url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://www.windows2universe.org?page=/glossary/gravity_defn.html&edu=high | 1,498,406,501,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320539.37/warc/CC-MAIN-20170625152316-20170625172316-00146.warc.gz | 968,802,809 | 12,994 | ## Shop Windows to the Universe
Ready, Set, SCIENCE!, by the National Research Council, focuses on K-8 science classsrooms. Check out the other publications in our online store, as well as classroom materials.
An artist's rendering of some of the forces of the universe. The apple falling is of course from the story of Isaac Newton discovering the law of gravity as an apple fell from a tree he was sitting underneath.
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# Gravity Definition Page
Gravity is one of the universal forces of nature. It is an attractive force between all matter, and is very weak as compared to the other forces of nature. The gravitational force between two objects is dependent on their masses, which is why we can only see gravity in action when at least one of the objects is very large (like the Earth).
Isaac Newton was the first scientist to define gravity mathematically when he formulated his law of universal gravitation, which says that the force of gravity between two objects (F) equals the mass of one object multiplied by the mass of the second object multiplied by the Gravitational constant (G, equal to 6.67 x 10^-11 Newton meters ^2/kg^2), all divided by the square of the distance between the two objects. This means that gravity is strongest between two very large objects, and gets much weaker as these objects get further apart.
One of the applications of this law is the concept of ‘escape velocity’, which is the velocity an object needs to achieve to escape the gravitational pull of another object (like the Earth). Escape velocity is equal to the square root of the quantity 2GM/R, where M is the mass of the object you’re trying to escape from, G is the Gravitational constant, and R is the radius of the object you’re escaping from. If we plug in the measurements we have for the planet Earth, we see that Earth’s escape velocity is about 11 km/s. This means that if you could throw a baseball at 11 km/s, it would never come down!
The concept of escape velocity is especially interesting when you consider black holes. These objects are extremely dense and very small, so M is huge and R is tiny. When we calculate the escape velocity for these objects, we find that the number is actually the speed of light, so not even light can get out! The size of these objects is almost unbelievable--if the Earth were to collapse so that it was this dense, it would end up being only about 2 cm in diameter!
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## Traveling Nitrogen Classroom Activity Kit
Check out our online store - minerals, fossils, books, activities, jewelry, and household items!...more
## Fundamental Forces
The interactions in the Universe are governed by four forces (strong, weak, electromagnetic and gravitational). Physicists are trying to derive a unified theory that would describe all the forces in nature...more
## What is mass?
Would it be more difficult to pull an elephant or a mouse? If you pulled each animal with the same amount of force, the elephant would respond less to pulling, even if he didn’t pull back at all. That’s...more
## Evidence of Evolution
Scientists use the word theory differently than nonscientists. "It’s just a theory," you hear people say. When speaking casually, people often use the word "theory" to mean a "guess" or a "feeling". In...more
## A Newly Discovered Solar System Contains Scaled-Down Versions of Saturn and Jupiter
A team of international astronomers reports in the Feb. 15 issue of Science the discovery of a solar system nearly 5,000 light years away containing scaled-down versions of Jupiter and Saturn. Their findings...more
## Small Planet, Small Star
Astronomers have discovered an extrasolar planet only three times more massive than our own, the smallest yet observed orbiting a normal star. The star itself is not large, perhaps as little as one twentieth...more
## Atmospheric Science Literacy - Essential Principle 1
Earth has a thin atmosphere that sustains life. Fundamental Concept 1.1 Earth's atmosphere is a mixture of gases with small, but important, quantities of liquid and solid particles. Fundamental Concept...more
## Earth Science Literacy - Big Idea 3
Earth is a complex system of interacting rock, water, air, and life. Big Idea 3.1 The four major systems of Earth are the geosphere, hydrosphere, atmosphere, and biosphere. The geosphere includes a metallic...more
Windows to the Universe, a project of the National Earth Science Teachers Association, is sponsored in part is sponsored in part through grants from federal agencies (NASA and NOAA), and partnerships with affiliated organizations, including the American Geophysical Union, the Howard Hughes Medical Institute, the Earth System Information Partnership, the American Meteorological Society, the National Center for Science Education, and TERC. The American Geophysical Union and the American Geosciences Institute are Windows to the Universe Founding Partners. NESTA welcomes new Institutional Affiliates in support of our ongoing programs, as well as collaborations on new projects. Contact NESTA for more information. | 1,096 | 5,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-26 | longest | en | 0.95913 |
https://www.paddlepaddle.org.cn/documentation/docs/en/api/paddle/tensor/linalg/cholesky_en.html | 1,618,308,712,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038072180.33/warc/CC-MAIN-20210413092418-20210413122418-00479.warc.gz | 1,043,222,558 | 20,380 | cholesky¶
`paddle.` `cholesky` ( x, upper=False, name=None ) [source]
Computes the Cholesky decomposition of one symmetric positive-definite matrix or batches of symmetric positive-definite matrice.
If upper is True, the decomposition has the form \(A = U^{T}U\) , and the returned matrix \(U\) is upper-triangular. Otherwise, the decomposition has the form \(A = LL^{T}\) , and the returned matrix \(L\) is lower-triangular.
Parameters
• x (Tensor) – The input tensor. Its shape should be [*, M, M], where * is zero or more batch dimensions, and matrices on the inner-most 2 dimensions all should be symmetric positive-definite. Its data type should be float32 or float64.
• upper (bool) – The flag indicating whether to return upper or lower triangular matrices. Default: False.
Returns
A Tensor with same shape and data type as x. It represents
triangular matrices generated by Cholesky decomposition.
Return type
Tensor
Examples
```import paddle
import numpy as np
a = np.random.rand(3, 3)
a_t = np.transpose(a, [1, 0])
x_data = np.matmul(a, a_t) + 1e-03 | 276 | 1,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.707757 |
https://gmatclub.com/forum/flavonoids-are-a-common-component-of-almost-all-plants-but-a-specific-245438.html?kudos=1 | 1,569,058,551,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574377.11/warc/CC-MAIN-20190921084226-20190921110226-00101.warc.gz | 486,087,495 | 144,282 | GMAT Question of the Day - Daily to your Mailbox; hard ones only
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# Flavonoids are a common component of almost all plants, but a specific
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Status: Greatness begins beyond your comfort zone
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Flavonoids are a common component of almost all plants, but a specific [#permalink]
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24 Jul 2017, 19:49
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Flavonoids are a common component of almost all plants, but a specific variety of flavonoid in apples has been found to be an antioxidant. Antioxidants are known to be a factor in the prevention of heart disease.
Which one of the following can be properly inferred from the passage?
(A) A diet composed largely of fruits and vegetables will help to prevent heart disease.
(B) Flavonoids are essential to preventing heart disease.
(C) Eating at least one apple each day will prevent heart disease.
(D) At least one type of flavonoid helps to prevent heart disease.
(E) A diet deficient in antioxidants is a common cause of heart disease.
_________________
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The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
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Re: Flavonoids are a common component of almost all plants, but a specific [#permalink]
### Show Tags
25 Jul 2017, 07:00
1
Skywalker18 wrote:
Flavonoids are a common component of almost all plants, but a specific variety of flavonoid in apples has been found to be an antioxidant. Antioxidants are known to be a factor in the prevention of heart disease.
Which one of the following can be properly inferred from the passage?
(A) A diet composed largely of fruits and vegetables will help to prevent heart disease.
(B) Flavonoids are essential to preventing heart disease.
(C) Eating at least one apple each day will prevent heart disease.
(D) At least one type of flavonoid helps to prevent heart disease.
(E) A diet deficient in antioxidants is a common cause of heart disease.
I'm choosing between C and D (A, B, and E are irrelevant)
Seems like "eating at least one apple each day will prevent heart disease" is for 100% sure (but we can't be 100% sure on that) whereas "at least one type of flavonoid helps to prevent heart disease" which is less intense in wording and correct according to the question stem.
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Joined: 23 Jul 2017
Posts: 2
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Concentration: General Management, Strategy
GMAT 1: 460 Q41 V13
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Re: Flavonoids are a common component of almost all plants, but a specific [#permalink]
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10 Aug 2017, 01:35
Chiru404 wrote:
Why cant be answer choice 'E'
Hi Chiru404,
E is out of scope since we don't talk about common causes of heart diseases.
We talk only about antioxidants being a factor in the prevention of heart disease.
Prevention is not equal cause.
A-->B doesn't mean B-->A
Hope this will help.
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Re: Flavonoids are a common component of almost all plants, but a specific [#permalink]
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30 Sep 2018, 21:21
D)
Since a specific variety of Flavonoids in apples are found to be an antioxidant and antioxidants are known to be a factor in prevention of heart disease.
Linking all the points above-At least one type of flavonoid helps to prevent heart disease.
Thanks
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Joined: 29 Oct 2015
Posts: 197
Flavonoids are a common component of almost all plants, but a specific [#permalink]
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Updated on: 13 Apr 2019, 03:29
Option A :- Too strong..."will help"..
Option B :- Too strong.."are essential"..
Option C :- Too strong..."will prevent"...
Option D :- moderate and correct option..."helps to prevent"...
Option E :- too generalised..."..a common cause.."
Give me kudos if you like my explanation...
Posted from my mobile device
Originally posted by sayan640 on 13 Apr 2019, 03:28.
Last edited by sayan640 on 13 Apr 2019, 03:29, edited 1 time in total.
Flavonoids are a common component of almost all plants, but a specific [#permalink] 13 Apr 2019, 03:28
Display posts from previous: Sort by | 1,300 | 5,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-39 | latest | en | 0.897004 |
https://www.luogu.com.cn/blog/xtb1228/solution-p1774 | 1,582,232,289,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145282.57/warc/CC-MAIN-20200220193228-20200220223228-00392.warc.gz | 713,402,448 | 3,012 | # henrytb 的博客
### 题解 P1774 【最接近神的人_NOI导刊2010提高（02）】
posted on 2018-12-23 15:19:01 | under 题解 |
//直接复制不能过哦
#include <cstdio>
#include <cstdlib>
#include <iostream>
#define Mid ((L+R)/2)
using namespace std;
const int N=1000005;
long long A[N],B[N],count=0,n;
void work(int L,int R){
if(L==R)return;
work(L,Mid);
work(Mid+1,R);
int tl=L,tr=Mid+1,t=L;
while(tl<=Mid&&tr<=R){
if(A[tl]<=A[tr])B[t++]=A[tl++];
else B[t++]=A[tr++],count+=tr-t;
}
while(tl<=Mid) B[t++]=A[tl++];
while(tr<=R) B[t++]=A[tr++];
for(int i=L;i<=R;i++)A[i]=B[i];
}
int main(){
scanf("%lld",&n);
for(int i=1;i<=n;i++)scanf("%lld",&A[i]);
work(1,n);
printf("%lld",count);
return 0;
}
//优秀的归并 | 273 | 660 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.16402 |
http://cr.openjdk.java.net/~bpb/7192954_4396272_7039391/src/share/classes/sun/misc/FloatingDecimal.java.sdiff.html | 1,656,804,959,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00585.warc.gz | 14,849,188 | 22,748 | ## src/share/classes/sun/misc/FloatingDecimal.java
Print this page
```rev 7487 : 7192954: Fix Float.parseFloat to round correctly and preserve monotonicity.
4396272: Parsing doubles fails to follow IEEE for largest decimal that should yield 0
7039391: Use Math.ulp in FloatingDecimal
Summary: Correct rounding and monotonicity problems in floats and doubles
Reviewed-by: martin
Contributed-by: Dmitry Nadezhin <dmitry.nadezhin@oracle.com>, Louis Wasserman <lowasser@google.com>```
``` `````` 32 * A class for converting between ASCII and decimal representations of a single 33 * or double precision floating point number. Most conversions are provided via 34 * static convenience methods, although a BinaryToASCIIConverter 35 * instance may be obtained and reused. 36 */ 37 public class FloatingDecimal{ 38 // 39 // Constants of the implementation; 40 // most are IEEE-754 related. 41 // (There are more really boring constants at the end.) 42 // 43 static final int EXP_SHIFT = DoubleConsts.SIGNIFICAND_WIDTH - 1; 44 static final long FRACT_HOB = ( 1L<String. 63 * 64 * @param d The double precision value. 65 * @return The value converted to a String. 66 */ 67 public static String toJavaFormatString(double d) { 68 return getBinaryToASCIIConverter(d).toJavaFormatString(); 69 } 70 71 /** 72 * Converts a single precision floating point value to a String. 73 * 74 * @param f The single precision value. 75 * @return The value converted to a String. 76 */ 77 public static String toJavaFormatString(float f) { `````` 985 } 986 987 /** 988 * A converter which can process an ASCII String representation 989 * of a single or double precision floating point value into a 990 * float or a double. 991 */ 992 interface ASCIIToBinaryConverter { 993 994 double doubleValue(); 995 996 float floatValue(); 997 998 } 999 1000 /** 1001 * A ASCIIToBinaryConverter container for a double. 1002 */ 1003 static class PreparedASCIIToBinaryBuffer implements ASCIIToBinaryConverter { 1004 final private double doubleVal; 1005 private int roundDir = 0; 1006 1007 public PreparedASCIIToBinaryBuffer(double doubleVal) { 1008 this.doubleVal = doubleVal; 1009 } 1010 1011 public PreparedASCIIToBinaryBuffer(double doubleVal, int roundDir) { 1012 this.doubleVal = doubleVal; 1013 this.roundDir = roundDir; 1014 } 1015 1016 @Override 1017 public double doubleValue() { 1018 return doubleVal; 1019 } 1020 1021 @Override 1022 public float floatValue() { 1023 return stickyRound(doubleVal,roundDir); 1024 } 1025 } 1026 1027 static final ASCIIToBinaryConverter A2BC_POSITIVE_INFINITY = new PreparedASCIIToBinaryBuffer(Double.POSITIVE_INFINITY); 1028 static final ASCIIToBinaryConverter A2BC_NEGATIVE_INFINITY = new PreparedASCIIToBinaryBuffer(Double.NEGATIVE_INFINITY); 1029 static final ASCIIToBinaryConverter A2BC_NOT_A_NUMBER = new PreparedASCIIToBinaryBuffer(Double.NaN); 1030 static final ASCIIToBinaryConverter A2BC_POSITIVE_ZERO = new PreparedASCIIToBinaryBuffer(0.0d); 1031 static final ASCIIToBinaryConverter A2BC_NEGATIVE_ZERO = new PreparedASCIIToBinaryBuffer(-0.0d); 1032 1033 /** 1034 * A buffered implementation of ASCIIToBinaryConverter. 1035 */ 1036 static class ASCIIToBinaryBuffer implements ASCIIToBinaryConverter { 1037 boolean isNegative; 1038 int decExponent; 1039 char digits[]; 1040 int nDigits; 1041 int roundDir = 0; // set by doubleValue 1042 1043 ASCIIToBinaryBuffer( boolean negSign, int decExponent, char[] digits, int n) 1044 { 1045 this.isNegative = negSign; 1046 this.decExponent = decExponent; 1047 this.digits = digits; 1048 this.nDigits = n; 1049 } 1050 1051 @Override 1052 public double doubleValue() { 1053 return doubleValue(false); 1054 } 1055 1056 /** 1057 * Computes a number that is the ULP of the given value, 1058 * for purposes of addition/subtraction. Generally easy. 1059 * More difficult if subtracting and the argument 1060 * is a normalized a power of 2, as the ULP changes at these points. 1061 */ 1062 private static double ulp(double dval, boolean subtracting) { 1063 long lbits = Double.doubleToLongBits(dval) & ~DoubleConsts.SIGN_BIT_MASK; 1064 int binexp = (int) (lbits >>> EXP_SHIFT); 1065 double ulpval; 1066 if (subtracting && (binexp >= EXP_SHIFT) && ((lbits & DoubleConsts.SIGNIF_BIT_MASK) == 0L)) { 1067 // for subtraction from normalized, powers of 2, 1068 // use next-smaller exponent 1069 binexp -= 1; 1070 } 1071 if (binexp > EXP_SHIFT) { 1072 ulpval = Double.longBitsToDouble(((long) (binexp - EXP_SHIFT)) << EXP_SHIFT); 1073 } else if (binexp == 0) { 1074 ulpval = Double.MIN_VALUE; 1075 } else { 1076 ulpval = Double.longBitsToDouble(1L << (binexp - 1)); 1077 } 1078 if (subtracting) { 1079 ulpval = -ulpval; 1080 } 1081 1082 return ulpval; 1083 } 1084 1085 /** 1086 * Takes a FloatingDecimal, which we presumably just scanned in, 1087 * and finds out what its value is, as a double. 1088 * 1089 * AS A SIDE EFFECT, SET roundDir TO INDICATE PREFERRED 1090 * ROUNDING DIRECTION in case the result is really destined 1091 * for a single-precision float. 1092 */ 1093 private strictfp double doubleValue(boolean mustSetRoundDir) { 1094 int kDigits = Math.min(nDigits, MAX_DECIMAL_DIGITS + 1); 1095 long lValue; 1096 double dValue; 1097 double rValue; 1098 1099 if (mustSetRoundDir) { 1100 roundDir = 0; 1101 } 1102 // 1103 // convert the lead kDigits to a long integer. 1104 // 1105 // (special performance hack: start to do it using int) 1106 int iValue = (int) digits[0] - (int) '0'; 1107 int iDigits = Math.min(kDigits, INT_DECIMAL_DIGITS); 1108 for (int i = 1; i < iDigits; i++) { 1109 iValue = iValue * 10 + (int) digits[i] - (int) '0'; 1110 } 1111 lValue = (long) iValue; 1112 for (int i = iDigits; i < kDigits; i++) { 1113 lValue = lValue * 10L + (long) ((int) digits[i] - (int) '0'); 1114 } 1115 dValue = (double) lValue; 1116 int exp = decExponent - kDigits; 1117 // 1118 // lValue now contains a long integer with the value of 1119 // the first kDigits digits of the number. 