-21 | latest | en | 0.824453 |
https://www.toppr.com/ask/question/define-electrical-circuit/ | 1,722,728,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640380725.7/warc/CC-MAIN-20240803214957-20240804004957-00792.warc.gz | 814,115,215 | 7,785 | 0
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Question
# Define electrical circuit.Electrical circuit is a path of electrical components through which the electric current flows.Electrical circuit is a closed path of electrical components through which the electric current flows.Electrical circuit is a combination of components which generate electromotive force.Electrical circuit is a system which stores electrical energy.
A
Electrical circuit is a closed path of electrical components through which the electric current flows.
B
Electrical circuit is a combination of components which generate electromotive force.
C
Electrical circuit is a path of electrical components through which the electric current flows.
D
Electrical circuit is a system which stores electrical energy.
Solution
Verified by Toppr
#### The correct option is B Electrical circuit is a closed path of electrical components through which the electric current flows.Electrical circuit is a closed path of wires and electrical components which allows a current through it on the application of potential difference between two points in the path.
16
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(a) supplies electricity? (b) has the power to stop the flow of electric current? (c) provides a path for the flow of electric current?
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A closed path through which electricity flows is called a circuit.
View Solution | 348 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.944001 |
http://www.ehow.com/about_6704652_social-benefits-military-retirement-pay.html | 1,503,511,597,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886123312.44/warc/CC-MAIN-20170823171414-20170823191414-00066.warc.gz | 525,658,955 | 18,897 | # Social Security Benefits & Military Retirement Pay
Save
Military pensions and Social Security benefits are both calculated using complex formulas. However, for Social Security purposes, military retirement income is calculated like any other income. This means you pay payroll tax on it as if you are an employee.
## Military Retirement Pay--Final Pay
Retirement pay for those who joined the military prior to September 8, 1980, is calculated by multiplying the number of military service years military by 2.5 percent, then multiplying that percentage by their last month's service. So, if someone earned \$5,000 in their last month after serving from 1975 to 2010, their monthly retirement pay would be \$4,375 (2.5 percent x 35 years = 87.5 percent x \$5,000).
## Military Retirement Pay--High 36
The High 36 plan applies to people who joined the military between September 8, 1980 and August 1, 1986. Rather than calculate from the last month's pay, the percentage is multiplied by the average of the final 36 months. If someone who joined the Army in 1985 made \$4,000 per month in 2003, 2004, and 2005, excepting December 2005 when they got a promotion and pay increase to \$5,000 a month, their pension will be 50 percent (20 years x 2.5) of \$4,027 a month, or \$2,013 a month.
## Military Retirement Pay--Career Status Bonus
Anyone who joined the military after 1986 can choose between the High 36 and the Career Status Bonus (CSB) program. Under the CSB program, retirees can choose between their last 36 months or their highest 36 months. Someone who joined in 1988, earned \$5,000 a month in 2003 and 2004, then was promoted to a position paying \$6,000 a month in 2005, and was then demoted in 2006 back to \$5,000 a month can choose to use 2005 in his retirement calculations in 2010.
What's more, the CSB program rewards service past 20 years. Every year is calculated at 2.5 percent up to 20 years. Then, every year after that is calculated at 3.5 percent. If our retiree chooses the CSB program, which gives him an average of \$5,333 from which to calculate retirement pay, for his first 20 years of service he receives 50 percent of his pay (2.5 x 20). For the following 2 years, he receives 7 percent (3.5 x 2). His retirement is 57 percent of \$5,333, or \$3,040 a month.
## Social Security
Social Security is calculated by determining a contributor's average wage over her entire working life, up to age 65. At 65, you will receive Social Security payments based on your retired income, in addition to continuing to receive retirement income.
## Changes to Social Security Payments
Earning income after retirement affects your Social Security payments. As of 2010, you lose \$1 from every \$3 you earn over the annual earnings limit, which was \$37,680 in 2010. Our CSB program retiree above is collecting \$3,040 a month, or \$36,480 a year--below the earnings threshold. His Social Security payments are unaffected. However, someone receiving \$5,000 a month in military retirement benefits collects \$60,000 a year, \$23,520 over the earnings limit: \$23,520 / 3=\$7,840. His annual Social Security benefits are reduced by \$7,840.
## References
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Is DIY in your DNA? Become part of our maker community. | 825 | 3,346 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-34 | latest | en | 0.970663 |
https://astronomy.stackexchange.com/questions/35882/how-to-make-projection-from-altitude-and-azimuth-to-screen-with-screen-coordinat | 1,718,767,651,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00637.warc.gz | 95,899,808 | 36,531 | # How to make projection from altitude and azimuth to screen with screen coordinate system?
Thanks to @ohoh for helped translate ra and dec into alt and az. Now that I have this data(altitude and azimuth), how can I plot it in something like this, with the coordinate system going like this?
• Commented Sep 11, 2020 at 18:23
The example sky map appears to be a stereographic projection; circles of right ascension and declination are projected as circular arcs, and their intersections are projected as right angles. The example has a 195° field of view, and the eastern and western horizons are mislabeled.
Let the celestial sphere be centered at the origin with radius 1, zenith at (0, 0, 1), northern horizon at (0, 1, 0), and eastern horizon at (1, 0, 0). Assuming azimuth 0° is north: \begin{align} x &= \sin \mathrm{Az} \cos \mathrm{Alt} \\ y &= \cos \mathrm{Az} \cos \mathrm{Alt} \\ z &= \sin \mathrm{Alt} \end{align}
To project from the nadir (0, 0, -1) onto the plane z = 0: $$(x', y') = \left( \frac{x}{z + 1}, \frac{y}{z + 1} \right)$$
To fit this between (0, 0) and (2R, 2R) on a canvas with northern horizon at (R, 0) and eastern horizon at (0, R), scale by -R and translate by (RR): $$(X, Y) = (R(1 - x'), R(1 - y'))$$
For a sky map centered overhead, if north is up, east should be on the left.
• This is so cool! I'm gonna give it a try..
– uhoh
Commented Apr 21, 2020 at 6:10
• Thanks you and uhoh for helping me out with this project. I owe you big time! Commented Apr 21, 2020 at 9:13
• Can I ask you last question? how can i calculate the coordinates for the grid as in the screenshot? Commented Apr 21, 2020 at 20:41
• @maisteRR Those appear to be at intervals of 10° in declination and 1h (15°) in right ascension. Process your synthetic grid points the same way as the star data. Commented Apr 22, 2020 at 3:43
• @MikeG, thanks, but I'm a little confused. This mean, that I must get alt and az for value "previous dec val + 10" and "previous RA + 15" every time when I getting coords the star? I know, most likely I'm wrong and should do it in another way. Can you give one more hint? Commented Apr 22, 2020 at 23:06 | 651 | 2,148 | {"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": 3, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-26 | latest | en | 0.884742 |
https://www.unidata.ucar.edu/mailing_lists/archives/visad/2012/msg00053.html | 1,566,694,177,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322160.92/warc/CC-MAIN-20190825000550-20190825022550-00363.warc.gz | 995,188,316 | 10,072 | # Re: [visad] Setting different colour to different range of values
```Hello Sir,
It's working.Thanks a lot.
Birinchi
On 4/19/12, Curtis Rueden <ctrueden@xxxxxxxx> wrote:
> Hi Birinchi,
>
> Please use "reply all" to respond to the entire list, rather than only to
> me.
>
>
> I examined that the colors got set according to dependent variable as
>> 0-20(red),21-40(blue),41-60(green),61-79(yellow),80-100(magenta).I
>> want to set the colors as
>> 0-15(red),16-25(blue),26-50(green),51-70(yellow),71-100(magenta).
>>
>
> You need to take the time to understand how the color table is generated
> and structured. Otherwise you will not be able to alter it according to
>
> Here is a version (untested) that does what you ask:
>
> int lutLen = 256;
> float[][] lut = new float[3][lutLen];
> int redBlueBorder = (int) (0.16 * lutLen);
> int blueGreenBorder = (int) (0.26 * lutLen);
> int greenYellowBorder = (int) (0.51 * lutLen);
> int yellowMagentaBorder = (int) (0.71 * lutLen);
> for (int i=0; i<redBlueBorder; i++) {
> lut[0][i] = Color.red.getRed() / 255f;
> lut[1][i] = Color.red.getGreen() / 255f;
> lut[2][i] = Color.red.getBlue() / 255f;
> }
> for (int i=redBlueBorder; i<blueGreenBorder; i++) {
> lut[0][i] = Color.blue.getRed() / 255f;
> lut[1][i] = Color.blue.getGreen() / 255f;
> lut[2][i] = Color.blue.getBlue() / 255f;
> }
> for (int i=blueGreenBorder; i<greenYellowBorder; i++) {
> lut[0][i] = Color.green.getRed() / 255f;
> lut[1][i] = Color.green.getGreen() / 255f;
> lut[2][i] = Color.green.getBlue() / 255f;
> }
> for (int i=greenYellowBorder; i<yellowMagentaBorder; i++) {
> lut[0][i] = Color.yellow.getRed() / 255f;
> lut[1][i] = Color.yellow.getGreen() / 255f;
> lut[2][i] = Color.yellow.getBlue() / 255f;
> }
> for (int i=yellowMagentaBorder; i<lutLen; i++) {
> lut[0][i] = Color.magenta.getRed() / 255f;
> lut[1][i] = Color.magenta.getGreen() / 255f;
> lut[2][i] = Color.magenta.getBlue() / 255f;
> }
> control.setTable(lut);
>
> Of course, generalizing this code over N colors in a loop, and/or
> encapsulating the three array assignments to a helper method, would make it
> less verbose and perhaps easier to understand.
>
>
> I genuinely hope that I am not irritating anyone.
>>
>
> The best approach to not irritate anyone is to ask questions in a smart
> way, which means making a best effort before leaning on someone else. See
> http://www.catb.org/~esr/faqs/smart-questions.html
>
>
> HTH,
> Curtis
>
>
> On Thu, Apr 19, 2012 at 1:39 AM, birinchi dutta
> <duttabirinchi@xxxxxxxxx>wrote:
>
>> I genuinely hope that I am not irritating anyone.I have applied the
>> color table of your example in the code below --
>>
>> import java.rmi.RemoteException;
>> import java.awt.*;
>> import javax.swing.*;
>> import java.io.*;
>>
>> public class color{
>>
>> private RealType longitude, latitude, temperature;
>> private RealTupleType domain_tuple;
>> private FunctionType func_dom_temp;
>> private Set domain_set;
>> private FlatField vals_ff;
>> private DataReferenceImpl data_ref;
>> private DisplayImpl display;
>> private ScalarMap latMap, lonMap, tempMap;
>>
>> @SuppressWarnings("empty-statement")
>> public color(String []args)
>>
>> latitude = RealType.getRealType("latitude");
>> longitude = RealType.getRealType("longitude");
>> domain_tuple = new RealTupleType(latitude, longitude);
>> temperature = RealType.getRealType("temperature");
>>
>> func_dom_temp = new FunctionType( domain_tuple,temperature);
>>
>>
>> int NCOLS = 10;
>> int NROWS = 10;
>>
>> domain_set = new Linear2DSet(domain_tuple, 0,9, NROWS,
>> 0,9,NCOLS);
>> float[][] temp_vals = new float[][]{{81,100,86,3,6,23,56,66,7,31},
>> {12,10,12,44,4,21,67,99,5,56},
>> {11,13,14,5,78,43,87,1,33,5},
>> {65,4,56,23,67,87,23,5,43,8},
>> {3,5,65,76,2,99,1,47,76,23},
>> {2,3,77,54,23,87,45,12,23,4},
>> {23,56,7,87,43,12,32,1,3,77},
>> {56,3,21,12,78,54,56,43,78,99},
>> {4,5,66,43,2,12,32,43,12,34},
>> {87,4,32,12,1,75,44,33,22,88}};
>>
>> float[][] set_samples = domain_set.getSamples( true );
>> float[][] flat_samples = new float[1][NCOLS * NROWS];
>> for(int c = 0; c < NCOLS; c++)
>> for(int r = 0; r < NROWS; r++)
>> {
>> flat_samples[0][ c * NROWS + r ] = temp_vals[r][c];
>>
>> }
>>
>>
>> vals_ff = new FlatField( func_dom_temp, domain_set);
>> vals_ff.setSamples( flat_samples);
>> display = new DisplayImplJ2D("display1");
>>
>> GraphicsModeControl dispGMC = (GraphicsModeControl)
>> display.getGraphicsModeControl();
>> dispGMC.setScaleEnable(true);
>>
>> latMap = new ScalarMap( latitude, Display.YAxis );
>> lonMap = new ScalarMap( longitude, Display.XAxis );
>> tempMap = new ScalarMap( temperature, Display.RGB);
>>
>>
>> ColorControl control = (ColorControl) tempMap.getControl();
>> float[][] lut = new float[3][256];
>> Color[] colors = {
>> Color.red, Color.blue, Color.green, Color.yellow, Color.magenta
>> };
>> int index = 0;
>> for (int c=0; c<colors.length; c++) {
>> int minIndex = c * 256 / colors.length;
>> int maxIndex = (c + 1) * 256 / colors.length;
>> for (int i=minIndex; i<maxIndex; i++) {
>> lut[0][i] = colors[c].getRed() / 255f;
>> lut[1][i] = colors[c].getGreen() / 255f;
>> lut[2][i] = colors[c].getBlue() / 255f;
>> }
>> }
>> control.setTable(lut);
>>
>> data_ref = new DataReferenceImpl("data_ref");
>> data_ref.setData( vals_ff );
>>
>> LabeledColorWidget widget =
>> new LabeledColorWidget(new ColorMapWidget(tempMap, false));
>>
>> JFrame jframe = new JFrame("DISPLAY");
>> jframe.getContentPane().setLayout(new FlowLayout());
>>
>>
>> jframe.setSize(300,550);
>> jframe.setVisible(true);
>>
>> }
>>
>>
>> public static void main(String[] args)
>> {
>> new color(args);
>> }
>>
>> }
>>
>> I examined that the colors got set according to dependent variable as
>> 0-20(red),21-40(blue),41-60(green),61-79(yellow),80-100(magenta).I
>> want to set the colors as
>> 0-15(red),16-25(blue),26-50(green),51-70(yellow),71-100(magenta).
>>
>> On 4/18/12, Curtis Rueden <ctrueden@xxxxxxxx> wrote:
>> > Hi Birinchi,
>> >
>> >
>> > Thank you for the reply.Its ok..but I want to set the color according
>> >> to intervals of values of the dependent variable like red for
>> >> dependent variable values ranging from 0 to 20,blue for 21 to 40
>> >> etc.Is it possible?
>> >>
>> >
>> > Sorry, I don't understand what you want. The code I sent does exactly
>> that.
>> > The ir_radiance RealType is a dependent variable, and it is colorized in
>> > stepwise fashion based on its value, as you say.
>> >
>> > If you need more guidance, you will have to be more specific in your
>> > question, or send your code (which we can compile and run) to illustrate
>> > what you mean.
>> >
>> > -Curtis
>> >
>> >
>> > On Wed, Apr 18, 2012 at 10:54 AM, birinchi dutta
>> > <duttabirinchi@xxxxxxxxx>wrote:
>> >
>> >> Thank you for the reply.Its ok..but I want to set the color according
>> >> to intervals of values of the dependent variable like red for
>> >> dependent variable values ranging from 0 to 20,blue for 21 to 40
>> >> etc.Is it possible?
>> >>
>> >> Birinchi
>> >>
>> >> On 4/12/12, Curtis Rueden <ctrueden@xxxxxxxx> wrote:
>> >> > Hi Birinchi,
>> >> >
>> >> >
>> >> > I have a problem.I want to set 5 colours to 5 different range
>> >> > of
>> >> >> values like 0-2,2-4,etc.How is it possible in VisAD?
>> >> >>
>> >> >
>> >> > You can set the lookup table of the ColorControl to have a stepwise
>> >> > color
>> >> > table.
>> >> >
>> >> > Here is an example:
>> >> > https://gist.github.com/2368231
>> >> >
>> >> > Regards,
>> >> > Curtis
>> >> >
>> >> >
>> >> > On Wed, Apr 11, 2012 at 4:16 AM, birinchi dutta
>> >> > <duttabirinchi@xxxxxxxxx>wrote:
>> >> >
>> >> >> Hello Sir,
>> >> >>
>> >> >> I have a problem.I want to set 5 colours to 5 different range
>> of
>> >> >> values like 0-2,2-4,etc.How is it possible in VisAD? Thanking you in
>> >> >>
>> >> >> Birinchi
>> >> >>
>> >> >> _______________________________________________
>> >> >> visad mailing list
>> >> >> For list information, to unsubscribe, visit:
>> >> >> http://www.unidata.ucar.edu/mailing_lists/
>> >> >>
>> >> >>
>> >> >
>> >>
>> >
>>
>
```
• 2012 messages navigation, sorted by:
• Search the `visad` archives: | 2,722 | 8,879 | {"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-2019-35 | latest | en | 0.462948 |
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# Problem 1226. Non-zero bits in 10^n.
Solution 197535
Submitted on 28 Jan 2013
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
%% n = 0; k_correct = 1; assert(isequal(num_ones(n),k_correct))
``` k = 1 ```
2 Pass
%% n = 1; k_correct = 2; assert(isequal(num_ones(n),k_correct))
``` k = 2 ```
3 Pass
%% n = 2; k_correct = 3; assert(isequal(num_ones(n),k_correct))
``` k = 3 ```
4 Pass
%% n = 5; k_correct = 6; assert(isequal(num_ones(n),k_correct))
``` k = 6 ```
5 Pass
%% n = 10; k_correct = 11; assert(isequal(num_ones(n),k_correct))
``` k = 11 ```
6 Pass
%% n = 15; k_correct = 20; assert(isequal(num_ones(n),k_correct))
``` k = 20 ```
7 Pass
%% n = 22; k_correct = 25; assert(isequal(num_ones(n),k_correct))
``` k = 25 ```
8 Fail
%% n = 23; k_correct = 27; assert(isequal(num_ones(n),k_correct))
```Error: Assertion failed. ```
9 Fail
%% n = 45; k_correct = 53; assert(isequal(num_ones(n),k_correct))
```Error: Assertion failed. ```
10 Fail
%% n = 100; k_correct = 105; assert(isequal(num_ones(n),k_correct))
```Error: Assertion failed. ``` | 418 | 1,207 | {"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-2017-22 | longest | en | 0.429207 |
http://www.nag.com/numeric/CL/nagdoc_cl23/html/D01/d01gyc.html | 1,369,206,989,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701459211/warc/CC-MAIN-20130516105059-00077-ip-10-60-113-184.ec2.internal.warc.gz | 615,445,664 | 3,872 | d01 Chapter Contents
d01 Chapter Introduction
NAG C Library Manual
# NAG Library Function Documentnag_quad_md_numth_coeff_prime (d01gyc)
## 1 Purpose
nag_quad_md_numth_coeff_prime (d01gyc) calculates the optimal coefficients for use by nag_quad_md_numth_vec (d01gdc), for prime numbers of points.
## 2 Specification
#include #include
void nag_quad_md_numth_coeff_prime (Integer ndim, Integer npts, double vk[], NagError *fail)
## 3 Description
The Korobov (1963) procedure for calculating the optimal coefficients ${a}_{1},{a}_{2},\dots ,{a}_{n}$ for $p$-point integration over the $n$-cube ${\left[0,1\right]}^{n}$ imposes the constraint that
(1)
where $p$ is a prime number and $a$ is an adjustable argument. This argument is computed to minimize the error in the integral
$3n∫01dx1⋯∫01dxn∏i=1n 1-2xi 2,$ (2)
when computed using the number theoretic rule, and the resulting coefficients can be shown to fit the Korobov definition of optimality.
The computation for large values of $p$ is extremely time consuming (the number of elementary operations varying as ${p}^{2}$) and there is a practical upper limit to the number of points that can be used. Function nag_quad_md_numth_coeff_2prime (d01gzc) is computationally more economical in this respect but the associated error is likely to be larger.
## 4 References
Korobov N M (1963) Number Theoretic Methods in Approximate Analysis Fizmatgiz, Moscow
## 5 Arguments
1: ndimIntegerInput
On entry: $n$, the number of dimensions of the integral.
Constraint: ${\mathbf{ndim}}\ge 1$.
2: nptsIntegerInput
On entry: $p$, the number of points to be used.
Constraint: ${\mathbf{npts}}$ must be a prime number $\text{}\ge 5$.
3: vk[ndim]doubleOutput
On exit: the $n$ optimal coefficients.
4: failNagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_ACCURACY
The machine precision is insufficient to perform the computation exactly. Try reducing npts: ${\mathbf{npts}}=〈\mathit{\text{value}}〉$.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{ndim}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ndim}}\ge 1$.
On entry, ${\mathbf{npts}}=〈\mathit{\text{value}}〉$.
Constraint: npts must be a prime number.
On entry, ${\mathbf{npts}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{npts}}\ge 5$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
## 7 Accuracy
The optimal coefficients are returned as exact integers (though stored in a double array).
The time taken is approximately proportional to ${p}^{2}$ (see Section 3).
## 9 Example
This example calculates the Korobov optimal coefficients where the number of dimensions is $4$ and the number of points is $631$.
### 9.1 Program Text
Program Text (d01gyce.c)
### 9.2 Program Data
Program Data (d01gyce.d)
### 9.3 Program Results
Program Results (d01gyce.r) | 934 | 3,059 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 25, "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-2013-20 | longest | en | 0.6466 |
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# Genius - nature or nurture watch
1. I've heard another saying that genius is 98% perspiration and 2% inspiration. The rest of us can have the 98% and get to be very good, but no matter how hard you try you cannot get the inspiration.
2. (Original post by Helenia)
I've heard another saying that genius is 98% perspiration and 2% inspiration. The rest of us can have the 98% and get to be very good, but no matter how hard you try you cannot get the inspiration.
Unfortunately, some us have 2% inspiration and 0% perspiration. Not that I like to blow my own trumpet or anything
3. (Original post by Helenia)
I've heard another saying that genius is 98% perspiration and 2% inspiration. The rest of us can have the 98% and get to be very good, but no matter how hard you try you cannot get the inspiration.
That was Edison I think, and he said 1% inspiration. He obviously wasn't a very inspired individual.
4. (Original post by Yannis)
teaching him the most complicated mathematics of the time is one thing. Becoming a complete genius in the middle of India on his own is another.
Tell me what great works of mathematics he developed on his own in India?
5. (Original post by ZJuwelH)
That was Edison I think, and he said 1% inspiration. He obviously wasn't a very inspired individual.
Dammit, someone misquoted to me then
At the moment I have about 0% anything. We have an entire dissection session tomorrow on a structure I still can't understand the point of.
6. (Original post by shiny)
Tell me what great works of mathematics he developed on his own in India?
Probably a formula for the number of grains of rice in the nth portion of lamb vindaloo...
7. (Original post by Dr. Blazed)
Unfortunately, some us have 2% inspiration and 0% perspiration. Not that I like to blow my own trumpet or anything
Hehe!
8. (Original post by ZJuwelH)
Probably a formula for the number of grains of rice in the nth portion of lamb vindaloo...
He did do some brilliant work on Bernoulli numbers in India but most of his ideas as a young man were too "raw". Hardy helped to shape him into an undisputable genius. I think that's the most important characteristic of a "genius" - someone who lives a lasting contribution to humanity.
I think without that stroke of luck that brought Ramanujan to England, and a partnership with Hardy, his genius would have been lost.
9. (Original post by shiny)
Tell me what great works of mathematics he developed on his own in India?
On his own, he developed the general formula for quartic equations, studied the Sum of 1/n series (aged 15-16) Later he gave lots of important results in elliptic modular functions and developed most of the theory of Bernoulli numbers. Those were only the most memorable of his achievements on his own before Hardy invited him over. Is that good enough?
10. (Original post by shiny)
He did do some brilliant work on Bernoulli numbers in India but most of his ideas as a young man were too "raw". Hardy helped to shape him into an undisputable genius. I think that's the most important characteristic of a "genius" - someone who lives a lasting contribution to humanity.
I think without that stroke of luck that brought Ramanujan to England, and a partnership with Hardy, his genius would have been lost.
there was also this bloke called Littlewood, whom everyone seems to neglect.
11. (Original post by Yannis)
there was also this bloke called Littlewood, whom everyone seems to neglect.
Poor bloke suffered from depression for most of his life.
Maybe you're right, maybe Ramanujan was a just a born genius!
I quite like the idea of using musicians as examples of genius though, i.e. Mozart! Can't really dispute the gene factor in that instance.
12. oh yea... coming back from a concert at 7 and reciting most of it from memory on the harpsicord... i agree. definite nature and no nurture.
13. many genii have the nurture factor, but I reckon they would still be genii without the education. It's just that neither they themselves, nor the rest of the world would ever realise that they are great
14. (Original post by shiny)
Maybe you're right, maybe Ramanujan was a just a born genius! .
Maybe? Maybe?! When Hardy found him he'd discovered, by himself, about half the proofs Western mathematics had taken 2,000 years to achieve. The guy was just about the nearest thing you could get to the definition of self-contained genius.
15. he did get some help at university level it must be noted as Madras wasn't the weakest educational establishment in the world, but nevertheless I agree. I think everyone agrees on that particular example now.
16. Outside of maths and music, what other areas of life can we find born geniuses?
I heard an argument that it is only in maths and music that we can truly find genius without nurture. Apparently these are the only areas in which one can readily find examples where children can exceed the performance of adults.
17. (Original post by Leeroy)
Did anyone just see that program about whether geniuses are genetically produced or whether its down to how theyre brought up as a child?
any thoughts?
After being in a rich prep school for several months - the ultimate 'nurture' environment - my views have changed to say its probably 70/30 to nurture. However, the really clever ones are probably born like that, because education doesn't teach you how to be a genius
18. yeah i agree that the truly exceptional geniuses are born.
did anyone see that program afterwards featuring people going to the university summer schools? eg Paul the best british mathematician, sum girl who's about 12 and learns about 6 languages etc...
a lot of repetition from the first program there, but i think it shows that for the lesser 'geniuses', hard work and nurturing is actually required..
19. (Original post by Leeroy)
Did anyone just see that program about whether geniuses are genetically produced or whether its down to how theyre brought up as a child?
any thoughts?
personally i think that there is no particular answer to this...nuture as if someone has the potential then they need to be bought up in a enviroment where they can release this potential. Nature, as what they have in the first place is needed in order to reach a full potentail xx
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 1,645 | 7,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-39 | longest | en | 0.967726 |
https://docs.data.world/documentation/sql/concepts/basic/BETWEEN.html | 1,721,904,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00694.warc.gz | 175,793,963 | 7,259 | # BETWEEN
An introduction to the keyword BETWEEN.
The keyword BETWEEN returns a range of data between two values. It can be used with numbers, text strings and dates, and is most commonly found in the WHERE clause.
If we wanted to know what months the animal center took in between 900 and 1000 animals, we could write the following query:
SELECT year,
month,
animal_type,
COUNT
FROM austin_animal_center_intakes_by_month
WHERE count BETWEEN 900 AND 2000
ORDER BY month
And the results would look like:
year month animal_type COUNT
2014 5 Cat 901
2014 5 Dog 966
2015 5 Cat 1009
2015 5 Dog 988
2016 5 Cat 921
2016 5 Dog 1020
2017 5 Cat 914
2015 6 Cat 1103
2015 6 Dog 1014
2014 7 Dog 926
2014 9 Dog 917
2017 9 Dog 943
2013 10 Dog 965
Though it isn’t clearly illustrated by the last query, note that BETWEEN returns everything between two values inclusive of the specified vales.
## Exercises (Continued from previous section)
There are two ways to do these exercises. The first is to use the “Try query” links to test your queries without saving them. The second is to create a data.world project and save your queries to it. If you are reading this documentation and completing the exercises as a tutorial, you will need to create your own project to save your work. Details and instructions are in the SQL tutorial which has instructions for setting up your project and links to all the current exercises.
#### Exercise 17
Using the patient, description and value columns in the observations table, write a query that returns all the patients expected to have between one and five quality years of life left. (Use the description column and “Quality adjusted life years”).
Try query
Hint
SELECT the patient, description, and value columns FROM the observations table WHERE the description column = “Quality adjusted life years” AND the value is BETWEEN 1 AND 5 years.
See solution | 454 | 1,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.843842 |
http://www.math.sci.kobe-u.ac.jp/OpenXM/1.2.3/doc/asir-contrib/html-ja/cman-ja_159.html | 1,695,890,121,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00717.warc.gz | 67,366,403 | 1,074 | Go to the first, previous, next, last section, table of contents.
### `sm1.ecartd_isSameIdeal_h`
sm1.ecartd_isSameIdeal_h(F)
: Here, F=[II,JJ,V]. It compares two ideals II and JJ in h[0,1](D)_alg.
Example:
```
input
II=[(1-x)^2*dx+h*(1-x)]\$ JJ = [(1-x)*dx+h]\$
V=[x]\$
sm1.ecartd_isSameIdeal_h([II,JJ,V]);
```
Go to the first, previous, next, last section, table of contents. | 136 | 381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2023-40 | latest | en | 0.475867 |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problems_7 | 1,603,812,341,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00702.warc.gz | 211,182,778 | 9,146 | # 2013 AMC 12A Problems/Problems 7
This article has been proposed for deletion. The reason given is: Wrong title, wrong problem (see 2013 AMC 12A Problems/Problem 7). Sysops: Before deleting this article, please check the article discussion pages and history.
x + (2/x) = y + (2/y)
((x^2)+2)/x = ((y^2)+2)/y
If we cross multiply,
y((x^2)+2) = x((y^2)+2)
(x^2)y + 2y = x(y^2) + 2x
(x^2)y - x(y^2) = 2x - 2y
xy(x-y) = 2(x-y)
xy = 2, which is D | 175 | 451 | {"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 | longest | en | 0.788527 |
http://perplexus.info/show.php?pid=9230&cid=53538 | 1,606,278,474,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00395.warc.gz | 74,377,584 | 4,304 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Union of squares (Posted on 2014-07-09)
For what number of non-overlapping unit squares can a figure be formed whose perimeter is numerically equal to the area?
No Solution Yet Submitted by Jer Rating: 3.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
set of numbers | Comment 9 of 10 |
Adding one square offset by 1/2 unit adds 1 unit area and 3 units perimeter so it subtracts 2 units from the quantity (A-P).
Adding on a unit square like this to a 4x5 rectangle makes A=P=21; and doing so to a 3x8 rectangle makes A=P=25.
Also, adding any number of 2x1 rectangles so that the side of length 2 is fully touching the wall of any existing structure (with length 1 sides not touching) adds 2 to both A and P.
I haven't ruled out smaller odd structures, but so far the allowable number of non-overlapping unit squares is:
16, 18, and all integers even and odd greater than or equal to 20
Posted by Larry on 2014-07-10 10:57:00
Search: Search body:
Forums (2) | 295 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-50 | latest | en | 0.851601 |
https://ncatlab.org/homotopytypetheory/show/diff/quotient+set+%3E+history | 1,670,540,758,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00318.warc.gz | 462,857,927 | 6,826 | Homotopy Type Theory quotient set > history (changes)
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Contents
Whenever editing is allowed on the nLab again, this article should be ported over there.
Definition
In set theory
Let $S$ be a set and let $\equiv$ be an equivalence relation on $S$. Let us define a quotient set algebra of $S$ and $\equiv$ to be a set $A$ with a function $\iota \in A^S$ such that
$\forall a \in S. \forall b \in S. (a \equiv b) \implies (\iota(a) = \iota(b))$
A quotient set algebra homomorphism of $S$ and $\equiv$ is a function $f \in B^A$ between two quotient set algebras $A$ and $B$ such that
$\forall a \in S. f(\iota_A(a)) = \iota_B(a)$
The category of quotient set algebras of $S$ and $\equiv$ is the category $QSA(S, \equiv)$ whose objects $Ob(QSA(S, \equiv))$ are quotient set algebras of $S$ and $\equiv$ and whose morphisms $Mor(QSA(S, \equiv))$ are quotient set algebra homomorphisms of $S$ and $\equiv$. The quotient set of $S$ and $\equiv$, denoted $S/\equiv$, is defined as the initial object in the category of quotient set algebras of $S$ and $\equiv$.
In homotopy type theory
Let $(T,\equiv)$ be a setoid. The quotient set $T/\equiv$ is inductively generated by the following:
• a function $\iota: T \to T/\equiv$
• a family of dependent terms
$a:T, b:T \vdash eq_{T/\equiv}(a, b): (a \equiv b) \to (\iota(a) =_{T/\equiv} \iota(b))$
• a family of dependent terms
$a:T/\equiv, b:T/\equiv \vdash \tau(a, b): isProp(a =_{T/\equiv} b)$ | 473 | 1,513 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 33, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-49 | latest | en | 0.814246 |
https://www.educator.com/mathematics/trigonometry/murray/pythagorean-identity.php | 1,721,805,048,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518157.20/warc/CC-MAIN-20240724045402-20240724075402-00795.warc.gz | 630,679,450 | 120,110 | Professor Murray
Pythagorean Identity
Slide Duration:
Section 1: Trigonometric Functions
Angles
39m 5s
Intro
0:00
Degrees
0:22
Circle is 360 Degrees
0:48
Splitting a Circle
1:13
2:08
2:31
2:52
Half-Circle and Right Angle
4:00
6:24
6:52
Coterminal, Complementary, Supplementary Angles
7:23
Coterminal Angles
7:30
Complementary Angles
9:40
Supplementary Angles
10:08
Example 1: Dividing a Circle
10:38
Example 2: Converting Between Degrees and Radians
11:56
Example 3: Quadrants and Coterminal Angles
14:18
Extra Example 1: Common Angle Conversions
-1
Extra Example 2: Quadrants and Coterminal Angles
-2
Sine and Cosine Functions
43m 16s
Intro
0:00
Sine and Cosine
0:15
Unit Circle
0:22
Coordinates on Unit Circle
1:03
Right Triangles
1:52
2:25
Master Right Triangle Formula: SOHCAHTOA
2:48
Odd Functions, Even Functions
4:40
Example: Odd Function
4:56
Example: Even Function
7:30
Example 1: Sine and Cosine
10:27
Example 2: Graphing Sine and Cosine Functions
14:39
Example 3: Right Triangle
21:40
Example 4: Odd, Even, or Neither
26:01
Extra Example 1: Right Triangle
-1
Extra Example 2: Graphing Sine and Cosine Functions
-2
Sine and Cosine Values of Special Angles
33m 5s
Intro
0:00
45-45-90 Triangle and 30-60-90 Triangle
0:08
45-45-90 Triangle
0:21
30-60-90 Triangle
2:06
Mnemonic: All Students Take Calculus (ASTC)
5:21
Using the Unit Circle
5:59
New Angles
6:21
9:43
Mnemonic: All Students Take Calculus
10:13
13:11
16:48
Example 3: All Angles and Quadrants
20:21
Extra Example 1: Convert, Quadrant, Sine/Cosine
-1
Extra Example 2: All Angles and Quadrants
-2
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D
52m 3s
Intro
0:00
Amplitude and Period of a Sine Wave
0:38
Sine Wave Graph
0:58
Amplitude: Distance from Middle to Peak
1:18
Peak: Distance from Peak to Peak
2:41
Phase Shift and Vertical Shift
4:13
Phase Shift: Distance Shifted Horizontally
4:16
Vertical Shift: Distance Shifted Vertically
6:48
Example 1: Amplitude/Period/Phase and Vertical Shift
8:04
Example 2: Amplitude/Period/Phase and Vertical Shift
17:39
Example 3: Find Sine Wave Given Attributes
25:23
Extra Example 1: Amplitude/Period/Phase and Vertical Shift
-1
Extra Example 2: Find Cosine Wave Given Attributes
-2
Tangent and Cotangent Functions
36m 4s
Intro
0:00
Tangent and Cotangent Definitions
0:21
Tangent Definition
0:25
Cotangent Definition
0:47
Master Formula: SOHCAHTOA
1:01
Mnemonic
1:16
Tangent and Cotangent Values
2:29
Remember Common Values of Sine and Cosine
2:46
90 Degrees Undefined
4:36
Slope and Menmonic: ASTC
5:47
Uses of Tangent
5:54
Example: Tangent of Angle is Slope
6:09
7:49
Example 1: Graph Tangent and Cotangent Functions
10:42
Example 2: Tangent and Cotangent of Angles
16:09
Example 3: Odd, Even, or Neither
18:56
Extra Example 1: Tangent and Cotangent of Angles
-1
Extra Example 2: Tangent and Cotangent of Angles
-2
Secant and Cosecant Functions
27m 18s
Intro
0:00
Secant and Cosecant Definitions
0:17
Secant Definition
0:18
Cosecant Definition
0:33
Example 1: Graph Secant Function
0:48
Example 2: Values of Secant and Cosecant
6:49
Example 3: Odd, Even, or Neither
12:49
Extra Example 1: Graph of Cosecant Function
-1
Extra Example 2: Values of Secant and Cosecant
-2
Inverse Trigonometric Functions
32m 58s
Intro
0:00
Arcsine Function
0:24
Restrictions between -1 and 1
0:43
Arcsine Notation
1:26
Arccosine Function
3:07
Restrictions between -1 and 1
3:36
Cosine Notation
3:53
Arctangent Function
4:30
Between -Pi/2 and Pi/2
4:44
Tangent Notation
5:02
Example 1: Domain/Range/Graph of Arcsine
5:45
Example 2: Arcsin/Arccos/Arctan Values
10:46
Example 3: Domain/Range/Graph of Arctangent
17:14
Extra Example 1: Domain/Range/Graph of Arccosine
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Computations of Inverse Trigonometric Functions
31m 8s
Intro
0:00
Inverse Trigonometric Function Domains and Ranges
0:31
Arcsine
0:41
Arccosine
1:14
Arctangent
1:41
Example 1: Arcsines of Common Values
2:44
Example 2: Odd, Even, or Neither
5:57
Example 3: Arccosines of Common Values
12:24
Extra Example 1: Arctangents of Common Values
-1
Extra Example 2: Arcsin/Arccos/Arctan Values
-2
Section 2: Trigonometric Identities
Pythagorean Identity
19m 11s
Intro
0:00
Pythagorean Identity
0:17
Pythagorean Triangle
0:27
Pythagorean Identity
0:45
Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity
1:14
Example 2: Find Angle Given Cosine and Quadrant
4:18
Example 3: Verify Trigonometric Identity
8:00
Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
-1
Extra Example 2: Find Angle Given Cosine and Quadrant
-2
Identity Tan(squared)x+1=Sec(squared)x
23m 16s
Intro
0:00
Main Formulas
0:19
Companion to Pythagorean Identity
0:27
For Cotangents and Cosecants
0:52
How to Remember
0:58
Example 1: Prove the Identity
1:40
Example 2: Given Tan Find Sec
3:42
Example 3: Prove the Identity
7:45
Extra Example 1: Prove the Identity
-1
Extra Example 2: Given Sec Find Tan
-2
52m 52s
Intro
0:00
0:09
How to Remember
0:48
Cofunction Identities
1:31
How to Remember Graphically
1:44
Where to Use Cofunction Identities
2:52
Example 1: Derive the Formula for cos(A-B)
3:08
Example 2: Use Addition and Subtraction Formulas
16:03
Example 3: Use Addition and Subtraction Formulas to Prove Identity
25:11
Extra Example 1: Use cos(A-B) and Cofunction Identities
-1
Extra Example 2: Convert to Radians and use Formulas
-2
Double Angle Formulas
29m 5s
Intro
0:00
Main Formula
0:07
How to Remember from Addition Formula
0:18
Two Other Forms
1:35
Example 1: Find Sine and Cosine of Angle using Double Angle
3:16
Example 2: Prove Trigonometric Identity using Double Angle
9:37
Example 3: Use Addition and Subtraction Formulas
12:38
Extra Example 1: Find Sine and Cosine of Angle using Double Angle
-1
Extra Example 2: Prove Trigonometric Identity using Double Angle
-2
Half-Angle Formulas
43m 55s
Intro
0:00
Main Formulas
0:09
Confusing Part
0:34
Example 1: Find Sine and Cosine of Angle using Half-Angle
0:54
Example 2: Prove Trigonometric Identity using Half-Angle
11:51
Example 3: Prove the Half-Angle Formula for Tangents
18:39
Extra Example 1: Find Sine and Cosine of Angle using Half-Angle
-1
Extra Example 2: Prove Trigonometric Identity using Half-Angle
-2
Section 3: Applications of Trigonometry
Trigonometry in Right Angles
25m 43s
Intro
0:00
Master Formula for Right Angles
0:11
SOHCAHTOA
0:15
Only for Right Triangles
1:26
Example 1: Find All Angles in a Triangle
2:19
Example 2: Find Lengths of All Sides of Triangle
7:39
Example 3: Find All Angles in a Triangle
11:00
Extra Example 1: Find All Angles in a Triangle
-1
Extra Example 2: Find Lengths of All Sides of Triangle
-2
Law of Sines
56m 40s
Intro
0:00
Law of Sines Formula
0:18
SOHCAHTOA
0:27
Any Triangle
0:59
Graphical Representation
1:25
Solving Triangle Completely
2:37
When to Use Law of Sines
2:55
ASA, SAA, SSA, AAA
2:59
SAS, SSS for Law of Cosines
7:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
8:44
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:30
Example 3: How Many Triangles Satisfy Conditions, Solve Completely
28:32
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely
-2
Law of Cosines
49m 5s
Intro
0:00
Law of Cosines Formula
0:23
Graphical Representation
0:34
Relates Sides to Angles
1:00
Any Triangle
1:20
Generalization of Pythagorean Theorem
1:32
When to Use Law of Cosines
2:26
SAS, SSS
2:30
Heron's Formula
4:49
Semiperimeter S
5:11
Example 1: How Many Triangles Satisfy Conditions, Solve Completely
5:53
Example 2: How Many Triangles Satisfy Conditions, Solve Completely
15:19
Example 3: Find Area of a Triangle Given All Side Lengths
26:33
Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
-1
Extra Example 2: Length of Third Side and Area of Triangle
-2
Finding the Area of a Triangle
27m 37s
Intro
0:00
Master Right Triangle Formula and Law of Cosines
0:19
SOHCAHTOA
0:27
Law of Cosines
1:23
Heron's Formula
2:22
Semiperimeter S
2:37
Example 1: Area of Triangle with Two Sides and One Angle
3:12
Example 2: Area of Triangle with Three Sides
6:11
Example 3: Area of Triangle with Three Sides, No Heron's Formula
8:50
Extra Example 1: Area of Triangle with Two Sides and One Angle
-1
Extra Example 2: Area of Triangle with Two Sides and One Angle
-2
Word Problems and Applications of Trigonometry
34m 25s
Intro
0:00
Formulas to Remember
0:11
SOHCAHTOA
0:15
Law of Sines
0:55
Law of Cosines
1:48
Heron's Formula
2:46
Example 1: Telephone Pole Height
4:01
Example 2: Bridge Length
7:48
Example 3: Area of Triangular Field
14:20
Extra Example 1: Kite Height
-1
Extra Example 2: Roads to a Town
-2
Vectors
46m 42s
Intro
0:00
Vector Formulas and Concepts
0:12
Vectors as Arrows
0:28
Magnitude
0:38
Direction
0:50
Drawing Vectors
1:16
Uses of Vectors: Velocity, Force
1:37
Vector Magnitude Formula
3:15
Vector Direction Formula
3:28
Vector Components
6:27
Example 1: Magnitude and Direction of Vector
8:00
Example 2: Force to a Box on a Ramp
12:25
Example 3: Plane with Wind
18:30
Extra Example 1: Components of a Vector
-1
Extra Example 2: Ship with a Current
-2
Section 4: Complex Numbers and Polar Coordinates
Polar Coordinates
1h 7m 35s
Intro
0:00
Polar Coordinates vs Rectangular/Cartesian Coordinates
0:12
Rectangular Coordinates, Cartesian Coordinates
0:23
Polar Coordinates
0:59
Converting Between Polar and Rectangular Coordinates
2:06
R
2:16
Theta
2:48
Example 1: Convert Rectangular to Polar Coordinates
6:53
Example 2: Convert Polar to Rectangular Coordinates
17:28
Example 3: Graph the Polar Equation
28:00
Extra Example 1: Convert Polar to Rectangular Coordinates
-1
Extra Example 2: Graph the Polar Equation
-2
Complex Numbers
35m 59s
Intro
0:00
Main Definition
0:07
Number i
0:23
Complex Number Form
0:33
Powers of Imaginary Number i
1:00
Repeating Pattern
1:43
Operations on Complex Numbers
3:30
3:39
Multiplying Complex Numbers
4:39
FOIL Method
5:06
Conjugation
6:29
Dividing Complex Numbers
7:34
Conjugate of Denominator
7:45
Example 1: Solve For Complex Number z
11:02
Example 2: Expand and Simplify
15:34
Example 3: Simplify the Powers of i
17:50
Extra Example 1: Simplify
-1
Extra Example 2: All Complex Numbers Satisfying Equation
-2
Polar Form of Complex Numbers
40m 43s
Intro
0:00
Polar Coordinates
0:49
Rectangular Form
0:52
Polar Form
1:25
R and Theta
1:51
Polar Form Conversion
2:27
R and Theta
2:35
Optimal Values
4:05
Euler's Formula
4:25
Multiplying Two Complex Numbers in Polar Form
6:10
Multiply r's Together and Add Exponents
6:32
Example 1: Convert Rectangular to Polar Form
7:17
Example 2: Convert Polar to Rectangular Form
13:49
Example 3: Multiply Two Complex Numbers
17:28
Extra Example 1: Convert Between Rectangular and Polar Forms
-1
Extra Example 2: Simplify Expression to Polar Form
-2
DeMoivre's Theorem
57m 37s
Intro
0:00
Introduction to DeMoivre's Theorem
0:10
n nth Roots
3:06
DeMoivre's Theorem: Finding nth Roots
3:52
Relation to Unit Circle
6:29
One nth Root for Each Value of k
7:11
Example 1: Convert to Polar Form and Use DeMoivre's Theorem
8:24
Example 2: Find Complex Eighth Roots
15:27
Example 3: Find Complex Roots
27:49
Extra Example 1: Convert to Polar Form and Use DeMoivre's Theorem
-1
Extra Example 2: Find Complex Fourth Roots
-2
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• Related Books
Pythagorean Identity
Main formulas:
• The Pythagorean theorem: The side lengths of a right triangle satisfy a2 + b2 = c2.
• The Pythagorean identity: For any angle x, we have sin2 x + cos2 x = 1.
Example 1:
Use the Pythagorean theorem to prove the Pythagorean identity.
Example 2:
If cosθ = 0.47 and θ is in the fourth quadrant, find sinθ .
Example 3:
Verify the following trigonometric identity :
1+cosθsinθ = sinθ1 − cosθ
Example 4:
Use the Pythagorean identity to prove the Pythagorean theorem.
Example 5:
If sinθ = − [5/13] and θ is in the third quadrant, find cosθ .
Pythagorean Identity
If cosθ = - 0.32 and θ is in the third quadrant, find sinθ.
• Use the Pythagorean Identity to solve for sinθ: sin2θ + cos2θ = 1
• sin2θ + (-0.32)2 = 1 ⇒ sin2θ + (0.1024) = 1 ⇒ sin2θ = 1 - 0.1024 ⇒ sin2θ = 0.8974 ⇒ sinθ = ±√{0.8974}
• sinθ = ± 0.9473 Remember the mnemonic ASTC. This tells which quadrant sine values will be positive. Sine is only positive in quadrants I and II
sinθ = - 0.9473 because θ is in quadrant III
If sinθ = - 0.47 and θ is in the fourth quadrant, find cosθ.
• Use the Pythagorean Identity to solve for cosθ: sin2θ + cos2θ = 1
• (-0.47)2 + cos2θ = 1 ⇒ (0.2209) + cos2θ = 1 ⇒ cos2θ = 1 - 0.2209 ⇒ cos2θ = 0.7791 ⇒ cosθ = ±√{0.7791}
• cosθ = ± 0.8827 Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV
cosθ = 0.8827 because θ is in quadrant IV
If cosθ = [12/13] and θ is in the second quadrant, find sinθ.
• Use the Pythagorean Identity to solve for sinθ: sin2θ + cos2θ = 1
• sin2θ + ([12/13])2 = 1 ⇒ sin2θ + ([144/169]) = 1 ⇒ sin2θ = [169/169] − [144/169] ⇒ sin2θ = [25/169] ⇒ sinθ = ±√{[25/169]}
• sinθ = ±[5/13] Remember the mnemonic ASTC. This tells which quadrant sine values will be positive. Sine is only positive in quadrants I and II
sinθ = [5/13] because θ is in quadrant II
If sinθ = [3/5] and θ is in the third quadrant, find cosθ.
• Use the Pythagorean Identity to solve for cosθ: sin2θ + cos2θ = 1
• ([3/5])2 + cos2θ = 1 ⇒ ([9/25]) + cos2θ = 1 ⇒ cos2θ = [25/25] - [9/25] ⇒ cos2θ = [16/25] ⇒ cosθ = ±√{[16/25]}
• cosθ = ±[4/5] Remember the mnemonic ASTC. This tells which quadrant cosine values will be positive. Cosine is only positive in quadrants I and IV
cosθ = − [4/5] because θ is in quadrant III
Verify the following identity: (1 + cosθ)(1 - cosθ) = sin2θ
• Try to get the left hand side to look like the right hand side because it is the more complicated side
• 1 - cos2θ = sin2θ by multiplying
sin2θ = sin2θ by the Pythagorean Identity: sin2θ + cos2θ= 1 or sin2θ = 1 - cos2θ
Verify the following identity: tanαcosα = sinα
• Try to get the left hand side to look like the right hand side because it is the more complicated side
• [(sinα)/(cosα)] · cosα = sinα because tanα = [(sinα)/(cosα)]
• [(sinαcosα)/(cosα)] = sinα by multiplying
sinα = sinα by simplifying
Verify the following identity: (1 + sinθ)(1 - sinθ) = cos2θ
• Try to get the left hand side to look like the right hand side because it is the more complicated side
• 1 - sin2θ = cos2θ by multiplying
cos2θ = cos2θ by the Pythagorean Identity: sin2θ + sin2θ = 1 or cos2θ = 1 - sin2θ
Verify the following identity: sin2θ - cos2θ = 2sin2θ - 1
• Try to get the left hand side to look like the right hand side because it is the more complicated side
• sin2θ - (1 - sin2θ) = 2sin2θ - 1 by substituting the Pythagorean Identity for cos2θ
• sin2θ - 1 + sin2θ = 2sin2θ - 1 by multiplying
2sin2θ - 1 = 2sin2θ - 1 by adding like terms
Verify the following identity: [(sinθ)/(cosθ)] + [(cosθ)/(sinθ)] = cscθsecθ
• Try to get the left hand side to look like the right hand side because it is the more complicated side
• [(sinθsinθ+ cosθcosθ)/(cosθsinθ)] = cscθsecθ by adding fractions
• [(sin2θ+ cos2θ)/(cosθsinθ)] = cscθsecθ by multiplying
• [1/(cosθsinθ)] = cscθsecθ by the Pythagorean Identity: sin2θ + cos2θ = 1
• [1/(cosθ)] · [1/(sinθ)] = cscθsecθ by separating fractions
cscθsecθ = cscθsecθ by the Reciprocal Identity
Verify the following identity: [1/(1 − sinθ)] + [1/(1 + sinθ)] = 2sec2θ
• [(1 + sinθ+ 1 − sinθ)/((1 − sinθ)(1 + sinθ))] = 2sec2θ by adding fractions
• [2/(1 − sin2θ)] = 2sec2θ by simplifying
• [2/(cos2θ)] = 2sec2θ by the Pythagorean Identity
2sec2θ = 2sec2θ by the Reciprocal Identity
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Pythagorean Identity
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Pythagorean Identity 0:17
• Pythagorean Triangle
• Pythagorean Identity
• Example 1: Use Pythagorean Theorem to Prove Pythagorean Identity 1:14
• Example 2: Find Angle Given Cosine and Quadrant 4:18
• Example 3: Verify Trigonometric Identity 8:00
• Extra Example 1: Use Pythagorean Identity to Prove Pythagorean Theorem
• Extra Example 2: Find Angle Given Cosine and Quadrant
Transcription: Pythagorean Identity
We are learning about the Pythagorean identity, what we are going to try now is to start with the Pythagorean identity and prove the Pythagorean theorem.0000
Remember what we did in the earlier example, was we started with the Pythagorean theorem and we proved the Pythagorean identity.0009
The point of this is to show that you can get from one to the other or from the other back to the first one.0015
And so that the two factor are equivalent even though one seems like geometric fact and one seems like a trigonometric fact.0022
Let us do that, remember that the Pythagorean identity says that sin2(x) + cos2(x) = 1.0033
The Pythagorean theorem, that is the theorem about right triangles, so let me set up a right triangle here.0045
I will label the sides a, b, and c, that is Pythagorean identity, now I’m going to use my SOHCAHTOA.0055
SOHCAHTOA tells us that if we have an angle here, I will this angle (x) in the corner here, the sin(x) is equal to the opposite/hypotenuse.0065
The opposite side here is b/c, the cos(x) is equal to a/c, the adjacent/hypotenuse.0090
I’m going to plug those in the Pythagorean identity because remember we are allowed to use the Pythagorean identity here.0105
That says the sin2(x) + cos2(x) = 1, if I plug those in I will get b/c2 + a/c2= 1.0112
Now I’m just going to do a little bit of algebraic manipulation b2/c2 + a2/c2 is equal to 1.0127
I’m going to multiply both sides there by c2 and that will clear my denominators on the left I got b2 + a 2 and on the right I have c2.0137
If I switch around the two terms here a2 + b2 is equal to c2.0151
That is the very familiar equation we approved the Pythagorean theorem.0160
We started out with the Pythagorean identity from trigonometry and we ended up proving the Pythagorean theorem from geometry a2 + b2=c2.0177
It is just a matter of writing down the right angle in one corner of the triangle and then working through the algebra, you end up with the Pythagorean theorem from geometry.0190
That shows that the Pythagorean identity and the Pythagorean theorem really are equivalent to each other.0200
You ought to be able to start with either one and prove the other one.0206
Let us try one more example, we are given that sin(theta) is -5/13, and (theta) is in the third quadrant and we want to find cos(theta).0000
Let me try graphing out what that might be.0010
Ok (theta) is in the third quadrant, that is down here and so its sum angle down there, I do not know exactly where it is but I will draw it down there.0022
What I’m given is that sin(theta) is -5/13 and I want to find cos(theta).0035
Well I have this Pythagorean identity that says sin2(theta) + cos2(theta) = 1.0040
I will plug in sin(theta ) that is -5/132 + cos2(theta) =10051
-5/13 when you square, the negative goes away so we get 25/169 + cos2(theta) is equal to0062
Well I’m going to have to subtract the 25/169 so I will write 1 as 169/169 then I will subtract 25/169 from both sides.0074
I got cos2(theta) is 144/169, then if I take the square root of both sides to solve for cos(theta), I get cos(theta) is equal to + or – square root of 144 is 12, square root of 169 is 13.0087
I know the my cos(theta) is equal to either positive or negative 12/13.0110
That is all I can get from the Pythagorean identity because it only told me what cos2(theta) is, I can not figure out from that whether cos(theta) is positive or negative.0116
But the problem gave us a little extra information, it says that (theta) is in the quadrant.0127
Knowing that (theta) is in the third quadrant, I looked down there and I remember that cos is equal to the x coordinate of my angle.0133
Cos is the x coordinate, remember all students take calculus, down there in the third quadrant tan are positive but nothing else is positive.0145
That means that cos is not positive, it is negative.0158
The cos(theta) is equal to -12/13 and must be the negative value because it is down in the third quadrant, that is where the x coordinate is negative.0162
The key to this problem is remembering the Pythagorean identity, sin2(theta) + cos2(theta)=1.0181
Then you plug the value you are given into the Pythagorean identity and you try to solve for cos(theta).0189
Once you work through the arithmetic you get the value for cos(theta) but you do not know if it is positive or negative.0197
Then you go over and look whether what quadrant the angle is in, it is in the quadrant and then you either remember the all students take calculus.0203
That tells you the plus or minus on the different functions or you just remember that in the third quadrant the x values are negative so the cos value has to be negative.0212
Either way you end up with cos(theta) is equal to -12/13.0222
That is the end of our set on the Pythagorean identity, this is www.educator.com.0229
Hello, this is the trigonometry lectures for educator.com and today we're going to learn about probably the single most important identity in all trigonometry which is the Pythagorean identity.0000
It says that sin2x + cos2x = 1.0011
This is known as the Pythagorean identity.0017
It takes its name from the Pythagorean theorem which you probably already heard of.0019
The Pythagorean theorem says that if you have a right triangle, very important that one of the angles be a right angle, then the side lengths satisfy a2 + b2 = c2.0024
The new fact for trigonometry class is that sin2x + cos2x = 1.0042
What we're going to learn is we work through the exercises for these lectures.0050
Is if these are really two different sides of the same coin, you should think of this as being sort of facts that come out of each other.0055
In fact, we're going to use each one of these facts to prove the other one.0064
These are really equivalent to each other.0069
Let's go ahead and start doing that.0071
In our first example, we are going to start with the Pythagorean theorem, remember that's a2 + b2 = c2.0074
We're going to try to prove the Pythagorean identity sin2x + cos2x = 1.0084
The way we'll do that is let x be an angle.0095
Let's draw x on the unit circle.0107
The reason I'm drawing it on the unit circle is because remember the definition of sine and cosine is the x and y coordinates of that angle.0112
If we draw x on the unit circle, the hypotenuse has length 1 and the x-coordinate of that point, remember, is the cos(x), and the y-coordinate is the sin(x).0124
Now, what we have here is a right triangle and we're allowed to use the Pythagorean theorem, we're given that and we're going to use that and try to prove the Pythagorean identity.0148
The Pythagorean theorem says that in a right triangle, by the Pythagorean theorem...0160
Let me draw my right triangle a little bigger, there's x, there's 1, this is cosx, this is sinx.0174
By the Pythagorean theorem, one side squared, let me write that first of all as cosine x squared plus the other side squared is equal to 12.0186
That's the length of the hypotenuse.0205
If we just do a little semantic cleaning up here, 12, of course, is just 1, cosine x squared, the common notation for that is cos2x + sin2x = 1.0207
We just derived an equation, and look this is the Pythagorean identity.0228
What we've done is we started by assuming the Pythagorean theorem and then we used the Pythagorean theorem to derive the Pythagorean identity.0246
Let's see an application of that in the next example.0256
We're given that θ is an angle whose cosine is 0.47, and θ is in the fourth quadrant.0260
We have to find sinθ.0266
Let me draw θ, θ is somewhere down there in the fourth quadrant.0272
I don't know exactly where it is but θ looks like that.0279
Here is what I know, by the Pythagorean identity, sin2θ + cos2θ = 1.0284
I'm going to fill in the one that I know, cosθ, cosθ is 0.47.0295
This is 0.472 = 1 + sin2θ.0302
Now, 0.47, that's not something I can easily find the square of, so I'll do that on my calculator.0310
0.472 = 0.2209, so that's +0.2209, sin2θ +0.2209 = 1, sin2θ = 1 - 0.2209, which is 0.7791.0317
Sinθ, if we take the square root of both sides, sinθ is equal to plus or minus the square root of 0.7791, which is approximately equal to 0.8827.0360
Now, it's plus or minus because I know that sine squared is this positive number, but I don't know whether this sine is a positive or negative.0382
Remember, sine is the y-coordinate, so the sine in the fourth quadrant is going to be negative because the y-coordinate is negative.0395
Because θ is in quadrant 4, sinθ is going to be negative, so we take the negative value, sinθ is approximately equal to -0.8827.0414
The whole key to doing this problem was to start with the Pythagorean identity sin2θ + cos2θ = 1.0446
Once you're given sine or cosine, you could plug those in and figure out the other one except that you can't figure out whether they're positive or negative.0457
Their identity doesn't tell you that so we had to get this little extra information about θ being in the fourth quadrant, that totals that the sinθ is negative and we were able to figure out that it was -0.8827.0463
Let's try another example of that.0480
We're going to verify a trigonometric identity.0483
This is a very common problem in trigonometry classes as you'll be given some kind of identity involving the trigonometric functions and you have to verify it.0486
For this one, what I want to do is start with the right hand side.0496
I'm going to label this RHS.0503
RHS stands for right-hand side.0504
The right-hand side here is equal to sinθ/(1 - cosθ).0510
Now, I'm going to do a little trick here which is very common when you have something plus something in the denominator, or something minus something in the denominator.0519
The trick is to multiply the conjugate of that thing.0529
Here I have 1 - cosθ in the denominator, I'm going to multiply by 1 + cosθ, and then, of course, I have to multiply the numerator by the same thing, 1 + cosθ).0533
The reason you do that, this is really an algebraic trick so you probably have learned about this in the algebra lectures.0547
The reason you do that is you want to take advantage of this formula, (a + b) × (a - b2.0554
That's often the way of simplifying things using that algebraic formula.0566
What we get here in the numerator is (sinθ) × (1 + cosθ), in the denominator, using this (a2 + b2) formula, we get (1 - cos2θ).0570
Now, let's remember the Pythagorean identity.0586
Pythagorean identity says sin2θ + cos2θ = 1.0590
That means 1 - cos2θ = sin2θ.0596
We can substitute that in into our work here, sinθ×(1 + cosθ).0603
The denominator, by the Pythagorean identity, turns into sin2θ.0613
We get some cancellation going on, the sine in the numerator cancels with one of the sines in the denominator leaving us just with (1 + cosθ)/sinθ in the denominator.0623
That's the same as the left-hand side that we started with.0639
We started with the right-hand side and we're able to work it all the way down and end up with the left-hand side verifying the trigonometric identity.0644
There were sort of two key steps there.0654
One was in looking at the denominator and recognizing that it was a good candidate to invoke this algebraic trick where you multiply by the conjugate.0657
If you have (a + b), you multiply by (a - b).0669
If you have (a - b), you multiply by (a + b).0672
Either way, you get to invoke this identity.0674
Here, we had (a - b), we multiplied by (a + b) and then we got to invoke the identity and get something nice on the bottom.0679
The second trick there was to remember the Pythagorean identity and notice that (1 - cos2θ) converts into sin2θ.0686
Once we did that, it was pretty to simplify it down to the left-hand side of the original identity.0697
We'll try some more examples later.0703
OR
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# Linear Regression by Hand
The basic idea behind linear regression is quite simple. In mathematical terms we want to predict a dependent variable Y using an independent variable X. By assuming that the two variables correlate in a linear fashion we can predict Y with a simple linear formula:
Linear equation by Author
(The wavy equal sign signifies “approximately”). Simply put, as soon as we know a bit about the relationship between the two coefficients, i.e. we have approximated the two coefficients **α **and β, we can (with some confidence) predict Y. Alpha α represents the intercept (value of y with f(x = 0)) and Beta β is the slope.
With the help of linear regression, we can answer a lot of questions; e.g.
• “Is the rise of sea level connected to rising temperatures?”,
• “How expensive will a house with 3 bedrooms be?”
• “How many items do we sell if we increase our marketing budget by 20%?”
#linear-regression #machine-learning-python #data-science #machine-learning #python
1594271340
## A Deep Dive into Linear Regression
Let’s begin our journey with the truth — machines never learn. What a typical machine learning algorithm does is find a mathematical equation that, when applied to a given set of training data, produces a prediction that is very close to the actual output.
Why is this not learning? Because if you change the training data or environment even slightly, the algorithm will go haywire! Not how learning works in humans. If you learned to play a video game by looking straight at the screen, you would still be a good player if the screen is slightly tilted by someone, which would not be the case in ML algorithms.
However, most of the algorithms are so complex and intimidating that it gives our mere human intelligence the feel of actual learning, effectively hiding the underlying math within. There goes a dictum that if you can implement the algorithm, you know the algorithm. This saying is lost in the dense jungle of libraries and inbuilt modules which programming languages provide, reducing us to regular programmers calling an API and strengthening further this notion of a black box. Our quest will be to unravel the mysteries of this so-called ‘black box’ which magically produces accurate predictions, detects objects, diagnoses diseases and claims to surpass human intelligence one day.
We will start with one of the not-so-complex and easy to visualize algorithm in the ML paradigm — Linear Regression. The article is divided into the following sections:
1. Need for Linear Regression
2. Visualizing Linear Regression
3. Deriving the formula for weight matrix W
4. Using the formula and performing linear regression on a real world data set
Note: Knowledge on Linear Algebra, a little bit of Calculus and Matrices are a prerequisite to understanding this article
Also, a basic understanding of python, NumPy, and Matplotlib are a must.
## 1) Need for Linear regression
Regression means predicting a real valued number from a given set of input variables. Eg. Predicting temperature based on month of the year, humidity, altitude above sea level, etc. Linear Regression would therefore mean predicting a real valued number that follows a linear trend. Linear regression is the first line of attack to discover correlations in our data.
Now, the first thing that comes to our mind when we hear the word linear is, a line.
Yes! In linear regression, we try to fit a line that best generalizes all the data points in the data set. By generalizing, we mean we try to fit a line that passes very close to all the data points.
But how do we ensure that this happens? To understand this, let’s visualize a 1-D Linear Regression. This is also called as Simple Linear Regression
#calculus #machine-learning #linear-regression-math #linear-regression #linear-regression-python #python
1592023980
## 5 Regression algorithms: Explanation & Implementation in Python
Take your current understanding and skills on machine learning algorithms to the next level with this article. What is regression analysis in simple words? How is it applied in practice for real-world problems? And what is the possible snippet of codes in Python you can use for implementation regression algorithms for various objectives? Let’s forget about boring learning stuff and talk about science and the way it works.
#linear-regression-python #linear-regression #multivariate-regression #regression #python-programming
1598352300
## Regression: Linear Regression
Machine learning algorithms are not your regular algorithms that we may be used to because they are often described by a combination of some complex statistics and mathematics. Since it is very important to understand the background of any algorithm you want to implement, this could pose a challenge to people with a non-mathematical background as the maths can sap your motivation by slowing you down.
In this article, we would be discussing linear and logistic regression and some regression techniques assuming we all have heard or even learnt about the Linear model in Mathematics class at high school. Hopefully, at the end of the article, the concept would be clearer.
**Regression Analysis **is a statistical process for estimating the relationships between the dependent variables (say Y) and one or more independent variables or predictors (X). It explains the changes in the dependent variables with respect to changes in select predictors. Some major uses for regression analysis are in determining the strength of predictors, forecasting an effect, and trend forecasting. It finds the significant relationship between variables and the impact of predictors on dependent variables. In regression, we fit a curve/line (regression/best fit line) to the data points, such that the differences between the distances of data points from the curve/line are minimized.
#regression #machine-learning #beginner #logistic-regression #linear-regression #deep learning
1602002446
## Hand Sanitizer in bulk - Get your effective hand sanitizer here
With the spread of various harmful virus globally causing immense distress and fatalities to human mankind, it has become absolutely essential for people to ensure proper and acute hygiene and cleanliness is maintained. To further add to the perennial hardship to save lives of people the recent pandemic of Covid-19 affected globally created the worst nightmare for people of all walks of life. Looking at the present crisis, it has become imperative for human beings to be encouraged to tackle this challenge with an everlasting strength to help protect oneself and their loved ones against the devastating effects of the virus. One thing that stands up between keeping all safe and vulnerable is by making sure that everybody attentively Hand wash periodically to help physically remove germs from the skin and getting rid of the live microbes.