1120 // dValue contains the (double) of the same. 1121 // 1122 1123 if (nDigits <= MAX_DECIMAL_DIGITS) { 1124 // 1125 // possibly an easy case. 1126 // We know that the digits can be represented 1127 // exactly. And if the exponent isn't too outrageous, 1128 // the whole thing can be done with one operation, 1129 // thus one rounding error. 1130 // Note that all our constructors trim all leading and 1131 // trailing zeros, so simple values (including zero) 1132 // will always end up here 1133 // 1134 if (exp == 0 || dValue == 0.0) { 1135 return (isNegative) ? -dValue : dValue; // small floating integer 1136 } 1137 else if (exp >= 0) { 1138 if (exp <= MAX_SMALL_TEN) { 1139 // 1140 // Can get the answer with one operation, 1141 // thus one roundoff. 1142 // 1143 rValue = dValue * SMALL_10_POW[exp]; 1144 if (mustSetRoundDir) { 1145 double tValue = rValue / SMALL_10_POW[exp]; 1146 roundDir = (tValue == dValue) ? 0 1147 : (tValue < dValue) ? 1 1148 : -1; 1149 } 1150 return (isNegative) ? -rValue : rValue; 1151 } 1152 int slop = MAX_DECIMAL_DIGITS - kDigits; 1153 if (exp <= MAX_SMALL_TEN + slop) { 1154 // 1155 // We can multiply dValue by 10^(slop) 1156 // and it is still "small" and exact. 1157 // Then we can multiply by 10^(exp-slop) 1158 // with one rounding. 1159 // 1160 dValue *= SMALL_10_POW[slop]; 1161 rValue = dValue * SMALL_10_POW[exp - slop]; 1162 1163 if (mustSetRoundDir) { 1164 double tValue = rValue / SMALL_10_POW[exp - slop]; 1165 roundDir = (tValue == dValue) ? 0 1166 : (tValue < dValue) ? 1 1167 : -1; 1168 } 1169 return (isNegative) ? -rValue : rValue; 1170 } 1171 // 1172 // Else we have a hard case with a positive exp. 1173 // 1174 } else { 1175 if (exp >= -MAX_SMALL_TEN) { 1176 // 1177 // Can get the answer in one division. 1178 // 1179 rValue = dValue / SMALL_10_POW[-exp]; 1180 if (mustSetRoundDir) { 1181 double tValue = rValue * SMALL_10_POW[-exp]; 1182 roundDir = (tValue == dValue) ? 0 1183 : (tValue < dValue) ? 1 1184 : -1; 1185 } 1186 return (isNegative) ? -rValue : rValue; 1187 } 1188 // 1189 // Else we have a hard case with a negative exp. 1190 // 1191 } 1192 } 1193 1194 // 1195 // Harder cases: 1196 // The sum of digits plus exponent is greater than 1197 // what we think we can do with one error. 1198 // 1199 // Start by approximating the right answer by, 1200 // naively, scaling by powers of 10. 1201 // 1202 if (exp > 0) { 1203 if (decExponent > MAX_DECIMAL_EXPONENT + 1) { 1204 // 1205 // Lets face it. This is going to be ``````1286 // Double.MIN_VALUE ). 1287 // 1288 t = dValue * 2.0; 1289 t *= TINY_10_POW[j]; 1290 if (t == 0.0) { 1291 return (isNegative) ? -0.0 : 0.0; 1292 } 1293 t = Double.MIN_VALUE; 1294 } 1295 dValue = t; 1296 } 1297 } 1298 1299 // 1300 // dValue is now approximately the result. 1301 // The hard part is adjusting it, by comparison 1302 // with FDBigInteger arithmetic. 1303 // Formulate the EXACT big-number result as 1304 // bigD0 * 10^exp 1305 // 1306 FDBigInteger bigD0 = new FDBigInteger(lValue, digits, kDigits, nDigits); 1307 exp = decExponent - nDigits; 1308 1309 final int B5 = Math.max(0, -exp); // powers of 5 in bigB, value is not modified inside correctionLoop 1310 final int D5 = Math.max(0, exp); // powers of 5 in bigD, value is not modified inside correctionLoop 1311 bigD0 = bigD0.multByPow52(D5, 0); 1312 bigD0.makeImmutable(); // prevent bigD0 modification inside correctionLoop 1313 FDBigInteger bigD = null; 1314 int prevD2 = 0; 1315 1316 correctionLoop: 1317 while (true) { 1318 // here dValue can't be NaN, Infinity or zero 1319 long bigBbits = Double.doubleToRawLongBits(dValue) & ~DoubleConsts.SIGN_BIT_MASK; 1320 int binexp = (int) (bigBbits >>> EXP_SHIFT); 1321 bigBbits &= DoubleConsts.SIGNIF_BIT_MASK; 1322 if (binexp > 0) { 1323 bigBbits |= FRACT_HOB; 1324 } else { // Normalize denormalized numbers. 1325 assert bigBbits != 0L : bigBbits; // doubleToBigInt(0.0) 1326 int leadingZeros = Long.numberOfLeadingZeros(bigBbits); 1327 int shift = leadingZeros - (63 - EXP_SHIFT); 1328 bigBbits <<= shift; 1329 binexp = 1 - shift; 1330 } 1331 binexp -= DoubleConsts.EXP_BIAS; 1332 int lowOrderZeros = Long.numberOfTrailingZeros(bigBbits); 1333 bigBbits >>>= lowOrderZeros; 1334 final int bigIntExp = binexp - EXP_SHIFT + lowOrderZeros; 1335 final int bigIntNBits = EXP_SHIFT + 1 - lowOrderZeros; 1336 1337 // 1338 // Scale bigD, bigB appropriately for 1339 // big-integer operations. 1340 // Naively, we multiply by powers of ten 1341 // and powers of two. What we actually do 1342 // is keep track of the powers of 5 and 1343 // powers of 2 we would use, then factor out 1344 // common divisors before doing the work. 1345 // 1346 int B2 = B5; // powers of 2 in bigB 1347 int D2 = D5; // powers of 2 in bigD 1348 int Ulp2; // powers of 2 in halfUlp. 1349 if (bigIntExp >= 0) { 1350 B2 += bigIntExp; 1351 } else { 1352 D2 -= bigIntExp; 1353 } 1354 Ulp2 = B2; 1355 // shift bigB and bigD left by a number s. t. 1356 // halfUlp is still an integer. 1357 int hulpbias; 1358 if (binexp <= -DoubleConsts.EXP_BIAS) { 1359 // This is going to be a denormalized number 1360 // (if not actually zero). 1361 // half an ULP is at 2^-(expBias+EXP_SHIFT+1) 1362 hulpbias = binexp + lowOrderZeros + DoubleConsts.EXP_BIAS; 1363 } else { 1364 hulpbias = 1 + lowOrderZeros; 1365 } 1366 B2 += hulpbias; 1367 D2 += hulpbias; 1368 // if there are common factors of 2, we might just as well 1369 // factor them out, as they add nothing useful. 1370 int common2 = Math.min(B2, Math.min(D2, Ulp2)); 1371 B2 -= common2; 1372 D2 -= common2; 1373 Ulp2 -= common2; 1374 // do multiplications by powers of 5 and 2 1375 FDBigInteger bigB = FDBigInteger.valueOfMulPow52(bigBbits, B5, B2); 1376 if (bigD == null || prevD2 != D2) { 1377 bigD = bigD0.leftShift(D2); 1378 prevD2 = D2; 1379 } 1380 // 1381 // to recap: ``````1405 Ulp2 -= 1; 1406 if (Ulp2 < 0) { 1407 // rats. Cannot de-scale ulp this far. 1408 // must scale diff in other direction. 1409 Ulp2 = 0; 1410 diff = diff.leftShift(1); 1411 } 1412 } 1413 } else if (cmpResult < 0) { 1414 overvalue = false; // our candidate is too small. 1415 diff = bigD.rightInplaceSub(bigB); // bigB is not user further - reuse 1416 } else { 1417 // the candidate is exactly right! 1418 // this happens with surprising frequency 1419 break correctionLoop; 1420 } 1421 cmpResult = diff.cmpPow52(B5, Ulp2); 1422 if ((cmpResult) < 0) { 1423 // difference is small. 1424 // this is close enough 1425 if (mustSetRoundDir) { 1426 roundDir = overvalue ? -1 : 1; 1427 } 1428 break correctionLoop; 1429 } else if (cmpResult == 0) { 1430 // difference is exactly half an ULP 1431 // round to some other value maybe, then finish 1432 dValue += 0.5 * ulp(dValue, overvalue); 1433 // should check for bigIntNBits == 1 here?? 1434 if (mustSetRoundDir) { 1435 roundDir = overvalue ? -1 : 1; 1436 } 1437 break correctionLoop; 1438 } else { 1439 // difference is non-trivial. 1440 // could scale addend by ratio of difference to 1441 // halfUlp here, if we bothered to compute that difference. 1442 // Most of the time ( I hope ) it is about 1 anyway. 1443 dValue += ulp(dValue, overvalue); 1444 if (dValue == 0.0 || dValue == Double.POSITIVE_INFINITY) { 1445 break correctionLoop; // oops. Fell off end of range. 1446 } 1447 continue; // try again. 1448 } 1449 1450 } 1451 return (isNegative) ? -dValue : dValue; 1452 } 1453 1454 /** 1455 * Takes a FloatingDecimal, which we presumably just scanned in, 1456 * and finds out what its value is, as a float. 1457 * This is distinct from doubleValue() to avoid the extremely 1458 * unlikely case of a double rounding error, wherein the conversion 1459 * to double has one rounding error, and the conversion of that double 1460 * to a float has another rounding error, IN THE WRONG DIRECTION, 1461 * ( because of the preference to a zero low-order bit ). 1462 */ 1463 @Override 1464 public strictfp float floatValue() { 1465 int kDigits = Math.min(nDigits, SINGLE_MAX_DECIMAL_DIGITS + 1); 1466 int iValue; 1467 float fValue; 1468 // 1469 // convert the lead kDigits to an integer. 1470 // 1471 iValue = (int) digits[0] - (int) '0'; 1472 for (int i = 1; i < kDigits; i++) { 1473 iValue = iValue * 10 + (int) digits[i] - (int) '0'; 1474 } 1475 fValue = (float) iValue; 1476 int exp = decExponent - kDigits; 1477 // 1478 // iValue now contains an integer with the value of 1479 // the first kDigits digits of the number. 1480 // fValue contains the (float) of the same. 1481 // 1482 1483 if (nDigits <= SINGLE_MAX_DECIMAL_DIGITS) { 1484 // 1485 // possibly an easy case. 1486 // We know that the digits can be represented 1487 // exactly. And if the exponent isn't too outrageous, 1488 // the whole thing can be done with one operation, 1489 // thus one rounding error. 1490 // Note that all our constructors trim all leading and 1491 // trailing zeros, so simple values (including zero) 1492 // will always end up here. 1493 // 1494 if (exp == 0 || fValue == 0.0f) { 1495 return (isNegative) ? -fValue : fValue; // small floating integer 1496 } else if (exp >= 0) { 1497 if (exp <= SINGLE_MAX_SMALL_TEN) { 1498 // 1499 // Can get the answer with one operation, 1500 // thus one roundoff. 1501 // 1502 fValue *= SINGLE_SMALL_10_POW[exp]; 1503 return (isNegative) ? -fValue : fValue; 1504 } 1505 int slop = SINGLE_MAX_DECIMAL_DIGITS - kDigits; 1506 if (exp <= SINGLE_MAX_SMALL_TEN + slop) { 1507 // 1508 // We can multiply dValue by 10^(slop) 1509 // and it is still "small" and exact. 1510 // Then we can multiply by 10^(exp-slop) 1511 // with one rounding. 1512 // 1513 fValue *= SINGLE_SMALL_10_POW[slop]; 1514 fValue *= SINGLE_SMALL_10_POW[exp - slop]; 1515 return (isNegative) ? -fValue : fValue; 1516 } 1517 // 1518 // Else we have a hard case with a positive exp. 1519 // 1520 } else { 1521 if (exp >= -SINGLE_MAX_SMALL_TEN) { 1522 // 1523 // Can get the answer in one division. 1524 // 1525 fValue /= SINGLE_SMALL_10_POW[-exp]; 1526 return (isNegative) ? -fValue : fValue; 1527 } 1528 // ``````1538 // First, finish accumulating digits. 1539 // Then convert that integer to a double, multiply 1540 // by the appropriate power of ten, and convert to float. 1541 // 1542 long lValue = (long) iValue; 1543 for (int i = kDigits; i < nDigits; i++) { 1544 lValue = lValue * 10L + (long) ((int) digits[i] - (int) '0'); 1545 } 1546 double dValue = (double) lValue; 1547 exp = decExponent - nDigits; 1548 dValue *= SMALL_10_POW[exp]; 1549 fValue = (float) dValue; 1550 return (isNegative) ? -fValue : fValue; 1551 1552 } 1553 // 1554 // Harder cases: 1555 // The sum of digits plus exponent is greater than 1556 // what we think we can do with one error. 1557 // 1558 // Start by weeding out obviously out-of-range 1559 // results, then convert to double and go to 1560 // common hard-case code. 1561 // 1562 if (decExponent > SINGLE_MAX_DECIMAL_EXPONENT + 1) { 1563 // 1564 // Lets face it. This is going to be 1565 // Infinity. Cut to the chase. 1566 // 1567 return (isNegative) ? Float.NEGATIVE_INFINITY : Float.POSITIVE_INFINITY; 1568 } else if (decExponent < SINGLE_MIN_DECIMAL_EXPONENT - 1) { 1569 // 1570 // Lets face it. This is going to be 1571 // zero. Cut to the chase. 1572 // 1573 return (isNegative) ? -0.0f : 0.0f; 1574 } 1575 1576 // 1577 // Here, we do 'way too much work, but throwing away 1578 // our partial results, and going and doing the whole 1579 // thing as double, then throwing away half the bits that computes 1580 // when we convert back to float. 1581 // 1582 // The alternative is to reproduce the whole multiple-precision 1583 // algorithm for float precision, or to try to parameterize it 1584 // for common usage. The former will take about 400 lines of code, 1585 // and the latter I tried without success. Thus the semi-hack 1586 // answer here. 1587 // 1588 double dValue = doubleValue(true); 1589 return stickyRound(dValue, roundDir); 1590 } 1591 1592 1593 /** 1594 * All the positive powers of 10 that can be 1595 * represented exactly in double/float. 1596 */ 1597 private static final double[] SMALL_10_POW = { 1598 1.0e0, 1599 1.0e1, 1.0e2, 1.0e3, 1.0e4, 1.0e5, 1600 1.0e6, 1.0e7, 1.0e8, 1.0e9, 1.0e10, 1601 1.0e11, 1.0e12, 1.0e13, 1.0e14, 1.0e15, 1602 1.0e16, 1.0e17, 1.0e18, 1.0e19, 1.0e20, 1603 1.0e21, 1.0e22 1604 }; 1605 1606 private static final float[] SINGLE_SMALL_10_POW = { 1607 1.0e0f, 1608 1.0e1f, 1.0e2f, 1.0e3f, 1.0e4f, 1.0e5f, 1609 1.0e6f, 1.0e7f, 1.0e8f, 1.0e9f, 1.0e10f ``````1918 // 1919 // We parsed everything we could. 1920 // If there are leftovers, then this is not good input! 1921 // 1922 if ( i < len && 1923 ((i != len - 1) || 1924 (in.charAt(i) != 'f' && 1925 in.charAt(i) != 'F' && 1926 in.charAt(i) != 'd' && 1927 in.charAt(i) != 'D'))) { 1928 break parseNumber; // go throw exception 1929 } 1930 if(isZero) { 1931 return isNegative ? A2BC_NEGATIVE_ZERO : A2BC_POSITIVE_ZERO; 1932 } 1933 return new ASCIIToBinaryBuffer(isNegative, decExp, digits, nDigits); 1934 } catch ( StringIndexOutOfBoundsException e ){ } 1935 throw new NumberFormatException("For input string: \"" + in + "\""); 1936 } 1937 1938 /** 1939 * Rounds a double to a float. 1940 * In addition to the fraction bits of the double, 1941 * look at the class instance variable roundDir, 1942 * which should help us avoid double-rounding error. 1943 * roundDir was set in hardValueOf if the estimate was 1944 * close enough, but not exact. It tells us which direction 1945 * of rounding is preferred. 1946 */ 1947 static float stickyRound( double dval, int roundDirection ){ 1948 if(roundDirection!=0) { 1949 long lbits = Double.doubleToRawLongBits( dval ); 1950 long binexp = lbits & DoubleConsts.EXP_BIT_MASK; 1951 if ( binexp == 0L || binexp == DoubleConsts.EXP_BIT_MASK ){ 1952 // what we have here is special. 1953 // don't worry, the right thing will happen. 1954 return (float) dval; 1955 } 1956 lbits += (long)roundDirection; // hack-o-matic. 1957 return (float)Double.longBitsToDouble( lbits ); 1958 } else { 1959 return (float)dval; 1960 } 1961 } 1962 1963 1964 private static class HexFloatPattern { 1965 /** 1966 * Grammar is compatible with hexadecimal floating-point constants 1967 * described in section 6.4.4.2 of the C99 specification. 1968 */ 1969 private static final Pattern VALUE = Pattern.compile( 1970 //1 234 56 7 8 9 1971 "([-+])?0[xX](((\\p{XDigit}+)\\.?)|((\\p{XDigit}*)\\.(\\p{XDigit}+)))[pP]([-+])?(\\p{Digit}+)[fFdD]?" 1972 ); 1973 } 1974 1975 /** 1976 * Converts string s to a suitable floating decimal; uses the 1977 * double constructor and sets the roundDir variable appropriately 1978 * in case the value is later converted to a float. 1979 * 1980 * @param s The String to parse. 1981 */ 1982 static ASCIIToBinaryConverter parseHexString(String s) { 1983 // Verify string is a member of the hexadecimal floating-point ``````2265 throw new AssertionError("Unexpected shift distance remainder."); 2266 // break; 2267 } 2268 2269 // Round is set; sticky might be set. 2270 2271 // For the sticky bit, it suffices to check the 2272 // current digit and test for any nonzero digits in 2273 // the remaining unprocessed input. 2274 i++; 2275 while (i < signifLength && !sticky) { 2276 currentDigit = getHexDigit(significandString, i); 2277 sticky = sticky || (currentDigit != 0); 2278 i++; 2279 } 2280 2281 } 2282 // else all of string was seen, round and sticky are 2283 // correct as false. 2284 2285 // Check for overflow and update exponent accordingly. 2286 if (exponent > DoubleConsts.MAX_EXPONENT) { // Infinite result 2287 // overflow to properly signed infinity 2288 return isNegative ? A2BC_NEGATIVE_INFINITY : A2BC_POSITIVE_INFINITY; 2289 } else { // Finite return value 2290 if (exponent <= DoubleConsts.MAX_EXPONENT && // (Usually) normal result 2291 exponent >= DoubleConsts.MIN_EXPONENT) { 2292 2293 // The result returned in this block cannot be a 2294 // zero or subnormal; however after the 2295 // significand is adjusted from rounding, we could 2296 // still overflow in infinity. 2297 2298 // AND exponent bits into significand; if the 2299 // significand is incremented and overflows from 2300 // rounding, this combination will update the 2301 // exponent correctly, even in the case of 2302 // Double.MAX_VALUE overflowing to infinity. 2303 2304 significand = ((( exponent + ``````2373 // Number Round(x) 2374 // x0.00 x0. 2375 // x0.01 x0. 2376 // x0.10 x0. 2377 // x0.11 x1. = x0. +1 2378 // x1.00 x1. 2379 // x1.01 x1. 2380 // x1.10 x1. + 1 2381 // x1.11 x1. + 1 2382 // 2383 boolean leastZero = ((significand & 1L) == 0L); 2384 if ((leastZero && round && sticky) || 2385 ((!leastZero) && round)) { 2386 significand++; 2387 } 2388 2389 double value = isNegative ? 2390 Double.longBitsToDouble(significand | DoubleConsts.SIGN_BIT_MASK) : 2391 Double.longBitsToDouble(significand ); 2392 2393 int roundDir = 0; 2394 // 2395 // Set roundingDir variable field of fd properly so 2396 // that the input string can be properly rounded to a 2397 // float value. There are two cases to consider: 2398 // 2399 // 1. rounding to double discards sticky bit 2400 // information that would change the result of a float 2401 // rounding (near halfway case between two floats) 2402 // 2403 // 2. rounding to double rounds up when rounding up 2404 // would not occur when rounding to float. 2405 // 2406 // For former case only needs to be considered when 2407 // the bits rounded away when casting to float are all 2408 // zero; otherwise, float round bit is properly set 2409 // and sticky will already be true. 2410 // 2411 // The lower exponent bound for the code below is the 2412 // minimum (normalized) subnormal exponent - 1 since a 2413 // value with that exponent can round up to the 2414 // minimum subnormal value and the sticky bit 2415 // information must be preserved (i.e. case 1). 2416 // 2417 if ((exponent >= FloatConsts.MIN_SUB_EXPONENT - 1) && 2418 (exponent <= FloatConsts.MAX_EXPONENT)) { 2419 // Outside above exponent range, the float value 2420 // will be zero or infinity. 2421 2422 // 2423 // If the low-order 28 bits of a rounded double 2424 // significand are 0, the double could be a 2425 // half-way case for a rounding to float. If the 2426 // double value is a half-way case, the double 2427 // significand may have to be modified to round 2428 // the the right float value (see the stickyRound 2429 // method). If the rounding to double has lost 2430 // what would be float sticky bit information, the 2431 // double significand must be incremented. If the 2432 // double value's significand was itself 2433 // incremented, the float value may end up too 2434 // large so the increment should be undone. 