The essence of apposite handwashing is based around time invested in washing and the amount of soap and water used. Technically, washing hands without soap is much less effective anyway. But incase a proper handwashing support system doesn’t become possible around, the usage of Effective Hand Sanitizer will certainly help fight to reduce the number of microbes on the surface of hands efficiently, eliminating most variants of harmful bacteria to settle.
The need has come about for Hand Sanitizer in bulk to save your daily life aptly maintaining a minimum of 60% alcohol - as per the CDC recommendations and approved by USFDA for its greater effectiveness. With the growing demand of people on the move the demand for easy to carry, small, and travel size worthy pouches that are also refillable once the product runs out is the need of the hour. To further make sure that human lives are well protected from these external viruses, it is mandatory for producer of effective Hand Sanitizer to evolve products circumspectly with ingredients that produce not just saving lives but with multiple benefits for people of all ages.
#hand sanitizer #hand sanitizer in bulk #hand sanitizer ingredient #hand sanitizer to alcohol #hand sanitizer travel size #hand sanitizer wholesale
1601431200
## Polynomial Regression — The “curves” of a linear model
The most glamorous part of a data analytics project/report is, as many would agree, the one where the Machine Learning algorithms do their magic using the data. However, one of the most overlooked part of the process is the preprocessing of data.
A lot more significant effort is put into preparing the data to fit a model on rather than tuning the model to fit the data better. One such preprocessing technique that we intend to disentangle is Polynomial Regression.
#data-science #machine-learning #polynomial-regression #regression #linear-regression | 1,756 | 8,832 | {"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-2023-14 | latest | en | 0.899818 |
http://asq.org/quality-progress/2013/11/expert-answers.html | 1,556,067,310,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578616424.69/warc/CC-MAIN-20190423234808-20190424020808-00174.warc.gz | 15,094,956 | 18,474 | ## 2019
### Reliable plan
Q: What are the most appropriate, non-industry specific sampling plans for reliability testing, particularly for ongoing reliability testing? The testing we are doing is destructive, so we cannot afford a large sample size. At the same time, we need to provide good customer protection.
A: Depending on the characteristics of the data being collected, there are many sampling plans that may apply. Three potential plans are:
1. Attribute testing: The test result is either pass or fail, and reliability is demonstrated by the likelihood of a pass. An example is go/no-go testing on the strength of a plastic tube, with the test being a weight hung from the end of the tube and the result being whether the tube breaks. The test purpose is to demonstrate that a percentage of the tubes meet the requirement at a given confidence level.
2. Variable testing: The numerical data are analyzed to determine the reliability demonstrated versus a specification. An example is tensile strength of the same plastic tube, with the purpose being to demonstrate that a percentage of the tubes meet a minimum requirement at a given confidence level.
3. Reliability testing: True product life data are being collected, and reliability is defined as the likelihood of surviving without failure under a given set of conditions. An example is an extension of the attribute testing earlier, in which the weight is hung from the tube until the tube breaks and the time to failure is recorded. Environmental conditions also may be involved.
For attributes, standard lot acceptance sampling is one place to start. The sampling plan should demonstrate a high probability of acceptance (for example, 0.95) at the acceptable quality level. Conversely, the sampling plan should have a low probability of acceptance (for example, 0.10) at the lot tolerance percent defective (LTPD). To minimize sample size in attribute testing, a plan that allows zero failures (C = 0 plan) provides the best consumer protection and the smallest sample size. The following simple equation can be used to calculate the required sample size (n) given a desired reliability (R) and a confidence level (CL):
n = ln (1 − CL) / ln (R).
To relate this equation to acceptance sampling terminology, a plan’s LTPD is equal to 1 − R.
Another way to reduce sample size is to make the testing requirements risk-based. Use the product’s risk analysis (hazard analysis or failure mode and effects analysis) to justify lower confidence and reliability requirements for testing. Table 1 is an example of a risk analysis severity rating scale and corresponding reliability/confidence requirements that can be used for attribute and variable testing.
To determine sample sizes for variable testing, additional parameters called "power" and "difference to detect" are needed. Statistical power is a measure of consumers’ protection. The higher, the better, and it also can be risk-based, as shown in Table 1. Lower risk implies lower required confidence and power, which leads to smaller sample sizes.
In addition to using risk, variable sample sizes also can be reduced by realistically increasing the difference to detect. If historical data on the product under test or on a similar product are available, the data can be used to estimate what difference to detect value would be so small that there would be insignificant risk if it were not detected. In other words, if the data infer that the requirement can likely be met comfortably, smaller sample sizes can be used.
Table 2 provides sample sizes for different powers and differences to detect, all with 95% confidence. Notice how much the difference to detect value can affect sample sizes, by entire orders of magnitude.
Table 2 applies to data that follow the normal distribution, so testing for normality is necessary before conducting the analysis. Alternate techniques such as data transformations and determining best-fit distributions are recommended if the data are not normally distributed.
For true reliability testing, the sample size calculation for a reliability demonstration test is straightforward, assuming the exponential distribution and its constant failure rate apply. Reliability software, such as ReliaSoft’s Weibull++ software and supporting literature, thoroughly explains the concepts that follow. The sample size is related to the total accumulated test time (T), which can be calculated from the equation:
T = (MTTF × X2CL, 2(r + 1)) / 2.
In this equation, MTTF is the mean time to failure that needs to be demonstrated, X2 is the Chi-squared distribution, CL is the confidence level and r is the number of failures.
After T is determined, you can calculate the sample size by dividing T by the available test time per unit. Note the tradeoff between sample size and test time per unit. Smaller sample sizes can be used, but the test time is longer. Also, as previously shown with attribute and variable testing, confidence level is a discretionary parameter. If the risk level allows, decreasing the confidence level can reduce the total test time needed, and therefore can decrease the sample size. The required MTTF also may have some discretion, although it is primarily driven by product requirements and a relevant safety margin. Decreasing the required MTTF can reduce sample size, again with caution and only if the risk level permits.
Other techniques to minimize the cost of testing or reduce sample size include:
1. Reduced inspection:1 For ongoing testing, in which production is at a steady rate, reduced inspection can be implemented based on previous acceptable results on a certain number of lots and units.
2. Double and multiple sampling: A smaller initial sample is drawn, and then the lot either accepted or rejected, with a third option being to take another sample. If the expected lot quality is very good or very bad, double and multiple sampling can be economical.
3. Combined environments or worst-case testing: In reliability testing, rather than testing the effects of two different conditions separately, such as temperature and vibration, combine them. The concept, which should be specifically validated by engineering expertise and the physics of the situation, is that subjecting the test specimen to more than one stress is more severe than applying the stresses separately. If the product passes the combined study, it is inferred that it would pass if exposed to the treatments separately.
4. Correlation with nondestructive evaluation (NDE): If product quality can be verified and the product is unharmed by the testing itself, NDE can be the way to go. For example, porosity or voids in a plastic tube can lead to reduced tensile strength. Ultrasonic inspection of the tubes can be used to qualitatively and quantitatively see the defects, and correlation with destructive test data can allow acceptance criteria to be set based on the ultrasonic scans.
5. Eliminate the need for testing by controlling the process: Use design of experiments and engineering knowledge to determine which manufacturing process parameters affect the product characteristics being tested. Then, control the key process parameters using statistical process control. Perform studies and validations to demonstrate that if the process is in control, the product will meet requirements.
6. Eliminate the part being tested: If you are testing a component of a more complex product, consider how you could redesign the overall product to eliminate the component or incorporate its function into another part of the design that does not require ongoing testing. This could be part of a design for reliability effort, or more specifically, design for testing.
These are a few of the ways in which sample sizes can be reduced to minimize cost without compromising the priority of protecting the users of the products from risks.
Scott A. Laman
Senior manager,
Quality engineering and risk management
Teleflex Inc. | 1,541 | 7,993 | {"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-2019-18 | latest | en | 0.921679 |
https://proofwiki.org/wiki/Group_is_Inverse_Semigroup_with_Identity | 1,679,585,914,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00591.warc.gz | 555,417,747 | 10,879 | # Group is Inverse Semigroup with Identity
## Theorem
A group is an inverse semigroup with an identity.
## Proof
Let $\struct {S, \circ}$ be a group.
Let $a \in S$.
Then:
$\ds e$ $=$ $\ds a \circ a^{-1}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds e \circ a$ $=$ $\ds a \circ a^{-1} \circ a$ $\ds \leadsto \ \$ $\ds a$ $=$ $\ds a \circ a^{-1} \circ a$ Definition of Identity Element
and
$\ds e$ $=$ $\ds a \circ a^{-1}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds a^{-1} \circ e$ $=$ $\ds a^{-1} \circ a \circ a^{-1}$ $\ds \leadsto \ \$ $\ds a^{-1}$ $=$ $\ds a^{-1} \circ a \circ a^{-1}$ Definition of Identity Element
Thus the criteria of an inverse semigroup are fulfilled.
$\blacksquare$ | 261 | 771 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-14 | latest | en | 0.220413 |
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ShowAll Questionssorted byDate Posted
apoorva pandey
# Can we write apex code to show reverse triangle (pattern)?
Best Answer chosen by apoorva pandey
Purushotham Yellanki
Hi Apoorva,
Try below Code in you Dev Console and see if this is what you are looking for. Basically reverse triangle logic is same in almost all OOP languages, it doesn't matter if it is Java or Apex all we are going to use is for loops!
String revTriangle = '';
Integer x = 1;
Integer y = 1;
for (x = 1; x <= 10; x++)
{
for (y = x; y <= 10; y++)
{
if (y<=x)
revTriangle +='*';
else
revTriangle+='*';
}
revTriangle+='\n';
}
System.debug('ReverserTriangle: \n' + revTriangle);
Thank you
Purushotham Yellanki
Hi Apoorva,
Try below Code in you Dev Console and see if this is what you are looking for. Basically reverse triangle logic is same in almost all OOP languages, it doesn't matter if it is Java or Apex all we are going to use is for loops!
String revTriangle = '';
Integer x = 1;
Integer y = 1;
for (x = 1; x <= 10; x++)
{
for (y = x; y <= 10; y++)
{
if (y<=x)
revTriangle +='*';
else
revTriangle+='*';
}
revTriangle+='\n';
}
System.debug('ReverserTriangle: \n' + revTriangle);
Thank you
This was selected as the best answer
apoorva pandey
Thank you Purushotham. This works! | 409 | 1,400 | {"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} | 2.71875 | 3 | CC-MAIN-2023-50 | longest | en | 0.806255 |
https://physicscalculations.com/what-is-decay-constant/ | 1,680,405,504,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950373.88/warc/CC-MAIN-20230402012805-20230402042805-00707.warc.gz | 499,889,884 | 98,091 | Home » What is Decay Constant?
# What is Decay Constant?
## Definition of Decay Constant
The decay constant is a measure of the probability that a given nucleus will decay per unit of time. The symbol for the decay constant is λ and is the ratio of the number of decays per unit of time to the total number of nuclei present
λ = (dN/dt)/N
where dN/dt is the rate of decay and N is the number of nuclei present. The units of λ are inverse time, such as per second (s-1) or per year (yr-1).
The decay constant is a fundamental property of each radioactive substance and is independent of external conditions, such as temperature or pressure. It is related to the half-life and means the lifetime of the substance, which are two other important quantities that describe the rate of decay.
## Understanding Radioactive Decay
Radioactive decay is a naturally occurring process by which unstable atomic nuclei spontaneously disintegrate, releasing radiation in the form of particles or waves. The rate at which this disintegration occurs is governed by the decay constant, which is a fundamental property of each radioactive substance. In this article, we will explore the concept of the decay constant in detail, including its definition, significance, calculation, and applications.
Radioactive decay refers to the spontaneous transformation of one type of nucleus into another by the emission of radiation. This phenomenon occurs because the nucleus is unstable and seeks to reach a more stable state. The process of decay can release various types of radiation, such as alpha particles, beta particles, or gamma rays, which can cause ionization of matter and other effects.
Radioactive decay is a random process, meaning that we cannot predict the exact moment of decay for any given nucleus. However, We can measure the overall rate of decay for a large number of nuclei of the same type. This is found to follow an exponential decay law.
## Types of Decay Constant
There are three types of decay constant that we commonly use to describe radioactive decay: half-life, mean lifetime, and activity.
### Half-Life
The half-life (t1/2) of a radioactive substance is the time it takes for half of the nuclei to decay. It is related to the decay constant by the equation:
t1/2 = ln(2)/λ = 0.693/λ
The half-life is a convenient way to describe the rate of decay because it is a fixed property of the substance, regardless of the amount present. For example, the half-life of carbon-14 is about 5,700 years, which means that half of the carbon-14 in a sample will decay in that time.
The mean lifetime (τ) of a radioactive substance is the average time that a nucleus will exist before decaying. It is related to the decay constant by the equation:
τ = 1/λ
The mean lifetime is a useful quantity for describing the decay process because it takes into account the fact that some nuclei may decay quickly, while others may last for a long time. It is also related to the energy spectrum of the emitted radiation.
### Activity
The activity (A) of a radioactive substance is the number of decays per unit time. It is related to the decay constant by the equation:
A = λN
where N is the number of nuclei present. The units of activity are typically Becquerels (Bq) or Curie (Ci). The activity of a substance decreases exponentially over time as the number of nuclei decreases due to decay.
## Significance of Decay Constant
The decay constant is a fundamental quantity in nuclear physics and plays a crucial role in understanding radioactive decay. It provides a measure of the stability of a nucleus and we can use it to predict the rate of decay for a given substance. Additionally, we use it to calculate the half-life, mean lifetime, and activity of a substance, which are important parameters in many applications.
Therefore, we can see that it is also related to the cross-section for nuclear reactions, which is a measure of the probability that a given reaction will occur. This is also important in understanding how nuclei interact with each other and with external particles, such as in nuclear reactors or particle accelerators.
## Calculation of Decay Constant
We can calculate the decay constant from experimental data by measuring the rate of decay and the number of nuclei present. We can do this by applying various techniques. These techniques include counting the number of emitted particles or measuring the intensity of emitted radiation.
Additionally, we can also calculate it theoretically by using quantum mechanics and the principles of nuclear physics. This involves modeling the nucleus as a system of protons and neutrons and calculating the probability of decay using the rules of quantum mechanics.
## Applications of Decay Constant
The decay constant has many important applications in various fields, such as radiometric dating, nuclear physics, and medical imaging.
Radiometric dating is a technique we use to determine the age of rocks and minerals based on the decay of radioactive isotopes. We use the decay constant to calculate the age of the sample by comparing the ratio of the parent isotope to the daughter isotope. The use of this technique heightened in geology, archaeology, and other fields to study the history of the Earth and its inhabitants.
### Nuclear Physics
We also apply it in nuclear physics to model the behavior of nuclei and predict the outcomes of nuclear reactions. We use it to calculate the cross-sections for various reactions, such as fission or fusion, and to design nuclear reactors and other devices.
### Medical Imaging
Another application is in medical imaging techniques. These techniques include PET (positron emission tomography), to image the distribution of radioactive tracers in the body. Thus, the decay of the tracer produces gamma rays, which can be detected and used to create an image of the body.
## Limitations and Challenges
There are several limitations and challenges associated with the concept of a decay constant. One of the main challenges is the measurement of the rate of decay. This is because external factors can affect it. These external factors can be temperature or pressure. Additionally, the decay constant can vary for different isotopes of the same element, which can complicate the analysis of complex systems.
Furthermore, another challenge is the prediction of decay pathways and outcomes. They are due to the influence of many factors. The factors are energy and momentum of the decaying nucleus. This requires sophisticated modeling techniques and experimental data to accurately predict the behavior of radioactive substances.
## Future Directions
Research into the decay constant and radioactive decay is ongoing, with many new discoveries and applications being explored. One area of interest is the study of exotic isotopes and their behavior in extreme conditions, such as in supernovae or neutron stars. Furthermore, this requires advanced experimental techniques and theoretical models to understand the fundamental nature of nuclear physics.
## Summary
Summarily, the decay constant is a fundamental property of radioactive substances. It describes the rate of decay and plays a crucial role in many applications. It is a measure of the stability of a nucleus. Additionally, it provides a way to predict the behavior of radioactive substances. Understanding the concept of decay constant is important for many fields, such as geology, nuclear physics, and medical imaging.
## Frequently Asked Questions (FAQs)
How is the decay constant calculated from experimental data?
We can calculate the decay constant from experimental data by measuring the rate of decay and the number of nuclei present. This can be done using various methods. These methods include counting the number of emitted particles or measuring the intensity of emitted radiation.
What is the relationship between decay constant and half-life?
Answer: The half-life of a substance is the time required for half of the original number of nuclei to decay. It is related to the decay constant by the formula t1/2 = ln(2)/λ, where t1/2 is the half-life and λ is the decay constant.
Can the decay constant change over time?
Answer: It is considered to be a constant for a given substance. It can be affected by external factors, such as temperature or pressure. In some cases, the decay constant may also vary for different isotopes of the same element.
What is the significance of the decay constant in radiometric dating?
Answer: It is used in radiometric dating to calculate the age of rocks and minerals. We compare the ratio of parent to daughter isotopes. This provides a way to determine the history of the Earth and its inhabitants over time.
What are some future directions in the study of decay constant and radioactive decay?
Answer: Research in this field is ongoing. The focus is on understanding the behavior of exotic isotopes and their role in astrophysical phenomena. There is also interest in developing new applications for radioactive isotopes in medicine and other fields.
You may also like to read:
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Share on social | 1,792 | 9,190 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-14 | longest | en | 0.941124 |
http://www.dailygk.in/2016/09/reasoning-quiz-65-for-ibps-po-clerk.html | 1,527,450,537,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00405.warc.gz | 360,302,481 | 39,684 | # REASONING QUIZ-65 FOR IBPS- PO & CLERK
REASONING QUIZ-65 FOR IBPS- PO & CLERK
Study the following information carefully to answer the given questions (1-5)
Avi, Bimal, Cruse, Deva, Edwen, Fatima, Gaurav and Heena are eight persons sitting around a circle facing outward in one arrangement and in a straight line facing north in another arrangement.
One of the immediate neighbour of Heena in straight line sits opposite Heena in the circle. Edwen sits third to the right of Bimal in the circle, while fourth to his left in the straight line. Fatima and Cruse are the immediate neighbour of Bimal in both the arrangements, but Cruse is not at the extreme ends of the row.
The one who sits on the extreme left end sits second to the right of Edwen in the circle. Heena is not on the immediate left of Fatima in both the arrangements. Gaurav sits on the immediate left of Heena in the circle, but both are not immediate neighbours of each other in the straight line. Dev sits third to the right of Fatima in the straight line. The one who sits on the immediate left of Bimal in straight line is sitting on the immediate right of Bimal in the circle
1. Which of the following pairs sits at the extreme ends of the row?
1. Gaurav,Avi
2. Dev,Gaurav
3. Dev,Fatima
4. Heena,Bimal
5. None of these
2. Who among the following sits on the immediate right of Fatima in the circle?
1. Avi
2. Bimal
3. Cruse
4. Edwen
5. None of these
3. The person sitting between Gaurav and Cruse in the circle is sitting what position in the straight line?
1. Between Cruse and Gaurav
2. Immediate right of Avi
3. Second from the right end
4. Third from the left end
5. None of these
4. The one who sitting in the right end of the row is sitting what position in the circle?
1. Between Bimal and Fatima
2. Immediate right of Gaurav
3. Between Edwen and Gaurav
4. Third to the right of Cruse
5. None of these
5. Who sits second to the left of Avi in the circle?
1. Edwen
2. Bimal
3. Cruse
4. Heena
5. None of these
Study the following information carefully to answer the questions given below (6-10).
Eight persons A,B,C,D,E,F,H and I are going to three different destinations Bihar, Assam, Rajasthan in three different cars- Swift, Avanti, Alto. Out of these three are females and also one in each car. At Least two persons are there in each car.
E, a male, is travelling with only I and they are not going to Assam.
A is travelling in Swift and going to Rajasthan. C is not travelling with B and H.
C and F are travelling together. H is not going to Assam. D is the sister of A and is travelling by Alto.
6. Members of which cars are going to Assam?
1) Avanti
2) Can’t be determined
3) Alto
4) Swift
7. In which car are four members travelling?
1) None
2) Alto
3) Avanti
4) Swift
8. Which of the following combinations represents the three female members?
1) ABC
2) CID
3) CFE
4) Can’t be determined
9. Who is travelling with H?
1) CF
2) SI
3) AB
4) Can’t be determined
10. Members of which of the following are travelling in Swift?
1) ACH
2) ABI
3) HAB
4) None of these
1. 2
2. 1
3. 4
4. 3
5. 2
6. 3
7. 1
8. 4
9. 3
10. 3 | 884 | 3,123 | {"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-2018-22 | latest | en | 0.898405 |
http://openstudy.com/updates/4f6f21cce4b0772daa08ed75 | 1,513,110,127,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948517917.20/warc/CC-MAIN-20171212192750-20171212212750-00264.warc.gz | 214,972,219 | 8,445 | • anonymous
Interpret the definite integral A=a~b dA=a~b L(x)dx as computing the area of a region in the xy-plane. Then one can think of the definite integral as: A. the antiderivative of L(x) B. "accumulating" all of the small segments of area "dA" from a to b. C. "accumulating" all of the small segments of area "L(x) x d(x)" from a tob, where L(x) represents the length of a rectangle at a particular xvalue, and dx the width. D. both "accumulating" all of the small segments of area "dA" from a to bAND "accumulating" all of the small segments of area "L(x) x d(x)" from a to b, where be back
Mathematics
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Not the answer you are looking for? Search for more explanations. | 416 | 1,527 | {"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 | 3 | CC-MAIN-2017-51 | latest | en | 0.485646 |
https://www.physicsforums.com/threads/a-bit-confused-about-polar-basis-vectors.767087/ | 1,508,503,856,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00728.warc.gz | 972,318,352 | 16,564 | # A Bit Confused About Polar Basis Vectors
1. Aug 22, 2014
### MrBillyShears
Let me say from the beginning I'm not talking about the non-coordinate unit vectors for polar coordinates. I'm talking about basis vectors. Let me just ask it as boldly as possible: how does one use these basis vectors in order to describe a vector? I know they are different at every point, so which point do you use? Is it completely arbitrary? Why is there different basis vectors at every point? And, I am new with this kind of stuff, so try to keep it as simple as possible in your explanation.
And also, if so, if you pick $\vec{r}$ to describe your vectors, would the "tails" of your vectors come from the origin, or from your point $\vec{r}$? And, if someone could give an example with numbers, that would be great.
Last edited: Aug 22, 2014
2. Aug 22, 2014
### Blazejr
To really understand this concept properly one needs to learn some basic introduction to manifolds and tangent spaces. But let's first clarify on the underlying idea, without going into details. Those vectors you are refering to are in a way "glued to a point", and if you are going to imagine them as arrows, they would come from points in space, not always from origin. We call them vectors tangent to a point. Space of all vectors tangent to a given point is obviously linear space. Space of all such vectors (at any point) is not, because it's meaningless to add vectors with tails at different points! Therefore, for every point we have a tangent space. You can have different basis in those spaces (from now on I assume that we chose one point). We can have, for example, basis related to cartesian cordinates. Those basis vectors are of unit length and point in direction of coordinate axes. You can decompose any vector in terms of those basis vectors and get cartesian components. Make no mistake though - those components are not coordinates of a vector. Vector is not a point in the original space. Now you can also have basis related to polar coordinates. One vector pointing in the direction of growing $r$ etc., three vectors, each parallel to each other. There is important difference between those and cartesian vectors. If you go to different point in space. There direction of growing r is different! At point $(3,0,0)$ in euclidean space cartesian coordinates of r vector would be just $(1,0,0)$. However, at point $(-3,0,0)$ has cartesian components $(-1,0,0)$. Both point radially outwards and are of unit length, but what it means to point radially outward clearly depends on where you are.
3. Aug 22, 2014
### WWGD
One issue is that, outside of R^n , there is rarely a natural isomorphism between vector spaces at different points. The differential quotient then takes tangent vectors at different points, and, like blazejr said, this difference --the whole expression-- is not well-defined. To "well-define" it , one uses connections, which are choices of vector-space isomorphisms between the tangent spaces.
4. Aug 22, 2014
### Fredrik
Staff Emeritus
I don't understand what you're saying here. "Non-coordinate unit vectors" sounds like something that has nothing to do with the coordinate system. "for polar coordinates" sounds like the exact opposite. Then the second sentence suggests that the vectors you were talking about in the first sentence aren't basis vectors. Every linearly independent set with two vectors is a basis.
I'm still assuming that you're talking about these guys:
\begin{align}
&\hat r=(\cos\varphi,\sin\varphi)\\
&\hat\varphi =(-\sin\varphi,\cos\varphi)
\end{align} For all $r,\varphi$, the vectors above are the ones that the polar coordidinate system associates with the tangent space at the point $(r\cos\varphi,r\sin\varphi)$. Since that tangent space is an identical copy of the vector space that your $\vec r$ is an element of, you can also use these vectors as a basis for that space.
Same way you use any other basis.
A point on a particle's trajectory at which you intend to calculate something, like that particle's centripetal acceleration.
They're defined by
\begin{align}
\hat r=\frac{\frac{d\vec r}{dr}}{\left|\frac{d\vec r}{dr}\right|},\qquad \hat\varphi=\frac{\frac{d\vec r}{d\varphi}}{\left|\frac{d\vec r}{d\varphi}\right|}
\end{align} The right-hand sides have different values at each point. If you don't find this useful, then you can use some other basis. But you will certainly find these bases useful when you describe circular motion, where the particle's position and velocity are respectively $r\hat r$ and $r\dot\varphi\hat\varphi$, at every point.
The position vector should be drawn from the origin, but the velocity vector from the particle's location. It makes sense to think of the position vector $\vec r$ as an element of your original copy of $\mathbb R^2$ ("the configuration space"), and the velocity vector as an element of a different copy of $\mathbb R^2$ ("the tangent space at $\vec r$") with its origin attached to the point $\vec r$. When you're dealing with a significantly less trivial manifold than $\mathbb R^2$, you pretty much have to think this way.
Last edited: Aug 22, 2014 | 1,233 | 5,151 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-43 | longest | en | 0.952817 |
https://www.instructables.com/Automatic-Streetlight-System/ | 1,721,634,090,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00096.warc.gz | 678,007,282 | 32,962 | Introduction: Automatic Streetlight System
This doesn't needs manual operation for switching ON and OFF. When there is a need of light
it automatically switches ON. When darkness rises to a certain level then sensor circuit gets activated and switches ON and when there is other source of light i.e. daytime, the street light gets OFF. The sensitiveness of the street light can also be adjusted.
Step 1: First Collect the Components...
1. A IC-555 (Timer IC)
2. A 470 Ohm Resistor
3. A LED (Red Preferable)
4. A 0-50K preset or Potentiometer for setting sensitivity of LDR (I have used a preset in my Project)
5. A LDR (Light dependent Resistor)-This acts as a sensor in our project
6. A 9 Volts Battery and a battery snap
7. A Toggle Switch
8. An Old CD case
9. A piece of straw
10. A white LED (This acts as a streetlight)
Step 2: Circuit Diagram and Working
When light falls on the LDR then its resistance decreases which results in increase of the voltage at pin 2 of the IC 555. IC 555 has got comparator inbuilt, which compares between the input voltage from pin2 and 1/3rd of the power supply voltage. When input falls below 1/3rd then output is set high otherwise it is set low. Since in brightness, input voltage rises so we obtain no positive voltage at output of pin 3 to drive relay or LED, besides in poor light condition we get output to energize.
Step 3: Working Principle...
*This circuit uses a popular timer I.C 555. I.C 555 is connected as comparator with pin-6 connected with positive rail, the output goes high(1) when the trigger pin 2 is at lower then 1/3rd level of the supply voltage. Conversely the output goes low (0) when it is above 1/3rd level. So small change in the voltage of pin-2 is enough to change the level of output (pin-3) from 1 to 0 and 0 to 1. The output has only two states high and low and can not remain in any intermediate stage. It is powered by a 6V battery for portable use. The circuit is economic in power consumption. Pin 4, 6 and 8 is connected to the positive supply and pin 1 is grounded. To detect the present of an object we have used LDR and a source of light.
*LDR is a special type of resistance whose value depends on the brightness of the light which is falling on it. It has resistance of about 1 mega ohm when in total darkness, but a resistance of only about 5k ohms when brightness illuminated. It responds to a large part of light spectrum. We have made a potential divider circuit with LDR and 100K variable resistance connected in series. We know that voltage is directly proportional to conductance so more voltage we will get from this divider when LDR is getting light and low voltage in darkness. This divided voltage is given to pin 2 of IC 555. Variable resistance is so adjusted that it crosses potential of 1/3rd in brightness and fall below 1/3rd in darkness.
*Sensitiveness can be adjusted by this variable resistance. As soon as LDR gets dark the voltage of pin 2 drops 1/3rd of the supply voltage and pin 3 gets high and LED or buzzer which is connected to the output gets activated. | 745 | 3,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.892467 |
https://aa.quae.nl/en/reken/juliaansedag.html | 1,656,531,430,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00732.warc.gz | 128,495,367 | 89,656 | $$\def\|{&}\DeclareMathOperator{\D}{\bigtriangleup\!} \DeclareMathOperator{\d}{\text{d}\!}$$
$$\DeclareMathOperator{\trunc}{trunc} \DeclareMathOperator{\Div}{div} \DeclareMathOperator{\gcd}{gcd} \DeclareMathOperator{\bdom}{dom} \def\floorratio#1#2{\left\lfloor \dfrac{#1}{#2} \right\rfloor} \def\ceilratio#1#2{\left\lceil \dfrac{#1}{#2} \right\rceil} \def\mod1ratio#1#2{\left\lfloor \dfrac{#1}{#2} \right\rceil_1} \def\dom1ratio#1#2{\left\lceil \dfrac{#1}{#2} \right\rfloor_1}$$
In astronomical formulas that contain a date, it is not convenient to write that date as a combination of years, months, and days, especially because not all years have the same number of days and not all months have the same number of days. It is much more convenient to measure the date as the number of days since some fixed day. The Julian Day Number (JDN) or Julian Date (JD) and its relatives are much used for this in astronomy. This page explains how you can translate a date from various calendars to the Julian Date, or the other way around.
The first part of this page provides algorithms for some modern and historical calendars. The second part (from Section 12) explains how I deduced those algorithms.
There are many different algorithms in use for converting between calendar dates and day numbers. The only thing that counts is that they provide the correct answers for all day numbers and calendar dates. If there is only a limited number of input values and possible outcomes (for example for calculating the month number and day-within-the-month number from the day-within-the-year number), then you can usually find many different algorithms that all give just the right answers for those particular values (but perhaps very different answers for all other input values), even if there is no logical connection between that algorithm and the calendar. In such a case there is often also little or no logical connection between the algorithm that converts from date to day number and the algorithm that converts from day number to date, but the algorithms may be shorter then.
For every algorithm, you should know for which input values it was designed to work. Some calendar algorithms only work well with positive numbers and give wrong results for negative years or negative day numbers. The algorithms I give below are designed to work for all years (including negative ones), all months in each year, all days in each month, and for all Julian Day Numbers (including negative ones).
Some of the fomulas express a choice: If a certain condition is met, then a certain action should be taken (for example, apply a correction to something), and if that condition is not met, then a different action or no action at all should be taken. Such alternative paths are easy to use in manual calculations, in applications that process one date at a time, and in applications written in a programming language that requires compilation before the program can be executed (for example, written in C or Fortran) but are not very convenient for use in applications that can handle many dates at once (in an array) and that are written in a programming language that can be executed immediately (such as Basic or Perl or IDL), for which the execution of an if-then statement per date is much slower than the execution of a single fixed calculation for each date in an array.
If the calendrical calculations below here involve such choices, then where possible I give a formula that circumvents if-then constructions and is therefore convenient for application to arrays of dates in direct-execution programming languagues. The accompanying text explains which choice is being worked around.
## 1. Different Kinds of Day Numbers
For astronomical calculations that depend on time it is useful to have a continuous time scale that uses only a single unit, and not three like most calendars (days, months, years), and that in addition uses a fixed time zone so that there is no confusion about which precise instant of time is meant.
The IAU has adopted the Julian Date (JD) for this purpose. This should not be confused with "a date in the Julian Calendar". There are various other timescales that look like JD. The next table describes a few of them.
Name Kind Begins From JD
JD Julian Date Fractional 12:00 UTC
JDN Julian Day Number Whole 12:00 UTC = ⌊JD⌋
CJD Chronological Julian Date Fractional 00:00 LT = JD + 0.5 + TZ
CJDN Chronological Julian Day Number Whole 00:00 LT = ⌊JD + 0.5 + TZ⌋
"Whole" means that that timescale uses whole numbers only and counts whole days only. "Fractional" means that that timescale uses fractional numbers (with a part after the decimal point) and indicates instants of time. The "Begins" column shows when each next calendar day begins. 12:00 UTC means that a new calendar day begins at 12:00 (noon) universal time, which is not 12:00 noon local time ― except if you happen to be in a time zone equal to UTC, such as in Great Britain in winter. 00:00 LT means that a new calendar day begins at midnight local time. The "From JD"-column shows how the other numbers can be calculated from JD. In that column, "TZ" stands for an adjustment for your local time zone; that adjustment is 0 for UTC.
In practice, "fractional" JDs and CJDs are usually written with at least one digit after the decimal marker (even if that digit is 0), and "whole" JDNs and CJDNs without digits after the decimal marker.
The zero point of JD (i.e., JD 0.0) corresponds to 12:00 UTC on 1 January −4712 in the Julian calendar. The zero point of CJD corresponds to 00:00 (midnight) local time on 1 January −4712. JDN 0 corresponds to the period from 12:00 UTC on 1 January −4712 to 12:00 UTC on 2 January −4712. CJDN 0 corresponds to 1 January −4712 (the whole day, in local time).
Let's see which instants or periods of time correspond in the Gregorian calendar to the number 2455772 in the timescales of the previous table, in a time zone in which the clock reads 2 hours later than UTC (for example, in Central Europe in summer).
JD 2.45577e+06 exactly at 14:00 hours on 29 July 2011 JDN 2.45577e+06 from 14:00 hours on 29 July 2011 until 14:00 hours on 30 July 2011 CJD 2.45577e+06 exactly at 00:00 hours (midnight) on 29 July 2011 CJDN 2.45577e+06 the whole calendar day of 29 July 2011
To avoid round-off errors it is best to do calendrical calculations with whole numbers only. In what follows we use CJDN for that.
## 2. Notation
### 2.1. Rounding
In the formulas given below, many numbers are rounded to a nearby whole number. It is important that those numbers get rounded in the right direction. Whole numbers are already rounded, so those don't change.
The types of rounding that we use here are downward rounding to the nearest whole number (for $$x$$ this is indicated by $$⌊x⌋$$) or upward rounding to the nearest whole number ($$⌈x⌉$$). The standard rounding function on calculators usually rounds to the nearest whole number (our notation for that is $$[x]$$) and the standard conversion from a fractional number to a whole number in a computer language is usually by rounding in the direction of zero (our notation for that is $$\trunc(x)$$). We do not want those kinds of rounding here! The below table shows these three different kinds of rounding for some values.
$${x}$$ $${⌊x⌋}$$ $${[x]}$$ $${⌈x⌉}$$ $${\trunc(x)}$$ −5 −5 −5 −5 −5 −4.9 −5 −5 −4 −4 −4.2 −5 −4 −4 −4 −4 −4 −4 −4 −4 −0.2 −1 0 0 0 0 0 0 0 0 0.2 0 0 1 0 4 4 4 4 4 4.2 4 4 5 4 4.9 4 5 5 4 5 5 5 5 5
### 2.2. Modular Arithmetic
If you divide the whole number $$y$$ by the whole number $$x$$ then you get a quotient $$q$$ and a remainder $$r$$. Sometimes we're interested in the quotient, and sometimes in the remainder. We choose to have the remainder never be negative. Then $$0 ≤ r \lt |x|$$ and
\begin{align} q \| = \floorratio{y}{x} \\ r \| = y \bmod x = y − x\floorratio{y}{x} \\ y \| = qx + r \end{align}
The notation $$y \bmod x$$ means: the non-negative remainder from the division of $$y$$ by $$x$$.
We find
$$\frac{y}{x} = \floorratio{y}{x} + \left( \frac{y}{x} \bmod 1 \right) = \floorratio{y}{x} + \frac{y \bmod x}{x}$$
and if $$x = 1$$ then
$$y = ⌊y⌋ + (y \bmod 1)$$
The notation $$x ≡ y \pmod{n}$$ means that $$x$$ and $$y$$ differ by some multiple of $$n$$. One says that $$x$$ and $$y$$ are congruent modulo $$n$$, and $$n$$ is called the modulus. The equation $$x ≡ y \pmod{n}$$ is called a congruence.
Modular arithmetic is arithmetic in which all calculations are modulo a fixed $$n$$. That means that you can subtract arbitrary multiples of $$n$$ from all terms and factors without changing the congruence. If $$x ≡ y \pmod{n}$$ and $$p$$ is a whole number, then also $$x + pn ≡ y \pmod{n}$$, and $$px ≡ py \pmod{n}$$, and also $$px ≡ py \pmod{pn}$$.
A well-known example of modular arithmetic is clock face arithmetic. The face of a typical clock can show hours between 1 and 12. If you see two pictures of a clock face and in the first picture it shows 3 o'clock and in the second picture it shows 7 o'clock, then the second picture might have been made 4 hours after the first one, but also 4 + 12 = 16 hours later, or 4 + 10×12 hours = 5 days and 4 hours later, or 4 hours plus any desired multiple of 12 hours later (or earlier). If $$t_1$$ is the time of the first picture and $$t_2$$ is the time of the second picture (measured in hours from a fixed moment), then all you know from the pictures is that $$t_2 ≡ t_1 + 4 \pmod{12}$$. And if the clock shows $$t$$ hours now, then after $$y$$ more hours it will show $$(t + y) \bmod 12$$ hours.
Note the difference in notation: $$z = x \bmod y$$ means that $$z$$ is equal to the remainder of dividing $$x$$ by $$y$$, and $$z ≡ x \pmod{y}$$ means that $$z$$ is equal to $$x$$ except for an arbitrary multiple of $$y$$. If $$z = x \bmod y$$ then also $$z ≡ x \pmod{y}$$, but the opposite need not be true,
## 3. The GregorianCalendar
See section 13.2.1 for the derivation of this algorithm.
### 3.1. From Gregorian Date to CJDN
One algorithm to calculate a CJDN $$J$$ from a Gregorian date (calendar year $$j$$, calendar month $$m$$, calendar day $$d$$) is:
\begin{align} c_0 \| = \floorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ \{x_3, x_2\} \| = \Div(x_4, 100) \\ x_1 \| = m − 12c_0 − 3 \\ J \| = \floorratio{146097x_3}{4} + \floorratio{36525x_2}{100} + \floorratio{153x_1 + 2}{5} + d + 1721119 \end{align}
Here $$J_1$$ is the number of days from the beginning of calculation year 0 until the beginning of the current 100-calculation-year-period, $$J_2$$ the number of days since the beginning of the current 100-calculation-year-period, and $$J_3$$ the number of days minus one from the beginning of the current calculation year until the beginning of the current month. Calculation years run from 1 March to 1 March.
For example: what is the CJDN for 7 September 2010 on the Gregorian calendar? Then $$j = 2010$$, $$m = 9$$, $$d = 7$$ so
\begin{align*} c_0 \| = \floorratio{9 − 3}{12} = 0 \\ x_4 \| = 2010 + 0 = 2010 \\ \{x_3, x_2\} \| = \Div(2010, 100) = \{20, 10\} \\ x_1 \| = 9 − 12×0 − 3 = 6 \\ J \| = \floorratio{146097×20}{4} + \floorratio{36525×10}{100} + \floorratio{153×6 + 2}{5} + 7 + 1721119 \\ \| = 730485 + 3652 + 184 + 7 + 1721119 = 2455447 \end{align*}
And here are the calculations for a few more dates:
calendar date $${c_0}$$ $${x_4}$$ $${x_3}$$ $${x_2}$$ $${x_1}$$ $${…x_3…}$$ $${…x_2…}$$ $${…x_1…}$$ $${J}$$
2000-02-29 −1 1999 19 99 11 693960 36159 337 2451604
2000-03-01 0 2000 20 0 0 730485 0 0 2451605
2001-02-28 −1 2000 20 0 11 730485 0 337 2451969
2001-03-01 0 2001 20 1 0 730485 365 0 2451970
2100-02-28 −1 2099 20 99 11 730485 36159 337 2488128
2100-03-01 0 2100 21 0 0 767009 0 0 2488129
### 3.2. From CJDN to Gregorian Date
The algorithm to calculate a Gregorian date (calendar year $$j$$, calendar month $$m$$, calendar day $$d$$) from a CJDN $$J$$ is:
\begin{align} \{x_3, r_3\} \| = \Div(4×J − 6884477, 146097) \\ \{x_2, r_2\} \| = \Div\left( 100\floorratio{r_3}{4} + 99, 36525 \right) \\ \{x_1, r_1\} \| = \Div\left( 5\floorratio{r_2}{100} + 2, 153 \right) \\ d \| = \floorratio{r_1}{5} + 1 \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \end{align}
Here $$x_3$$ is the number of calculation centuries since that time, $$x_2$$ the number of calculation years since that time, and $$x_1$$ the number of months since the beginning of the current calculation year.
For example, which date in the Gregorian calendar corresponds to CJDN 2452827? Then $$J = 2452827$$ and then
\begin{align*} k_3 \| = 4×(2452827 − 1721120) + 3 = 2926831 \\ x_3 \| = \floorratio{2926831}{146097} = 20 \\ k_2 \| = 100\floorratio{2926831 \bmod 146097}{4} + 99 = 100\floorratio{4891}{4} + 99 = 100×1222 + 99 = 122299 \\ x_2 \| = \floorratio{122299}{36525} = 3 \\ k_1 \| = 5\floorratio{122299 \bmod 36525}{100} + 2 = 5×\floorratio{12724}{100} + 2 = 5×127 + 2 = 637 \\ x_1 \| = \floorratio{k_1}{153} = \floorratio{637}{153} = 4 \\ c_0 \| = \floorratio{4 + 2}{12} = 0 \\ j \| = 100×20 + 3 + 0 = 2003 \\ m \| = 4 − 12×0 + 3 = 7 \\ d \| = \floorratio{637 \bmod 153}{5} + 1 = \floorratio{25}{5} + 1 = 5 + 1 = 6 \end{align*}
The date is 6 July 2003.
And here are the calculations for the same dates as before.
$${J}$$ $${x_3}$$ $${x_2}$$ $${x_1}$$ $${c_0}$$ $${j}$$ $${m}$$ $${d}$$ calendar date
2451604 19 99 11 1 2000 2 29 2000-02-29
2451605 20 0 0 0 2000 3 1 2000-03-01
2451969 20 0 11 1 2001 2 28 2001-02-28
2451970 20 1 0 0 2001 3 1 2001-03-01
2488128 20 99 11 1 2100 2 28 2100-02-28
2488129 21 0 0 0 2100 3 1 2100-03-01
## 4. The Milanković Calendar
See section 13.3 for the derivation of this algorithm.
### 4.1. From Milanković Date to CJDN
Some Eastern Orthodox churches use a calendar invented by Milutin Milanković, that differs from the Gregorian calendar only in the rules for which century years are leap years. One algorithm to calculate a CJDN $$J$$ from a Milanković date (calendar year $$j$$, calendar month $$m$$, calendar day $$d$$) is:
\begin{align} c_0 \| = \floorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ x_3 \| = \floorratio{x_4}{100} \\ x_2 \| = x_4 \bmod 100 \\ x_1 \| = m − 12c_0 − 3 \\ J \| = \floorratio{328718x_3 + 6}{9} + \floorratio{36525x_2}{100} + \floorratio{153x_1 + 2}{5} + d + 1721119 \end{align}
### 4.2. From CJDN to Milanković Date
An algorithm to convert CJDN $$J$$ to year $$j$$, month $$m$$, day $$d$$ in the Milanković calendar is:
\begin{align} k_3 \| = 9×(J − 1721120) + 2 \\ x_3 \| = \floorratio{k_3}{328718} \\ k_2 \| = 100\floorratio{k_3 \bmod 328718}{9} + 99 \\ x_2 \| = \floorratio{k_2}{36525} \\ x_1 \| = \floorratio{5\floorratio{k_2 \bmod 36525}{100} + 2}{153} \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = \floorratio{k_1 \bmod 153}{5} + 1 \end{align}
## 5. The Julian Calendar
See section 13.1 for the derivation of this algorithm.
### 5.1. From Julian Date to CJDN
The algorithm to calculate a CJDN $$J$$ from a Julian date (calendar year $$j$$, calendar month $$m$$, calendar day $$d$$) is:
\begin{align} J_0 \| = 1721117 \\ c_0 \| = \floorratio{m − 3}{12} \\ J_1 \| = \floorratio{1461×(j + c_0)}{4} \\ J_2 \| = \floorratio{153m − 1836c_0 − 457}{5} \\ J \| = J_1 + J_2 + d + J_0 \end{align}
Here $$J_1$$ is the number of days between the beginning of calculation year 0 and the beginning of the current calculation year, and $$J_2$$ the number of days between the beginning of the current calculation year and the beginning of the current month.
For example: what is the CJDN for 7 September 2010 on the Julian calendar? Then $$j = 2010$$, $$m = 9$$, $$d = 7$$
\begin{align*} c_0 \| = \floorratio{9 − 3}{12} = 0 \\ J_1 \| = \floorratio{1461×(2010 + 0)}{4} = 734152 \\ J_2 \| = \floorratio{153×9 − 1836×0 − 457}{5} = 184 \\ J \| = 734152 + 184 + 7 + 1721117 = 2455460 \end{align*}
And here are the calculations for a few more dates:
calendar date $${c_0}$$ $${J_1}$$ $${J_2}$$ $${J}$$
2000-02-29 −1 730134 337 2451617
2000-03-01 0 730500 0 2451618
2001-02-28 −1 730500 337 2451982
2001-03-01 0 730865 0 2451983
2100-02-28 −1 766659 337 2488141
2100-02-29 −1 766659 337 2488142
2100-03-01 0 767025 0 2488143
### 5.2. From CJDN to Julian Date
The algorithm to calculate a Julian date (calendar year $$j$$, calendar month $$m$$, calendar day $$d$$) from a CJDN $$J$$ is:
\begin{align} y_2 \| = J − 1721118 \\ k_2 \| = 4y_2 + 3 \\ k_1 \| = 5\floorratio{k_2 \bmod 1461}{4} + 2 \\ x_1 \| = \floorratio{k_1}{153} \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = \floorratio{k_2}{1461} + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = \floorratio{k_1 \bmod 153}{5} + 1 \end{align}
Which Julian date corresponds to CJDN 2451893? Then $$J = 2451893$$ so
\begin{align*} y_2 \| = 2451893 − 1721118 = 730775 \\ k_2 \| = 4×730775 + 3 = 2923103 \\ k_1 \| = 5\floorratio{2923103 \bmod 1461}{4} + 2 = 5\floorratio{1103}{4} + 2 = 5×275 + 2 = 1377 \\ x_1 \| = \floorratio{1377}{153} = 9 \\ c_0 \| = \floorratio{9 + 2}{12} = \floorratio{11}{12} = 0 \\ j \| = \floorratio{2923103}{1461} + 0 = 2000 + 0 = 2000 \\ m \| = 9 − 12×0 + 3 = 12 \\ d \| = \floorratio{1377 \bmod 153}{5} + 1 = \floorratio{0}{5} + 1 = 1 \end{align*}
The date is 1 December 2000.
And here are the calculations for the same dates as before.
$${J}$$ $${y_2}$$ $${k_2}$$ $${k_1}$$ $${x_1}$$ $${c_0}$$ $${j}$$ $${m}$$ $${d}$$ calendar date
2451617 730499 2921999 1827 11 1 2000 2 29 2000-02-29
2451618 730500 2922003 2 0 0 2000 3 1 2000-03-01
2451982 730864 2923459 1822 11 1 2001 2 28 2001-02-28
2451983 730865 2923463 2 0 0 2001 3 1 2001-03-01
2488141 767023 3068095 1822 11 1 2100 2 28 2100-02-28
2488142 767024 3068099 1827 11 1 2100 2 29 2100-02-29
2488143 767025 3068103 2 0 0 2100 3 1 2100-03-01
## 6. The Islamic Calendar
See section 13.8 for the derivation of this algorithm.
### 6.1. From Islamic Date to CJDN
The religious Islamic calendar depends on observations and so cannot be caught in formulas. The administrative calendar has fixed rules so it can be caught in formulas. The difference between the religious and administrative calendars should usually be no more than 1 day.
The CJDN can be derived from the year number $$j$$, month number $$m$$, and day number $$d$$ in the most commonly used administrative Islamic calendar (used by al-Fazārī, al-Khwārizmī, al-Battānī, and in the Toledan and Alfonsine Tables) as follows:
$$J = \floorratio{10631j − 10617}{30} + \floorratio{325m − 320}{11} + d + 1948439$$
For example, which CJDN $$J$$ corresponds to the Islamic date 29-08-1432? Then $$j = 1432$$, $$m = 8$$, $$d = 29$$, so
$\begin{split} J \| = \floorratio{10631×1432 − 10617}{30} + \floorratio{325×8 − 320}{11} + 29 + 1948440 \\ \| = \floorratio{15212975}{30} + \floorratio{2280}{11} + 29 + 1948440 \\ \| = 507099 + 207 + 29 + 1948439 \\ \| = 2455774 \end{split}$
The CJDN is 2455774, which corresponds to 31 July 2011 in the Gregorian calendar.
And how about the first day of the first month of the first year of the Islamic calendar? Then $$j = 1$$, $$m = 1$$, $$d = 1$$, so
$\begin{split} J \| = \floorratio{10631×1 − 10617}{30} + \floorratio{325×1 − 320}{11} + 1 + 1948439 \\ \| = \floorratio{14}{30} + \floorratio{5}{11} + 1 + 1948440 = 0 + 0 + 1 + 1948439 = 1948440 \end{split}$
which corresponds to 16 July 622 in the Julian calendar.
### 6.2. From CJDN to Islamic Date
The following formulas calculate the year number $$j$$, month number $$m$$, and day number $$d$$ in the most commonly used administrative Islamic calendar from the CJDN $$J$$:
\begin{align} k_2 \| = 30×(J − 1948440) + 15 \\ k_1 \| = 11\floorratio{k_2 \bmod 10631}{30} + 5 \\ j \| = \floorratio{k_2}{10631} + 1 \\ m \| = \floorratio{k_1}{325} + 1 \\ d \| = \floorratio{k_1 \bmod 325}{11} + 1 \end{align}
For example, which Islamic date corresponds to CJDN 2455774? Then $$J = 2455774$$, and then
\begin{align*} k_2 \| = 30×(2455774 − 1948440) + 15 = 15220035 \\ k_1 \| = 11×\floorratio{15220035 \bmod 10631}{30} + 5 = 11×\floorratio{7074}{30} + 5 = 11×235 + 5 = 2590 \\ j \| = \floorratio{15220035}{10631} + 1 = 1431 + 1 = 1432 \\ m \| = \floorratio{2590}{325} + 1 = 7 + 1 = 8 \\ d \| = \floorratio{2590 \bmod 325}{11} + 1 = \floorratio{315}{11} + 1 = 29 \end{align*}
so $$j = 1432$$, $$m = 8$$, $$d = 29$$ which means 29 Shaban 1432.
## 7. The Babylonian Calendar
See section 13.5 for the derivation of this algorithm.
### 7.1. From Babylonian Date to CJDN
The Babylonians had a lunisolar calendar in which the beginning of each month was determined from observations of the Moon, but the number of months in each year followed a fixed 19-year pattern. The below formulas for an administrative Babylonian calendar uses that same 19-year pattern, and should usually deviate by at most one day from the calendar that the Babylonians used that depended on observations.
If $$j$$ is the year number in the era of Seleukos, and $$m$$ is the month number (beginning at 1) in the current year, and $$d$$ is the day number (beginning at 1) in the current month, then the CJDN $$J$$ can be found as follows:
\begin{align} y_1 \| = \floorratio{235j − 241}{19} + m \\ J \| = \floorratio{6940y_1}{235} + d + 1607557 \end{align}
For example, which CJDN $$J$$ corresponds to year 3, month 9, day 27 of the era of Seleukos? Then $$j = 3$$, $$m = 9$$, $$d = 27$$, so
\begin{align*} x_1 \| = 2 \\ z_1 \| = 8 \\ z_2 \| = 26 \\ c_1 \| = \floorratio{235×2 + 13}{19} = \floorratio{483}{19} = 25 \\ x_2 \| = y_1 = 25 + 8 = 33 \\ c_2 \| = \floorratio{6940×33}{235} = \floorratio{229020}{235} = 974 \\ y_2 \| = 974 + 26 = 1000 \\ J \| = 1000 + 1607558 = 1607658 \end{align*}
### 7.2. From CJDN to Babylonian Date
From CJDN $$J$$, we find the Babylonian year number $$j$$, month number $$m$$, and day number $$d$$ as follows:
\begin{align} k_2 \| = 235×(J − 1607558) + 234 \\ k_1 \| = 19\floorratio{k_2}{6940} + 5 \\ j \| = \floorratio{k_1}{235} + 1 \\ m \| = \floorratio{k_1 \bmod 235}{19} + 1 \\ d \| = \floorratio{k_2 \bmod 6940}{235} + 1 \end{align}
For example, which date in the Babylonian calendar corresponds to CJDN $$1608558$$? Then $$J = 1608558$$, so
\begin{align*} k_2 \| = 235×(1608558 − 1607558) + 234 = 235234 \\ k_1 \| = 19×\floorratio{235234}{6940} + 5 = 19×33 + 5 = 632 \\ j \| = \floorratio{632}{235} + 1 = 3 \\ m \| = \floorratio{632 \bmod 235}{19} + 1 = \floorratio{162}{19} + 1 = 9 \\ d \| = \floorratio{235234 \bmod 6940}{235} + 1 = \floorratio{6214}{235} + 1 = 27 \end{align*}
which means day 27 of month 9 of year 3.
## 8. The Jewish Calendar
See section 13.6 for the derivation of this algorithm.
### 8.1. From Jewish Date to CJDN
The Jewish calendar is a lunisolar calendar with complicated rules, and repeats itself only after 251,827,457 days = 35,975,351 weeks = 8,527,680 months = 689,472 years. The algorithm to transform a Jewish date into CJDN is long, and it features very large intermediate results. The algorithm that I provide here uses only whole numbers (to avoid round-off error), and assumes that all input values, intermediate results, and output values cannot be greater than 231 = 2,147,483,648.
In the Jewish calendar, a year has 12 or 13 months, and a month has 29 or 30 days. New Year is the first day of the 7th month, Tishri. The months in a year of 12 months are: 1 = Nisan (נִיסָן), 2 = Iyar (אִיָּר / אייר), 3 = Sivan (סִיוָן / סיוון), 4 = Tammuz (תַּמּוּז), 5 = Av (אָב), 6 = Elul (אֱלוּל), 7 = Tishri (תִּשׁרִי), 8 = Ḥeshvan (מַרְחֶשְׁוָן / מרחשוון), 9 = Kislev (כִּסְלֵו / כסליו), 10 = Tevet (טֵבֵת), 11 = Shevat (שְׁבָט), 12 = Adar (אֲדָר). In a year of 13 months, embolismic month Adar Ⅰ (אֲדָר א׳) is inserted between Shevat and regular Adar, and that regular Adar is then temporarily renamed to Adar Ⅱ (אֲדָר ב׳). In a year of 13 months, we count Adar Ⅰ as month number 12, and Adar Ⅱ as month number 13.
One algorithm to transform calendar year $$j$$, calendar month $$m$$ (from 1 through 12 or 13), and calendar day $$d$$ (from 1 through 29 or 30) into CJDN is then:
\begin{align} c_0 \| = \floorratio{13 − m}{7} \\ x_1 \| = j − 1 + c_0 \\ x_3 \| = m − 1 \\ z_4 \| = d − 1 \\ c_1(x_1) \| = \floorratio{235x_1 + 1}{19} \\ q(x_1) \| = \floorratio{c_1(x_1)}{1095} \\ r(x_1) \| = c_1(x_1) \bmod 1095 \\ υ_1(x_1) \| = 32336q(x_1) + \floorratio{15q(x_1) + 765433r(x_1) + 12084}{25920} \\ υ_2(x_1) \| = υ_1(x_1) + \left(\floorratio{6×(υ_1(x_1) \bmod 7)}{7} \bmod 2\right) \end{align}
Calculate in the same way also $$υ_2(x_1 + 1)$$. Then
\begin{align} L_2(x_1) \| = υ_2(x_1 + 1) − υ_2(x_1) \\ L_2(x_1 − 1) \| = υ_2(x_1) − υ_2(x_1 − 1) \\ v_3 \| = 2\left(\floorratio{L_2(x_1) + 19}{15} \bmod 2\right) \\ v_4 \| = \floorratio{L_2(x_1 − 1) + 7}{15} \bmod 2 \\ c_2(x_1) \| = υ_2 + v_3 + v_4 \end{align}
Calculate in the same way also $$c_2(x_1 + 1)$$ (which means you need to calculate $$υ_2(x_1 + 2)$$, too). Then
\begin{align} L \| = c_2(x_1 + 1) − c_2(x_1) \\ c_8 \| = \floorratio{L + 7}{2} \bmod 15 \\ c_9 \| = −\left( \floorratio{385 − L}{2} \bmod 15 \right) \\ c_3 \| = \floorratio{384x_3 + 7}{13} + c_8\floorratio{x_3 + 4}{12} + c_9\floorratio{x_3 + 3}{12} \\ c_4(x_1,x_3) \| = c_2(x_1) + c_3(x_3) \\ J \| = J_0 − 177 + c_4(x_1,x_3) + z_4 = 347821 + c_2(x_1) + c_3 + z_4 \end{align}
As an example, we'll calculate which CJDN corresponds to 18 Sivan A.M. 4682. Then $$j = 4682$$, $$m = 3$$, $$d = 18$$. And then
\begin{align*} c_0 \| = \floorratio{13 − m}{7} \\ \| = \floorratio{13 − 3}{7} \\ \| = \floorratio{10}{7} = 1 \\ x_1 \| = j − 1 + c_0 = 4682 − 1 + 1 = 4682 \\ x_3 \| = m − 1 = 3 − 1 = 2 \\ z_4 \| = d − 1 = 18 − 1 = 17 \\ c_1(4681) \| = c_1(x_1 − 1) = \floorratio{235×(x_1 − 1) + 1}{19} \\ \| = \floorratio{235×4681 + 1}{19} \\ \| = \floorratio{1100036}{19} = 57896 \\ c_1(4682) \| = c_1(x_1) = \floorratio{235×x_1 + 1}{19} \\ \| = \floorratio{235×4682 + 1}{19} \\ \| = \floorratio{1100270 + 1}{19} \\ \| = \floorratio{1100271}{19} = 57909 \\ c_1(4683) \| = c_1(x_1 + 1) = \floorratio{235×(x_1 + 1) + 1}{19} \\ \| = \floorratio{235×4683 + 1}{19} \\ \| = \floorratio{1100506}{19} = 57921 \\ c_1(4684) \| = c_1(x_1 + 2) = \floorratio{235×(x_1 + 2) + 1}{19} \\ \| = \floorratio{235×4684 + 1}{19} \\ \| = \floorratio{1100741}{19} = 57933 \\ q(4681) \| = q(x_1 − 1) = \floorratio{c_1(x_1 − 1)}{1095} \\ \| = \floorratio{57896}{1095} = 52 \\ r(4681) \| = r(x_1 − 1) = c_1(x_1 − 1) \bmod 1095 \\ \| = 57896 \bmod 1095 = 956 \\ υ_1(4681) \| = υ_1(x_1 − 1) = 32336×q(x_1 − 1) \\ \| + \floorratio{15×q(x_1 − 1) + 765433×r(x_1 − 1) + 12084}{25920} \\ \| = 32336×52 + \floorratio{15×52 + 765433×956 + 12084}{25920} \\ \| = 1681472 + \floorratio{731766812}{25920} \\ \| = 1681472 + 28231 = 1709703 \\ υ_2(4681) \| = υ_2(x_1 − 1) = υ_1(x_1 − 1) + \left( \floorratio{6×(υ_1(x_1 − 1) \bmod 7)}{7} \bmod 2\right) \\ \| = 1709703 + \left( \floorratio{6×(1709703 \bmod 7)}{7} \bmod 2 \right) \\ \| = 1709703 + \left( \floorratio{6×2}{7} \bmod 2 \right) \\ \| = 1709703 + \left( \floorratio{12}{7} \bmod 2 \right) \\ \| = 1709703 + (1 \bmod 2) = 1709703 + 1 = 1709704 \\ q(4682) \| = q(x_1) = \floorratio{c_1(x_1)}{1095} \\ \| = \floorratio{57909}{1095} = 52 \\ r(4682) \| = r(x_1) = c_1(x_1) \bmod 1095 = 57909 \bmod 1095 = 969 \\ υ_1(4682) \| = υ_1(x_1) = 32336×q(x_1) \\ \| + \floorratio{15×q(x_1) + 765433×r(x_1) + 12084}{25920} \\ \| = 32336×52 + \floorratio{15×52 + 765433×969 + 12084}{25920} \\ \| = 1681472 + \floorratio{741717441}{25920} \\ \| = 1681472 + 28615 = 1710087 \\ υ_2(4682) \| = υ_2(x_1) = υ_1(x_1) + \left( \floorratio{6×(υ_1(x_1) \bmod 7)}{7} \bmod 2 \right) \\ \| = 1710087 + \left( \floorratio{6×(1710087 \bmod 7)}{7} \bmod 2 \right) \\ \| = 1710087 + \left( \floorratio{1}{7} \bmod 2 \right) = 1710087 + (0 \bmod 2) \\ \| = 1710087 + 0 = 1710087 \\ q(4683) \| = q(x_1 + 1) = \floorratio{c_1(x_1 + 1)}{1095} \\ \| = \floorratio{57921}{1095} = 52 \\ r(4683) \| = r(x_1 + 1) = c_1(x_1 + 1) \bmod 1095 \\ \| = 57921 \bmod 1095 = 981 \\ υ_1(4683) \| = υ_1(x_1 + 1) = 32336×q(x_1 + 1) \\ \| + \floorratio{15×q(x_1 + 1) + 765433×r(x_1 + 1) + 12084}{25920} \\ \| = 332336×52 + \floorratio{15×52 + 765433×981 + 12084}{25920} \\ \| = 17281472 + \floorratio{750902637}{25920} \\ \| = 17281472 + 28970 = 17310442 \\ υ_2(4683) \| = υ_2(x_1 + 1) = υ_1(x_1 + 1) \\ \| + \left( \floorratio{6×(υ_1(x_1 + 1) \bmod 7}){7} \bmod 2 \right) \\ \| = 1710442 + \left( \floorratio{6×(1710442 \bmod 7)}{7} \bmod 2 \right) \\ \| = 1710442 + \left( \floorratio{6×6}{7} \bmod 2 \right) \\ \| = 1710442 + \left( \floorratio{36}{7} \bmod 2 \right) \\ \| = 1710442 + (5 \bmod 2) = 1710442 + 1 = 1710443 \\ q(4684) \| = q(x_1 + 2) = \floorratio{c_1(x_1 + 2)}{1095} \\ \| = \floorratio{57933}{1095} = 52 \\ r(4684) \| = r(x_1 + 2) = c_1(x_1 + 2) \bmod 1095 \\ \| = 57933 \bmod 1095 = 993 \\ υ_1(4684) \| = υ_1(x_1 + 2) = 32336×q(x_1 + 2) \\ \| + \floorratio{15×q(x_1 + 2) + 765433×r(x_1 + 2) + 12084}{25920} \\ \| = 32336×52 + \floorratio{15×52 + 765433×993 + 12084}{25920} \\ \| = 1681472 + \floorratio{760087833}{25920} \\ \| = 1681472 + 29324 = 1710796 \\ υ_2(4684) \| = υ_2(x_1 + 2) = υ_1(x_1 + 2) + \left( \floorratio{6×(υ_1(x_1 + 2) \bmod 7)}{7} \bmod 2 \right) \\ \| = 1710796 + \left( \floorratio{6×(1710796 \bmod 7)}{7} \bmod 2 \right) \\ \| = 1710796 + \left( \floorratio{6×3}{7} \bmod 2 \right) \\ \| = 1710796 + \left( \floorratio{18}{7} \bmod 2 \right) = 1710796 + (2 \bmod 2) \\ \| = 1710796 + 0 = 1710796 \\ L_2(4681) \| = L_2(x_1 − 1) = υ_2(x_1) − υ_2(x_1 − 1) \\ \| = 1710087 − 1709704 = 383 \\ L_2(4682) \| = L_2(x_1) = υ_2(x_1 + 1) − υ_2(x_1) \\ \| = 1710443 − 1710087 = 356 \\ L_2(4683) \| = L_2(x_1 + 1) = υ_2(x_1 + 2) − υ_2(x_1 + 1) \\ \| = 1710796 − 1710443 = 353 \\ v_3(4682) \| = v_3(x_1) = 2×\left( \floorratio{L_2(x_1) + 19}{15} \bmod 2 \right) \\ \| = 2×\left( \floorratio{356 + 19}{15} \bmod 2 \right) \\ \| = 2×\left( \floorratio{375}{15} \bmod 2 \right) \\ \| = 2×(25 \bmod 2) = 2 \\ v_4(4682) \| = v_4(x_1) = \floorratio{L_2(x_1 − 1) + 7}{15} \bmod 2 \\ \| = \floorratio{383 + 7}{15} \bmod 2 \\ \| = \floorratio{390}{15} \bmod 2 = 26 \bmod 2 = 0 \\ c_2(4682) \| = c_2(x_1) = υ_2(x_1) + v_3(x_1) + v_4(x_1) \\ \| = 1710087 + 2 + 0 = 1710089 \end{align*}
\begin{align*} v_3(4683) \| = v_3(x_1 + 1) = 2×\left( \floorratio{L_2(x_1 + 1) + 19}{15} \bmod 2 \right) \\ \| = 2×\left( \floorratio{353 + 19}{15} \bmod 2 \right) \\ \| = 2×\left( \floorratio{372}{15} \bmod 2 \right) \\ \| = 2×(24 \bmod 2) = 0 \\ v_4(4683) \| = v_4(x_1 + 1) = \floorratio{L_2(x_1) + 7}{15} \bmod 2 \\ \| = \floorratio{356 + 7}{15} \bmod 2 \\ \| = \floorratio{363}{15} \bmod 2 = 24 \bmod 2 = 0 \\ c_2(4683) \| = c_2(x_1 + 1) = υ_2(x_1 + 1) + v_3(x_1 + 1) + v_4(x_1 + 1) = 1710443 \\ L \| = y_2(x_1 + 1) − y_2(x_1) = 1710443 − 1710089 = 354 \\ c_8 \| = \floorratio{L + 7}{2} \bmod 15 \\ \| = \floorratio{354 + 7}{2} \bmod 15 \\ \| = \floorratio{361}{2} \bmod 15 = 180 \bmod 15 = 0 \\ c_9 \| = −\left( \floorratio{385 − L}{2} \bmod 15 \right) \\ \| = −\left( \floorratio{385 − 354}{2} \bmod 15 \right) \\ \| = −\left( \floorratio{31}{2} \bmod 15 \right) \\ \| = −(15 \bmod 15) = 0 \\ c_3 \| = \floorratio{384×x_3 + 7}{13} + c_8×\floorratio{x_3 + 4}{12} + c_9×\floorratio{x_3 + 3}{12} \\ \| = \floorratio{384×2 + 7}{13} + 0×\floorratio{2 + 4}{12} + 0×\floorratio{2 + 3}{12} \\ \| = \floorratio{775}{13} + 0 + 0 = 59 \\ c_4 \| = c_2 + c_3 = 1710089 + 59 = 1710148 \\ J \| = J_0 − 177 + c_4 + z_4 = 347821 + 1710148 + 17 = 2057986 \end{align*}
This date corresponds to 17 June 922 in the Julian calendar.
### 8.2. From CJDN to Jewish Calendar
From CJDN $$J$$ we go to Jewish year $$j$$, month $$m$$, and day $$d$$:
\begin{align} y_4 \| = J − 347821 \\ q \| = \floorratio{y_4}{1447} \\ r \| = y_4 \bmod 1447 \\ y_1' \| = 49q + \floorratio{23q + 25920r + 13835}{765433} \\ γ_1 \| = y_1' + 1 \\ ξ_1 \| = \floorratio{19γ_1 + 17}{235} \\ μ_1 \| = γ_1 − \floorratio{235ξ_1 + 1}{19} \end{align}
Calculate $$c_{41}' = c_4(ξ_1, μ_1)$$ in the way described in the previous section. Then
\begin{align} ζ_1 \| = y_4 − c_{41}' \\ γ_2 \| = γ_1 + \floorratio{ζ_1}{33} \\ ξ_2 \| = \floorratio{19γ_2 + 17}{235} \\ μ_2 \| = γ_2 − \floorratio{235ξ_2 + 1}{19} \end{align}
Calculate $$c_{42}' = c_4(ξ_2, μ_2)$$ in the way described in the previous section. Then
\begin{align} ζ_2 \| = y_4 − c_{42}' \\ γ_3 \| = γ_2 + \floorratio{ζ_1}{33} \\ x_1 \| = ξ_3 = \floorratio{19γ_3 + 17}{235} \\ x_3 \| = μ_3 = γ_3 − \floorratio{235ξ_3 + 1}{19} \end{align}
Calculate $$c_{43}' = c_4(ξ_3, μ_3)$$ in the way described in the previous section. Then
\begin{align} z_4 \| = ζ_3 = y_1 − c_{43}' \\ c \| = \floorratio{12 − x_3}{7} \\ j \| = x_1 + 1 − c \\ m \| = x_3 + 1 \\ d \| = z_4 + 1 \end{align}
For $$J = 2057986$$ we find
\begin{align*} y_4 \| = J − 347821 = 2057986 − 347821 = 1710165 \\ q \| = \floorratio{y_4}{1447} = \floorratio{1710165}{1447} = 1181 \\ r \| = y_4 \bmod 1447 = 1710165 \bmod 1447 = 1258 \\ y_1' \| = 49q + \floorratio{23q + 25920r + 13835}{765433} \\ \| = 49×1181 + \floorratio{23×1181 + 25920×1258 + 13835}{765433} \\ \| = 57869 + \floorratio{32648358}{765433} = 57869 + 42 = 57911 \\ γ_1 \| = y_1' + 1 = 57911 + 1 = 57912 \\ ξ_1 \| = \floorratio{19γ_1 + 17}{235} \\ \| = \floorratio{19×57912 + 17}{235} \\ \| = \floorratio{1100345}{235} = 4682 \\ μ_1 \| = γ_1 − \floorratio{235ξ_1 + 1}{19} \\ \| = 57912 − \floorratio{235×4682 + 1}{19} \\ \| = 57912 − \floorratio{1100271}{19} = 57912 − 57909 = 3 \end{align*}
In the way described in the previous section we calculate that
$c_{41}' = c_4(ξ_1,μ_1) = c_4(4682,3) = 1710178$
Then
\begin{align*} ζ_1 \| = y_4 − c_{41}' = 1710165 − 1710178 = −13 \\ γ_2 \| = γ_1 + \floorratio{ζ_1}{33} = 57912 + \floorratio{−13}{33} = 57912 + −1 = 57911 \\ ξ_2 \| = \floorratio{19γ_2 + 17}{235} = \floorratio{19×57911 + 17}{235} = \floorratio{1100326}{235} = 4682 \\ μ_2 \| = γ_2 − \floorratio{235ξ_2 + 1}{19} = 57911 − \floorratio{235×4682 + 1}{19} \\ \| = 57911 − \floorratio{1100271}{19} = 57911 − 57909 = 2 \end{align*}
In the way described in the previous section we calculate that
$c_{42} = c_4(ξ_2,μ_2) = c_4(4682,2) = 1710148$
Then
\begin{align*} ζ_2 \| = y_4 − c_{42}' = 1710165 − 1710148 = 17 \\ γ_3 \| = γ_2 + \floorratio{ζ_2}{33} = 57911 + \floorratio{17}{33} = 57911 + 0 = 57911 \end{align*}
Because in this example $$ζ_3 = ζ_2$$, also $$ξ_3 = ξ_2$$, $$μ_3 = μ_2$$, $$c_4(ξ_3,μ_3) = c_4(ξ_2,μ_2)$$, and $$ζ_3 = ζ_2$$, so in this example you need not calculate $$ξ_3$$, $$μ_3$$, $$c_4(ξ_3,μ_3)$$, $$ζ_3$$ explicitly, but you can still do so if that is more convenient.
\begin{align*} x_1 \| = ξ_3 = \floorratio{19γ_3 + 17}{235} = \floorratio{19×57911 + 17}{235} = \floorratio{1100326}{235} = 4682 \\ x_3 \| = μ_3 = γ_3 − \floorratio{235ξ_3 + 1}{19} = 57911 − \floorratio{235×4682 + 1}{19} \\ \| = 57911 − \floorratio{1100271}{19} = 57911 − 57909 = 2 \end{align*}
In the way described in the previous section we calculate that
$c_{43}' = c_4(ξ_3,μ_3) = c_4(4682,2) = 1710148$
Then
\begin{align*} z_4 \| = ζ_3 = y_4 − c_{43}' = 1710165 − 1710148 = 17 \\ c \| = \floorratio{12 − x_3}{7} = \floorratio{12 − 2}{7} = \floorratio{10}{7} = 1 \\ j \| = x_1 + 1 − c = 4682 + 1 − 1 = 4682 \\ m \| = x_3 + 1 = 2 + 1 = 3 \\ d \| = z_4 + 1 = 17 + 1 = 18 \end{align*}
so the desired date in the Jewish calendar is day 18 of month 3 (Sivan) of year 4682: 18 Sivan A.M. 4682.
## 9. The Egyptian Calendar
The ancient Egyptians had a very simple calendar indeed, without leap years and with 30 days to every month, except that the last month had 5 days. In the calendar according to the Era of Nabonassar the first day (1 Thoth of year 1) was equivalent to 26 February −746 in the Julian calendar.
### 9.1. From Egyptian Date to CJDN
The algorithm to translate an Egyptian date (calendar year $$j$$, calendar month $$m$$, calendar day $$d$$) to CJDN $$J$$ is as follows:
$$J = 365 j + 30 m + d + 1448242$$
For example, which CJDN corresponds to day 7 of month 5 of year 218 in the Egyptian calendar? Then we find
\begin{align*} J \| = 365j + 30m + d + 1448242 \\ \| = 365×218 + 30×5 + 7 + 1448242 \\ \| = 79570 + 150 + 7 + 1448242 = 1527969 \end{align*}
so the answer is CJDN 1527969.
### 9.2. From CJDN to Egyptian Date
In the opposite direction, the calculations are simple as well.
\begin{align} y_2 \| = J − 1448638 \\ x_2 \| = \floorratio{y_2}{365} \\ y_1 \| = y_2 \bmod 365 \\ j \| = x_2 + 1 \\ m \| = \floorratio{y_1}{30} + 1 \\ d \| = y_1 − 30m + 31 = (y_1 \bmod 30) + 1 \end{align}
What date in the Egyptian calendar corresponds to CJDN 1527969? Then
\begin{align*} y_2 \| = J − J_0 = 1527969 − 1448638 = 79331 \\ x_2 \| = \floorratio{y_2}{365} = \floorratio{79331}{365} = 217 \\ y_1 \| = y_2 − 365x_2 = 79331 − 365×217 = 79331 − 79205 = 126 \\ j \| = x_2 + 1 = 217 + 1 = 218 \\ m \| = \floorratio{y_1}{30} + 1 = \floorratio{126}{30} + 1 = 4 + 1 = 5 \\ d \| = y_1 − 30m + 31 = 126 − 30×5 + 31 = (y_1 \bmod 30) + 1 = (126 \bmod 30) + 1 = 6 + 1 = 7 \end{align*}
so it is day 7 of month 5 of year 218.
## 10. The Maya Calendar
### 10.1. From Maya Calendar to CJDN
The Maya from Anahuac (Central America) used three different calendars, of which two (the Tzolkin and Haab) were periodic with fairly short periods, and the third one (the Long Count) might have been intended to be periodic but has some periods that are so long that it can be considered to be continuous rather than repeating.
Different areas in Central America had slightly different versions of these calendars, with different names for days and months, and different ways to indicate years, and sometimes days were counted from 0 rather than from 1. Below we describe the calendars from the city of Tikal.
#### 10.1.1. From Haab to CJDN
The Haab has a day number and a month, but no year number. There are 18 months of 20 days, plus a 19th month of 5 days, which makes 365 days in total, which we call a Haab year. There are no leap years. The months have a name, and the days have a number beginning at 0.
We write a date in the Haab as $$\{h_d,h_m\}$$ where $$h_d$$ is the day number (from 0 through 19) and $$h_m$$ is the month number (from 1 through 19).
There is no year number in the Haab, so a particular Haab date $$\{h_d,h_m\}$$ returns every 365 days. We need to apply additional information to pick the right one. We can find the last CJDN $$J$$ on or before a particular CJDN $$J_0$$ that corresponds to this Haab date, as follows:
\begin{align} H \| = h_d + 20×(h_m − 1) \\ J \| = J_0 − ((J_0 − H + 65) \bmod 365) \end{align}
What is the last CJDN in the year 1965 that corresponds to Haab date $$\{5,13\}$$? Then $$h_d = 5$$, $$h_m = 13$$, $$J_0 = 2439126$$ (which corresponds to 31 December 1965), so
\begin{align*} H \| = 5 + 20×(13 − 1) = 245 \\ J \| = J_0 − ((J_0 − H + 65) \bmod 365) \\ \| = 2439126 − (2438946 \bmod 365) \\ \| = 2439126 − 16 = 2439110 \end{align*}
#### 10.1.2. From Tzolkin to CJDN
The Tzolkin has a period of 20 days (the venteina) with a name for each day, and a period of 13 days (the trecena) with a number for each day (beginning at 1).
We write a date in the Tzolkin as $$\{t_t,t_v\}$$, where $$t_t$$ is the day number in the trecena (from 1 through 13) and $$t_v$$ is the day number in the venteina (from 1 through 20).
The trecena and venteina increase simultaneously, so after day $$\{4,7\}$$ comes day $$\{5,8\}$$, and then $$\{6,9\}$$, and so on. Because 13 and 20 are relatively prime, all possible combinations of venteina and trecena occur in this calendar, so after 13×20 = 260 days the dates repeat again.
There is no year number in the Tzolkin so a particular Tzolkin date returns every 260 days. We can find the last CJDN $$J$$ that corresponds to Tzolkin date $$\{t_t,t_v\}$$ on or before a particular CJDN $$J_0$$ as follows:
$$J = J_0 − ((J_0 − 40 t_t + 39 t_v + 97) \bmod 260)$$
The last Tzolkin date $$\{4,7\}$$ that occurs on or before 31 December 1965 (= CJDN 2439126) is
\begin{align*} J \| = 2439126 − ((2439126 − 40×4 + 39×7 + 97) \bmod 260) \\ \| = 2439126 − (2439336 \bmod 260) \\ \| = 2439126 − 16 = 2439110 \end{align*}
For the number of days $$T$$ since the last $$\{1,1\}$$ we have
$$T = (40 t_t + 221 t_v − 1) \bmod 260$$
For CJDN 2439110 (with $$\{t_t,t_v\} = \{4,7\}$$) we find
\begin{align*} T \| = (40×4 + 221×7 − 1) \bmod 260 \\ \| = 1706 \bmod 260 = 146 \end{align*}
so the last day before then that had $$\{1,1\}$$ was CJDN $$J = 2439110 − 146 = 2438964$$, and that fits, because $$t_t = ((J + 5) \bmod 13) + 1 = (2438969 \bmod 13) + 1 = 1$$ and $$t_v = ((J + 16) \bmod 20) + 1 = (2438980 \bmod 20) + 1 = 1$$.
#### 10.1.3. From Tzolkin and Haab to CJDN
Sometimes a date is indicated in both Tzolkin and Haab. The CJDN $$J$$ of the last Tzolkin/Haab date $$\{ h_d, h_m, t_t, t_v \}$$ on or before CJDN $$J_0$$ is:
\begin{align} H \| = (h_d + 20 × (h_m − 1)) \bmod{365} \\ T \| = (40t_t + 221 t_v − 1) \bmod{260} \\ J \| = J_0 − ((J_0 − 365 T + 364 H − 7600) \bmod{18980}) \end{align}
The Tzolkin and Haab together have a period of 18980 days, which is approximately 52 years.
Which CJDN is the last one before the end of 1965 that corresponds to $$\{t_t,t_v,h_d,h_m\} = \{4,7,5,13\}$$? Then $$J_0 = 2439126$$ and
\begin{align*} H \| = (5 + 20×12) \bmod{365} = 245 \bmod{365} = 245 \\ T \| = (40×4 + 221×7 − 1) \bmod{260} = 1706 \bmod{260} = 146 \\ J \| = 2439126 − ((2439126 − 365×146 + 364×245 − 7600) \bmod{18980}) \\ \| = 2439126 − (2467416 \bmod{18980}) \\ \| = 2439126 − 16 = 2439110 \end{align*}
Beware! Not all possible combinations of $$\{ h_d, h_m, t_t, t_v \}$$ occur in the calendar. You can fill in any combination you like in the preceding formulas, but you'll get useful results only for combinations that really occur in the calendar. Those are combinations for which $$H − T ≡ 65 − 96 ≡ 4 \pmod{5}$$.
In the example that we've been using we have $$H = 245$$ and $$T = 146$$, so $$H − T = 99 ≡ 4 \pmod{5}$$, so this combination indeed occurs in this calendar.
#### 10.1.4. From the Long Count to CJDN
The Long Count counts days and has a series of increasingly longer periods. The smallest period is 20 days, the next one is 18 times longer (i.e., 360 days), and after that each next period is 20 times longer than the previous one. The number for each period begins at 0. Often dates in the Long Count are given with five numbers, but even longer periods are known, up to nine numbers. The longest known period corresponds to about 63 million years. Here we assume that the Long Count has five numbers, and that the fifth number can get arbitrarily long.
The epoch (0.0.0.0.0) of the Long Count probably corresponds to CJDN $$J_0 = 584283$$ (6 September −3113 in the Julian calendar).
Translating a Long Count to CJDN goes as follows:
$$\begin{split} J \| = l_1 + 20×(l_2 + 18×(l_3 + 20×(l_4 + 20×l_5))) + J_0 \\ \| = l_1 + 20×l_2 + 360×l_3 + 7200×l_4 + 144000×l_5 \end{split}$$
Which CJDN corresponds to Long Count $$L = 12.17.12.5.7$$? Then
$J = 7 + 20×(5 + 18×(12 + 20×(17 + 20×12))) + 584283 = 2439110$
### 10.2. From CJDN to the Maya Calendar
#### 10.2.1. From CJDN to Haab
We calculate Haab date $$\{h_d,h_m\}$$ from CJDN $$J$$ as follows:
\begin{align} H \| = (J + 65) \bmod 365 \\ h_m \| = \floorratio{H}{20} + 1 \\ h_d \| = H \bmod 20 \end{align}
$$H$$ is the number of days since the beginning of the current Haab year (with $$H = 0$$ for the first day of the Haab year).
For example, which Haab date corresponds to 15 December 1965 = CJDN 2439110? Then $$J = 2439110$$ and
\begin{align*} H \| = (J + 65) \bmod 365 = 2439175 \bmod 365 = 245 \\ h_m \| = \floorratio{245}{20} + 1 = 13 \\ h_d \| = 245 \bmod 20 = 5 \end{align*}
so that day is 245 days since the beginning of the current Haab year and corresponds to the 5th day of the 13th month, i.e., $$\{5,13\}$$.
#### 10.2.2. From CJDN to Tzolkin
We translate CJDN $$J$$ to a Tzolkin date as follows:
\begin{align} t_t \| = ((J + 5) \bmod{13}) + 1 \label{eq:trecena} \\ t_v \| = ((J + 16) \bmod{20}) + 1 \label{eq:venteina} \end{align}
What is the Tzolkin date corresponding to CJDN 2439110? Then
\begin{align*} t_t \| = ((2439110 + 5) \bmod{13}) + 1 = 3 + 1 = 4 \\ t_v \| = ((2439110 + 16) \bmod{20}) + 1 = 6 + 1 = 7 \end{align*}
so that date is $$\{4,7\}$$.
#### 10.2.3. From CJDN to the Long Count
Translating CJDN $$J$$ to Long Count $$L ≡ l_5.l_4.l_3.l_2.l_1$$ goes as follows:
\begin{align} x_5 \| = J − J_0 \\ l_5 \| = \floorratio{x_5}{144000} \\ x_4 \| = x_5 \bmod 144000 \\ l_4 \| = \floorratio{x_4}{7200} \\ x_3 \| = x_4 \bmod 7200 \\ l_3 \| = \floorratio{x_3}{360} \\ x_2 \| = x_3 \bmod 360 \\ l_2 \| = \floorratio{x_2}{20} \\ l_1 \| = x_2 \bmod 20 \end{align}
Which Long Count $$L$$ corresponds to CJDN $$J = 2439110$$? Then
\begin{align*} x_5 \| = J − J_0 = 2439110 − 584283 = 1854827 \\ l_5 \| = \floorratio{1854827}{144000} = 12 \\ x_4 \| = 1854827 \bmod 144000 = 126827 \\ l_4 \| = \floorratio{126827}{7200} = 17 \\ x_3 \| = 126827 \bmod 7200 = 4427 \\ l_3 \| = \floorratio{4427}{360} = 12 \\ x_2 \| = 4427 \bmod 360 = 107 \\ l_2 \| = \floorratio{107}{20} = 5 \\ l_1 \| = 107 \bmod 20 = 7 \end{align*}
The answer is $$L = 12.17.12.5.7$$.
### 10.3. A LunarCalendar With Many Fixed Month Lengths
I've designed a lunisolar calendar in which nearly all months have the same number of days in every calendar year. The calendar follows the seasons by means of the Cycle of Meton, which lasts 19 years. Each year contains 12 or 13 months. The "shortest" years of 12 months contain 354 days. The "short" years of 12 months contain 355 days. The "long" years of 13 months contain 384 days. The year lengths are:
Year Months Days Type
1 12 354 shortest
2 12 355 short
3 13 384 long
4 12 354 shortest
5 12 354 shortest
6 13 384 long
7 12 355 short
8 13 384 long
9 12 354 shortest
10 12 354 shortest
11 13 384 long
12 12 355 short
13 12 354 shortest
14 13 384 long
15 12 354 shortest
16 12 355 short
17 13 384 long
18 12 354 shortest
19 13 384 long
All together there are 235 months (125 long, 110 short) and 6940 days in the cycle of 19 years.
All odd months (except for the 13th) contain 30 days, and all even months contain 29 days, except that in short and long years (of 355 and 384 days) the 12th month contains 30 days instead of 29.
354 355 384
Month Days
1 30 30 30
2 29 29 29
3 30 30 30
4 29 29 29
5 30 30 30
6 29 29 29
7 30 30 30
8 29 29 29
9 30 30 30
10 29 29 29
11 30 30 30
12 29 30 30
13 29
#### 10.3.1. From LunarCalendar to Running Day Number
Given year $$j$$, month $$m$$, and day $$d$$ in this lunar calendar, the running day number $$y_2$$ can be calculated as follows:
$$y_2 = 354j + 29m + 30\floorratio{7j − 6}{19} + \floorratio{4j + 8}{19} + \floorratio{7m}{13} + d − 384$$
For example, which running day number corresponds to day 27 of month 9 of year 3 in this calendar? Then
\begin{align*} j \| = 3 \\ m \| = 9 \\ d \| = 27 \\ y_2 \| = 354×3 + 29×9 + 30\floorratio{7×3 − 6}{19} + \floorratio{4×3 + 8}{19} + \floorratio{7×9}{13} + 27 − 384 \\ \| = 1323 + 30\floorratio{15}{19} + \floorratio{20}{19} + \floorratio{63}{13} − 357 \\ \| = 1323 + 0 + 1 + 4 − 357 = 971 \end{align*}
#### 10.3.2. From Running Day Number to LunarCalendar
Given running day number $$y_2$$, we calculate year $$j$$, month $$m$$, and day $$d$$ in this lunar calendar as follows:
\begin{align} x' \| = \floorratio{19y_2 + 546}{6940} \\ x_2 \| = x' + \floorratio{y_2 − 354x' − 30\floorratio{7x' + 1}{19} − \floorratio{4x' + 12}{19}}{385} \\ k_1 \| = 13×\left( y_2 − 354x_2 − 30\floorratio{7x_2 + 1}{19} − \floorratio{4x_2 + 12}{19} \right) + 5 \\ j \| = x_2 + 1 \\ m \| = \floorratio{k_1}{384} + 1 \\ d \| = \floorratio{k_1 \bmod 384}{13} + 1 \end{align}
For example, which day in this calendar corresponds to running day number 971? Then $$y_2 = 971$$ and then
\begin{align*} x' \| = \floorratio{19×971 + 546}{6940} = \floorratio{18995}{6940} = 2 \\ x_2 \| = 2 + \floorratio{971 − 354×2 − 30 \floorratio{7×2 + 1}{19} − \floorratio{4×2 + 12}{19}}{385} \\ \| = 2 + \floorratio{263 − 30 \floorratio{15}{19} − \floorratio{20}{19}}{385} \\ \| = 2 + \floorratio{263 − 0 − 1}{385} = 2 \\ k_1 \| = 13×\left( 971 − 354×2 − 30\floorratio{7×2 + 1}{19} − \floorratio{4×2 + 12}{19} \right) + 9 \\ \| = 13×\left( 263 − 30\floorratio{15}{19} − \floorratio{20}{19} \right) + 9 \\ \| = 13×(263 − 0 − 1) + 5 = 3411 \\ j \| = 2 + 1 = 3 \\ m \| = \floorratio{3411}{384} + 1 = 9 \\ d \| = \floorratio{3411 \bmod 384}{13} + 1 = \floorratio{339}{13} + 1 = 27 \end{align*}
so the date is day 27 of month 9 of year 3.
## 11. The Chinese Calendar
The Chinese calendar is a lunisolar calendar. The date of Chinese New Year depends on the position of the Sun and the Moon, but a precise method for calculating those positions has not been fixed. Here I discuss a few characteristics of the Chinese calendar.
### 11.1. The Year of the Monkey
In the traditional Chinese calendar, each year has a name from a cycle of 10 celestial stems and a name from a cycle of 12 terrestrial branches. For each next year both the celestial stem and terrestrial branch advance by one. The (untranslatable) celestial stems are 1. jiǎ, 2. yǐ, 3. bǐng, 4. dīng, 5. wù, 6. jǐ, 7. gēng, 8. xīn, 9. rén, 10. guǐ. The terrestrial branches are 1. zǐ (rat), 2. chǒu (ox), 3. yín (tiger), 4. mǎo (rabbit), 5. chén (dragon), 6. sì (snake), 7. wǔ (horse), 8. wèi (goat), 9. shēn (monkey), 10. yǒu (rooster), 11. xū (dog), 12. hài (pig).
You can calculate the celestial stem $$s$$ (1 through 10) and terrestrial branch $$b$$ (1 through 12) from the (astronomical) Gregorian year number $$j$$ as follows:
\begin{align} s \| = ((j + 6) \bmod 10) + 1 \label{eq:jnaarchinees} \\ b \| = ((j + 8) \bmod 12) + 1 \end{align}
The number $$n$$ in the combined cycle of 60 years is
$$n = ((j + 56) \bmod 60) + 1$$
For example, for the year 2000 in the Gregorian calendar:
\begin{align*} s \| = ((2000 + 6) \bmod 10) + 1 = 6 + 1 = 7 \\ b \| = ((2000 + 8) \bmod 12) + 1 = 4 + 1 = 5 \\ n \| = ((2000 + 56) \bmod 60) + 1 = 16 + 1 = 17 \end{align*}
so early in the Gregorian year 2000 a Chinese Year of the Metal Dragon began, of the gēng branch (gēngchén).
The years that have a particular celestial stem $$s$$ or terrestrial branch $$b$$ or combination $$n$$ are
\begin{align} j \| ≡ 1983 + s \pmod{10} \\ j \| ≡ 1983 + b \pmod{12} \\ j \| ≡ 1983 + n \pmod{60} \end{align}
The years that have celestial stem $$s$$ as well as terrestrial branch $$b$$ are
$$j ≡ 1983 + 25b − 24s \pmod{60} \label{eq:chineesnaarj}$$
For example, for $$s = 7$$ we find
$j ≡ 1983 + 7 = 1990 \pmod{10}$
so 1990 and every 10 years earlier or later, so 2000 fits, too.
For $$b = 5$$ we find
$j ≡ 1983 + 5 = 1988 \pmod{12}$
so 1988 and every 12 years earlier or later, so 2000 fits, too.
For $$s = 7$$ and $$b = 5$$ we find
$j ≡ 1983 + 25×5 − 24×7 = 1940 \pmod{60}$
so 1940 and every 60 years earlier or later, so 2000 fits, too.
Beware! Not every combination of $$s$$ and $$b$$ is valid! Only combinations for which $$s ≡ b \pmod{2}$$ are valid, so the difference between $$s$$ and $$b$$ must be even.
For example, the combination $$s = 3$$ and $$b = 6$$ does not occur in this calendar, so is not valid. If you enter that combination into equation \ref{eq:chineesnaarj} then you get an answer that looks reasonable: $$j ≡ 1983 + 25×6 − 24×3 = 2061 \pmod{60}$$, but if you enter $$j = 2061$$ into equations \ref{eq:jnaarchinees}ff then you get $$s = 8$$ and $$b = 6$$, a different (and valid) combination than the one that you started with.
### 11.2. HYSN
The HYSN system is a special way to assign a number to a year. The conversion of a Gregorian (or Julian proleptic) year number to a HYSN year number goes as follows:
\begin{align} \{ H − 1, j_2 \} \| = \Div(j + 67016, 10800) \\ \{ Y − 1, j_3 \} \| = \Div(j_2, 360) \\ \{ S − 1, N − 1 \} \| = \Div(j_3, 30) \end{align}
For example, which HYSN year corresponds approximately to our year 2016? Then
\begin{align*} j \| = 2016 \\ \{ H − 1, j_2 \} \| = \Div(2016 + 67016, 10800) = \Div(69032, 10800) = \{ 6, 4232 \} \\ H \| = 7 \\ j_2 \| = 4232 \\ \{ Y − 1, j_3 \} \| = \Div(4232, 360) = \{ 11, 272 \} \\ Y \| = 12 \\ j_3 \| = 272 \\ \{ S − 1, N − 1 \} \| = \Div(272, 30) = \{ 9, 2 \} \\ S \| = 10 \\ N \| = 3 \end{align*}
so the HYSN year number is 0712-1003.
In the other direction things are simpler. Given $$H$$, $$Y$$, $$S$$, and $$N$$, the Gregorian year is
$$j = H×10800 + Y×360 + S×30 + N − 78207$$
For example, HYSN 0712-1003 means
\begin{align*} H \| = 7 \\ Y \| = 12 \\ S \| = 10 \\ N \| = 3 \\ j \| = 7×10800 + 12×360 + 10×30 + 3 − 78207 \\ \| = 75600 + 4320 + 300 + 3 − 78207 = 2016 \end{align*}
## 12. Derivation of the General Algorithms
We can derive many calendar formulas by finding straight lines that, with rounding, yield the correct results.
### 12.1. Notation
We use special notation for some things that are useful later.
$$C(x)$$ has value 1 if test $$x$$ is satisfied, and 0 otherwise.
$$x∥y$$ asserts that $$x$$ is a multiple of $$y$$.
$$x⊥y$$ asserts that $$x$$ is not a multiple of $$y$$.
$$⌊x⌋$$ is the floor function, which returns the nearest whole number not exceeding $$x$$ (i.e., in the direction of $$−∞$$).
$$⌊x⌉_y ≡ x \bmod y ≡ x − y\floorratio{x}{y}$$
is the modulus function that yields the non-negative difference between $$x$$ and the nearest multiple of $$y$$ not exceeding $$x$$. Always
\begin{align} x \| = ⌊x⌋ + ⌊x⌉_1 \\ x \| = y\floorratio{x}{y} + ⌊x⌉_y \\ ⌊x⌉_y \| = y\left\lfloor \frac{x}{y} \right\rceil_1 \end{align}
$$⌈x⌉$$ is the ceiling function that yields the nearest whole number not less than $$x$$ (i.e., in the direction of $$+∞$$).
$$⌈x⌋_y ≡ x \bdom y ≡ y\ceilratio{x}{y} − x$$
is the domulus function that yields the non-negative difference between $$x$$ and the nearest multiple of $$y$$ not less than $$x$$. I made up the notation $$\bdom$$, which is $$\bmod$$ spelled backwards. And I invented the name "domulus", like "modulus" but with "dom" instead of "mod". Always
\begin{align} x \| = ⌈x⌉ − ⌈x⌋_1 \\ x \| = y\ceilratio{x}{y} − ⌈x⌋_y \\ ⌈x⌋_y \| = y\left\lceil \frac{x}{y} \right\rfloor_1 \end{align}
$$\{ m, n \} = \Div(x,y)$$ is a function that yields two values $$m, n$$ such that $$x = my + n$$ with $$m$$ a whole number, $$0 ≤ n \lt y > 0$$, and
\begin{align*} m \| = \floorratio{x}{y} \\ n \| = ⌊x⌉_y = x \bmod y \end{align*}
A few useful relationships:
\begin{align} ⌈x⌉ − ⌊x⌋ \| = C(x⊥1) \\ ⌈x⌋_p + ⌊x⌉_p \| = pC(x⊥p) \\ ⌊−x⌋ \| = −⌈x⌉ = −⌊x⌋ − C(x⊥1) \\ ⌈−x⌉ \| = −⌊x⌋ = −⌈x⌉ + C(x⊥1) \\ ⌊−x⌉_p \| = ⌈x⌋_p = −⌊x⌉_p + pC(x⊥p) \\ ⌈−x⌋_p \| = ⌊x⌉_p = −⌈x⌋_p + pC(x⊥p) \\ ⌊x + y⌋ \| = ⌊x⌋ + ⌊y⌋ + C(⌊x⌉_1 + ⌊y⌉_1 ≥ 1) \\ ⌊x + y⌉_p \| = ⌊x⌉_p + ⌊y⌉_p − pC(⌊x⌉_p + ⌊y⌉_p ≥ p) \\ ⌊x − y⌋ \| = ⌊x⌋ − ⌊y⌋ − C(⌊x⌉_1 \lt ⌊y⌉_1) \\ ⌊x − y⌉_p \| = ⌊x⌉_p − ⌊y⌉_p + pC(⌊x⌉_p \lt ⌊y⌉_p) \\ ⌈x + y⌉ \| = ⌈x⌉ + ⌈y⌉ − C(⌈x⌋_1 + ⌈y⌋_1 ≥ 1) \\ ⌈x + y⌋_p \| = ⌈x⌋_p + ⌈y⌋_p − pC(⌈x⌋_p + ⌈y⌋_p ≥ p) \\ ⌈x − y⌉ \| = ⌈x⌉ − ⌈y⌉ + C(⌈x⌋_1 \lt ⌈y⌋_1) \\ ⌈x − y⌋_p \| = ⌈x⌋_p − ⌈y⌋_p + pC(⌈x⌋_p \lt ⌈y⌋_p) \end{align}
$${x}$$ $${⌊x⌉_3}$$ $${⌈x⌋_3}$$ −4 2 1 −3 0 0 −2 1 2 −1 2 1 0 0 0 1 1 2 2 2 1 3 0 0 4 1 2
### 12.2. Large Intermediate Results
Many of the calendar calculations that are described below are of the type
\begin{align} y \| = \floorratio{fx + t}{d} \label{eq:template} \\ e \| = (fx + t) \bmod d \label{eq:template-e} \end{align}
i.e.,
$\{y, e\} = \Div(fx + t, d)$
with
$$fx + t = dy + e$$
where $$f$$, $$t$$, $$d$$ are fixed whole numbers with $$f, d \gt 1$$ and $$0 ≤ e \lt d$$ and $$y$$, $$x$$ are variable but whole numbers. The intermediate result $$fx + t$$ is then usually much greater in magnitude than $$x$$ and than the end result. This can give trouble on calculators and in computer programs where whole numbers cannot exceed a certain size that we'll refer to as $$w$$. (Floating-point numbers like 1.234 × 1020 can be much greater, but have limited accuracy, and then you have to worry about round-off errors.)
Intermediate results greater than $$w$$ give trouble, so to avoid trouble we'd need roughly $$|fx| ≤ w$$, so $$|x| ≤ \frac{w}{f}$$, which is much less than $$w$$ itself.
If $$|t| ≥ d$$, then we can rewrite equation \eqref{eq:template}ff to
\begin{align} fx + t \| = fx + \floorratio{t}{d} d + (t \bmod d) \\ y \| = \floorratio{t}{d} + \floorratio{fx + (t \bmod d)}{d} \\ e \| = (fx + (t \bmod d)) \bmod d \end{align}
where the second term of the last two equations has the same form as equation \eqref{eq:template}, if you rename $$t \bmod d$$ to a new $$t$$. Below, we assume that that has been done, which means that $$|t| \lt d$$.
If $$f \gt d$$, then we can rewrite equation \eqref{eq:template}ff to
\begin{align} fx + t \| = \left( \floorratio{f}{d} d + (f \bmod d) \right) x + t \\ y \| = \floorratio{f}{d} x + \floorratio{(f \bmod d) x + t}{d} \label{eq:template2} \\ e \| = ((f \bmod d)x + t) \bmod d \end{align}
Then $$|x|$$ may be nearly as great as $$\frac{w}{f \bmod d}$$ before we run into trouble.
Equations \eqref{eq:template2}ff again have the form of equations \eqref{eq:template}ff, if you rename $$f \bmod d$$ to a new $$f$$. Below, we assume that this has been done, so that $$f \lt d$$.
If we could first do the division by $$d$$ and then the multiplication by $$f$$, then the intermediate results would stay smaller than $$x$$ itself. We'd like to calculate something like $$f\floorratio{x}{d}$$ which is roughly equal to $$y$$, and then apply a correction to get the real $$y$$.
We can rewrite equation \eqref{eq:template}ff to
\begin{align} fx + t \| = f\left( \floorratio{x}{d} d + (x \bmod d) \right) + t \\ y \| = f\floorratio{x}{d} + \floorratio{f(x \bmod d) + t}{d} \label{eq:detour1} \\ e \| = (f(x \bmod d) + t) \bmod d \end{align}
Now the greatest intermediate result is no longer $$fx + t$$, which can get as large as $$fw$$, but $$f(x \bmod d) + t$$, which cannot exceed $$fd$$.
Equations \eqref{eq:detour1}ff again have the form of equations \eqref{eq:template}ff, if you rename $$x \bmod d$$ to a new $$x$$.
For example, let $$d = 12345$$, $$f = 8432$$, $$t = 871$$, and $$w = 2^{31} − 1 = 2,147,483,647$$, and suppose we want to calculate $$y$$, $$e$$ for $$x = 300,000$$.
We use the detour of equation \eqref{eq:detour1}. We find
\begin{align*} \floorratio{x}{d} \| = \floorratio{300000}{12345} = 24 \\ x \bmod d \| = 300000 \bmod 12345 = 3720 \\ y = \| f\floorratio{x}{d} + \floorratio{f(x \bmod d) + t}{d} \\ \| = 8432×24 + \floorratio{8432×3720 + 871}{12345} \\ \| = 202368 + \floorratio{31367911}{12345} \\ \| = 202368 + 2540 = 204908 \\ e \| = 31367911 \bmod 12345 = 11611 \end{align*}
The greatest intermediate result is 31,367,911, which is much less than $$w$$.
If we could have used arbitrarily large intermediate results, then we'd have found
\begin{align*} y \| = \floorratio{fx + t}{d} = \floorratio{8432×300000 + 871}{12345} = \floorratio{2529600871}{12345} = 204908 \\ e \| = 2529600871 \bmod 12345 = 11611 \end{align*}
which are the same results, but with a greatest intermediate result of 2,529,600,871 which is greater than $$w$$.
Sometimes an intermediate result of (nearly) $$fd$$ is still too great, because $$fd$$ can be much greater than $$w$$ even if $$d$$ and $$f$$ are each much less than $$w$$. If $$fd$$ must remain less than $$w$$, then $$d$$ and $$f$$ cannot both be greater than $$\sqrt{w}$$, which is a lot less than $$w$$ itself. For example, for 32-bit numbers, $$w = 2^{31} − 1 = 2,147,483,647$$ and $$\sqrt{w} = 46341$$, so even if $$d$$ and $$f$$ are both just a bit greater than 46431, which isn't really very large, then $$fd$$ is already too great to fit into a 32-bit number.
Suppose we pick a $$P \gt 0$$ and then calculate once
\begin{align} \{Q, R\} \| = \Div(fP, d) \\ Q \| = \floorratio{fP}{d} \\ R \| = fP \bmod d \end{align}
so
$$fP = dQ + R$$
Then, for each desired $$x$$, calculate
\begin{align} \{q, r\} \| = \Div(x, P) \\ q \| = \floorratio{x}{P} \\ r \| = x \bmod P \end{align}
so
$$x = qP + r$$
and then
\begin{align} fx + t \| = fqP + fr + t = qdQ + qR + fr + t \\ y \| = \floorratio{qdQ + qR + fr + t}{d} = qQ + \floorratio{qR + fr + t}{d} \label{eq:detour2} \\ e \| = (qdQ + qR + fr + t) \bmod d = (qR + fr + t) \bmod d \end{align}
Now the greatest intermediate result is $$m ≡ qR + fr + t$$. We know that $$r \lt P$$, so $$fr \lt fP$$. To stay out of trouble we need $$|m| ≤ w$$. We have
$|m| ≤ \left| \floorratio{x}{P} \right| R + fP + |t| ≤ |x|\frac{R}{P} + fP + |t|$
so we need
$|x|\frac{R}{P} + fP + |t| ≤ w$
If $$R = 0$$, then we need $$fP + |t| ≤ w$$, so then
$$P ≤ \frac{w − |t|}{f}$$
If $$R \gt 0$$ then we need
$$|x| ≤ X ≡ (w − fP − |t|) \frac{P}{R}$$
We must have $$0 \lt P \lt \frac{w − |t|}{f} ≤ \frac{w}{f}$$, otherwise certainly $$X ≤ 0$$.
We want to make the range of $$x$$ as large as possible, so we seek $$P$$ and $$R$$ such that $$X$$ becomes as great as possible, given $$w$$, $$f$$, $$t$$. This means that $$R$$ should preferably be small.
If we use equations \eqref{eq:detour1}ff then it is sufficient if $$X ≥ d$$ to be able to handle all values $$|x| \lt w$$, because then we can use equations \eqref{eq:detour1}ff to first make $$|x| \lt d$$. If we do not use equations \eqref{eq:detour1}ff, then we must have $$X ≥ w$$ to be able to handle all values $$|x| \lt w$$.
There can be multiple combinations of $$P$$, $$Q$$, $$R$$ that yield $$X ≥ w$$ or $$X ≥ d$$, and those are all good enough. Within that group, we prefer small $$P$$, $$Q$$, $$R$$, because that makes for easier calculations.
The smallest possible positive value of $$R$$ is equal to the greatest common divisor $$c$$ of $$d$$ and $$f$$, and the corresponding $$P, Q \gt 0$$ can be found using the Extended Algorithm of Euclid ― or by trying all $$d$$ or $$w/f$$ (pick the smallest) possibilities for $$P$$.
The Extended Algorithm of Euclid produces a relation of the form
$$φf − δd = c$$
for $$c \gt 0$$ (if you find $$c \lt 0$$, then multiply $$φ$$, $$δ$$, and $$c$$ by −1). If $$φ$$ and $$δ$$ are positive, then $$P_1 = φ$$ and $$Q_1 = δ$$, but $$φ$$ and $$δ$$ can be negative. In general, calculate $$P_1$$, $$Q_1$$, $$R_1$$ as follows:
\begin{align*} P_1 \| = φ \bmod \frac{d}{c} \\ Q_1 \| = δ \bmod \frac{f}{c} \\ R_1 \| = c \end{align*}
If that $$P = P_1$$, $$Q = Q_1$$, $$R = R_1$$ fit together, then
$$fP_1 − dQ_1 = R_1$$
Then also $$P = kP_1$$, $$Q = kQ_1$$, $$R = kR_1$$ belong together, for whole $$0 \lt k ≤ \floorratio{d}{R_1}$$, because $$R$$ must be less than $$d$$. To be able to handle greater values of $$k$$, it is therefore better to calculate $$Q$$ and $$R$$ from $$P$$ according to their definition.
\begin{align*} P \| = kP_1 \\ Q \| = \floorratio{fP}{d} = \floorratio{kfP_1}{d} = \floorratio{k(dQ_1 + R_1)}{d} = kQ_1 + \floorratio{kR_1}{d} \\ R \| = fP \bmod d = k(dQ_1 + R_1) \bmod d = kR_1 \bmod d \end{align*}
so the curtailing of $$R$$ to values less than $$d$$ leads to an increase in $$Q$$ so that $$fP − dQ = R$$ is still satisfied. If $$kR_1 \lt d$$, then $$P = kP_1$$, $$Q = kQ_1$$, and $$R = kR_1$$ as expected.
It is not useful to allow $$P ≥ d$$, because otherwise the greatest intermediate result can be as great or greater than $$fP = fd$$ which can become greater than $$w$$, because if it couldn't exceed $$w$$ then we could have used equation \eqref{eq:detour1} and would not need the current more complicated method at all. We seek $$P \lt d$$.
If $$P_2 = P + md/c$$, with $$c$$ the greatest common divisor of $$f$$ and $$d$$, then
\begin{align*} Q_2 \| = \floorratio{fP_2}{d} = \floorratio{fP + m\dfrac{fd}{c}}{d} = \floorratio{fP}{d} + m\frac{f}{c} = Q + m\dfrac{f}{c} \\ R_2 \| = fP_2 − dQ_2 = fP + m\dfrac{fd}{c} − \left( dQ + m \dfrac{df}{c} \right) = fP − dQ = R \end{align*}
so if we add or subtract a multiple of $$d/c$$ from $$P$$, then $$Q$$ shifts by the corresponding multiple of $$f/c$$, and $$R$$ remains the same. In particular, we can take
\begin{align} P \| = kP_1 \bmod d \\ Q \| = kQ_1 \bmod f \\ R \| = kR_1 \bmod d \end{align}
for $$k \lt d$$ because there are not more than that number of unique values of $$P \lt d$$.
For example, let $$d = 765433$$, $$f = 25920$$, $$t = 13835$$, $$w = 2^{31} − 1 = 2147483647$$. Then $$\frac{w}{f} = \frac{2^{31} − 1 }{25920} ≈ 82850$$, so we seek a $$P \lt 82850 \lt 765433$$, and an $$R$$ as small as possible.
The Extended Algorithm of Euclid (or a search) for this $$d$$ and $$f$$ yields that $$99902f − 3383d = 1$$, so $$P_1 = 99902$$, $$Q_1 = 3383$$, $$R_1 = 1$$. Then
\begin{align*} P \| = kP_1 \bmod d \\ Q \| = kQ_1 \bmod f \\ R \| = kR_1 \bmod d \end{align*}
The following table shows a couple of combinations for small $$R$$.
$${R}$$ $${P}$$ $${Q}$$ $${Q′}$$
1 99902 3383 3383
2 199804 6766 6766
7 699314 23681 23681
8 33783 1144 1144
9 133685 4527 4527
The next table shows the combinations with the greatest $$X$$.
$${R}$$ $${P}$$ $${Q}$$ $${X}$$
8 33783 1144 5.4 × 1012
16 67566 2288 1.7 × 1012
31 35230 1193 1.4 × 1012
54 36677 1242 8.1 × 1011
39 69013 2337 6.3 × 1011
All of these $$X$$ are comfortably greater than $$w ≈ 2.15×10^{9}$$. $$X ≥ w$$ for 1498 values of $$R$$ between 8 and 20312, and $$X ≥ d$$ for 82576 values of $$R$$ between 8 and 765426. We found solutions for $$X ≥ w$$ so we do not need equations \eqref{eq:detour1}ff. Within the group with $$X ≥ w$$ we search for the smallest values of $$PQR$$. We find
$${R}$$ $${P}$$ $${Q}$$ $${PQR}$$ $${X}$$
23 1447 49 1.6 × 106 1.33 × 1011
46 2894 98 1.3 × 107 1.30 × 1011
544 945 32 1.6 × 107 3.69 × 109
69 4341 147 4.4 × 107 1.28 × 1011
92 5788 196 1.0 × 108 1.26 × 1011
We choose $$P = 1447$$, then $$R = 23$$ and $$Q = 49$$.
Now we calculate $$y$$, $$e$$ for $$x = 1710321$$. We find
\begin{align*} q \| = \floorratio{x}{P} = \floorratio{1710321}{1447} = 1181 \\ r \| = x \bmod P = 1710321 \bmod 1447 = 1414 \\ y \| = qQ + \floorratio{qR + fr + t}{d} \\ \| = 1181×49 + \floorratio{1181×23 + 25920×1414 + 13835}{765433} \\ \| = 57869 + \floorratio{36691878}{765433} \\ \| = 57869 + 47 = 57916 \\ e \| = 36691878 \bmod 765433 = 716527 \end{align*}
The greatest intermediate result is 36,691,878, which is considerably less than $$w$$.
Had we been allowed arbitrarily large intermediate results, then we'd have found
\begin{align*} y \| = \floorratio{fx + t}{d} = \floorratio{25920×1710321 + 13835}{765433} = \floorratio{44331534155}{765433} = 57916 \\ e \| = 44331534155 \bmod 765433 = 716527 \end{align*}
which are the same results, but with the much greater intermediate result of 44,331,534,155.
Another example: $$d = 146097$$, $$f = 4800$$, $$t = 15793$$, $$w = 2^{31} − 1 = 2147483647$$. Then $$\frac{w}{f} = \frac{2^{31} − 1}{4800} ≈ 447392$$, so we seek a $$P \lt 146097 \lt 447392$$ and an $$R$$ as small as possible.
The Extended Algorithm of Euclid (or a search) yields that $$−15188f + 499d = 3$$, and there is no sum of a multiple of $$f$$ and a multiple of $$d$$ that is closer to 0. It follows that
\begin{align*} φ \| = −15188 \\ δ \| = −499 \\ c \| = 3 \\ P_1 \| = −15188 \bmod \frac{146097}{3} = −15188 \bmod 48699 = 33511 \\ Q_1 \| = −499 − \floorratio{−15188×3}{146097} \frac{4800}{3} = −499 − (−1)×1200 = 1101 \\ R_1 \| = 3 \end{align*}
Let's check:
$fP_1 − dQ_1 = 4800×33511 − 146097×1101 = 3 = R_1$
For small $$R$$ we find the following combinations:
$${k}$$ $${R}$$ $${P}$$ $${Q}$$
1 3 33511 1101
2 6 67022 2202
3 9 100533 3303
4 12 134044 4404
There are 57132 combinations for which $$X ≥ w$$. The combinations with the smallest $$PQR$$ are:
$${R}$$ $${P}$$ $${Q}$$ $${PQR}$$ $${X}$$
48 487 16 3.74 × 105 2.18 × 1010
9 3135 103 2.91 × 106 7.43 × 1011
96 974 32 2.99 × 106 2.17 × 1010
375 761 25 7.13 × 106 4.35 × 109
144 1461 48 1.01 × 107 2.17 × 1010
We choose $$P = 487$$, then $$Q = 16$$ and $$R = 48$$. We calculate $$y$$, $$e$$ for $$x = 731767$$. Then
\begin{align*} q \| = \floorratio{731767}{487} = 1502 \\ r \| = 731767 \bmod 487 = 293 \\ y \| = 1502×16 + \floorratio{1502×48 + 4800×293 + 15793}{146097} \\ \| = 24032 + \floorratio{1494289}{146097} = 24032 + 10 = 24042 \\ e \| = 1494289 \bmod 146097 = 33319 \end{align*}
The greatest intermediate result is 1.494.289, which is comfortably smaller than $$w$$. With the original formula we'd have found
\begin{align*} y \| = \floorratio{4800×731767 + 15793}{146097} = \floorratio{3512497393}{146097} = 24042 \\ e \| = 3512497393 \bmod 146097 = 33319 \end{align*}
which are the same results as before but with the greatest intermediate result equal to 3,512,497,393, which is considerably larger than $$w$$.
I've investigated for all $$w$$ from 3 through 1024 what the greatest values of $$f$$ and $$d$$ are for which solutions can be found with $$X ≥ d$$ or $$X ≥ w$$, and what the greatest $$P$$, $$Q$$, $$R$$, and $$X$$ are among those solutions, and what the greatest number $$c$$ of such solutions is, and the total count $$N$$ of combinations of $$f$$ and $$d$$ for which there is at least one solution. I find
\begin{align} \max(f_w) \| ≈ \floorratio{w}{4} \\ \max(f_d) \| ≈ \floorratio{w}{3} \\ \max(d_w) \| ≈ w − 5 + (w \bmod 2) \\ \max(d_d) \| ≈ w − 1 \\ \max(c_w) \| ≈ \floorratio{w+8}{16} \\ \max(c_d) \| ≈ \floorratio{w−1}{2} \\ \max(P_w) \| ≈ \floorratio{w}{2} − 2 + (w \bmod 2) \\ \max(P_d) \| ≈ \floorratio{w}{2} − 1 \\ \max(Q) \| ≈ \floorratio{2\sqrt{w − 1} − 3}{2} \\ \max(R_w) \| ≈ \floorratio{\sqrt{w + 1} − 1}{2} \\ \max(R_d) \| ≈ \frac{\sqrt{w(w + 1100)}}{25} \\ \max(X) \| ≈ \floorratio{w^2}{8} \\ N_w \| ≈ \frac{1}{5}w^{5/3} \\ N_d \| ≈ \frac{1}{3}w^{5/3} \end{align}
where the variables with subscript $$w$$ apply only to $$X \gt w$$, those with subscript $$d$$ apply only to $$X \gt d$$, and those without a subscript apply to both. The formulas for $$\max(R_d)$$, $$N_w$$, and $$N_d$$ are approximations; the other formulas are exact for $$10 ≤ w ≤ 1024$$.
For example, for $$w = 1023$$ I find
\begin{align*} \max(f_w) \| ≈ \floorratio{1023}{4} = 255 \\ \max(f_d) \| ≈ \floorratio{1023}{3} = 341 \\ \max(d_w) \| ≈ 1023 − 5 + (1023 \bmod{2}) = 1019 \\ \max(d_d) \| ≈ 1023 − 1 = 1022 \\ \max(c_w) \| ≈ \floorratio{1023 + 8}{16} = 64 \\ \max(c_d) \| ≈ \floorratio{1023 − 1}{2} = 511 \\ \max(P_w) \| ≈ \floorratio{1023}{2} − 2 + (1023 \bmod{2}) = 510 \\ \max(P_d) \| ≈ \floorratio{1023}{2} − 1 = 510 \\ \max(Q) \| ≈ \floorratio{2×\sqrt{1023 − 1} − 3}{2} ≈ ⌊30.47⌋ = 30 \\ \max(R_w) \| ≈ \floorratio{\sqrt{1023 + 1} − 1}{2} ≈ ⌊15.50⌋ = 15 \\ \max(R_d) \| ≈ \frac{\sqrt{1023 × (1023 + 1100)}}{25} ≈ 58.95 \\ \max(X) \| ≈ \floorratio{1023^2}{8} = 130816 \\ N_w \| ≈ \frac{1}{5}1023^{5/3} ≈ 20772.52 \\ N_d \| ≈ \frac{1}{3}1023^{5/3} ≈ 34620.87 \end{align*}
These values are all correct, except that $$R_d = 59$$, $$N_w = 20014$$, and $$N_d = 33937$$.
If we extrapolate the above formules to $$w = 2^{31} − 1$$ (for values with a width of 32 bits), then we find, approximately
\begin{align*} \max(f_w) \| ≈ 5.4×10^8 \\ \max(f_d) \| ≈ 7.2×10^8 \\ \max(d) \| ≈ 2.1×10^9 \\ \max(c_w) \| ≈ 1.3×10^8 \\ \max(c_d) \| ≈ 1.1×10^9 \\ \max(P) \| ≈ 1.1×10^9 \\ \max(Q) \| ≈ 46340 \\ \max(R_w) \| ≈ 23170 \\ \max(R_d) \| ≈ 8.6×10^7 \\ \max(X) \| ≈ 5.8×10^{17} \\ N_w \| ≈ 7.1×10^{14} \\ N_d \| ≈ 1.2×10^{15} \end{align*}
so if we seek a solution for $$X ≥ w$$ (so that we do not need equations \eqref{eq:detour1}ff), then we need try only up to 23170 different values of $$R$$. If such a solution does not exist, then we seek a solution for $$X ≥ d$$ (for which we do need to use equations \eqref{eq:detour1}) and then we must try up to 86 million values of $$R$$, which is still much less than the up to 1100 million values of $$P$$.
For a given $$w$$ there are many combinations of $$d$$ and $$f$$ for which there are no solutions with $$X ≥ w$$ or $$X ≥ d$$. For example, for $$w = 499$$ there are only 6554 combinations of $$d$$ and $$f$$ with at least one solution with $$X ≥ w$$, and 10560 combinations with at least one solution with $$X ≥ d$$, but there are $$\frac{1}{2} (w − 2)×(w − 3) = 123256$$ combinations with $$2 ≤ f \lt d \lt w$$. The greater $$f$$ is, the smaller is the chance that there is a solution for the combination of that $$f$$ and an arbitrary $$d \gt f$$.
However, when $$f$$ and $$d$$ are considerably less than $$w$$, then there is a good chance of a solution. For example, for $$w = 2^{31} − 1$$ and for $$f$$ and $$d$$ near $$f ≈ 25920$$ and $$d ≈ 765433$$ there is no solution with $$X ≥ d$$ for only 1 out of about 9100 combinations of $$f$$ and $$d$$, and no solution with $$X ≥ w$$ for only 1 out of about 54 combinations.
### 12.3. The Simple Calendar
For convenience, we'll work with a calendar that has only "days" and "months", where the months are longer than the days, but the formulas also work for other units of time. Later, we'll discuss calendars with more than two time periods.
In the simplest case, the average length of the month is equal to $$p$$ days, and all months are either $$⌊p⌋$$ or $$⌊p⌋ + 1$$ days long. The running day number $$y$$ depends on the month number $$x$$ and the day number $$z$$ within the current month as follows:
\begin{align} c(x) \| ≡ ⌊px + r⌋ \\ y \| ≡ c(x) + z \end{align}
where $$x$$, $$y$$, $$c(x)$$, $$z$$ are whole numbers, and $$p ≥ 1$$ and $$r$$ are fixed values that depend on the calendar. The first month has $$x = 0$$ and the first day of the month has $$z = 0$$, for easier calculation.
The first day ($$z = 0$$) of the first month ($$x = 0$$) has running day number $$y = 0$$, so
$$⌊r⌋ = 0 ⇔ 0 ≤ r \lt 1$$
The length $$L(x)$$ of month $$x$$ is
$$\begin{split} L(x) \| = c(x + 1) − c(x) \\ \| = ⌊px + r + p⌋ − ⌊px + r⌋ \\ \| = ⌊px + r⌋ + ⌊p⌋ + C(⌊px + r⌉_1 + ⌊p⌉_1 ≥ 1) − ⌊px + r⌋ \\ \| = ⌊p⌋ + C(⌊px + r⌉_1 ≥ 1 − ⌊p⌉_1) \end{split}$$
All months are $$⌊p⌋$$ or $$⌊p⌋ + 1$$ days long. Month $$x$$ is a long month (with $$⌊p⌋ + 1$$ days) if $$⌊px + r⌉_1 ≥ 1 − ⌊p⌉_1$$. On average there is a long month after every
$$Q = \frac{1}{⌊p⌉_1} \label{eq:Q}$$
months, but that isn't always a whole number, so in practice the number of months between two successive long months is equal to $$⌊Q⌋$$ or $$⌈Q⌉$$.
With $$r = 0$$ we get a certain pattern of long and short months. Suppose that that pattern would be good if only it were shifted by $$a$$ months so that month $$x_0$$ in the calendar with $$r = 0$$ corresponded with month $$x = x_0 + a$$ in the new calendar. Then
$$\begin{split} c(x) \| = c(x_0 + a) − c(a) \\ \| = ⌊p(x_0 + a)⌋ − ⌊pa⌋ \\ \| = ⌊px_0 + pa − ⌊pa⌋⌋ \\ \| = ⌊px_0 + ⌊pa⌉_1⌋ \end{split}$$
so
$$r = ⌊pa⌉_1$$
The subtracting of $$c(a)$$ ensures that $$c(0) = 0$$.
Then we go from running month number $$x$$ and day number $$z$$ in the month to running day number $$y$$ through:
$$y = ⌊px + r⌋ + z \label{eq:xnaary}$$
How do we go from $$y$$ back to $$x$$?
\begin{eqnarray} y = ⌊px + r⌋ + z \| = \| px + r − ⌊px + r⌉_1 + z \\ x + \frac{z − ⌊px + r⌉_1}{p} \| = \| \frac{y − r}{p} \\ x + \frac{z + 1 − ⌊px + r⌉_1}{p} \| = \| \frac{y + 1 − r}{p} \end{eqnarray}
We have
$$0 \lt 1 − ⌊px + r⌉_1 ≤ 1$$
for all $$x$$, so if $$z ≥ 0$$ then
$$z + 1 − ⌊px + r⌉_1 \gt 0$$
For the last day of month $$x$$ we have
$$z = L(x) − 1 = ⌊px + r + p⌋ − ⌊px + r⌋$$
so if $$z ≤ L(x) − 1$$ then
$$\begin{split} z + 1 − ⌊px + r⌉_1 \| ≤ ⌊px + r + p⌋ − ⌊px + r⌋ − ⌊px + r⌉_1 \\ \| = ⌊px + r + p⌋ − px − r \\ \| = px + r + p − ⌊px + r + p⌉_1 − px − r \\ \| = p − ⌊px + r + p⌉_1 ≤ p \end{split}$$
Combined, this means, for $$0 ≤ z ≤ L(x) − 1$$,
\begin{eqnarray} 0 \| \lt \| z + 1 − ⌊px + r⌉_1 ≤ p \\ 0 \| \lt \| \dfrac{z + 1 − ⌊px + r⌉_1}{p} ≤ 1 \end{eqnarray}
so
\begin{eqnarray} x + 1 \| = \| \left\lceil x + \frac{z + 1 − ⌊px + r⌉_1}{p} \right\rceil = \ceilratio{y + 1 − r}{p} \\ x \| = \| \ceilratio{y + 1 − r}{p} − 1 \label{eq:ξ} \end{eqnarray}
And then we can also find $$z$$
$$\begin{split} z \| = y − ⌊px + r⌋ \\ \| = y − \left\lfloor p \left( \ceilratio{y + 1 − r}{p} − 1 \right) + r \right\rfloor \\ \| = y − \left\lfloor p \left( \dfrac{y + 1 − r}{p} + \left\lceil \dfrac{y + 1 − r}{p} \right\rfloor_1 − 1 \right) + r \right\rfloor \\ \| = y − \left\lfloor y + 1 − r + p \left\lceil \dfrac{y + 1 − r}{p} \right\rfloor_1 − p + r \right\rfloor \\ \| = −\left\lfloor 1 + ⌈y + 1 − r⌋_p − p \right\rfloor \\ \| = ⌈p − 1 − ⌈y + 1 − r⌋_p⌉ \\ \| = ⌈p − ⌈y + 1 − r⌋_p⌉ − 1 \end{split}$$
So, all together,
\begin{eqnarray} x \| = \| \ceilratio{y + 1 − r}{p} − 1 \label{eq:ynaarx2} \\ z \| = \| ⌈p − ⌈y + 1 − r⌋_p⌉ − 1 \end{eqnarray}
We can rewrite this so we can use the $$\Div$$ function. If
\begin{eqnarray} k \| = \| y + 1 − r \\ \{ m,n \} \| = \| \Div(−k, p) \end{eqnarray}
then
\begin{eqnarray} x \| = \| −m − 1 \\ z \| = \| −⌊n − p⌋ − 1 \end{eqnarray}
An example, with $$p = 30.6$$ and $$r = 0$$. We then find, for the beginning of the first few months
$${x}$$ 0 1 2 3 4 5 6 7 8 9 10 $${px}$$ 0 30.6 61.2 91.8 122.4 153 183.6 214.2 244.8 275.4 306 $${c(x)}$$ 0 30 61 91 122 153 183 214 244 275 306 $${L(x)}$$ 0 30 31 30 31 31 30 31 30 31 31
Because $$5p = 153$$ is a whole number, the pattern of month lengths repeats itself after 5 months. We see here the pattern 31-30-31-30-31 that in the Gregorian calendar describes the months March - July and August - December.
Now we go in the opposite direction, for a few running day numbers:
$${y}$$ −2 −1 0 1 120 121 122 123 $${k}$$ 1 0 −1 −2 −121 −122 −123 −124 $${m}$$ 0 0 −1 −1 −4 −4 −5 −5 $${n}$$ 1 0 29.6 28.6 1.4 0.4 30 29 $${x}$$ −1 −1 0 0 3 3 4 4 $${z}$$ 29 30 0 1 29 30 0 1
### 12.4. With Whole Numbers Only
Calculating machines often work with a limited number of decimals behind the decimal separator, so you can get into trouble with round-off errors. If $$(y + 1)/p = 6.99999999998$$ but because of a very small round-off error your calculating machine thinks that $$(y + 1)/p = 7.00000000001$$, then your calculating machine thinks that $$⌈(y + 1)/p⌉ − 1 = 7$$ rather than 6, and then you find the wrong month.
Converting equations with $$⌈•⌉$$ or $$⌈•⌋_1$$ into equations with $$⌊•⌋$$ or $$⌊•⌉_1$$ or $$\Div$$ is easier when only whole numbers are used, because for arbitrary whole numbers $$v$$ and $$w$$ ($$w \gt 0$$)
\begin{align} \ceilratio{v}{w} \| = \floorratio{v − 1}{w} + 1 \\ ⌈v⌋_w \| = w − 1 − ⌊v − 1⌉_w \end{align}
If the average length $$p$$ is a ratio (of whole numbers), then we can avoid such round-off errors by rewriting the formulas based on ratios. We define
$$p = \frac{f}{g}$$
with $$f$$, $$g$$ whole numbers greater than zero. Then we find, for an arbitrary value $$q$$,
$$\begin{split} ⌊q⌋_p \| = p\left\lfloor \dfrac{q}{p} \right\rceil_1 \| = \dfrac{f}{g} \left\lfloor \dfrac{qg}{f} \right\rceil_1 \\ \| = \dfrac{1}{g} ⌊qg⌉_f \end{split}$$
Then
\begin{eqnarray} r \| = \| ⌊pa⌉_1 = \left\lfloor \frac{fa}{g} \right\rceil_1 = \frac{⌊fa⌉_g}{g} ≡ \frac{s}{g} \label{eq:r} \\ x \| = \| \ceilratio{y + 1 − r}{p} − 1 \notag \\ \| = \| \ceilratio{y + 1 − (s/g)}{f/g} − 1 \notag \\ \| = \| \ceilratio{gy + g − s}{f} − 1 \notag \\ \| = \| \floorratio{gy + g − s − 1}{f} \\ z \| = \| \left\lceil \dfrac{f}{g} − ⌈y + 1 − (s/g)⌋_p \right\rceil − 1 \notag \\ \| = \| \ceilratio{f − ⌈gy + g − s⌋_f}{g} − 1 \notag \\ \| = \| \ceilratio{f − (f − 1 − ⌊gy + g − s − 1⌉_f)}{g} − 1 \notag \\ \| = \| \ceilratio{f − f + 1 + ⌊gy + g − s − 1⌉_f}{g} − 1 \notag \\ \| = \| \ceilratio{1 + ⌊gy + g − s − 1⌉_f}{g} − 1 \notag \\ \| = \| \floorratio{⌊gy + g − s − 1⌉_f}{g} \end{eqnarray}
Summarizing, from running month number $$x$$ and day number $$z$$ in the month to running day number $$y$$:
$$y = \floorratio{fx + s}{g} + z$$
where
$$s = fa \bmod g$$
and from $$y$$ to $$x$$ and $$z$$:
\begin{align} k \| = gy + g − 1 − s \\ x \| = \floorratio{k}{f} \label{eq:ynaarxr} \\ z \| = \floorratio{k \bmod f}{g} \end{align}
Now you can do all calculations with ratios, which have no numbers after the decimal mark and hence no round-off errors.
For the same calendar as before we have $$p = 30.6 = 153/5$$, so $$f = 153$$, $$g = 5$$. Also $$r = 0$$ so $$a = 0$$. We then find, for the beginning of the first few months
$${x}$$ 0 1 2 3 4 5 6 7 8 9 10 $${⌊fx/g⌋}$$ 0 30 61 91 122 153 183 214 244 275 306 $${⌈fx⌋_g}$$ 0 3 1 4 2 0 3 1 4 2 0 $${c(x)}$$ 0 30 61 91 122 153 183 214 244 275 306 $${L(x)}$$ 0 30 31 30 31 31 30 31 30 31 31
Now we go in the opposite direction, for a few running day numbers:
$${y}$$ −2 −1 0 1 120 121 122 123 $${k}$$ −6 −1 4 9 604 609 614 619 $${x}$$ −1 −1 0 0 3 3 4 4 $${n}$$ 147 152 4 9 145 150 2 7 $${z}$$ 29 30 0 1 29 30 0 1
### 12.5. Very Unequal Months
So far we have assumed that a long month was only one day longer than a short month, but that difference can be larger. Suppose that short months are $$q$$ days long and that long months are $$q + d$$ days long, with $$q$$, $$d$$ whole numbers greater than 0, and that a fraction $$ψ$$ (between 0 and 1) of all months is long. Then the average length of a month is equal to
$$p = q + dψ$$
days. The formula to calculate the running day number $$y$$ from month number $$x$$ and day number $$z$$ in the current month is then
$$\begin{split} y \| = c + z = qx + d⌊ψ(x + a)⌋ − d⌊ψa⌋ + z \\ \| = qx + d⌊ψx + ψa − ⌊ψa⌋⌋ + z = qx + d⌊ψx + σ⌋ + z \end{split} \label{eq:xnaary3}$$
$$σ ≡ ψa − ⌊ψa⌋ = ψa \bmod 1$$
$$c = qx + d⌊ψx + σ⌋$$
How do we go in the opposite direction? To figure that out we compare the running day number $$c$$ that comes from equation \eqref{eq:xnaary3} for the first day ($$z = 0$$) of month $$x$$ with the running day number (for convenience renamed to $$c_s$$) that comes from equation \eqref{eq:xnaary} for the same $$x$$ and the same average month length $$p$$ (but with $$d = 1$$ and $$a = 0$$).
$$\begin{split} c_s(x) \| ≡ ⌊px⌋ \\ \| = ⌊qx + dψx⌋ \\ \| = qx + ⌊dψx⌋ \end{split}$$
Then we find
$$c − c_p = d⌊ψx + σ⌋ − ⌊dψx⌋ = d(ψx + σ − δ) − (dψx − ε) = dσ − dδ + ε$$
for certain $$δ$$, $$ε$$ satisfying
$$0 ≤ \{δ,ε\} \lt 1$$
so
$$dσ − d \lt c − c_p \lt dσ + 1$$
$$dσ − d + 1 ≤ c − c_p ≤ dσ$$
so the beginning of month $$x$$ in the target calendar is between $$dσ + 1 − d$$ and $$dσ$$ days after the beginning of month $$x$$ in the simpler calendar. Because $$0 ≤ σ \lt 1$$, we also have $$−d \lt c − c_p \lt d$$, so the difference between the beginning of a particular month in the target calendar and the same month in the simpler calendar is always less than $$d$$ days (in either direction).