2435 // 2436 if ((significand & 0xfffffffL) == 0x0L) { 2437 // For negative values, the sign of the 2438 // roundDir is the same as for positive values 2439 // since adding 1 increasing the significand's 2440 // magnitude and subtracting 1 decreases the 2441 // significand's magnitude. If neither round 2442 // nor sticky is true, the double value is 2443 // exact and no adjustment is required for a 2444 // proper float rounding. 2445 if (round || sticky) { 2446 if (leastZero) { // prerounding lsb is 0 2447 // If round and sticky were both true, 2448 // and the least significant 2449 // significand bit were 0, the rounded 2450 // significand would not have its 2451 // low-order bits be zero. Therefore, 2452 // we only need to adjust the 2453 // significand if round XOR sticky is 2454 // true. 2455 if (round ^ sticky) { 2456 roundDir = 1; 2457 } 2458 } else { // prerounding lsb is 1 2459 // If the prerounding lsb is 1 and the 2460 // resulting significand has its 2461 // low-order bits zero, the significand 2462 // was incremented. Here, we undo the 2463 // increment, which will ensure the 2464 // right guard and sticky bits for the 2465 // float rounding. 2466 if (round) { 2467 roundDir = -1; 2468 } 2469 } 2470 } 2471 } 2472 } 2473 return new PreparedASCIIToBinaryBuffer(value,roundDir); 2474 } 2475 } 2476 } 2477 2478 /** 2479 * Returns s with any leading zeros removed. 2480 */ 2481 static String stripLeadingZeros(String s) { 2482 // return s.replaceFirst("^0+", ""); 2483 if(!s.isEmpty() && s.charAt(0)=='0') { 2484 for(int i=1; iBinaryToASCIIConverter 35 * instance may be obtained and reused. 36 */ 37 public class FloatingDecimal{ 38 // 39 // Constants of the implementation; 40 // most are IEEE-754 related. 41 // (There are more really boring constants at the end.) 42 // 43 static final int EXP_SHIFT = DoubleConsts.SIGNIFICAND_WIDTH - 1; 44 static final long FRACT_HOB = ( 1L<String. 65 * 66 * @param d The double precision value. 67 * @return The value converted to a String. 68 */ 69 public static String toJavaFormatString(double d) { 70 return getBinaryToASCIIConverter(d).toJavaFormatString(); 71 } 72 73 /** 74 * Converts a single precision floating point value to a String. 75 * 76 * @param f The single precision value. 77 * @return The value converted to a String. 78 */ 79 public static String toJavaFormatString(float f) { `````` 987 } 988 989 /** 990 * A converter which can process an ASCII String representation 991 * of a single or double precision floating point value into a 992 * float or a double. 993 */ 994 interface ASCIIToBinaryConverter { 995 996 double doubleValue(); 997 998 float floatValue(); 999 1000 } 1001 1002 /** 1003 * A ASCIIToBinaryConverter container for a double. 1004 */ 1005 static class PreparedASCIIToBinaryBuffer implements ASCIIToBinaryConverter { 1006 final private double doubleVal; 1007 final private float floatVal; 1008 1009 public PreparedASCIIToBinaryBuffer(double doubleVal, float floatVal) { 1010 this.doubleVal = doubleVal; 1011 this.floatVal = floatVal; 1012 } 1013 1014 @Override 1015 public double doubleValue() { 1016 return doubleVal; 1017 } 1018 1019 @Override 1020 public float floatValue() { 1021 return floatVal; 1022 } 1023 } 1024 1025 static final ASCIIToBinaryConverter A2BC_POSITIVE_INFINITY = new PreparedASCIIToBinaryBuffer(Double.POSITIVE_INFINITY, Float.POSITIVE_INFINITY); 1026 static final ASCIIToBinaryConverter A2BC_NEGATIVE_INFINITY = new PreparedASCIIToBinaryBuffer(Double.NEGATIVE_INFINITY, Float.NEGATIVE_INFINITY); 1027 static final ASCIIToBinaryConverter A2BC_NOT_A_NUMBER = new PreparedASCIIToBinaryBuffer(Double.NaN, Float.NaN); 1028 static final ASCIIToBinaryConverter A2BC_POSITIVE_ZERO = new PreparedASCIIToBinaryBuffer(0.0d, 0.0f); 1029 static final ASCIIToBinaryConverter A2BC_NEGATIVE_ZERO = new PreparedASCIIToBinaryBuffer(-0.0d, -0.0f); 1030 1031 /** 1032 * A buffered implementation of ASCIIToBinaryConverter. 1033 */ 1034 static class ASCIIToBinaryBuffer implements ASCIIToBinaryConverter { 1035 boolean isNegative; 1036 int decExponent; 1037 char digits[]; 1038 int nDigits; 1039 1040 ASCIIToBinaryBuffer( boolean negSign, int decExponent, char[] digits, int n) 1041 { 1042 this.isNegative = negSign; 1043 this.decExponent = decExponent; 1044 this.digits = digits; 1045 this.nDigits = n; 1046 } 1047 1048 /* 1049 * Takes a FloatingDecimal, which we presumably just scanned in, 1050 * and finds out what its value is, as a double. 1051 * 1052 * AS A SIDE EFFECT, SET roundDir TO INDICATE PREFERRED 1053 * ROUNDING DIRECTION in case the result is really destined 1054 * for a single-precision float. 1055 */ 1056 @Override 1057 public double doubleValue() { 1058 int kDigits = Math.min(nDigits, MAX_DECIMAL_DIGITS + 1); 1059 // 1060 // convert the lead kDigits to a long integer. 1061 // 1062 // (special performance hack: start to do it using int) 1063 int iValue = (int) digits[0] - (int) '0'; 1064 int iDigits = Math.min(kDigits, INT_DECIMAL_DIGITS); 1065 for (int i = 1; i < iDigits; i++) { 1066 iValue = iValue * 10 + (int) digits[i] - (int) '0'; 1067 } 1068 long lValue = (long) iValue; 1069 for (int i = iDigits; i < kDigits; i++) { 1070 lValue = lValue * 10L + (long) ((int) digits[i] - (int) '0'); 1071 } 1072 double dValue = (double) lValue; 1073 int exp = decExponent - kDigits; 1074 // 1075 // lValue now contains a long integer with the value of 1076 // the first kDigits digits of the number. 1077 // dValue contains the (double) of the same. 1078 // 1079 1080 if (nDigits <= MAX_DECIMAL_DIGITS) { 1081 // 1082 // possibly an easy case. 1083 // We know that the digits can be represented 1084 // exactly. And if the exponent isn't too outrageous, 1085 // the whole thing can be done with one operation, 1086 // thus one rounding error. 1087 // Note that all our constructors trim all leading and 1088 // trailing zeros, so simple values (including zero) 1089 // will always end up here 1090 // 1091 if (exp == 0 || dValue == 0.0) { 1092 return (isNegative) ? -dValue : dValue; // small floating integer 1093 } 1094 else if (exp >= 0) { 1095 if (exp <= MAX_SMALL_TEN) { 1096 // 1097 // Can get the answer with one operation, 1098 // thus one roundoff. 1099 // 1100 double rValue = dValue * SMALL_10_POW[exp]; 1101 return (isNegative) ? -rValue : rValue; 1102 } 1103 int slop = MAX_DECIMAL_DIGITS - kDigits; 1104 if (exp <= MAX_SMALL_TEN + slop) { 1105 // 1106 // We can multiply dValue by 10^(slop) 1107 // and it is still "small" and exact. 1108 // Then we can multiply by 10^(exp-slop) 1109 // with one rounding. 1110 // 1111 dValue *= SMALL_10_POW[slop]; 1112 double rValue = dValue * SMALL_10_POW[exp - slop]; 1113 return (isNegative) ? -rValue : rValue; 1114 } 1115 // 1116 // Else we have a hard case with a positive exp. 1117 // 1118 } else { 1119 if (exp >= -MAX_SMALL_TEN) { 1120 // 1121 // Can get the answer in one division. 1122 // 1123 double rValue = dValue / SMALL_10_POW[-exp]; 1124 return (isNegative) ? -rValue : rValue; 1125 } 1126 // 1127 // Else we have a hard case with a negative exp. 1128 // 1129 } 1130 } 1131 1132 // 1133 // Harder cases: 1134 // The sum of digits plus exponent is greater than 1135 // what we think we can do with one error. 1136 // 1137 // Start by approximating the right answer by, 1138 // naively, scaling by powers of 10. 1139 // 1140 if (exp > 0) { 1141 if (decExponent > MAX_DECIMAL_EXPONENT + 1) { 1142 // 1143 // Lets face it. This is going to be ``````1224 // Double.MIN_VALUE ). 1225 // 1226 t = dValue * 2.0; 1227 t *= TINY_10_POW[j]; 1228 if (t == 0.0) { 1229 return (isNegative) ? -0.0 : 0.0; 1230 } 1231 t = Double.MIN_VALUE; 1232 } 1233 dValue = t; 1234 } 1235 } 1236 1237 // 1238 // dValue is now approximately the result. 1239 // The hard part is adjusting it, by comparison 1240 // with FDBigInteger arithmetic. 1241 // Formulate the EXACT big-number result as 1242 // bigD0 * 10^exp 1243 // 1244 if (nDigits > MAX_NDIGITS) { 1245 nDigits = MAX_NDIGITS + 1; 1246 digits[MAX_NDIGITS] = '1'; 1247 } 1248 FDBigInteger bigD0 = new FDBigInteger(lValue, digits, kDigits, nDigits); 1249 exp = decExponent - nDigits; 1250 1251 long ieeeBits = Double.doubleToRawLongBits(dValue); // IEEE-754 bits of double candidate 1252 final int B5 = Math.max(0, -exp); // powers of 5 in bigB, value is not modified inside correctionLoop 1253 final int D5 = Math.max(0, exp); // powers of 5 in bigD, value is not modified inside correctionLoop 1254 bigD0 = bigD0.multByPow52(D5, 0); 1255 bigD0.makeImmutable(); // prevent bigD0 modification inside correctionLoop 1256 FDBigInteger bigD = null; 1257 int prevD2 = 0; 1258 1259 correctionLoop: 1260 while (true) { 1261 // here ieeeBits can't be NaN, Infinity or zero 1262 int binexp = (int) (ieeeBits >>> EXP_SHIFT); 1263 long bigBbits = ieeeBits & DoubleConsts.SIGNIF_BIT_MASK; 1264 if (binexp > 0) { 1265 bigBbits |= FRACT_HOB; 1266 } else { // Normalize denormalized numbers. 1267 assert bigBbits != 0L : bigBbits; // doubleToBigInt(0.0) 1268 int leadingZeros = Long.numberOfLeadingZeros(bigBbits); 1269 int shift = leadingZeros - (63 - EXP_SHIFT); 1270 bigBbits <<= shift; 1271 binexp = 1 - shift; 1272 } 1273 binexp -= DoubleConsts.EXP_BIAS; 1274 int lowOrderZeros = Long.numberOfTrailingZeros(bigBbits); 1275 bigBbits >>>= lowOrderZeros; 1276 final int bigIntExp = binexp - EXP_SHIFT + lowOrderZeros; 1277 final int bigIntNBits = EXP_SHIFT + 1 - lowOrderZeros; 1278 1279 // 1280 // Scale bigD, bigB appropriately for 1281 // big-integer operations. 1282 // Naively, we multiply by powers of ten 1283 // and powers of two. What we actually do 1284 // is keep track of the powers of 5 and 1285 // powers of 2 we would use, then factor out 1286 // common divisors before doing the work. 1287 // 1288 int B2 = B5; // powers of 2 in bigB 1289 int D2 = D5; // powers of 2 in bigD 1290 int Ulp2; // powers of 2 in halfUlp. 1291 if (bigIntExp >= 0) { 1292 B2 += bigIntExp; 1293 } else { 1294 D2 -= bigIntExp; 1295 } 1296 Ulp2 = B2; 1297 // shift bigB and bigD left by a number s. t. 1298 // halfUlp is still an integer. 1299 int hulpbias; 1300 if (binexp <= -DoubleConsts.EXP_BIAS) { 1301 // This is going to be a denormalized number 1302 // (if not actually zero). 1303 // half an ULP is at 2^-(DoubleConsts.EXP_BIAS+EXP_SHIFT+1) 1304 hulpbias = binexp + lowOrderZeros + DoubleConsts.EXP_BIAS; 1305 } else { 1306 hulpbias = 1 + lowOrderZeros; 1307 } 1308 B2 += hulpbias; 1309 D2 += hulpbias; 1310 // if there are common factors of 2, we might just as well 1311 // factor them out, as they add nothing useful. 1312 int common2 = Math.min(B2, Math.min(D2, Ulp2)); 1313 B2 -= common2; 1314 D2 -= common2; 1315 Ulp2 -= common2; 1316 // do multiplications by powers of 5 and 2 1317 FDBigInteger bigB = FDBigInteger.valueOfMulPow52(bigBbits, B5, B2); 1318 if (bigD == null || prevD2 != D2) { 1319 bigD = bigD0.leftShift(D2); 1320 prevD2 = D2; 1321 } 1322 // 1323 // to recap: ``````1347 Ulp2 -= 1; 1348 if (Ulp2 < 0) { 1349 // rats. Cannot de-scale ulp this far. 1350 // must scale diff in other direction. 1351 Ulp2 = 0; 1352 diff = diff.leftShift(1); 1353 } 1354 } 1355 } else if (cmpResult < 0) { 1356 overvalue = false; // our candidate is too small. 1357 diff = bigD.rightInplaceSub(bigB); // bigB is not user further - reuse 1358 } else { 1359 // the candidate is exactly right! 1360 // this happens with surprising frequency 1361 break correctionLoop; 1362 } 1363 cmpResult = diff.cmpPow52(B5, Ulp2); 1364 if ((cmpResult) < 0) { 1365 // difference is small. 1366 // this is close enough 1367 break correctionLoop; 1368 } else if (cmpResult == 0) { 1369 // difference is exactly half an ULP 1370 // round to some other value maybe, then finish 1371 if ((ieeeBits & 1) != 0) { // half ties to even 1372 ieeeBits += overvalue ? -1 : 1; // nextDown or nextUp 1373 } 1374 break correctionLoop; 1375 } else { 1376 // difference is non-trivial. 1377 // could scale addend by ratio of difference to 1378 // halfUlp here, if we bothered to compute that difference. 1379 // Most of the time ( I hope ) it is about 1 anyway. 1380 ieeeBits += overvalue ? -1 : 1; // nextDown or nextUp 1381 if (ieeeBits == 0 || ieeeBits == DoubleConsts.EXP_BIT_MASK) { // 0.0 or Double.POSITIVE_INFINITY 1382 break correctionLoop; // oops. Fell off end of range. 1383 } 1384 continue; // try again. 1385 } 1386 1387 } 1388 if (isNegative) { 1389 ieeeBits |= DoubleConsts.SIGN_BIT_MASK; 1390 } 1391 return Double.longBitsToDouble(ieeeBits); 1392 } 1393 1394 /** 1395 * Takes a FloatingDecimal, which we presumably just scanned in, 1396 * and finds out what its value is, as a float. 1397 * This is distinct from doubleValue() to avoid the extremely 1398 * unlikely case of a double rounding error, wherein the conversion 1399 * to double has one rounding error, and the conversion of that double 1400 * to a float has another rounding error, IN THE WRONG DIRECTION, 1401 * ( because of the preference to a zero low-order bit ). 1402 */ 1403 @Override 1404 public float floatValue() { 1405 int kDigits = Math.min(nDigits, SINGLE_MAX_DECIMAL_DIGITS + 1); 1406 // 1407 // convert the lead kDigits to an integer. 1408 // 1409 int iValue = (int) digits[0] - (int) '0'; 1410 for (int i = 1; i < kDigits; i++) { 1411 iValue = iValue * 10 + (int) digits[i] - (int) '0'; 1412 } 1413 float fValue = (float) iValue; 1414 int exp = decExponent - kDigits; 1415 // 1416 // iValue now contains an integer with the value of 1417 // the first kDigits digits of the number. 1418 // fValue contains the (float) of the same. 1419 // 1420 1421 if (nDigits <= SINGLE_MAX_DECIMAL_DIGITS) { 1422 // 1423 // possibly an easy case. 1424 // We know that the digits can be represented 1425 // exactly. And if the exponent isn't too outrageous, 1426 // the whole thing can be done with one operation, 1427 // thus one rounding error. 1428 // Note that all our constructors trim all leading and 1429 // trailing zeros, so simple values (including zero) 1430 // will always end up here. 1431 // 1432 if (exp == 0 || fValue == 0.0f) { 1433 return (isNegative) ? -fValue : fValue; // small floating integer 1434 } else if (exp >= 0) { 1435 if (exp <= SINGLE_MAX_SMALL_TEN) { 1436 // 1437 // Can get the answer with one operation, 1438 // thus one roundoff. 1439 // 1440 fValue *= SINGLE_SMALL_10_POW[exp]; 1441 return (isNegative) ? -fValue : fValue; 1442 } 1443 int slop = SINGLE_MAX_DECIMAL_DIGITS - kDigits; 1444 if (exp <= SINGLE_MAX_SMALL_TEN + slop) { 1445 // 1446 // We can multiply fValue by 10^(slop) 1447 // and it is still "small" and exact. 1448 // Then we can multiply by 10^(exp-slop) 1449 // with one rounding. 1450 // 1451 fValue *= SINGLE_SMALL_10_POW[slop]; 1452 fValue *= SINGLE_SMALL_10_POW[exp - slop]; 1453 return (isNegative) ? -fValue : fValue; 1454 } 1455 // 1456 // Else we have a hard case with a positive exp. 1457 // 1458 } else { 1459 if (exp >= -SINGLE_MAX_SMALL_TEN) { 1460 // 1461 // Can get the answer in one division. 1462 // 1463 fValue /= SINGLE_SMALL_10_POW[-exp]; 1464 return (isNegative) ? -fValue : fValue; 1465 } 1466 // ``````1476 // First, finish accumulating digits. 1477 // Then convert that integer to a double, multiply 1478 // by the appropriate power of ten, and convert to float. 1479 // 1480 long lValue = (long) iValue; 1481 for (int i = kDigits; i < nDigits; i++) { 1482 lValue = lValue * 10L + (long) ((int) digits[i] - (int) '0'); 1483 } 1484 double dValue = (double) lValue; 1485 exp = decExponent - nDigits; 1486 dValue *= SMALL_10_POW[exp]; 1487 fValue = (float) dValue; 1488 return (isNegative) ? -fValue : fValue; 1489 1490 } 1491 // 1492 // Harder cases: 1493 // The sum of digits plus exponent is greater than 1494 // what we think we can do with one error. 1495 // 1496 // Start by approximating the right answer by, 1497 // naively, scaling by powers of 10. 1498 // Scaling uses doubles to avoid overflow/underflow. 1499 // 1500 double dValue = fValue; 1501 if (exp > 0) { 1502 if (decExponent > SINGLE_MAX_DECIMAL_EXPONENT + 1) { 1503 // 1504 // Lets face it. This is going to be 1505 // Infinity. Cut to the chase. 1506 // 1507 return (isNegative) ? Float.NEGATIVE_INFINITY : Float.POSITIVE_INFINITY; 1508 } 1509 if ((exp & 15) != 0) { 1510 dValue *= SMALL_10_POW[exp & 15]; 1511 } 1512 if ((exp >>= 4) != 0) { 1513 int j; 1514 for (j = 0; exp > 0; j++, exp >>= 1) { 1515 if ((exp & 1) != 0) { 1516 dValue *= BIG_10_POW[j]; 1517 } 1518 } 1519 } 1520 } else if (exp < 0) { 1521 exp = -exp; 1522 if (decExponent < SINGLE_MIN_DECIMAL_EXPONENT - 1) { 1523 // 1524 // Lets face it. This is going to be 1525 // zero. Cut to the chase. 1526 // 1527 return (isNegative) ? -0.0f : 0.0f; 1528 } 1529 if ((exp & 15) != 0) { 1530 dValue /= SMALL_10_POW[exp & 15]; 1531 } 1532 if ((exp >>= 4) != 0) { 1533 int j; 1534 for (j = 0; exp > 0; j++, exp >>= 1) { 1535 if ((exp & 1) != 0) { 1536 dValue *= TINY_10_POW[j]; 1537 } 1538 } 1539 } 1540 } 1541 fValue = Math.max(Float.MIN_VALUE, Math.min(Float.MAX_VALUE, (float) dValue)); 1542 1543 // 1544 // fValue is now approximately the result. 1545 // The hard part is adjusting it, by comparison 1546 // with FDBigInteger arithmetic. 1547 // Formulate the EXACT big-number result as 1548 // bigD0 * 10^exp 1549 // 1550 if (nDigits > SINGLE_MAX_NDIGITS) { 1551 nDigits = SINGLE_MAX_NDIGITS + 1; 1552 digits[SINGLE_MAX_NDIGITS] = '1'; 1553 } 1554 FDBigInteger bigD0 = new FDBigInteger(iValue, digits, kDigits, nDigits); 1555 exp = decExponent - nDigits; 1556 1557 int ieeeBits = Float.floatToRawIntBits(fValue); // IEEE-754 bits of float candidate 1558 final int B5 = Math.max(0, -exp); // powers of 5 in bigB, value is not modified inside correctionLoop 1559 final int D5 = Math.max(0, exp); // powers of 5 in bigD, value is not modified inside correctionLoop 1560 bigD0 = bigD0.multByPow52(D5, 0); 1561 bigD0.makeImmutable(); // prevent bigD0 modification inside correctionLoop 1562 FDBigInteger bigD = null; 1563 int prevD2 = 0; 1564 1565 correctionLoop: 1566 while (true) { 1567 // here ieeeBits can't be NaN, Infinity or zero 1568 int binexp = ieeeBits >>> SINGLE_EXP_SHIFT; 1569 int bigBbits = ieeeBits & FloatConsts.SIGNIF_BIT_MASK; 1570 if (binexp > 0) { 1571 bigBbits |= SINGLE_FRACT_HOB; 1572 } else { // Normalize denormalized numbers. 