For that simpler calendar we have formula \eqref{eq:ynaarx2} to calculate the corresponding month number for any of the $$\lfloor p\rfloor$$ or $$⌈p⌉$$ days in that month, so if running day number $$y$$ corresponds to one of the first $$\lfloor p\rfloor$$ days of month $$x$$ in the target calendar, then we'll have for that $$x$$
\begin{align} x_\text{lower} \| ≤ x ≤ x_\text{upper} \label{eq:ondbov3} \\ x_\text{lower} \| ≡ \ceilratio{y − dσ + 1}{p} − 1 \\ x_\text{upper} \| ≡ \ceilratio{y − dσ + d}{p} − 1 \label{eq:ynaarx3} \end{align}
In the target calendar, a month can be up to $$q + d$$ days long, while in the corresponding simpler calendar a month cannot be longer than $$⌈p⌉$$ days, which is less than $$q + d$$. A number (less than $$d$$) of days beyond day $$\lfloor p \rfloor$$ of a month can still belong to that same month in the target calendar, but already belong to a later month in the simpler calendar. For such days, formula \eqref{eq:ynaarx3} can yield a month number that is too great, but then formula \eqref{eq:ondbov3} still holds.
If $$d ≤ p$$ then the difference between $$x_\text{upper}$$ and $$x_\text{lower}$$ is at most 1. In that case
\begin{align} c_\text{upper} \| = qx_\text{upper} + d⌊ψx_\text{upper} + σ⌋ \\ z_\text{upper} \| = y − c_\text{upper} \\ ζ \| = \floorratio{z_\text{upper}}{q + d} \\ x \| = x_\text{upper} + ζ \label{eq:xζ3} \\ z \| = y − c = y − qx − d⌊ψx + σ⌋ \end{align}
If $$f = 1001$$, $$g = 165$$, $$d = 2$$, $$q = 5$$, $$a = 77$$, then
\begin{align*} p \| = \frac{f}{g} = \frac{1001}{165} = 6 \frac{11}{165} \\ ψ \| = \frac{176}{330} \\ σ \| = ψa \bmod 1 = \frac{13552}{330} \bmod 1 = \frac{22}{330} \end{align*}
Then also $$d ≤ p$$, so equation \eqref{eq:xζ3} can be used.
The running day number $$y$$ of the first day of month number $$x = 37$$ is then (from equation \eqref{eq:xnaary3})
$y = qx + d⌊ψx + σ⌋ = 5×37 + 2×\left\lfloor \frac{37×176}{330} + \frac{22}{330} \right\rfloor = 185 + 2×\floorratio{6534}{330} = 223$
Now we calculate the month that corresponds to running day number $$y = 223$$. Then
\begin{align*} x_\text{lower} \| = \ceilratio{y + 1}{p} − 1 = \ceilratio{224×165}{1001} − 1 = 36 \\ x_\text{upper} \| = \ceilratio{y + d}{p} − 1 = \ceilratio{225×165}{1001} − 1 = 37 \end{align*}
so the month number can be 36 or 37. For $$x_\text{upper}$$ we find
\begin{align*} c_\text{upper} \| = 5×37 + 2×\floorratio{176×37}{330} + \frac{22}{330} = 223 \\ z_\text{upper} \| = y − c_\text{upper} = 223 − 223 = 0 \\ ζ \| = 0 \\ x \| = x_\text{upper} = 37 \\ z \| = z_\text{upper} = 0 \end{align*}
And here is another example. With $$f = 829$$, $$g = 235$$, $$d = 2$$, $$q = 3$$, $$ψ = 124/470$$, $$a = 0$$ we calculate which month corresponds to running day number $$y = 1448$$. Then
\begin{align*} x_\text{lower} \| = \ceilratio{y + 1}{p} − 1 = \ceilratio{1449×235}{829} − 1 = 410 \\ x_\text{upper} \| = \ceilratio{y + d}{p} − 1 = \ceilratio{1450×235}{829} − 1 = 411 \end{align*}
so the month number can be 410 or 411. For $$x_\text{upper}$$ we find
\begin{align*} c_\text{upper} \| = 3×411 + 2×\floorratio{411×124}{470} = 1449 \\ z_\text{upper} \| = y − c_\text{upper} = 1448 − 1449 = −1 \\ ζ \| = −1 \\ x \| = x_\text{upper} + ζ = 410 \\ z \| = y − c = 1448 − 3×410 − 2×\floorratio{411×124}{470} = 2 \end{align*}
If $$p = f/g = q + dh/g$$ is a ratio of whole numbers ($$f$$, $$g$$, $$q$$, $$d$$, $$h$$ are whole numbers), then we find
\begin{align} y \| = c + z = qx + d\floorratio{hx + s}{g} + z \label{eq:xnaary3r} \\ s \| ≡ ha − g\floorratio{ha}{g} = ha \bmod g \\ x_\text{upper} \| = \floorratio{gy + dg − ds − 1}{f} \label{eq:ynaarx3r} \\ c_\text{upper} \| = qx_\text{upper} + d\floorratio{hx_\text{upper} + s}{g} \\ z_\text{upper} \| = y − c_\text{upper} \\ ζ \| = \floorratio{z_\text{upper}}{q + d} \\ x \| = x_\text{upper} + ζ \\ z \| = y − c = y − qx − d\floorratio{hx + s}{g} \end{align}
### 12.6. Many Kinds of Unequal Months
We now allow there to be more than two kinds of months with different month lengths. In equation \eqref{eq:xnaary3}, $$⌊ψ(x + a)⌋$$ gave the pattern of long months (with a shift as desired). We now allow an arbitrary number of such pattern, each with its own length difference $$d_i$$ (which may now also be negative), relative frequency $$0 \lt ψ_i \lt 1$$, and pattern shift $$a_i$$. Then we find
$$\begin{split} y \| = c + z = qx + ∑_{i}d_i ⌊ψ_i(x + a_i)⌋ − ∑_{i}d_i ⌊ψ_{i}a_i⌋ + z \\ \| = qx + ∑_{i}d_i ⌊ψ_{i}x + ψ_{i}a_i − ⌊ψ_{i}a_i⌋⌋ + z \\ \| = qx + ∑_{i}d_i ⌊ψ_{i}x + σ_i⌋ + z \label{eq:xnaary4} \end{split}$$
$$c = qx + ∑_{i}d_i ⌊ψ_{i}x + σ_i⌋$$
$$σ_i ≡ ψ_{i}a_i − ⌊ψ_{i}a_i⌋ = ψ_{i}a_i \bmod 1$$
The derivation of the formulas for the opposite direction goes analogous to that of equation \eqref{eq:ynaarx3}, but now we have to take into account possibly negative $$d_i$$. We find
\begin{align} p \| = q + ∑_{i}d_{i}ψ_{i} \\ c_s \| = ⌊px⌋ = qx + ⌊∑_{i}d_{i}ψ_{i}⌋ \end{align}
$$\begin{split} c − c_s \| = ∑_{i}d_i⌊ψ_{i}x + σ_i⌋ − ⌊∑_{i}d_{i}ψ_{i}x⌋ \\ \| = ∑_{i}d_i (ψ_{i}x + σ_i − δ_i) − \left( ∑_{i}d_{i}ψ_{i}x − ε \right) \\ \| = ∑_{i}d_{i}σ_i − ∑_{i}d_{i}δ_i + ε \end{split}$$
for
$$0 ≤ δ_i, ε \lt 1$$
If there is at least one $$d_i \gt 0$$, then
$$c − c_s \gt ∑_{i}d_{i}σ_i − ∑_{\gt0}d_i$$
(where $$∑_{\gt0}d_i$$ is the sum of all $$d_i$$ that are greater than 0) and otherwise
$$c − c_s ≥ ∑_{i}d_{i}σ_i$$
Also
$$c − c_s \lt ∑_{i}d_{i}σ_i − ∑_{\lt0}d_i + 1$$
Together this yields
$$∑_{i}d_{i}σ_i − ∑_{\gt0}d_i + 1 ≤ y − y_p ≤ ∑_{i}d_{i}σ_i − ∑_{\lt0}d_i$$
if there are positive $$d_i$$, and otherwise
$$∑_{i}d_{i}σ_i ≤ c − c_s ≤ ∑_{i}d_{i}σ_i − ∑_{i}d_i$$
We can combine this to
$$∑_{i}d_{i}σ_i − ∑_{\gt0}d_i ≤ c − c_s ≤ ∑_{i}d_{i}σ_i − ∑_{\lt0}d_i$$
regardless of whether there are positive $$d_i$$. With that,
$$c − ∑_{i}d_{i}σ_i + ∑_{\lt0}d_i ≤ c_s ≤ c − ∑_{i}d_{i}σ_i + ∑_{\gt0}d_i$$
and so
\begin{align} x_\text{lower} \| = \ceilratio{y − ∑_{i}d_{i}σ_i + ∑_{\lt0}d_i}{p} − 1 \\ x_\text{upper} \| = \ceilratio{y − ∑_{i}d_{i}σ_i + ∑_{\gt0}d_i}{p} − 1 \label{eq:ynaarx4} \end{align}
If $$∑_{\gt0}d_i − ∑_{\lt0}d_i \lt p$$ then $$x_\text{lower}$$ and $$x_\text{upper}$$ differ by at most 1. Then
\begin{align} c_\text{upper} \| = qx_\text{upper} + ∑_{i}d_i⌊ψ_{i}x_\text{upper} + σ_i⌋ \\ z_\text{upper} \| = y − c_\text{upper} \\ ζ \| = \floorratio{z_\text{upper}}{q + ∑_{\gt0}d_i} \\ x \| = x_\text{upper} + ζ \label{eq:xζ4} \\ z \| = y − c − y − qx − ∑_{i}d_i⌊ψ_{i}x + σ_i⌋ \end{align}
If all $$ψ_i = h_i/g$$ are ratios of whole numbers with a common denominator $$g$$, then we find
\begin{align} f \| ≡ qg + ∑_{i}d_{i}h_{i} \\ p \| = \frac{f}{g} \\ y \| = c + z = qx + ∑_{i}d_i\floorratio{h_{i}x + s_i}{g} + z \label{eq:xnaary4r} \\ s_i \| ≡ h_{i}a_i − g\floorratio{h_{i}a_i}{g} = h_{i}a_i \bmod g \\ x_\text{upper} \| = \floorratio{gy + g∑_{\gt0}d_i − ∑_{i}d_{i}s_i − 1}{f} \label{eq:ynaarx4r} \\ c_\text{upper} \| = qx_\text{upper} + ∑_{i}d_i\floorratio{h_{i}x_\text{upper} + s_i}{g} \\ z_\text{upper} \| = y − c_\text{upper} \\ ζ \| = \floorratio{z_\text{upper}}{q + ∑_{\gt0}d_i} \\ x \| = x_\text{upper} + ζ \\ z \| = y − c = y − qx − ∑_{i}d_i\floorratio{h_{i}x + s_i}{g} \end{align}
$$g$$ is the least common multiple of the denominators of all $$ψ_i$$, or is a multiple of that. You can take the product of the denominators of all $$ψ_i$$ for $$g$$, but sometimes a smaller value can be used.
Suppose we want a calendar in which each month has 17 days, with each 3rd month getting 2 days extra, and each 5th month getting 3 days extra. Then the first couple of months contain the following number of days: 17, 17, 19, 17, 20, 19, 17, 17, 19, 20, 17, 19, 17, 17, 22.
Then we have
$${i}$$ 1 2 $${d_i}$$ 2 3 $${ψ_i}$$ 1/3 = 5/15 1/5 = 3/15 $${a_i}$$ 0 0 $${h_i}$$ 5 3 $${s_i}$$ 0 0
for $$g = 15$$ (the product of denominators 3 and 5). Then $$p = (17×15 + 2×5 + 3×3)/15 = 268/15 = 17 + 13/15$$, so $$f = 268$$ and $$y = 17x + 2⌊x/3⌋ + 3⌊x/5⌋$$ and $$x_\text{upper} = ⌊(15y + 15×5 − 0 − 1)/268⌋ = ⌊(15y + 74)/268⌋$$.
The following table shows the results that you get when you calculate for certain calendar dates $$x$$, $$z$$ what the corresponding running day number $$y$$ is, and then calculate from that $$y$$ what the month number $$x = x_\text{upper} + ζ$$ is.
$${x}$$ 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 $${z}$$ 0 16 0 16 0 18 0 16 0 19 0 18 0 16 0 16 0 18 0 19 0 16 0 18 0 16 0 16 0 21 0 16 $${y}$$ 0 16 17 33 34 52 53 69 70 89 90 108 109 125 126 142 143 161 162 181 182 198 199 217 218 234 235 251 252 273 274 290 $${x_\text{upper}}$$ 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 $${y_\text{upper}}$$ 0 17 17 34 34 53 53 70 70 90 90 109 109 126 126 143 143 162 162 182 182 199 199 218 218 235 235 252 252 274 274 291 $${ζ}$$ 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 0 −1 $${x_\text{upper} + ζ}$$ 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15
### 12.7. Very Unequal Month Lengths
The preceding methods for calculating the calendar date from the running day number for calendars with very unequal month lengths assumed that the variation in month lengths is less than the average month length: $$d ≤ p$$ when there are two different month lengths, or $$∑_{\gt0}d_i − ∑_{\lt0}d_i \lt p$$ if there are more month lengths. What to do if that condition is not met?
In that case you'll have to find the correct month by searching for it. First try month number $$x = x_\text{upper}$$ and calculate the corresponding $$z = y − c$$ and $$ζ$$. Is that $$ζ \lt 0$$? Then add $$ζ$$ to $$x$$ and try again, until $$ζ = 0$$: then you have found the correct $$x$$.
Let $$q = 3$$, $$d = 7$$, $$ψ = 1/3$$, then $$h = 1$$, $$g = 3$$, $$f = 16$$, $$p = q + dψ = 3 + 5/3 = 5 \frac{2}{3}$$ so the condition $$d ≤ p$$ is not satisfied. In this calendar the first three months have length $$3, 3, 10$$ days, and that pattern repeats every three months. For this calendar, we have
\begin{align*} x_\text{upper} \| = \floorratio{3y + 20}{16} \\ x_\text{lower} \| = \floorratio{3y + 2}{16} \\ c \| = 3x + 7\floorratio{x}{3} \end{align*}
We find
$${y}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 $${x_\text{upper}}$$ 1 1 1 1 2 2 2 2 2 2 3 3 3 3 4 4 4 $${c_\text{upper}}$$ 3 3 3 3 6 6 6 6 6 6 16 16 16 16 19 19 19 $${z_0}$$ −3 −2 −1 0 −2 −1 0 1 2 3 −6 −5 −4 −3 −5 −4 −3 $${ζ_0}$$ −1 −1 −1 0 −1 −1 0 0 0 0 0 0 0 0 −1 −1 0 $${c_1}$$ 0 0 0 3 3 6 6 6 6 16 16 16 $${z_1}$$ 0 1 2 1 2 4 5 6 7 −2 −1 0 $${ζ_1}$$ 0 0 0 0 0 0 0 0 0 −1 −1 0 $${c_2}$$ 6 6 $${z_2}$$ 8 9 $${ζ_2}$$ 0 0 $${x}$$ 0 0 0 1 1 1 2 2 2 2 2 2 2 2 2 2 3 $${z}$$ 0 1 2 0 1 2 0 1 2 3 4 5 6 7 8 9 10
For example: for $$y = 11$$ we find
\begin{align*} x_\text{upper} \| = \floorratio{3×11 + 20}{16} = 3 \\ c_\text{upper} \| = 3×3 + 7\floorratio{3}{3} = 16 \\ z_0 \| = y − c_\text{upper} = 11 − 16 = −5 \end{align*}
That $$z_0$$ is negative, so we decrement $$x$$ by one ($$x = 3 − 1 = 2$$) and try again. With the new $$x$$ we find
\begin{align*} c_1 \| = 3×2 + 7\floorratio{2}{3} = 6 \\ z_1 \| = y − c_1 = 11 − 6 = 5 \end{align*}
which is no longer negative, so we're done. For $$y = 11$$ we find $$x = 2$$ and $$z = 5$$.
If you want to, then you can use this procedure also if $$d ≤ p$$.
The greatest number of months that you may have to try before you find the right one is equal to the greatest value that $$x_\text{upper} − x_\text{lower}$$ can attain, plus 1. For calendars with two different month lengths we have
$$x_\text{upper} − x_\text{lower} = \ceilratio{y − dσ + d}{p} − \ceilratio{y − dσ + 1}{p}$$
The greatest value that this can attain is $$⌈(d − 1)/p⌉$$, so the number $$N$$ of month that you have to try at most is equal to
$$N = \ceilratio{d − 1}{p} + 1$$
For calendars with more than two different month lengths we have
$$x_\text{upper} − x_\text{lower} = \ceilratio{y − ∑_{i}d_{i}σ_i + ∑_{\gt0}d_i}{p} − \ceilratio{y − ∑_{i}d_{i}σ_i + ∑_{\lt0}d_i}{p}$$
which has for its greatest possible value $$⌈(∑_{\gt0}d_i − ∑_{\lt0}d_i)/p⌉$$ so for such a calendar
$$N = \ceilratio{∑_{\gt0}d_i − ∑_{\lt0}d_i}{p} + 1$$
### 12.8. Month Lengths Without Internal Patterns
Previously we looked at calendars of which the month lengths can be expressed as the sum of various patterns, for example with an extra day every second month and two additional days every third month, so that each sixth month gets three extra days in total. But what if the months repeat after some time but show no other discernable pattern?
As an example we use a calendar with the following month lengths, that keep repeating themselves indefinitely: 7, 13, 5, 11, 4 days. The "year" is 7 + 13 + 5 + 11 + 4 = 40 days long. The relationship between running day number $$y$$, running month number $$x$$, and day number $$z$$ in the current month (all starting at 0) is then
$${y}$$ 0 1 … 6 7 … 19 20 … 24 25 … 35 36 … 39 $${x}$$ 0 0 … 0 1 … 1 2 … 2 3 … 3 4 … 4 $${z}$$ 0 1 … 6 0 … 12 0 … 4 0 … 10 0 … 3
The calendar has $$f$$ days in $$g$$ months, for an average month length of $$p = f/g$$ days. The relationship between the running day number $$y$$ and the running month number $$x$$ is a step function: after a (varying) number of days, the month number increases by one. Such a step of 1 that occurs just before day $$a$$ and repeats itself every $$f$$ days can be obtained through the formula
$\floorratio{y − a}{f}$
For the example calendar we have $$f = 40$$ and $$g = 5$$ and there is (amongst others) a step just before day $$a = 7$$. We get that with
$x = \floorratio{y − 7}{40}$
$${y}$$ 0 1 … 6 7 8 … 46 47 48 $${y−7}$$ −7 −6 −1 0 1 39 40 41 $${x}$$ −1 −1 −1 0 0 0 1 1
When there are $$g$$ months, each with its own $$a_i$$ (for $$i$$ from 1 through $$g$$), then their combined effect is
$$x = g + ∑_i\floorratio{y − a_i}{f}$$
If the successive month lengths are $$m_i$$, then the first step is at $$a_1 = 0$$, the second one is at $$a_2 = a_1 + m_1 = m_1$$, the third one is at $$a_3 = a_2 + m_2 = m_1 + m_2$$, and in general
$$a_i = ∑_{j\lt i}m_j$$
Then
$$x = g + ∑_i\floorratio{y − ∑_{j\lt i}m_j}{f} \label{eq:willekeurigynaarx}$$
For the example calendar we have $$f = 40$$, $$g = 5$$, and
$${i}$$ 1 2 3 4 5 $${m_i}$$ 7 13 5 11 4 $${a_i}$$ 0 7 20 25 36
and so
\begin{align*} x \| = \floorratio{y + 40 − 0}{40} + \floorratio{y + 40 − 7}{40} + \floorratio{y + 40 − 20}{40} + \floorratio{y + 40 − 25}{40} + \floorratio{y + 40 − 36}{40} \\ \| = \floorratio{y + 40}{40} + \floorratio{y + 33}{40} + \floorratio{y + 20}{40} + \floorratio{y + 15}{40} + \floorratio{y + 4}{40} \end{align*}
For example, if $$y = 21$$ then
$x = \floorratio{61}{40} + \floorratio{54}{40} + \floorratio{41}{40} + \floorratio{36}{40} + \floorratio{25}{40} = 1 + 1 + 1 + 0 + 0 = 3$
If we go in the other direction then we also have a staircase, but now days and months exchange roles as length and height of the steps. With that, the formula for the running day number $$c$$ of the first day of month $$x$$ is:
$$c = ∑_{i}m_i\floorratio{x + g − i}{g}$$
With again the same calendar, we find
$c = 7 \floorratio{x + 4}{5} + 13 \floorratio{x + 3}{5} + 5 \floorratio{x + 2}{5} + 11 \floorratio{x + 1}{5} + 4 \floorratio{x}{5}$
$${x}$$ 0 1 2 3 4 5
$${7⌊(x+4)/5⌋}$$ 0 7 7 7 7 7
$${{13⌊(x+3)/5⌋}}$$ 0 0 13 13 13 13
$${5⌊(x+2)/5⌋}$$ 0 0 0 5 5 5
$${11⌊(x+1)/5⌋}$$ 0 0 0 0 11 11
$${4⌊x/5⌋}$$ 0 0 0 0 0 4
$${c}$$ 0 7 20 25 36 40
The formula to go from running month number $$x$$ and day number $$z$$ in the current month to running day number $$y$$ (all beginning at 0) is then (with $$i$$ running from 1 through $$g$$)
$$y = c + z = ∑_{i}m_i\floorratio{x + g − i}{g} + z \label{willekeurigxnaary}$$
These formulas are, if you write out the summation, a lot longer than the formulas that we found earlier for calendars with internal patterns, so it is convenient if you recognize such patterns for a calendar, but not every calendar has such patterns.
The simple calendar from section 12.3 has $$y = ⌊px + σ⌋ + z$$. To have that calendar begin month $$x$$ at day $$y$$ we need $$y = ⌊px + σ⌋$$, from which follows $$y ≤ px + σ \lt y + 1$$, hence (for $$x \gt 0$$) $$(y − σ)/x ≤ p \lt (y + 1 − σ)/x$$. We know that $$0 ≤ σ \lt 1$$, so
$\frac{y − 1}{x} \lt \frac{y − σ}{x} ≤ p \lt \frac{y + 1 − σ}{x} \lt \frac{y + 1}{x}$
so
$$\frac{y − 1}{x} \lt p \lt \frac{y + 1}{x} \label{eq:beperkp}$$
Not every $$p$$ that meets these restrictions yields a simple calendar, but if a $$p$$ does not meet these restrictions for at least one of its months, then there is certainly no simple calendar for these month lengths.
Our example calendar does not meet the restrictions that are necessary for the simple formulas from section 12.3 to apply, because there are more than two different month lengths, even if we leave the last month out of consideration. We can also see this using equation \eqref{eq:beperkp}. The first day of month $$x = 1$$ has $$y = 7$$, so we must have $$6 \lt p \lt 8$$. The first day of month $$x = 2$$ has $$y = 20$$ so we need $$9\frac{1}{2} = 19/2 \lt p \lt 21/2 = 10\frac{1}{2}$$, and already we're in trouble, because to have the beginning of month $$x = 1$$ in the right place $$p$$ must be less than 8, but to have the beginning of month $$x = 2$$ in the right place $$p$$ must be greater than 9½, and those restrictions cannot be met at the same time.
### 12.9. Combinations of straight lines
It is rare for a calendar to be fully defined by just one period, so usually you have to combine several periods. Let's assume that we have two straight lines for two periods:
\begin{align} y_1 \| = c_1(x_1) + z_1 \\ y_2 \| = c_2(x_2) + z_2 \end{align}
and in the other direction
\begin{align} x_1 \| = c_1'(y_1) \\ z_1 \| = y_1 − c_1(x_1) \\ x_2 \| = c_2'(y_2) \\ z_2 \| = y_2 − c_2(x_2) \end{align}
where $$c'(y)$$ is the inverse of $$c(x)$$: $$c'(c(x)) = x$$ and $$c(c'(y)) = y$$.
In simple cases we have
\begin{eqnarray} \{ x_1, r_1 \} \| = \Div(g_1y_1 + g_1 − 1 − s_1, f_1) \\ z_1 \| = \floorratio{r_1}{g_1} \\ \{ x_2, r_2 \} \| = \Div(g_2y_2 + g_2 − 1 − s_2, f_2) \\ z_2 \| = \floorratio{r_2}{g_2} \end{eqnarray}
We assume that the first line is for the smaller period and the second line is for the larger period. There are two ways to combine these: We can equate $$y_1$$ to $$z_2$$ or to $$x_2$$. In the first case, the two periods have the same leap unit, for example days. In the second case, the two periods have different leap units (for example, sometimes an extra day for the first period, and sometimes an extra month for the second period). If the larger period needs no leap rules at all, then you can choose which combination method to use.
Let's call the first case the "flat" combination, because the leap units remain the same. Let's call the second case the "stepped" combination, because the leap unit goes a step higher.
The combination of months and years in most (perhaps all) solar calendars (such as the Gregorian calendar, the Julian calendar, and the Egyptian calendar) is flat, i.e. of the first kind. A month can be a day shorter or longer than another month (and exactly one month can be a lot shorter), and a year can be a day shorter or longer than another year. The number of days in a year does not depend on the months, because the rules to calculate the length of the year depend on the year number but not on the month number.
The combination of months and years in a lunisolar calendar (such as the Hebrew and Babylonian calendars) can be flat with very unequal year lengths ($$d \gt 1$$) or stepped (then usually $$d = 1$$). A month can be a day longer or shorter than another month, and a year can be a month shorter or longer than another month. (For the flat combination, one month of the year can be much shorter.) In this way you can follow two separate (astronomical) cycles: the motion of the Sun (with the year) and the motion of the Moon (with the month). If the combination is stepped, then the length of the year depends on the length of the months, because the year is then defined in terms of a fixed number of months, not a fixed number of days.
Lunar calendars that are not lunisolar (such as the administrative Islamic calendar) usually do not have any leap rules, so then both methods can be used.
If a calendar has more than two important large periods with leap rules (for example, not just for the month and the year, but also for the century), then it is possible that some combinations are flat and others are stepped.
If we have to deal with more than one period, then it is important for translating a running day number into a calendar date to have the running numbers begin at 0 for each of those periods.
As an example we'll take for the first calendar a simple one with $$p_1 = 7/3 = 2 \frac{2}{3}$$ and for the second calendar a simple one with $$p_2 = 37/5 = 7 \frac{2}{5}$$. Then we have
\begin{align*} \\ y_1 \| = \floorratio{7x_1}{3} + z_1 = c_1 + z_1 \\ y_2 \| = \floorratio{37x_2}{5} + z_2 = c_2 + z_2 \\ x_1 \| = \floorratio{3y_1 + 2}{7} \\ z_1 \| = \floorratio{(3y_1 + 2) \bmod 7}{3} \\ x_2 \| = \floorratio{5y_2 + 4}{37} \\ z_2 \| = \floorratio{(5y_2 + 4) \bmod 37}{5} \end{align*}
or
\begin{align*} \{x_1, r_1\} \| = \Div(3y_1 + 2, 7) \\ z_1 \| = \floorratio{r_1}{3} \\ \{x_2, r_2\} \| = \Div(5y_2 + 4, 37) \\ z_2 \| = \floorratio{r_2}{5} \end{align*}
#### 12.9.1. Flat combination
In this case $$z_2 = y_1$$, so
$$y_2 = c_2(x_2) + c_1(x_1) + z_1 \label{eq:vlak}$$
and in the other direction
\begin{align} x_2 \| = c_2'(y_2) \label{eq:vlakr} \\ y_1 \| = z_2 = y_2 − c_2(x_2) \\ x_1 \| = c_1'(y_1) \\ z_1 \| = y_1 − c_1(x_1) \end{align}
For simple cases
\begin{align} \{x_2, r_2\} \| = \Div(g_2y_2 + g_2 − 1 − s_2, f_2) \\ \{x_1, r_1\} \| = \Div\left( g_1\floorratio{r_2}{g_2} + g_1 − 1 − s_1, f_1 \right) \\ z_1 \| = \floorratio{r_1}{g_1} \end{align}
Here (for example) $$y_2$$ is the running day number, $$x_2$$ is the year number, $$y_1 = z_2$$ is the day number in the current year, $$x_1$$ is the month number in the current year (with 0 for the first month), and $$z_1$$ is the day number in the current month (with 0 for the first day).
To calculate the calendar date from the running day number, we must first calculate the larger period ($$x_2$$, $$z_2$$), and then the smaller period ($$x_1$$, $$z_1$$).
The two example calendars then yield, for the first 20 days,
$${y_2}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
$${x_2}$$ 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2
$${r_2}$$ 4 9 14 19 24 29 34 2 7 12 17 22 27 32 0 5 10 15 20 25
$${z_2}$$ 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5
$${y_1}$$ 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5
$${x_1}$$ 0 0 1 1 2 2 2 0 0 1 1 2 2 2 0 0 1 1 2 2
$${r_1}$$ 2 5 1 4 0 3 6 2 5 1 4 0 3 6 2 5 1 4 0 3
$${z_1}$$ 0 1 0 1 0 1 2 0 1 0 1 0 1 2 0 1 0 1 0 1
Running day number $$y_2 = 16$$ corresponds to day $$z_1 = 0$$ of month $$x_1 = 1$$ of year $$x_2 = 2$$.
#### 12.9.2. Stepped combination
In this case $$x_2 = y_1$$, so
$$y_2 = c_2(c_1(x_1) + z_1) + z_2$$
and in the other direction
\begin{align} y_1 \| = c_2'(y_2) \\ z_2 \| = y_2 − c_2(y_1) \\ x_1 \| = c_1'(y_1) \\ z_1 \| = y_1 − c_1(x_1) \end{align}
For simple cases
\begin{align} \{y_1, r_2\} \| = \Div(g_2y_2 + g_2 − 1 − s_2, f_2) \\ z_2 \| = \floorratio{r_2}{g_2} \\ \{x_1, r_1\} \| = \Div(g_1y_1 + g_1 − 1 − s_1, f_1) \\ z_1 \| = \floorratio{r_1}{g_1} \end{align}
Here (for example) $$y_2$$ is the running day number, $$x_1$$ is the year number, $$x_2 = y_1$$ is the running month number, $$z_1$$ is the month number in the current year (with 0 for the first month), and $$z_2$$ is the day number in the current month (with 0 for the first day).
With our two example calendars we find
$${y_2}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
$${x_2}$$ 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2
$${r_2}$$ 4 9 14 19 24 29 34 2 7 12 17 22 27 32 0 5 10 15 20 25
$${z_2}$$ 0 1 2 3 4 5 6 0 1 2 3 4 5 6 0 1 2 3 4 5
$${y_1}$$ 0 0 0 0 0 0 0 1 1 1 1 1 1 1 2 2 2 2 2 2
$${x_1}$$ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1
$${r_1}$$ 2 2 2 2 2 2 2 5 5 5 5 5 5 5 1 1 1 1 1 1
$${z_1}$$ 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0
Running day number $$y_2 = 16$$ corresponds to day $$z_2 = 2$$ of month $$z_1 = 0$$ of year $$x_1 = 1$$.
### 12.10. Simultaneous Cycles
Most calendars have higher and lower periods, and the number of a higher period changes much less often than the number of a lower period. After the 7th day of the 3rd month follows the 8th day of the 3rd month − the month number changes less often than the day number.
Some calendars have different periods that change simultaneously. The most common calendars often have such a case, too, in the form of weekdays and days of the month. When the next day arrives, then the day number of the month changes, but the weekday changes, too. After Monday the 7th we get Tuesday the 8th − the weekday and the day number change equally fast.
We don't need the weekday to be able to point to a unique day (30 August 2011 indicates exactly one day; for that we do not need to know that that day was a Tuesday), so the weekday does not usually play a role in calendar calculations. However, there are calendars (for example the calendars of Central America) which do use simultaneous cycles to point at particular days. We look at this type of calendar below.
Suppose that a calendar uses simultaneous periods $$p_i$$ for $$i$$ from 1 through $$n$$. The day number in each period begins at 0. We write a date in that calendar as $$\{x\} = \{x_1,x_2,…,x_n\}$$, where $$x_i$$ is the day number from period $$i$$. If $$x_i$$ is equal to $$p_i − 1$$, then the next day has $$x_i = 0$$ again.
For example, in a calendar with $$p_1 = 13$$ and $$p_2 = 20$$, we get after day $$\{11,8\}$$ the following days: $$\{12,9\}$$, $$\{0,10\}$$, $$\{1,11\}$$, …, $$\{9,19\}$$, $$\{10,0\}$$, $$\{11,1\}$$.
#### 12.10.1. From Running Day Number to Date
To translate running day number $$J$$ into $$\{x\}$$ we can use the following formula:
$$x_i = (J + a_i) \bmod p_i \label{eq:cycli}$$
Here $$a_i$$ is the value of $$x_i$$ when $$J = 0$$. That value depends on the calendar.
Suppose that in the calendar from the previous example $$J = 0$$ corresponds to $$\{11,8\}$$. Then $$a_1 = 11$$ and $$a_2 = 8$$, and then
\begin{align*} x_1 = (J + 11) \bmod 13 \\ x_2 = (J + 8) \bmod 20 \end{align*}
Day $$J = 11$$ then corresponds to $$x_1 = (11 + 11) \bmod 13 = 22 \bmod 13 = 9$$ and $$x_2 = (11 + 8) \bmod 20 = 19 \bmod 20 = 19$$, so $$\{9,19\}$$.
#### 12.10.2. From Date to Running Day Number
It is a lot more difficult to go in the other direction. Then we have to find a $$J$$ for which equation \eqref{eq:cycli} is satisfied for all $$i$$, i.e.,
$$J ≡ x_i − a_i \pmod{p_i}$$
We define
$$c_i = x_i − a_i$$
We first look at the case $$n = 2$$. Then we need to solve $$J$$ from
\begin{align} J \| ≡ c_1 \pmod{p_1} \label{eq:c1} \\ J \| ≡ c_2 \pmod{p_2} \label{eq:c2} \end{align}
All solutions to equation \eqref{eq:c1} are of the kind
$$J = c_1 + kp_1$$
when $$k$$ runs through all whole numbers. Substituting into formula \eqref{eq:c2} yields
$$kp_1 ≡ c_2 − c_1 \pmod{p_2} \label{eq:kp_1}$$
If we could find a number $$r_2$$ such that $$r_2p_1 ≡ 1 \pmod{p_2}$$, then we could solve the preceding formula for $$k$$. Such a number $$r_2$$ only exists if $$p_1$$ and $$p_2$$ are relative prime, which means that they have no common divisors (greater than 1).
If the greatest common divisor of $$p_1$$ and $$p_2$$ is equal to $$g$$ (greater or equal to 1), and
$$q_i = \frac{p_i}{g}$$
then we can rewrite formula \eqref{eq:kp_1} to
$$kgq_1 ≡ c_2 − c_1 \pmod{gq_2}$$
If $$c_2 − c_1$$ is evenly divisible by $$g$$, then we can divide by $$g$$ throughout and then we find
$$kq_1 = \frac{c_2 − c_1}{g} \pmod{q_2} \label{eq:red}$$
For dates from calendars based on formula \eqref{eq:cycli} $$c_2 − c_1$$ is evenly divisible by $$g$$, so we make that assumption.
Because $$q_1$$ and $$q_2$$ are relative prime, there is a number $$r_2$$ such that $$r_2q_1 ≡ 1 \pmod{q_2}$$. With that we find from equation \eqref{eq:red}
$$k ≡ kq_1r_2 ≡ \frac{c_2 − c_1}{g} r_2 \pmod{q_2}$$
and then
\begin{align} kp_1 \| ≡ q_1r_2\frac{c_2 − c_1}{g} \pmod{q_2p_1} \\ J \| = c_1 + kp_1 ≡ c_1 + q_1r_2(c_2 − c_1) ≡ c_1 (1 − r_2q_1) + c_2r_2q_1 \pmod{q_2p_1} \end{align}
thus
$$J ≡ (x_1 − a_1) (1 − r_2q_1) + (x_2 − a_2)r_2q_1 \pmod{p_2q_1} \label{eq:cyclitoj}$$
Summarizing:
1. Calculate the greatest common divisor $$g$$ of $$p_1$$ and $$p_2$$:
$$g = \gcd(p_1, p_2)$$
2. Find the smallest multiple $$s$$ of $$p_1/g$$ for which
$s ≡ 1 \pmod{p_2/g}$
3. Then
$$J ≡ sx_2 + (1 − s)x_1 − (sa_2 + (1 − s)a_1) \pmod{p_1p_2/g}$$
How can you find a $$r_2$$ such that $$s = r_2q_1 ≡ 1 \pmod{q_2}$$? Because you only need to find that $$r_2$$ or $$s$$ once for a given calendar, it is usually the least amount of work to just try successive values until you find the correct one. You need to check at most $$q_2$$ successive values. If you need to calculate such $$r_2$$ often, then you can use the Extended Euclidean Algorithm.
You can find the greatest common divisor of $$p_1$$ and $$p_2$$ by using the regular Euclidean Algorithm.
We again use the calendar from the previous example, with $$p_1 = 13$$, $$p_2 = 20$$, $$a_1 = 11$$ and $$a_2 = 8$$. The greatest common divisor of 13 and 20 is 1, so $$g = 1$$. Now we seek $$s$$. That is the smallest multiple of $$p_1/g = 13/1 = 13$$ for which $$s ≡ 1 \pmod{20}$$, which is equivalent to $$s \bmod 20 = 1$$.
$${k}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 $${s = 13k}$$ 0 13 26 39 52 65 78 91 104 117 130 143 156 169 182 195 208 221 234 247 260 $${s \bmod 20}$$ 0 13 6 19 12 5 18 11 4 17 10 3 16 9 2 15 8 1 14 7 0
We find $$s = 17×13 = 221$$, because $$221 ≡ 1 \pmod{20}$$.
Then
\begin{align*} J \| ≡ 221x_2 − 220x_1 − (221×8 − 220×11) \\ \| ≡ 221x_2 − 220x_1 + 652 \\ \| ≡ 221x_2 + 40x_1 + 132 \\ \| ≡ 40x_1 + 221x_2 + 132 \pmod{13×20 = 260} \end{align*}
For date $$\{9,19\}$$ we then find
\begin{align*} J \| ≡ 40×9 + 221×19 + 132 \\ \| ≡ 360 + 4199 + 132 \\ \| ≡ 100 + 39 + 132 ≡ 271 ≡ 11 \pmod{260} \end{align*}
so day $$J = 11$$ and every 260 days earlier or later correspond to date $$\{9,19\}$$.
The order of the two congruence relations that we began with is not important, so we can switch the roles of $$p_1$$ and $$p_2$$ (and of $$a_1$$ and $$a_2$$). We seek the smallest multiple $$s$$ of $$p_2/g = 20/1 = 20$$ for which $$s ≡ 1 \pmod{13}$$, which means $$s \bmod 13 = 1$$. We find $$s = 40$$ because $$40 = 3×13 + 1 ≡ 1 \pmod{13}$$. Then
\begin{align*} J \| ≡ 40x_1 ― 39x_2 − (40×11 − 39×8) \\ \| ≡ 40x_1 ― 39x_2 − 128 \\ \| = 40x_1 + 221x_2 + 132 \pmod{260} \end{align*}
which is the same as before.
Warning! Formula \eqref{eq:cyclitoj} only gives useful results for $$\{x\}$$ that really occur in the calendar. If $$g$$ is not equal to 1, then not all possible $$\{x\}$$ can occur. If you enter an $$\{x\}$$ that does not occur in the calendar, then you get a reasonable-looking result from the formula, but that result then still won't be valid.
Now we look at a calendar for which the periods are not relatively prime, with $$p_1 = 10$$, $$p_2 = 15$$, $$a_1 = a_2 = 0$$. Then
\begin{align*} x_1 = J \bmod 10 \\ x_2 = J \bmod 15 \end{align*}
The largest common divisor of $$p_1$$ and $$p_2$$ is $$g = 5$$. We seek the smallest multiple $$s$$ of $$p_1/g = 10/5 = 2$$ for which $$s ≡ 1 \pmod{15/5 = 3}$$. That is $$s = 4$$, because $$4 = 2×2 = 3 + 1 ≡ 1 \pmod{3}$$. With that,
\begin{align*} J \| ≡ 4x_2 − 3x_1 − 0 ≡ 27x₁ + 4x₂ \pmod{10×15/5 = 30} \end{align*}
For $$J$$ from 8 to 17 we then find for $$\{x\}$$: $$\{8,8\}$$, $$\{9,9\}$$, $$\{0,10\}$$, $$\{1,11\}$$, $$\{2,12\}$$, $$\{3,13\}$$, $$\{4,14\}$$, $$\{5,0\}$$, $$\{6,1\}$$, $$\{7,2\}$$.
For that $$\{x\}$$ we find:
$${x_1}$$ 8 9 0 1 2 3 4 5 6 7 $${x_2}$$ 8 9 10 11 12 13 14 0 1 2 $${27x_1 + 4x_2}$$ 248 279 40 71 102 133 164 135 166 197 $${J \bmod 30}$$ 8 9 10 11 12 13 14 15 16 17
Because $$g$$ is not equal to 1, many combinations of $$x_1$$ and $$x_2$$ do not occur in this calendar. Only combinations for which $$x_1 − x_2$$ is evenly divisible by 5 occur in this calendar. For example, $$\{3,7\}$$ does not occur. If we apply the equation for $$J$$ to that date, then we find $$J ≡ 27×3 + 4×7 ≡ 19 \pmod{30}$$, but for $$J = 19$$ we find $$x_1 = 9$$ and $$x_2 = 4$$, so $$\{9,4\}$$ and not $$\{3,7\}$$. For impossible $$\{x\}$$ we still get nice-looking $$J$$ out of the formula.
#### 12.10.3. More than Two Periods
Using equation \eqref{eq:cyclitoj}, $$J ≡ x_1 − a_1 \pmod{p_1}$$ and $$J ≡ x_2 − a_2 \pmod{p_2}$$ lead to another equation $$J ≡ C_2 \pmod{P_2}$$ with
\begin{align} C_2 \| ≡ (x_1 − a_1) (1 − s_2) + (x_2 − a_2) s_2 \\ P_2 \| ≡ p_1p_2/g_1 \end{align}
That formula has the same form as the two formulas that we started with, so we can combine that formula in the same manner with $$J ≡ x_3 − a_3 \pmod{p_3}$$, and so on until we've handled all $$p_i$$. We end up with a formula that again has the same form $$J ≡ C_n \pmod{P_n}$$, and that is the solution of the $$n$$ equations.
The procedure is then as follows. Given $$x_i$$, $$p_i$$, $$a_i$$ for $$i$$ from 1 through $$n$$,
1. Assign $$C_1 = x_1 − a_1$$ and $$P_1 = p_1$$.
2. For $$i$$ from 2 through $$n$$:
3. Determine $$g_i = \gcd(P_{i−1}, p_i)$$, the greatest common divisor of $$P_{i−1}$$ and $$p_i$$.
4. Find the smallest $$s_i$$ that is a multiple of $$P_{i−1}/g_i$$ and that satisfies $$s_i ≡ 1 \pmod{p_i/g_i}$$.
5. Assign $$C_i = C_{i−1} (1 − s_i) + (x_i − a_i) s_i$$ and $$P_i = P_{i−1}p_i/g_i$$.
6. The solution is $$J ≡ C_n \pmod{P_n} \label{eq:cycle}$$
Let's look at a calendar with the following characteristics:
$${i}$$ 1 2 3 $${p_i}$$ 4 5 6 $${a_i}$$ 3 1 2
You translate running day number $$J$$ to calendar date $$\{x\}$$ as follows:
\begin{align*} x_1 \| = (J + 3) \bmod 4 \\ x_2 \| = (J + 1) \bmod 5 \\ x_3 \| = (J + 2) \bmod 6 \end{align*}
In the opposite direction we find:
\begin{align*} C_1 \| = x_1 − 3 \\ P_1 \| = 4 \\ g_2 \| = \gcd(4, 5) = 1 \\ s_2 \| = 16 = 4×4 = 3×5 + 1 ≡ 1 \pmod{5} \\ C_2 \| = C_1×(1 − 16) + (x_2 − 1)×16 \\ \| = −15×(x_1 − 3) + 16×(x_2 − 1) \\ \| = −15x_1 + 16x_2 + 29 \\ P_2 \| = 4×5/1 = 20 \\ g_3 \| = \gcd(20, 6) = 2 \\ s_3 \| = 10 = 5×2 = 3×3 + 1 ≡ 1 \pmod{3} \\ C_3 \| = C_2×(1 − 10) + (x_3 − 2)×10 \\ \| = −9×(−15x_1 + 16x_2 + 29) + 10×(x_3 − 2) \\ \| = 135x_1 − 144x_2 + 10x_3 − 281 \\ P_3 \| = 20×6/2 = 60 \end{align*}
The solution is $$J ≡ C_3 ≡ 15x_1 + 36x_2 + 10x_3 + 19 \pmod{60}$$.
Let's check: if $$J = 654$$, then $$x_1 = (654 + 3) \bmod 4 = 1$$, $$x_2 = (654 + 1) \bmod 5 = 0$$, $$x_3 = (654 + 2) \bmod 6 = 2$$. And from $$\{x\}$$ to $$J$$ we find $$J ≡ 15×1 + 36×0 + 10×2 + 19 ≡ 54 \pmod{60}$$, which is correct, because $$54 ≡ 654 \pmod{60}$$.
#### 12.10.4. To One Solution
Because the date $$\{x\}$$ is given as a collection of positions in repeating periods, there is an infinite number of days that fit that date. For the running day number $$J$$ that corresponds to date $$\{x\}$$ we find formula \eqref{eq:cycle}, which gives a solution $$\bmod P_n$$. If you have found a running day number $$J$$ that fits date $$\{x\}$$, then you can add or subtract arbitrary multiples of $$P_n$$ and then you find another $$J$$ that belongs to date $$\{x\}$$. How do we pick the right one?
To pick the right solution out of an infinite collection of solutions, we need to use additional information. One appropriate way is to pick the solution that is closest to a date $$J_0$$ chosen by us. One can interpret "closest to" in at least four different ways:
1. the last one at or before $$J_0$$
2. the first one at or after $$J_0$$
3. the last one before $$J_0$$
4. the first one after $$J_0$$
We investigate these cases one by one, for $$J ≡ C_n \pmod{P_n}$$.
##### 12.10.4.1. The last one at or before
We seek the last $$J$$ at or before $$J_0$$ that satisfies $$J ≡ C_n \pmod{P_n}$$. Then, with $$k$$ a whole number,
\begin{align} J_0 − P_n \| \lt J ≤ J_0 \\ J_0 − P_n \| \lt C_n + kP_n ≤ J_0 \\ \frac{J_0 − C_n}{P_n} − 1 \| \lt k ≤ \frac{J_0 − C_n}{P_n} \end{align}
There is only one whole number $$k$$ that satisfies this inequality, and that is
$$k = \floorratio{J_0 − C_n}{P_n}$$
so
$$\begin{split} J \| = C_n + P_n \floorratio{J_0 − C_n}{P_n} \\ \| = C_n + P_n \left( \frac{J_0 − C_n}{P_n} − \mod1ratio{J_0 − C_n}{P_n} \right) \\ \| = C_n + (J_0 − C_n) − P_n \mod1ratio{J_0 − C_n}{P_n} \\ \| = J_0 − ⌊J_0 − C_n⌉_{P_n} \\ \| = J_0 − ((J_0 − C_n) \bmod P_n) \end{split} \label{eq:nearestle}$$
For example, what is the last day at or before $$J_0 = 700$$ that corresponds to date $$\{1,0,2\}$$ in the calendar from the previous example?
There we found $$C_n = 15x_1 + 36x_2 + 10x_3 + 19$$ and $$P_n = 60$$, so $$C_n = 15×1 + 36×0 + 10×2 + 19 = 54$$. With formula \eqref{eq:nearestle} we then find $$J = 700 − ((700 − 54) \bmod 60) = 700 − (646 \bmod 60) = 700 − 46 = 654$$ which is correct: 654 + 60 = 714 is greater than 700, so 654 is the last $$J$$ that is less than or equal to $$J_0$$ and that fits date $$\{1,0,2\}$$.
We should find $$J = 654$$ for all $$J_0$$ from 654 through 654 + 60 − 1 = 713, and we do:
$${J_0}$$ $${J_0 − C_n}$$ $${(J_0 − C_n) \bmod P_n}$$ $${J}$$ 653 599 59 594 654 600 0 654 712 658 58 654 713 659 59 654 714 660 0 714
##### 12.10.4.2. The first at or after
We seek the first $$J$$ at or after $$J_0$$ that satisfies $$J ≡ C_n \pmod{P_n}$$. Then, with $$k$$ a whole number,
\begin{align} J_0 \| ≤ J \lt J_0 + P_n \\ J_0 \| ≤ C_n + kP_n \lt J_0 + P_n \\ \frac{J_0 − C_n}{P_n} \| ≤ k \lt \frac{J_0 − C_n}{P_n} + 1 \end{align}
There is only one whole number $$k$$ that satisifies this inequality, and that number is
$$k = \ceilratio{J_0 − C_n}{P_n}$$
so
$$\begin{split} J \| = C_n + P_n \ceilratio{J_0 − C_n}{P_n} \\ \| = C_n + P_n \left( \frac{J_0 − C_n}{P_n} + \dom1ratio{J_0 − C_n}{P_n} \right) \\ \| = C_n + (J_0 − C_n) + P_n \dom1ratio{J_0 − C_n}{P_n} \\ \| = J_0 + ⌈J_0 − C_n⌋_{P_n} \\ \| = J_0 + ⌊C_n − J_0⌉_{P_n} \\ \| = J_0 + ((C_n − J_0) \bmod P_n) \end{split} \label{eq:nearestge}$$
For example, what is the first day at or after $$J_0 = 700$$ that corresponds to date $$\{1,0,2\}$$ in the calendar from the previous example?
We had $$C_n = 54$$. With formula \eqref{eq:nearestge} we then find $$J = 700 + ((54 − 700) \bmod 60) = 700 + ((−646) \bmod 60) = 700 + 14 = 714$$ which is correct: 714 − 60 = 654 is less than 700, so 714 is the first $$J$$ that is greater than or equal to $$J_0$$ and that fits date $$\{1,0,2\}$$.
We should find $$J = 714$$ for all $$J_0$$ from 714 − 60 + 1 = 655 through 714, and we do:
$${J_0}$$ $${J_0 − C_n}$$ $${(C_n − J_0) \bmod P_n}$$ $${J}$$ 654 600 0 654 655 601 59 714 656 602 58 714 713 659 1 714 714 660 0 714 715 661 59 774
##### 12.10.4.3. The last before
The last $$J$$ before $$J_0$$ is one $$P_n$$ before the first $$J$$ at or after $$J_0$$, so
$$J = J_0 − P_n − ((J_0 − C_n) \bmod P_n)$$
##### 12.10.4.4. The first after
The first $$J$$ after $$J_0$$ is one $$P_n$$ after the last $$J$$ at or before $$J_0$$, so
\begin{align} J \| = J_0 + P_n + ⌈J_0 − C_n⌋_{P_n} \\ \| = J_0 + P_n + ((C_n − J_0) \bmod P_n) \end{align}
### 12.11. Summary
This summary is in terms of days and months, but holds also for other units of time. We distinguish four different degrees of difficulty: from 1 to 4 the formulas get more complicated but also more capable.
1. The simplest calendar has two different month lengths, with the greatest being one greater than the smallest, and allows no shifting of the pattern of month lengths. Month 0 is always a short month, and month $$g − 1$$ (the last month before everything repeats itself) is always a long month.
2. Calendar type 2 is equal to type 1, except that shifting of the pattern of month lengths is allowed.
3. Calendar type 3 is equal to type 2, but now the difference between the month lengths can be more than one.
4. Calendar type 4 is equal to type 3, but allows more than two month lengths.
The following table summarizes the calendar formulas, and uses the following definitions:
• $$q$$ is the basic number of days in each month. For calendar types 1 through 3, $$q$$ is the number of days in short months. $$q$$ is a whole number greater than 0.
• $$p$$ is the average length of the infinite number of months that this calendar provides (even if in practice the calendar is used in a flat combination with another calendar so that only a limited number of months actually get used). $$p$$ is greater than zero, but is usually not a whole number.
• $$f$$, $$g$$ are the numerator and denominator from $$p = f/g$$. They are whole numbers greater than zero.
• $$ψ$$, $$ψ_i$$ is the fraction of the infinite number of months that has the non-basic month length (unequal to $$q$$).
• $$d$$, $$d_i$$ is the difference between a month length and the basic month length $$q$$. They are whole numbers. For calendar type 3, $$d$$ must be greater than zero. For calendar type 4 $$d_i$$ may be greater or less than zero.
• $$h$$, $$h_i$$ are the numerator and denominator from $$ψ = h/g$$ or $$ψ_i = h_i/g$$. They are whole numbers greater than zero.
• $$a$$, $$a_i$$ indicate by how many months the pattern of month lengths should be shifted compared to the simplest case. They are a whole numbers. Only the values from 0 through $$g − 1$$ are important: value $$a + g$$ has the same effect as value $$a$$.
• $$x$$ is the running month number. It is a whole number, and can be negative, zero, or positive.
• $$c$$ is the running day number of the beginning of month $$c$$. It is a whole number, and can be negative, zero, or positive.
• $$ζ$$ is a correction to be added to month number $$x$$ to get a better estimate for the month number. This is only relevant for calendar types 3 and 4.
• $$z$$ is the number of days since the beginning of the current month, $$z = y − c$$.
#
$${p}$$ $${f}$$ $${c}$$ $${c'}$$ $${s}$$ $${ζ}$$
1 $${q+ψ}$$ $${qg+h}$$ $${\floorratio{fx}{g}}$$ $${\floorratio{gy+g−1}{f}}$$ $${0}$$ $${0}$$
2 $${q+ψ}$$ $${qg+h}$$ $${\floorratio{fx+s}{g}}$$ $${\floorratio{gy+g−s−1}{g}}$$ $${fa \bmod g}$$ $${0}$$
3 $${q+dψ}$$ $${qg+dh}$$ $${qx+d\floorratio{hx+s}{g}}$$ $${\floorratio{gy+dg−ds−1}{g}}$$ $${ha \bmod g}$$ $${\floorratio{z}{q + d}}$$
4 $${q+∑d_{i}ψ_i}$$ $${qg+∑d_{i}h_i}$$ $${qx+∑d_i\floorratio{h_{i}x+s_i}{g}}$$ $${\floorratio{gy+g∑_{\gt0}d_i−∑d_{i}s_i−1}{f}}$$ $${h_{i}a_i \bmod g}$$ $${\floorratio{z}{q + ∑_{\gt0}d_i}}$$
For calendar types 3 and 4, the formula for $$x$$ yields an upper boundary for the month number. The general procedure for finding the right month number is then:
Try the $$x$$ from the formula in the table. Calculate the corresponding $$c$$ and then $$z = y − c$$ and $$ζ$$. If $$ζ$$ is equal to 0, then you're done. Otherwise, add $$ζ$$ (which is negative) to $$x$$ and try again.
If $$d ≤ p$$ (for calendar type 3) or $$∑_{\gt0}d_i − ∑_{\lt0}d_i \lt p$$ (for calendar type 4), then you need to calculate $$ζ$$ at most once. Otherwise you may need to do it more often, up to $$⌈(d − 1)/p⌉ + 1$$ (for calendar type 3) or $$⌈(∑_{\gt0}d_i − ∑_{\lt0}d_i)/p⌉ + 1$$ (for calendar type 4) times.
If there are more that two relevant units of time, then different calendar levels must be combined. From calendar date in the direction of running day number we have
$$y_i = c_i(x_i) + z_i$$
and from running day number in the direction of calendar date
\begin{align} x_i \| = c'_i(y_i) \\ z_i \| = y_i − c_i(x_i) \end{align}
If the smallest unit of time in two calendars is the same (for example, days), then those calendars should be combined in a flat manner, then $$z_{i+1} = y_i$$. If the smallest unit of time from calendar $$i + 1$$ is equal to the largest unit of time from calendar $$i$$, then those calendars should be combined in a stepped manner, then $$x_{i+1} = y_i$$.
## 13. Derivation for Specific Calendars
Almost all calendars distinguish between three basic periods: days, months, and years. One calendar level (straight line) links two periods, so at least two calendar levels are needed to link three periods.
### 13.1. The Julian Calendar
#### 13.1.1. From Julian Date to CJDN (1)
In the Julian calendar, every three regular years of 365 days are followed by a leap year of 366 days. This period of four years contains 3×365 + 366 = 1461 days. Each period of four years has the same sequence of months with their month lengths.
All months have 30 or 31 days, except for February which has 28 or 29 days. That extra-short month February is a bit of a problem; because of it, we have to deal not with two but with four different month lengths. We can work around the problem by regarding February as the last month of the calculation year. Then March is month 0 of the calculation year, and February is month 11 of the same calculation year.
For month $$m$$ of year $$j$$ we then find calculation month $$m_2$$ and calculation year $$j_2$$:
\begin{align} \{ c, m_2 \} \| = \Div(m − 3, 12) \\ c \| = \floorratio{m − 3}{12} \\ m_2 \| = m − 12c − 3 = (m ― 3) \bmod 12 \\ j_2 \| = j + c \end{align}
$${m}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 $${c}$$ −1 −1 −1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 $${m_2}$$ 9 10 11 0 1 2 3 4 5 6 7 8 9 10 11 0 1 2 $${j_2}$$ $${j−1}$$ $${j−1}$$ $${j−1}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j}$$ $${j+1}$$ $${j+1}$$ $${j+1}$$
and in the opposite direction
\begin{align} \{ c, m − 1 \} \| = \Div(m_2 + 2, 12) \\ c \| = \floorratio{m_2 + 2}{12} \\ m \| = m_2 − 12c + 3 \\ j \| = j_2 + c \end{align}
$${m_2}$$ 0 1 2 3 4 5 6 7 8 9 10 11 $${c}$$ 0 0 0 0 0 0 0 0 0 0 1 1 $${m}$$ 3 4 5 6 7 8 9 10 11 12 1 2 $${j}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2}$$ $${j_2+1}$$ $${j_2+1}$$
If we use a flat combination of the calendar levels, then the lower calendar (days-months) starts over for every year in the upper calendar (days-years). The last month of the calculation year is then cut short by the upper calendar, so then it does not matter if that last month would contain too many days if it weren't cut short. Then we can give all calendar months 30 or 31 days, which makes the calculations a lot simpler.
The month lengths from March through January are 31 - 30 - 31 - 30 - 31 - 31 - 30 - 31 - 30 - 31 - 31. We get this sequence of month lengths if we use calendar type 2 from section 12.11 with $$f = 153$$, $$g = 5$$, $$a = 4$$ so $$s = fa \bmod g = 153×4 \bmod 5 = 2$$. In that calendar, February gets 30 days, but that is cut short to 28 or 29 days by the upper calendar.
That upper calendar should yield 365 - 365 - 365 - 366 days and repeat that every 4 years. We get that sequence of year lengths if we use calendar type 1 with $$f = 1461$$, $$g = 4$$.
The first day of calculation year 0, i.e. 1 March of the year 0 in the Julian calendar, has CJDN $$J_0 = 1721118$$.
The steps to get from the Julian calendar to the CJDN are now:
1. Calculate calculation year number $$x_2$$, calculation month number $$x_1$$, and calculation day number $$z_1$$ from the year number $$j$$, month number $$m$$, and day number $$d$$. The first time period for each unit of time (calculation year, calculation month, calculation day) must have number 0, otherwise the calculations don't give the right results.
\begin{align} \{ c_0, x_1\} \| = \Div(m ― 2, 12) \\ c_0 \| = \floorratio{m − 3}{12} \\ x_1 \| = m − 12c_0 − 3 = (m ― 3) \bmod 12 \\ x_2 \| = j + c_0 \\ z_1 \| = d − 1 \end{align}
2. We combine the two calendar levels in the flat manner, so $$z_2 = y_1$$:
\begin{align} c_1 \| = \floorratio{153x_1 + 2}{5} \\ y_1 \| = c_1 + z_1 \\ z_2 \| = y_1 \\ c_2 \| = \floorratio{1461x_2}{4} \\ y_2 \| = c_2 + z_2 \end{align}
3. Add the CJDN of running day number 0:
$$J = y_2 + J_0 = y_2 + 1721118$$
For example, what CJDN corresponds to Julian date 6 July 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$, and thus
\begin{align*} c_0 \| = \floorratio{7 − 3}{12} = 0 \\ x_1 \| = 7 − 12×0 − 3 = 4 \\ x_2 \| = 2003 + 0 = 2003 \\ z_1 \| = 6 − 1 = 5 \\ c_1 \| = \floorratio{153×4 + 2}{5} = \floorratio{614}{5} = 122 \\ y_1 \| = 122 + 5 = 127 \\ z_2 \| = 127 \\ c_2 \| = \floorratio{1461×2003}{4} = \floorratio{2926383}{4} = 731595 \\ y_2 \| = 731595 + 127 = 731722 \\ J \| = 731722 + 1721118 = 2452840 \end{align*}
Another example. What CJDN corresponds to Julian date 1 December 2000? Then $$j = 2000$$, $$m = 12$$, $$d = 1$$, so
\begin{align*} c_0 \| = \floorratio{12 − 3}{12} = 0 \\ x_1 \| = 12 − 12×0 − 3 = 9 \\ x_2 \| = 2000 + 0 = 2000 \\ z_1 \| = 1 − 1 = 0 \\ c_1 \| = \floorratio{153×9 + 2}{5} = \floorratio{1379}{5} = 275 \\ y_1 \| = 275 + 0 = 275 \\ z_2 \| = 275 \\ c_2 \| = \floorratio{1461×2000}{4} = \floorratio{2922000}{4} = 730500 \\ y_2 \| = 730500 + 275 = 730775 \\ J \| = 730775 + 1721118 = 2451893 \end{align*}
We can combine and squeeze these steps into:
\begin{align} c_0 \| = \floorratio{m − 3}{12} \\ J \| = \floorratio{1461×(j + c_0)}{4} + \floorratio{153m − 1836c_0 − 457}{5} + d + 1721117 \end{align}
For the same example as above we find $$j = 2003$$, $$m = 7$$, $$d = 6$$, so
\begin{align*} c_0 \| = \floorratio{7 − 3}{12} = 0 \\ J \| = \floorratio{1461×(2003 + 0)}{4} + \floorratio{153×7 − 1836×0 − 457}{5} + 6 + 1721117 \\ \| = \floorratio{2926383}{4} + \floorratio{614}{5} + 6 + 1721117 \\ \| = 731595 + 122 + 6 + 1721117 = 2452840 \end{align*}
#### 13.1.2. From CJDN to Julian Date (1)
We run through the opposite procedure from the one described above, with the same calendar levels as above. Calendar level 1 is of calendar type 2 with $$f = 153$$, $$g = 5$$, $$a = 4$$ (so $$s = fa \bmod g = 153×4 \bmod 5 = 2$$), and calendar level 2 is of calendar type 1 with $$f = 1461$$, $$g = 4$$.
First we calculate the running day number from the Julian Day Number, by subtracting the Julian Day Number of running day number 0:
$$y_2 = J − J_0 = J − 1721118$$
Then we calculate the calculation year, calculation month, and calculation day from the running day number, in the flat manner:
\begin{align} \{x_2, r_2\} \| = \Div(4y_2 + 3, 1461) \\ z_2 \| = \floorratio{r_2}{4} \\ \{x_1, r_1\} \| = \Div(5z_2 + 2, 153) \\ z_1 \| = \floorratio{r_1}{5} \end{align}
And lastly we translate these calculation values to calendar values:
\begin{align} c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 = ((x_1 + 2) \bmod 12) + 1 \\ d \| = z_1 + 1 \end{align}
For example, which Julian Date corresponds to Julian Day number 2452840? Then $$J = 2452840$$ so
\begin{align*} y_2 \| = 2452840 − 1721118 = 731722 \\ \{x_2, r_2\} \| = \Div(4×731722 + 3, 1461) = \Div(2926891, 1461) = \{2003, 508\} \\ z_2 \| = \floorratio{508}{4} = 127 \\ y_1 \| = 127 \\ \{x_1, r_1\} \| = \Div(5×127 + 2, 153) = \Div(637, 153) = \{4, 25\} \\ z_1 \| = \floorratio{25}{5} = 5 \\ c_0 \| = \floorratio{4 + 2}{12} = 0 \\ j \| = 2003 + 0 = 2003 \\ m \| = 4 − 12×0 + 3 = 7 \\ d \| = 5 + 1 = 6 \end{align*}
The date is July 6th, 2003.
Which Julian date corresponds to CJDN 2451893? Then $$J = 2451893$$ so
\begin{align*} y_2 \| = 2451893 − 1721118 = 730775 \\ \{x_2, r_2\} \| = \Div(4×730775, 1461) = \Div(2923103, 1461) = \{2000, 1103\} \\ z_2 \| = \floorratio{1103}{4} = 275 \\ y_1 \| = z_2 = 275 \\ \{x_1, r_1\} \| = \Div(5×275 + 2, 153) = \Div(1377, 153) = \{9, 0\} \\ z_1 \| = \floorratio{0}{4} = 0 \\ c_0 \| = \floorratio{9 + 2}{12} = \floorratio{11}{12} = 0 \\ j \| = 2000 + 0 = 2000 \\ m \| = 9 − 12×0 + 3 = 12 \\ d \| = 0 + 1 = 1 \end{align*}
The date is 1 December 2000.
And which Julian date corresponds to CJDN 173? Then $$J = 173$$ and
\begin{align*} y_2 \| = 173 − 1721118 = −1720945 \\ \{x_2, r_2\} \| = \Div(4×−1720945, 1461) = \Div(−6883777, 1461) = \{−4712, 455\} \\ z_2 \| = \floorratio{455}{4} = 113 \\ y_1 \| = 113 \\ \{x_1, r_1\} \| = \Div(5×113 + 2, 153) = \Div(567, 153) = \{3, 108\} \\ z_1 \| = \floorratio{108}{5} = 21 \\ c_0 \| = \floorratio{3 + 2}{12} = 0 \\ j \| = −4712 + 0 = −4712 \\ m \| = 3 − 12×0 + 3 = 6 \\ d \| = 21 + 1 = 22 \end{align*}
The date is 22 June −4712.