1573 assert bigBbits != 0 : bigBbits; // floatToBigInt(0.0) 1574 int leadingZeros = Integer.numberOfLeadingZeros(bigBbits); 1575 int shift = leadingZeros - (31 - SINGLE_EXP_SHIFT); 1576 bigBbits <<= shift; 1577 binexp = 1 - shift; 1578 } 1579 binexp -= FloatConsts.EXP_BIAS; 1580 int lowOrderZeros = Integer.numberOfTrailingZeros(bigBbits); 1581 bigBbits >>>= lowOrderZeros; 1582 final int bigIntExp = binexp - SINGLE_EXP_SHIFT + lowOrderZeros; 1583 final int bigIntNBits = SINGLE_EXP_SHIFT + 1 - lowOrderZeros; 1584 1585 // 1586 // Scale bigD, bigB appropriately for 1587 // big-integer operations. 1588 // Naively, we multiply by powers of ten 1589 // and powers of two. What we actually do 1590 // is keep track of the powers of 5 and 1591 // powers of 2 we would use, then factor out 1592 // common divisors before doing the work. 1593 // 1594 int B2 = B5; // powers of 2 in bigB 1595 int D2 = D5; // powers of 2 in bigD 1596 int Ulp2; // powers of 2 in halfUlp. 1597 if (bigIntExp >= 0) { 1598 B2 += bigIntExp; 1599 } else { 1600 D2 -= bigIntExp; 1601 } 1602 Ulp2 = B2; 1603 // shift bigB and bigD left by a number s. t. 1604 // halfUlp is still an integer. 1605 int hulpbias; 1606 if (binexp <= -FloatConsts.EXP_BIAS) { 1607 // This is going to be a denormalized number 1608 // (if not actually zero). 1609 // half an ULP is at 2^-(FloatConsts.EXP_BIAS+SINGLE_EXP_SHIFT+1) 1610 hulpbias = binexp + lowOrderZeros + FloatConsts.EXP_BIAS; 1611 } else { 1612 hulpbias = 1 + lowOrderZeros; 1613 } 1614 B2 += hulpbias; 1615 D2 += hulpbias; 1616 // if there are common factors of 2, we might just as well 1617 // factor them out, as they add nothing useful. 1618 int common2 = Math.min(B2, Math.min(D2, Ulp2)); 1619 B2 -= common2; 1620 D2 -= common2; 1621 Ulp2 -= common2; 1622 // do multiplications by powers of 5 and 2 1623 FDBigInteger bigB = FDBigInteger.valueOfMulPow52(bigBbits, B5, B2); 1624 if (bigD == null || prevD2 != D2) { 1625 bigD = bigD0.leftShift(D2); 1626 prevD2 = D2; 1627 } 1628 // 1629 // to recap: 1630 // bigB is the scaled-big-int version of our floating-point 1631 // candidate. 1632 // bigD is the scaled-big-int version of the exact value 1633 // as we understand it. 1634 // halfUlp is 1/2 an ulp of bigB, except for special cases 1635 // of exact powers of 2 1636 // 1637 // the plan is to compare bigB with bigD, and if the difference 1638 // is less than halfUlp, then we're satisfied. Otherwise, 1639 // use the ratio of difference to halfUlp to calculate a fudge 1640 // factor to add to the floating value, then go 'round again. 1641 // 1642 FDBigInteger diff; 1643 int cmpResult; 1644 boolean overvalue; 1645 if ((cmpResult = bigB.cmp(bigD)) > 0) { 1646 overvalue = true; // our candidate is too big. 1647 diff = bigB.leftInplaceSub(bigD); // bigB is not user further - reuse 1648 if ((bigIntNBits == 1) && (bigIntExp > -FloatConsts.EXP_BIAS + 1)) { 1649 // candidate is a normalized exact power of 2 and 1650 // is too big (larger than Float.MIN_NORMAL). We will be subtracting. 1651 // For our purposes, ulp is the ulp of the 1652 // next smaller range. 1653 Ulp2 -= 1; 1654 if (Ulp2 < 0) { 1655 // rats. Cannot de-scale ulp this far. 1656 // must scale diff in other direction. 1657 Ulp2 = 0; 1658 diff = diff.leftShift(1); 1659 } 1660 } 1661 } else if (cmpResult < 0) { 1662 overvalue = false; // our candidate is too small. 1663 diff = bigD.rightInplaceSub(bigB); // bigB is not user further - reuse 1664 } else { 1665 // the candidate is exactly right! 1666 // this happens with surprising frequency 1667 break correctionLoop; 1668 } 1669 cmpResult = diff.cmpPow52(B5, Ulp2); 1670 if ((cmpResult) < 0) { 1671 // difference is small. 1672 // this is close enough 1673 break correctionLoop; 1674 } else if (cmpResult == 0) { 1675 // difference is exactly half an ULP 1676 // round to some other value maybe, then finish 1677 if ((ieeeBits & 1) != 0) { // half ties to even 1678 ieeeBits += overvalue ? -1 : 1; // nextDown or nextUp 1679 } 1680 break correctionLoop; 1681 } else { 1682 // difference is non-trivial. 1683 // could scale addend by ratio of difference to 1684 // halfUlp here, if we bothered to compute that difference. 1685 // Most of the time ( I hope ) it is about 1 anyway. 1686 ieeeBits += overvalue ? -1 : 1; // nextDown or nextUp 1687 if (ieeeBits == 0 || ieeeBits == FloatConsts.EXP_BIT_MASK) { // 0.0 or Float.POSITIVE_INFINITY 1688 break correctionLoop; // oops. Fell off end of range. 1689 } 1690 continue; // try again. 1691 } 1692 1693 } 1694 if (isNegative) { 1695 ieeeBits |= FloatConsts.SIGN_BIT_MASK; 1696 } 1697 return Float.intBitsToFloat(ieeeBits); 1698 } 1699 1700 1701 /** 1702 * All the positive powers of 10 that can be 1703 * represented exactly in double/float. 1704 */ 1705 private static final double[] SMALL_10_POW = { 1706 1.0e0, 1707 1.0e1, 1.0e2, 1.0e3, 1.0e4, 1.0e5, 1708 1.0e6, 1.0e7, 1.0e8, 1.0e9, 1.0e10, 1709 1.0e11, 1.0e12, 1.0e13, 1.0e14, 1.0e15, 1710 1.0e16, 1.0e17, 1.0e18, 1.0e19, 1.0e20, 1711 1.0e21, 1.0e22 1712 }; 1713 1714 private static final float[] SINGLE_SMALL_10_POW = { 1715 1.0e0f, 1716 1.0e1f, 1.0e2f, 1.0e3f, 1.0e4f, 1.0e5f, 1717 1.0e6f, 1.0e7f, 1.0e8f, 1.0e9f, 1.0e10f ``````2026 // 2027 // We parsed everything we could. 2028 // If there are leftovers, then this is not good input! 2029 // 2030 if ( i < len && 2031 ((i != len - 1) || 2032 (in.charAt(i) != 'f' && 2033 in.charAt(i) != 'F' && 2034 in.charAt(i) != 'd' && 2035 in.charAt(i) != 'D'))) { 2036 break parseNumber; // go throw exception 2037 } 2038 if(isZero) { 2039 return isNegative ? A2BC_NEGATIVE_ZERO : A2BC_POSITIVE_ZERO; 2040 } 2041 return new ASCIIToBinaryBuffer(isNegative, decExp, digits, nDigits); 2042 } catch ( StringIndexOutOfBoundsException e ){ } 2043 throw new NumberFormatException("For input string: \"" + in + "\""); 2044 } 2045 2046 private static class HexFloatPattern { 2047 /** 2048 * Grammar is compatible with hexadecimal floating-point constants 2049 * described in section 6.4.4.2 of the C99 specification. 2050 */ 2051 private static final Pattern VALUE = Pattern.compile( 2052 //1 234 56 7 8 9 2053 "([-+])?0[xX](((\\p{XDigit}+)\\.?)|((\\p{XDigit}*)\\.(\\p{XDigit}+)))[pP]([-+])?(\\p{Digit}+)[fFdD]?" 2054 ); 2055 } 2056 2057 /** 2058 * Converts string s to a suitable floating decimal; uses the 2059 * double constructor and sets the roundDir variable appropriately 2060 * in case the value is later converted to a float. 2061 * 2062 * @param s The String to parse. 2063 */ 2064 static ASCIIToBinaryConverter parseHexString(String s) { 2065 // Verify string is a member of the hexadecimal floating-point ``````2347 throw new AssertionError("Unexpected shift distance remainder."); 2348 // break; 2349 } 2350 2351 // Round is set; sticky might be set. 2352 2353 // For the sticky bit, it suffices to check the 2354 // current digit and test for any nonzero digits in 2355 // the remaining unprocessed input. 2356 i++; 2357 while (i < signifLength && !sticky) { 2358 currentDigit = getHexDigit(significandString, i); 2359 sticky = sticky || (currentDigit != 0); 2360 i++; 2361 } 2362 2363 } 2364 // else all of string was seen, round and sticky are 2365 // correct as false. 2366 2367 // Float calculations 2368 int floatBits = isNegative ? FloatConsts.SIGN_BIT_MASK : 0; 2369 if (exponent >= FloatConsts.MIN_EXPONENT) { 2370 if (exponent > FloatConsts.MAX_EXPONENT) { 2371 // Float.POSITIVE_INFINITY 2372 floatBits |= FloatConsts.EXP_BIT_MASK; 2373 } else { 2374 int threshShift = DoubleConsts.SIGNIFICAND_WIDTH - FloatConsts.SIGNIFICAND_WIDTH - 1; 2375 boolean floatSticky = (significand & ((1L << threshShift) - 1)) != 0 || round || sticky; 2376 int iValue = (int) (significand >>> threshShift); 2377 if ((iValue & 3) != 1 || floatSticky) { 2378 iValue++; 2379 } 2380 floatBits |= (((((int) exponent) + (FloatConsts.EXP_BIAS - 1))) << SINGLE_EXP_SHIFT) + (iValue >> 1); 2381 } 2382 } else { 2383 if (exponent < FloatConsts.MIN_SUB_EXPONENT - 1) { 2384 // 0 2385 } else { 2386 // exponent == -127 ==> threshShift = 53 - 2 + (-149) - (-127) = 53 - 24 2387 int threshShift = (int) ((DoubleConsts.SIGNIFICAND_WIDTH - 2 + FloatConsts.MIN_SUB_EXPONENT) - exponent); 2388 assert threshShift >= DoubleConsts.SIGNIFICAND_WIDTH - FloatConsts.SIGNIFICAND_WIDTH; 2389 assert threshShift < DoubleConsts.SIGNIFICAND_WIDTH; 2390 boolean floatSticky = (significand & ((1L << threshShift) - 1)) != 0 || round || sticky; 2391 int iValue = (int) (significand >>> threshShift); 2392 if ((iValue & 3) != 1 || floatSticky) { 2393 iValue++; 2394 } 2395 floatBits |= iValue >> 1; 2396 } 2397 } 2398 float fValue = Float.intBitsToFloat(floatBits); 2399 2400 // Check for overflow and update exponent accordingly. 2401 if (exponent > DoubleConsts.MAX_EXPONENT) { // Infinite result 2402 // overflow to properly signed infinity 2403 return isNegative ? A2BC_NEGATIVE_INFINITY : A2BC_POSITIVE_INFINITY; 2404 } else { // Finite return value 2405 if (exponent <= DoubleConsts.MAX_EXPONENT && // (Usually) normal result 2406 exponent >= DoubleConsts.MIN_EXPONENT) { 2407 2408 // The result returned in this block cannot be a 2409 // zero or subnormal; however after the 2410 // significand is adjusted from rounding, we could 2411 // still overflow in infinity. 2412 2413 // AND exponent bits into significand; if the 2414 // significand is incremented and overflows from 2415 // rounding, this combination will update the 2416 // exponent correctly, even in the case of 2417 // Double.MAX_VALUE overflowing to infinity. 2418 2419 significand = ((( exponent + ``````2488 // Number Round(x) 2489 // x0.00 x0. 2490 // x0.01 x0. 2491 // x0.10 x0. 2492 // x0.11 x1. = x0. +1 2493 // x1.00 x1. 2494 // x1.01 x1. 2495 // x1.10 x1. + 1 2496 // x1.11 x1. + 1 2497 // 2498 boolean leastZero = ((significand & 1L) == 0L); 2499 if ((leastZero && round && sticky) || 2500 ((!leastZero) && round)) { 2501 significand++; 2502 } 2503 2504 double value = isNegative ? 2505 Double.longBitsToDouble(significand | DoubleConsts.SIGN_BIT_MASK) : 2506 Double.longBitsToDouble(significand ); 2507 2508 return new PreparedASCIIToBinaryBuffer(value, fValue); 2509 } 2510 } 2511 } 2512 2513 /** 2514 * Returns s with any leading zeros removed. 2515 */ 2516 static String stripLeadingZeros(String s) { 2517 // return s.replaceFirst("^0+", ""); 2518 if(!s.isEmpty() && s.charAt(0)=='0') { 2519 for(int i=1; i | 17,725 | 53,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.55109 |
http://superuser.com/questions/529549/choose-maximum-from-two-excel-cells | 1,467,054,047,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396106.25/warc/CC-MAIN-20160624154956-00066-ip-10-164-35-72.ec2.internal.warc.gz | 278,072,393 | 16,114 | Choose maximum from two Excel cells
We have two pricing models for an item.
• `Price 1` = Cell 1
• `Price 2` = Cell 2
Is there a formula that I can use where the price for the customer would be the whichever is higher among `Price 1` and `Price 2`?
-
• Put 27 in Cell B2
• Put 35 in Cell B3
• Put `=MAX(B2:B3)` in B4
When done, B4 better be 35
It also works on rectangular ranges
• Put 27 in Cell B2
• Put 35 in Cell B3
• Put 37 in Cell C2
• Put 31 in Cell C3
• Put `=MAX(B2:C3)` in B4
When done, B4 better be 37
To be honest, both question and answer are trivial.
Using the B4 Cell, when you type =MAX, a little help screen pops up to give you the option to add cells/values.
I am using Excel 2010, BTW
Click on the `File` tab, look to the far right and you will see the Greek Letter Sigma.
Click on the Sigma's DropDown Arrow and Choose More Functions. You will get a list of Functions available in Excel to Insert into the Cell.
-
Thank you so much! – user1953920 Jan 7 '13 at 21:34 | 305 | 1,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | latest | en | 0.839589 |
https://forum.gamsworld.org/viewtopic.php?f=15&p=28811&sid=2a15ee8031c0abba9313901af8f16e74 | 1,627,116,846,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150134.86/warc/CC-MAIN-20210724063259-20210724093259-00542.warc.gz | 273,773,658 | 7,843 | ## how to use .lp file with GAMS?
Moderator: aileen
Forum rules
abhosekar
Moderator
Posts: 190
Joined: 9 months ago
### how to use .lp file with GAMS?
To generate lp file from GAMS, you can use writelp option. An example is shown here using the rotdk model from GAMS model library where the lines before the solve statement demonstrate how to use the option writelp.
Code: Select all
``````\$title Robust Optimization (ROTDK,SEQ=185)
\$onText
Robust Optimization.
Laguna, M, Applying Robust Optimization to Capacity Expansion of
One Location in Telecommunications with Demand Uncertainty.
Management Science 44, 11 (1998), 101-110.
Keywords: mixed integer linear programming, robust optimization, capacity expansion,
time-dependent knapsack problem
\$offText
Set
s 'scenarios' / 1*1000 /
t 'time periods' / t1*t12 /
j 'components' / C001*C010 /;
Alias (t,tt);
Parameter
di(s,t) 'increment'
D(t,s) 'demand'
c(j) 'capacity size'
p(j) 'capacity cost'
mu 'mean capacity parameter'
sigma 'std capacity parameter';
mu = 100;
sigma = 10;
c(j) = round(uniform(1,mu));
p(j) = round(mu + c(j) + uniform(-sigma,sigma));
di(s,t)\$(ord(s) <= 0.25*card(s)) = round(normal( 50,10));
di(s,t)\$(ord(s) > 0.25*card(s) and ord(s) <= 0.75*card(s)) = round(normal(100,20));
di(s,t)\$(ord(s) > 0.75*card(s)) = round(normal(150,40));
d(t,s) = sum(tt\$(ord(tt) <= ord(t)), di(s,tt));
* display c, p, di, d;
Parameter
dis(t) 'discount factor'
w 'shortage penalty';
dis(t) = power(.86,ord(t) - 1);
w = 5;
Variable
x(j,t) 'expansion'
z(s) 'max capacity shortage'
cap(t) 'installed capacity'
obj;
Integer Variable x;
Positive Variable z;
Equation
capbal(t) 'capacity balance'
dembal(t,s) 'demand balance'
objdef;
objdef.. obj =e= sum((j,t), dis(t)*p(j)*x(j,t)) + w/card(s)*sum(s, z(s));
capbal(t).. cap(t) =e= cap(t-1) + sum(j, c(j)*x(j,t));
dembal(t,s).. cap(t) + z(s) =g= d(t,s);
Model rotdk / all /;
\$onEcho > cplex.opt
lazyconstraints 1
dembal.lazy(t,s)\$(ord(t) eq 1) 1
dembal.lazy 1
dembal.lazy('t1','s1') 1
\$offecho
rotdk.optfile=1;
option limCol = 0, limRow = 0;
* do not reset optcr if already set to a nondefault value
if{(.1 = %gams.optCr%), option optCr = 0.05;};
\$onecho > cplex.opt
writelp rotdk.lp
\$offecho
rotdk.optfile = 1;
solve rotdk min obj using mip;
``````
Running this code will generate rotdk.lp file. To use this model from lp file instead of using it from GAMS, one can use option probread. This can be done by replacing the option writelp from the above code fragment with probread as follows.
Code: Select all
``````\$onecho > cplex.opt | 860 | 2,624 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-31 | latest | en | 0.506685 |
https://www.physicsforums.com/threads/red-shift-question-from-another-thread.709299/ | 1,708,806,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474544.15/warc/CC-MAIN-20240224180245-20240224210245-00770.warc.gz | 969,353,071 | 17,288 | # Red-shift question from another thread.
• Jufro
In summary, the conversation is about a question regarding the acceleration of a source based on its redshift. The individual is working through the equation and taking the derivative, but has encountered an issue with the sign of the result. They are asking for clarification and guidance on how to proceed.
## Homework Statement
This is not a HW questions but from another thread.
The statement I made was that if z increases for a source, then it is accelerating away. Or, it could that if z is constant then the sources moves with constant velocity away from an observer.
I just need to know where my logic is breaking down and what direction I can try to find the right answer.
## Homework Equations
I took the equation:
1+z = $\sqrt{\frac{1+v/c}{1-v/c}}$
## The Attempt at a Solution
So taking d/dt on both sides I end up with:
dz/dt = 1/2 $\frac{1+v/c}{1-v/c}-1/2$*$\frac{(1/c dv/dt)*(1-v/c)-(1/c dv/dt)*(1+v/c)}{(1-v/c)2}$
Re-writting this:
dz/dt= $\frac{-v* dv/dt}{c2*(z+1)*(1-v/c)2}$
Since the first term is negative (from the minus sign) and v is positive, then a positive dz/dt would require that dv/dt is negative or that the sources acceleration is radially inward when I had suspected outward.
This may be a result of me using the flat-spacetime (Minkowski metric) but I am not sure. Can anyone point me in the right direction.
Please and thank you.
Jufro said:
## Homework Statement
This is not a HW questions but from another thread.
The statement I made was that if z increases for a source, then it is accelerating away. Or, it could that if z is constant then the sources moves with constant velocity away from an observer.
I just need to know where my logic is breaking down and what direction I can try to find the right answer.
## Homework Equations
I took the equation:
$$1+z = \sqrt{\frac{1+v/c}{1-v/c}}$$
## The Attempt at a Solution
So taking d/dt on both sides I end up with:
$$\frac{dz}{dt} = \frac{1}{2} \left(\frac{1+v/c}{1-v/c}\right)^{-1/2} \frac{(1/c\ dv/dt)(1-v/c)-(1/c\ dv/dt)(1+v/c)}{(1-v/c)^2}.$$ Re-writing this:
$$\frac{dz}{dt}= \frac{-v\ dv/dt}{c^2(z+1)(1-v/c)^2}.$$ Since the first term is negative (from the minus sign) and v is positive, then a positive dz/dt would require that dv/dt is negative or that the sources acceleration is radially inward when I had suspected outward.
This may be a result of me using the flat-spacetime (Minkowski metric) but I am not sure. Can anyone point me in the right direction.
Please and thank you.
You dropped a sign when you applied the quotient rule. The numerator should end up as a sum, not a difference.
Ah, I used the wrong derivative in the second half of my quotient rule. Thanks for that.
## What causes red-shift?
Red-shift is caused by the Doppler effect, which is the perceived change in frequency of a wave due to the relative motion between the source of the wave and the observer. In the case of red-shift, the source of the wave (usually a galaxy) is moving away from the observer, causing the wavelength of the light to stretch and appear more red.
## How is red-shift measured?
Red-shift is measured using spectroscopy, which involves splitting the light from an object into its component wavelengths. By analyzing the wavelengths of light emitted by a galaxy, scientists can determine how much the light has been stretched due to red-shift and calculate the object's velocity and distance from Earth.
## What is the significance of red-shift in astronomy?
Red-shift is a fundamental concept in astronomy and is used to measure the distance and speed of objects in the universe. It also provides evidence for the expansion of the universe and the Big Bang theory.
## Can red-shift be used to determine the age of the universe?
Yes, red-shift can be used to estimate the age of the universe. By measuring the red-shift of objects at different distances, scientists can calculate the rate at which the universe is expanding and estimate the age of the universe to be around 13.8 billion years.
## Are there different types of red-shift?
Yes, there are two types of red-shift: cosmological red-shift and gravitational red-shift. Cosmological red-shift is caused by the expansion of the universe, while gravitational red-shift is caused by the gravitational pull of massive objects, such as black holes. | 1,055 | 4,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-10 | latest | en | 0.944762 |
http://blog.wolframalpha.com/2011/06/01/wolframalpha-performs-vector-operations/ | 1,394,510,603,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011118294/warc/CC-MAIN-20140305091838-00072-ip-10-183-142-35.ec2.internal.warc.gz | 20,967,703 | 12,544 | # Wolfram|Alpha Performs Vector Operations
June 1, 2011
Posted by
What do you get when you cross a mountain climber with a mosquito? Nothing—you can’t cross a scalar with a vector!
But what do you get when you cross two vectors? Wolfram|Alpha can tell you. For example:
And in fact, Wolfram|Alpha can give lots of information on vectors. A vector is commonly defined as a quantity with both magnitude and direction and is often represented as an arrow. The direction of the arrow matches the direction of the vector, while the length represents the magnitude of the vector. Wolfram|Alpha can now plot vectors with this arrow representation in 2D and 3D and return many other properties of the vector.
Suppose you know only the point in R^n corresponding to your vector and you want to know its magnitude and direction. You can query Wolfram|Alpha for the vector’s length to find its magnitude:
And to find the direction, you can ask for the angles between the vector and the coordinate axes:
If you want to find both the magnitude and direction, you can represent the vector in polar or spherical coordinates. The radius gives you the magnitude of your vector, while the angles specify its direction.
So whether your studies are in algebra, calculus, or physics, Wolfram|Alpha can be your resource for learning about vectors.
Posted by J Januszewski June 1, 2011 at 3:38 pm
Lovely! Just asking: why do I keep getting error 101 when trying to compute something at W|A, and that I cannot use your calculus or W|A app?
Posted by Santiago Sinisterra June 1, 2011 at 5:54 pm
Are we ever going to have Mathematica for iPad?
Posted by Libby June 1, 2011 at 10:29 pm
This is excellent! Thanks…..I will definitely be playing with this new toy!
Posted by fizixx June 10, 2011 at 8:29 am | 424 | 1,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2014-10 | longest | en | 0.93041 |
https://educatorpages.com/site/brownsprealgebra/pages/94830 | 1,627,525,433,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00079.warc.gz | 235,870,237 | 7,828 | # Classifying & Ordering Real Numbers
Real Number System
You can divide the real numbers into two major groups: Rational & Irrational.
Irrational numbers are numbers that can be made into a decimal that goes on forever, but it does not repeat or terminate(end). For example, an irrational number can look like this 3.56469882224689....
All other real numbers are classified as rational.
Rational numbers can be subgrouped into natural numbers, whole numbers, and integers.
Natural numbers are the counting numbers. 1,2,3,....
Whole numbers are natural numbers plus zero. 0,1,2,3...
Integers are whole numbers plus their opposites. ...,-2,-1,0,1,2,...
All of those subgroups are rational numbers, but rational numbers can be decimals too (as long as they repeat or terminate). For example, some rational numbers are 3.45498, 2.3333333..., 8.5
Here is a chart that may help you visualize classifying real numbers: | 220 | 922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-31 | latest | en | 0.85038 |
https://www.gkseries.com/computer-based-optimization/multiple-choice-questions-and-answers-on-computer-based-optimization-3 | 1,669,555,167,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00126.warc.gz | 852,842,040 | 11,528 | # Computer Based Optimization Techniques - Questions & Answers for Competitive Exams | GkSeries
Questions
25. _____________is used to reduce the game.
• [A] Transportation Problem
• [B] Assignment Problem
• [C] Dual simplex method
• [D] Graphical Method
26. _____is a mathematical technique used to solve the problem of allocating limited resource among the competing activities
• [A] Linear Programming problem
• [B] Assignment Problem
• [C] Replacement Problem
• [D] Non linear Programming Problem
27. Traveling salesman problem will have a total of _____different sequences.
• [A] n!
• [B] n-1
• [C] (n-a)!.
• [D] n
28. If there exists a saddle point for a given problem it, implies that the players are using ______strategies.
• [A] Pure
• [B] Mixed
• [C] Optimal
• [D] Pure and Mixed
29. FCFS refers to _______in queueing theory.
• [A] First Come First Serve
• [B] Last In First Out
• [C] Service In Random Order
• [D] Fastest Server Rule
30. SIRO refers to _______in queueing theory.
• [A] First In First Out
• [B] Last In First Out
• [C] Service In Random Order
• [D] Fastest Server Rule
31. If a customers decides not to enter the queue since it is too long is called_________.
• [A] balking
• [B] reneging
• [C] jockeying
• [D] queue
32. When there are two or more parallel queues and customer move from one queue to another queue it is called___________
• [A] balking
• [B] reneging
• [C] jockeying
• [D] system capacity
33. If the operating characteristic vary within time is called_____.
• [A] balking
• [B] reneging
• [C] transient state
34. The __________ time for an activity can be reduced by using increased resources.
• [A] normal
• [B] optimistic
• [C] pessimistic
• [D] most likely
35. Graphical method of linear programming is useful when the number of decision variable are _______
• [A] 2
• [B] 3
• [C] 4
• [D] 1
36. The activity cost corresponding to the crash time is called the _____.
• [A] critical time
• [B] normal time
• [C] cost slope
• [D] crash cost | 580 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.811603 |
https://rdrr.io/cran/rmutil/man/Burr.html | 1,544,603,270,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00341.warc.gz | 740,230,132 | 13,338 | # Burr: Burr Distribution In rmutil: Utilities for Nonlinear Regression and Repeated Measurements Models
## Description
These functions provide information about the Burr distribution with location parameter equal to `m`, dispersion equal to `s`, and family parameter equal to `f`: density, cumulative distribution, quantiles, log hazard, and random generation.
The Burr distribution has density
f(y) = f s (y/m)^(s-1)/(m (1+(y/m)^s)^(f+1))
where m is the location parameter of the distribution, s is the dispersion, and f is the family parameter.
## Usage
```1 2 3 4``` ```dburr(y, m, s, f, log=FALSE) pburr(q, m, s, f) qburr(p, m, s, f) rburr(n, m, s, f) ```
## Arguments
`y` vector of responses. `q` vector of quantiles. `p` vector of probabilities `n` number of values to generate `m` vector of location parameters. `s` vector of dispersion parameters. `f` vector of family parameters. `log` if TRUE, log probabilities are supplied.
J.K. Lindsey
## Examples
```1 2 3 4``` ```dburr(2, 5, 1, 2) pburr(2, 5, 1, 2) qburr(0.3, 5, 1, 2) rburr(10, 5, 1, 2) ```
rmutil documentation built on May 27, 2018, 5:03 p.m. | 350 | 1,127 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-51 | longest | en | 0.678256 |
https://www.aopa.org/news-and-media/all-news/2010/august/01/test-pilot-(2) | 1,723,572,198,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00439.warc.gz | 507,935,927 | 25,929 | # Test Pilot
General 1. From reader Patty Schmidt: What was the first jet-powered airplane to be piloted by an American woman? 2.
#### General
1. From reader Patty Schmidt: What was the first jet-powered airplane to be piloted by an American woman?
2. There are six variables in the classic wind triangle: wind speed, wind direction, true course, groundspeed, true heading, and airspeed. Given any four of these variables, it is possible to find the other two. Assume that wind speed = 20 knots, true heading = 100 degrees, true course = 110 degrees, and groundspeed = 100 knots. What is the wind direction and true airspeed?
3. From reader George Shanks: Almost every pilot has heard of Lockheed’s high-flying U–2 spy plane, but what is or was the U–1?
4. From reader Patrick Mathews: Estimate within 2,000 the number of Army Air Force aircraft lost in training accidents within the continental United States during America’s involvement in World War II (December 1941 to August 1945).
5. Name a well-known American aircraft manufacturer that built only multiengine airplanes. Although it no longer makes aircraft, the company is still in business.
6. From reader Ken Thompson: Explain how it is possible for a private pilot flying an airplane to carry passengers at night over the contiguous United States without being night current and without violating the Federal Aviation Regulations.
7. From reader John Schmidt: There are six airports in the United States that are served by air carriers and are named after U.S. presidents. How many of them can you name?
8. Who probably is best known for having said, “Another thousand feet, and I’ll be on top?”
#### Multiple Choice
9. To be type-certificated under Part 23 of the Federal Aviation Regulations, a lightplane must be demonstrated capable of landing with a crosswind component of no less than _____ Vso.
a. 0.20; b. 0.25; c. 0.30; d. No minimum cross- wind component is required to be demonstrated.
10. The following is a list of the four most common violations of the Federal Aviation Regulations committed by VFR pilots. Arrange them in order of their occurrence, the most common deviation first and the least common last.
a. airspace violations; b. airspeed violations; c. flying VFR into IFR conditions; d. low-level flight
11. From reader Alasdair Halleron: Henry Ford took his first airplane ride in
a. a Fokker Trimotor.; b. a Ford Tri-Motor.; c. the Spirit of St. Louis.; d. Henry Ford never flew in an airplane.
#### True or False
12. The Cessna Aircraft Company never produced a piston-powered, single-engine, low-wing monoplane subsequent to the end of World War II.
13. From reader Hans Strömberg: The British navy had an aircraft carrier that used a sail to help keep the ship headed into the wind.
14. From reader Dan Stroud: A person can climb a ladder inside the vertical stabilizer of a Lockheed C–5 Galaxy and step outside the aircraft at the top of the fin.
1. America’s first turbojet fighter, the Bell P–59A Airacomet was first flown on October 1, 1942. It was flown a year later by Ann Baumgartener Carl, a WASP pilot who authored the book, A WASP Among Eagles: A Woman Military Test Pilot in World War II.
2. There are two possible solutions: If the wind is from 040 degrees, the true airspeed is 108 knots; if the wind is from 340 degrees, the true airspeed is 88 knots. The solutions are easier to obtain by plotting a wind triangle on a chart or on graph paper than with an E6B computer.
3. U is the military designation for utility aircraft, and the U–1 was a de Havilland Otter. The U–3 was the Cessna 310.
4. 13,873 aircraft were lost (approximately 10 per day).
5. Ford built only various models of the Ford Tri-Motor. There might be other such manufacturers.
6. Part 1.1 defines “night” as beginning at the end of evening civil twilight. This is typically 30-40 minutes after sunset. According to Part 61.57, however, a pilot may not carry passengers beginning one hour after sunset unless he is night current. Therefore, he may carry passengers at night from the end of twilight until one hour after sunset (a 20- to 30-minute period) without being night current. (We do not recommend this, however.)
7. John F. Kennedy International Airport (New York City), Bush Intercontinental Airport (Houston), Reagan Washington National Airport (Washington, D.C.), Abraham Lincoln Capital Airport (Springfield, Illinois), Gerald R. Ford International Airport (Grand Rapids, Michigan), and Theodore Roosevelt Regional Airport (Dickinson, North Dakota).
8. Mercury Astronaut John H. Glenn Jr. during his flight in Friendship 7 when he became the first American to orbit Earth (February 20, 1962).
9. (a) VSO is the 1-G stall speed of the aircraft with the engine idling and the landing gear and flaps fully extended.
10. The four choices are listed in the order requested: a, b, c, and d. Airspeed violations are typically committed by pilots of turbine aircraft.
11. Henry Ford experienced his first flight when Charles Lindbergh took him aloft for 10 minutes in the Spirit of St. Louis on August 11, 1927.
12. False. Production of the Cessna 188 AG Wagon (aka AGwagon), an agricultural sprayer, ended in 1981 after 1,589 such aircraft had been built.
13. True. The steam-powered HMS Royal Ark was used by the Royal Navy beginning in 1914 to launch floatplanes on dollies.
14. True. This should not be attempted during flight. | 1,269 | 5,441 | {"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} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.926957 |
https://ccssmathanswers.com/mcgraw-hill-my-math-grade-5-chapter-5-review-answer-key/ | 1,718,550,283,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00265.warc.gz | 138,714,482 | 57,164 | All the solutions provided in McGraw Hill My Math Grade 5 Answer Key PDF Chapter 5 Review will give you a clear idea of the concepts.
Vocabulary Check
Use the words in the words bank to complete each sentence.
Associative Property
Commutative Property
Identity Property
inverse operations
estimate
difference
place value
regroup
Question 1.
The _______________ states that you can add numbers in any orders.
Commutative Property
Explanation:
The Commutative Property states that you can add numbers in any order.
Question 2.
When you round a number, you find its _________________.
Estimate
Explanation:
When you round a number, you find its Estimate.
Question 3.
The value given to a digit by its position in a number is its ________________.
Place value.
Explanation:
The value given to a digit by its position in a number is its place value.
Question 4.
To ______________ is to form into a new or restructured group or grouping.
Regroup.
Explanation:
To regroup is to form into a new or restructured group or grouping.
Question 5.
Subtracting results in finding the ___________________ between two numbers.
Difference.
Explanation:
Subtracting results in finding the difference between two numbers.
Question 6.
The _____________ states that the sum of any number and 0 equals the number.
Identity property.
Explanation:
The Identity property states that the sum of any number and 0 equals the number.
Question 7.
The _____________ states that the way in which numbers are grouped does not change the sum or product.
Associative Property.
Explanation:
The Associative Property states that the way in which numbers are grouped does not change the sum or product.
Question 8.
When computing using actual numbers, you find a(n) ______________.
Explanation:
When computing using actual numbers, you find a(n) exact answer.
Concept Check
Round each decimal to the place indicated.
Question 9.
0.781; hundredths _______________
0.78
Explanation:
By rounding off the decimal number to the place indicated is 0.78.
Question 10.
17.93; ones _________________
18
Explanation:
By rounding off the decimal number to the place indicated is 18.
Round each decimal to the nearest tenth. Then add or subtract.
Question 11.
17.16 – 9.01 = ______________
8.2
Explanation:
Given,
17.16 – 9.01
Now round off each decimal to the nearest tenth.
17.2 – 9 = 8.2
Question 12.
62.84 + 74.2 = _______________
137
Explanation:
Given,
62.84 + 74.2
Now round off each decimal to the nearest tenth.
63 + 74 = 137.
Question 13.
7.41
Explanation:
Question 14.
$5.95 Explanation: Question 15. 4.15 + 20.68 = _______________ Answer: 24.83 Explanation: The addition of the given two decimal numbers is 24.83. Question 16. Use properties of addition to find 37 + 48 mentally. Show your steps and identify the properties that you used. 37 + 48 = ______________ Answer: 85 Explanation: 37 + 48 = (30 + 7) + (40 + 8) = 30 + 40 + 7 + 8 Commutative property = (30 + 40) + (7 + 8) Associative property = 70 + 15 Add = 85 Add Subtract. Use addition to check your answer. Question 17. Answer: 2.86 Explanation: The subtraction for the given decimal numbers is 2.86. Question 18. Answer:$8.37
Explanation:
The subtraction for the given decimal numbers is $8.37. Question 19. 8.3 – 4.7 = _______________ Answer: 3.6 Explanation: The subtraction for the given decimal numbers is 3.6. Problem Solving For Exercises 20 and 21, determine whether you need an estimate or an exact answer. Then solve. Question 20. A family is staying in a hotel for 2 nights. The hotel costs$91.35 a night. About how much will they pay for the hotel?
$91 +$91 = $182. Explanation: Given, Cost of hotel =$91.35 a night.
The family is staying in a hotel for 2 nights.
$91 +$91 = $182 They will pay$182 for the hotel.
Question 21.
Jesse had three long jumps during a track meet of 17 feet, 18 feet, and 15 feet. Andrew had three long jumps during a track meet of 17 feet, 15 feet, and 18 feet. Overall, who had a greater total jump distance? Which addition property can you use to find the answer?
Commutative property.
Explanation:
The jump distances were equal. So the addition property used to find the answer is called the Commutative property.
Question 22.
You have to be 48 inches tall to ride the roller coaster. Eva is 45 inches tall. If she grows 2 inches each year, will she be able to ride the roller coaster in 2 years?
45 + 2 + 2 = 49 inches.
Explanation:
Given,
Eva is 45 inches tall.
If she grows 2 inches each year.
To find:Â will she be able to ride the roller coaster in 2 years?
45 + 2 + 2 = 49 inches.
Test Practice
Question 23.
Micah had $10.52 after he left the book store. If he bought a book for$6.39, how much money did he have before he went to the book store?
(A) $4.13 (B)$8.23
(C) $14.91 (D)$16.91
$16.91 Explanation: Given, Micah had$10.52.
If he bought a book for $6.39 Total amount =$10.52 + $6.39 =$16.91.
Reflect
Use what you learned about adding and subtracting decimals to complete the graphic organizer. | 1,310 | 5,017 | {"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} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.830555 |
http://engineering-references.sbainvent.com/fluid-mechanics/hydrostatic-pressure.php | 1,369,145,083,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700074077/warc/CC-MAIN-20130516102754-00072-ip-10-60-113-184.ec2.internal.warc.gz | 93,787,989 | 6,827 | Fluid Mechanics: Hydrostatic Pressure
PLAY
Hydrostatic Pressure
Length: 12 Minutes 31 Seconds
Static Pressure
When dealing with static pressure it can be assumed that the pressure will be same around a specific point or element of interest. This is due to Pascal's Law "The pressure at a point in a fluid at rest, or in motion, is independent of direction as long as there are no shearing stress present." Refer to the equation below.
(1)
Hydrostatic Pressure
Hydrostatic pressure has a relation to the depth of fluid. This is also known as fluid head. The relationship between height and pressure of a fluid can be seen in equations 2-4.
(2)
(3)
(4)
Notice from equations 2-4 the x and y plane has no influence on hydrostatic pressure. Instead the entire influence is due to the height of the fluid. This means the container's cross-sectional area has no influence on hydro static pressure.
Hydrostatic Pressure of an Incompressibile Fluid
A fluid can be considered incompressible if it takes a large amount of pressure to compress the fluid. Liquids can be considered incompressible fluids. To determine the hydrostatic pressure of an incompressible fluid refer to equation 5.
(5)
Hydrostatic Pressure of a Compressibile Fluid
A compressible fluid easily changes its density and volume under different pressures. To calculate the hydrostatic pressure of a compressible fluid the ideal gas law would be used.
(6)
Hydrostatic Force on a Plane
When calculating hydrostatic force on a plane you have to realize that the pressure along the plane can change based on its relationship to the fluids depth. To calculate the resultant force equation 7 would be used. Also, to find the location of the resultant force on the plane equations 8-9 would be used.
(7)
(8)
(9)
Hydrostatic Force on a Curved Surface
There are also some cases where the hydrostatic force on a curved surface is of interest. To calculate the hydrostatic force on a curved surface refer to the equations below.
(10)
(11)
(12)
(13)
(14)
F1 in equation 12 represents the force above the curve. | 463 | 2,088 | {"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} | 3.765625 | 4 | CC-MAIN-2013-20 | longest | en | 0.926714 |
http://www.instructables.com/id/Servo-Motor-Control-by-using-Microcontroller-PIC16/ | 1,529,936,800,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867885.75/warc/CC-MAIN-20180625131117-20180625151117-00417.warc.gz | 413,992,630 | 12,025 | # Servo Motor Control by Using Microcontroller PIC16F877A
75,379
37
3
Published
## Introduction: Servo Motor Control by Using Microcontroller PIC16F877A
PIC16F877A is an easy micro controller for the beginners
Here in project:
1) How PIC switches servo motor forward and back ward
When we push one switch it goes to positive 90 degree , on second switch it moves backward to negative 90 degree
a. MickroC PRO // for programming of MICRO CONTROLLER
b. Proteous 8 professional //For simulation of CKT and Design
BEST OF LUCK.
Eng. Mahmoud Abuzayed
## Step 1: R/C Servo Control Theory
Servos designed for use in radio-controlled airplanes, cars, and boats can be easily interfaced to a PIC MCU. They are often used for robots and applications where simple mechanical movement is required. This might be surprising because a positional servo is considered to be an analog device. The output of an RC servo is usually a wheel that can be rotated from 0 to 90 degrees. (some servos can turn from 0 to 180 and others have very high torque outputs for special applications). Typically, they only require +5 V, ground, and an input signal.
An RC servo is an analog device; the input is a PWM signal at digital voltage level. This pulse is between 1.0 and 2.0 ms long and repeats every 20 ms .The length of the PWM pulse determines the position of the servo's wheel. A 1.0-ms pulse will cause the wheel to go to 0 degrees and a 2.0-ms pulse will cause the wheel to go to 90 degrees.
Pulse Width Modulation (PWM)
Pulse Width Modulation (PWM) is one of the powerful techniques used in control systems today. They are not only employed in wide range of control application which includes: speed control, power control, measure and communication.