We can combine and compress these a bit, to
\begin{align} y_2 \| = J − 1721118 \\ k_2 \| = 4y_2 + 3 \\ k_1 \| = 5\floorratio{k_2 \bmod 1461}{4} + 2 \\ x_1 \| = \floorratio{k_1}{153} \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = \floorratio{k_2}{1461} + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = \floorratio{k_1 \bmod 153}{5} + 1 \end{align}
or, with use of $$\Div()$$,
\begin{align} \{x_2, r_2\} \| = \Div(4J − 6884469, 1461) \\ \{x_1, r_1\} \| = \Div\left( 5\floorratio{r_2}{4} + 2, 153 \right) \\ d \| = \floorratio{r_1}{5} + 1 \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \end{align}
Which Julian date corresponds to CJDN 2451893? Then $$J = 2451893$$ so
\begin{align*} y_2 \| = 2451893 − 1721118 = 730775 \\ k_2 \| = 4×730775 + 3 = 2923103 \\ k_1 \| = 5\floorratio{2923103 \bmod 1461}{4} + 2 = 5\floorratio{1103}{4} + 2 = 5×275 + 2 = 1377 \\ x_1 \| = \floorratio{1377}{153} = 9 \\ c_0 \| = \floorratio{9 + 2}{12} = \floorratio{11}{12} = 0 \\ j \| = \floorratio{2923103}{1461} + 0 = 2000 + 0 = 2000 \\ m \| = 9 − 12×0 + 3 = 12 \\ d \| = \floorratio{(1377 \bmod 153}{5} + 1 = \floorratio{0}{5} + 1 = 1 \end{align*}
or
\begin{align} \{x_2, r_2\} \| = \Div(4×2451893 − 6884469, 1461) = \Div(2923103, 1461) = \{2000, 1103\} \\ \{x_1, r_1\} \| = \Div\left( 5\floorratio{1103}{4} + 2, 153 \right) = \Div(5×275 + 2, 153) = \Div(1377, 153) = \{9, 0\} \\ d \| = \floorratio{0}{5} + 1 = 1 \\ c_0 \| = \floorratio{9 + 2}{12} = \floorratio{11}{12} = 0 \\ j \| = 2000 + 0 = 2000 \\ m \| = 9 − 12×0 + 3 = 12 \end{align}
The date is 1 December 2000.
#### 13.1.3. With the Standard Algorithms
With the standard algorithms for calendar levels we can describe this conversion as follows:
$${q}$$ $${g}$$ $${d}$$ $${h}$$ $${s}$$ $${x}$$ $${z}$$
⇒
$${y}$$ description
1 1 1 0 −1 $${m}$$ 0
⇒
$${y_1}$$ month index (0…11)
12 1 1 0 0 $${j}$$ $${y_1}$$
⇒
$${y_2}$$ running month index
1 1 1 0 −2 $${y_2}$$ 0
⇒
$${y_3}$$ running calculation month index
12 1 1 0 0 $${y_4}$$ $${z_4}$$
⇐
$${y_3}$$ calculation year, calculation month
1 1 1 0 −1 $${d}$$ 0
⇒
$${y_5}$$ calculation day of the month
30 5 1 3 2 $${z_4}$$ $${y_5}$$
⇒
$${y_6}$$ running day number in the year
365 4 1 1 0 $${y_4}$$ $${y_6}$$
⇒
$${y_7}$$ running day number
1 1 1 0 1721118 $${y_7}$$ 0
⇒
$${J}$$ CJDN
If you transform from date to CJDN then you go from top to bottom and follow the arrows, and if you transform frm CJDN to date then you go from bottom to top and against the arrows.
For $$j = 2003, m = 7, d = 6$$ we find
$${q}$$ $${g}$$ $${d}$$ $${h}$$ $${s}$$ $${x}$$ $${z}$$
↔
$${y}$$ description
1 1 1 0 −1 7 0
→
$${y_1}$$ 6 month index (0…11)
12 1 1 0 0 2003 6
→
$${y_2}$$ 24042 running month index
1 1 1 0 −2 24042 0
→
$${y_3}$$ 24040 running calculation month index
12 1 1 0 0 $${y_4}$$ 2003 $${z_4}$$ 4
←
24040 calculation year, calculation month
1 1 1 0 −1 6 0
→
$${y_5}$$ 5 calculation day of the month
30 5 1 3 2 4 5
→
$${y_6}$$ 127 running day number in the year
365 4 1 1 0 2003 127
→
$${y_7}$$ 731722 running day number
1 1 1 0 1721118 731722 0
→
$${J}$$ 2452840 CJDN
#### 13.1.4. From Julian Date to CJDN (2)
We can also use a stepped combination of calendars; then we need a calendar level between a running month number $$x$$ and a running day number $$y$$, taking into account leap years. In that case there is no longer a higher calendar level that will shorten a too-long last month, so the formulas must be correct also for very large running month numbers.
First we calculate the running month number $$y_1$$ from the year $$x_1$$ and the month number $$z_1$$ within the year. The formula is very simple:
$$y_1 = 12x_1 + z_1$$
At the second calendar level, we must calculate a running day number from the running month number. Of the 12 months of a year, 7 are long, with 31 days each, so we need a formula $$⌊7×(x_2 + a)/12⌋$$ (which repeats itself every 12 months), but what should be the value of $$a$$?
If $$a = 0$$, then $$⌊7x_2/12⌋$$ yields the values (for $$x_2$$ from 0 through 12): 0 0 1 1 2 2 3 4 4 5 5 6 7. Each time that the value increases by one, the preceding month was a long one, so these values mean that the sequence of short (S) and long (L) months was: S L S L S L L S L S L L, but we're searching for L S L S L S L L S L S L. We get that if we shift the pattern by −1 month (one to the left), so $$a = −1$$. Then $$s = fa \bmod g = −7 \bmod 12 = 5$$, so we find $$⌊(7x_2 + 5)/12⌋$$.
Then all months are good, except for February ($$x_2 = 1$$), because that month gets 30 days but should have only 28 days in a regular year or 29 days in a leap year. We subtract 2 days from every February using the formula $$−2⌊(x_2 + 10)/12⌋$$. We add 1 day back in for the first of every four years by using $$⌊(x_2 + 46)/48⌋$$. All in all we find
$$c_2 = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48}$$
This is calendar type 4, but for calculating in the opposite direction it is better if all fractions use the same numerator. The least common multiple of the numerators 12, 12, and 48 is 48, so we use $$g = 48$$. Then
$$c_2 = 30x_2 + \floorratio{28x_2 + 20}{48} − 2\floorratio{4x_2 + 40}{48} + \floorratio{x_2 + 46}{48}$$
which means calendar type 4 with $$q = 30$$, $$g = 48$$, $$\{d_1,h_1,s_1\} = \{1,28,20\}$$, $$\{d_2,h_2,s_2\} = \{−2,4,40\}$$, $$\{d_3,h_3,s_3\} = \{1,1,46\}$$.
Then the procedure is
1. Translate the year $$j$$, month $$m$$, day $$d$$ to calculation year $$x_1$$, calculation month $$z_1$$, and calculation day $$z_2$$. The number of the first calculation year/month/day must be 0, otherwise the calculations don't give the right results.
\begin{align} x_1 \| = j \\ z_1 \| = m − 1 \\ z_2 \| = d − 1 \end{align}
The CJDN $$J_0$$ that corresponds to $$\{x_1,z_1,z_2\} = \{0,0,0\}$$, i.e., 1 January of the year 0 (in the Julian calendar) is 1721058.
2. Combine the calendar levels in the stepped fashion, i.e., $$x_2 = y_1$$:
\begin{align} c_1 \| = 12x_1 \\ y_1 \| = c_1 + z_1 \\ x_2 \| = y_1 \\ c_2 \| = 30x_2 + \floorratio{28x_2 + 20}{48} − 2\floorratio{4x_2 + 40}{48} + \floorratio{x_2 + 46}{48} \\ y_2 \| = c_2 + z_2 \end{align}
3. Add the CJDN of running day number 0:
$$J = y_2 + J_0 = y_2 + 1721058$$
For example, what CJDN corresponds to Julian date July 6th, 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$, so $$x_1 = 2003$$, $$z_1 = 6$$, $$z_2 = 5$$. Then we find
\begin{align*} c_1 \| = 12×2003 = 24036 \\ y_1 \| = 24036 + 6 = 24042 \\ x_2 \| = 24042 \\ c_2 \| = 30×24042 + \floorratio{28×24042 + 20}{48} − 2×\floorratio{4×24042 + 40}{48} + \floorratio{24042 + 46}{48} \\ \| = 721260 + \floorratio{673196}{48} − 2×\floorratio{96208}{48} + \floorratio{24088}{48} \\ \| = 721260 + 14024 − 2×2004 + 501 = 731777 \\ y_2 \| = 731777 + 5 = 731782 \\ J \| = 731782 + 1721058 = 2452840 \end{align*}
We can combine the above formulas to
\begin{align} c \| = 12j + m \\ J \| = 30c + \floorratio{7c − 2}{12} − 2\floorratio{c + 9}{12} + \floorratio{c + 45}{48} + d + 1721027 \end{align}
For example, what CJDN corresponds to Julian date July 6th, 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$, so
\begin{align*} c \| = 12×2003 + 7 = 24043 \\ J \| = 30×24043 + \floorratio{7×24043 − 2}{12} − 2\floorratio{24043 + 9}{12} + \floorratio{24043 + 45}{48} + 6 + 1721027 \\ \| = 721290 + \floorratio{168299}{12} − 2\floorratio{24052}{12} + \floorratio{24088}{48} + 6 + 1721027 \\ \| = 721290 + 14024 − 2×2004 + 501 + 6 + 1721027 = 2452840 \end{align*}
#### 13.1.5. From Julian Day Number to Julian Date (2)
We run through the above procedure in the opposite direction.
1. Calculate the running day number $$y_2$$ by subtracting the CJDN of running day number 0 from the Julian Day number $$J$$:
$$y_2 = J − J_0 = J − 1721058$$
2. With $$q = 30$$, $$g = 48$$, $$\{d_1,h_1,s_1\} = \{1,28,20\}$$, $$\{d_2,h_2,s_2\} = \{−2,4,40\}$$, $$\{d_3,h_3,s_3\} = \{1,1,46\}$$ we have $$f = qg + ∑d_{i}h_i = 30×48 + 1×28 − 2×4 + 1×1 = 1461$$, so the average month length is $$f/g = 1461/48 = 487/16 = 30 + 7/16$$ days. Also $$∑d_{i}s_i = 1×20 − 2×40 + 1×46 = −14$$ and $$∑_{\gt0}d_i = 2$$ and so
$$g∑_{\gt0}d_i − ∑_i d_i s_i − 1 = 48×2 − (−14) − 1 = 109$$
Translate to calculation year $$x_1$$, calculation month $$z_1$$, and calculation day $$z_2$$:
\begin{align} x_2 \| = c'_2(y_2) = \floorratio{48y_2 + 109}{1461} \\ \| \begin{split} c_2 \| = 30x_2 + \floorratio{28x_2 + 20}{48} − 2\floorratio{4x_2 + 40}{48} + \floorratio{x_2 + 46}{48} \\ \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \end{split} \\ z_2 \| = y_2 − c_2 \\ ζ \| = \floorratio{z_2}{32} \\ x_2 \| ↤ x_2 + ζ \\ \| \begin{split} c_2 \| ↤ 30x_2 + \floorratio{28x_2 + 20}{48} − 2\floorratio{4x_2 + 40}{48} + \floorratio{x_2 + 46}{48} \\ \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \end{split} \\ z_2 \| ↤ y_2 − c_2 \\ y_1 \| = x_2 \\ x_1 \| = c'_1(y_1) = \floorratio{y_1}{12} \\ c_1 \| = 12x_1 \\ z_1 \| = y_1 − c_1 \end{align}
3. Then translate to calendar year $$j$$, calendar month $$m$$, and calendar day $$d$$:
\begin{align} j \| = x_1 \\ m \| = z_1 + 1 \\ d \| = z_2 + 1 \end{align}
For example, what Julian Date corresponds to CJDN 2452840? Then $$J = 2452840$$, so
\begin{align*} y_2 \| = 2452840 − 1721058 = 731782 \\ x_2 \| = \floorratio{48×731782 + 81}{1461} = \floorratio{35125617}{1461} = 24042 \\ c_2 \| = 30×24042 + \floorratio{28×24042 + 20}{48} − 2×\floorratio{4×24042 + 40}{48} + \floorratio{24042 + 46}{48} \\ \| = 721260 + \floorratio{673196}{48} − 2×\floorratio{96208}{48} + \floorratio{24088}{48} \\ \| = 721260 + 14024 − 4008 + 501 = 731777 \\ z_2 \| = 731782 − 731777 = 5 \\ ζ \| = \floorratio{5}{32} = 0 \end{align*}
Because $$ζ = 0$$, the $$x_2$$, $$c_2$$, $$z_2$$ remain unchanged.
\begin{align*} y_1 \| = 24042 \\ x_1 \| = \floorratio{24042}{12} = 2003 \\ c_1 \| = 12×2003 = 24036 \\ z_1 \| = 24042 − 24036 = 6 \\ j \| = 2003 \\ m \| = 6 + 1 = 7 \\ d \| = 5 + 1 = 6 \end{align*}
so the date is July 6th, 2003.
### 13.2. The GregorianCalendar
#### 13.2.1. From Gregorian Date to CJDN (1)
The Gregorian calendar is equal to the Julian calendar, except for the rules that say when the month of February has 29 days instead of 28 days. In the Julian calendar, February has 29 days in every fourth year, when the year number is evenly divisible by 4. In the Gregorian calendar the same rule holds, except that February has 28 days when the year number is evenly divisible by 100, except if the year number is evenly divisible by 400, when February has 29 days after all.
To approximate Gregorian years with a straight line with a single pattern, the leap values have to return after every $$⌊Q⌋$$ or $$⌈Q⌉$$ periods (for a well-chosen $$Q$$; see equation \eqref{eq:Q}). Unfortunately, the leap year rules of the Gregorian Calendar do not meet that demand, because they mean that the time between two leap years is sometimes 4 years and sometimes 8 years.
The following $$p$$ might be considered:
• 146097 days in 400 years → $$p = f/g = 146097/400$$. This yields 85 sequences of a leap year every 4 years, but also 12 sequences of a leap year every 5 years. That does not fit the Gregorian Calendar, so this straight line is not useful for calculating the Gregorian Calendar.
• 146097 days in 100 periods of 4 years → $$p = f/g = 146097/100$$. This yields two sequences of 33 periods (of 4 years) with one fewer leap day at the end (1460 days instead of 1461 days), and one sequence of 34 periods. The Gregorian Calendar has two sequences of 25 periods (of 4 years) with one fewer leap day at the end, and then a sequence of 50 periods with one fewer leap day at the end. This one is not suitable, either.
• 36524 days in 100 years → $$p = f/g = 36524/100$$. This yields 20 sequences of a leap year every 4 years, but also 4 sequences of a leap year every 5 years. This one is not suitable.
• 146097 days in 4 periods of 100 years → $$p = f/g = 146097/4$$. This yields 3 sequences of 36524 days and 1 sequence of 36525 days, and that fits the Gregorian Calendar.
• 36525 days in 100 years → $$p = f/g = 36525/100$$. This yields 25 sequences with a leap year every 4 years. That fits, except that we sometimes need to stop already after 36524 days.
We can now use the 146097/4 line to find the 100-year periods, and then use the 36525/100 line to find the years within the 100-year period, and finally (just like for the Julian Calendar) use the 153/5 line to find the months within the year. 1 March of year 0 in the Gregorian Calendar corresponds to $$J_0 = 1721120$$. We then use three calendar levels, with the following values:
$${i}$$ $${f_i}$$ $${g_i}$$ $${a_i}$$ $${s_i}$$
1 153 5 4 2
2 36525 100 0 0
3 146097 4 0 0
The steps to get from the Gregorian calendar to the CJDN are now:
1. Calculate calculation century number $$x_3$$, calculation year number $$x_2$$, calculation month number $$x_1$$, and calculation day number $$z_1$$ from the year number $$j$$, month number $$m$$, and day number $$d$$
\begin{align} c_0 \| = \floorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ \{x_3, x_2\} \| = \Div(x_4, 100) \\ x_1 \| = m − 12c_0 − 3 \\ z_1 \| = d − 1 \end{align}
2. We combine the two calendar levels in the flat manner, so $$z_2 = y_1$$ and $$z_3 = y_2$$:
\begin{align} c_1 \| = \floorratio{153x_1 + 2}{5} \\ y_1 \| = c_1 + z_1 \\ z_2 \| = y_1 \\ c_2 \| = \floorratio{36525x_2}{100} \\ y_2 \| = c_2 + z_2 \\ z_3 \| = y_2 \\ c_3 \| = \floorratio{146097x_3}{4} \\ y_3 \| = c_3 + z_3 \end{align}
3. Add the CJDN of running day number 0:
$$J = y_3 + J_0 = y_3 + 1721120$$
Combined and condensed, this yields:
\begin{align} c_0 \| = \floorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ \{x_3, x_2\} \| = \Div(x_4, 100) \\ x_1 \| = m − 12c_0 − 3 \\ J \| = \floorratio{146097x_3}{4} + \floorratio{36525x_2}{100} + \floorratio{153x_1 + 2}{5} + d + 1721119 \label{eq:gregtocjdn1} \end{align}
For example, which Julian Day number corresponds to Gregorian date July 6th, 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$ and then
\begin{align*} c_0 \| = \floorratio{7 − 3}{12} = 0 \\ x_4 \| = 2003 + 0 = 2003 \\ \{x_3, x_2\} \| = \Div(2003, 100) = \{20, 3\} \\ x_1 \| = 7 − 12×0 − 3 = 4 \\ J \| = \floorratio{146097×20}{4} + \floorratio{36525×3}{100} + \floorratio{153×4 + 2}{5} + 6 + 1721119 \\ \| = 730485 + 1095 + 122 + 6 + 1721119 = 2452827 \end{align*}
#### 13.2.2. From CJDN to Gregorian Date (1)
Now we go in the opposite direction again, with the same three calendar levels as before.
First we subtract the CJDN of running day number 0 from the CJDN to get the running day number:
$$y_3 = J − J_0 = J − 1721120$$
Then we translate the running day number in the flat manner into a calculation century, calculation year, calculation month, and calculation day:
\begin{align} \{x_3, r_3\} \| = \Div(4y_3 + 3, 146097) \\ z_3 \| = \floorratio{r_3}{4} \\ y_2 \| = z_3 \\ \{x_2, r_2\} \| = \Div(100y_2 + 99, 36525) \\ z_2 \| = \floorratio{r_2}{100} \\ y_1 \| = z_2 \\ \{x_1, r_1\} \| = \Div(5y_1 + 2, 153) \\ z_1 \| = \floorratio{r_1}{5} \end{align}
And lastly we translate to day-month-year:
\begin{align} c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = z_1 + 1 \end{align}
For example, which date in the Gregorian calendar corresponds to CJDN 2452827? Then $$J = 2452827$$ and then
\begin{align*} y_3 \| = J − 1721120 = 731707 \\ \{x_3, r_3\} \| = \Div(4×731707, 146097) = \Div(2926831, 146097) = \{20, 4891\} \\ z_3 \| = \floorratio{4891}{4} = 1222 \\ y_2 \| = 1222 \\ \{x_2, r_2\} \| = \Div(100×1222 + 99, 36525) = \Div(122299, 36525) = \{3, 12724\} \\ z_2 \| = \floorratio{12724}{100} = 127 \\ y_1 \| = 127 \\ \{x_1, r_1\} \| = \Div(5×127 + 2, 153) = \Div(637, 153) = \{4, 25\} \\ z_1 \| = \floorratio{25}{5} = 5 \\ c_0 \| = \floorratio{4 + 2}{12} = 0 \\ j \| = 100×20 + 3 + 0 = 2003 \\ m \| = 4 − 12×0 + 3 = 7 \\ d \| = 5 + 1 = 6 \end{align*}
The date is July 6th, 2003.
This condenses a little bit, to:
\begin{align} \{x_3, r_3\} \| = \Div(4J − 6884477, 146097) \\ \{x_2, r_2\} \| = \Div\left( 100\floorratio{r_3}{4} + 99, 36525 \right) \\ \{x_1, r_1\} \| = \Div\left( 5\floorratio{r_2}{100} + 2, 153 \right) \\ d \| = \floorratio{r_1}{5} + 1 \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \end{align}
For example, which date in the Gregorian calendar corresponds to CJDN 2452827? Then $$J = 2452827$$ and then
\begin{align*} \{x_3, r_3\} \| = \Div(4×2452827 − 6884477, 146097) = \Div(2926831, 146097) = \{20, 4891\} \\ \{x_2, r_2\} \| = \Div\left( 100\floorratio{4891}{4} + 99, 36525 \right) = \Div(100×1222 + 99, 36525) = \Div(122299, 36525) = \{3, 12724\} \\ \{x_1, r_1\} \| = \Div\left( 5\floorratio{12724}{100} + 2, 153 \right) = \Div(5×127 + 2, 153) = \Div(637, 153) = \{4, 25\} \\ d \| = \floorratio{25}{5} + 1 = 5 + 1 = 6 \\ c_0 \| = \floorratio{4 + 2}{12} = 0 \\ j \| = 100×20 + 3 + 0 = 2003 \\ m \| = 4 − 12×0 + 3 = 7 \end{align*}
The date is July 6th, 2003.
#### 13.2.3. From Gregorian Date to CJDN (2)
We can make do with just two calendar levels combined in the flat manner if we use calendar type 4 for the upper level, with multiple month lengths. Then, for that upper level,
\begin{align*} q \| = 365 \\ g \| = 400 \\ \{d_1,h_1,s_1\} \| = \{1,100,0\} \\ \{d_2,h_2,s_2\} \| = \{−1,4,0\} \\ \{d_3,h_3,s_3\} \| = \{1,1,0\} \end{align*}
The lower calendar level is the same as before, so
\begin{align} c_3 \| = \floorratio{m − 3}{12} \\ x_2 \| = j + c_3 \\ x_1 \| = m − 12c_3 − 3 \\ z_1 \| = d − 1 \\ c_1 \| = \floorratio{153x_1 + 2}{5} \\ y_1 \| = c_1 + z_1 \\ z_2 \| = y_1 \\ \| \begin{split} c_2 \| = 365x_2 + \floorratio{100x_2}{400} − \floorratio{4x_2}{400} + \floorratio{x_2}{400} \\ \| = 365x_2 + \floorratio{x_2}{4} − \floorratio{x_2}{100} + \floorratio{x_2}{400} \end{split} \\ y_2 \| = c_2 + z_2 \\ J \| = y_2 + J_0 = y_2 + 1721120 \end{align}
For example, which CJDN corresponds to Gregorian date July 6th, 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$, so
\begin{align*} c_3 \| = \floorratio{7 − 3}{12} = 0 \\ x_1 \| = 7 − 12×0 − 3 = 4 \\ x_2 \| = 2003 + 0 = 2003 \\ z_1 \| = 6 − 1 = 5 \\ c_1 \| = \floorratio{153×4 + 2}{5} = \floorratio{614}{5} = 122 \\ y_1 \| = 122 + 5 = 127 \\ z_2 \| = 127 \\ c_2 \| = 365×2003 + \floorratio{100×2003}{400} − \floorratio{4×2003}{400} + \floorratio{2003}{400} \\ \| = 731095 + \floorratio{200300}{400} − \floorratio{8012}{400} + 5 \\ \| = 731095 + 500 − 20 + 5 = 731580 \\ y_2 \| = 731580 + 127 = 731707 \\ J \| = 731707 + 1721120 = 2452827 \end{align*}
This condenses to:
\begin{align} c_3 \| = \floorratio{m − 3}{12} \\ x_2 \| = j + c_3 \\ x_1 \| = m − 12c_3 − 3 \\ J \| = 365x_2 + \floorratio{x_2}{4} − \floorratio{x_2}{100} + \floorratio{x_2}{400} + \floorratio{153x_1 + 2}{5} + d + 1721119 \end{align}
#### 13.2.4. From CJDN to Gregorian Date (2)
With the upper calendar of type 4, we go in the opposite direction, again with
$$q = 365$$, $$g = 400$$, $$\{d_1,h_1,s_1\} = \{1,100,0\}$$, $$\{d_2,h_2,s_2\} = \{−1,4,0\}$$, $$\{d_3,h_3,s_3\} = \{1,1,0\}$$
Then
$$f = qg + ∑d_{i}h_i = 365×400 + 1×100 − 1×4 + 1×1 = 146097$$
so the average length of the year is $$f/g = 146097/400 = 365 \frac{97}{400}$$ days.
First we calculate the running day number from the CJDN, by subtracting the CJDN of running day number 0:
$$y_2 = J − J_0 = J − 1721120$$
Then we calculate the calculation year, calculation month, and calculation day from the running day number, in the flat manner:
\begin{align} x_2 \| = c'_2(y_2) = \floorratio{400y_2 + 799}{146097} \\ \| \begin{split} c_2 \| = 365x_2 + \floorratio{100x_2}{400} − \floorratio{4x_2}{400} + \floorratio{x_2}{400} \\ \| = 365x_2 + \floorratio{x_2}{4} − \floorratio{x_2}{100} + \floorratio{x_2}{400} \end{split} \\ z_2 \| = y_2 − c_2 \\ ζ \| = \floorratio{z_2}{367} \\ x_2 \| ↤ x_2 + ζ \\ \| \begin{split} c_2 \| ↤ 365x_2 + \floorratio{100x_2}{400} − \floorratio{4x_2}{400} + \floorratio{x_2}{400} \\ \| = 365x_2 + \floorratio{x_2}{4} − \floorratio{x_2}{100} + \floorratio{x_2}{400} \end{split} \\ z_2 \| ↤ y_2 − c_2 \\ y_1 \| = z_2 \\ x_1 \| = c'_1(y_1) = \floorratio{5y_1 + 2}{153} \\ c_1 \| = \floorratio{153x_1 + 2}{5} \\ z_1 \| = y_1 − c_1 \end{align}
And lastly we translate these calculation values to calendar values:
\begin{align} c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = z_1 + 1 \end{align}
For example, which date in the Gregorian calendar corresponds to CJDN 2452827? Then $$J = 2452827$$ and then
\begin{align*} y_2 \| = J − 1721120 = 731707 \\ x_2 \| = \floorratio{400×731707 + 799}{146097} = \floorratio{292683599}{146097} = 2003 \\ c_2 \| = 365×2003 + \floorratio{100×2003}{400} − \floorratio{4×2003}{400} + \floorratio{2003}{400} \\ \| = 731095 + \floorratio{200300}{400} − \floorratio{8012}{400} + 5 = 731095 + 500 − 20 + 5 = 731580 \\ z_2 \| = 731707 − 731580 = 127 \\ ζ \| = \floorratio{127}{367} = 0 \end{align*}
$$ζ = 0$$, so $$x_2$$, $$c_2$$, $$z_2$$ remain the same. Then
\begin{align*} y_1 \| = 127 \\ x_1 \| = \floorratio{5×127 + 2}{153} = \floorratio{637}{153} = 4 \\ c_1 \| = \floorratio{153×4 + 2}{5} = \floorratio{614}{5} = 122 \\ z_1 \| = 127 − 122 = 5 \\ c_0 \| = \floorratio{4 + 2}{12} = 0 \\ j \| = 2003 + 0 = 2003 \\ m \| = 4 − 12×0 + 3 = 7 \\ d \| = 5 + 1 = 6 \end{align*}
The date is July 6th, 2003.
The numerator of $$x_2$$ can get very large in this algorithm: In the example it was equal to 292,683,599 for a date in the year 2003. If you use this algorithm in a computer program that uses values of 32 bits large, then $$x_2$$ will get too large for dates more than 14,698 years from Julian Day number 0, i.e., from the year 9986. Section 12.2 explains how to find a solution for this. Such a solution is to replace formula
$x_2 = \floorratio{400y_2 + 799}{146097}$
with
\begin{align} q \| = \floorratio{y_2}{1461} \\ r \| = y_2 \bmod 1461 \\ x_2 \| = 4q + \floorratio{12q + 400r + 799}{146097} \end{align}
For example, when $$y_2 = 9,000,000$$, then
$x_2 = \floorratio{400×9000000 + 799}{146097} = \floorratio{3600000799}{146097} = 24641$
The greatest intermediate result is 3,600,000,799, which exceeds the greatest value (2,147,483,648) that fits a 32-bit value. With the alternative, we find
\begin{align*} q \| = \floorratio{9000000}{1461} = 6160 \\ r \| = 9000000 \bmod 1461 = 240 \\ x_2 \| = 4×6160 + \floorratio{12×6160 + 400×240 + 799}{146097} = 24640 + \floorratio{170719}{146097} = 24640 + 1 = 24641 \end{align*}
Now the greatest intermediate result was 9000000, which comfortably fits a 32-bit value.
#### 13.2.5. From Gregorian Date to CJDN (3)
Just like for the Julian calendar we can use a stepped combination of calendar levels. Compared to the Julian calendar we have to add the leap year rules for each 100 years and for each 400 years:
$$\begin{split} c_2 \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \\ \| − \floorratio{x_2 + 1198}{1200} + \floorratio{x_2 + 4798}{4800} \end{split}$$
Here, too, we want to have all denominators be equal; then we get
$$\begin{split} c_2 \| = 30x_2 + \floorratio{2800x_2 + 2000}{4800} − 2\floorratio{400x_2 + 4000}{12} \\ \| + \floorratio{100x_2 + 4600}{4800} − \floorratio{4x_2 + 4792}{4800} \\ \| + \floorratio{x_2 + 4798}{4800} \end{split}$$
which means calendar type 4 with $$q = 30$$, $$g = 4800$$, $$\{d_1,h_1,s_1\} = \{1,2800,2000\}$$, $$\{d_2,h_2,s_2\} = \{−2,400,4000\}$$, $$\{d_3,h_3,s_3\} = \{1,100,4600\}$$, $$\{d_4,h_4,s_4\} = \{−1,4,4792\}$$, $$\{d_5,h_5,s_5\} = \{1,1,4798\}$$.
The CJDN $$J_0$$ that corresponds to $$\{x_1,z_1,z_2\} = \{0,0,0\}$$, i.e., January 1st of year 0 (in the Gregorian calendar) is 1721060.
For the rest things are the same as for the Julian calendar. We then find
\begin{align} x_1 \| = j \\ z_1 \| = m − 1 \\ z_2 \| = d − 1 \\ c_1 \| = 12x_1 \\ y_1 \| = c_1 + z_1 \\ x_2 \| = y_1 \\ \| \begin{split} c_2 \| = 30x_2 + \floorratio{2800x_2 + 2000}{4800} − 2\floorratio{400x_2 + 4000}{4800} \\ \| + \floorratio{100x_2 + 4600}{4800} − \floorratio{4x_2 + 4792}{4800} + \floorratio{x_2 + 4798}{4800} \\ \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \\ \| − \floorratio{x_2 + 1198}{1200} + \floorratio{x_2 + 4798}{4800} \end{split} \\ y_2 \| = c_2 + z_2 \\ J \| = y_2 + J_0 = y_2 + 1721060 \end{align}
For example, which CJDN corresponds to Gregorian date July 6th, 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$, so $$x_1 = 2003$$, $$z_1 = 7 − 1 = 6$$, $$z_2 = 6 − 1 = 5$$, and then
\begin{align*} c_1 \| = 12×2003 = 24036 \\ y_1 \| = 24036 + 6 = 24042 \\ x_2 \| = 24042 \\ c_2 \| = 30×24042 + \floorratio{2800×24042 + 2000}{4800} − 2×\floorratio{400×24042 + 4000}{4800} \\ \| + \floorratio{100×24042 + 4600}{4800} − \floorratio{4×24042 + 4792}{4800} + \floorratio{24042 + 4798}{4800} \\ \| = 721260 + \floorratio{67319600}{4800} − 2×\floorratio{9620800}{4800} + \floorratio{2408800}{4800} − \floorratio{100960}{4800} + \floorratio{28840}{4800} \\ \| = 721260 + 14024 − 4008 + 501 − 21 + 6 = 731762 \\ y_2 \| = 731762 + 5 = 731767 \\ J \| = 731767 + 1721060 = 2452827 \end{align*}
We can combine the above formulas into
\begin{align} c \| = 12j + m \\ \| \begin{split} J \| = 30c + \floorratio{7c − 2}{12} − 2\floorratio{c + 9}{12} + \floorratio{c + 45}{48} \\ \| − \floorratio{c + 1197}{1200} + \floorratio{c + 4797}{4800} + d + 1721029 \end{split} \end{align}
#### 13.2.6. From CJDN to Gregorian Date (3)
We run through the same procedure as in the previous section, but in the opposite direction.
1. Calculate the running day number $$y_2$$ by subtracting the CJDN of running day number 0 from CJDN $$J$$.
$$y_2 = J − J_0 = J − 1721060$$
2. With $$q = 30$$, $$g = 4800$$, $$\{d_1,h_1,s_1\} = \{1,2800,2000\}$$, $$\{d_2,h_2,s_2\} = \{−2,400,4000\}$$, $$\{d_3,h_3,s_3\} = \{1,100,4600\}$$, $$\{d_4,h_4,s_4\} = \{−1,4,4792\}$$, $$\{d_5,h_5,s_5\} = \{1,1,4798\}$$ we have
$f = qg + ∑d_{i}h_i = 30×4800 + 1×2800 − 2×400 + 1×100 − 1×4 + 1×1 = 146097$
so the average month length is $$f/g = 146097/4800 = 30 + 699/1600$$ days. Also,
$∑d_{i}s_i = 1×2000 − 2×4000 + 1×4600 − 1×4792 + 1×4798 = −1394$
Translate to calculation year $$x_1$$, calculation month $$z_1$$, and calculation day $$z_2$$:
\begin{align} x_2 \| = c'_2(y_2) = \floorratio{4800y_2 + 15793}{146097} \label{eq:g3} \\ \| \begin{split} c_2 \| = 30x_2 + \floorratio{2800x_2 + 2000}{4800} − 2\floorratio{400x_2 + 4000}{4800} + \floorratio{100x_2 + 4600}{4800} \\ \| − \floorratio{4x_2 + 4792}{4800} + \floorratio{x_2 + 4798}{4800} \\ \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \\ \| − \floorratio{x_2 + 1198}{1200} + \floorratio{x_2 + 4798}{4800} \end{split} \\ z_2 \| = y_2 − c_2 \\ ζ \| = \floorratio{z_2}{33} \\ x_2 \| ↤ x_2 + ζ \\ \| \begin{split} c_2 \| ↤ 30x_2 + \floorratio{2800x_2 + 2000}{4800} − 2\floorratio{400x_2 + 4000}{4800} + \floorratio{100x_2 + 4600}{4800} \\ \| − \floorratio{4x_2 + 4792}{4800} + \floorratio{x_2 + 4798}{4800} \\ \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \\ \| − \floorratio{x_2 + 1198}{1200} + \floorratio{x_2 + 4798}{4800} \end{split} \\ z_2 \| ↤ y_2 − c_2 \\ y_1 \| = x_2 \\ x_1 \| = c'_1(y_1) = \floorratio{y_1}{12} \\ c_1 \| = 12x_1 \\ z_1 \| = y_1 − c_1 \end{align}
3. Then translate to year $$j$$, month $$m$$, and day $$d$$:
\begin{align} j \| = x_1 \\ m \| = z_1 + 1 \\ d \| = z_2 + 1 \end{align}
For example, which date in the Gregorian calendar corresponds to CJDN 2452827? Then $$J = 2452827$$ and then
\begin{align*} y_2 \| = J − 1721060 = 731767 \\ x_2 \| = \floorratio{4800×731767 + 15793}{46097} = \floorratio{3512497393}{146097} = 24042 \\ c_2 \| = 30×24042 + \floorratio{2800×24042 + 2000}{4800} − 2×\floorratio{400×24042 + 4000}{4800} \\ \| + \floorratio{100×24042 + 4600}{4800} − \floorratio{4×24042 + 4792}{4800} + \floorratio{24042 + 4798}{4800} \\ \| = 721260 + \floorratio{67319600}{4800} − 2×\floorratio{9620800}{4800} + \floorratio{2408800}{4800} − \floorratio{100960}{4800} + \floorratio{28840}{4800} \\ \| = 721260 + 14024 − 4008 + 501 − 21 + 6 = 731762 \\ z_2 \| = 731767 − 731762 = 5 \\ ζ \| = \floorratio{5}{33} = 0 \end{align*}
$$ζ = 0$$ so $$x_2$$, $$c_2$$, $$z_2$$ remain the same. Then
\begin{align*} y_1 \| = 24042 \\ x_1 \| = \floorratio{24042}{12} = 2003 \\ c_1 \| = 12×2003 = 24036 \\ z_1 \| = 24042 − 24036 = 6 \\ j \| = 2003 \\ m \| = 6 + 1 = 7 \\ d \| = 5 + 1 = 6 \end{align*}
The date is July 6th, 2003.
This can be condensed a bit to
\begin{align} y_2 \| = J − J_0 = J − 1721060 \\ x_2 \| = \floorratio{4800y_2 + 15793}{146097} \\ \| \begin{split} c_2 \| = 30x_2 + \floorratio{7x_2 + 5}{12} − 2\floorratio{x_2 + 10}{12} + \floorratio{x_2 + 46}{48} \\ \| − \floorratio{x_2 + 1198}{1200} + \floorratio{x_2 + 4798}{4800} \end{split} \\ z_2 \| = y_2 − c_2 \\ ζ \| = \floorratio{z_2}{33} \\ u \| = x_2 + ζ \\ \| \begin{split} d \| = y_2 − 30u − \floorratio{7u + 5}{12} + 2\floorratio{u + 10}{12} − \floorratio{u + 46}{48} \\ \| + \floorratio{u + 1198}{1200} − \floorratio{u + 4798}{4800} + 1 \end{split} \\ j \| = \floorratio{u}{12} \\ m \| = u − 12j + 1 \end{align}
In the calculation of $$x_2$$, the numerator can get very large. In a computer program that uses 32-bit values, the calculation of $$x_2$$ will yield wrong results when $$|y_2|$$ exceeds about 447,000 (days), which corresponds to only 1224 years. Section 12.2 explains how to find a solution for this. Such a solution is to replace the formula for $$x_2$$ with
\begin{align} q \| = \floorratio{y_2}{487} \\ r \| = y_2 \bmod 487 \\ x_2 \| = 16q + \floorratio{48q + 4800r + 15793}{146097} \end{align}
For example, if $$y_2 = 730,000$$, then $$x_2 = \floorratio{4800×730000 + 15793}{146097} = \floorratio{3504015793}{146097} = 23984$$, so then the greatest intermediate result is equal to 3,504,015,793, which is much greater than the greatest value (2,147,483,648) that fits in memory of 32 bits wide. With the alternative method, we find
\begin{align*} q \| = \floorratio{730000}{487} = 1498 \\ r \| = 730000 \bmod 487 = 474 \\ x_2 \| = 1498×16 + \floorratio{1498×48 + 4800×474 + 15793}{146097} = 23968 + \floorratio{2362897}{146097} = 23968 + 16 = 23984 \end{align*}
which is the same result as before, but now with the greatest intermediate result equal to 2,362,897, which is comfortably small enough to fit in 32 bits.
For the Gregorian calendar the formulas (for days and years) from the sections with "(2)" and "(3)" in the title aren't easier than the formulas from the section with "(1)" in the title, so we recommend the formulas from the sections with "(1)" in the title for general use for this calendar. In general, it is not always possible to find such formulas, so then it is nice to also have more complicated but also more powerful formulas available.
### 13.3. The Milanković Calendar
#### 13.3.1. From Milanković Date to CJDN
Some Eastern Orthodox Churches have for some time (since 1923) used a calendar invented by Milutin Milanković that only differs from the Gregorian Calendar in the rule about which century years are leap years. In the Gregorian Calendar those are all years whose number is not evenly divisible by 400. In the Milanković Calendar those are are years that when dividing by 900 leave a remainder equal to 200 or 600. The following table shows for the century years from 1500 through 2900 which of them are leap years in the Gregorian and Milanković Calendars.
Year Gregorian Milanković 1500 no yes 1600 yes no 1700 no no 1800 no no 1900 no no 2000 yes yes 2100 no no 2200 no no 2300 no no 2400 yes yes 2500 no no 2600 no no 2700 no no 2800 yes no 2900 no yes
Between the years 1601 and 2799, the Milanković- and Gregorian calendars give the same results. The years 1600 and 2800 are leap years in the Gregorian calendar but not in the Milanković calendar.
With these rules, the Milanković calendar has a period of 365×900 + 900/4 − (900/100)×(7/9) = 328718 days. The conversion of a date to CJDN is nearly the same for the Milanković calendar as for the Gregorian calendar (see Section 13.2.1):
\begin{align} c_0 \| = \floorratio{m − 3}{12} \\ x_4 \| = j + c_0 \\ x_3 \| = \floorratio{x_4}{100} \\ x_2 \| = x_4 \bmod 100 \\ x_1 \| = m − 12c_0 − 3 \\ J \| = \floorratio{328718x_3 + 6}{9} + \floorratio{36525x_2}{100} + \floorratio{153x_1 + 2}{5} + d + 1721119 \end{align}
For example, which CJDN corresponds to Milanković date July 6th, 2003? Then $$j = 2003$$, $$m = 7$$, $$d = 6$$ hence
\begin{align*} c_0 \| = \floorratio{7 − 3}{12} = 0 \\ x_4 \| = 2003 + 0 = 2003 \\ x_3 \| = \floorratio{2003}{100} = 20 \\ x_2 \| = 2003 − 100×20 = 3 \\ x_1 \| = 7 − 12×0 − 3 = 4 \\ z_1 \| = 6 − 1 = 5 \\ c_1 \| = \floorratio{153×4 + 2}{5} = \floorratio{614}{5} = 122 \\ y_1 \| = 122 + 5 = 127 \\ z_2 \| = 127 \\ c_2 \| = \floorratio{36525×3}{100} = \floorratio{109575}{100} = 1095 \\ y_2 \| = 1095 + 127 = 1222 \\ z_3 \| = 1222 \\ c_3 \| = \floorratio{328718×20 + 6}{9} = \floorratio{6574366}{9} = 730485 \\ y_3 \| = 730485 + 1222 = 731707 \\ J \| = 731707 + 1721120 = 2452827 \end{align*}
#### 13.3.2. From CJDN to Milanković Date
Also in the opposite direction, we do the same as for the Gregorian calendar (see Section 13.2.2), except that we use different formulas to calculate $$x_3$$ and $$k_3$$:
\begin{align} \{x_3, r_3\} \| = \Div\left( 9×(J − 1721120) + 2, 328718 \right) \\ \{x_2, r_2\} \| = \Div(100r_3 + 99, 36525) \\ \{x_1, r_1\} \| = \Div(5r_2 + 2, 153) \\ c_0 \| = \floorratio{x_1 + 2}{12} \\ j \| = 100x_3 + x_2 + c_0 \\ m \| = x_1 − 12c_0 + 3 \\ d \| = \floorratio{k_1 \bmod 153}{5} + 1 \end{align}
For example, which date in the Gregorian calendar corresponds to CJDN 2452827? Then $$J = 2452827$$ and then
\begin{align*} \{x_3, r_3\} \| = \Div( 9×(2452827 − 1721120) + 2, 328718 ) \\ \| = \Div(6585365, 328718) = \{20, 11005\} \\ \{x_2, r_2\} \| = \Div\left( 100\floorratio{11005}{9} + 99, 36525 \right) \\ \| = \Div(100×1222 + 99, 36525) \\ \| = \Div(122299, 36525) = \{3, 12724\} \\ \{x_1, r_1\} \| = \Div\left( 5\floorratio{12724}{100} + 2, 153 \right) \\ \| = \Div(5×127 + 2, 153) \\ \| = \Div(637, 153) = \{4, 25\} \\ d \| = \floorratio{25}{5} + 1 = 6 \\ c_0 \| = \floorratio{4 + 2}{12} = 0 \\ j \| = 100×20 + 3 + 0 = 2003 \\ m \| = 4 − 12×0 + 3 = 7 \end{align*}
The date is July 6th, 2003.
### 13.4. The Egyptian Calendar
#### 13.4.1. From Egyptian Date to CJDN
The ancient Egyptians had a very simple calendar indeed, without leap years and with 30 days to every month, except that the last month had 5 days. In the calendar according to the Era of Nabonassar the first day (1 Thoth of year 1) was equivalent to 26 February −746 in the Julian calendar, i.e., to JD 1448638 ($$= J_0$$).
At the higher calendar level, each year has 365 days. At the lower level, each month has 30 days. The last month of the year actually has only 5 days, but we get that for free from the higher calendar level. We combine the two levels in the flat way, with $$z_2 = y_1$$.
For day $$d$$ of month $$m$$ of year $$j$$ we then find
\begin{align} x_2 \| = j − 1 \\ x_1 \| = m − 1 \\ z_1 \| = d − 1 \\ c_1 \| = 30x_1 \\ y_1 \| = c_1 + z_1 \\ z_2 \| = y_1 \\ c_2 \| = 365x_2 \\ y_2 \| = c_2 + z_2 \\ J \| = y_2 + J_0 \end{align}
This condenses to
$$J = 365×(j − 1) + 30×(m − 1) + d − 1 + J_0 = 365j + 30m + d + 1448242$$
For example, which CJDN corresponds to day 7 of month 5 of year 218 in the Egyptian calendar? Then we find
\begin{align*} x_2 \| = j − 1 = 218 − 1 = 217 \\ x_1 \| = m − 1 = 5 − 1 = 4 \\ z_1 \| = d − 1 = 7 − 1 = 6 \\ c_1 \| = 30x_1 = 30×4 = 120 \\ z_2 \| = y_1 = c_1 + z_1 = 120 + 6 = 126 \\ c_2 \| = 365x_2 = 365×217 = 79205 \\ y_2 \| = c_2 + z_2 = 79205 + 126 = 79331 \\ J \| = y_2 + J_0 = 79331 + 1448638 = 1527969 \end{align*}
so the answer is CJDN 1527969.
With the condensed formula we find
\begin{align*} J \| = 365j + 30m + d + 1448242 \\ \| = 365×218 + 30×5 + 7 + 1448242 \\ \| = 79570 + 150 + 7 + 1448242 = 1527969 \end{align*}
which is the same answer as before.
#### 13.4.2. From CJDN to Egyptian Date
In the opposite direction, the calculations are simple as well.
\begin{align} y_2 \| = J − J_0 \\ \{x_2, r_2\} \| = \Div(y_2, 365) \\ \{x_1, r_1\} \| = \Div(r_2, 30) \\ j \| = x_2 + 1 \\ m \| = x_1 + 1 \\ d \| = r_1 + 1 \end{align}
What date in the Egyptian calendar corresponds to CJDN 1527969? Then
\begin{align*} y_2 \| = J − J_0 = 1527969 − 1448638 = 79331 \\ \{x_2, r_2\} \| = \Div(79331, 365) = \{217, 126\} \\ \{x_1, r_1\} \| = \Div(126, 30) = \{4, 6\} \\ j \| = x_2 + 1 = 217 + 1 = 218 \\ m \| = x_1 + 1 = 4 + 1 = 5 \\ d \| = r_1 + 1 = 6 + 1 = 7 \end{align*}
so it is day 7 of month 5 of year 218.
### 13.5. The Babylonian Calendar
#### 13.5.1. From Babylonian Date to Julian Day Number
The Metonic Cycle says that 235 (synodical) months are equal to 19 (tropical) years of 12 or 13 months each, with 125 months of 30 days and 110 months of 29 days, so 6940 days in total. The long years (with 13 months) are the 1st, 4th, 7th, 9th, 12th, 15th, and 18th year of each cycle. In the 18th year, month 6 is doubled; in the other long years month 12 is doubled. Day 1 of month 1 (Nisannu) of year 1 of the era of Seleukos corresponds to 3 April −310 in the Julian Calendar, or CJDN 1607558.
The Babylonians determined the beginning of each month by looking for a particular phase of the Moon. They did not give the first month of each year the same number of days, and likewise for the other months, but did vary the number of months in each year according to the method described above. The distribution of months into years ran independently from the distribution of days into months.
If a calendar is (partly) based on direct observations, then it is a bit unpredictable and cannot be completely caught in formulas. The calendar then depends on things like the weather. If there happen to be many clouds in the sky one day, then the calendar officials may not see the moon at all that night, and then the official start of the new month may be delayed until after they see the moon the next night.
We derive a calendar that looks like the one of the Babylonians, but is entirely predictable. The difference with the historical calendar of the Babylonians should be at most 1 day.
That means a calendar with a stepped combination of a calendar level between day and month and a calendar level between month and year. We go from year number $$y$$ (the first year has number 1), month number $$m$$ (the first month has number 1), and day number $$d$$ (the first day has number 1) to running day number $$y_2$$ as follows:
\begin{align} x_1 \| = j − 1 \\ z_1 \| = m − 1 \\ z_2 \| = d − 1 \\ c_1 \| = \floorratio{235x_1 + 13}{19} \\ y_1 \| = c_1 + z_1 \\ x_2 \| = y_1 \\ c_2 \| = \floorratio{6940x_2}{235} \\ y_2 \| = c_2 + z_2 \\ J \| = y_2 + 1607558 \end{align}
This can be compressed to
\begin{align} y_1 \| = \floorratio{235j − 241}{19} + m \\ J \| = \floorratio{6940y_1}{235} + d + 1607557 \end{align}
The number of years since the beginning of the current Metonic cycle is equal to $$y_1 \bmod 19$$.
For example, which CJDN $$J$$ corresponds to year 3, month 9, day 27 of the era of Seleukos? Then $$j = 3$$, $$m = 9$$, $$d = 27$$, so
\begin{align*} y_1 \| = \floorratio{235×3 − 241}{19} + 9 = \floorratio{464}{19} + 9 = 24 + 9 = 33 \\ J \| = \floorratio{6940×33}{235} + 27 + 1607557 = \floorratio{229020}{235} + 1607584 = 974 + 1607584 = 1608558 \end{align*}
#### 13.5.2. From CJDN to Babylonian Date
Now we go in the opposite direction, from Julian Day number $$J$$ to day $$d$$, month $$m$$, year $$j$$ in the era of Seleukos.
\begin{align} y_2 \| = J − 1607558 \\ \{x_2, r_2\} \| = \Div(235y_2 + 234, 6940) \\ z_2 \| = \floorratio{r_2}{235} \\ y_1 \| = x_2 \\ \{x_1, r_1\} \| = \Div(19y_1 + 5, 235) \\ z_1 \| = \floorratio{r_1}{19} \\ j \| = x_1 + 1 \\ m \| = z_1 + 1 \\ d \| = z_2 + 1 \end{align}
For example, which date in the Babylonian calendar corresponds to CJDN $$1608558$$? Then $$J = 1608558$$, so
\begin{align*} y_2 \| = 1608558 − 1607558 = 1000 \\ \{x_2, r_2\} \| = \Div(235×1000 + 234, 6940) = \Div(235234, 6940) = \{33, 6214\} \\ z_2 \| = \floorratio{6214}{235} = 26 \\ \{x_1, r_1\} \| = \Div(19×33 + 5, 235) = \Div(632, 235) = \{2, 162\} \\ z_1 \| = \floorratio{162}{19} = 8 \\ j \| = 2 + 1 = 3 \\ m \| = 8 + 1 = 9 \\ d \| = 26 + 1 = 27 \end{align*}
which means day 27 of month 9 of year 3.
This can be compressed to
\begin{align} \{x_2, r_2\} \| = \Div(235×(J − 1607558) + 234, 6940) \\ \{x_1, r_1\} \| = \Div(19x_2 + 5, 235) \\ j \| = x_1 + 1 \\ m \| = \floorratio{r_1}{19} + 1 \\ d \| = \floorratio{r_2}{235} + 1 \end{align}
For example, which date in the Babylonian calendar corresponds to CJDN $$1608558$$? Then $$J = 1608558$$, so
\begin{align*} \{x_2, r_2\} \| = \Div(235×(1608558 − 1607558) + 234, 6940) = \Div(235234, 6940) = \{33, 6214\} \\ \{x_1, r_1\} \| = \Div(19×33 + 5, 235) = \Div(632, 235) = \{2, 162\} \\ j \| = 2 + 1 = 3 \\ m \| = \floorratio{162}{19} + 1 = 9 \\ d \| = \floorratio{6214}{235} + 1 = 27 \end{align*}
which means day 27 of month 9 of year 3.
### 13.6. The Jewish Calendar
#### 13.6.1. From Jewish Date to CJDN
The Jewish calendar is a lunisolar calendar, like the Babylonian calendar. The Jewish calendar is a lot more complicated than the Babylonian one, because
• the Jewish calendar has four rules that can delay the start of a new year, to ensure that certain calendar days do not fall on certain days of the week.
• the calendar year in the Jewish calendar can have six different lengths.
• the Jewish calendar has three different places where leap days can be inserted.
• the Jewish New Year does not fall at the beginning of the first month but at the beginning of the seventh month.
When it is necessary to indicate explicitly that a year is given according to the Jewish calendar, then A.M. (Anno Mundi, Year of the World) is used for that.
A calendar month has 29 or 30 days, a calendar year has 12 or 13 months, and a calendar day begins at 6 o'clock in the evening.
There are 6 different lengths of calendar years, namely 353, 354, 355, 383, 384, or 385 days. The years with 354 or 384 days are called regular. Years with one day more are called complete. Years with a day less are called deficient. Years with 355 or fewer days are called common, and years with more days are called embolismic.
New Year is the day at which the calendar year number increases by one. In the Jewish calendar, this is not the first day of month number 1, but the first day of month number 7, which is the month of Tishri.
##### 13.6.1.1. Calculation Year, Calculation Month, Calculation Day
For the calendrical calculations, it is convenient when each calendar period begins counting at 0. It is therefore inconvenient that the year number changes at the beginning of the 7th month. We calculate with a calculation year that begins with the 1st month (corresponding to calculation month number $$x_3 = 0$$). Months $$m$$ = 1 through 6 of the preceding calendar year $$j − 1$$ and months 7 through 12 or 13 of the current calendar year $$j$$ are part of calculation year $$x_1 = j − 1$$ as calculation months $$x_3$$ = 0 through 11 or 12. The first day of each month has calculation day number $$z_3 = 0$$. New Year's Day of calendar year $$j$$ corresponds to $$x_1 = j − 1$$, $$x_3 = 6$$, and $$z_4 = 0$$.
$${m}$$ 1 2 3 4 5 6 7 8 9 10 11 12 13 $${x_3}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 $${x_1 − j}$$ 0 0 0 0 0 0 −1 −1 −1 −1 −1 −1 −1
For converting from year $$j$$, month $$m$$, day $$d$$ in the Jewish calendar to calculation year $$x_1$$, calculation month $$x_3$$, and calculation day $$z_4$$, we find
\begin{align} c_0 \| = \floorratio{13 − m}{7} \\ x_1 \| = j − 1 + c_0 \\ x_3 \| = m − 1 \\ z_4 \| = d − 1 \end{align}
For example, after the last day of month 13 of year 4682 in the Jewish calendar comes the first day of month 1 of year 4682. A few months later, after the last day of month 6 of year 4682 comes the first day of month 7 of year 4683. That day is New Year's Day. The following table shows a few correspondences.
$${j}$$ $${m}$$ $${d}$$ $${c_0}$$ $${x_1}$$ $${x_3}$$ $${z_4}$$
4682 13 29 0 4681 12 28
4682 1 1 1 4682 0 0
4682 6 29 1 4682 5 28
4683 7 1 0 4682 6 0
4683 12 30 0 4682 11 29
##### 13.6.1.2. Calendar Levels
Transforming a date from the Jewish calendar into CJDN requires combining four calendar levels:
1. First we calculate the running calculation month number $$c_1$$ of the first month (calculation month number $$z_1 = 0$$) of calculation year $$x_1$$.
2. Then we calculate the running day number $$c_2$$ of New Year (calculation day number 0, calculation month number 6) of calculation year $$x_1$$ from the running calculation month number $$c_1$$. With that, we can calculate the length of the year, and from that the length of all of the months.
3. Then we calculate $$y_3$$, the day number in the year, from month number $$x_3$$ in the calculation year.
4. Then we calculate the running day number $$y_4$$ of the desired date from the running day number $$c_2$$ of New Year and from the day number $$y_3$$ in the current calculation year and from the calculation day number $$z_4$$ in the calculation month.
5. And finally we add the CJDN of the day with $$y_4 = 0$$ to get the CJDN of the desired date.
##### 13.6.1.3. First CalendarLevel
At the first calendar level, $$y_1 = c_1(x_1) + z_1$$, where $$y_1$$ is the running month number, $$x_1$$ is the calculation year number, $$z_1$$ is the calculation month number in the calculation year, and $$c_1$$ is the running month number of the first month ($$z_1 = 0$$) of the calculation year.
The leap years (calendar years with 13 months) are the 3rd, 6th, 8th, 11th, 14th, 17th, and 19th year in each cycle of 19 years, which holds 235 months. The running month number $$c_1$$ of the first calculation month ($$z_1 = 0$$) of year $$x_1$$ is
$$c_1 = \floorratio{235x_1 + 1}{19}$$
As an example we look at the beginning of the Jewish year A.M. 4682. Then $$x_1 = 4681$$ so
$c_1 = \floorratio{235×4681 + 1}{19} = \floorratio{1100036}{19} = 57896$
New Year of A.M. 4682 is 57896 months after New Year of A.M. 1.
If you're working in a calculation environment where whole numbers cannot be greater than a certain value $$w$$, then (to avoid overflow) $$|x_1|$$ in the preceding formula must not exceed $$w/235$$, instead of $$w$$. For $$w = 2^{31}$$, $$w/235$$ years correspond to 9,138,228 years. If that is too few for you, then you can find alternatives in the manner of Section 12.2. One alternative is, if $$w \gt 133$$
$$c_1 = 12x_1 + \floorratio{7x_1 + 1}{19}$$
for which $$|x_1|$$ must not exceed $$w/7$$. Another alternative is, for $$w = 2^{31} = 2147483648$$
\begin{align} q \| = \floorratio{x_1}{3} \\ r \| = x_1 \bmod 3 \\ c_1 \| = 37q + \floorratio{2q + 235r + 1}{19} \end{align}
for which $$|x_1|$$ may have any value up to $$w$$.
##### 13.6.1.4. SecondCalendarLevel
At the second calendar level, $$y_2 = c_2(x_2) + z_2$$, where $$x_2 = c_1$$ is the running calculation month number of the first calculation month of the year, $$z_2$$ is the calculation day number in the current month of the first day of that month, and $$y_2$$ is the running day number of the first day of that month.
The epoch of the calendar (the beginning of 1 Tishri of year A.M. 1) is at 6 o'clock at night on Sunday 6 October −3760 in the Julian proleptic calendar, which corresponds to CJD 347997.75. For correspondences between calendar dates from different calendars we look at the situation at noon. Noon of the first day (New Year) of the Jewish calendar was on Monday 7 October −3760 in the Julian proleptic calendar (CJDN 347998 = $$J_0$$).
The beginning of the year is determined by the mean conjunction between the Sun and the Moon, where the length of a synodical month is set at 29 and 13753/25920 days, which is 765433/25920 days. The New Moon at the beginning of the month of Tishri (the New Year's month) of year A.M. 1 was at 5 hours, 11 minutes, and 20 seconds (= 5604/25920 days) after the beginning of the first day of that month (and that began at 6 o'clock at night).
The time $$μ$$ in days that elapsed since the beginning of the first day of the calendar and the New Moon at the beginning of the month with running month number $$c_1$$ is equal to
$$μ = \frac{5604 + 765433c_1}{25920} \label{eq:μ}$$
The running day number $$υ_0$$ of the calendar day on which that New Moon occurs is
$$υ_0 = ⌊μ⌋$$
We again look at New Year of the Jewish year A.M. 4682. We earlier found that $$c_1 = 57896$$. Then
\begin{align*} μ \| = \frac{5604 + 765433×57896}{25920} = \frac{44315514572}{25920} = 1709703 \frac{12812}{25920} \\ υ_0 \| = 1709703 \end{align*}
If you're working in a calculation environment where whole numbers cannot exceed a certain value $$w$$, then (to avoid overflow) in equation \eqref{eq:μ} $$|c_1|$$ must not exceed $$w/765433$$, which is far less than $$w$$. For 32-bit numbers this corresponds to only about 226 jaar − far too few to be practical. Using the methods of Sec. 12.2 we can find better alternatives. One alternative, when $$w \gt 356,477,760$$, is
$$μ = 29c_1 + \frac{5604 + 13753c_1}{25920}$$
for which $$|c_1|$$ must not exceeed $$w/13753$$, which for 32-bit numbers corresponds to about 156146 months, or about 12624 years. Another alternative, for $$w = 2^{31} = 2147483648$$, is
\begin{align} q \| = \floorratio{c_1}{1095} \\ r \| = c_1 \bmod 1095 \\ μ \| = 32336q + \frac{15q + 765433r + 5604}{25920} \end{align}
for which $$|c_1|$$ may have any value up to $$w$$, which corresponds to about 173 million years.
The following table shows some results.
$${x_1}$$ $${c_1}$$ $${q}$$ $${r}$$ $${μ}$$ $${υ_0}$$
4681 57896 52 956 $${1709703 \frac{12812}{25920}}$$ 1709703
4682 57909 52 969 $${1710087 \frac{10161}{25920}}$$ 1710087
4683 57921 52 981 $${1710441 \frac{19677}{25920}}$$ 1710441
5516 68224 62 334 $${2014695 \frac{12196}{25920}}$$ 2014695
5517 68236 62 346 $${2015049 \frac{21712}{25920}}$$ 2015049
5518 68249 62 359 $${2015433 \frac{19061}{25920}}$$ 2015433
The running day number $$υ_0$$ can be modified by four delays.
• The first delay occurs when the mean conjunction falls at or after noon and before the end of the calendar day at 6 o'clock at night. In that case, New Year is delayed by 1 day, the current year gets 1 day shorter, and the preceding year gets 1 day longer.
The time of day follows from $$μ \bmod 1$$, which is equal to 0 at the beginning of the calendar day at 6 o'clock at night, and is equal to 3/4 at noon. So if $$μ \bmod 1 ≥ 3/4$$, then New Year is delayed by one day. If $$μ \bmod 1 ≥ 3/4$$ then $$(μ \bmod 1) + 1/4 ≥ 1$$ so $$\left\lfloor μ + \frac{1}{4} \right\rfloor = ⌊μ⌋ + 1$$, so the running day number $$υ_1$$ of the first day of the month, including the effect of the first delay, is
$$υ_1 = \left\lfloor μ + \frac{1}{4} \right\rfloor = \left\lfloor μ + \frac{6480}{25920} \right\rfloor = 29c_1 + \floorratio{12084 + 13753c_1}{25920} \label{eq:υ_1}$$
For A.M. 4682 we find
$υ_1 = \left\lfloor 1709703 + \frac{12812 + 6480}{25920} \right\rfloor = 1709703 + \floorratio{19292}{25920} = 1709703 = υ_0$
so that year's beginning is unaffected by the first delay.
If you're working in a calculation environment where whole numbers must not exceed some value $$w$$, then the same limitations hold here as in the previous example. You can then use the same alternatives as in the previous example, if you replace 5604 by 12084 there.
• The second delay occurs when the mean conjunction (after taking into account the first delay) falls on a Sunday, Wednesday, or Friday. In that case, New Year is delayed by 1 day, the current year gets one day shorter, and the preceding year gets one day longer.
$$υ_0 = υ_1 = 0$$ corresponds to CJDN 347998 which was a Monday, because $$347998 \bmod 7 = 0$$ and 0 corresponds to Monday. The second delay occurs when
$$δ = υ_1 \bmod 7$$
is equal to 2, 4, or 6.
We seek a formula for $$v_2(δ)$$ that returns 0 if $$δ$$ is equal to 0, 1, 3, or 5, and that returns 1 if $$δ$$ is equal to 2, 4, or 6. Because we only want 0 or 1 to be returned, something involving $$\bmod 2$$ seems to be needed. That we only want whole numbers as results points at something looking like $$⌊...δ...⌋ \bmod 2$$. The simplest formula that might work is then $$v_2(δ) = ⌊qδ⌋ \bmod 2$$ for a suitable factor $$q$$. Some searching shows that we get the desired outcomes when $$5/6 \lt q \lt 1$$. A few ratios with small denominators that are suitable for $$q$$ are 6/7, 7/8, 8/9, 9/10, 10/11, 11/12. The ratio with denominator 7 has the smallest denominator and has as additional bonus that it returns the desired values also if one calculates $$v_2(υ_1)$$ instead of $$v_2(δ) = v_2(υ_1 \bmod 7)$$: that saves one modulus calculation. With that, the running day number $$υ_2$$ of the first day of the month, including the first two delays, is equal to
$$υ_2 = υ_1 + \left( \floorratio{6υ_1}{7} \bmod 2 \right) \label{eq:υ_2}$$
If you're working in a calculation environment where whole numbers must not exceed some value $$w$$, then $$υ_1$$ must not exceed $$w/6$$. For 32-bit numbers this corresponds to about 979,914 years. If that is not enough, then you can use $$v_2(υ_1 \bmod 7)$$ after all:
$$υ_2 = υ_1 + \left( \floorratio{6×(υ_1 \bmod 7)}{7} \bmod 2 \right)$$
• The third delay is needed if a year, after treatment of the first two delays, would get a length of 356 days, which is deemed unacceptably long. By delaying New Year by 2 days, the length of the year is reduced to 354 days, which is acceptable. The preceding year (which was 353 or 383 days long) then gets 2 days longer (i.e., 355 or 385 days).
The length $$L_2(x_1)$$ of year $$x_1$$ (including the effects of the first two delays) is equal to
$$L_2(x_1) = υ_2(x_1 + 1) − υ_2(x_1)$$
so New Year of year $$x_1$$ is delayed by 2 days if $$L_2(x_1) = 356$$. We achieve the desired effect with formula
$$v_3 = 2\left( \floorratio{L_2 + 19}{15} \bmod 2 \right)$$
$${L_2}$$ 353 354 355 356 382 383 384 385 $${v_3}$$ 0 0 0 2 0 0 0 0
• The fourth delay is needed when the preceding year (after treatment of the first two delays) would get a length of 382 days, which is deemed unacceptably short. By delaying New Year of the current year by 1 day the preceding year becomes 383 days long, which is acceptable. The current year (which was 355 days long) then gets one day shorter (i.e., 354 days).
So, New Year of year $$x_1$$ is delayed by 1 day when $$L_2(x_1 − 1) = 382$$. We get the desired effect with formula
$$v_4 = \floorratio{L_2(x_1 − 1) + 7}{15} \bmod 2$$
$${L_2}$$ 353 354 355 356 382 383 384 385 $${v_4}$$ 0 0 0 0 1 0 0 0
Years do not exist for which $$L_2(x_1) = 356$$ and at the same time $$L_2(x_1 − 1) = 382$$, so the third and fourth delays are never active at the same time.
The running day number $$c_2$$ of New Year, taking into account all four delays, is
$$c_2 = υ_2 + v_3 + v_4$$
##### 13.6.1.5. The Third CalendarLevel
Some month lengths depend on the length of the year. With all delays taken into account, years can only have the following lengths $$L_4$$: 353, 354, 355, 383, 384, or 385 days.