Basic Principal of PWM
PWM is achieved with the help of a square wave whose duty cycle is changed to get a varying voltage output as a result of average value of waveform. A mathematical explanation of this is given below:
Consider a square wave shown in the figure above. Ton is the time for which the output is high and Toff is the time for which output is low. Let Ttotal be time period of the wave such that,
Duty cycle of a square wave is defined as:
The output voltage varies with duty cycle as:
So you can see from the final equation the output voltage can be directly varied by varying the Ton value.
If Ton is 0, Vout is also 0.
If Ton is Ttotal then Vout is Vin or say maximum.
## Step 2: PIC PWM Module
In Pulse Width Modulation mode, the CCP1(RC2) pin produces up to a 10-bit resolution PWM output. Since the CCP1 pin is multiplexed with the PORTC data latch, the TRISC<2> bit must be cleared to make the CCP1pin an output.
Figure below shows a simplified block diagram of the CCP module in PWM mode.
A PWM output has a time-base (period) and a time that the output stays high (duty cycle). The frequency of the PWM is the inverse of the period (1/period).
## Step 3: PWM PERIOD & DUTY CYCLE
The PWM period is specified by writing to the PR2 register. The PWM period can be calculated using the following formula:
PWM period = [(PR2) + 1] • 4 • TOSC • (TMR2 prescale value)
PWM frequency is defined as 1 / [PWM period]. When TMR2 is equal to PR2, the following three events occur on the next increment cycle:
TMR2 is cleared
The CCP1 pin is set (exception: if PWM duty cycle = 0%, the CCP1 pin will not be set)
The PWM duty cycle is latched from CCPR1L into CCPR1H
The PWM duty cycle is specified by writing to the CCPR1L register and to the CCP1CON<5:4> bits. Up to 10-bit resolution is available. The CCPR1L contains the eight MSbs and the CCP1CON<5:4> contains the two LSbs. This 10-bit value is represented by CCPR1L:CCP1CON<5:4>. The following equation is used to calculate the PWM duty cycle in time:
PWM duty cycle =(CCPR1L:CCP1CON<5:4>) •TOSC • (TMR2 prescale value)
SETUP FOR PWM OPERATION
The following steps should be taken when configuring the CCP module for PWM operation:
1. Set the PWM period by writing to the PR2 register.
2. Set the PWM duty cycle by writing to the CCPR1L register and CP1CON<5:4> bits.
3. Make the CCP1 pin an output by clearing the TRISC<2> bit.
4. Set the TMR2 prescale value and enable Timer2 by writing to T2CON.
5. Configure the CCP1 module for PWM operation.
Circuit Diagram
## Step 5: Code for PIC 16F877A
void main()
{
//int x = 125;
TRISC = (1<<3)|(1<<4) ; // configure RC3 as input (1 for input and 0 for output)
PR2 = 0xFA; //oxFA for 250 (used to set time period)
CCPR2L = 125; // (used to set duty cycle)
CCP2CON |= (1<<2)|(1<<3); // select for PWM mode
T2CON |= (1<<2)|(1<<1); // enabling timer 2, prescalar 16
while(1)
{
if(PORTC.F3==1) // forward
{
// x=x+1
CCPR2L = 250+125; // (used to set duty cycle)
T2CON |= (1<<2)|(1<<1); // enabling timer 2, prescalar 16
}
if(PORTC.F4==1) // back ward
{
// x=x-1
CCPR2L = 250-245; // (used to set duty cycle)
T2CON |= (1<<2)|(1<<1); // enabling timer 2, prescalar 16
}
delay_ms(100);
}
}
## Step 6: OUT PUT
output of the code :
## Recommendations
606 Enrolled
• ### Casting Contest
We have a be nice policy.
## Questions
sir how can i control 2 servos bidirectional with different buttons
Frequecy of the crystal oscillator is not shown in the schematic?
If the PIC is running at 20MHz crystal, I dont think it is possible to generate a low frequency PWM (20ms cycle time or 50Hz) using the CCP module | 1,497 | 5,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-26 | latest | en | 0.874532 |
http://back2cloud.com/percent-error/percent-error-calculation.php | 1,527,149,649,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866107.79/warc/CC-MAIN-20180524073324-20180524093324-00598.warc.gz | 29,150,628 | 5,344 | Home > Percent Error > Percent Error Calculation
# Percent Error Calculation
## Contents
Please enter a valid email address. Logga in och gör din röst hörd. When you calculate the density using your measurements, you get 8.78 grams/cm3. This is also called the accepted, experimental or true value.Note due to the absolute value in the actual equation (above) there are two value. have a peek at these guys
You might also enjoy: Sign up There was an error. Jumeirah College Science 68 533 visningar 4:33 Scientific Notation: Multiplication and Division - Längd: 5:30. Imaging the Universe A lab manual developed by the University of Iowa Department of Physics and Astronomy Site Navigation[Skip] Home Courses Exploration of the Solar System General Astronomy Stars, Galaxies, and VisningsköKöVisningsköKö Ta bort allaKoppla från Läser in ... http://astro.physics.uiowa.edu/ITU/glossary/percent-error-formula/
## How To Calculate Percent Error In Chemistry
Co-authors: 11 Updated: Views:260,694 70% of people told us that this article helped them. Example: Sam does an experiment to find how long it takes an apple to drop 2 meters. For example, you would not expect to have positive percent error comparing actual to theoretical yield in a chemical reaction.[experimental value - theoretical value] / theoretical value x 100%Percent Error Calculation A stopwatch has a circular dial divided into 120 divisions.time interval of 10 oscillation of a simple pendulum is measure as 25 sec.by using the watch Max. % error in the
Tips Some teachers like the percent error to be rounded to a certain point; most people will be satisfied with the percent error rounded to three significant digits. Percentage Difference Percentage Index Search :: Index :: About :: Contact :: Contribute :: Cite This Page :: Privacy Copyright © 2014 MathsIsFun.com About.com Autos Careers Dating & Relationships Education en Thank you,,for signing up! Negative Percent Error Läser in ...
Answer this question Flag as... What Is A Good Percent Error Create an account EXPLORE Community DashboardRandom ArticleAbout UsCategoriesRecent Changes HELP US Write an ArticleRequest a New ArticleAnswer a RequestMore Ideas... Flag as... Ex: 10 - 9 = 1 3 Divide the result by the real number.
## Percent Error Calculator
Whether error is positive or negative is important.
This can give a positive or negative result, which may be useful to know. How To Calculate Percent Error In Chemistry You measure the dimensions of the block and its displacement in a container of a known volume of water. Percent Error Definition Todd also writes many of the example problems and general news articles found on the site.
Learn more You're viewing YouTube in Swedish. http://back2cloud.com/percent-error/percent-error-calculation-used-determine.php But Sam measures 0.62 seconds, which is an approximate value. |0.62 − 0.64| |0.64| × 100% = 0.02 0.64 × 100% = 3% (to nearest 1%) So Sam was only The absolute value of the error is divided by an accepted value and given as a percent.|accepted value - experimental value| \ accepted value x 100%Note for chemistry and other sciences, The difference between the actual and experimental value is always the absolute value of the difference. |Experimental-Actual|/Actualx100 so it doesn't matter how you subtract. Can Percent Error Be Negative
AJ Design☰ MenuMath GeometryPhysics ForceFluid MechanicsFinanceLoan Calculator Percent Error Equations Calculator Math Physics Chemistry Biology Formulas Solving for percent error. Please select a newsletter. Visa mer Läser in ... check my blog Arbetar ...
Copper's accepted density is 8.96 g/cm3. Percent Error Worksheet The result of the difference is positive and therefore the percent error is positive. This will convert the answer into percent form.
## It is used in chemistry and other sciences to report the difference between a measured or experimental value and a true or exact value.
Example: I estimated 260 people, but 325 came. 260 − 325 = −65, ignore the "−" sign, so my error is 65 "Percentage Error": show the error as a percent of Flag as... Share it. Percent Error Definition Chemistry Flag as...
Läser in ... View all posts by Todd Helmenstine → Post navigation ← Direct Image Of Exoplanet Sets New Record Using Stem Cells and Herpes To Fight Brain Cancer → 3 thoughts on “Calculate Flag as... http://back2cloud.com/percent-error/percent-error-calculation-online.php EditRelated wikiHows How to Calculate Modulus How to Calculate Confidence Interval How to Calculate Variance How to Calculate P Value How to Calculate Uncertainty How to Calculate Probability How to Calculate
This means that you should subtract the real value from the estimated value. We can also use a theoretical value (when it is well known) instead of an exact value. Did you mean ? You look up the density of a block aluminum at room temperature and find it to be 2.70 g/cm3.
Flag as... Visningskö Kö __count__/__total__ Ta reda på varförStäng Error and Percent Error Tyler DeWitt PrenumereraPrenumerantSäg upp277 864277 tn Läser in ... In many situations, the true values are unknown. About this wikiHow 190reviews Click a star to vote Click a star to vote Thanks for voting!
Logga in 595 29 Gillar du inte videoklippet? The percentage error in measurement of time period "T"and length "L" of a simple pendulum are 0.2% and 2% respectively ,the maximum % age error in LT2 is? Solve for percent error Solve for the actual value. | 1,234 | 5,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-22 | longest | en | 0.768279 |
http://aidmeditechclients.com/utilize-the-bmi-calculator-to-find-your-recommended-weight5809/ | 1,516,403,676,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888302.37/warc/CC-MAIN-20180119224212-20180120004212-00628.warc.gz | 13,133,996 | 14,430 | # Utilize The Bmi Calculator To Find Your Recommended Weight5809
Written by
You understand you have to shed weight but do you know exactly how much you need to lose? Do you have a goal to snap for? Like the majority of things we want to accomplish, shedding pounds is most effective when we have a goal. Usually, how can you know when you have dropped adequate weight? For desired goals to be effective, they have to be certain. You should use the Body Mass Index calculator that will help you set up the healthful weight focus on you must satisfy. body mass index calculator
• You realize you have to slim down but have you any idea simply how much you have to get.
• In case you are very little proficient at maths so you feel your eyesight.
• body mass index calculator.
To figure out if we are underweight or overweight the Body mass index looks at our bodyweight with regards to our size. This then slips within a a number of collection, which is regarded as healthful. Outside that variety is regarded as under bodyweight or obese and not healthy. For instance, an individual who is half a dozen toes taller (182cm) is regarded as healthy if how much they weigh tumbles involving 145 and 182 weight (67-83 kg). An individual 5 toes 8 " (172cm) should weigh up in between 131 and 161 pounds (62-74 kg).
## Underweight or overweight the Body mass index
You can determine your own BMI or use one of the many sites by using a calculator. If you want to do-it-yourself, the BMI formulation can be your body weight separated by you elevation squared, then multiplied by 703, should you be employed in pounds. This means for a person considering 140 pounds who is 5 ft . 10 in . (70 ins) tall, the computation will be 140 divided by 4900 (70x70), which could equivalent .028. You grow this by 703, which supplies you with a BMI of 19.68. To perform the calculations in kgs you simply divide your weight through your level squared and end up forgetting the 703. For instance, somebody who weighs 65 kilograms and is also 1.78 m taller would have a Body mass index of 20.51 (65/1.78x1.78). calculate bmi
### Be divided by x which could equivalent
When you are little efficient at maths so you really feel your vision glazing over whenever you check out the numbers, just go more than them once more and you will probably see they are not way too hard to sort out. healthy bmi
### Them once more and you
The World Wellness Organization has defined categories of BMI according to evidence of improved likelihood of persistent sickness and mortality. A Body mass index of 25-29.9 is incorporated in the category of heavy, 30-39.9 is being overweight as well as over 40 is morbid obesity. It could be a little distress to find you go with one of these brilliant classes, however it is some thing we should are all aware so that we can perform some thing regarding it. This means we must be seeking to attain a Bmi of less than 25 if we should remain in the healthier excess weight class.
### Persistent sickness
There may be some discussion as to how powerful the numerous methods of gauging recommended weight are, nevertheless the BMI has grown to be broadly approved. The Entire World Well being Organisation and other health government bodies apply it like a normal in their self-help guide to healthy weight. Several life coverage firms utilize it as strategies for over weight and underweight inside their underwriting selections. It really is a sensible strategy to allow us to handle our excess weight in order that we stay in that healthier weight variety and get away from many of the modern day life-style conditions. You can use the Body mass index formulation to create your ideal weight focus on and also to go on monitor while you are attaining it. healthy bmi
#### For over
• You already know you will need to lose weight but have you figured out simply how much you.
• Should you be very little proficient at maths and you also truly feel your eyes glazing.
• There is some dialogue regarding how powerful the various methods for.
• To ascertain once we are underweight.
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Technology | 894 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-05 | longest | en | 0.951015 |
https://www.jiskha.com/display.cgi?id=1305505177 | 1,516,674,139,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891705.93/warc/CC-MAIN-20180123012644-20180123032644-00725.warc.gz | 934,528,072 | 3,880 | # calculus
posted by .
Decompose 58-x/x^2-6x-16 into partial fractions.
• calculus -
If I wrote
(58 - x)/(x^2 - 6x - 16) as
= (58-x)/((x-8)(x+2))
can you follow the procedure I showed you in your post above this one?
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More Similar Questions | 715 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-05 | latest | en | 0.865348 |
thecorridortheory.blogspot.com | 1,553,329,849,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202781.33/warc/CC-MAIN-20190323080959-20190323102959-00062.warc.gz | 198,692,759 | 18,777 | ## August 12, 2011
### Missing dollar riddle
I was asked the following question last thursday:
You want to buy a T-shirt that costs RM97. You have no money.
You borrow RM50 each from your brother and your sister and with your current RM100, you purchase said T-shirt.
You hold in your hands the balance (RM3). You then decide to return RM1 to your brother, RM1 to your sister, and keep the remainder.
Now that each of them has RM1 back, the total owed is RM98. If you have RM1, what happened to the remaining RM1?
This riddle illustrates problems of confusion and misdirection can foil a person's clear understanding of the problem.
The original problem is as follows:
Three guests check into a hotel room. The clerk says the bill is \$30, so each guest pays \$10. Later the clerk realizes the bill should only be \$25. To rectify this, he gives the bellhop \$5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest \$1 and keep \$2 for himself.
Now that each of the guests has been given \$1 back, each has paid \$9, bringing the total paid to \$27. The bellhop has \$2. If the guests originally handed over \$30, what happened to the remaining \$1?
I shall attempt to solve the original question.
Explanation 1:
The initial payment of \$30 is accounted for as the clerk takes \$25, the bellhop takes \$2, and the guests get a \$3 refund. It adds up. After the refund has been applied, we only have to account for a payment of \$27. Again, the clerk keeps \$25 and the bellhop gets \$2. This also adds up.
There is no reason to add the \$2 and \$27 – the \$2 is contained within the \$27 already. Thus the addition is meaningless. Instead the \$2 should be subtracted from the \$27 to get the revised bill of \$25.
This becomes clearer when the initial and net payments are written as simple equations. The first equation shows what happened to the initial payment of \$30:
\$30 (initial payment) = \$25 (to clerk) + \$2 (to bellhop) + \$3 (refund)
The second equation shows the net payment after the refund is applied (subtracted from both sides):
\$27 (net payment) = \$25 (to clerk) + \$2 (to bellhop)
Both equations make sense, with equal totals on either side of the equal sign. The correct way to get the bellhop's \$2 and the guests \$27 on the same side of the equal sign ("The bellhop has \$2, and the guests paid \$27, how does that add up?") is to subtract, not add:
\$27 (final payment) - \$2 (to bellhop) = \$25 (to clerk)
Explanation 2(or the TL;DR version):
The question attempts to confuse you by adding up loaned money. Simply put, the \$2 is contained within the \$27, and should not be added together.
Why?
An analogous situation would be adding 99 empty pails and 1 full pail of water. The two quantities are distinct from each other and it is meaningless to call the total "100 full pails of water" nor "100 empty pails of water".
The wikipedia entry elaborates further:
### Misdirection
The "paradox" cleverly sets its room rates so that when we add the two terms \$27 and \$2, we nearly get \$30. If not for this "near-miss", we would be more inclined to ask if those two terms have to add up to \$30 when we break down the situation this way (and to realize that they do not).
With different prices, the illusion would vanish. Say the clerk initially accepted \$30 but then learned that rooms are only \$10 no matter how many people are in them, and sends back a refund of \$20 via the bellhop. Again, the bellhop, seeing that \$20 doesn't evenly divide, gives each guest \$6 (for a total of \$18) and keeps the leftover \$2 for himself. Therefore each of the three guests paid \$4, bringing the total paid to \$12; add that to the bellhop's 2 dollars to get a total of \$14. So where did the other \$16 go?
With this setup it is more clear that the guests' new total amount paid (\$12) is only the bellhop's \$2 away from the actual room price of \$10, not the original room price of \$30. The target price to account for is the new \$10 bill, not the old \$30 one. In the original riddle it is only the "near-miss" with \$30 that makes \$30 seem like the correct target of the operation.
The riddle involves the phenomenon of 'suspension of disbelief' inherent in storytelling and its power over the human imagination. If one were to make the story a bit more complex and compelling the illusion is almost guaranteed to work in the moment of its telling and can be a good illustration for the explanation of the anomaly, although not a perfect one because there is an explanation. The more points added to the story cause the listener to pause and try to compute what each element may signify.
There are dozens of variations to the riddle.
Have a nice day.
http://en.wikipedia.org/wiki/Ignoratio_elenchi
Source:
1. Pretty awesome stuff, I like breaking my head on these things... thanks!
2. Honestly: I stopped reading in one particular moment... The payment examples resembles the case of arrow and the turtle. It's too much philosophy for me, yet I have to admit: misdirection is a very interesting idea/problem/case/ whatever one could call it. It's me who simply isn't into that kind of philosophical debates.
3. Pretty cool I got caught by this one
4. Portugal should hire you :D
5. That was intense, lol... great brain exercises... +followed
6. I feel stupid. lol
7. I'm no good at math, and this proved it haha
8. Mindfuck, now in a text-mode for your pleasure.
Nah but honestly, great subject :). Original and fun !
9. I got it! Its a simmple but good illusion :D
Nice post! want more of these!
Following!
greetings
10. There is no dollar missing. The question is itself mis leading us.
The answer to this misleading question is pretty simple rather very simple if you understand the question. | 1,412 | 5,918 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.96317 |
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Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 1029 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
Rs. 450.00 or
## Question number: 372
MCQ▾
### Question
A grocer purchased 70 kg. Of sugar at Rs. 11.00 per kg and mixed it with 110 kg sugar at Rs. 15 per kg. At what rate should he sell the mixture (per kg) to gain 15%?
### Choices
Choice (4) Response
a.
20.5 Rs.
b.
14.30 Rs.
c.
20.00 Rs.
d.
15.46 Rs.
## Question number: 373
MCQ▾
### Question
The value of the product and at is________.
### Choices
Choice (4) Response
a.
15
b.
18
c.
20
d.
17
## Question number: 374
MCQ▾
### Question
Find the missing term in the following problem.
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 375
MCQ▾
### Question
The quotient when is divided by is________.
### Choices
Choice (4) Response
a.
b.
3q
c.
d.
## Question number: 376
MCQ▾
### Question
Find the missing term in the following problem.
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 377
MCQ▾
### Question
The value of the product and at a = 3 is________.
### Choices
Choice (4) Response
a.
15
b.
24
c.
26
d.
18
## Question number: 378
MCQ▾
### Question
The perimeter of a triangle is and two of its sides are and. Find the third side of the triangle.
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 379
MCQ▾
### Question
On dividing by , we get quotient________.
### Choices
Choice (4) Response
a.
7
b.
2
c.
d.
f Page | 515 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-26 | longest | en | 0.629357 |
https://www.studyadda.com/question-bank/atomic-molecular-and-equivalent-masses_q39/1316/94227 | 1,601,004,947,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00213.warc.gz | 1,072,817,465 | 18,530 | • # question_answer The weight of a molecule of the compound ${{C}_{60}}{{H}_{122}}$is [AIIMS 2000] A) $1.4\times {{10}^{-21}}$g B) $1.09\times {{10}^{-21}}$g C) $5.025\times {{10}^{23}}$g D) $16.023\times {{10}^{23}}$g
Molecular weight of ${{C}_{60}}{{H}_{122}}$$=12\times 60+122\times 1$ $=720+122=842$ $\because$ $6\times {{10}^{23}}$molecule ${{C}_{60}}{{H}_{122}}$has mass = 842gm $\therefore$ 1 molecule ${{C}_{60}}{{H}_{122}}$has mass $\frac{842}{6\times {{10}^{23}}}$ $=140.333\times {{10}^{-23}}gm$ $=1.4\times {{10}^{-21}}gm$. | 261 | 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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2020-40 | latest | en | 0.405426 |
http://www.physicsmynd.com/?p=1103 | 1,529,639,781,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864343.37/warc/CC-MAIN-20180622030142-20180622050142-00052.warc.gz | 486,136,047 | 19,413 | # IIT JEE Main / Advanced Physics 2018 Tips & Trends – Young’s Double Slit Experiment [YDSE]
The experiment by the English scientist Thomas Young in 1801 demonstrated the inseparability of the wave and particle nature of light or wave – particle duality.Morever , he was able to determine the wavelength of light by this experiment . The importance of this experiment was such that , Richard Feyman , one of the greatest physicts opined that all of quantum mechanics could be gleaned from carefully thinking through the implications of this single experiment. In this post , we will look at the key theoretical aspects of YDSE and problems related to it.