For $$x_1 = 4681$$ we found earlier that $$υ_0 = 1709703$$. Equation \eqref{eq:υ_1} then yields $$υ_1 = 1709703 = υ_0$$, so the first delay does not affect that year. Equation \eqref{eq:υ_2} yields
\begin{align*} υ_2 \| = υ_1 + \left( \floorratio{6υ_1}{7} \bmod 2 \right) \\ \| = 1709703 + \left( \floorratio{6×1709703}{7} \bmod 2 \right) \\ \| = 1709703 + \left(\floorratio{10258218}{7} \bmod 2 \right) \\ \| = 1709703 + (1465459 \bmod 2) = 1709703 + 1 = 1709704 \end{align*}
so the second delay does have effect. The length of year $$x_1 = 4681$$ taking into account the first two delays is $$L_2(4681) = υ_2(4682) − υ_2(4681) = 1710087 − 1709704 = 383$$ days. The year with $$x_1 = 4682$$ therefore has a length (taking into account the first two delays) of 356 days, so the third delay applies, and New Year of that year is delayed by 2 days. The results for this and some other years are listed in the following table. The column with title "v" mentions the numbers of the delays that applied.
$${x_1}$$ $${υ_0}$$ $${υ_1}$$ $${υ_2}$$ $${L_2}$$ $${c_2}$$ $${L_4}$$ v
4681 1709703 1709703 1709704 383 1709704 385 2
4682 1710087 1710087 1710087 356 1710089 354 3
4683 1710441 1710442 1710443 353 1710443 353 1, 2
5516 2014695 2014695 2014696 355 2014696 355 2
5517 2015049 2015050 2015051 382 2015051 383 1, 2
5518 2015433 2015433 2015433 355 2015434 354 4
The first delay depends on the time of day, and the second delay depends on the day of the week, so those two delays combined depend on the time since the beginning of the week. The leap years depend on the position of the year in the current cycle of 19 years = 235 months. The calendar as a whole repeats itself when a whole number of weeks coincides with a whole number of leap cycles. The leap cycle has a length of 235×765433/25920 = 35975351/5184 days, so each period of 5184 cycles covers a whole number of days, namely 35975351. This number of days is not a multiple of 7, so the calendar only repeats itself after 7×35975351 = 251827457 days = 35975351 weeks = 689472 years = 8527680 months.
The distribution of the calendar year lengths in that period of 689472 years is listed in the following table, including how many of those years have that length if you take into account different delays.
353 354 355 356 382 383 384 385
$${υ_0}$$ 275583 159873 25984 228032
$${υ_1}$$ 275583 159873 25984 228032
$${υ_2}$$ 78738 140946 192933 22839 3712 116288 36288 97728
$${υ_4}$$ 69222 167497 198737 106677 36288 111051
The next table shows of how many years from that big cycle of 689472 years the New Year is affected by the delays mentioned in the column with title "v".
v
#
%
−
268937 39.0
1 98496 14.3
2 221616 32.1
3 22839 3.3
4 3712 0.5
1 + 2 73872 10.7
rest 0 0
The next table shows how often a year with a certain length follows a year with that same or another length. The columns are for the current year, and the rows are for the following year. For example, that a year of 354 days is followed by a year of 383 days occurs 40000 times in the great cycle of 689472 years.
353 354 355 383 384 385 353 0 16404 16404 0 0 36414 354 13776 0 54468 53354 0 45899 355 9516 54491 16381 53323 36288 28738 383 0 40000 66677 0 0 0 384 29965 0 13323 0 0 0 385 22965 56602 31484 0 0 0
We now know the running day number $$c_2$$ of New Year (1 Tishri) of calculation year $$x_1$$.
The length $$L$$ of calendar year $$x_1$$ is then equal to
$$L = c_2(x_1 + 1) − c_2(x_1)$$
The month lengths are as follow, in years with different lengths:
$${m}$$ 353 354 355 383 384 385
1 Nisan 30 30 30 30 30 30
2 Iyar 29 29 29 29 29 29
3 Sivan 30 30 30 30 30 30
4 Tammuz 29 29 29 29 29 29
5 Av 30 30 30 30 30 30
6 Elul 29 29 29 29 29 29
7 Tishri 30 30 30 30 30 30
8 Ḥeshvan 29 29 30 29 29 30
9 Kislev 29 30 30 29 30 30
10 Tevet 29 29 29 29 29 29
11 Shevat 30 30 30 30 30 30
12 Adar Ⅰ 0 0 0 30 30 30
13 Adar Ⅱ 29 29 29 29 29 29
In a year of 13 months, the embolismic (extra) month Adar Ⅰ is inserted between Shevat and the original Adar, which is then renamed to Adar Ⅱ.
The lengths of the months do not fit a simple calendar (of types 1 or 2 from Sec. 12.11), because the number of successive months with the same length varies by more than one. In the first seven months, the number of successive months with the same length is equal to 1, but after that it is sometimes equal to 3. For example, months 8 through 10 in a year of 353 or 383 days are all 29 days long, and months 7 through 9 in a year of 355 or 385 days are all 30 days long.
The lengths that the months have in a year of 384 days long fit the equation
$$y_3 = \floorratio{384x_3 + 7}{13} + z_3$$
where $$x_3$$ is the calculation month number in the year, $$y_3$$ is the running day number in the year, $$z_3$$ is the running day number in the month, and all of them begin at 0 for the first month and day.
This formula can also be used for month lengths in a year of 354 days, when the length of the year terminates the 12th month at 29 days.
A correction to this formula is needed for years with a length $$L$$ unequal to 354 or 384 days. In a year with 353 or 383 days, month 9 must be one day shorter. In a year with 355 or 385 days, month 8 must be one day longer. The corrections $$c_8$$ for month 8 and $$c_9$$ for month 9 as a function of year length $$L$$ are
$${L}$$ 353 354 355 383 384 385 $${c_8}$$ 0 0 1 0 0 1 $${c_9}$$ −1 0 0 −1 0 0
We find the following formula for the month−8 correction
$$c_8 = \floorratio{L + 353}{2} \bmod 15 = \floorratio{L + 7}{2} \bmod 15$$
and for the month−9 correction
$$c_9 = −\left( \floorratio{385 − L}{2} \bmod 15 \right)$$
We calculate the running day number of a date in months 1 through 6 ($$x_3$$ = 0 through 5) from the following New Year (the first day of month 7), and the length of those months is the same in every year, so the month-8 and month-9 corrections do not affect such dates. We calculate the running day number of a date in months 7 through 13 ($$x_3$$ = 6 through 12) from the preceding New Year. The month-8 correction applies to dates in months 9 through 13, and the month-9 correction to dates in months 10 through 13. In the following table, $$f_8$$ shows (with value 1) to which months the month-8 correction applies, and $$f_9$$ likewise for the month-9 correction.
$${x_3}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 $${f_8}$$ 0 0 0 0 0 0 0 0 1 1 1 1 1 $${f_9}$$ 0 0 0 0 0 0 0 0 0 1 1 1 1
We get that using the equations
\begin{align} f_8 \| = \floorratio{x_3 + 4}{12} \\ f_9 \| = \floorratio{x_3 + 3}{12} \end{align}
With that, the formula to go from month number $$x_3$$ in the calculation year to day number $$c_3$$ in the calculation year of the first day of that month is
$$c_3 = \floorratio{384x_3 + 7}{13} + c_8\floorratio{x_3 + 4}{12} + c_9\floorratio{x_3 + 3}{12}$$
and we add the day number $$z_3$$ in the month to get that day's day number $$y_3$$ in the calculation year
$$y_3 = c_3 + z_3$$
##### 13.6.1.6. The Fourth CalendarLevel
The running day number $$c_4$$ (since 1 Tishri of A.M. 1) of the first day of the desired month is
$$c_4 = c_2 + c_3 \label{eq:c_4}$$
The first six months (from Nisan through Elul) together always have 177 days, regardless of the length of the calendar year, so the running day number $$c_2$$ of 1 Tishri of the calculation year is 177 days greater than the running day number of 1 Nisan of that same calculation year. So, the running day number $$y_4$$ (since 1 Nisan of A.M. 1) of the desired date is
$$y_4 = c_4 + z_3 − 177$$
The epoch of the calendar is 1 Tishri of the year A.M. 1 and corresponds to CJDN 347998 = $$J_0$$. So, the CJDN $$J$$ of the desired date is
$$J = J_0 + y_4 = J_0 + c_2 + c_3 + z_3 − 177 = 347821 + c_2 + c_3 + z_3$$
#### 13.6.2. From CJDN to Jewish Calendar
Now we go in the opposite direction.
1. Transform CJDN $$J$$ into running day number $$y_4$$ since 1 Tishri A.M. 1:
$$y_4 = J − J_0 + 177 = J − 347821$$
As examples we'll calculate the Hebrew calendar dates that correspond to the following CJDN: 2000087, 2001327, 2057986. For $$J = 2057986$$ we find $$y_4 = 2057986 − 347821 = 1710165$$. All in all we find
$${J}$$ $${y_4}$$
2000087 1652266
2001327 1653506
2057986 1710165
119311997 118964176
2. We need to take into account the effect of all delays to determine in which month the day with a certain running day number belongs, but the effect of the third and fourth delays can only be calculated for a specific month in a specific year, and that month and year are the very items we're trying to figure out. So, we need to first estimate a month and year without taking into account the third and fourth delays, and then calculate what those delays are, and then see if our estimate was good or bad.
Using the inverse of equation \eqref{eq:υ_1} we can calculate the provisional running calculation month number $$y_1'$$ for running day number $$y_4$$ as if the second, third, and fourth delays do not exist:
$$y_1' = \floorratio{25920y_4 + 25920 − 12084 − 1}{765433} = \floorratio{25920y_4 + 13835}{765433} \label{eq:y_1p}$$
For $$y_4 = 1710165$$ we find $$y_1' = \floorratio{25920×1710165 + 13835}{765433} = \floorratio{44327490635}{765433} = 57911$$.
If you're working in a calculation environment in which whole numbers must not exceed a certain value $$w$$, then (to avoid overflow) in equation \eqref{eq:y_1p} $$|y_4|$$ must not exceed $$w/25920$$, which is much less than $$w$$. For 32-bit numbers this corresponds to about 226 years − far too few to be practical. Using Sec. 12.2 we can find better alternatives. One alternative, when $$w = 2^{31} = 2147483648$$, is
\begin{align} q \| = \floorratio{y_4}{1447} \\ r \| = y_4 \bmod 1447 \\ y_1' \| = 49q + \floorratio{23q + 25920r + 13835}{765433} \end{align}
for which $$|y_4|$$ may have any value up to $$w$$, which corresponds to about 5.9 million years.
For $$y_4 = 1710165$$ we find
\begin{align*} q \| = \floorratio{1710165}{1447} = 1181 \\ r \| = 1710165 \bmod 1447 = 1258 \\ y_1' \| = 49×1181 + \floorratio{23×1181 + 25920×1258 + 13835}{765433} \\ \| = 57869 + \floorratio{32648358}{765433} = 57869 + 42 = 57911 \end{align*}
just like before.
For all example CJDN we find
$${J}$$ $${q}$$ $${r}$$ $${y_1'}$$
2000087 1141 1239 55951
2001327 1142 1032 55992
2057986 1181 1258 57911
119311997 82214 518 4028506
If we calculate the running day number $$c_4(y_1')$$ of the first day of that running calculation month with all applicable delays included, and then calculate the provisional day-number-in-the-month $$y_4 − c_4(y_1')$$, then we find results that vary between −2 en 29. The days for which this day number is negative belong in the previous month (with $$y_1 = y_1' − 1$$). There are 5963491 such days in each great calendar cycle of 251827457 days, which means that for one in about every 42 days $$y_1'$$ is one too great.
The days for which this day number is not less than 0 and not greater than 28 certainly belong in the current month (with $$y_1 = y_1'$$), because all months have at least 29 days. The days for which this day number is equal to 29 (which indicates the 30th day of the month, because the first day of the month has calculation day number 0) can belong in the current month, but can also belong in the next month, because some months have only 29 days.
If we calculate the day number in the month compared to the next month (i.e., $$y_4 − c_4(y_1' + 1)$$), then we find results that vary between −32 and 0. The days for which this day number is equal to 0 belong in this next month (with $$y_1 = y_1' + 1$$); There really are days for which the provisional running month number $$y_1'$$ is too small. There are 163258 such days in each great calendar cycle of 251827457 days, so for roughly one out of 1543 days (about one day out of every 52 months) $$y_1'$$ is one too small. For 97.6% of days, $$y_1'$$ is already the correct month.
Unfortunately, there is no simple relationship between the day number relative to month $$y_1'$$ and that relative to month $$y_1' + 1$$, so we must calculate both day numbers to figure out the month into which the desired day falls.
The minimal calculation is as follows: Calculate $$y_4 − c_4(y_1')$$. If that value is negative, then $$y_1 = y_1' − 1$$. If that value is not negative and is not greater than 28, then $$y_1 = y_1'$$. If that value is greater than 28, then also calculate $$y_4 − c_4(y_1' + 1)$$. If that second value is negative, then $$y_1 = y_1'$$. If that second value is not negative, then $$y_1 = y_1' + 1$$.
The number of calculations is then not the same for all dates. If it is advantageous to have the same number of calculations for all dates, then calculate as described below.
3. Calculate the provisional calculation year number $$ξ_1$$ and calculation month number $$μ_1$$ in the calculation year, for running calculation month number $$γ_1$$:
\begin{align} γ_1 \| = y_1' + 1 \\ κ_1 \| = 19γ_1 + 17 \\ ξ_1 \| = \floorratio{κ_1}{235} \\ μ_1 \| = \floorratio{κ_1 \bmod 235}{19} \end{align}
For our first example CJDN we had $$y_1' = 57911$$. Then
\begin{align*} γ_1 \| = y_1' + 1 = 57912 \\ κ_1 \| = 19×57912 + 17 = 1100345 \\ ξ_1 \| = \floorratio{1100345}{235} = 4682 \\ μ_1 \| = \floorratio{1100345 \bmod 235}{19} = \floorratio{75}{19} = 3 \end{align*}
For all example CJDNs we find
$${J}$$ $${y_1'}$$ $${γ_1}$$ $${κ_1}$$ $${ξ_1}$$ $${μ_1}$$
2000087 55951 55952 1063105 4523 10
2001327 55992 55993 1063884 4527 2
2057986 57911 57912 1100345 4682 3
119311997 4028506 4028507 76541650 325709 1
4. Calculate the running day number $$c_4'(ξ_1,μ_1)$$ of the beginning of calculation month $$μ_1$$ of calculation year $$ξ_1$$ in the way described in the previous section, and then the day number $$ζ_1$$ of the month, relative to $$c_4'(ξ_1,μ_1)$$:
$$ζ_1 = y_4 − c_4'(ξ_1,μ_1)$$
Apply a correction in case $$ζ_1$$ turns out to be negative:
$$γ_2 = γ_1 + \floorratio{ζ_1}{33}$$
If $$ζ_1$$ is not negative, then the correction is equal to 0.
For calculation year 4682 we found earlier that it has a length of $$L = 354$$ days, and that its New Year has for its running day number $$c_2' = 1710089$$. For calculation month $$μ_1 = 3$$ the month-8 and month-9 corrections $$c_8$$, $$c_9$$ are equal to 0 and we find
\begin{align*} c_3' \| = \floorratio{384μ_1 + 7}{13} + c_8×\floorratio{μ_1 + 4}{12} + c_9×\floorratio{μ_1 + 3}{12} \\ \| = \floorratio{384×3 + 7}{13} = \floorratio{1159}{13} = 89 \\ c_4' \| = c_2' + c_3' = 1710089 + 89 = 1710178 \\ ζ_1 \| = y_4 − c_4' = 1710165 − 1710178 = −13 \end{align*}
so the target day does not belong to calculation month 3 of calculation year 4682. Then
$γ_2 = γ_1 + \floorratio{ζ_1}{33} = 57912 + \floorratio{−13}{33} = 57912 + (−1) = 57911$
For all example CJDNs we find
$${J}$$ $${y_4}$$ $${γ_1}$$ $${ξ_1}$$ $${μ_1}$$ $${c_4'}$$ $${ζ_1}$$ $${⌊ζ_1/33⌋}$$ $${γ_2}$$
2000087 1652266 55952 4523 10 1652296 −30 −1 55951
2001327 1653506 55993 4527 2 1653506 0 0 55993
2057986 1710165 57912 4682 3 1710178 −13 −1 57911
119311997 118964176 4028507 325709 1 118964207 −31 −1 4028506
so $$y_1'$$ was one too small for $$J = 2001327$$.
5. Calculate in the same fashion the calculation year number $$ξ_2$$, calculation month number $$μ_2$$, running day number $$c_4'(ξ_2,μ_2)$$ of the first day of that month, day-number-in-the-month $$ζ_2$$, running month number $$γ_3$$, calculation year number $$ξ_3$$, calculation month number $$μ_3$$, running day number $$c_4'(ξ_3,μ_3)$$ of the first day of that month, and day-number-in-the-month $$ζ_3$$. The final calculation year number is then
$$x_1 = ξ_3$$
and the final calculation month number is
$$x_3 = μ_3$$
and the final calculation day number is
$$z_4 = ζ_3$$
Running month number $$γ_2 = 57911$$ corresponds to
\begin{align*} ξ_2 \| = \floorratio{19×57911 + 17}{235} = \floorratio{1100326}{235} = 4682 \\ μ_2 \| = γ_2 − \floorratio{235ξ_2 + 1}{19} \\ \| = 57911 − \floorratio{235×4682 + 1}{19} \\ \| = 57911 − \floorratio{1100271}{19} = 57911 − 57909 = 2 \end{align*}
For calculation year 4682 we earlier found that it has a length of $$L = 354$$ days, and that its New Year has for its running day number $$c_2' = 1710089$$. For calculation month $$μ_2 = 2$$ the month-8 and month-9 corrections $$c_8$$, $$c_9$$ are equal to 0, and we find
\begin{align*} c_3' \| = \floorratio{384μ_2 + 7}{13} + c_8\floorratio{μ_2 + 4}{12} + c_9\floorratio{μ_2 + 3}{12} \\ \| = \floorratio{384×2 + 7}{13} \\ \| = \floorratio{775}{13} = 59 \\ c_4' \| = c_2' + c_3' = 1710089 + 59 = 1710148 \\ ζ_2 \| = y_4 − c_4' = 1710165 − 1710148 = 17 \end{align*}
This is not negative and also not greater than 28, so we have found the correct month.
For all example CJDNs we find
$${J}$$ $${y_4}$$ $${γ_1}$$ $${γ_2}$$ $${ξ_2}$$ $${μ_2}$$ $${c_4'}$$ $${ζ_2}$$ $${γ_3}$$ $${x_1=ξ_3}$$ $${x_3=μ_3}$$ $${c_4'}$$ $${z_4=ζ_3}$$
2000087 1652266 55952 55951 4523 9 1652267 −1 55950 4523 8 1652237 29
2001327 1653506 55993 55993 4527 2 1653506 0 55993 4527 2 1653506 0
2057986 1710165 57912 57911 4682 2 1710148 17 57911 4682 2 1710148 17
119311997 118964176 4028507 4028506 325709 0 118964177 −1 4028505 325708 12 118964148 28
so $$y_1'$$ was correct for $$J = 2057986$$, and $$y_1'$$ was one too great for $$J = 2000087$$.
6. Transform calculation year $$x_1$$, calculation month $$x_3$$, and calculation day $$z_4$$ into calendar year $$j$$, calendar month $$m$$, and calendar day $$d$$:
\begin{align} c \| = \floorratio{12 − x_3}{7} \\ j \| = x_1 + 1 − c \\ m \| = x_3 + 1 \\ d \| = z_4 + 1 \end{align}
For calculation year 4682, calculation month 2, calculation day 17 we find
\begin{align*} c \| = \floorratio{12 − 2}{7} = \floorratio{10}{7} = 1 \\ j \| = x_1 + 1 − c = 4682 + 1 − 1 = 4682 \\ m \| = 2 + 1 = 3 \\ d \| = 17 + 1 = 18 \end{align*}
so the date is day 18 of calendar month 3 (Sivan) of year 4682.
For all example-CJDNs we find
$${J}$$ $${x_1}$$ $${x_3}$$ $${z_4}$$ $${c}$$ $${j}$$ $${m}$$ $${d}$$
2000087 4523 8 29 0 4524 9 30
2001327 4527 2 0 1 4527 3 1
2057986 4682 2 17 1 4682 3 18
119311997 325708 12 28 0 325709 13 29
### 13.7. A LunarCalendar With Many Fixed Month Lengths
If the distribution of days across months is independent from the distribution of months across years, then a particular month of the year won't have the same length in all years. For example, the first month of year 1 of the Babylonian calendar has 29 days, but the first month of year 2 has 30 days. For doing calendar calculations in your head, it would be convenient if most months have the same length in each year. Can we construct a lunisolar calendar that has that?
With the characteristics of the Cycle of Meton it is not possible to make each month have the same length in every year that contains it. If we make a fixed number of the first 12 months of every year long (with 30 days), then the corresponding number of long months in the 19-year cycle is a multiple of 19. If the 13th month were a short one, then the total number of long months would be a multiple of 19, and if the 13th month were a long one, then the total number would be 7 (the number of years in the 19-year cycle that have 13 months) plus a multiple of 19, but the Cycle of Meton requires 125 = 6×19 + 11 long months, which fits neither of the two possibilities. We'll have to vary the length of some of the first 12 months of every year, too, if we want most calendar months in each year to have a fixed length. We want to use only months of 29 or 30 days.
If we want to use the calendar calculation techniques discussed before, then it is most convenient if we need only a single calendar level for the calculations between months and days, but then we need both short and long years to use the same straight line and thus to have the same month lengths, apart from the shortening at the very end of the year.
We saw before that 6 long months in the first 12 months of the year is not enough, so we must use a formula that yields 7 long months in the first 12 months. With 7 long months in the first 12 months of every year, we have 7×12 = 133 long months in the 19-year cycle, and that is too many, because the Cycle of Meton only has 125 long months in 19 years. So, the 13th month must be a short month, and also we have to change 8 long months (of 30 days) into short months (of 29 days) in the short years (of 12 months). We can do that by making short years one day shorter, but then month 12 must be a long month, otherwise month 12 would have only 28 days in short years.
We can arrange all of that with the following three types of years in the 19-year cycle:
• 7 years of 13 months with 384 days (7 long months, 6 short months).
• 4 years of 12 months with 355 days (7 long months, 5 short months).
• 8 years of 12 months with 354 days (6 long months, 6 short months).
Then there are in total 7×13 + 4×12 + 8×12 = 235 months, which is as expected. There are then 7×7 + 4×7 + 8×6 = 125 long months, which is correct.
The straight line must yield a long month 12 (with 30 days) and a short month 13 (with 29 days), and that 7 of the first 12 months are long months.
The following formula provides this (with $$x_1$$ the number of months since the first month, and $$c_1$$ the number of days from the first day of the year to the first day of month $$x_1$$):
$$c_1 = 29x_1 + \floorratio{7x_1 + 7}{13} = \floorratio{384x_1 + 7}{13} \label{eq:metonmaand}$$
For example, if $$x_1 = 4$$, then equation \eqref{eq:metonmaand} yields
$c_1 = \floorratio{384×4 + 7}{13} = \floorratio{1543}{13} = 118$
so the first day of the fifth month ($$x_1 = 4$$) is the 118th day since the first day of the year.
For the first day of the first 14 months, equation \eqref{eq:metonmaand} gives the following results (with $$L$$ the length of the month):
$${m}$$ 1 2 3 4 5 6 7 8 9 10 11 12 13 14 $${x_1}$$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 $${c_1}$$ 0 30 59 89 118 148 177 207 236 266 295 325 355 384 $${L}$$ 30 29 30 29 30 29 30 29 30 29 30 30 29
Now we need a second calendar level to get the right number of days in each year. We can use the following formula for that, where $$x_2$$ is the number of years since the first year, and $$c_2$$ is the running day number of the first day of the year:
$$c_2 = 354x_2 + 30\floorratio{7x_2 + 1}{19} + \floorratio{4x_2 + 12}{19} \label{eq:metonjaar}$$
For example, if $$x_2 = 8$$, then equation \eqref{eq:metonjaar} yields
$c_2 = 354×8 + 30\floorratio{7×8 + 1}{19} + \floorratio{4×8 + 12}{19} = 2832 + 30×3 + 2 = 2837$
so the first day of year $$x_2 = 9$$ has running day number 2837.
This is calendar type 4, with $$q = 354$$, $$g = 19$$, $$\{d_1,h_1,s_1\} = \{30,7,1\}$$, $$\{d_2,h_2,s_2\} = \{1,4,12\}$$.
The complete algorithm is then, from year $$j$$, month $$m$$, day $$d$$ to running day number $$y_2$$:
\begin{align} x_2 \| = j − 1 \\ x_1 \| = m − 1 \\ z_1 \| = d − 1 \\ c_1 \| = \floorratio{384x_1 + 7}{13} \\ y_1 \| = c_1 + z_1 \\ z_2 \| = y_1 \\ c_2 \| = 354x_2 + 30\floorratio{7x_2 + 1}{19} + \floorratio{4x_2 + 12}{19} \\ y_2 \| = c_2 + z_2 \end{align}
For example, which running day number corresponds to day 27 of month 9 of year 3 in this calendar? Then
\begin{align*} j \| = 3 \\ m \| = 9 \\ d \| = 27 \\ x_2 \| = 3 − 1 = 2 \\ x_1 \| = 9 − 1 = 8 \\ z_1 \| = 27 − 1 = 26 \\ c_1 \| = \floorratio{384×8 + 7}{13} = \floorratio{3079}{13} = 236 \\ y_1 \| = 236 + 26 = 262 \\ z_2 \| = 262 \\ c_2 \| = 354×2 + 30×\floorratio{7×2 + 1}{19} + \floorratio{4×2 + 12}{19} \\ \| = 708 + 30×\floorratio{15}{19} + \floorratio{20}{19} = 708 + 0 + 1 = 709 \\ y_2 \| = 709 + 262 = 971 \end{align*}
This can be compressed to
$$y_2 = 354j + 29m + 30\floorratio{7j − 6}{19} + \floorratio{4j + 8}{19} + \floorratio{7m}{13} + d − 384$$
For the same date as before:
\begin{align*} y_2 \| = 354×3 + 29×9 + 30\floorratio{7×3 − 6}{19} + \floorratio{4×3 + 8}{19} + \floorratio{7×9}{13} + 27 − 384 \\ \| = 1323 + 30\floorratio{15}{19} + \floorratio{20}{19} + \floorratio{63}{13} − 357 \\ \| = 1323 + 0 + 1 + 4 − 357 = 971 \end{align*}
In the opposite direction, the formulas are
\begin{align} x_2 \| = \floorratio{19y_2 + 546}{6940} \\ c_2 \| = 354x_2 + 30\floorratio{7x_2 + 1}{19} + \floorratio{4x_2 + 12}{19} \\ z_2 \| = y_2 − c_2 \\ ζ \| = \floorratio{z_2}{385} \\ x_2 \| ↤ x_2 + ζ \\ c_2 \| ↤ 354x_2 + 30\floorratio{7x_2 + 1}{19} + \floorratio{4x_2 + 12}{19} \\ z_2 \| ↤ y_2 − c_2 \\ y_1 \| = z_2 \\ k_1 \| = 13y_1 + 5 \\ x_1 \| = \floorratio{k_1}{384} \\ z_1 \| = \floorratio{k_1 \bmod 384}{13} \\ j \| = x_2 + 1 \\ m \| = x_1 + 1 \\ d \| = z_1 + 1 \end{align}
For example, which day in this calendar corresponds to running day number 971? Then $$y_2 = 971$$ and then
\begin{align*} x_2 \| = \floorratio{19×971 + 546}{6940} = \floorratio{18995}{6940} = 2 \\ c_2 \| = 354×2 + 30×\floorratio{7×2 + 1}{19} + \floorratio{4×2 + 12}{19} \\ \| = 708 + 30×\floorratio{15}{19} + \floorratio{20}{19} = 708 + 0 + 1 = 709 \\ z_2 \| = 971 − 709 = 262 \\ ζ \| = \floorratio{262}{385} = 0 \end{align*}
$$ζ = 0$$ so $$x_2$$, $$c_2$$, $$z_2$$ remain the same. Then
\begin{align*} y_1 \| = 262 \\ k_1 \| = 13×262 + 5 = 3411 \\ x_1 \| = \floorratio{3411}{384} = 8 \\ z_1 \| = \floorratio{3411 \bmod 384}{13} = \floorratio{339}{13} = 26 \\ j \| = 2 + 1 = 3 \\ m \| = 8 + 1 = 9 \\ d \| = 26 + 1 = 27 \end{align*}
so the date is day 27 of month 9 of year 3.
This can be compressed a little, to
\begin{align} x' \| = \floorratio{19y_2 + 546}{6940} \\ x_2 \| = x' + \floorratio{y_2 − 354x' − 30\floorratio{7x' + 1}{19} − \floorratio{4x' + 12}{19}}{385} \\ k_1 \| = 13×\left( y_2 − 354x_2 − 30\floorratio{7x_2 + 1}{19} − \floorratio{4x_2 + 12}{19} \right) + 5 \\ j \| = x_2 + 1 \\ m \| = \floorratio{k_1}{384} + 1 \\ d \| = \floorratio{k_1 \bmod 384}{13} + 1 \end{align}
For example, which day in this calendar corresponds to running day number 971? Then $$y_2 = 971$$ and then
\begin{align*} x' \| = \floorratio{19×971 + 546}{6940} = \floorratio{18995}{6940} = 2 \\ x_2 \| = 2 + \floorratio{971 − 354×2 − 30 \floorratio{7×2 + 1}{19} − \floorratio{4×2 + 12}{19}}{385} \\ \| = 2 + \floorratio{263 − 30 \floorratio{15}{19} − \floorratio{20}{19}}{385} \\ \| = 2 + \floorratio{263 − 0 − 1}{385} = 2 \\ k_1 \| = 13×\left( 971 − 354×2 − 30\floorratio{7×2 + 1}{19} − \floorratio{4×2 + 12}{19} \right) + 5 \\ \| = 13×\left( 263 − 30\floorratio{15}{19} − \floorratio{20}{19} \right) + 5 \\ \| = 13×(263 − 0 − 1) + 5 = 3411 \\ j \| = 2 + 1 = 3 \\ m \| = \floorratio{3411}{384} + 1 = 9 \\ d \| = \floorratio{3411 \bmod 384}{13} + 1 = \floorratio{339}{13} + 1 = 27 \end{align*}
so the date is day 27 of month 9 of year 3.
### 13.8. The Islamic Calendar
#### 13.8.1. From Islamic Date to CJDN
The religious Islamic calendar depends on observations and so cannot be caught in formulas. The administrative calendar has fixed rules so it can be caught in formulas. In the administrative calendar, the odd months have 30 days and the even months have 29 days, except that the last month sometimes has 29 days and sometimes has 30 days. This gives a year 354 or 355 days. Of each 30 years, 11 are long (with 355 days) and the remainder are short (with 354 days). A complete cycle contains 11×355 + 19×354 = 10631 days.
Rob van Gent (at //www.staff.science.uu.nl/~gent0113/islam/islam_tabcal.htm) mentions four different ways that are used to distribute the 11 leap years across each cycle of 30 years. Table 1 lists those ways and their leap year distributions, and shows the corresponding $$r$$ (as in equation \eqref{eq:r}).
Table 1: Leap Years in Islamic Calendar
Type Leap Years $${r}$$ Origin/Use
I 2 5 7 10 13 15 18 21 24 26 29 15 Kūshyār ibn Labbān, Ulugh Beg, "Kuwaiti Algorithm"
II 2 5 7 10 13 16 18 21 24 26 29 14 al-Fazārī, al-Khwārizmī, al-Battānī, Toledan and Alfonsine Tables
III 2 5 8 10 13 16 19 21 24 27 29 11 Fātimids, Misri, or Bohra Calendar
IV 2 5 8 11 13 16 19 21 24 27 30 9 Habash al-Hāsib, al-Bīrūnī, Elias of Nisibis
The number of leap years between the beginning of year 0 and the beginning of year $$j$$ is
$$N = \floorratio{11j + r}{30}$$
There are two variants for each pattern shift: the astronomical epoch "a" according to which the first day of the calendar (1 Muharram of year 1) began on the evening of Wednesday 14 July 622 (approximately CJD 1948438.75), and the civil epoch "c" according to which the calendar began on the evening of Thursday 15 July 622 (approximately CJD 1948439.75). The most commonly used administrative calendar is IIc.
For the lowest calendar level we seek a straight line that alternates between 30 and 29 days except that the 12th month has 30 days, and for short years (with 354 days) we cut away the last day. In a calendar based on a straight line for 355/12, the alternation between 30 and 29 days is already broken after 5 or 7 months, but we need that alternation to continue for at least 11 months. It turns out that we need a $$p$$ for that of at most 29 + 6/11, and 355/12 = 29 + 7/12 is greater than that. We use $$p = 29 \frac{6}{11} = 325/11$$.
In the Islamic Calendar the calendar days run from sunset to sunset. The CJDN calendar days run from midnight to midnight. For the calculations, we make the connection at noon. The CJDN that according to the following formulas is connected with a particular Islamic calendar date, actually corresponds to that Islamic calendar date only between midnight and sunset. Between sunset and midnight, the next calendar day has already begun according to the Islamic calendar, but not yet according to the CJDN.
For example, CJDN 2455774 corresponds to Gregorian date 31 July 2011 (31-07-2011) and runs from midnight to midnight local time. The below formulas say (for type IIc) that CJDN 2455774 corresponds to Islamic date 29 Shaban 1432 (29-08-1432) ― this means that those two dates correspond to each other between midnight and sunset. From sunset to midnight, it is still 31 July in the Gregorian calendar and still CJDN 2455774, but in the Islamic calendar it is already the next day (i.e., 1 Ramaḍān).
Type $${r}$$ $${J_0}$$ Ia 15 1948439 Ic 15 1948440 IIa 14 1948439 IIc 14 1948440 IIIa 11 1948439 IIIc 11 1948440 IVa 9 1948439 IVc 9 1948440
\begin{align} x_2 \| = j − 1 \\ x_1 \| = m − 1 \\ z_1 \| = d − 1 \\ c_1 \| = \floorratio{325x_1 + 5}{11} \\ y_1 \| = c_1 + z_1 \\ z_2 \| = y_1 \\ c_2 \| = \floorratio{10631x_2 + r}{30} \\ y_2 \| = c_2 + z_2 \\ J \| = y_2 + J_0 \end{align}
For example, which CJDN corresponds to the Islamic date of 29 Shaban 1432 according to calendar type IIc? Then
\begin{align*} j \| = 1432 \\ m \| = 8 \\ d \| = 29 \\ r \| = 14 \\ J_0 \| = 1948440 \\ x_2 \| = 1432 − 1 = 1431 \\ x_1 \| = 8 − 1 = 7 \\ z_1 \| = 29 − 1 = 28 \\ c_1 \| = \floorratio{325×7 + 5}{11} = \floorratio{2280}{11} = 207 \\ y_1 \| = 207 + 28 = 235 \\ z_2 \| = 235 \\ c_2 \| = \floorratio{10631×1431 + 14}{30} = \floorratio{15212975}{30} = 507099 \\ y_2 \| = 507099 + 235 = 507334 \\ J \| = 507334 + 1948440 = 2455774 \end{align*}
This can be condensed to
$$J = \floorratio{10631j − 10631 + r}{30} + \floorratio{325m − 320}{11} + d − 1 + J_0$$
i.e.,
Ia $${J = \floorratio{10631j − 10616}{30} + \floorratio{325m − 320}{11} + d + 1948438}$$ Ic $${J = \floorratio{10631j − 10616}{30} + \floorratio{325m − 320}{11} + d + 1948439}$$ IIa $${J = \floorratio{10631j − 10617}{30} + \floorratio{325m − 320}{11} + d + 1948438}$$ IIc $${J = \floorratio{10631j − 10617}{30} + \floorratio{325m − 320}{11} + d + 1948439}$$ IIIa $${J = \floorratio{10631j − 10620}{30} + \floorratio{325m − 320}{11} + d + 1948438}$$ IIIc $${J = \floorratio{10631j − 10620}{30} + \floorratio{325m − 320}{11} + d + 1948439}$$ IVa $${J = \floorratio{10631j − 10622}{30} + \floorratio{325m − 320}{11} + d + 1948438}$$ IVc $${J = \floorratio{10631j − 10622}{30} + \floorratio{325m − 320}{11} + d + 1948439}$$
For the same Islamic date as before we now find
\begin{align*} J \| = \floorratio{10631×1432 − 10617}{30} + \floorratio{325×8 − 320}{11} + 29 + 1948439 \\ \| = \floorratio{15212975}{30} + \floorratio{2280}{11} + 29 + 1948439 \\ \| = 507099 + 207 + 29 + 1948439 = 2455774 \end{align*}
which is the same answer as before.
#### 13.8.2. From CJDN to Islamic Date
In the opposite direction we find:
\begin{align} y_2 \| = J − J_0 \\ \{x_2, r_2\} \| = \Div(30y_2 + 29 − r, 10631) \\ z_2 \| = \floorratio{r_2}{30} \\ y_1 \| = z_2 \\ \{x_1, r_1\} \| = \Div(11y_1 + 5, 325) \\ z_1 \| = \floorratio{r_1}{11} \\ j \| = x_2 + 1 \\ m \| = x_1 + 1 \\ d \| = z_1 + 1 \end{align}
For example, which Islamic date (calendar type IIc) corresponds to CJDN 2455774? Then $$J = 2455774$$, $$r = 14$$, $$J_0 = 1948440$$, and then
\begin{align*} y_2 \| = 2455774 − 1948440 = 507334 \\ \{x_2, r_2\} \| = \Div(30×507334 + 15, 10631) = \Div(15220035, 10631) = \{1431, 7074\} \\ z_2 \| = \floorratio{7074}{30} = 235 \\ y_1 \| = 235 \\ \{x_1, r_1\} \| = \Div(11×235 + 5, 325) = \Div(2590, 325) = \{7, 315\} \\ z_1 \| = \floorratio{315}{11} = 28 \\ j \| = 1432 \\ m \| = 8 \\ d \| = 29 \end{align*}
which means 29 Shaban 1432.
This can be condensed to
\begin{align} \{x_2, r_2\} \| = \Div(30(J − J_0) + 29 − r, 10631) \\ \{x_1, r_1\} \| = \Div\left( 11\floorratio{r_2}{30} + 5, 325 \right) \\ j \| = x_2 + 1 \\ m \| = x_1 + 1 \\ d \| = \floorratio{r_1}{11} + 1 \end{align}
For example, which Islamic date (calendar type IIc) corresponds to CJDN 2455774? Then $$J = 2455774$$, $$r = 14$$, $$J_0 = 1948440$$, and then
\begin{align*} \{x_2, r_2\} \| = \Div(30×(2455774 − 1948440) + 29 − 14, 10631) \\ \| = \Div(15220035, 10631) = \{1431, 7074\} \\ \{x_1, r_1\} \| = \Div\left( 11×\floorratio{7074}{30} + 5, 325 \right) \\ \| = \Div(11×235 + 5, 325) = \Div(2590, 325) = \{7, 315\} \\ j \| = 1431 + 1 = 1432 \\ m \| = 7 + 1 = 8 \\ d \| = \floorratio{315}{11} + 1 = 29 \end{align*}
which means 29 Shaban 1432, which is the same as in the previous example.
### 13.9. The Maya Calendar
#### 13.9.1. Between CJDN and Maya Calendar
The Maya from Anahuac (Central America) used three different calendars, of which two (the Tzolkin and Haab) were periodic with fairly short periods, and the third one (the Long Count) might have been intended to be periodic but has some periods that are so long that it can be considered to be continuous rather than repeating.
Different areas in Central America had slightly different versions of these calendars, with different names for days and months, and different ways to indicate years, and sometimes days were counted from 0 rather than from 1. Below we describe the calendars from the city of Tikal.
#### 13.9.2. The Haab
The Haab has a day number and a month, but no year number. There are 18 months of 20 days, plus a 19th month of 5 days, which makes 365 days in total, which we call a Haab year. There are no leap years. The months have a name, and the days have a number beginning at 0.
We write a date in the Haab as $$\{h_d,h_m\}$$ where $$h_d$$ is the day number (from 0 through 19) and $$h_m$$ is the month number (from 1 through 19).
To translate a CJDN $$J$$ to a Haab date $$\{h_d,h_m\}$$ we must first calculate $$H$$, the number of days since the beginning of the current Haab year (with $$H = 0$$ for the first day of the Haab year).
\begin{align} H \| = (J + 65) \bmod 365 \\ h_m \| = \floorratio{H}{20} + 1 \\ h_d \| = H \bmod 20 \end{align}
For example, which Haab date corresponds to 15 December 1965 = CJDN 2439110? Then $$J = 2439110$$ and
\begin{align*} H \| = (J + 65) \bmod 365 = 2439175 \bmod 365 = 245 \\ h_m \| = \floorratio{245}{20} + 1 = 13 \\ h_d \| = 245 \bmod 20 = 5 \end{align*}
so that day is 245 days since the beginning of the current Haab year and corresponds to the 5th day of the 13th month, i.e., $$\{5.13\}$$.
To translate a Haab date $$\{h_d,h_m\}$$ to a CJDN we first calculate $$H$$.
$$H = h_d + 20×(h_m − 1)$$
We know that
$$J ≡ H − 65 \pmod{365}$$
Because there is no year number in the Haab, this Haab date returns every 365 days. We can use equation \eqref{eq:laatste} to find the last $$J$$ that corresponds to this Haab date, on or before a specific $$J_0$$:
$$J = J_0 − ((J_0 − H + 65) \bmod 365)$$
Which CJDN corresponds to Haab date $$\{5,13\}$$? Then $$h_d = 5$$, $$h_m = 13$$, so $$H = 5 + 20×(13 − 1) = 245$$ and then $$J ≡ 245 − 65 ≡ 180 \pmod{365}$$. That fits the original $$J$$ that we began with in the previous example (2439110), because $$2439110 = 6682×365 + 180 ≡ 180 \pmod{365}$$.
What is the last CJDN in the year 1965 that corresponds to Haab date $$\{5,13\}$$? We seek the last $$J ≡ 180 \pmod{365}$$ on or before CJDN $$J_0 = 2439126$$ (which corresponds to 31 December 1965), using equation \eqref{eq:laatste}. We find
\begin{align*} J \| = J_0 − ((J_0 − 180) \bmod 365) \\ \| = 2439126 − (2438946 \bmod 365) \\ \| = 2439126 − 16 = 2439110 \end{align*}
#### 13.9.3. The Tzolkin
The Tzolkin has a period of 20 days (the venteina) with a name for each day, and a period of 13 days (the trecena) with a number for each day (beginning at 1).
We write a date in the Tzolkin as $$\{t_t,t_v\}$$, where $$t_t$$ is the day number in the trecena (from 1 through 13) and $$t_v$$ is the day number in the venteina (from 1 through 20).
The trecena and venteina increase simultaneously, so after day $$\{4,7\}$$ comes day $$\{5,8\}$$, and then $$\{6,9\}$$, and so on. Because 13 and 20 are relatively prime, all possible combinations of venteina and trecena occur in this calendar, so after 13×20 = 260 days the dates repeat again.
We translate CJDN $$J$$ to a Tzolkin date as follows:
\begin{align} t_t \| = ((J + 5) \bmod{13}) + 1 \label{eq:trecena2} \\ t_v \| = ((J + 16) \bmod{20}) + 1 \label{eq:venteina2} \end{align}
What is the Tzolkin date corresponding to CJDN 2439110? Then
\begin{align*} t_t \| = ((2439110 + 5) \bmod 13) + 1 = 3 + 1 = 4 \\ t_v \| = ((2439110 + 16) \bmod 20) + 1 = 6 + 1 = 7 \end{align*}
so that date is $$\{4,7\}$$.
To translate a Tzolkin date $$\{t_t,t_v\}$$ to a CJDN we need to solve the two simultaneous congruences of equations \eqref{eq:trecena2} and \eqref{eq:venteina2}, i.e.,
\begin{align} J \| ≡ t_t − 6 \pmod{13} \\ J \| ≡ t_v − 17 \pmod{20} \end{align}
We combine those congruences in the way of section 12.10.3. We have $$x_1 = t_t$$, $$x_2 = t_v$$, $$p_1 = 13$$, $$p_2 = 20$$, $$a_1 = 6$$, $$a_2 = 17$$. Then
\begin{align} C_1 \| = x_1 − a_1 = t_t − 6 \\ P_1 \| = p_1 = 13 \end{align}
The greatest common divisor of 13 and 20 is 1:
\begin{align} g_2 \| = 1 \\ Q_1 \| = \frac{P_1}{g_2} = \frac{13}{1} = 13 \\ q_2 \| = \frac{p_2}{g_2} = \frac{20}{1} = 20 \end{align}
We now seek a $$r_2$$ that solves $$r_2Q_1 ≡ 1 \pmod{q_2}$$, i.e., $$13r_2 ≡ 1 \pmod{20}$$, and find $$r_2 = 17$$, because $$13×17 = 221 ≡ 1 \pmod{20}$$.
\begin{align} \| \begin{split} C_2 \| = C_1 (1 − r_2Q_1) + (x_2 − a_2) r_2Q_1 \\ \| = (t_t − 6)×(1 − 17×13) + (t_v − 17)×17×13 \\ \| = 221 t_v − 220 t_t − 2437 \end{split} \\ P_2 \| = p_2Q_1 = 20×13 = 260 \end{align}
so the solution is
$$J ≡ 221 t_v − 220 t_t − 2437 = −39 t_v + 40 t_t − 97 \pmod{260}$$
Which days correspond to Tzolkin date $$\{4,7\}$$? Those days have $$J ≡ −39×7 + 40×4 − 97 ≡ 50 \pmod{260}$$. That fits with $$J = 2439110$$ because $$2439110 = 9381×260 + 50 ≡ 50 \pmod{260}$$.
Because there is no year number in the Tzolkin, this Tzolkin date returns every 260 days. We can use equation \eqref{eq:laatste} to find the last $$J$$ on or before a particular $$J_0$$ that corresponds with this Tzolkin date:
$$J = J_0 − ((J_0 − 40 t_t + 39 t_v + 97) \bmod 260)$$
The last Tzolkin date $$\{4,7\}$$ that occurs on or before 31 December 1965 (= CJDN 2439126) is $$J = 2439126 − ((2439126 − 50) \bmod 260) = 2439126 − 16 = 2439110$$.
For the number of days $$T$$ since the last $$\{1,1\}$$ we have
$$\begin{split} T \| ≡ 40 (t_t − 1) − 39 (t_v − 1) \\ \| ≡ 40 t_t − 39 t_v − 1 \\ \| ≡ 40 t_t + 221 t_v − 1 \pmod{260} \end{split}$$
so
$$T ≡ (40 t_t + 221 t_v − 1) \bmod 260$$
For CJDN 2439110 (with $$\{t_t,t_v\} = \{4,7\}$$) we find $$T = (40×4 + 221×7 − 1) \bmod 260 = 1706 \bmod 260 = 146$$, so the last day before then that had $$\{1,1\}$$ was CJDN $$J = 2439110 − 146 = 2438964$$, and that fits, because $$t_t = ((J + 5) \bmod 13) + 1 = (2438969 \bmod 13) + 1 = 1$$ and $$t_v = ((J + 16) \bmod 20) + 1 = (2438980 \bmod 20) + 1 = 1$$.
#### 13.9.4. Tzolkin and Haab
Sometimes a date is indicated in both Tzolkin and Haab. From CJDN to Tzolkin and Haab is described above. From a date with Tzolkin and Haab to CJDN is a bit more complicated. We need to solve simultaneously:
\begin{align} J \| ≡ H − 65 \pmod{365} \\ J \| ≡ T − 96 \pmod{260} \end{align}
We found the first formula for the Haab alone, and the second formula for the Tzolkin alone. These formulas again have the form from section 12.10.3, with $$x_1 = H$$, $$x_2 = T$$, $$p_1 = 365$$, $$p_2 = 260$$, $$a_1 = 65$$, $$a_2 = 96$$. Then
\begin{align} C_1 \| = x_1 − a_1 = H − 65 \\ P_1 \| = p_1 = 365 \end{align}
The greatest common divisor of 365 and 260 is 5, so
\begin{align} g_2 \| = 5 \\ Q_1 \| = \frac{P_1}{g_1} = \frac{365}{5} = 73 \\ q_2 \| = \frac{p_2}{g_1} = \frac{260}{5} = 52 \end{align}
We now seek $$r_2$$ that satisfies $$r_2Q_1 ≡ 1 \pmod{q_2}$$, i.e., $$73r_2 ≡ 1 \pmod{52}$$, and find $$r_2 = 5$$, because $$73×5 = 365 = 7×52 + 1 ≡ 1 \pmod{52}$$. Then
\begin{align} \| \begin{split} C_2 \| = C_1 (1 − r_2Q_1) + (x_2 − a_2) r_2Q_1 \\ \| = (H − 65)×(1 − 5×73) + (T − 96)×5×73 \\ \| = 365 T − 364 H − 11380 \end{split} \\ P_2 \| = p_2Q_1 = 365×52 = 18980 \end{align}
so the solution is
$$J ≡ 365 T − 364 H − 11380 ≡ 365 T − 364 H + 7600 \pmod{18980}$$
The Tzolkin and Haab together have a period of 18980 days, which is approximately 52 years.
Which CJDN corresponds to $$\{t_t,t_v,h_d,h_m\} = \{4,7,5,13\}$$? Then $$H = 245$$ and $$T = 146$$, so $$J ≡ 365×146 − 364×245 + 7600 ≡ −28290 ≡ 9670 \pmod{18980}$$, and that fits with the original $$J = 2439110$$ that we began with, because $$2439110 = 128×18980 + 9670 ≡ 9670 \pmod{18980}$$.
Because $$g_2$$ is not equal to 1, not all possible combinations of $$H$$ and $$T$$ occur in the calendar. The preceding formula yields a $$J$$ for every combination of $$H$$ and $$T$$, but that $$J$$ is only useful for combinations of $$H$$ and $$T$$ that actually occur in the calendar. Those are combinations for which $$H − T ≡ 65 − 96 ≡ 4 \pmod{5}$$.
In the example that we've been using we have $$H = 245$$ and $$T = 146$$, so $$H − T = 99 ≡ 4 \pmod{5}$$, so this combination indeed occurs in this calendar.
#### 13.9.5. The Long Count
The Long Count counts days and has a series of increasingly longer periods. The smallest period contains 20 days, the next one is 18 times longer (i.e., 360 days), and after that each next period is 20 times longer than the previous one. The number for each period begins at 0. Often dates in the Long Count are given with five numbers, but even longer periods are known, up to nine numbers. The longest known period corresponds to about 63 million years. Here we assume that the Long Count has five numbers, and that the fifth number can get arbitrarily long.
The epoch (0.0.0.0.0) of the Long Count probably corresponds to CJDN $$J_0 = 584283$$ (6 September −3113 in the Julian calendar).
Translating CJDN $$J$$ to Long Count $$L ≡ l_5.l_4.l_3.l_2.l_1$$ goes as follows:
\begin{align} y \| = J − J_0 \\ \{l_5, r_5\} \| = \Div(y, 144000) \\ \{l_4, r_4\} \| = \Div(r_5, 7200) \\ \{l_3, r_3\} \| = \Div(r_4, 360) \\ \{l_2, l_1\} \| = \Div(r_3, 20) \end{align}
and translating a Long Count to CJDN goes as follows:
$$\begin{split} J \| = l_1 + 20×(l_2 + 18×(l_3 + 20×(l_4 + 20×l_5))) + J_0 \\ \| = l_1 + 20×l_2 + 360×l_3 + 7200×l_4 + 144000×l_5 \end{split}$$
Which Long Count $$L$$ corresponds to CJDN $$J = 2439110$$? Then
\begin{align*} y \| = J − J_0 = 2439110 − 584283 = 1854827 \\ \{l_5, r_5\} \| = \Div(1854827, 144000) = \{12, 126827\} \\ \{l_4, r_4\} \| = \Div(126827, 7200) = \{17, 4427\} \\ \{l_3, r_3\} \| = \Div(4427, 360) = \{12, 107\} \\ \{l_2, l_1\} \| = \Div(107, 20) = \{5, 7\} \end{align*}
The answer is $$L = 12.17.12.5.7$$.
And now in the opposite direction: $$J = 7 + 20×(5 + 18×(12 + 20×(17 + 20×12))) + 584283 = 2439110$$.
### 13.10. Algorithm with a Curved Line
We are not limited to using straight lines. The method that we derived above for using a straight line can be expanded to more complicated functions. If we have functions $$v(x)$$ and $$w(y)$$ such that
\begin{align*} v'(x) \| ≥ 1 \\ v(w(x)) \| = w(v(x)) = x \end{align*}
(so $$v$$ and $$w$$ are each other's inverse functions) then the calendar formulas become
\begin{align} y \| = ⌊v(x)⌋ + z \\ x \| = ⌈w(y + 1)⌉ − 1 = −⌊−w(y + 1)⌋ − 1 \end{align}
We get the case of the straight line if we set $$v(x) = px = fx/g$$; to this corresponds $$w(y) = y/p = gy/f$$.
The derivation goes similar to the case with the straight line.
With a curved line we can make a calendar in which the average length of the month or the year does not stay constant in the long run. In this way we can keep the calendar better in step with the Sun and the Moon than for a calendar that has fixed average lengths.
A practical problem for this is that not every function $$v(x)$$ has a (reasonably simple to calculate) corresponding inverse function $$w(y)$$. You cannot use just any $$v(x)$$.
If you use a function $$v(x)$$ that (for whole numbers $$x$$) does not always yield a whole number or a ratio, or if you do the calculations with floating-point numbers, then you should worry about round-off errors. If you calculator or program makes a round-off error and yields $$v(x) = 6.9999$$ instead of $$v(x) = 7.0000$$, then you get $$⌊v(x)⌋ = 6$$ instead of $$⌊v(x)⌋ = 7$$, and so a wrong day number. To prevent round-off errors, you should pick a function $$v(x)$$ that allows calculating with ratios and that has an inverse function $$w(y)$$ that also allows calculating with ratios, so that all of the calculations can be done with whole numbers only (separately in numerators and denominators).
## 14. Validity of the Algorithms When Used in Computer Programs
The above-mentioned algorithms are in principle valid for all days, but computer programs usually work with numbers of a limited size, so in practice the algorithms won't work anymore for dates that are too far from the begin date of the calendar, or too far from CJDN 0. If the computer program works with whole numbers of $$n$$ bits wide, then the greatest number that it can handle is equal to $$w = 2^{n − 1}$$. If one or more of the intermediate results of the algorithms gets greater than $$w$$ (in positive or negative direction), then they won't fit anymore and then the program yields wrong results.
For translating a calendar date into a running day number, that running day number is usually the greatest (intermediate) result, so algorithms that translate a calendar date into a running day number will give valid results in computer programs if the running day number remains clearly less (in absolute sense) than $$w$$. The number of days since or before CJDN 0 must also remain less than $$w$$. The value of $$w$$ for oft-used number widths is shown in table 2.
$${n}$$ $${w}$$ $${w/365.25}$$
2 32 768 89
4 2 147 483 648 5.8 × 106
8 9 223 372 036 854 775 808 2.5 × 1016
For example, if the computer program uses numbers of 32 bits wide, then it can translate calendar dates into running day numbers for dates up to at least 5 million years from the beginning of the calendar (both to the future and to the past).
While translating a running day number into a calendar date, the running day number $$y$$ (since the beginning of the calendar) is entered into a formula like equation \eqref{eq:ynaarxr}, i.e., multiplied by a factor $$g$$ (a whole number greater than 1), then added to a fixed (usually small) whole number, and then divided by a divisor $$f$$ (a whole number greater than $$g$$). We can indicate that calculation schematically like this:
$$x = \floorratio{gy + t}{f}$$
If we do the calculations in the listed order, then we get an intermediate result $$gy + t$$ which is much greater than $$y$$ itself. Intermediate results greater than $$w$$ give trouble, so to avoid trouble we'd need roughly $$|gy| \lt w$$, so $$|y| \lt w/g$$, which is much less than $$w$$ itself.
For example, the "CJDNGregorian (3)" algorithm has $$g = 4800$$, so then we'd need $$|y| \lt w/4800$$. For numbers of 32 bits wide, this algorithm would be useful as long as approximately $$|y| \lt 447392$$, i.e., for 447,392 days or 1224 years from the beginning of the calendar. This is far too small to cover the whole period since the end of prehistory.
Section 12.2 describes how we can rearrange the calculation to avoid this problem.
As an example we look at equation \eqref{eq:g3}, which we repeat here:
$$x_2 = \floorratio{4800y_2 + 15793}{146097}$$
Then $$g = 4800$$, $$t = 15793$$, $$f = 146097$$. Suppose that $$y_2 = 730,000$$ (we calculate for a date that is 730,000 days, approximately 2000 years, after the beginning of the calendar) and $$n = 4$$ so $$w = 2^{31} = 2,147,483,648$$.
Then $$4800y_2 + 15793 = 3,504,015,793$$, which is greater than $$w$$, and $$3504015793/146097 = 23984 + 25345/146097$$, so $$x_2 = 23984$$. If we calculate $$x_2$$ in this way then we get an intermediate result that is too great to fit into a number of 32 bits wide.
Now we use the alternative method. Then
$x_2 = 4800\floorratio{y_2}{146097} + \floorratio{4800(y_2 \bmod 146097) + 15793}{146097}$
We find
\begin{align*} \floorratio{y_2}{146097} \| = \floorratio{730000}{146097} = 4 \\ y_2 \bmod 146097 \| = y_2 − 146097\floorratio{y_2}{146097} = 145612 \\ x_2 \| = 4800×4 + \floorratio{4800×145612 + 15793}{146097} \\ \| = 19200 + \floorratio{698953393}{146097} = 19200 + 4784 = 23984 \end{align*}
This gives the same end result as above, but now the greatest intermediate result was 698,953,393, which is less than $$w$$.
With this alternative method, the greatest possible intermediate result for this calendar is equal to $$η = fg − f + t = f×(g − 1) + t = 4799×146097 + 15793 = 701,135,296$$, which fits into a number of 32 bits wide.
With this alternative we can, by doing more calculations, translate running day numbers into calendar dates for about the same range of dates as the translation of calendar dates into running day numbers, if $$fg − f + t \lt w$$.
Table 3 shows, for all algorithms described above, the relevant value of $$g$$ and (approximately) the greatest allowed value of $$y$$ (measured in days and years) for $$n$$ equal to 16, 32, and 64 ― the width in bits of the most commonly used types of whole numbers in computer programs. This shows the limitations of the use of the algorithms if the alternative way of calculation that was discussed above is not used. The last columns shown $$t$$, $$f$$, the greatest possible intermediate result $$η$$ (apart from $$y$$) if the alternative method is used, and the minimum width $$n_{min}$$ in bits that is required for the alternative method to work well in the computer program.
Calendar $${g}$$ $${y}$$ $${y/365.25}$$ $${t}$$ $${f}$$ $${η}$$ $${n_{min}}$$
16 32 64 16 32 64
CJDN → Julian (1) 4 8192 536 × 106 2305 × 1015 22 1469872 6313 × 1012 3 1461 4386 2
CJDN → Julian (2) 48 682 44 × 106 192 × 1015 1 122489 526 × 1012 109 1461 68776 3
CJDNGregorian (1) 4 8192 536 × 106 2305 × 1015 22 1469872 6313 × 1012 3 146097 438294 2
CJDNGregorian (2) 400 81 5 × 106 23 × 1015 0 14698 63 × 1012 799 146097 58293502 4
CJDNGregorian (3) 4800 6 447392 1.9 × 1015 0 1224 5 × 1012 15793 146097 701135296 4
CJDN → Babylonian (1) 235 139 9 × 106 39 × 1015 0 25019 10 × 1012 234 6940 1624194 3
CJDN → Babylonian (2) 19 1724 113 × 106 485 × 1015 4 309446 1329 × 1012 546 6940 125466 3
CJDN → Islamic 30 1092 71 × 106 307 × 1015 3 195983 841 × 1012 15 10631 308314 3
The "CJDNGregorian (3)" algorithm is, for whole numbers of 32 bits wide, valid for dates up to 1224 years from the beginning of the calendar. If you use that algorithm with that number width for dates near today, then you'll get wrong answers. How the answers are wrong depends on what your computer does with intermediate results that don't fit, and that can be different for each computer. If you want to use this algorithm in practice, then your computer program should either use whole numbers of 64 bits wide (then the algorithm is valid for about 5,000,000,000,000 years), or use whole numbers of 32 bits wide and also use the alternative method explained above (then the algorithm is valid for about 6,000,000 years).
## 15. General Algorithms for Calculating Between Calendar Date and Running Day Number
### 15.1. From Calendar Date to Running Day Number
Below is an algorithm (in pseudo-code) for calculating a running day number from a calendar date. The meaning of the various function arguments is:
date[i,j] is a two-dimensional table, in which i selects the different calendar units (day, month, and so on; there are n + 1 of them) and j selects the different dates (there are t of them).
q[i] is the q (base period) of calendar level i.
g[i] is the g (denominator) of calendar level i.
b[i] indicates the calendar type (1 through 4) of calendar level i.
d[i][k] is the d (deviation from the base period) of period k (there are m of these) of calendar level i.
h[i][k] is the h (numerator) of period k of calendar level i.
s[i][k] is the s (pattern shift) of period k of calendar level i.
y[j] is the array of running day numbers.
The function DIV(x,y) corresponds to ⌊x/y⌋. The function MOD(x,y) corresponds to x mod y = x − y*⌊x/y⌋.
Here is the algorithm:
function CAL2DAY(date, q, g, b, d, h, s)
for j from 0 through t − 1:
set y[j] to date[0,j]
for i from 0 through n − 1:
if MOD(b[i],2) = 1 then:
set z[j] to y[j]
set x[j] to date[i+1,j]
else:
set z[j] to date[i+1,j]
set x[j] to y[j]
set c[j] to q[i]*x[j]
for k from 0 through m − 1:
set c[j] to c[j] + MOD(h[i][k]*x[k,j] + s[i][k],g[i])*d[i][k]
set y[j] to c[j] + z[j]
return y
end function
### 15.2. From Running Day Number to Calendar Date
Here is a general algorithm (in pseudo-code) for calculating a calendar date from a running day number for all four calendar types, in similar fashion as above.
function DAY2CAL(day, q, g, b, d, h, s)
for i from 0 through n − 1:
set dp[i] to 0
set s2[i] to 0
set f[i] to q[i]*g[i]
for k from 0 through m − 1:
set f[i] to f[i] + d[i][k]*h[i][k]
if d[i][k] gt 0 then:
set dp[i] to dp[i] + d[i][k]
set s2[i] to d[i][k]*s[i][k]
set s2[i] to g[i]*dp[i] − s2[i] − 1
set zz[i] to q[i] + dp[i]
for j from 0 through t − 1:
set y[j] to day[j]
for i from n − 1 through 0:
set x[j] to DIV(g[i]*y[j] + s2[i],f[i])
set c[j] to q[i]*x[j]
for k from 0 through m − 1:
set c[j] to c[j] + DIV(h[i][k]*x[j] + s[i][k],g[i])*d[i,k]
set z[j] to y[j] − c[j]
if b[i] gt 2 then:
while z[j] lt 0:
set x[j] to x[j] + DIV(z[j],zz[i])
set c[j] to q[i]*x[j]
for k from 0 through m − 1:
set c[j] to c[j] + DIV(h[i][k]*x[j] + s[i][k],g[i])*d[i,k]
set z[j] to y[j] − c[j]
if MOD(b[i],2) = 1 then:
set y[j] = z[j]
set date[i + 1,j] = x[j]
else:
set date[i + 1,j] = z[j]
set y[j] = x[j]
set date[0,j] = y[j]
return date
end function
### 15.3. Parameters for Various Calendars
The below table shows for each of the calendars which calendar units in which order should be provided to CAL2DAY as date, or that are returned by DAY2CAL. d, m, y, c indicate day, month, year, and century, respectively. A * indicates the use of calculation months/years/centuries where the calculation year begins with the month of March.
Calendar Order
Julian (1) dm*y*
Julian (2) ymd
Gregorian (1) dm*y*c*
Gregorian (2) dm*y*
Gregorian (3) ymd
Babylonian (1) ymd
Babylonian (2) dmy
Islamic dmy
The parameters that are needed to use the above algorithms for the various calendars are shown in the following table.
Julian Gregorian Babylonian Islamic
(1) (2) (1) (2) (3) (1) (2)
J₀1721117.5 1721057.5 1721119.5 1721119.5 1721059.5 1607557.6 1607557.5 1948440
q₁, g₁, b₁30, 5, 1 12, 1, 4 30, 5, 1 30, 5, 3 12, 1, 4 12, 19, 2 29, 13, 1 29, 11, 1
d, h, s1, 3, 2 1, 0, 0 1, 3, 2 1, 3, 2 1, 0, 0 1, 7, 13 1, 7, 3 1, 6, 5
q₂, g₂, b₂365, 4, 1 30, 48, 4 365, 100, 1 365, 400, 3 30, 4800, 4 29, 235, 2 354, 19, 3 354, 30, 1
d, h, s1, 1, 0 1, 28, 20 1, 25, 0 1, 100, 0 1, 2800, 2000 1, 125, 0 30, 7, 1 1, 11, 14
−2, 4, 40 −1, 4, 0 −2, 400, 4000 1, 4, 12
1, 1, 46 1, 1, 0 1, 100, 4600
−1, 4, 4792
1, 1, 4798
q₃, g₃, b₃ 36524, 4, 1
d, h, s 1, 1, 0
### 15.4. Julian Calendar
If you want to use this function for the Julian (1) calendar, then you should use the following function arguments:
1. q = [30, 365]
2. g = [5, 4]
3. b = [1, 1]
4. d = [1, 1]
5. h = [3, 1]
6. s = [2, 0]
q should contain the following elements: [ [ 1461, 4, 0 ], [ 153, 5, 4 ] ]. x[0,j] is then the calculation year number, x[1,j] is the month number within the calculation year (March is month number 0!), and x[2,j] is the day number within the month (the first day has day number 0!).
If you want to use the function for the Gregorian calendar, then q should contain: [ [ 146097, 4, 0 ], [ 36525, 100, 0 ], [ 153, 5, 4 ] ]. Then x[0,j] is the calculation century number, x[1,j] the calculation year within the calculation century, x[2,j] the month number within the calculation year (March has month number 0!), and x[3,j] the day number within the month (the first day has number 0!).