Young’s double slit experiment
Points to note
• The light used for this experiment is usually ( for best results ) collimated – light rays are distinct , parallel to each other and doesn’t mix with each other , monochromatic – having a single wavelength [ to validate the results] and coherent – having a single phase.
• The single slit ensures that light from only a single direction falls on the double slits.
• The light passing through S1 and S2 will be coherent as their source is the same.
• A bright fringe is created directly opposite the mid-point between the two slits. Here , the distances between the fringe and the two slits ( l1 and l2 ) are equal and waves contain the same no of wave lenghts – thus resulting in constructive interference . In addition , constructive interference produces more bright fringes on both sides of the middle , wherever the difference between l1 and l2 is an integer number of wave lenghts – λ , 2λ, 3λ etc.
• For dark fringes, l2 is larger than l1 by exactly one half a wavelength. Additional dark fringes are created at points where the difference between l1 and l2 equals an odd integer no of half wavelengths like 1[λ/2] , 3[λ/2]..etc . Dark fringes indicate destructive interference.
• Angle θ for maximum interference ( constructive) is given by sin θ = m λ/d where m = 0,1,2,3.. and d is the distance between S1 and S2.
• Angle θ for minimum interference ( destructive ) is given by sin θ = [ m + ½ ] λ/d where m = 0,1,2,3….
Conditions for Constructive & Destructive Interference
Consider a point on a distant screen , such that the distance D from the slits is D >> d .Now , the small arc of the circle from the chosen point is almost a straight line and the path difference , Δx = d sin θ .
• The condition for constructive interference is Δ x = ± nλ ( n = o,1,2,3..) and for maxima , d sin θ = ± nλ .
• The condition for destructive interference is Δ x = ± [ n – ½ ] λ ( n = 1,2,3..) and for minima d sin θ = ± [ n – ½]λ .
• When D >> d , then sin θ = tan θ = θ = y/D .
• Position of nth bright fringe is yn = nλ [ D/d ] .
• Position of the nth dark fringe is yn = [ n – ½ ] λ D/d .
Fringe width
Fringe width ( β ) is defined as the distance between two sucessive maxima or minima. It is given by β = λD/d .
• β is independent of n ( fringe order) as long as d and θ are small , i.e fringes are evenly spaced.
• For red light β will be higher as β is proportional to λ .
• If this experiment is done in a medium other than air , say water with a refractive index of μ , then , β’ = β/μ .
Maximum order for Interference Fringes
• When n << d/λ , y/D = nλ / d .
• When n ≈ d /λ , then n = d sin θ /λ .
• Therefore n max = [ d/λ ] and n min = [ d/λ + ½ ] .
Variations
1 . When the rays are not parallel to the Principal axis
• Δ x = d sin θ – yd/D
2 . When the source is placed beyond the central line
• Maxima : Δ x = nλ
• Minima : Δ x = ( 2n – 1 ) λ / 2
3 . When a transparent glass slab of thickness t and refractive index μ is placed in one of the incoming wave paths , due to the increase of the path by ( μ – 1 ) t , the fringe pattern undergoes a shift , s given by
• s = D/d ( μ – 1 ) t
Now on to some problems ……..
1. In a Young’s experiment, the upper slit is covered by a thin glass pate of refractive index 1.4, while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen, where the central maximum(n = 0) fall before the glass plates were inserted, now has ¾ the original intensity. It is further observed that what used to be the fifth maximum earlier lies below the point P while the sixth minima lies above P. Calculate the thickness of glass plate. (Absorption of light by glass plate may be neglected). ( IIT JEE 1997 )
Solution –
From the given data ,μ1 = 1.4 and μ2 = 1.7 . Let t be the thickness of each glass plates. Path difference at O, due to insertion of glass plates will be
Δx = ( μ2– μ1) t = (1.7 – 1.4) t = 0.3 t …(1) . The point to deduce here is that the path dfference would be lying between 5 λ and 5λ + λ/2 , since the 5th maxima and the 6th minima lies on both sides of O .
Hence , Δx is taken as 5λ + Δ …(2) . Here , Δ < λ/2 . Now , the phase difference at O will be –
Φ = 2π /λ . Δx = 2π /λ ( 5λ + Δ ) = ( 10 π + 2π/λ Δ ) …… (3)
Now , we know that I (Φ ) = Imax cos² [ Φ / 2 ] and the intensity at O is given as 3/4 I max . Therefore ,
3/4 I max = I (Φ ) = Imax cos² [ Φ / 2 ] or 3/4 = cos² [ Φ / 2 ] ….. (4)
Substituting the value of eq (3 ) in eq. (4) and solving , we get Δ = λ /6 .
Hence , Δx = 5λ + λ /6 = 31 λ / 6 . Now from eq. (1) Δx = 0.3 t . Therefore ,
t = 31 λ/6 x 0.3 = 9.3 x 10-6 m = 9.3 μm .
2.A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 m from it as shown in the figure. a ) If the incident beam falls normally on the double slit apparatus, find the y-coordinates of all the interference minima on the screen. b ) If the incident beam makes an angle of 30o with the x-axis (as in the dotted arrow shown in figure), find the y-coordinates of the first minima on either side of the central maximum. ( IIT JEE 1998 )
Solution –
normal path of incident rays
Refer [variations.2] above . For minimum intensity , d sin θ = ( 2n – 1 ) λ /2 so , sin θ = ( 2n – 1 ) λ /2d . Subtituting the values from the given data,
sin θ = 2n – 1 /4 . The point to note – as sin θ ≤ 1 , ( 2n – 1 ) λ /2 ≤ 1 , means that n ≤ 2.5 . Hence n can either be 1 or 2 .
If n = 1 , sin θ1 = 1/4 and tan θ1 = 1 /√15 . Conversely, if n = 2 , then sin θ2 = 3/4 and tan θ2 = 3 /√7 .
When D = 1m , y = tan θ . Hence , y1 = 1/√15 m = 0.26 m .Similarily , y2 = 3 / √7 m = 1.13 m .
As there can be minima on either side of O , there ideally should be 4 minima at positions ± 0.26 m and ± 1.13 m .
Part b ) If α = 30° , then Δx1 = d sin 30 = 0.5 mm which is equal to the given λ . Hence , the path difference betwwen the rays is λ .
At central maxima , the net path difference is zero . Hence , d sin θ = d sin α and θ = α = 30° or tan θ = 1/√3 = y0/D . Therefore , y0 = 0.58m.
Let X1 and X2 be the minimas flanking the central maxima. Now , considering X2 –
Δx2 – Δx1 = λ /2 therefore , Δx2 = 3λ /2 as Δx1 = λ . Thus , d sin θ2 = 3λ /2 = ¾ .
tan θ2 = y2/D = 3/√7 , hence , y2 = 1.13m . Working out in the same manner for X1 , tan θ1 = y1/D = 1/√15 , hence , y1 = 0.26m.
3. In a Young’s double slit experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again? Take D/d = 103. Symbols have their usual meanings. ( IIT JEE 2004 )
Taking n1 as the bright fringe of wavelength λ1 which is given as 500 nm and n2 as that of the bright fringe λ2 given as 700nm , then –
n1 . λ1D/d = n2 . λ2 D/d . Therefore , n1/n2 = λ2/λ1 = 7/5 . The inference here is that the 7th maxima of λ1 will coincide with the 5th maxima of λ2 . Hence , the minimum distance = n1λ1D/d = 7 x 5 x 10 -7 x 10³ = 3.5 mm | 2,596 | 8,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-26 | latest | en | 0.905567 |
https://www.geeksforgeeks.org/length-of-the-longest-substring-with-equal-1s-and-0s/?ref=rp | 1,660,204,920,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571246.56/warc/CC-MAIN-20220811073058-20220811103058-00049.warc.gz | 657,050,725 | 31,314 | # Length of the longest substring with equal 1s and 0s
• Difficulty Level : Medium
• Last Updated : 20 Jul, 2022
Given a binary string. We need to find the length of the longest balanced substring. A substring is balanced if it contains an equal number of 0 and 1.
Examples:
```Input : input = 110101010
Output : Length of longest balanced
sub string = 8
Input : input = 0000
Output : Length of longest balanced
sub string = 0```
A simple solution is to use two nested loops to generate every substring. And a third loop to count number of 0s and 1s in current substring. Time complexity of this would be O(n3)
Below is the implementation of the above approach:
## C++
`// C++ program to find the length of``// the longest balanced substring``#include``using` `namespace` `std;` `// Function to check if a string contains``// equal number of one and zeros or not``bool` `isValid(string p)``{`` ``int` `n = p.length();`` ``int` `c1 = 0, c0 = 0;`` ` ` ``for``(``int` `i =0; i < n; i++)`` ``{`` ``if``(p[i] == ``'0'``)`` ``c0++;`` ``if``(p[i] == ``'1'``)`` ``c1++;`` ``}`` ` ` ``return` `(c0 == c1) ? ``true` `: ``false``;``}` `// Function to find the length of``// the longest balanced substring``int` `longestSub(string s)``{`` ``int` `max_len = 0;`` ``int` `n = s.length();`` ` ` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``for``(``int` `j = i; j < n; j++)`` ``{`` ``if``(isValid(s.substr(i, j - i + 1)) && max_len < j - i + 1)`` ``max_len = j - i + 1;`` ``}`` ` ` ``}`` ``return` `max_len;`` ` `}` `// Driver code``int` `main()``{`` ``string s = ``"101001000"``;`` ` ` ``// Function call`` ``cout << longestSub(s);`` ` ` ``return` `0;``}`
## Java
`// Java program to find the length of ``// the longest balanced substring``import` `java.io.*;``import` `java.util.*;``class` `GFG``{` ` ``// Function to check if a string contains`` ``// equal number of one and zeros or not`` ``public` `static` `boolean` `isValid(String p)`` ``{`` ``int` `n = p.length();`` ``int` `c1 = ``0``, c0 = ``0``;`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``{`` ``if``(p.charAt(i) == ``'0'``)`` ``{`` ``c0++;`` ``}`` ``if``(p.charAt(i) == ``'1'``)`` ``{`` ``c1++;`` ``}`` ``}`` ``if``(c0 == c1)`` ``{`` ``return` `true``;`` ``}`` ``else`` ``{`` ``return` `false``;`` ``}`` ``}`` ``// Function to find the length of `` ``// the longest balanced substring`` ``public` `static` `int` `longestSub(String s)`` ``{`` ``int` `max_len = ``0``;`` ``int` `n = s.length();`` ``for``(``int` `i = ``0``; i < n; i++)`` ``{`` ``for``(``int` `j = i; j < n; j++)`` ``{`` ``if``(isValid(s.substring(i, j + ``1``)) && max_len < j - i + ``1``)`` ``{`` ``max_len = j - i + ``1``;`` ``}`` ``}`` ``}`` ``return` `max_len;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main (String[] args)`` ``{`` ``String s = ``"101001000"``;` ` ``// Function call`` ``System.out.println(longestSub(s));`` ``}``}` `// This code is contributed by avanitrachhadiya2155`
## Python3
`# Python3 program to find the length of``# the longest balanced substring` `# Function to check if a contains``# equal number of one and zeros or not``def` `isValid(p):` ` ``n ``=` `len``(p)`` ``c1 ``=` `0`` ``c0 ``=` `0` ` ``for` `i ``in` `range``(n):`` ``if` `(p[i] ``=``=` `'0'``):`` ``c0 ``+``=` `1`` ``if` `(p[i] ``=``=` `'1'``):`` ``c1 ``+``=` `1` ` ``if` `(c0 ``=``=` `c1):`` ``return` `True`` ``else``:`` ``return` `False` `# Function to find the length of``# the longest balanced substring``def` `longestSub(s):`` ` ` ``max_len ``=` `0`` ``n ``=` `len``(s)` ` ``for` `i ``in` `range``(n):`` ``for` `j ``in` `range``(i, n):`` ``if` `(isValid(s[i : j ``-` `i ``+` `1``]) ``and`` ``max_len < j ``-` `i ``+` `1``):`` ``max_len ``=` `j ``-` `i ``+` `1` ` ``return` `max_len` `# Driver code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``s ``=` `"101001000"` ` ``# Function call`` ``print``(longestSub(s))` `# This code is contributed by mohit kumar 29`
## C#
`// C# program to find the length of ``// the longest balanced substring``using` `System;` `class` `GFG{`` ` `// Function to check if a string contains``// equal number of one and zeros or not``static` `bool` `isValid(``string` `p)``{`` ``int` `n = p.Length;`` ``int` `c1 = 0, c0 = 0;`` ` ` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``if` `(p[i] == ``'0'``)`` ``{`` ``c0++;`` ``}`` ``if` `(p[i] == ``'1'``)`` ``{`` ``c1++;`` ``}`` ``}`` ``if` `(c0 == c1)`` ``{`` ``return` `true``;`` ``}`` ``else`` ``{`` ``return` `false``;`` ``}``}` `// Function to find the length of ``// the longest balanced substring``public` `static` `int` `longestSub(``string` `s)``{`` ``int` `max_len = 0;`` ``int` `n = s.Length;`` ` ` ``for``(``int` `i = 0; i < n; i++)`` ``{`` ``for``(``int` `j = i; j < n; j++)`` ``{`` ``if` `(isValid(s.Substring(i, j - i + 1)) &&`` ``max_len < j - i + 1)`` ``{`` ``max_len = j - i + 1;`` ``}`` ``}`` ``}`` ``return` `max_len;``}` `// Driver code``static` `public` `void` `Main()``{`` ``string` `s = ``"101001000"``;`` ` ` ``// Function call`` ``Console.WriteLine(longestSub(s));``}``}` `// This code is contributed by rag2127`
## Javascript
``
Output
`6`
An efficient solution is to use hashing.
1. Traverse string and keep track of counts of 1s and 0s as count_1 and count_0 respectively.
2. See if current difference between two counts has appeared before (We use hashing to store all differences and first index where a difference appears). If yes, then substring from previous appearance and current index has same number of 0s and 1s.
Below is the implementation of above approach.
## C++
`// C++ for finding length of longest balanced``// substring``#include``using` `namespace` `std;` `// Returns length of the longest substring``// with equal number of zeros and ones.``int` `stringLen(string str)``{`` ``// Create a map to store differences`` ``// between counts of 1s and 0s.`` ``map<``int``, ``int``> m;`` ` ` ``// Initially difference is 0.`` ``m[0] = -1; `` ` ` ``int` `count_0 = 0, count_1 = 0;`` ``int` `res = 0;`` ``for` `(``int` `i=0; i
## Java
`// Java Code for finding the length of``// the longest balanced substring` `import` `java.io.*;``import` `java.util.*;` `public` `class` `MAX_LEN_0_1 {`` ``public` `static` `void` `main(String args[])``throws` `IOException`` ``{`` ``String str = ``"101001000"``;`` ` ` ``// Create a map to store differences`` ``//between counts of 1s and 0s.`` ``HashMap map = ``new` `HashMap();`` ` ` ``// Initially difference is 0;`` ``map. put(``0``, -``1``);`` ``int` `res =``0``;`` ``int` `count_0 = ``0``, count_1 = ``0``;`` ``for``(``int` `i=``0``; i
## Python3
`# Python3 code for finding length of``# longest balanced substring` `# Returns length of the longest substring``# with equal number of zeros and ones.``def` `stringLen( ``str` `):` ` ``# Create a python dictionary to store`` ``# differences between counts of 1s and 0s.`` ``m ``=` `dict``()`` ` ` ``# Initially difference is 0.`` ``m[``0``] ``=` `-``1`` ` ` ``count_0 ``=` `0`` ``count_1 ``=` `0`` ``res ``=` `0`` ``for` `i ``in` `range``(``len``(``str``)):`` ` ` ``# Keeping track of counts of`` ``# 0s and 1s.`` ``if` `str``[i] ``=``=` `'0'``:`` ``count_0 ``+``=` `1`` ``else``:`` ``count_1 ``+``=` `1`` ` ` ``# If difference between current`` ``# counts already exists, then`` ``# substring between previous and`` ``# current index has same no. of`` ``# 0s and 1s. Update result if`` ``# this substring is more than`` ``# current result.`` ``if` `m.get(count_1 ``-` `count_0)!``=``None``:`` ``res ``=` `max``(res, i ``-` `m[count_1 ``-` `count_0])`` ` ` ``# If current difference is`` ``# seen first time.`` ``else``:`` ``m[count_1 ``-` `count_0] ``=` `i`` ``return` `res` `# driver code``str` `=` `"101001000"``print``(``"Length of longest balanced"`` ``" sub string = "``,stringLen(``str``))` `# This code is contributed by "Sharad_Bhardwaj"`
## C#
`// C# Code for finding the length of``// the longest balanced substring``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``public` `static` `void` `Main(``string``[] args)``{`` ``string` `str = ``"101001000"``;` ` ``// Create a map to store differences`` ``//between counts of 1s and 0s.`` ``Dictionary<``int``,`` ``int``> map = ``new` `Dictionary<``int``,`` ``int``>();`` ` ` ``// Initially difference is 0;`` ``map[0] = -1;`` ``int` `res = 0;`` ``int` `count_0 = 0, count_1 = 0;`` ``for` `(``int` `i = 0; i < str.Length;i++)`` ``{`` ``// Keep track of count of 0s and 1s`` ``if` `(str[i] == ``'0'``)`` ``{`` ``count_0++;`` ``}`` ``else`` ``{`` ``count_1++;`` ``}` ` ``// If difference between current counts`` ``// already exists, then substring between`` ``// previous and current index has same`` ``// no. of 0s and 1s. Update result if this`` ``// substring is more than current result.`` ``if` `(map.ContainsKey(count_1 - count_0))`` ``{`` ``res = Math.Max(res, (i - map[count_1 -`` ``count_0]));`` ``}` ` ``// If the current difference is`` ``// seen first time.`` ``else`` ``{`` ``map[count_1 - count_0] = i;`` ``}` ` ``}` ` ``Console.WriteLine(``"Length of longest balanced"` `+`` ``" sub string = "` `+ res);``}``}` `// This code is contributed by Shrikant13`
## Javascript
``
Output
`Length of longest balanced sub string = 6`
Time Complexity: O(n)
Extended Problem: Largest subarray with equal number of 0s and 1s
This article is contributed by Shivam Pradhan (anuj_charm) and ASIPU PAWAN KUMAR. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
My Personal Notes arrow_drop_up | 3,921 | 11,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-33 | latest | en | 0.514004 |
https://economics.stackexchange.com/questions/26256/keynesian-cross-and-balanced-budget-multiplier | 1,718,732,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00186.warc.gz | 189,487,374 | 39,305 | # Keynesian Cross and Balanced Budget Multiplier
Hi! I am trying to understand the problem above, and was wondering if someone could help me with the last question. I think I am fine with all other questions.
Here is my attempt:
(i) The fiscal multiplier is $$\frac{1}{1-c} \Delta G$$, as it is the infinite sum $$\Delta G + c\Delta G + c^2\Delta G + ...$$.
Depends on c because the increase in government spending first leads to an increase in Y, which leads to an increase in C, which then leads to another increase in Y, but this time the increase in only $$cY$$, as part of the income is saved.
(ii) Balanced budget multiplier is 1, as $$\Delta Y = \frac{1}{1-c} \Delta G + \frac{-c}{1-c} \Delta T$$ with $$\Delta T = \Delta G$$
This means that an increase in government spending that is matched by an identical increase in lump sum taxed increases output by the same amount as the increase in G (and T).
(iii) $$\Delta Y = \frac{\Delta X}{0.01} = 100$$
This means that giving me an additional unit of currency will increase total income by 100, as it feeds back into income and consumption infinitely.
(iv) This is the question where I would like some guidance. Here is my work so far:
We have: $$C= a + c(Y-T) - d\alpha Y$$
The multiplier associated with an increase in G should now be: $$\frac{1}{1-c} + \frac{1}{1-d\alpha}$$ The first part is the "old multiplier, the second is the effect of the new part of the consumption function? I am not sure that this is correct, happy to receive comments!
This would mean that $$c=d\alpha$$.
And then, the multiplier associated with an increase in T:
• 1st round: Y increases by $$c\Delta T$$
• 2nd round: C increases by $$c^2\Delta T + d\alpha \Delta T = 2 c^2\Delta T$$
• Then Y increases by $$2 c^2\Delta T$$
• 3rd round: C increases by $$4 c^3\Delta T$$
Does this mean that the multiplier is this sum: $$2^0C^1\Delta T + 2^1C^2\Delta T + 2^2C^2\Delta T + ...$$
And then the balanced budget multiplier would be this last result + 1 (the other multiplier, from the increase in G)?