Here is pseudocode to calculate a CJDN from a Julian calendar date x:
function JUL2CJDN(x)
set q to [ [ 1461, 4, 0 ], [ 153, 5, 4 ] ]
for i from 0 through n−1:
set x[1,i] to x[1,i] − 3
set x[2,i] to x[2,i] − 1
if x[1,i] is less than 0:
set x[1,i] to x[1,i] + 12
set x[0,i] to x[0,i] − 1
set d to CAL2DAY(x,q)
for j from 0 through t−1:
set y[j] to y[j] + 1721118
return y
end function
Here is pseudocode to calculate a CJDN from a Gregorian calendar date c:
function GREG2CJDN(x)
set q to [ [ 146097, 4, 0 ], [ 36525, 100, 0 ], [ 153, 5, 4 ] ]
for i from 0 through n−1:
set x[1,i] to x[1,i] − 3
if x[1,i] is less than 0:
set x[1,i] to x[1,i] + 12
set x[0,i] to x[0,i] − 1
set b[0,i] to DIV[x[0,i],100]
set b[1,i] to x[0,i] − 100*b[0,i]
set b[2,i] to x[1,i]
set b[3,i] to x[2,i] − 1
set d to CAL2DAY(b,q)
for j from 0 through t−1:
set y[j] to y[j] + 1721120
return y
end function
Here is pseudocode to calculate Julian calendar dates from CJDN y:
function CJDN2JUL(y)
set q to [ [ 1461, 4, 0 ], [ 153, 5, 4 ] ]
for j from 0 through t−1:
set y[j] to y[j] − 1721118
set c to DAY2CAL(y,q)
for i from 0 through n−1:
set x[1,i] to x[1,i] + 3
set x[2,i] to x[2,i] + 1
if x[1,i] is greater than 12:
set x[1,i] to x[1,i] − 12
set x[0,i] to x[0,i] + 1
return x
end function
And here is pseudocode to calculate Gregorian calendar dates from CJDN y:
function CJDN2GREG(y)
set q to [ [ 146097, 4, 0 ], [ 36525, 100, 0 ], [ 153, 5, 4 ] ]
for j from 0 through t−1:
set y[j] to y[j] − 1721120
set b to DAY2CAL(y,q)
for i from 0 through n−1:
set x[0,i] to 100*b[0,i] + b[1,i]
set x[1,i] to b[2,i] + 3
set x[2,i] to b[3,i] + 1
if x[1,i] is greater than 12:
set x[1,i] to x[1,i] − 12
set x[0,i] to x[0,i] + 1
return x
end function
`
## 16. Fast Estimate of the Number of Days Between Two Dates
If you want to know how many days there are between two dates, then you can calculate that exactly by subtracting the CJDNs of the two dates. If you are satisfied with a possible error of a few days, then you can also estimate it from the difference measured in calendar years, calendar months, and calendar days.
Such information (the number of calendar years, months, and days) is not sufficient to calculate the exact number of days for all calendars, because not every year contains the same number of days as every other year, and not every month contains the same number of days as every other month (in all calendars).
Let's assume that the second date is $$j$$ calendar years and $$m$$ calendar months and $$d$$ calendar days later than the first date (both in the same calendar, the Gregorian Calendar or the Jewish Calendar or the administrative Islamic Calendar), where you just subtract the year number, month number, and day number of the second date from those of the first date (which must be later). The variable $$j$$ may be positive or zero, but not negative. The variables $$m$$ and $$d$$ may be positive or zero or negative, but $$m$$ must not correspond to more than one calendar year, and $$d$$ must not correspond to more than one calendar month. For the Jewish calendar we define Tishri to be month number 1 and define Adar Ⅱ to belong to month number 6 together with Adar Ⅰ (so that month number 6 may contain up to 59 days). For the Egyptian calendar we count the last 5 days of the year as a 13th month. Then you can estimate the number $$n$$ of days between those two dates using the formula
$$n = ⌊a j + b m + c d⌋$$
with $$a$$, $$b$$, and $$c$$ from the following table:
$${a}$$ $${b}$$ $${c}$$ $${|Δ|_\text{max}}$$ $${σ}$$ $${p_0}$$
Gregorian 365.24 30.4 1 4 1.1 35 %
Jewish 365.25 31.0 0.9 37 12.9 3 %
Islamic 354.367 29.5 1 2 0.6 62 %
Egyptian 365 30 1 0 0 100 %
The $$|Δ|_\text{max}$$ from the table shows the greatest error (measured in days) that the formula for that calendar yields (based on 10,000 random test dates) for dates that are not more than 400 years apart. The $$σ$$ shows the standard deviation (say, the average error, also measured in days). The $$p_0$$ shows the probability that the estimated number $$n$$ of days between the two dates is exactly right.
It is clear that the Egyptian calendar is the most predictable one, because it yields the least mistakes (namely zero). Then comes the administrative Islamic calendar, then the Gregorian calendar (with more variation in the month lengths than the Islamic calendar has), en lastly the Jewish calendar (with variation in the number of months in the year).
An example: How many days are there between 2003-05-25 and 2017-01-17 in the three calendars? Those dates differ by 14 years, −4 months, and −8 days. In the Gregorian calendar this yields $$n = ⌊365.24×14 + 30.4×−4 − 8⌋ = ⌊4983.76⌋ = 4983$$ days. The real number of days between those two dates is 4986.
In the Jewish calendar this yields $$n = ⌊365.25×14 + 31.0×−4 + 0.9×−8⌋ = ⌊4982.3⌋ = 4983$$ days, which happens to be equal to the real number of days between those two dates in the Jewish calendar.
In the administrative Islamic calendar this yields $$n = ⌊354.367×14 + 29.5×−4 − 8⌋ = ⌊4835.138⌋ = 4835$$ days, which is equal to the real number of days between those two dates in the administrative Islamic calendar.
In the Egyptian calendar this yields $$n = ⌊365×14 + 30×−4 − 8⌋ = ⌊4982⌋ = 4982$$ days, which is equal to the real number of days between those two dates in the Egyptian calendar.
languages: [en] [nl]
//aa.quae.nl/en/reken/juliaansedag.html;
Last updated: 2022-03-16 | 99,011 | 251,962 | {"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": 127, "equation": 149, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-27 | latest | en | 0.876465 |
http://www.srcmini.com/70307.html | 1,632,671,269,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057882.56/warc/CC-MAIN-20210926144658-20210926174658-00337.warc.gz | 118,243,127 | 14,440 | # Python逻辑运算符用法示例指南
## 逻辑运算符
and 逻辑AND：如果两个操作数都为true, 则为true x和y
or 逻辑或：如果任何一个操作数为true, 则为true x或y
not 逻辑非：如果操作数为假, 则为真 不是x
#### 逻辑AND运算子
# Python program to demonstrate
# logical and operator
a = 10
b = 10
c = - 10
if a> 0 and b> 0 :
print ( "The numbers are greater than 0" )
if a> 0 and b> 0 and c> 0 :
print ( "The numbers are greater than 0" )
else :
print ( "Atleast one number is not greater than 0" )
The numbers are greater than 0
Atleast one number is not greater than 0
# Python program to demonstrate
# logical and operator
a = 10
b = 12
c = 0
if a and b and c:
print ( "All the numbers have boolean value as True" )
else :
print ( "Atleast one number has boolean value as False" )
Atleast one number has boolean value as False
#### 逻辑或运算符
# Python program to demonstrate
# logical or operator
a = 10
b = - 10
c = 0
if a> 0 or b> 0 :
print ( "Either of the number is greater than 0" )
else :
print ( "No number is greater than 0" )
if b> 0 or c> 0 :
print ( "Either of the number is greater than 0" )
else :
print ( "No number is greater than 0" )
Either of the number is greater than 0
No number is greater than 0
# Python program to demonstrate
# logical and operator
a = 10
b = 12
c = 0
if a or b or c:
print ( "Atleast one number has boolean value as True" )
else :
print ( "All the numbers have boolean value as False" )
Atleast one number has boolean value as True
#### 逻辑非运算符
# Python program to demonstrate
# logical not operator
a = 10
if not a:
print ( "Boolean value of a is True" )
if not (a % 3 = = 0 or a % 5 = = 0 ):
print ( "10 is not divisible by either 3 or 5" )
else :
print ( "10 is divisible by either 3 or 5" )
10 is divisible by either 3 or 5
## 逻辑运算符的评估顺序
# Python program to demonstrate
# order of evaluation of logical
# operators
def order(x):
print ( "Method called for value:" , x)
return True if x> 0 else False
a = order
b = order
c = order
if a( - 1 ) or b( 5 ) or c( 10 ):
print ( "Atleast one of the number is positive" )
Method called for value: -1
Method called for value: 5
Atleast one of the number is positive
• 回顶 | 718 | 2,100 | {"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} | 3.84375 | 4 | CC-MAIN-2021-39 | latest | en | 0.838151 |
https://physics.stackexchange.com/questions/154019/why-isnt-length-contraction-permanent-even-though-time-dilation-is/154110#154110 | 1,638,608,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362952.24/warc/CC-MAIN-20211204063651-20211204093651-00286.warc.gz | 476,426,629 | 42,557 | # Why isn't length contraction permanent even though time dilation is?
It's my understanding that when something is going near the speed of light in reference to an observer, time dilation occurs and time goes slower for that fast-moving object. However, when that object goes back to "rest", it has genuinely aged compared to the observer. It's not like time goes slow for a while, and then speeds back to "normal," so that the age of the observer once again matches the object. The time dilation is permanent. Why wouldn't the same thing happen with length contraction? Since the two are so related, you'd think if one is permanent, the other would be also. And from everything I've read so far, length contraction is not permanent. An object will be at rest touching an observer, go far away near light speed, return to touching the observer, and be the same length it was at the beginning. It shortens, and then grows long again, as if its shrinkage was an illusion the whole time. Did I just not read the right things or what? Were my facts gathered incorrectly?
• The relative movement of time (which is what we call time dilation) is just as observer dependent as the relative distances in space. When the velocity between the observed object and the observer goes to zero, both effects disappear equally. The actual time dilation between two clocks in the same reference frame is zero, but because time itself is cumulative, we retain a memory of the relative state of motion in the eigentime of the object, i.e. in the total amount of time that has passed according to a clock in the object's frame of reference. Dec 19 '14 at 2:32
• Time dilation is not permanent. If you bring a moving clock to rest then it is no longer time dilated. Dec 19 '14 at 3:20
• An object will be at rest with a clock moving at the same speed as an observer, go far away near light speed, return to the observer, and the clocks will still be moving at the same speed. Dec 19 '14 at 5:25
• The length of space passed is in fact permanently contracted. Without loss of generality, consider a particle moving at C. For it the space passed contracted to 0 and the time passed was 0. If the space travelled were not permanently contracted, the particle would not be permanently at the new location after 0 time. Note that this does not imply that the distance can be traversed again in its contracted state. Dec 19 '14 at 12:42
• Or, to summarize things in an easy to remember way: The effect of length contraction has the same kind of permanency on your odometer as time dilation does on your clock. Dec 19 '14 at 14:56
Time dilation is a comparison of rates. When an object is moving fast with respect to you, it's clock rate is slow, and when it comes to rest with respect to you its clock rate returns to normal. The time difference between the two clocks at this time is due to the accumulation due to these different time rates. That is the leftover effect of the time dilation but not the time dilation itself.
Length contraction, like time dilation, exists when there is relative motion and goes away when there is no relative motion, but there isn't any "accumulation" with length contraction, so there is nothing to be "left over".
The way I see it, time dilation is the real effect here.
Length contraction (in SR) is just a consequence of the fact that the "length" of a rod is the distance between simultaneous positions of the rod's endpoints. But two observers with different velocities will have different ideas about what simultaneous is, and this means they measure different lengths.
• +1 Best answer. Time can be measured at a single location using a clock, but length is a measurement between two separate locations, which depends on simultaneity. Welcome to StackExchange! Dec 20 '14 at 8:19
It's my understanding that when something is going near the speed of light in reference to an observer, time dilation occurs and time goes slower for that fast-moving object.
According to the 'something', it is the observer's clock that runs slower and it is the observer's rulers that are contracted. That is to say, the time dilation and length contraction are symmetrical. Neither clock can be objectively said to be running slow and neither ruler can objectively said to be contracted.
However, when that object goes back to "rest",
Now the symmetry is lost; the object's accelerometer registered non-zero acceleration for some time while the observer remained inertial. This means that there is now an objective difference between the object (non-inertial) and observer (inertial) and, thus, an objective difference in elapsed times.
Why wouldn't the same thing happen with length contraction?
In fact, in SR, acceleration of an extended object must be handled with great care. If an object is to not to stretch or compress during acceleration, different parts of the object must have different (proper) acceleration.
Length contraction effects may be permanent in the same way as time dilation! You just have to choose the right example.
Example: An astronaut is traveling at v=0,99 c to an exoplanet, according to Earth frame he is traveling 198 light years in 200 years. According to his frame (reciprocal gamma = 0,141) he is traveling 27,9 light years in 28,2 years. After his arrival on the exoplanet, he is permanently younger than (and outliving) his twin brother on Earth, and he is permanently at a distance of 198 light years from Earth, a distance which he could never have traveled without length contraction.
• "A distance which he could never have traveled without length contraction" -- But there is a perfectly consistent analysis of this situation in the Earth/exoplanet rest frame which doesn't involve length contraction at all, just time dilation which causes him to age 28.2 years over the 200 years of coordinate time it took him to cross 198 light years of coordinate distance. Whereas with proper time, all frames agree that the the accelerated twin in the twin paradox aged less than the inertial one. Dec 19 '14 at 15:52
• @Hypnosifl - No, time dilation and length contraction are separate effects, each of them increasing the autonomy of the astronaut. You may not mix up Earth frame and astronaut's frame because the result would be a velocity faster than speed of light (198 light years in 28,2 years), t=28,2 years does not belong to the Earth frame. Dec 19 '14 at 16:37
• What statement of mine are you saying "no" to? I didn't claim they weren't separate effects, nor did I mix up different frames. I just said that the age of the astronaut on arriving at the exoplanet can be accounted for in different ways depending on what frame you use. If you use only the Earth/exoplanet frame, the coordinate distance is 198 l.y. and the coordinate time is 200 y, but the astronaut's clock only elapses 28.2 y due to time dilation. If you use only the astronaut frame, the coordinate distance is 27.9 l.y. due to length contraction, so he takes 28.2 y to get there. Dec 19 '14 at 16:41
There is actually an equivalent to "total elapsed proper time" along time-like curves in spacetime (which can represent the worldlines of particles moving slower than light), and that is the "proper distance" along a space-like curve (which cannot be any real particle's worldline). See the spacetime wikipedia article for more on time-like vs. space-like, particular the basic concepts section dealing with different types of intervals in special relativity, and the spacetime in general relativity generalizing that discussion.
The simplest physical interpretation of proper time on a time-like curve is just the total elapsed time on an ideal clock that has that curve as its worldline. But just as an arbitrary curve can be approximated as a polygonal shape consisting of a series of straight segments connected at their endpoints, so an arbitrary time-like curve can be approximated as a series of short inertial segments, which could represent bits of the worldlines of a bunch of different inertial clocks which cross paths with one another at the point the segments join. Then if you add the time elapsed by each inertial clock on each segment, this is approximately the proper time on the whole curve. Analogously, an arbitrary space-like curve can be approximated by a series of space-like segments, and the endpoints of each segment can be events at either end of a short inertial ruler which is moving at just the right velocity so that its plane of simultaneity is parallel to the the segment. Then the total proper distance is just the sum of the proper length of the rulers for all the segments. But this will probably only make sense if you already have a decent familiarity with spacetime diagrams in special relativity.
To give a mathematical example, suppose we are dealing with curves in SR which can be described in the coordinates of some inertial frame, and suppose the curves only vary their position along the x-axis so we can ignore the y and z space coordinates, and just describe the curves by some x(t) function. Then a time-like curve is one where $\frac{dx}{dt} < c$ everywhere, and a space-like curve is one where $\frac{dx}{dt} > c$ everywhere. If the time-like curve is approximated by a "polygonal" path made up of a series of inertial segments that each have a constant velocity $v$ over a time-interval $\Delta t$ in the inertial frame, then the elapsed proper time on each segment is $\sqrt{1 - v^2/c^2} \Delta t$ (this is just the time dilation equation), and the total proper time along the whole polygonal path is the sum or the proper time for each each segment. In the limit as the time-intervals become infinitesimal, this sum becomes an integral, and in this limit the error in the polygonal approximation goes to zero, so the actual proper time along the curve is $\int \sqrt{1 - v(t)^2/c^2} \, dt$.
Similarly, the space-like curve can be approximated by a polygonal path made up of a series of space-like segments whose endpoints have a spatial interval of $\Delta x$ between them, and with each segment having a constant value of $v^{\prime} = \frac{dx}{dt}$, where $v^{\prime} > c$. Each segment will be parallel to the simultaneity plane of a ruler moving at a slower-than-light speed $v = \frac{c^2}{v^{\prime}}$, and if the ruler's ends line up with the endpoints of the spacelike segment, that means the ruler has a contracted length of $\Delta x$ in the inertial frame we're using, which means the ruler's proper length is $\frac{1}{\sqrt{1 - v^2/c^2}} \Delta x$ (this is just the length contraction equation). So the total proper distance along the polygonal path is just the sum of the proper length for each ruler, and in the limit as the rulers' proper lengths become infinitesimal the sum becomes an integral and the error goes to zero, so the actual proper distance along the curve is $\int \frac{1}{\sqrt{1 - v(t)^2/c^2}} \, dx$.
So, you can see that in the first integral for proper time the factor in the integral is the same one that appears in the time dilation equation $dt_{proper} = \sqrt{1 - v^2/c^2} \, dt$, and in the second integral for proper distance the factor in the integral is the same one that appears in the length contraction equation $dx_{proper} = \frac{1}{\sqrt{1 - v^2/c^2}} \, dx$.
Putting CuriousOne's comment into an answer,
In the theory of relativity, time dilation is an actual difference of elapsed time between two events as measured by observers either moving relative to each other or differently situated from gravitational masses. Wikipedia
I see that such a definition might be misleading as it talks about time dilation in an "elapsed time" sense. Although I can't say it is technically wrong, perhaps a better way to understand it would be in terms of speed that time goes at for observers moving relative to each other.
Just like length contraction, time dilation, interpreted as difference of speed of time flow also disappears as the observers again come at rest relative to each other. But, the elapsed time is a cumulative quantity. That difference can't be restored. Total time or any such concept may not be covered by General Relativity or any such present theory, as far as my limited knowledge goes.
• Would this mean that time and space have different properties since time has a cumulative property that space apparently lacks? If so, why is there so much literature uniting them as space-time? I'm not trying to win an argument or anything, I'm just thoroughly confused. So far the only thing I've seen justifying space-time as a united thing is a four-dimensional coordinate system. Dec 19 '14 at 3:09
• Total time is more tied with entropy rather than space. Its flow however can be united with space into the 4-vector. Dec 19 '14 at 3:12
As you say, time dilation and length contraction occur when two frames of reference (observer and observed) travel at two different speeds. Both of these effects "go away" if the two frames of reference subsequently travel at the same speed; that is, time will pass at the same rate and two yardsticks will have the same length.
But the EFFECTS of these relativistic effects are permanent in BOTH cases. For time dilation it's easy to imagine (i.e. the "old twin" scenario that you mentioned). So here's an example for length contraction:
Imagine that there's an immense opaque disc between you (on Earth) and a big stellar nebula. The disc is big, but not so big that it completely obscures the nebula. Some of the photons coming from the nebula are blocked by the disc.
Now imagine that the same scenario, but the disc is travelling very fast tangentially to you and the nebula. In other words, it's moving across your field of vision. Now, at such immense distances, you won't be able to easily see the disc moving, but it WILL be length-contracted. So fewer photons from the nebula will be blocked by the disc, allowing you to see more of the nebula.
This is a permanent effect! Those extra photons that slipped by the foreshortened disc are now streaming out into the universe, interacting with things and hitting retinas (maybe yours) long after the disc slows down (relative to you).
Nothing that happens in the universe every really "goes away". I didn't even mention the increased mass of the disc, which will distort the paths of those photons and everything else around it. Every distortion is "permanent" in that respect.
• maybe it would be better to say, "Every distortion leaves permanent effects?" Dec 20 '14 at 19:54
• Good recommendation. I clarified my answer. Dec 22 '14 at 20:22
No, your facts weren't gathered incorrectly, your reasoning is just incorrect. It doesn't even take knowledge of physics to answer the question, just logical reasoning. (Don't take my language as a personal insult, I'm just trying to be clear.)
"It's not like time goes slow for a while, and then speeds back to "normal," so that the age of the observer once again matches the object."
Well, actually, yes, time does go back to "normal" when the moving clock comes back to rest. (All relative to the observer, of course). Once the clock comes to rest, it will once again tick at its normal rate, which is faster than the rate it ticked at when it was moving.
The clock will be behind, however, because it spent some time being a slowpoke. Your idea that once the clock begins to tick at its normal rate it will somehow "catch up" with the other clock is incorrect.
That's like saying if one marathon runner spends an hour walking, while his competitor runs, once he starts running again, he'll immediately catch up to the other guy. No, he'll be behind because of the time he spent walking while the other guy was running. Same idea with the clocks.
Time dilation does disappear as relative velocity approaches zero. The things experienced during the time experienced do not disappear; cells which have died remain dead and second hands which have ticked ahead do not reverse direction. To undo those things would require time reversal.
Sine we as humans only perceive time in one direction, time reversal is irrelevant: if an object is travelling in direction a at 1m/-s we would perceive and record it as travelling in direction -a at 1m/s, or direction a at -1m/s. We always record and perceive time as forward moving, but it can as easily be seen in the other direction.
Time dilation and length contraction are effects that occur when two observers move relative to each other. They both are zero when the observers are in the same inertial reference frame. While they exist, the effects are entirely reciprocal, so both observers see lengths contracted and time running slowly in the other frame. Neither is permanent- they both vanish if the observers stop moving relative to each other.
The permanent effect on the respective ages of the twins in the twin paradox is not a result of time dilation, but of the changing reference frames of the travelling twin, which shifts the traveller's plane of simultaneity. On both legs of the journey the travelling twin sees the stationary twin's clock running more slowly than their own.
Length contraction is caused by the two ends of a rod accelerating in different ways. If a one-meter rod starts moving to the right, with the left end and the right end accelerating in exactly the same way (that is, with the acceleration $$a(t)$$ the same for each end at each time $$t$$) then the rod does not contract (in its original rest frame). If it decelerates in the same way, then it does not change length at that end either.
To take an extreme but entirely illustrative example, if both ends of the rod instantaneously jump from speed 0 to speed .9, then the entire rod is going to travel at speed .9, and it is going (according to any observer on earth) to be the same length while traveling that it always was. If it decelerates just as uniformly, then when it stops it will still have that same length.
Things look different in the frame of the moving rod, because the accelerations of the two ends, if they are simulataneous in the earth frame, cannot be simulataneous in the moving-rod frame (and vice versa). If both ends accelerate simulataneously in the earth frame, then the right end accelerates before the left end in the moving frame, causing the rod to stretch in that frame. When it comes to a halt, if the deceleration is simulataneous at both ends in the earth frame, the left end decelerates before the right end, causing the rod to contract and return to its original length.
If the accelerations are such that the movng rod maintains its original length in the moving frame, then it must be shrunk in the earth frame. The ratio of the two lengths is always determined by the Lorentz factor, but the Lorentz factor does not tell you you anything at all about the ratio of either of these lengths to the length of the rod before it was moving.
So your question is based on a false premise. When a rod starts to move, it might (in the earth frame) either contract, expand, or maintain its original length. When it stops, the same is true. What happens depends on the details of the acceleration/deceleration. As to "why" it expands or contracts, the answer, in its entirety is that it expands or contracts because (and only if) one end starts moving before the other. Likewise when it stops.
You don't need relativity to see that this has to be what happens in the earth frame. You only need relativity to predict what happens in the moving-rod frame --- but your question is not about that!
Any change in time at all is only "permanent" because of the second law of thermodynamics and the resulting arrow of time.
• This should be the correct answer. In the spatial dimensions, we can invert the direction of propagation, in the time dimension we cannot. Imagine a simple 1-D dimensional spacetime. In special relativity, nothing holds us back to go back in time. Only when thermodynamics is added, does OP's problem happen. Oct 17 '19 at 22:51 | 4,393 | 20,024 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-49 | latest | en | 0.955278 |
https://www.tutorialcup.com/leetcode-solutions/bus-routes-leetcode-solution.htm | 1,657,166,405,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00184.warc.gz | 1,089,221,061 | 43,737 | Bus Routes Leetcode Solution
Difficulty Level Hard
Array Breadth First Search HashingViews 69
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Problem Statement
The Bus Routes LeetCode Solution – “Bus Routes” states that you’re given an array of routes where routes[i] is a bus route such that ith bus repeats the route forever. We’ll be given a bus stop source and we want to reach the bus stop target. We can travel between bus stops using buses only.
We need to find the least number of buses that are needed to travel from source to target. Return -1 if this is not possible.
Example:
Input: routes = [[1,2,7],[3,6,7]], source = 1, target = 6
Output: 2
Explanation:
• We can switch from the first bus to the second bus when we reach 7 with bus1, and then we will move to target = 6 from the second bus.
• Hence, our answer is 2.
Input: routes = [[7,12],[4,5,15],[6],[15,19],[9,12,13]], source = 15, target = 12
Output: -1
Explanation:
• There are no bus routes that can travel from source = 15 to target = 12, hence we return -1.
Approach
Idea:
1. The main idea to solve this problem is to use Breadth-First Search.
2. Store the mapping of each route[i] to the bus number and perform the breadth-first search for the buses.
3. We’ll start from the source route position and push all those routed that we can travel from this source into the queue. All such routes can be easily obtained with the help of adjacent lists.
4. Each time, pop the front element of the queue and iterate in the adjacency lists to find all the next routes that we can go from the current state with the count of a number of buses as a current number of buses + 1.
5. Finally, we’ll end up with the least number of buses required to reach the target route.
Code
Bus Routes Leetcode C++ Solution:
class Solution {
public:
const int N = 1e6;
int numBusesToDestination(vector<vector<int>>& routes, int source, int target) {
if(source==target){
return 0;
}
int n = routes.size();
for(int i=0;i<n;i++){
for(int j=0;j<routes[i].size();j++){
}
}
queue<pair<int,int>> q;
for(auto& route:routes[bus]){
q.push({route,1});
}
buses[bus] = true;
}
while(!q.empty()){
int route = q.front().first,amount = q.front().second;
q.pop();
if(route==target){
return amount;
}
if(!buses[bus]){
buses[bus] = true;
for(auto& x:routes[bus]){
q.push({x,amount+1});
}
}
}
}
return -1;
}
};
Bus Routes Leetcode Java Solution:
class Solution {
public int numBusesToDestination(int[][] routes, int S, int T) {
int n = routes.length;
HashMap<Integer, HashSet<Integer>> to_routes = new HashMap<>();
for (int i = 0; i < routes.length; ++i) {
for (int j : routes[i]) {
if (!to_routes.containsKey(j))
to_routes.put(j, new HashSet<Integer>());
}
}
Queue<int[]> bfs = new ArrayDeque();
bfs.offer(new int[] {S, 0});
HashSet<Integer> seen = new HashSet<>();
boolean[] seen_routes = new boolean[n];
while (!bfs.isEmpty()) {
int stop = bfs.peek()[0], bus = bfs.peek()[1];
bfs.poll();
if (stop == T) return bus;
for (int i : to_routes.get(stop)) {
if (seen_routes[i]) continue;
for (int j : routes[i]) {
if (!seen.contains(j)) {
bfs.offer(new int[] {j, bus + 1});
}
}
seen_routes[i] = true;
}
}
return -1;
}
}
Complexity Analysis for Bus Routes Leetcode Solution
Time Complexity
The time complexity of the above code is as the Breadth-first Search of the normal graph.
Space Complexity
The space complexity of the above code is O(N), for storing the adjacency lists, where N = max route seen.
Translate » | 984 | 3,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-27 | longest | en | 0.883419 |
https://getrevising.co.uk/revision-tests/hess_law_quiz | 1,529,282,297,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267859904.56/warc/CC-MAIN-20180617232711-20180618012711-00190.warc.gz | 609,120,721 | 13,445 | # Hess' Law Quiz
HideShow resource information
## 1. For a given process at constant pressure, enthalpy change is negative. This means that the process is _______.
• Endothermic
• Exothermic
• Equithermic
• A state function
1 of 5
## 2. Hess' Law states...
• if a reaction can take place by more than one route and the initial and final conditions are the same, the total enthalpy change is the same for each route
• if the reaction can take place by more than one route and the initial and final Hess' are the same, the total enthalpy change is the same for each route
• if the reaction can take place by more than one route and the initial and final temperatures are the same, the total enthalpy change is the same for each route
• if the reaction can take place by more than one route and the initial and final masses are the same, the total enthalpy change is the same for each route
## 3. Which of the following is an example of an exothermic situation?
• An ice cube melting in your hand
• A log burning in the fireplace
• A chemical cold pack placed over an injured knee
• True
• False
## 5. Hess' Law provides a method for finding an enthalpy change...
• More accurately
• Faster
• Indirectly
• Directly | 300 | 1,221 | {"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} | 2.578125 | 3 | CC-MAIN-2018-26 | latest | en | 0.882966 |
https://www.nagwa.com/en/videos/692187679745/ | 1,576,493,941,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541319511.97/warc/CC-MAIN-20191216093448-20191216121448-00344.warc.gz | 817,857,190 | 6,551 | # Video: Solving Problems Involving Adjacent Angles
Find the sum of the two adjacent angles from the given angles in the diagram.
03:31
### Video Transcript
Find the sum of the two adjacent angles from the given angles in the diagram.
In the diagram, we can see a whole turn. And it’s been split into six different angles. But there are only three of those six angles that we’re given. This angle here is labelled as 22 degrees. The size of this angle is 88 degrees. And the only other angle that we know is this one here, and it’s 64 degrees. The first keyword that we come across in our problem is the word “sum,” find the sum, although sometimes you might hear people using the words sum to mean calculation — I’m going to work out my sums. Really, we shouldn’t use it like this. It means something very definite in maths. And it means to add together. So to find the sum of something is to find the total.
So what do we need to find a total of? We need to find the sum of the two adjacent angles. And we’re told that these are the two adjacent angles from the given angles in the diagram. As we’ve just seen, we only have three given angles. So two of them must be adjacent. And we need to add those two adjacent angles together. Let’s remind ourselves what makes two angles adjacent. Well, the word adjacent means next door, too. But we can recognise adjacent angles because they have two things in common. Firstly, they share a common vertex. We know the vertex in an angle is the point from which the rays of the sides come from. When we have two angles that are adjacent, all the sides come from the same point.
Now, if we look at our diagram, we can see that the three labelled angles, in fact, all of the angles, do come from the same point. They all have a common vertex. So we can’t separate our three angles out that way. But as we’ve said already, there’s also something else that adjacent angles have in common. And this has got to do with the fact that they’re next door to each other. Adjacent angles also share a common side. In other words, one of the sides in the angles is shared between both angles. It’s part of them both. This is what makes them next door to each other or adjacent.
And if we remember this fact, we can spot which two angles are going to be adjacent out of the three that are labelled. It’s 64 and 88. They share a common vertex but also a common side. So to solve the problem, we need to find the sum of these adjacent angles. What is 64 plus 88? Four plus eight equals 12. Six tens plus eight tens equals 14 tens plus the 10 we’ve exchanged equals 15 tens. And because our answer is the total of two angles, we need to write it using a degrees symbol. Out of the given angles in the diagram, we know that the two adjacent angles are 64 degrees and 88 degrees. We know this because they share a common vertex and a common side. And so the sum of these two adjacent angles is 152 degrees. | 675 | 2,938 | {"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-2019-51 | longest | en | 0.94352 |
https://www.studypool.com/discuss/209878/how-many-weeks-would-it-take-gabe-to-pay-his-dad-back-with-the-doubling-payback?free | 1,481,431,491,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698544097.11/warc/CC-MAIN-20161202170904-00479-ip-10-31-129-80.ec2.internal.warc.gz | 1,013,907,555 | 13,886 | ##### how many weeks would it take gabe to pay his dad back with the doubling payback?
Mathematics Tutor: None Selected Time limit: 1 Day
on the first day he owes dad \$.1, on the second day \$.02, and on the third day \$.04, and so on for 14 days. How many weeks would it take Gabe to pay his dad back with the doubling payback?
Aug 14th, 2014
in order to pay back double the amount, it would take 15 days..........
Aug 14th, 2014
...
Aug 14th, 2014
...
Aug 14th, 2014
Dec 11th, 2016
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check_circle | 171 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2016-50 | longest | en | 0.95931 |
https://www.studymode.com/essays/Econ-151A-Midterm-1-1832034.html | 1,716,770,874,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058984.87/warc/CC-MAIN-20240526231446-20240527021446-00766.warc.gz | 883,700,164 | 15,291 | Econ 151a Midterm 1
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Econ 151a Midterm 1
Midterm #1 Exam Answer Key ECN 151A Summer 2013 Eschker (3pts each) Version A 1. Ans: E 2. Question deleted; Everyone gets 3 points 3. Ans: D 4. Ans: A 5. Ans: D 6. Ans: B 7. Ans: C 8. Ans: C 9. Ans: B 10. Ans: E Version B 1. Ans: C 2. Ans: A 3. Ans: B 4. Ans: D 5. Ans: E 6. Ans: C 7. Ans: B 8. Ans: E 9. Question deleted; Everyone gets 3 points 10. Ans: D
Part II. Answer the following. Show your work for all answers. 10 points each. 1. After watching the TV show American Idol, Blake thinks that people prefer younger performers over older performers. Maybe this is because younger singers have a “lighter” voice that sounds more pleasant. Thus, Blake thinks younger voters will receive more votes. To answer his question, he gets super-secret telephone voting data from the show and regresses the log of Votes for the singer on singer Age and obtains the following regression output (standard error in parentheses):
Log Votes = 5.0 - 0.03 Age (0.01)
R-squared=0.25
a. (3pts) Plot the Log Votes on the vertical axis and Age on the horizontal axis. Indicate vertical intercept and slope.
Age
b. (4pts) Interpret the value of the coefficient in the above output. Does the regression coefficient support Kathleen’s hypothesis? Be sure to consider tests of significance. = -0.03, which means that for every one year that Age rises, votes are 3% lower Yes, his hypothesis is supported, since the coefficient is negative, and he expected a negative relationship. The significance is also high, with a t-stat= -0.03/0.01 = 3 (absolute value), which is greater than 2. The R-squared is also somewhat low, however, but this is hard to interpret without comparison to another model specification
c. (3pts) After watching another season of American Idol, Blake now thinks that singing too long can hurt the voice and make the voice sound “harsh.” He believes that singers with too much experience will have damaged voices and will receive fewer votes.
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# hw3 - both 1 2 Find the most probable number of boys and...
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Homework 3 Probability Theory (MATH 235A, Fall 2007) 1. Zero-one law. Let A 1 ,A 2 ,... be independent events. Prove that the events lim sup n A n and lim inf n A n each have probabilities either zero or one. 2. Bounding random variables. Let X be a random variable. Prove that for every ε > 0 there exists K such that P ( | X | > K ) < ε. (Hint: Using Theorem 3.2, show that the probabilities of the events A n = {| X | > n } converge to zero as n → ∞ .) 3. Approximation by bounded random variables. Let X be a random variable. Show that for every ε > 0 there exists a bounded random variable X ε such that P ( X negationslash = X ε ) < ε. (Hint: use Problem 2). 4. Absolute value Show that if X is a random variable then so is | X | . 5. Family There are 4 children in a family. The probabilities of a boy or a girl are
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Unformatted text preview: both 1 / 2. Find the most probable number of boys and girls in the family, and the corresponding probability. 6. Composition of measurable maps. Prove that a composition of two measurable maps is measurable. 7. Random variables induce probability measures on R Let X be a random variable on a probability space (Ω , F , P ). Prove that X induces a probability measure on R in the following sense. For every Borel subset A of R , de±ne P ( A ) := P ( X ∈ A ) = P ( ω ∈ Ω : X ( ω ) ∈ A ) . Prove that ( R , R ,P ) is a probability space....
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Ask a homework question - tutors are online | 471 | 1,731 | {"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} | 3.84375 | 4 | CC-MAIN-2018-13 | latest | en | 0.853776 |
https://usaco.org/current/data/sol_prob1_gold_jan23.html | 1,716,463,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00213.warc.gz | 521,038,175 | 4,468 | (Analysis by Andi Qu)
A natural data structure to represent the string $S$ is a binary tree where each leaf node contains an individual character and an operation involves replacing some leaf nodes with subtrees. For instance, the tree corresponding to the example input would evolve like so:
This data structure is helpful for printing out the answer because we can easily and efficiently prune large branches off the tree until we're only left with what we want. We can do this by storing the size of each subtree in its root node.
However, the final string may be upwards of $2^{100000}$ characters long, so constructing the tree explicitly is out of the question. It helps to take a step back and ask, how can the string get so long? In this case, $S$ gets long when it contains many occurrences of one letter that get replaced with a longer string $s$. This results in $s$ appearing many times in $S$ and, as a result, many identical subtrees being appended to the tree.
Storing all of these identical subtrees explicitly would be extremely inefficient; a better idea would be to store only one instance of the subtree and then replace the affected leaves with pointers to that instance. We can illustrate this using the example input again:
This solves the problem of fitting everything into the available memory, but we still have another problem to deal with. Iterating through each occurrence of some letter in $S$ and pointing each to the new subtree is also quite inefficient and difficult to implement.
To address this problem, we can try to construct the final tree from the leaves up. To do this, we must iterate through the operations in reverse order, as strange as that may seem. Instead of starting with a single tree representing the string "$\texttt{a}$", we store $26$ trees. The $i$-th tree represents the final string if we started with only the $i$-th lowercase character. This allows us to merge large trees together efficiently when processing each operation.
Using these two ideas, we can construct an efficient, linear-time solution. The analysis of the runtime is as follows:
Constructing all the trees can be done in linear time because merging two trees happens in constant time, and we merge two trees each time we process a single character in the input. Let $M = r - l + 1$ and $N$ be the number of operations. When we traverse the binary tree, there are three categories of nodes we encounter:
1. Nodes whose entire subtree lies in $S_{l\dots r}$.
2. Nodes whose subtree partially contains $S_{l\dots r}$.
3. Nodes whose subtree doesn't contain any of $S_{l\dots r}$.
The number of nodes in a binary tree is equal to twice the number of leaves minus 1. This means the union of the subtrees of nodes in the first category contains at most $2M - 1$ nodes. The contribution from those nodes is then $\mathcal O(M)$. There are at most 2 nodes in the second category at each depth of the tree, and the tree has depth $\mathcal O(N)$. Similarly, there are at most 2 nodes in the third category at each depth of the tree. The contribution from those nodes is then $\mathcal O(N)$. The overall runtime is, therefore, $\mathcal O(N + M)$.
To prevent integer overflow in languages like C++ and Java, we need to cap subtree sizes at $10^{18}$. This works because we never care about the contents of $S$ beyond the $10^{18}$-th character.
My C++ code:
#include <algorithm>
#include <iostream>
#include <string>
#include <utility>
typedef long long ll;
using namespace std;
const ll INF = 1e18;
struct Node {
char value;
ll size;
Node *l, *r;
void print_substring(ll start, ll end) {
start = max(start, 1ll);
end = min(end, size);
if (start > end) {
return;
}
if (value != '.') {
cout << value;
} else {
l->print_substring(start, end);
r->print_substring(start - l->size, end - l->size);
}
}
};
Node* current[26];
pair<char, string> operations[200000];
int main() {
cin.tie(0)->sync_with_stdio(0);
ll l, r;
int n;
cin >> l >> r >> n;
for (int i = n - 1; i >= 0; i--) {
cin >> operations[i].first >> operations[i].second;
}
for (char c = 'a'; c <= 'z'; c++) {
current[c - 'a'] = new Node{c, 1};
}
for (int i = 0; i < n; i++) {
Node* result = nullptr;
for (char c : operations[i].second) {
Node* to_merge = current[c - 'a'];
if (result == nullptr) {
result = to_merge;
} else {
result = new Node{
'.',
min(INF, result->size + to_merge->size),
result,
to_merge
};
}
}
current[operations[i].first - 'a'] = result;
}
current[0]->print_substring(l, r);
cout << '\n';
return 0;
}
Note that it's very important that leaf nodes only contain individual letters and not empty strings. Changing the one line
Node* result = nullptr;
to
Node* result = new Node();
results in an $\mathcal O(NM)$ solution instead because the empty string gets repeated many times in $S$ but doesn't count toward its length.
Danny Mittal's Java code:
import java.io.BufferedReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;
public class FindAndReplace {
public static final long MAXVAL = 1000000000000000000L;
public static void main(String[] args) throws IOException {
long l = Long.parseLong(tokenizer.nextToken());
long r = Long.parseLong(tokenizer.nextToken());
int amtOperations = Integer.parseInt(tokenizer.nextToken());
List<Operation> operations = new ArrayList<>();
for (; amtOperations > 0; amtOperations--) {
char before = tokenizer.nextToken().charAt(0);
char[] after = tokenizer.nextToken().toCharArray();
}
BigString[] currs = new BigString[26];
for (char chara = 'a'; chara <= 'z'; chara++) {
currs[chara - 'a'] = new BigString(true, chara, null, 1);
}
Collections.reverse(operations);
for (Operation operation : operations) {
if (operation.after.length == 1) {
currs[operation.before - 'a'] = currs[operation.after[0] - 'a'];
} else {
BigString[] elements = new BigString[operation.after.length];
long length = 0;
for (int j = 0; j < elements.length; j++) {
elements[j] = currs[operation.after[j] - 'a'];
length += elements[j].length;
length = Math.min(length, MAXVAL);
}
currs[operation.before - 'a'] = new BigString(false, '\0', elements, length);
}
}
StringBuilder out = new StringBuilder();
currs[0].append(l, r, out);
System.out.println(out);
}
static class Operation {
final char before;
final char[] after;
Operation(char before, char[] after) {
this.before = before;
this.after = after;
}
}
static class BigString {
final boolean isSingleton;
final char chara;
final BigString[] elements;
final long length;
BigString(boolean isSingleton, char chara, BigString[] elements, long length) {
this.isSingleton = isSingleton;
this.chara = chara;
this.elements = elements;
this.length = length;
}
void append(long from, long to, StringBuilder builder) {
from = Math.max(from, 1);
to = Math.min(to, length);
if (from <= to) {
if (isSingleton) {
builder.append(chara);
} else {
long curr = 0;
for (BigString element : elements) {
element.append(from - curr, to - curr, builder);
curr += element.length;
curr = Math.min(curr, MAXVAL);
}
}
}
}
}
} | 1,743 | 7,088 | {"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} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.936691 |
http://www.jiskha.com/display.cgi?id=1201488140 | 1,495,947,453,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609598.5/warc/CC-MAIN-20170528043426-20170528063426-00083.warc.gz | 684,680,493 | 4,806 | # science
posted by on .
I did this experiment and I found these validity issues: they never specified if temperature to be measured in celcius or faren., they never mentioned that the buckets were to be the same, stirring until your arms get tired is not a good idea, but I need three more validity issues?????
DO you consider the companys conclusion valid? How could them improve?
To prevent future brownfield sites the DEP's concer lies in the fact that the chemical in the wastewater eventually reaches natural waterways and affects drinking water. It decided to test the temperatureeffects of a chemial to the natural water that could be contaminating.
Their experiment was based on this statement: We are trying to determine if adding chamiall X to stream water causes the stream water to rise in temperature.
HEre is the procedure:
1. Take two buckets and fill each one about halfway with tap water.
2. Record the temperature of each bucket of water.
3. Put a shovel full of chemical into bucket number 1 and stir until arms get tired. Record the temperature.
4. Put 2 shovel fulls in bucket number two and record temperature.
So it again for accurate results.
Graph and make conclusion
In one bucket the temperature increased from 15 to 35 degrees and in the other bucket the temperature increased from 15 to 37 degrees. They concluded the chemical was responsible for temperature rise.
• science - ,
In 1, not only do they not specify the size of the buckets BUT they also say "FILL ABOUT HALFWAY WITH TAP WATER." The exact amount of water should be stated. Second, is tap water the same as "stream water."
In 2. T in C or F you have.
In 3. A shovel full of chemical is not precise enough. The researcher should specify the amount of chemical added. The researcher should specify over what period of time the chemical is added. Also, a specific amount of time should be given. Stirring until arms get tired is not good enough since that would vary depending upon whose arms we were using.
I would add one more to the steps above. Since the initial statement was that they were trying to determine if chemical X caused a rise in the termperature of stream water, my answer would be, "So what?" Whether the chemical raises the temperature of stream water has nothing to do with the wastewater eventually reaching streams. Why? One could argue that even if adding a chemical to waste water raised the temperature of the wastewater, letting that wastewater cool ( in air) before dumping it into a stream would not raise the temperature of the stream. My point in the last argument is that this experiment will NOT tell them anything they want to know. | 556 | 2,666 | {"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-2017-22 | latest | en | 0.947072 |
https://courses.cs.northwestern.edu/325/readings/pat-match.html | 1,721,344,246,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00591.warc.gz | 159,629,043 | 3,579 | ## Overview
Matching (of all kinds) is used for
• inferencing (if A implies B, and C matches A, then B)
• classification (A matches B therefore A is a kind of B)
• similarity judgment (A partially matches B, so A is similar to B)
• case-based reasoning (A is the best partial match to B that can be found in memory, so use the information attached to A when processing B)
What can be matched?
• patterns -- can match many different instances or, with matchers called unifiers, can match other patterns
• instances (or constants) -- can match other instances, albeit often only partially
Graham gives a simple but not very well-designed unifier. A better version is covered in the reading on deductive data retrieval.
Here we're talking about a simpler pattern matcher that takes one pattern and one object that is not a pattern. A pattern can be
• an atom, like 12 or A
• a variable, which is any symbol beginning with a question mark, e.g., `?X` or ?PERSON
• a list of patterns
Note the recursive nature of this definition. Here are some patterns:
```(like mary jon)
(like ?x ?y)
(?fn ?arg1 ?arg2)
(?fn (null ?x))```
Here's what matches what and why:
PatternInputMatches?
`mary` `mary` Yes, they're equal
`(like mary jon)` `(like mary jon)` Yes, they're equal
`(like mary jon)` `(like jon mary)` No, they're not equal
`(like ?x ?y)` `(like jon mary)` Yes, and `?x` is `jon` and `?y` is `mary`
`(like ?x ?x)` `(like jon mary)` No, `?x` can't be `jon` and `mary` at the same time
`(like ?x ?x)` `(like jon jon)` Yes, `?x` is `jon`
`(like ?x ?y)` `(like jon jon)` Yes, `?x` and `?y` are both `jon`
## Basic Pattern Match Rules
The basic rule for pattern matching is simple. Given a pattern P and an S-expression S, P matches S if
• P is equal to S, or
• P is a variable that can be "bound" to S
• P and S are both lists and their corresponding CAR's and CDR's match
A variable P can be "bound" to S if
• P has never been matched against anything before, or
• P was previously matched against something that matches S
The matcher returns
• `NIL` if the match fails
• a list of the list of bindings if the match succeeds
## Binding Lists
A binding list is simply a list of the variables in a pattern, paired with the items those variables matched in the input S-expression. For example, the binding list
`((?x . jon) (?y . mary))`
says that the variable `?x` is bound to `jon` and the variable `?y` is bound to `mary`.
Why does the pattern matcher return a list of binding lists? For two reasons:
• to distinguish successful matches with no bindings in a simple fashion
• to allow pattern matches to return more than one binding list
The "success but no bindings" problem: Consider matching `(likes mary jon)`with `(likes mary jon)`. This clearly matches but equally clearly generates no bindings. If our matcher returned just a binding list, it would return `NIL` in this case, but that looks like the match failed.
The multiple bindings problem: Multiple binding lists can't arise in our simple matcher, but they arise as soon as we make any extensions, such as "match ?x to A or B". Or, for a more complex example, assume the pattern `(A ?* C)` matches any list that starts with A and ends with C, where `?X` matches zero or more intermediate elements. Here are some different examples of using `?*`:
PatternInputBindings
`(like ?* ?x)` `(like mary jon)` `?x` is bound to `jon`
`(like ?x ?*)` `(like mary jon)` `?x` is bound to `mary`
`(like ?* ?x ?*)` `(like mary jon)` `?x` is bound to `mary` OR
`?x` is bound to `jon`
The last case is the important one. There are two equally valid bindings.
### Other approaches
Many authors treat the "success but no bindings" case specially and return `T`. While intuitive, this means that every function that calls the pattern matcher, including the internal recursive calls of the matcher itself, have to check for three possible values: `NIL`, `T`, or a list. The simple recursive calling pattern found in match.lisp is lost when you do this. Furthermore, it doesn't handle multiple binding lists.
Other authors, including Graham in Chapter 15, return multiple values: the binding list and a flag indicating whether the match succeeded or failed. Again, you lose the simple recursive calling pattern and you still don't handle multiple binding lists.
Our matcher, while less intuitive at first glance, has a very simple semantics for its return value:
Matching returns a list of all possible binding lists.
If the list is `NIL`, there were no possible binding lists and the match must have failed.
If the list is `(NIL)`, that's the list of one binding list which happens to be the empty binding list.
There are no special cases with this approach. All code that calls the matcher can simply iterate through the list returned (which will usually have 0 or 1 lists in it) to process each binding list. No special checks. | 1,207 | 4,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.902205 |
http://sabrinamanin.it/decaying-exponential-fit-matlab.html | 1,606,967,522,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141718314.68/warc/CC-MAIN-20201203031111-20201203061111-00011.warc.gz | 61,962,438 | 22,288 | ## Decaying Exponential Fit Matlab
When b is negative, an exponential function decreases, flattening out as it approaches the t axis. Half-life calculation by non-linear least squares curvefitting In order to improve the statistical rigor and accuracy of the half-life calculation, the data for each RNA was fitted to an exponential function (of the form A = A 0 e kt) using a non-linear least squares algorithm implented in MATLAB (function nlinfit in the statistics toolbox which uses the Gauss-Newton method). Exponential curve fitting matlab Exponential curve fitting matlab. e forexponential", it computes the exponentially weighted moving average. Includes: Graphs of linear functions|Domain and range|Quadratic equations|Exponential growth and decay|Direct and inverse variation|Systems of linear inequalities. DA: 30 PA: 77 MOZ Rank: 12. Gaussian curve fitting to ADC histograms (11) was performed with the MATLAB curve-fitting tool (Method: nonlinear least squares. Learn how to solve exponential equations including growth and The math form f(x) = ax defines Exponential Functions of base a if a is a positive real number. Smoothing Data Points The Curve After you enable Flash, refresh this page and the presentation should play. Thanks for responding. the equation used has two double exponential parameters and other 2 parameters. Question: This Is A Matlab Exercise. In other words, the formula gives recent prices more weight than past prices. The three types of cavities are shown here. This is the basic introduction to Matlab. 5 * params[1] * np. Matlab Online gives you complete knowledge about Matlab, which will be useful while implementing your project. If you're seeing this message, it means we're having trouble loading external resources on our website. 1 Recommendation. In the spectral domain, the harmonic peaks which can represent most of the energy of BPW spectrum were used to fit the exponential decay function in the following: (1) P = A e α f where P is the harmonic peak value of BPW spectrum, A is the coefficient of the exponential decay function, α (< 0) is the exponent of the exponential decay. Just click the green Download button above to start. Example 4: The Pareto distribution has been used in economics as a model for a density function with a slowly decaying tail = λxe−λ x!. I want to curve fit a set of data points using an exponential decay function. Robust: On. The general […]. The remaining IVIM component was fit to the standard mono-exponential IVIM model (1), ( ) ( ), [3] where. Select the type of fit from the STAT CALC menu ii. For c) and d). The doc string for it even has an example of fitting an exponential decay in it which I'll copy here: >>> import numpy as np >>> from scipy. Fitting to a model of the type,. Shown below is the result from a python program using Padé-Laplace to curve-fit a noisy 3-exponential decay with decay constants 5, 1, and 0. The lifetime -of the disturbed donor was obtained using a two-exponential model for the fit performed on N2A cells transfected with EGFP and mRFP. Kurtosis parameter characterizing the deviation from the exponential decay. During the 1980s the population of a certain city went from 100,000 to 205,000. In this chapter, a will always be a positive number. of 10 K (Figure S2). For over 30 years, NCSS, LLC has been dedicated to providing researchers, investigators, academics, scientists, and other professionals with quality statistical software that is comprehensive and accurate but still intuitive and easy to use. This example fits two poorly resolved Gaussian peaks on a decaying exponential background using a general (nonlinear) custom model. An exponential continuous random variable. Before look at the problems, if you like to learn about exponential growth and decay, Please click here. Smooth surface cavities occur on the smooth sides of your teeth, while root cavities develop on the surface over the roots. To fit a functional form (1) take the logarithm of both sides (2) The best-fit values are then (3) (4) where and. I just want fo fit an exponential curve to my XY data. For example:. Ich habe KEINE Toolbox für Kurvenanpassung oder -optimierung. In a number of studies, it has been observed that diffusion signal decay in breast tumors is better fitted by bi-exponential function with smaller residual errors (14-16). I thing that will work. The relative predictive power of an exponential model is denoted by R^2. I want to plot an exponential function that approaches 1, then decays to zero. In this lesson you will learn how to write an exponential equation by finding a pattern in a table. Hi, I am trying to do a custom fit in matlab's curve fitting toolbox, but am having some problems. Requirements Be familiar with fundamental MATLAB concepts such as if statements, looping, logicals,user-defined functions, matrices Must have access to MATLAB - student versions are avaialble at the MathWorks website Knowledge of calculus topics like derivatives, polynomials, and integration. Choose one or two terms to fit exp1 or exp2. Diktat Kuliah Pak Sujantoko ST. Y = [iy1, iy2, iy3, x. For each additional decay mode, you add another exponential term to the model. You feed sigmaplot10 with it and using the simple equation y=a*exp(-bx)+c*exp(-dx) it gives you a nice fit. Exponential decay extrap to y=0. Ewma Matlab - mrn. Compared to the normal, it has a stronger peak, more rapid decay, and heavier tails. Noting that we are only fitting two variables, and starting from the beginning: M = ¦ ¸ ¹. Exponential smoothing schemes weight past observations using exponentially decreasing weights This is a very popular scheme to produce a smoothed Time Series. I've used this resource here as a base for building my program. Spline Fitting. curve_fit function. Please nd the MLE of the parameter λ. See full list on ece. (Optional) Click Fit Options to specify coefficient starting values and constraint bounds appropriate for your data, or change algorithm settings. This file was created by the Typo3 extension sevenpack version 0. x축은 t1, y 축은 P1이라는 변수로 선언을 합니다. dexp gives the density, pexp gives the distribution function, qexp gives the quantile function, and rexp generates random deviates. Fitting a curve in the form y = Ae^(-t/tau) (Equation 2) is simple, because the equation can be linearized and you can use linear least squares to find A and tau. Batch gradient descent also doesn't allow us to update our model online, i. Volume 13 | Number 3 | 2013 Lab on a Chip Pages 313–476 1473-0197(2013)13:3;1-2 ISSN 1473-0197 Lab on a Chip Miniaturisation for chemistry, physics, biology, materials science and bioengineering. The integrated differential equation for the process that created your data would require values for all three parameters. Inverse application: scheme 1 with a fitting function that is a polynomial is 1/x. The behavior of the absorbed power is exponentially decaying sinusoidal wave. Wolfram Community forum discussion about Use FindFit with an exponential function?. For example, this is an exponential decay of the form: a*(1-2*exp(-x/b)). In the exponential growth of f(x), the function doubles every time you add one to its input x. e forexponential", it computes the exponentially weighted moving average. The Matlab command randn generates samples of a Gaussian distributed random variable with mean 0 and variance 1. We have started our service. • Removing a trend • ts=detrend(ts); • subtracts best fitting line • detrend can be used to subtract mean: detrend(ts,'constant'). I thought it should work with my old code, but apparently, I am doing something wrong, but I don't see my mistake Excel retuns an exponential function of 150e-0. 5*x) + 4*exp (-3*x) + 2*exp (-2*x); % calculate integrals. Activity of a sample of Barium 137 was measured as a function of time, and the results are shown in file 'decay. MATLAB: An Introduction with Applications, 3rd Edition. Matlab Code For Logistic Growth Model. Nieuwe en tweedehands goederen, auto's en diensten, kopen en verkopen op Marktplaats. The model that resulted in the lowest overall least squares fit was two exponentials as given by (1) where A is the amplitude and T. Its domain includes complex numbers. A Gaussian minus exponential distribution has been suggested for modelling option prices. As we need to calculate the gradients for the whole dataset to perform just one update, batch gradient descent can be very slow and is intractable for datasets that don't fit in memory. The double exponential is a symmetric distribution. outer(x, y)**2). I'd like to plot a decaying exponential function similar to this plot, while being able to vary the spike time and rate of decay as well as the amplitude. Before we dive into algorithms and problem solving, we need to gain some basic familiarity with MATLAB. A demonstration of several scatter point styles. of Exponential Decay. 먼저 위와 같은 코드를 통해 데이터를 불러오겠습니다. For real numbers c and d, a function of the form. optimize import curve_fit >>> def func(x, a, b, c):. Bi-T 2* decay was found in some, but not all, selected pixels: 32 out of 110 pixels (29%) were determined to have a bi-T 2* decay. Matlab Fit - cocr. I'm looking for best methods to fit experimental data to a series of exponentials, either 3 or 4 depending on sample temperature. In this chapter, a will always be a positive number. There might even be an excellent open-source software library that can't be included in Matlab, but can be used within Matlab. outer(x, y)**2). DOEpatents. We also computed the value of the exponential function at x = 1. Kurtosis parameter characterizing the deviation from the exponential decay. # Import modules import numpy as np # Exponential Recovery xdata = np. How do you fit non-negative exponential decay that is biased with non-uniform noise over time? If anyone can clarify John's last response or help in replying to my last comment, I would greatly appreciate it. 15-06-0302-02-003c Submission Sheet3. Thanks everyone in advance. The equation for the. Sometime it is useful to take data from a real life situation and plot the points on a graph. In this % case it is recommended that the user either modify the code to % add parameters for the background (e. Exponentials are often used when the rate of change of a quantity is proportional to the initial amount of the quantity. The theory needed to understand this lecture is explained in the lecture entitled Maximum likelihood. The inverse of an exponential function is a logarithm function. The number b is often called the base. Least squares fit is a method of determining the best curve to fit a set of points. We exploit the fminsearch function in Matlab to fit essential parameters for curve fitting, namely d (maximum kill rate) and r (tumor growth rate). Here are two ways to do this. Instructions: Use this step-by-step Exponential Function Calculator, to find the function that describe the exponential function that passes through two given points in the plane XY. MATLAB Builder EX (for Microsoft Excel) MATLAB Builder JA (for Java language) MATLAB Builder NE (for Microsoft. The following estimators have built-in variable selection fitting procedures, but any estimator using a L1 or elastic-net penalty also performs variable selection: typically SGDRegressor or SGDClassifier with an appropriate penalty. If two decay modes exist, then you must use the two-term exponential model. A method for estimating modal decay parameters in the presence of noise is given in [118,220,221]. To construct an Optimizer you have to give it an iterable containing the parameters (all should be Variable s) to optimize. Curve Fitting in Matlab. Fit to exponential decay Suppose you have data, and want to fit a model of the form yi = a1 + a2 exp (– a3xi) + εi. However, someone might guess on the basis of previous knowledge of the system under study. 905 X), that is, a = 0. the equation used has two double exponential parameters and other 2 parameters. It does not attempt to model market conditional heteroskedasticity any more than UWMA does. Students who have not used MATLAB before should go to the. Push ZOOM and ZoomStat to see the graph of the data h. Fit Zwei-Term-Exponential-Modell in Matlab - Matlab, exponentiell. ELUs tested on CIFAR-10 and CIFAR-100. 959 and b = -0. For this example, the nonlinear function is the standard exponential decay curve y ( t ) = A exp ( - λ t ) , where y ( t ) is the response at time t , and A and λ are the parameters to fit. %----- % Demonstrate nonlinear regression -- quick & dirty example with "fminsearch" % Fit y=alpha*exp(beta*x) % Programming Note: call "fminsearch" % Instructor: Nam. % Uses fitnlm() to fit a non-linear model (an exponential decay curve, Y = a * exp(-b*x)) through noisy data. Exponential. Volume 13 | Number 3 | 2013 Lab on a Chip Pages 313–476 1473-0197(2013)13:3;1-2 ISSN 1473-0197 Lab on a Chip Miniaturisation for chemistry, physics, biology, materials science and bioengineering. There might even be an excellent open-source software library that can't be included in Matlab, but can be used within Matlab. Compile the source into a code or AST object. If you had a straight line, then n=1, and the equation would be: f(x) = a0x + a1. It's not perfect because the exponential envelope doesn't have a continuous gradient at 0 (i. f(x) = c x,. We will discuss in this lesson three of the most common applications: population growth, exponential decay, and compound interest. How to fit an exponential decay to find damping Learn more about exponential decay, nonlinear square least. I've used this resource here as a base for building my program. It has the mathematical form lr = lr0 * e^(−kt), where lr, k are hyperparameters and t is the iteration number. After measuring the voltage and time data points of the exponential decay in charge experienced by the capacitor bank, Matlab is used to t the data. MATLAB, the language of technical computing, is a programming environment for algorithm development, data analysis, visualization, and numeric computation. Next: Exponential growth and decay Up: Background Previous: Background. Fit the data using this equation. Matlab has a function called polyfit. txt) or view presentation slides online. After measuring the voltage and time data points of the exponential decay in charge experienced by the capacitor bank, Matlab is used to t the data. Thanks for responding. Combine multiple words with dashes(-), and seperate tags You must enter a body with at least 15 characters. Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. Exponential Line Of Best Fit Calculator. (power-like) than an exponential decay. Cubic splines are a common class of functions which are used in many interpolation applications. doc 2 of 2 e. call: par = T2fit(x,y); [par,delta]=gaussfit(x,y); [par,delta]=gaussfit(x,y,p); [par,delta]=gaussfit(x,y,p,out); par = fitted parameters like p delta = error estimate (optional) in abs. Example of MATLAB Exponential Function. For f(x) in the previous example, the function doubles every time we add to x. Step 3, Save the script. Carrier recombination dynamics exhibit both single exponential (dashed bold lines, fit) and bi-exponential (solid bold lines, fit) decay from different spatial regions. The behavior of the absorbed power is exponentially decaying sinusoidal wave. of the form: A + B e ^ {C x}. Measuring rates of decay Mean lifetime. The function approaches 1 but does not decay. ilcanzonieredisansevero. We'll just look at the simplest possible example of this. That means it will take that long for 100 grams of uranium-238 to turn into 50 grams of uranium-238 (the other 50 grams will have turned into another element). The bi-exponential fitting produced a short T 2* of 5ms and a long T 2* of 34. Fitting Exponential Decay. Extend your MATLAB program to fit this resulting data to an exponential function for the decay of material 2. It concentrates on Matlab fundamentals and gives examples of its application to a wide range of current bioengineering problems in computational biology, molecular biology, bio-kinetics, biomedicine, bioinformatics, and. This method is useful because it is straightforward and does not require a computer,. It has literally hundreds of built-in functions for a wide variety of computations and many toolboxes designed for specific research disciplines, including statistics. The code generates xdata from 100 independent samples of an exponential distribution with mean 2. The equation for a polynomial line is: Here, the coefficients are the a0, a1, and so on. To run this Monte Carlo study you will need to download the OnScale and MATLAB files. This example shows how to perform noncompartmental analysis to calculate NCA parameters and estimate the tumor growth model parameters from experimental data using nonlinear regression in the SimBiology Model Analyzer app. This is a feature of exponential functions, indicating how fast they grow or decay. It provides command-line functions and a basic graphical user interface for interactive selection of the data. Word Problems with Exponential Functions How to solve word problems involving exponential functions? Examples: 1. It's complete Iso offline installer of MathWorks MATLAB R2019b Download Tested 100% working link. I know you have said that MATLAB 2013b run ~10 times slower than with previous releases. That is it represents 5. Therefore, the equation for the exponential curve of best fit through the given points is: y = 2. e forexponential", it computes the exponentially weighted moving average. The code generates xdata from 100 independent samples of an exponential distribution with mean 2. Derived Parameters. In mathematics, an exponential function is a function of the form. Random Exponential Sequence. Below is a section of matlab code which performs linear regression: function [slope,offset] = fitline(x,y); %FITLINE fit line 'y = slope * x + offset' % to column vectors x and y. The integrated differential equation for the process that created your data would require values for all three parameters. Use exp for the element-by-element exponential. Therefore, the equation for the exponential curve of best fit through the given points is: y = 2. We can use the command >> x= 0:0. To test an example execute in Matlab: Example_1. 25+ years serving the scientific and engineering community Log In Watch Videos Try Origin for Free Buy. Solar system fault detection. However, problems arise when using standard algorithms to fit this function: we have observed that different initializations result in distinct fitted parameters. In order to compute this information using just MATLAB, you need to do a lot of typing. An exponential model is a mathematical model in which the variable is in the exponent. The exponential can be used to model lifetimes or the time between unlikely events. MATLAB still enforces the rules of linear algebra so paying attention to the details of vector creation and manipulation is always. But I think I did not frame my question well. Many important systems follow exponential patterns of growth and decay. Fitting Exponential Decay Exponential decay is a very common process. Example 4: The Pareto distribution has been used in economics as a model for a density function with a slowly decaying tail = λxe−λ x!. Fit decays using a Gaussian distribution of donor-acceptor (D-A) distances. Word Problems with Exponential Functions How to solve word problems involving exponential functions? Examples: 1. % The code does nonlinear least-square fitting with % covariance matrix to the exponential 'a*exp(-(x/b))', % given following things: % X=XDATA=control parameter, N. Exponential time decay and laser scanning instrument. Aug 28, 2020 computing qualitatively correct approximations of balance laws exponential fit well balanced and asymptotic preserving sema simai springer series Posted By Karl MayLtd. I have extracted data from a florescence decay graph. Half-life calculation by non-linear least squares curvefitting In order to improve the statistical rigor and accuracy of the half-life calculation, the data for each RNA was fitted to an exponential function (of the form A = A 0 e kt) using a non-linear least squares algorithm implented in MATLAB (function nlinfit in the statistics toolbox which uses the Gauss-Newton method). An exponential function f is defined by. Exponential decay, on the other hand, is a similar idea, but formed around C e − k t instead, for some constants c and k. • In Excel, you can create an XY (Scatter) chart and add a best-fit “trendline” based on the exponential function. you're not simply seeking an optimal superposition. Jan 08, 2017 · Performing a Chi-Squared Goodness of Fit Test in Python. Include in your ticket your institutional affiliation and a brief statement confirming that you will use MATLAB only for non-commercial, academic purposes. To construct an Optimizer you have to give it an iterable containing the parameters (all should be Variable s) to optimize. T he exponent x is any real number and f is called an exponential function. If the coefficient associated with b and/or d is negative, y represents exponential decay. The exponential can be used to model lifetimes or the time between unlikely events. Since the data usually has measurement errors, the measured data from an exponential decay will usually contain an error term. Requirements Be familiar with fundamental MATLAB concepts such as if statements, looping, logicals,user-defined functions, matrices Must have access to MATLAB - student versions are avaialble at the MathWorks website Knowledge of calculus topics like derivatives, polynomials, and integration. • In Excel, you can create an XY (Scatter) chart and add a best-fit "trendline" based on the exponential function. Another common schedule is exponential decay. Matlab has two functions, polyfit and polyval, which can quickly and easily fit a set of data points with a polynomial. Matlab package, which is available on NuGet as separate package and not included in the basic distribution. It does not attempt to model market conditional heteroskedasticity any more than UWMA does. The formulas below are used in calculations involving the exponential decay of, for example, radioactive materials Let’s code a Matlab function to calculate the final amount of a substance, given the elapsed time, half-life and initial amount. However if you are sure that the data is of some exponential decay you can try taking logarithm of the data first and then using the polyfit function. This is due the random fluctuations in the data introduced by the statistical nature. The key concept that makes this possible is the fact that a sine wave of arbitrary phase can be represented by the sum of a sin wave and a cosine wave. Bi-T 2* decay was found in some, but not all, selected pixels: 32 out of 110 pixels (29%) were determined to have a bi-T 2* decay. I tried to use the explicit expression for the Gaussian and nlinfit, but the sigmoidal shape of the Gaussian disappears (it behaves like an exponential decay function). y 다음 MATLAB 명령에 해당하는 링크를 클릭했습니다. fig, ax = plt. If the exponential decay constant of each peak is expected to be different and you need to measure those values, use shapes 31 or 39, but the decay constant of all the peaks is expected to be the same, use shape 5, and determine the decay constant by fitting an isolated peak. Decaying Exponential Fit Matlab. Aug 28, 2020 computing qualitatively correct approximations of balance laws exponential fit well balanced and asymptotic preserving sema simai springer series Posted By Karl MayLtd. If b r is below a certain threshold, the mono-exponential decay model is considered an adequate representation of the data, and we proceed with the extraction of the fit parameters from the data, including the scattering mean-free path, l s, as the global parameter or quantitative measure of transparency. Exponential Modelling and Curve Fitting. classdef Appl < matlab. Massive underwater coral 'skyscraper' discovered in the Great Barrier Reef. 02 however using the fittype and fit to replicate this in MATLAB I get the following results:. Look in the Results pane to see the model terms, the values of the coefficients, and the goodness-of-fit statistics. The detected signal is a convolution of IRF and the model, in this case an exponential decay. I want to plot an exponential function that approaches 1, then decays to zero. The decay of this device type is consistent with earlier findings for ITO/PEDOT:PSS/MEHPPV/Al homojunctions [9]. Mathematical Curves. A double-exponential decay corresponding to two lifetimes. NASA Technical Reports Server (NTRS) Stein, Robert F. Smooth surface cavities occur on the smooth sides of your teeth, while root cavities develop on the surface over the roots. Step 2 Create an FDG model. Accessing elements within a vector. eps Accuracy of floating-point precision. pdf), Text File (. In a number of studies, it has been observed that diffusion signal decay in breast tumors is better fitted by bi-exponential function with smaller residual errors (14-16). This is meant for experimental data, specifically optically switching transmission data from glass. A pplicAtion This problem appears as Exercise 69 in Section 5. Link to download MATLAB 2020A: Download and install MATLAB 2020A for free #Day61 #100DaysChallenge- Matlab Loops| Palindrome or Not #Day61-Palindrom or Not Task: Write a code to find if the given vector is palindrome or not x=[0,2,0,2,2,0,2,0] Palindrome. If the decaying quantity, N(t), is the number of discrete elements in a certain set, it is possible to compute the average length of time that an element remains in the set. Figure 1 presents the decay of short circuit current density of the cell measured within ~ 6 hours and a double exponential fit to the curve for guiding the eye. %exponential wave t=0:0. I'm looking for best methods to fit experimental data to a series of exponentials, either 3 or 4 depending on sample temperature. Step 3, Save the script. That means it will take that long for 100 grams of uranium-238 to turn into 50 grams of uranium-238 (the other 50 grams will have turned into another element). A 2-D sigma should contain the covariance matrix of errors in ydata. Graph exponential growth and decay functions. One can see that after 20 minutes of. Exponential growth and decay: a differential equation This little section is a tiny introduction to a very important subject and bunch of ideas: solving differential equations. Is this consistent with parameters from your curve fit? ## Exercise 3: Exponential fit Barium 137 is radioactive. Least Squares Fitting--Exponential. In mathematics, an exponential function is a function of the form. pdf), Text File (. ft=fittype('exp1'); cf=fit(time,data,ft) This is when time and data are your data vectors; time is the independent variable and data is the dependent variable. Again, I have to fit exponential data and get the coefficients. RANDN Normally distributed random numbers. MATLAB® is presented with examples and applications to various school-level and advanced life science / bioengineering problems - from growing populations of microorganisms and population dynamics, reaction kinetics and reagent concentrations, predator-prey models, to data fitting and time series analysis. An exponential least-squares fit (solid line) applied to a noisy data set (points) in order to estimate the decay constant,. iy1 = cumtrapz (x, y); iy2 = cumtrapz (x, iy1); iy3 = cumtrapz (x, iy2); % get exponentials lambdas. 1 Recommendation. | Recommend：Exponential Curve Fitting in Matlab. Does anyone knows the code that can do 'exp' function in MATLAB? You might use the Taylor series, has someone already did Writing a taylor series function for e^x - MATLAB Answers - MATLAB Central[^]. Below is a section of matlab code which performs linear regression: function [slope,offset] = fitline(x,y); %FITLINE fit line 'y = slope * x + offset' % to column vectors x and y. Creating the script will help to store your work in a program and will allow reusability. The behavior of the absorbed power is exponentially decaying sinusoidal wave. You feed sigmaplot10 with it and using the simple equation y=a*exp(-bx)+c*exp(-dx) it gives you a nice fit. An exponential function written as f(x) = 4^x is read as "four to the x power. In a number of studies, it has been observed that diffusion signal decay in breast tumors is better fitted by bi-exponential function with smaller residual errors (14-16). exp(-xdata/p1) # Sum of squared of the residuals sum_sqr_res = lambda y_observed, y_predicted: np. Gaussian Fitting with an Exponential Background. To obtain a mean other than zero, just 3 Matlab Help on randn. The problem is, no matter what the x-value I put in is, the y-value ALWAYS comes up as 1. 5ms, 30% shorter than the long T 2* component. The symbols in this and the following figures show the recorded currents (mean ± SE for n = 10), and the lines show the currents fit with an optimal exponential decay constant using MATLAB (The MathWorks, Natick, MA). y You clicked a link that corresponds to this MATLAB command:. Implement your own decay models (only MATLAB version). Numerical methods for partial differential equations (PDEs). perfettoshop. with new examples on-the-fly. An exponential decay function is For a system whose behavior can be defined by exponential decay, the parameters for the decay function can be found using least-squares. Fitting Exponential Decay Exponential decay is a very common process. A Gaussian minus exponential distribution has been suggested for modelling option prices. 5 signal_observed. We have started our service. The echoes/pulses in the first and exponential decaying in the second parts of APDP are clearly seen (also observed in the measurements data). LaTeX can automate this task by abstracting objects such as tables, pictures, etc. However, the decay function given by Eq. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. Cavities are decayed areas of your teeth that develop into tiny openings or holes. Simple fit: exponential decay First plot some data, say, an exponential decay. This might become an even bigger problem if one wants to apply this algorithm to retrieve a larger number of time constants. % The code does nonlinear least-square fitting with % covariance matrix to the exponential 'a*exp(-(x/b)^c)', % given following things: % X=XDATA=control parameter, N by 1; % Y=YDATA=mean observed data, N by 1; % CM=Covariant_Matrix=correlations in data, N by N; % N_s=number of samples used to construct CM (Y's are means of N_s samples) % Init=initial guess on two fit parameters (a,b), 1 by 3. It appears to me that excel only curve fits exponential growth data (the data increases slowly before increasing sharply (data look. I assume that this time-varying mean has an exponential decay at a single rate with a single frequency. As we need to calculate the gradients for the whole dataset to perform just one update, batch gradient descent can be very slow and is intractable for datasets that don't fit in memory. I am reasonably new to matlab and I am unsure about how to fit an exponential curve to the plot that reaches y=0 to fit a non. Heat Transfer Matlab Project Interseasonal Heat Transfer provides sustainable energy using a new form of on site renewable energy that channels naturally occurring heat from the sun down to the ground in summer and back to buildings in winter to heat buildings without burning fossil fuels. 5 * params[1] * np. %First read. If b r is below a certain threshold, the mono-exponential decay model is considered an adequate representation of the data, and we proceed with the extraction of the fit parameters from the data, including the scattering mean-free path, l s, as the global parameter or quantitative measure of transparency. The function approaches 1 but does not decay. The lifetime -of the disturbed donor was obtained using a two-exponential model for the fit performed on N2A cells transfected with EGFP and mRFP. However the concern is the deconvolution of the Instrument response function(IRF, due to device) from the experimental decay data before fitting it. f(x) = λ {e}^{- λ x} for x ≥ 0. values p = (optional) starting values for fit and control on fitting order (default =3) see below out = creates output plot for being different from 0 (optional, default =0) 1 = linear plot, 2 = logarithmic y-axis length of p: 0-3 monoexponential = A-(A-B)*exp(-t/T1) 6 biexponential = Aa-(Aa-Ba)*exp(-t/T1a. text and: In the exponential, processes give education of their happy patients. However, I am not able to get a fit. Gaussian Fitting with an Exponential Background. To determine which pulse model was the best fit, the time constants (and line slope), and the pulse amplitude were swept until the least squared error was found for each pulse. Heat Transfer Matlab Project Interseasonal Heat Transfer provides sustainable energy using a new form of on site renewable energy that channels naturally occurring heat from the sun down to the ground in summer and back to buildings in winter to heat buildings without burning fossil fuels. Readers who feel the Python examples are too hard to follow will benefit from reading a tutorial, e. Inf Infinity. Fitting a curve in the form y = Ae^(-t/tau) (Equation 2) is simple, because the equation can be linearized and you can use linear least squares to find A and tau. {\displaystyle {\frac {dN}{dt}}=-\lambda N. ELUs tested on CIFAR-10 and CIFAR-100. Y = exp(X) returns the exponential for each element of X. Fit decays using the combination of a Gaussian D-A distribution plus an imported D-A distribution, e. As we need to calculate the gradients for the whole dataset to perform just one update, batch gradient descent can be very slow and is intractable for datasets that don't fit in memory. However, an exponential can approximate a second order polynomial, over a small range, well enough. Mathematical optimization problems may include equality constraints. Prerequisite: Polynomial Regression in MATLAB. MATLAB Function Reference. is also an exponential function, since it can be rewritten as. Discover the innovative world of Apple and shop everything iPhone, iPad, Apple Watch, Mac, and Apple TV, plus explore accessories, entertainment, and expert device support. The discrepancy between the models reflects real-life denoising in which the true signal decay is typically unknown and is not fully described by the available models used for denoising. If you want to get more accurate, you could model the background as an exponential decay -- seems to be approximately 800*10-x/100, but it's best to do a least-squares regression for a more accurate fit. doc 2 of 2 e. Previous posts explained how numerical solutions work and how Matlab will perform the calculations for you automatically. 30421 t, and therefore, the estimated half-life is t = log(2)/0. Exponential. It is a fully automatic system that detects 66 landmark points on the face and estimates the rough 3D head pose. See full list on graphpad. Since the wavefunction penetration effectively. Exponential Line Of Best Fit Calculator. Exponential Growth: y = a e bx, b > 0. of Equation & Graph of Exponential Decay Function. I got your point. Start studying Exponential Functions, Exponential Functions. MATLAB can be used to optimize parameters in a model to best fit data, increase profitability of a potential engineering design, or meet some other type of objective that can be described mathematically with variables and equations. For example, uranium-238 is a slowly decaying radioactive element with a half-life of about 4. For example, the matlab function polyfit can be used for this purpose (where the requested polynomial order is set. The following is an example of a "sampled" decaying exponential. There can be numerous. The three types of cavities are shown here. A pplicAtion This problem appears as Exercise 69 in Section 5. Exponentials are often used when the rate of change of a quantity is proportional to the initial amount of the quantity. 6k86cw1vy85v zeycw6yv2okl7 8ny1xq29lll 4pn0z1y2f9cuhj 2636sy232xqi0p j4jd773186bgi few8lg8xnyjld7g kvav88064x 3j560i0mqaqd2j4 4naga4u5d87xug5 qdm3kc8u653ggyf 2go19ypnlb8ka 5m990b43prb87 2p0c33n59c idu859zt7bw9 32gyfm97w9f2r 7jt23b7ko4 tefzm23ia4k xp8y2578buga4fh pi4en8ys2lbd jq211xzh3lseihp s7q9x79v0ucyn x05h64qb5m8i2 5523qxoryrcv 1rtf1we069vr6nw gbmnhfmesb2c. This represents the decay of certain variables. An exponential function written as f(x) = 4^x is read as "four to the x power. The Pareto distribution, named after the Italian civil engineer, economist, and sociologist Vilfredo Pareto, (US: / p ə ˈ r eɪ t oʊ / pə-RAY-toh), is a power-law probability distribution that is used in description of social, quality control, scientific, geophysical, actuarial, and many other types of observable phenomena. Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. [1] X Research sourceStep 2, Type commands 'clc' and 'clear all' in the command window. ^2, x, ones (size (x))]; A = pinv (Y)*y;. INTRODUCTION In this document, we briefly describe the model [22, 23] adopted by the 15. If you had a straight line, then n=1, and the equation would be: f(x) = a0x + a1. Choose one or two terms to fit exp1 or exp2. The doc string for it even has an example of fitting an exponential decay in it which I'll copy here: >>> import numpy as np >>> from scipy. Drag force effect on a skydiver free fall (integra-tion of rational functions and use of integration tables) 3. Equation 2 has the vertical offset C and the exponential term makes it impossible to isolate A, tau and C when deriving the least squares relations. 15-06-0302-02-003c Submission Sheet3. the equation is in the matlab code. will search through all of the H1 lines in all MATLAB functions for this word and will return information on that function if autocorrelation is found. For a graph to display exponential decay, either the exponent is "negative" or else the base is between 0 and 1. In this case, the optimized function is chisq = sum((r / sigma) ** 2). it Matlab Fit. However, the decay function given by Eq. I have extracted data from a florescence decay graph. The reactant concentration according to a decaying exponential function I_r(t)=I_0*exp(-k_0*t) where I_r(t) is the concentration at a given time point, I_0 is the initial reactant concentration, k_0 is the reactant decay rate constant, and t is the time. This abstraction is called a float. python,numpy,scipy,curve-fitting,data-fitting Say I want to fit two arrays x_data_one and y_data_one with an exponential function. Exponential. Choose one or two terms to fit exp1 or exp2. I'd like to plot a decaying exponential function similar to this plot, while being able to vary the spike time and rate of decay as well as the amplitude. Start studying Exponential Functions, Exponential Functions. 5 * params[1] * np. In this lesson you will learn how to write an exponential equation by finding a pattern in a table. % This is the same as % N(t) = N(0) exp(k t) for a constant k. exp(-xdata/p1) # Sum of squared of the residuals sum_sqr_res = lambda y_observed, y_predicted: np. Many sites have the solution to the case where one wants to fit an exponential of the form A e ^ {B x}. Here is the code I used: function expo = exponential(b,x) %EXPONENTIAL. call: par = T2fit(x,y); [par,delta]=gaussfit(x,y); [par,delta]=gaussfit(x,y,p); [par,delta]=gaussfit(x,y,p,out); par = fitted parameters like p delta = error estimate (optional) in abs. The units are: ml/ 100 ml of tissue / min to k 1, k 2 e k 3; sec-1 to 1 and ; Hz to and degrees to th [1]. GaussConvDemo. Exponential Modelling and Curve Fitting. 1 are, respectively, the exponential amplitude and the decay rate. DA: 30 PA: 77 MOZ Rank: 12. A method for estimating modal decay parameters in the presence of noise is given in [126,235,236,125]. The toolbox calculates optimized start points for exponential fits, based on the. Exponential Growth. 115x, so I took this as starting values for the coefficients p. Does anyone knows the code that can do 'exp' function in MATLAB? You might use the Taylor series, has someone already did Writing a taylor series function for e^x - MATLAB Answers - MATLAB Central[^]. optimization module provides: An optimizer with weight decay fixed that can be used to fine-tuned models, and. f(x) = λ {e}^{- λ x} for x ≥ 0. I assume that this time-varying mean has an exponential decay at a single rate with a single frequency. introduction to Matlab programming language for engineering,vector ,matrix,statistics,mathematics,control loops,plotting 44. The common term regression line is used for a first-degree polynomial. (power-like) than an exponential decay. The diffusion component was extrapolated for b-values < b lim and subtracted from ( ). 1 INTRODUCTION Many processes in nature have exponential dependencies. The berfit function is intended for curve fitting or interpolation, not extrapolation. Choose one or two terms to fit exp1 or exp2. Exponential data fitting 1. An exponential function is a function of the form where b is a positive constant and x is any real number. Basically I want MATLAB to find the exact end of the decay period. algebra review exponential growth and decay, graphing exponential growth and decay montescience org, login river mill academy, 6 1 exponential growth and decay functions, exponential functions quiz review kyrene org, ixl exponential growth and decay word problems algebra, ixl exponential growth and decay word problems algebra, exponential growth and decay worksheet for 11th grade, infinite pre. Lesson 4 Exploring Exponential Functions 1 2 Constructing exponential functions mathbitsnotebook a2 ccss math function tables desmos write and graph an exponential function by examining a table function tables desmos. {\displaystyle {\frac {dN}{dt}}=-\lambda N. •To solve this equation using matlab, we first rewrite the diff eq in normal form: •We then write an m-file that, when given t and y, returns the value of the rhs of this equation y' y e t2sin(t); y(0) 1. Exponential regression is probably one of the simplest nonlinear regression models. First, the response of a SDOF to white noise is simulated in the time domain using [3]. sample from an exponential distribution with the. Exponential functions in Maple. doc 2 of 2 e. , are temporarily saved in FitOutput waiting for acceptance or rejection. Otherwise, the mono-exponential decay model may not be appropriate and we perform a fit by an extended exponential decay, as follows. In these cases, stimulated echoes confound the signal decay and preclude the use a mono- or multi-exponential fit. Step 3, Save the script. % This is the same as % N(t) = N(0) exp(k t) for a constant k. By taking data and plotting a curve, scientists are in a better position to make predictions. At each harmonic frequency, -the nearest-partial decay-rate gives the desired loop-filter gains, -the nearest-partial peak-frequency give the desired loop-filter phase delay. The present submission contains: - a function RDT. August, 2004 IEEE P802. In an earlier report the authors developed an initial value algorithm for one class of exponential sum least squares fitting problems. The goal is to find parameters a ˆ i, i = 1, 2, 3, for the model that best fit the data. Mathematical Curves. I'm trying to put an exponential decay curve onto some vehicle data I have. Since the wavefunction penetration effectively. Excel returns "y = 2814. We will discuss in this lesson three of the most common applications: population growth, exponential decay, and compound interest. 5_Word Problems. uxmk0js6176ls3f n26o11eaue zyx9juwsyi9nwo mmj37oeb40uh lyqs3oc85wpqx 0rasoijsvplsw 9b4sf1ob9xb orbv6txj2mh7 hqct8wclhus 82foxzv7omvvym4 zzmwediqdkl4efg lrrjk703593g5. (D) Time course of inhibition by 1mM Mg2+. Step 2 Create an FDG model. 43 X 10-8, compared to 3. In this % case it is recommended that the user either modify the code to % add parameters for the background (e. Figure % UI Figure. Therefore, a simple method of estimating the exponential decay time-constant for each overtone frequency is to fit a straight line to its amplitude envelope and use the slope of the fitted line to compute the decay time-constant. % Uses fitnlm() to fit a non-linear model (an exponential decay curve) through noisy data. For the finite potential well, the solution to the Schrodinger equation gives a wavefunction with an exponentially decaying penetration into the classicallly forbidden region. Note, among other criteria, that the variance is only 5. We analyze the average ACF of different BAN/BBN channels where we fit the single term exponential and power series models to the ACF decay in MATLAB, which uses the trust-region algorithm with nonlinear least-square method. Its domain is all real numbers and its range is all. After measuring the voltage and time data points of the exponential decay in charge experienced by the capacitor bank, Matlab is used to t the data. par) Anelva AGS-7000 (DOS)(*. The table below only lists parameters in addition to location and scale. Not only does that convey the intent of the code better, but it also won't have the frame-rate-dependency and floating-point rounding issues that I mentioned above. Tools for bi- multiple exponential decay of a MRI pixel use *fittype* to create your fittype object, and pass that to the *fit* command. The coming programming examples assumes familiarity with variables, for loops, lists, arrays, functions, positional arguments, and keyword (named) arguments. Mathematical Curves. 3 Observe using graphs and tables that a. f(x) = c x,. The inverse of an exponential function is a logarithm function. 0ww8ue7znr0csl gdwy51g1vtp cgafyfuz4pr lq901vacw87k 51kgk1jhq1d c7zulakmpcs8 ovkdbjfa2c z2l7sbpcxewvg3 9cwoizunjnlg 02wlaamd0w 9ityk3vjiclum k77l9r7p69 zjb6lkmv9tpnbj. Ask Question Asked 5 years, 3 months ago. Exponential Growth and Decay (Section 3. I could be able to fit the exponential decay data with exponential function. It must be one term of the form c x n. That is it represents 5. command window에 cftool이라고 입력하거나 그림 2와 같이 Curve Fitting Toolbox를 실행시킵니다. Is it more like the bars in the top plot (like an exponential decay) or in the bottom plot (like a log-normal or Poisson)? You say: " is there a way to get MatLab to identify an offset term in examining a data set for distribution fitting?". In a number of studies, it has been observed that diffusion signal decay in breast tumors is better fitted by bi-exponential function with smaller residual errors (14-16). In order to define the problem n and solve it execute the following in Matlab. This Account has been suspended. Notice that, while the the training loss is going down with each epoch, the validation loss is increasing! This suggests that we are training our model too long, and it's over-fitting on the training data. Double-exponential equations can be tried when easier forms like straight lines, parabolas, hyperbolas, and single-exponential equations are not satisfactory [1-3]. Basic statistics and also linear algebra. I'm just thinking that with all Matlab's Curve Fitting Toolbox can do, there has to be a way to produce an uncertainty for the coefficients. %----- % Demonstrate nonlinear regression -- quick & dirty example with "nlinfit" % Fit y=alpha*exp(beta*x) % Programming Note: call "nlinfit" % Instructor: Nam Sun. “decay time. In this week's lab we will generate some data that should follow this law, and you will have to fit exponential data at least twice more this quarter. I thought it should work with my old code, but apparently, I am doing something wrong, but I don't see my mistake Excel retuns an exponential function of 150e-0. m that includes both a rate constant and an initial condition. ans Most recent answer. Click here for the earlier JSim 1. , equation 3 ): f ( y) = a × e x p ( − S × x) + K. Exponential Modelling and Curve Fitting. Today, we’ll put that knowledge to good use. Note, among other criteria, that the variance is only 5. However the concern is the deconvolution of the Instrument response function(IRF, due to device) from the experimental decay data before fitting it. The Poisson distribution is related to the exponential distribution. Ich möchte ein abklingendes Exponential an die aufgezeichneten Daten anpassen. Using the non-linear least-square fitting for FLIM data analysis, the approach utilized by the user typically encompasses the following sequential steps: Establish a decay model such as the N value in Eq. relevant in that it governs the behavior of a heating and cooling, radioactive decay of materials, absorption of drugs in the body, the charging of a capacitor, and population growth just to name a few. And creating the different types of 3D plots with its function. The exponential funtion is very useful to describe important functions,like population growth ,decay. minimize python constraints hope curve-fitting signal sympy decay decay-rate dissipation-fit Updated Mar 18, 2017. Euler's formula relates its values at purely imaginary arguments to. 5 TOPIC FOCUS I can… Identify exponential growth and decay Graph exponential functions. % The code does nonlinear least-square fitting with % covariance matrix to the exponential 'a*exp(-(x/b))', % given following things: % X=XDATA=control parameter, N. Look in the Results pane to see the model terms, the values of the coefficients, and the goodness-of-fit statistics. These results confirm that the decay function in form of Eq. In MATLAB®, the fft function computes the Fourier transform using a fast Fourier transform algorithm. This example fits two poorly resolved Gaussian peaks on a decaying exponential background using a general (nonlinear. An exponential moving average (EMA) is a type of moving average that places a greater weight and significance on the most recent data points. employ an exponential decay model to account for the sigmoidal behavior of the tumor growth when in interaction with the drugs. org Measuring rates of decay Mean lifetime. I use a sine function as an example, but this method can be extended for use in. 959 exp(- 0. Log InorSign Up. You should expect to need to be able to identify the type of exponential equation from the graph. Evaluates polynomial and generates error estimates. , equation 3 ): f ( y) = a × e x p ( − S × x) + K. The general […]. e forexponential", it computes the exponentially weighted moving average. Exponential functions and power functions are compared interactively, using an applet. Problem Statement: Consider 6 points in a two-dimensional space: (1, 2), (2, 3), (1,−1), (−1, 3), (1,−2), (0,−1) Build a MATLAB figure in which the points are represented with their linear and quadratic regression functions. What would you like to see on the Matlab blog? Hi everyone! We recently passed the 100-post mark on the blog! What would you like to see as future blog posts? Leave a comment here! % categories: miscellaneous. It's written in a standard exponential form. We’re going to experiment in Matlab with this type of functions. I want to fit a decaying exponential to the plotted data. The two types of exponential functions are exponential growth and exponential decay. The second method uses switch-case statements, and the third method uses indices to define different sections of the domain. The toolbox calculates optimized start points for exponential fits, based on the. However, I am not able to get a fit. mexpfit: Multi-exponential Fitting in pracma: Practical Numerical Math Functions. Curve Fitting. The frequency domain instrument is a lifetime fluorometer (ChronosFD, ISS, Champaign, IL) using a laser diode centred at 470 nm (90099, ISS) with the modulation frequency between 8 to 200 MHz. You can perform least squares fit with or without the Symbolic Math Toolbox. Jacobian Multiply Function with Linear Lea. We'll just look at the simplest possible example of this. If the decaying quantity, N(t), is the number of discrete elements in a certain set, it is possible to compute the average length of time that an element remains in the set. Least Squares Fitting--Exponential. 47 billion years. I want to fit an exponential curve with a DC shift. NASA Technical Reports Server (NTRS) Stein, Robert F. All available built-in curve fitting functions are listed here. Here are two ways to do this. Another common schedule is exponential decay. In other words, the formula gives recent prices more weight than past prices. Warning: Do not confuse the int function in Matlab with the integer (int) data type in C or the int8, int16, int32 data types in Matlab. Learn more about fit, curve fitting, decay, function, exponential MATLAB Answers. For the Cavendish experiment, we'll need to fit our data to a sinusoidal curve with exponential decay. Wolfram Community forum discussion about Use FindFit with an exponential function?. I would really appreciate it if someone could please take a look at my code and give me suggestions to fit the data. In the exponential decay of $g(x)$, the function shrinks in half every time you add one to its input $x$. The TA data analysis is performed using a specialized data class Exponen-tial_Decay, which includes a range of different multi-exponential models. % This is the same as % N(t) = N(0) exp(k t) for a constant k. exp(-xdata/p1) # Sum of squared of the residuals sum_sqr_res = lambda y_observed, y_predicted: np. (27) (circles). View Decision Tree. I'd like to plot a decaying exponential function similar to this plot, while being able to vary the spike time and rate of decay as well as the amplitude. Exponential Growth. * * Submission * * Submission Alexei Davydov (Intel Corporation) doc. Creation of vectors is included with a few basic operations. This method is useful because it is straightforward and does not require a computer,. • Problem: Regarding the fitted curve for Excel's Exponential Trendline,. Lesson 4 Exploring Exponential Functions 1 2 Constructing exponential functions mathbitsnotebook a2 ccss math function tables desmos write and graph an exponential function by examining a table function tables desmos. ELUs tested on CIFAR-10 and CIFAR-100. | 12,490 | 54,834 | {"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} | 3.53125 | 4 | CC-MAIN-2020-50 | latest | en | 0.86819 |
http://www.ck12.org/book/CK-12-Algebra-I-Second-Edition/r3/section/6.4/ | 1,493,334,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122629.72/warc/CC-MAIN-20170423031202-00155-ip-10-145-167-34.ec2.internal.warc.gz | 503,745,027 | 36,798 | # 6.4: Absolute Value Equations and Inequalities
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Solve an absolute value equation.
• Analyze solutions to absolute value equations.
• Graph absolute value functions.
• Solve absolute value inequalities.
• Rewrite and solve absolute value inequalities as compound inequalities.
• Solve real-world problems using absolute value equations and inequalities.
## Introduction
Timmy is trying out his new roller skates. He’s not allowed to cross the street yet, so he skates back and forth in front of his house. If he skates 20 yards east and then 10 yards west, how far is he from where he started? What if he skates 20 yards west and then 10 yards east?
The absolute value of a number is its distance from zero on a number line. There are always two numbers on the number line that are the same distance from zero. For instance, the numbers 4 and -4 are each a distance of 4 units away from zero.
represents the distance from 4 to zero, which equals 4.
represents the distance from -4 to zero, which also equals 4.
In fact, for any real number :
if is not negative, and if is negative.
Absolute value has no effect on a positive number, but changes a negative number into its positive inverse.
Example 1
Evaluate the following absolute values.
a)
b)
c)
d)
e)
Solution
a) Since 25 is a positive number, the absolute value does not change it.
b) Since -120 is a negative number, the absolute value makes it positive.
c) Since -3 is a negative number, the absolute value makes it positive.
d) Since 55 is a positive number, the absolute value does not change it.
e) Since is a negative number, the absolute value makes it positive.
Absolute value is very useful in finding the distance between two points on the number line. The distance between any two points and on the number line is or .
For example, the distance from 3 to -1 on the number line is .
We could have also found the distance by subtracting in the opposite order: . This makes sense because the distance is the same whether you are going from 3 to -1 or from -1 to 3.
Example 2
Find the distance between the following points on the number line.
a) 6 and 15
b) -5 and 8
c) -3 and -12
Solution
Distance is the absolute value of the difference between the two points.
a)
b)
c)
Remember: When we computed the change in and the change in as part of the slope computation, these values were positive or negative, depending on the direction of movement. In this discussion, “distance” means a positive distance only.
## Solve an Absolute Value Equation
We now want to solve equations involving absolute values. Consider the following equation:
This means that the distance from the number to zero is 8. There are two numbers that satisfy this condition: 8 and -8.
When we solve absolute value equations we always consider two possibilities:
1. The expression inside the absolute value sign is not negative.
2. The expression inside the absolute value sign is negative.
Then we solve each equation separately.
Example 3
Solve the following absolute value equations.
a)
b)
Solution
a) There are two possibilities: and .
b) There are two possibilities: and .
## Analyze Solutions to Absolute Value Equations
Example 4
Solve the equation and interpret the answers.
Solution
We consider two possibilities: the expression inside the absolute value sign is nonnegative or is negative. Then we solve each equation separately.
and are the solutions.
The equation can be interpreted as “what numbers on the number line are 5 units away from the number 4?” If we draw the number line we see that there are two possibilities: 9 and -1.
Example 5
Solve the equation and interpret the answers.
Solution
Solve the two equations:
The equation can be re-written as: . We can interpret this as “what numbers on the number line are 2 units away from -3?” There are two possibilities: -5 and -1.
Example 6
Solve the equation and interpret the answers.
Solution
Solve the two equations:
The interpretation of this problem is clearer if the equation is divided by 2 on both sides to get . Because is nonnegative, we can distribute it over the absolute value sign to get . The question then becomes “What numbers on the number line are 3 units away from ?” There are two answers: and .
## Graph Absolute Value Functions
Now let’s look at how to graph absolute value functions.
Consider the function . We can graph this function by making a table of values:
-2
-1
0
1
2
3
4
You can see that the graph of an absolute value function makes a big “V”. It consists of two line rays (or line segments), one with positive slope and one with negative slope, joined at the vertex or cusp.
We’ve already seen that to solve an absolute value equation we need to consider two options:
1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.
Combining these two options gives us the two parts of the graph.
For instance, in the above example, the expression inside the absolute value sign is . By definition, this expression is nonnegative when , which is to say when . When the expression inside the absolute value sign is nonnegative, we can just drop the absolute value sign. So for all values of greater than or equal to 1, the equation is just .
On the other hand, when — in other words, when — the expression inside the absolute value sign is negative. That means we have to drop the absolute value sign but also multiply the expression by -1. So for all values of less than 1, the equation is , or .
These are both graphs of straight lines, as shown above. They meet at the point where — that is, at .
We can graph absolute value functions by breaking them down algebraically as we just did, or we can graph them using a table of values. However, when the absolute value equation is linear, the easiest way to graph it is to combine those two techniques, as follows:
1. Find the vertex of the graph by setting the expression inside the absolute value equal to zero and solving for .
2. Make a table of values that includes the vertex, a value smaller than the vertex, and a value larger than the vertex. Calculate the corresponding values of using the equation of the function.
3. Plot the points and connect them with two straight lines that meet at the vertex.
Example 7
Graph the absolute value function .
Solution
Step 1: Find the vertex by solving . The vertex is at .
Step 2: Make a table of values:
-8
-5
-2
Step 3: Plot the points and draw two straight lines that meet at the vertex:
Example 8
Graph the absolute value function:
Solution
Step 1: Find the vertex by solving . The vertex is at .
Step 2: Make a table of values:
0
4
8
Step 3: Plot the points and draw two straight lines that meet at the vertex.
## Solve Real-World Problems Using Absolute Value Equations
Example 9
A company packs coffee beans in airtight bags. Each bag should weigh 16 ounces, but it is hard to fill each bag to the exact weight. After being filled, each bag is weighed; if it is more than 0.25 ounces overweight or underweight, it is emptied and repacked. What are the lightest and heaviest acceptable bags?
Solution
The weight of each bag is allowed to be 0.25 ounces away from 16 ounces; in other words, the difference between the bag’s weight and 16 ounces is allowed to be 0.25 ounces. So if is the weight of a bag in ounces, then the equation that describes this problem is .
Now we must consider the positive and negative options and solve each equation separately:
The lightest acceptable bag weighs 15.75 ounces and the heaviest weighs 16.25 ounces.
We see that and . The answers are 0.25 ounces bigger and smaller than 16 ounces respectively.
The answer you just found describes the lightest and heaviest acceptable bags of coffee beans. But how do we describe the total possible range of acceptable weights? That’s where inequalities become useful once again.
## Absolute Value Inequalities
Absolute value inequalities are solved in a similar way to absolute value equations. In both cases, you must consider the same two options:
1. The expression inside the absolute value is not negative.
2. The expression inside the absolute value is negative.
Then you must solve each inequality separately.
## Solve Absolute Value Inequalities
Consider the inequality . Since the absolute value of represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero is less than or equal to 3. The following graph shows this solution:
Notice that this is also the graph for the compound inequality .
Now consider the inequality . Since the absolute value of represents the distance from zero, the solutions to this inequality are those numbers whose distance from zero are more than 2. The following graph shows this solution.
Notice that this is also the graph for the compound inequality or .
Example 1
Solve the following inequalities and show the solution graph.
a)
b)
Solution
a) represents all numbers whose distance from zero is less than 5.
This answer can be written as “”.
b) represents all numbers whose distance from zero is more than or equal to 2.5
This answer can be written as “ or ”.
## Rewrite and Solve Absolute Value Inequalities as Compound Inequalities
In the last section you saw that absolute value inequalities are compound inequalities.
Inequalities of the type can be rewritten as “”.
Inequalities of the type can be rewritten as “ or .”
To solve an absolute value inequality, we separate the expression into two inequalities and solve each of them individually.
Example 2
Solve the inequality and show the solution graph.
Solution
Re-write as a compound inequality:
Write as two separate inequalities: and
Solve each inequality: and
Re-write solution:
The solution graph is
We can think of the question being asked here as “What numbers are within 7 units of 3?”; the answer can then be expressed as “All the numbers between -4 and 10.”
Example 3
Solve the inequality and show the solution graph.
Solution
Re-write as a compound inequality:
Write as two separate inequalities: and
Solve each inequality: and
and
Re-write solution:
The solution graph is
Example 4
Solve the inequality and show the solution graph.
Solution
Re-write as a compound inequality: or
Solve each inequality: or
The solution graph is
Example 5
Solve the inequality and show the solution graph.
Solution
Re-write as a compound inequality: or
Solve each inequality: or
or
The solution graph is
## Solve Real-World Problems Using Absolute Value Inequalities
Absolute value inequalities are useful in problems where we are dealing with a range of values.
Example 6
The velocity of an object is given by the formula , where the time is expressed in seconds and the velocity is expressed in feet per second. Find the times for which the magnitude of the velocity is greater than or equal to 60 feet per second.
Solution
The magnitude of the velocity is the absolute value of the velocity. If the velocity is feet per second, then its magnitude is feet per second. We want to find out when that magnitude is greater than or equal to 60, so we need to solve for .
First we have to split it up: or
Then solve: or
or
The magnitude of the velocity is greater than 60 ft/sec for times less than 0.8 seconds and for times greater than 5.6 seconds.
When . The magnitude of the velocity is 60 ft/sec. (The negative sign in the answer means that the object is moving backwards.)
When .
To find where the magnitude of the velocity is greater than 60 ft/sec, check some arbitrary values in each of the following time intervals: and .
Check
Check
Check
You can see that the magnitude of the velocity is greater than 60 ft/sec only when or when .
## Further Resources
For a multimedia presentation on absolute value equations and inequalities, see http://www.teachertube.com/viewVideo.php?video_id=124516.
## Lesson Summary
• The absolute value of a number is its distance from zero on a number line.
• if is not negative, and if is negative.
• An equation or inequality with an absolute value in it splits into two equations, one where the expression inside the absolute value sign is positive and one where it is negative. When the expression within the absolute value is positive, then the absolute value signs do nothing and can be omitted. When the expression within the absolute value is negative, then the expression within the absolute value signs must be negated before removing the signs.
• Inequalities of the type can be rewritten as “.”
• Inequalities of the type can be rewritten as “ or .”
## Review Questions
Evaluate the absolute values.
Find the distance between the points.
1. 12 and -11
2. 5 and 22
3. -9 and -18
4. -2 and 3
Solve the absolute value equations and interpret the results by graphing the solutions on the number line.
Graph the absolute value functions.
Solve the following inequalities and show the solution graph.
1. How many solutions does the inequality have?
2. How about the inequality ?
1. A company manufactures rulers. Their 12-inch rulers pass quality control if they are within of the ideal length. What is the longest and shortest ruler that can leave the factory?
2. A three month old baby boy weighs an average of 13 pounds. He is considered healthy if he is at most 2.5 lbs. more or less than the average weight. Find the weight range that is considered healthy for three month old boys.
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Subjects: | 3,003 | 13,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2017-17 | latest | en | 0.920077 |
https://math.stackexchange.com/questions/549302/showing-a-function-is-bijective-and-finding-its-inverse | 1,726,647,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00816.warc.gz | 352,536,047 | 40,531 | # Showing a function is bijective and finding its inverse
The function $$f: \mathbb{R}^2 \rightarrow\mathbb{R}^2$$ is defined by $$f(x,y)=(2x+3y,x+2y)$$. Show that $$f$$ is bijective and find its inverse.
I've got so far: Bijective = $$1-1$$ and onto.
$$1-1$$ if $$(2x_1+3y_1,x_1+2y_1)=(2x_2+3y_2,x_2+2y_2)$$
Then $$2x_1+3y_1=2x_2+3y_2 \qquad (1)$$ $$x_1+2y_1=x_2+2y_2 \qquad (2)$$
$$(1)-(2)$$
$$x_1+y_1=x_2+y_2$$
$$x_1=x_2+y_2-y_1$$
Substituting into equation 1:
$$2(x_2+y_2-y_1)+3y_1=2x_2+3y_2$$
$$y_1=3y_2-2y_2$$
$$y_1=y_2$$
Substituting into equation 2:
$$x_1+2y_1=x_2+2y_1$$
$$x_1=x_2$$
$$(2x_1+3y_1,x_1+2y_1)=(2x_2+3y_2,x_2+2y_2)$$
Thus 1-1
Onto
if $$(u,v) \in \mathbb{R}^2$$ (codomain) we want $$(x,y)$$ with $$f(x,y)=(u,v)$$
$$(2x+3y,x+2y)=(u,v)$$
$$2x+3y=u$$
$$x+2y=v$$
Eliminating x:
$$y=2v-u$$
Substituting:
$$2x+3(2v-u)=u$$
$$2x+6v-3u=u$$
$$2x=4u-6v$$
$$x=2u-3v$$
Therefore:
$$f(2u-3v,2v-u)=(u,v)$$
Now how do you find it’s inverse? And is that correct what I have done?
• Do you know about matrices? Commented Nov 2, 2013 at 19:24
• Looks right. And you have found the inverse transformation. It sends $(u,v)$ to $(2u-3v, 2v-u)$. If you like, you can change the variables to $x$ and $y$. Commented Nov 2, 2013 at 19:28
• So u to be x and v to be y? Commented Nov 2, 2013 at 19:29
• Yes, if you want to use those names for variables. To me, $u$ and $v$ are just as good, but they may be expecting you to use $x$ and $y$. Commented Nov 2, 2013 at 19:31
• so the inverse is literally: (2-3y,2y-x)? Commented Nov 2, 2013 at 19:41
Superhint: If you had started with the second part ("and find its inverse") you could have "coimpletely" avoided the first part calculations.
You have $f(x,y)=(2x+3y,x+2y)$. This function is what we call linear, because it satisfies the following property: $f(\lambda v+w) = \lambda f(v)+f(w)$ for $v,w\in \Bbb R^2$ as you can check.
Now, linear functions are much easier do deal with than general functions. Why? Because they can be (at least in spaces like $\mathbb{R}^2$ that are what we call finite dimensional) be expressed in terms of matrices. Indeed, look that you have the following:
$$f(x,y)=xf(1,0)+yf(0,1)$$
Exactly because of that property of linearity. In that case, if you know about matrix multiplication, you'll see that this is equal to saying that
$$f(x,y)=\begin{pmatrix}f(1,0) & f(0,1)\end{pmatrix}\begin{pmatrix}x \\y\end{pmatrix},$$
where we consider elements of $\mathbb{R}^2$ as columns. In that case, you have
$$f(x,y)=\begin{pmatrix}2 & 3\\ 1 & 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix},$$
if we call that matrix $A$, then $f(v)=A v$ where $v=(x,y)$. Now, Look that because of that, things get easier. Indeed, to show that $f$ is injective, is equivalent to showing that $f(v)=f(w)$ implies $v=w$, but $f(v)=f(w)$ is the same as $Av=Aw$, or $A(v-w)=0$. This implies $v=w$, if and only if the homegenous system of linear equations $Av=0$ has only the trivial solution. That is, if and only if $\det A \neq 0$. But you can check that $\det A = 1$, so that $A$ is invertible, hence, $A(v-w)=0$ implies $v=w$, and hence $f$ is injective.
To show surjectivity, notice that $A$ being invertible, already implies that, because you can multiply the following
$$f(x,y)=A\begin{pmatrix}x\\y\end{pmatrix}$$
on the left by $A^{-1}$ to find $(x,y)$. It is easy, because
$$A^{-1}=\begin{pmatrix}-2 & \phantom{-}1 \\ \phantom{-}3 & -2\end{pmatrix},$$
and so we can take the inverse to be $g(v)=A^{-1}v$. In that case, $f(g(v))=A(A^{-1}v)=v$ and similarly $g(f(v))=A^{-1}(A(v))=v$, so $g$ is inverse to $f$ and hence $f$ is bijective.
Hint: You want a function $g(x,y)$ such that $g(2x+3y,x+2y) = (x,y)$. In the work you've already shown, you set $u = 2x+3y$ and $v = x+2y$ and then solved for $x$ and $y$ in terms of $u$ and $v$. These expressions for $u$ and $v$ are somehow related to $g$.
That $f(2u-3v,2v-u)=(u,v)$ means that with $g(u,v)=(2u-v,2v-u)$ you have $f\circ g=\mathrm{id}$.
Had you not already shown that $f$ is injective, you could do so by checking that $g\circ f=\mathrm{id}$. (Indeed linear algebra tells us that “for dimension reasons” this follows automatically in this case.) | 1,623 | 4,204 | {"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": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.780798 |
https://math.stackexchange.com/questions/2728830/showing-a-theory-has-aleph-0-structures | 1,566,333,684,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315618.73/warc/CC-MAIN-20190820200701-20190820222701-00113.warc.gz | 549,717,800 | 32,606 | Showing a theory has $\aleph_0$ structures
Consider a first order language with a single binary relation $R$ and consider the theory of a set $A$, Th($A$), where $A$ is the set containing a single axiom describing that there are exactly three elements: $A = \{ \exists x \exists y \exists z (x \neq y \land x \neq z \land y \neq z \land \forall w(w = x \lor w = y \lor w = z))\}$.
I'm trying to show that i. $T$ has countably infinite many structures up to isomorphism and ii. to find a decision procedure for $T$.
Here are my thoughts: i. I understand how to find a class of countably many structures, e.g., the class of structures whose domains consist of $i$, $i+1$, $i+2$ for any $i \in \mathbb{N}$. We see in my example any two models are isomorphic, but I'm unsure of how to show that there can't be a larger class or isomorphic models of any sort.
ii. Since $A$ contains a single axiom and is therefore enumerable, then its consequences, are enumerable and hence $T$ is enumerable. We may write out all strings representing proofs and by Los Vaught Test (we showed in part i. that $T$ is $\aleph_0$-categorical), we know $T$ is complete so for every formula we must reach at some point a proof of either it or its negation.
I'd appreciate any insight as to whether I am on the right track and hints of any sort would be much appreciated!
• (i) is false, as it is rather obvious that there are only finitely many structures up to isomorphism (there are only finitely many binary relations on any 3-element set)... – Eric Wofsey Apr 9 '18 at 5:06
• @EricWofsey but what about my example of structures with domains of a different set of three natural numbers? These structures are isomorphic and there are infinite of them. – Raton Apr 9 '18 at 5:10
• "Up to isomorphism" means you consider isomorphic structures to be the same. In any case, you haven't actually named any examples, since you never said what the binary relations are. – Eric Wofsey Apr 9 '18 at 5:12
• @EricWofsey and also, this complicates part ii since we can't claim that Th($A$) is complete any more by the Los Vaught Test... – Raton Apr 9 '18 at 5:19
• @EricWofsey I understand how we would come up with a bijective mapping, call it $h$, between domains of any two structures of $T$, but how would we show that $h(R(a,b)) = R(h(a),h(b))$? – Raton Apr 9 '18 at 5:40
1 Answer
There are two key observations here.
The first is purely combinatorial:
Fix a three-element set $\{a, b, c\}$. There are only finitely many binary relations on $\{a, b, c\}$. That is, there are only finitely many models of your theory with domain $\{a, b, c\}$.
The second observation is:
The statement we want to prove is $$\mbox{"Each model of T is isomorphic to some structure with domain \{a, b, c\}."}$$
Since this will let us count the models of $T$ up to isomorphism, by counting the models of $T$ with domain $\{a, b, c\}$ up to isomorphism.
So suppose $(\{x, y, z\}; R)$ is a model of $T$. Pick a bijection $f:\{x,y,z\}\rightarrow\{a, b, c\}$; do you see a way to use $f$ to define a relation $S$ on $\{a, b, c\}$ such that $f$ is an isomorphism from $(\{x, y, z\}; R)$ to $(\{a, b, c\}; S)$?
HINT: you're going to define $S$ in terms of $R$ and $h$. Suppose for example that $R(x, y)$ holds; using $h$, does this suggest some fact about the $S$ we want to build?
• what is $h$? Do you mean $f$? I understand that we have $8$ isomorphic classes of structures ($\{a\}, \{b\}, \{a,b\}, \cdots$). Why is it important to define $S$ in terms of $R$ and do you have any hints as to how to go about it? – Raton Apr 9 '18 at 23:30
• also, regarding ii. if my decision procedure is to look at one structure in each of the 8 isomorphic classes and decide whether the formula holds in each. If the formula fails in at least one structure, we reject and otherwise accept. Is this correct reasoning and what do you recommend I add to make it more rigorous? Once again, thank you. – Raton Apr 9 '18 at 23:33
• I meant $2^9$ iso classes – Raton Apr 10 '18 at 0:59 | 1,175 | 4,023 | {"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} | 2.984375 | 3 | CC-MAIN-2019-35 | latest | en | 0.942597 |
http://forums.holdemmanager.com/manager-general/14949-roi-calculation.html | 1,368,926,216,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696383081/warc/CC-MAIN-20130516092623-00067-ip-10-60-113-184.ec2.internal.warc.gz | 106,546,276 | 16,085 | # Thread: ROI Calculation
1. ## ROI Calculation
How exactly is ROI calculated in HEM.
I was doing this manually before recently using HEM, and the figures are different...
I use: ROI = (Winnings - Buyins) / Buyins * 100
In this case I cite of 1 particular tournament from an online league, I get 6.83 and HEM say 106.8%???
Winnings: \$102.24
Games: 29
Buyin: \$3 + 30c
It seems HEM is adding 100???
2. I'll forward it to Fabio for an answer.
3. I am a tourney noob, but I am pretty adept at math (or at least I was 20 years ago in High-School), so hopefully I am explaining this right. This is my laymen's understanding of ROI in HM.
You should not be multiplying your divisor (BuyIns) by 100
The return on investment formula:
Code:
``` (Gain Form Investment - Cost of Investment)
ROI = ---------------------------------------------------------
Cost of Investment```
Your cost of investment was \$95.70
That money is gone, and no longer your own.
Your return on that investment was \$102.24
You got your entire investment back (100%) + the profits, for an ROI of >100%
4. Well, thanks for the reply, but I will correct you on 1 thing. The formula i cited doesn't multiply the divisor, it multiplies the result of the division (multiplication and division have the same precedence, hence are simple performed left to right)
ie:
Code:
```winnings - buyins
----------------- x 100
and you need to multiply by 100 to get a % figure...
OK, so by HEM ROI anythin < 100 is loosing money?
ROI should be standardised across all poker software imo...
Yet looking atound i see a number of formulae for it.
5. I'd like to know how ROI is calculated in HM. Do you compute ROI with the real \$\$\$, or do you compute ROI in buyins? E.g. I've paid 5 buyins, took 2 first places, other 3 SNGs out of the money. So ROI in this case would be (10-5)/5=100%
It differs whether \$\$\$ or buyins are used, if you play different buyins simultaniously (as you haven't fixed it yet, that HM recognizes what buyin you are currently playing at. It takes all SNGs as \$11 SNGs)
6. Yes, that is the best I can figure HEM calculated ROI that anything < 100 is losing.
But the figures you cite should show an ROI of 200% as you have got back twice your investment.
in the above formula buyins is the total amount spens on buyining into tournaments, so if you have played 5 SnGs with a total buyin af \$1 each, then yes this would be 5.
Everything is in total dollars...
Seems the real formula HEM uses is winnings / buyins x 100...
the \$11 figure for buyins only refers to one site, which doesnt record the amount in the HH files, but other sites (PS, Tilt etc) do, so you dont need to enter the amount for those, although you can filter some out of the report.
Overall ROI is irrespective of buyin level if you buyin at diferent levels, it just totals your winnings, and you buyins and divides to two...
Hope this helps...
7. I forgot to mention that I'm playing at a Casino in the ongame network, where HM doesn't recognize the different buyins.
8. I'll forward it to Fabio again...
9. I've got an question about ROI calculation as well. It's a more general question. But since ya'll are such good at math, i think at least one of you probably knows this one:
If i want to know what my average ROI % per game should be if i want to make x amount of money with x amount of games to play. Lets say:
• I play: \$2.25 stakes STTSNG;
• i want to play 600 games;
• I want to make a profit of \$750.
Is it elegible to say:
Over 600 \$2.25 games i invest \$1350 (2.25*600). So that is the total investment. What i want to make is \$750,00. So how do we get that? -\$1350 + ??payback?? = \$750. Well: (\$750--\$1350)=\$2100 payback. So we got to make a profit of \$2100 in total. Here is the equasion:
ROI = [ (\$2100 – \$1350)/\$1350)]*100 = 55%.
ROI = 55%
Only thing is that an ROI of 55% would be insane and fairly impossible imo lol | 1,049 | 3,943 | {"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} | 3.953125 | 4 | CC-MAIN-2013-20 | latest | en | 0.942059 |
http://www.enotes.com/homework-help/find-distance-between-6-2-0-10-units-226411 | 1,462,533,346,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861754936.62/warc/CC-MAIN-20160428164234-00001-ip-10-239-7-51.ec2.internal.warc.gz | 503,908,657 | 11,477 | # Find a if the distance between (6.2) and (0,a) is 10 units
Posted on
Given the distance between the points (0, 6) and ( 2,a) is 10 units.
We will use the distance between two points formula to solve.
We know that:
D (AB) = sqrt[ ( xA-xB)^2 + ( yA-yB)^2 ].
Then, we will substitute.
==> D = sqrt( 6-0)^2 + ( a-2)^2 = 10
==> sqrt(6^2) +(a-2)^2 = 10
==> sqrt(36+(a-2)^2] = 10
Now we will square both sides.
==> 36 + (a-2)^2 = 100.
Now we will subtract 36 from both sides.
==> (a-2)^2 = 64.
Now we will take the root of both sides.
==> (a-2) = +- 8
==> (a-2) = 8 ==> a= 10
==> (a-2) = -8 ==> a= -6
Then we have two possible values for a.
==> a1= 10
==> a2= -6
Then, the points ( 0, -6) and ( 0, 10) are located 10 units from the point (6,2). | 299 | 761 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-18 | longest | en | 0.845061 |
https://nl.mathworks.com/matlabcentral/cody/problems/48-making-change/solutions/1830258 | 1,585,978,144,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370520039.50/warc/CC-MAIN-20200404042338-20200404072338-00188.warc.gz | 588,879,089 | 15,615 | Cody
# Problem 48. Making change
Solution 1830258
Submitted on 29 May 2019 by josselin LEDENT
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 = [257.68]; b = [2 1 0 0 1 1 0 1 0 1 1 3]; out = makingChange(a); assert(isequal(out(:), b(:)))
2 Pass
a = [135.01]; b = [1 0 1 1 1 0 0 0 0 0 0 1]; out = makingChange(a); assert(isequal(out(:), b(:)))
3 Pass
a = [10035.99]; b = [100 0 1 1 1 0 0 1 1 2 0 4]; out = makingChange(a); assert(isequal(out(:), b(:))) | 230 | 582 | {"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-2020-16 | latest | en | 0.657369 |
www.pinzhi.org | 1,713,653,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817688.24/warc/CC-MAIN-20240420214757-20240421004757-00697.warc.gz | 873,052,143 | 3,602 | Stat > Basic Statistics > Correlation
A correlation coefficient measures the extent to which two variables tend to change together. Minitab offers two different correlation analyses:
· Pearson product moment correlation - The Pearson correlation evaluates the linear relationship between two continuous variables. A relationship is linear when a change in one variable is associated with a proportional change in the other.
For example, you might use a Pearson correlation to evaluate whether increases in temperature at your production facility are associated with decreasing thickness of your chocolate coating.
· Spearman rank-order correlation (also called Spearman's rho) - The Spearman correlation evaluates the monotonic relationship between two continuous or ordinal variables. In a monotonic relationship, the variables tend to change together, but not necessarily at a constant rate. The Spearman correlation coefficient is based on the ranked values for each variable rather than the raw data.
Spearman correlation is often used to evaluate relationships involving ordinal variables. For example, you might use a Spearman correlation to evaluate whether the order in which employees complete a test exercise is related to the number of months they have been employed.
There are a few points to keep in mind when performing or interpreting a correlational analysis:
· It is always a good idea to examine the relationship between variables with a scatterplot. Correlation coefficients only measure linear (Pearson) or monotonic (Spearman) relationships. Other relationships are possible.
· It is never appropriate to conclude that changes in one variable cause changes in another based on a correlation alone. Only properly controlled experiments allow you to determine if a relationship is causal.
· The Pearson correlation coefficient is very sensitive to extreme values. A single value that is very different from the others in a data set can change the value of the coefficient a great deal.
## Dialog box items
Variables: Choose the columns containing the variables you want to correlate. When you list two columns, Minitab calculates the correlation coefficient for the pair. When you list more than two columns, Minitab calculates the correlation for every possible pair, and displays the lower triangle of the correlation matrix (in blocks if there is insufficient room to fit across a page).
Method
Pearson correlation: Calculate the linear correlation coefficient for each pair of variables.
Spearman rho: Calculate the rank-order correlation coefficient for each pair of variables.
Display p-values: Check to display p-values for the hypothesis test. For a coefficient, r, the hypothesis are: H0: r = 0 versus H1: r ≠ 0.
Store matrix (display nothing): Check to store the correlation matrix. Minitab does not display the correlation matrix when you choose this option. To display the matrix, choose Data > Display Data. | 560 | 2,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-18 | latest | en | 0.904822 |
sabkablogger.blogspot.com | 1,503,474,950,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00629.warc.gz | 374,782,639 | 16,890 | ## 28 Dec 2012
### Special Theory of Relativity - Full Lecture
Before i start my topic i will tell you about frame reference What is frame of reference? you will find much detail about this but here i will tell you a little. Let first define it "A frame of reference is any coordinate system relative to which measurements are taken". For example suppose some measurements for an experiment are taken in laboratory then laboratory is the frame of reference. Another example is suppose a body is palced in car then car is a frame of refence. Lets move further. There are two types of frame of reference.
1) Inertial frame of reference
2) Non inertaial frame of reference
In above example if the body is at rest or moves with consatnt velocity so that no acceleration is produced then it is called inertial frame of reference.
And if the body moves with some variable velocity so that acceleration is produced then it is called non inertial frame of reference.
Special theory of relativity treats the problems involving inertial frames of reference and do you know? there is another theory called general theory of relativity that treat the problems involving non inertial frames of refernce.
In 1905, a paper entitled " on the electrodynamics of moving bodies" Einstein offered the following postulates which are the basis of the special theory of relativity.
The two postulates are:
#### 1) Principle of relativity: The laws of physics are same in all inertial frames of reference.2) The constancy of the speed of light: Speed of light in free space(vacum) is constant(has same value c) in all inertial frames of reference.
Now we will discuss the above two postulates.
The first postulate declares that the laws of physics that hold for an observer in inertial frame of reference cannot be violated for an observer in an other inertial frame of reference.
To accept the second postulate we consider the three observers at rest in three different inertial frames of reference.
The flash of light emitted by observer A is measured by him to be travel at speed c. If the frame of
observer B is moving away from A at a speed of c/4, then Galiliean kinematics predicts that B
measures the c-c/4=3c/4 for the speed of light emitted by A. If the observer C is moving towards A
with a speed of c/4, then the Galilean kinematics predicts that C measures the value c+c/4=5c/4 for
the speed of light emitted by A.
BUT
Second postulate tells that all three observers measures the same speed c for the speed of light.
However, some experiments are conducted which satisfies second postulate. This is a big problem
and confusion which must be solved in future.
Now we will discuss some consequences(results) of special theory of realativity. There are many
consequences but here i willl discuss only three consquences which are important.
These are
i) Time Dilation
ii) Length Contraction
iii) Mass Variation
Here i will more focus on time dilation. Lets discuss these consequences one by one.
1) Time Dilation:
From centuries it is considered that time is an absolute quantity. Time is running, running and running.
But
Einstein said time is not an absolute quantity it depends on velocity, it depends on motion of frame of
reference.
Now suppose an observer is stationary in an inertial frame of reference. He observes the time interval
between the two events in this frame let it be t0 . This is called proper time. If the observer is moving
with velocity v with respect to frame of events, then the time measured by observer would not be it
will be t0
given as
As the quantity of right hand side is less than 1, so t is always greater than t0.
There are many verses in Holy Book Quran that tells that time is not an absolute quantity. Time is not
same for all. It is different for diffferent persons.
In Quran,
"The angel and spirit ascend to Him in day whose length is fifty thousand years"
(Chap.29, Surah Al-Maarij 70:4)
AND
"They ask you to hasten the punishment God will not break His promise. A day with your Lord is
equivalent to a thousand years in the way you count" (Al-Quran 22:47)
Now focus on this picture.
Here people are working together and time is passing on Earth which is t0 and time passing on
spaceship is t. Focus on above two clocks you can see that
More time pass on Earth
Less time pass in spaceship
So from above verses and example it is clear that time is not an absolute quantity.
2) Length Contraction:
Suppose an observer in spaceship is at rest in space. The observer measured the length of a rod
(say 1 meter), which is ℓ0 and called proper length. And if the observer moves with spaceship in a
space then the length measured by the observer will not be 1 meter it will be less than 1 meter, denoted
by ℓ and
given as
3) Mass Variation:
Suppose an observer in a spaceship is at rest in space. The mass of an object meaured by the
observers m(say 1kg), called proper mass. Now, if the observer moves at high speed, then the
time maesured by observer will not be 1 kg, it will be greater than 1 kg. denoted by m and
given as | 1,193 | 5,058 | {"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-2017-34 | longest | en | 0.918529 |
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Enter formulas
# Test 2, question 3
If you copy the formula =C4*\$D\$9 from cell C4 to cell C5, what will the formula be in cell C5? (Pick one answer.)
1.
=C5*\$D\$9
2.
=C4*\$D\$9
3.
=C5*\$E\$10
Document views 244 Page views 244 Page last viewed Sat Dec 03 14:57:36 UTC 2016 Pages 65 Paragraphs 433 Words 3509 | 133 | 374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2016-50 | longest | en | 0.735192 |
https://www.physicsforums.com/threads/gravity-inside-the-earth.207148/ | 1,485,309,797,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285337.76/warc/CC-MAIN-20170116095125-00499-ip-10-171-10-70.ec2.internal.warc.gz | 968,827,441 | 20,341 | # Gravity inside the Earth
1. Jan 4, 2008
### PlasmaSphere
I am looking for any data on the weight of objects as they decend into the earth. I cant seem to find any, which i find surprising, that would be an ideal way to prove gravity being exclusively due to the attraction of mass.
I think i am correct in saying that as you decend the weight should decrease uniformly, as there is more mass above you pulling you back, and less below you pulling you towards the core. And when you get to the centre you would be completely weightless, as gravity cancels out in every direction. That is what Newtons Shell theorem suggests; http://en.wikipedia.org/wiki/Shell_theorem
is there any data on weights at different depths inside the earth? surely there must be some, i cant seem to find any.
Last edited: Jan 4, 2008
2. Jan 4, 2008
### Fizzicist
This is only true if you assume that the density of the earth is uniform, which is not the case. The earth is made of several types of rock, all with different densities. You are correct in that the amount of mass pulling you back does increase as you descend to the center of the earth, but because of the earth's variable desnity the change in the amount of mass pulling you back relative to the amount of mass pulling you to the core as you descend will not result in a uniform decrease in weight.
Last edited: Jan 4, 2008
3. Jan 4, 2008
### Andre
Been there:
Unfortunately the last link is broken. That domain is gone. I see if I can dig up the excel sheet with the calculations again.
Last edited: Jan 4, 2008
4. Jan 4, 2008
### PlasmaSphere
that is very true, it would not be uniform. It should still derease to zero as you get to the centre though, as at the COG it cancels out in every direction.
I just find it odd that there is no data on this. There are plenty of deep caves and pipelines about the earth, some that go over thirty miles down, so i would have thought that at least some experiments would have been carried out to see what gravity is doing inside the earth. there must be some experiments that have been conducted. can anyone find any? i certainly cant
A fine piece of work! i had not thought about this before, and to be honest the maths was beyond me, until i saw that.
i would like to see that graph, it doesn't seem to work.
That surprises me somewhat. I would have thought it was stongest at the surface, as the further down you go the more mass cancels out. hmmm..... i'm missing something.
Last edited: Jan 4, 2008
5. Jan 4, 2008
### Fizzicist
Precisely.
I don't really find it odd. We can only make rough estimates of the density of each layer of the earth, which means that we can only make very crude approximations of the gravitational field inside the earth. These approximations would essentially be useless.
If the earth were uniformly dense, the weight would be given by (GMem/Re^3)r, where Me is the mass of the earth, m is the mass of the object, Re is the radius of the earth and r is the distance between the CoM of the earth and of the object. This would mean that the gravitational force is only dependent on and directly proportional to r, and so weight would be greatest at the surface of the earth. However, the earth is not uniformly dense so this isn't true. The density of the earth increases as the distance to the core decreases. To find the gravitational force at any point within the earth, you take the sphere of the same radius as the distance from that point to the center of the earth and ignore the rest of the mass, and then you use the equation F=GMm/r^2 where M is the mass of this sphere. Because the density is greater nearer the earth's core, this means that enough mass is concentrated there that M is not much smaller than the mass of the earth while the value of r^2 is much smaller. It shouldn't be surprising, then, that at some point within the earth this expression is greater than it is at the earth's surface.
6. Jan 4, 2008
### PlasmaSphere
I thought that from seismology the density of each layer was fairly well established?
I suppose that since the drop in gravitational field strength would be negligable over 20 miles it would be a fairly pointless experiment too. but you could still do it at multiple points on the earth and work with the avergaes, to see if the value is as predicted. Maybe i'm going to have do this myself as a small science project, although there certainly wouldn't be anything small about it
I see now, i was thinking that the graphs of g/r in my old physics textbook were literally what happened, but i remember now that they specifically noted that it was for a uniform spherical mass. The just drew the line straight from the radius to the origin, like the one on this site; http://www.saburchill.com/physics/chapters/0007.html
I wonder what the graph would look like if you took varying density of the earth into account. It would be a smoother transition for a start, past that I'm not sure what exact shape it would take.
7. Jan 4, 2008
### Andre
I hope I can find that spreadsheat back.
I seem to remember that going down the (light) crust g increases some procent, then in the upper mantle g is near constant. In the lower mantle g goes up a bit sharper towards the core-mantle boundary where it's the highest. In the core the decrease is indeed as expected, near linear to zero.
Last edited: Jan 5, 2008
8. Jan 4, 2008
### Andre
got it. The zip file contains the spreadsheet
File size:
25.8 KB
Views:
92
File size:
9.6 KB
Views:
236
9. Jan 4, 2008
### billiards
You wanted data. Well it has been published, the best I have to hand is in this preliminary reference earth model (PREM) table. If you wanted, you could plot it up on excel next to Andre's to see how they match... might be interesting to see.
#### Attached Files:
• ###### PREMTable.pdf
File size:
148.7 KB
Views:
124
10. Jan 4, 2008
### Andre
Seems to be reasonable within the error margin that I mentioned in the old thread.
11. Jan 5, 2008
### ZapperZ
Staff Emeritus
Actually, this can be worked out easily even if the earth doesn't have a uniform density. All one has to assume is that it only has a radial dependence of density, which isn't that unreasonable. If one goes by that, then one can apply the Gauss's Law equivalence to calculate the gravitational force because one has the radial symmetry to construct a simple gaussian surface.
Zz.
Last edited: Jan 5, 2008
12. Jan 5, 2008
### Fizzicist
I suppose if you had the density values at various depths within the earth you could plot them and do a curve fit to derive an equation (to find the radial dependence of density). Then you could use this equation with Gauss's Law to solve for the gravitational force at any point. However, since the values we have for the density of the earth at various depths are not very accurate, our density equation probably wouldn't be very accurate and therefore our results wouldn't be accurate.
13. Jan 5, 2008
### billiards
Note that the PREM assumes spherical symmetry (and also isotropy if you're interested in the seismic aspect).
Note also that if you're looking for more physical understanding about how g can be greater somewhere in the middle of the earth than it is at the surface you need to understand a bit more about what g actually is. g is actually defined as the (maximum negative) rate of change of the "gravitational potential" with respect to position. g=-grad(gravitational potential scalar). So if you have some amount of gravitational energy in a gravitational field and you change position in the direction of g, the amount of gravitational energy you lose corresponds to the magnitude of g - if you lose a lot of energy for a very small change in position then g is big, conversely if you lose a little gravitational potential energy over a large positional change then g is small.
Bringing it back to the earth, this simply means that you lose more gravitational potential energy if you move some small distance at around 2890 km depth (i.e. at the core mantle boundary), than you do if you move the same small distance at the surface of the earth. Note that this fact is related to the fact that the density contrast at the core mantle boundary is greater even than the density contrast at the surface of the earth!
Density within the earth is reasonably well constrained by things such as normal modes, birch's law, and moment of inertia. Temperature is a lot more difficult, and the errors in the earth's temperature profile are more significant. But I don't think the errors in density are too bad. | 1,979 | 8,603 | {"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-2017-04 | longest | en | 0.962456 |
http://wiigamesdownload.tk/puzor/pile-design-calculation-xls-1711.php | 1,513,470,099,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948592202.83/warc/CC-MAIN-20171217000422-20171217022422-00730.warc.gz | 302,525,062 | 4,242 | # Pile design calculation xls
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