I'd be super grateful to anyone who could help me with question (iv)!
• This question would be improved by converting the picture at the beginning into text. Commented Jan 4, 2019 at 18:01
• Sorry about this, I will definitely covert pictures into text next time I post a question. Thank you for the feedback.
– Dila
Commented Jan 5, 2019 at 19:06
• The balanced budget multiplier is 1 if you assume that all of the increase in income becomes disposable income but that is often not the case, due to taxes, transfers, as well as depreciation payments. If we assume that the proportion of the increase in income that results in the increase in disposable income is $d_0$, then the balanced budget multiplier is $\frac{1 - c}{1 - cd_0}$ where c is the marginal propensity to consume. Commented May 25, 2020 at 2:58
The fiscal multiplier in (iv) would arise from
$$Y = C+I+G$$
$$Y = a + c(Y-T)-d\alpha Y +G+I.$$
This can be rearranged as
$$Y(1-c+d\alpha)=a-cT+G+I,$$
and, therefore, the fiscal multiplier would be
$$\frac{1}{1-c+d\alpha}.$$
In order for this to equal $$1$$, you just need $$d\alpha=c.$$
Note that setting $$d\alpha=c$$ in your "multiplier" does not yield 1:
$$\frac{1}{1-c}+\frac{1}{1-d\alpha}=\frac{1}{1-c}+\frac{1}{1-c}=\frac{2}{1-c}.$$
• Thank you! This is very helpful. As for the balanced budget multiplier, am I correct to think that it is equal to 1 (the fiscal multiplier) + c (the multiplier associated with an increase in T)? That is, balanced budget multiplier is 1 + c?
– Dila
Commented Jan 5, 2019 at 19:02 | 1,024 | 3,597 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.943165 |
https://blog.csdn.net/cmershen/article/details/51548744 | 1,521,765,150,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648103.60/warc/CC-MAIN-20180322225408-20180323005408-00048.warc.gz | 549,110,283 | 11,816 | # Middle-题目123：335. Self Crossing
You are given an array x of n positive numbers. You start at point (0,0) and moves x[0] metres to the north, then x[1] metres to the west, x[2] metres to the south, x[3] metres to the east and so on. In other words, after each move your direction changes counter-clockwise.
Write a one-pass algorithm with O(1) extra space to determine, if your path crosses itself, or not.
Example 1:
Given x = [2, 1, 1, 2],
Return true (self crossing)
Example 2:
Given x = [1, 2, 3, 4],
Return false (not self crossing)
Example 3:
Given x = [1, 1, 1, 1],
Return true (self crossing)
public class Solution {
public boolean isSelfCrossing(int[] x) {
for(int i=3;i<x.length;i++){
if(i>=3&&x[i]>=x[i-2]&&x[i-1]<=x[i-3])
return true;
if(i>=4&&x[i-1]==x[i-3]&&x[i-2]<=x[i]+x[i-4])
return true;
if(i>=5&&x[i-2]>x[i-4]&&x[i-1]<=x[i-3]&&x[i-1]+x[i-5]>=x[i-3]&&x[i]+x[i-4]>=x[i-2])
return true;
}
return false;
}
}
0ms，beats 43.49%，众数0ms,56.51%
Cmershen的碎碎念：
• 本文已收录于以下专栏：
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https://visualfractions.com/unit-converter/convert-100-pclmb-to-nclmb/ | 1,653,092,620,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00035.warc.gz | 682,801,853 | 6,287 | # Convert 100 pclmb to nclmb
So you want to convert 100 picocoulombs into nanocoulombs? If you're in a rush and just need the answer, the calculator below is all you need. The answer is 0.1 nanocoulombs.
## How to convert picocoulombs to nanocoulombs
We all use different units of measurement every day. Whether you're in a foreign country and need to convert the local imperial units to metric, or you're baking a cake and need to convert to a unit you are more familiar with.
Luckily, converting most units is very, very simple. In this case, all you need to know is that 1 pclmb is equal to 0.001 nclmb.
Once you know what 1 pclmb is in nanocoulombs, you can simply multiply 0.001 by the total picocoulombs you want to calculate.
So for our example here we have 100 picocoulombs. So all we do is multiply 100 by 0.001:
100 x 0.001 = 0.1
## What is the best conversion unit for 100 pclmb?
As an added little bonus conversion for you, we can also calculate the best unit of measurement for 100 pclmb.
What is the "best" unit of measurement? To keep it simple, let's say that the best unit of measure is the one that is the lowest possible without going below 1. The reason for this is that the lowest number generally makes it easier to understand the measurement.
For 100 pclmb the best unit of measurement is picocoulombs, and the amount is 100 pclmb. | 352 | 1,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2022-21 | latest | en | 0.919145 |
http://www.docstoc.com/docs/126658826/HDPE-Irradiation | 1,427,552,688,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131297587.67/warc/CC-MAIN-20150323172137-00108-ip-10-168-14-71.ec2.internal.warc.gz | 465,189,320 | 32,312 | Documents
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• pg 1
``` HDPE Berm Pipe
Michael Geelhoed
In the spring of 2011, the berm pipe from enclosure G2 to NM1 developed a significant
leak. Vacuum pumps were destroyed when ground water infiltrated the quarter inch thick steel
wall pipe. After months of troubleshooting and surveying, the idea to sleeve this pipe using
High Density Polyethylene, or HDPE, pipe is considered.
One aspect in deciding to use HDPE is to consider the radiation damage and its impact
on the ability of the material to hold vacuum well enough to transport beam. The radiation
dose to the HDPE pipe is calculated for the expected normal running beam losses and tested for
accidental direct hit beam pulses. Based on vacuum readings of adjacent berm pipes, we
consider looking at the vacuum at 1E-2 Torr as our base line for acceptable vacuum readings.
With the normal running conditions of 1E+13 protons per pulses delivered at one pulse
a minute, which delivers 5.256E+18 protons per year, for a 1.27 inch thick and 760 feet long
pipe at 1E-2 Torr pressure, it is calculated to receive an equivalent of 3.15E+13 interacting
protons each year it is in service. A multiwire, with Gold plated Tungsten wires, will be placed
just upstream of the HDPE pipe causing an equivalent of 6.62E+14 protons scattered to interact
with the pipe.
Combined lost estimate is 6.94E+14 protons per year, this equates to 1.67E5 Rads per
year based on the pipe’s mass. Radiation damage to HDPE below 1E+7 Rads will not
compromise its mechanical properties which are deemed as always satisfactory1, and is often
satisfactory2 for 5E+7 Rads3. The calculation of the dose incurred by the HDPE pipe is presented
below.
Equation 1 Calculation for yearly protons lost to Rads per year, and calculations for years for HDPE material to become
Always and Often Satisfactory
1
Always Satisfactory is no changes to materials properties i.e. tensile strength, elongation, etc.
2
Often Satisfactory is the change is first detected in one or more properties
3
REIC Report 21 and Addendum August 31, 1964
Accidental loss is replicated using the same material blind flanges. Four flanges were
procured, three were irradiated as one remained as the control. Each HDPE Blind flange is one
inch thick and 21 inches in diameter. From the manufacturer they have a bolt-hole pattern for a
back collar, and we specially machined the front face for a smooth finish and an o-ring groove
to achieve a good vacuum seal before it was irradiated.
Individually each flange was placed in front of the Switchyard absorber line and was
irradiated, with increasing amounts of pulses. Each flange had approximately a factor of 10
more than the previous flange. The spot size of the beam varied slightly from pulse to pulse as
did the intensity but on average was consistent. The flange with 1 pulse had a half hour cool-
down period and the remaining flanges had an hour cool -down before we retrieved the flange.
Each flange irradiated was frisked and surveyed to be Class 1, less than 1mR/hr4, only flange number
four showed contact residual dose at 150 counts per minute, the others had no evidence of contact
residual dose. Shown in Table 1 is the amount of integrated beam delivered to each flange,
average spot size of the beam on the flange and the vacuum leak rate after irradiation.
Flange Number Average Integrated Average 1 Average 1 Vacuum Leak Rate
of pulse Protons Sigma Sigma
pulses intensity Horizontal Vertical
(mm) (mm)
1 0 0 0 N/A N/A 3.2E-10
2 1 6.94E+11 6.94E+11 2.521 2.915 3.2E-10
3 10 7.50E+11 7.50E+12 2.484 2.782 3.9E-10
4 124 5.81E+11 7.20E+13 3.161 2.627 3.3E-10
Table 1: Flange leak rate vs. integrated protons
Under normal running conditions the calculated lifetime of the HDPE pipe exceeds the
lifetime of the current experiment, (E-906 SeaQuest) and covers far into the future by a safety
factor of greater than 12. With accidental proton beam losses on the pipe, it has the budget of
at least 124 direct beam hits of 5.81E11 protons per pulse on the HDPE pipe (7.2E13 protons),
without compromising the integrity of the HDPE beam pipe.
4
FRCM Chapter 4 Part 1
```
To top | 1,161 | 4,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-14 | longest | en | 0.922726 |
https://socratic.org/questions/how-do-you-divide-6-15-4 | 1,623,754,731,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00214.warc.gz | 481,617,591 | 5,642 | # How do you divide 6.15/4?
I found: $1.5375$
I would multiply top and bottom by $100$:
$\frac{6.15 \cdot 100}{4 \cdot 100} = \frac{615}{400}$ | 58 | 143 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.788686 |
https://lagevilcei.web.app/1021.html | 1,722,735,033,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640388159.9/warc/CC-MAIN-20240804010149-20240804040149-00338.warc.gz | 292,805,524 | 5,045 | # Node analysis problems pdf
Nodal analysis examples nodal equations by inspection method. Circuit analysis with dependent sources anode equations c. Assumed states for analysis of ideal diode circuits example 3. Chapter 3 nodal and mesh equations circuit theorems. Kcl analysis of a circuit the node in the center of this circuit involves four currents. Express the currents in terms of the voltages v1, v2, and vo and other node voltages as necessary. Pan 10 nfundamental loop analysis is based on a systematic application of kvl to the. Use nodal analysis to compute the voltage across the 18 a current source in the circuit of figure 3. Note that the ground node does not necessarily represent an actual connection to ground, it is just a device to make the analysis simpler. This is the same where we use supernode circuit analysis instead of node or nodal circuit analysis to simplify such a network where the assign supernode, fully enclosing the voltage source.
Nodal circuit analysis using kcl most useful for when we have mostly current sources node analysis uses kcl to establish the currents procedure 1 choose one node as the common or datum node number label the nodes designate a voltage for each node number each node voltage is with respect to the common or datum node. For example, if a node has a voltage of 5 volts, then the voltage drop between that node and the ground node. Supermesh or supermesh analysis is a better technique instead of using mesh analysis to analysis such a complex electric circuit or network, where two meshes have a current source as a common element. Supernode analysis circuit diagram equation examples. Pan 9 nnodal analysis is based on a systematic application of kcl and is a general method.
So that, we may minimize the number of of equations. In the figure below, means the voltage of the node a with respect to node r. There are a few general guidelines that we need to remember as we make the. Nodal analysis of electronic circuits is based on assigning nodal voltages at various nodes of the circuit with respect to a reference and then finding these nodal voltages to analyze the circuit. This app teaches the nodal analysis method of solving dc electric networks. The node method or the node voltage method, is a very powerful approach for circuit analysis and it is based on the application of kcl, kvl and ohms law. Circuit analysis i with matlab applications 357 orchard publications exercises problems 1. In our case we have this voltage here, is the node voltage on node one, well call it. Node voltage method steps 1 to 4 video khan academy. The field is the domain of interest and most often represents a. The finite element method fem, or finite element analysis fea, is a computational technique used to obtain approximate solutions of boundary value problems in engineering. Summary of supernode analysis step by step redraw the circuit if possible. Node voltages, in this context, are the independent voltages in the circuit. Once you have done this you can easily work out anything else you need.
Nodal analysis of ideal operational amplifier circuits ee. Solving by nodal analysis circuit with four nodes solved problems. Nodal analysis with dependent sources utilized kirchhoffs current law with algebra and ohms law to substitute an unknown voltage for a node and to find other circuit values. Chapter 3 solving for voltages and currents in circuits. In nodal analysis, the unknowns will be node voltages.
Boundary value problems are also called field problems. E40m solving circuits using nodal analysis, part ii and. This section contains analysis steps that can be performed to resolve sap hana inter node communication issues. In general, in a n node circuit, one of the nodes is chosen as reference or datum node, then it is possible to write n 1 nodal analysis examples by assuming n 1 node. A circuit with four nodes solved using the nodal analysis. Basic procedure node voltage analysis is a systematic method for finding the voltage at all nodes in an electrical circuit. Mesh analysis meshcurrent method these methods are based on the systematic application of kirchhoffs laws kvl and kcl. Weve picked a reference node to be node three, down here. Remember that in an ideal op amp, currents i1 and i2 are zero. Nodal analysis and dependent sources technical articles. This module demonstrates physical resistive circuits and introduces several systematic ways to solve circuit problems. Use nodal analysis to compute the voltage across the 18 a current source in the circuit of figure. Each tree branch defines a unique cutset or super node.
Useful nodes are the ones which make the problem easier to understand and solve. And just as weve seen in previous problems, we notice that there is a 6 volt source between node three and the ground node. Video created by georgia institute of technology for the course linear circuits 1. Using nikon 18 55 repair manual pdf mesh analysis or kirchhoffs laws, determine the values of i, i1 and i2. Using nodal method, find the current through resistor r2 figure 1. The procedure for analyzing a circuit with the node method is based on the following steps. The ground node should be chosen carefully for convenience. Nodal analysis example with solution electronics tutorials. Write the admittance matrix y of the network where. Boser department of electrical engineering and computer sciences university of california, berkeley 1. The sum of the two currents entering the node the current from elements a and d are equal to the two currents leaving the node towards elements b and c. Determine the node voltages, vt1 and vt2, and the mesh currents, it1 and it2, for this circuit.
Electronicsnodal analysis wikibooks, open books for an. The basic idea is that applying kirchhoffs current law at the various nodes of the network yields a system of equations whose simultaneous solution is the values of the node voltages. Analyzing the center node of this circuit shows kirchhoffs voltage law kvl. This may be the node with the greatest number of branches. Determine the number of nodes n, label them and pick the reference. In analyzing a circuit using kirchhoffs circuit laws, one can either do nodal analysis using kirchhoffs current law.
In electric circuits analysis, nodal analysis, node voltage analysis, or the branch current method is a method of determining the voltage potential difference between nodes points where elements or branches connect in an electrical circuit in terms of the branch currents. Use nodal analysis to compute the voltage in the circuit of figure 3. If we consider the node r as a reference node, then is simply written as, because the voltage of the reference node is zero. By taking the time to carefully label the nodes, by identifying the proper node voltages and polarities, problem solving is made easier and can avoid mistakes. If youre seeing this message, it means were having. Linear constant coefficient differential equations.
The rules for modified nodal analysis are given by. Supermesh circuit analysis step by step with solved example. Nodal analysis examples we discussed simple circuits containing only two nodes, including the reference node. Circuit analysis with dependent sources a node equations bequivalent sources camplifier parameters. Introduction to finite element analysis fea or finite. Procedure if communication issues occur between different nodes within an sap hana scaleout environment, usually the sap hana tracefiles contain corresponding errors. And again, noting that for nodal analysis what were ultimately looking for are the nodal voltages, we can find for this problem that we can directly calculate or find the nodal voltage for node three from this. Solved question on nodal analysis topics discussed. Selective a reference node usually ground and name the remaining n1 nodes. To address this problem, after we describe nodal analysis we will provide a number of techniques that allow you to simplify the network you need to solve. This symbol on a schematic indicates the reference node. Find the efficient method to determine the voltage in a circuit using the nodal analysis method by applying kcl at a given node.
Chapter 3 nodal and mesh equations circuit theorems 352 3. The basic procedure for solving nodal analysis equations is as follows. To apply the node voltage method to a circuit with n nodes with m voltage sources, perform the following steps after decarlolin. Critical path analysis solved example milestonetask. Clearly label all circuit parameters and distinguish the unknown parameters from the known. The circuit has three current sources and three resistors. So, a node voltage is measured between a node and the reference node. Simple representation of nodal voltages shown below. Node analysis in its simplest form only has current sources present.
1213 763 543 577 634 1421 574 312 1051 1503 1160 1495 1647 166 544 1222 547 939 1262 1262 106 170 339 1363 618 1240 1419 706 4 883 1466 3 1483 296 350 | 1,872 | 9,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.881424 |
http://stackoverflow.com/questions/2722124/word-jumble-algorithm | 1,394,817,760,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678694108/warc/CC-MAIN-20140313024454-00092-ip-10-183-142-35.ec2.internal.warc.gz | 135,533,671 | 16,195 | # Word Jumble Algorithm
Given a word jumble (i.e. ofbaor), what would be an approach to unscramble the letters to create a real word (i.e. foobar)? I could see this having a couple of approaches, and I think I know how I'd do it in .NET, but I curious to see what some other solutions look like (always happy to see if my solution is optimal or not).
This isn't homework or anything like that, I just saw a word jumble in the local comics section of the paper (yes, good ol' fashioned newsprint), and the engineer in me started thinking.
edit: please post some pseudo code or real code if you can; it's always nice to try and expand language knowledge by seeing examples like this.
-
Have a dictionary that's keyed by the letters of each word in sorted order. Then take you jumble an sort the letters - look up all the words in the dictionary by that sorted-letter string.
So, as an example, the words 'bear' and 'bare' would be in the dictionary as follows:
``````key word
----- ------
aber bear
aber bare
``````
And if you're given the jumble, 'earb', you'd sort the letters to 'aber' and be able to look up both possible words in the dictionary.
-
Create all the permutations of the string, and look them up in a dictionary.
You can optimize by looking up shorter strings that begin words, and if there are no words of suitable length in the dictionary that start with those strings, eliminating permutations starting with those letters from further consideration.
-
CodeProject has a couple of articles here and here. The second uses recursion. Wikipedia also outlines a couple of algorithms here. The Wikipedia article also mentions a program called Jumbo that uses a more heuristic approach that solves the problem like a human would. There seem to be a few approaches to the problem.
-
Depending on the length of the string WhirlWind's approach could be faster, but an alternative way of doing it which has more or less O(n) complexity is instead of creating all the permutations of the string and looking them up, you go through all the words in the dictionary and see if each can be built from the input string.
A really smart algorithm that knows the number of words in the dictionary in advance could do something like this:
``````sort letters in jumbled string
if factorial(length of jumbled string) > count of words in dictionary:
for each word in dictionary:
sort letters in dictionary word
if sorted letters in dictionary word == sorted letters in jumbled string:
append original dictionary word to acceptable word list
else:
create permutation list of jumbled letters
for each permutation of letters:
search in dictionary for permutation
if permutation in dictionary:
add word to acceptable word list
``````
- | 593 | 2,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2014-10 | latest | en | 0.944985 |
https://uk.news.yahoo.com/yue-yuen-industrial-holdings-limited-005746979.html | 1,585,707,561,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505359.23/warc/CC-MAIN-20200401003422-20200401033422-00051.warc.gz | 753,302,413 | 96,002 | # Is Yue Yuen Industrial (Holdings) Limited (HKG:551) Trading At A 26% Discount?
Today we will run through one way of estimating the intrinsic value of Yue Yuen Industrial (Holdings) Limited (HKG:551) by estimating the company's future cash flows and discounting them to their present value. I will use the Discounted Cash Flow (DCF) model. Don't get put off by the jargon, the math behind it is actually quite straightforward.
Companies can be valued in a lot of ways, so we would point out that a DCF is not perfect for every situation. Anyone interested in learning a bit more about intrinsic value should have a read of the Simply Wall St analysis model.
Check out our latest analysis for Yue Yuen Industrial (Holdings)
### Is Yue Yuen Industrial (Holdings) fairly valued?
As Yue Yuen Industrial (Holdings) operates in the luxury sector, we need to calculate the intrinsic value slightly differently. Instead of using free cash flows, which are hard to estimate and often not reported by analysts in this industry, dividends per share (DPS) payments are used. This often underestimates the value of a stock, but it can still be good as a comparison to competitors. The 'Gordon Growth Model' is used, which simply assumes that dividend payments will continue to increase at a sustainable growth rate forever. The dividend is expected to growth at an annual growth rate equal to the 10-year government bond rate of 1.6%. We then discount this figure to today's value at a cost of equity of 7.1%. Relative to the current share price of HK\$19.6, the company appears a touch undervalued at a 26% discount to where the stock price trades currently. Remember though, that this is just an approximate valuation, and like any complex formula - garbage in, garbage out.
Value Per Share = Expected Dividend Per Share / (Discount Rate - Perpetual Growth Rate)
= US\$0.2 / (7.1% – 1.6%)
= HK\$26.5
### The assumptions
We would point out that the most important inputs to a discounted cash flow are the discount rate and of course the actual cash flows. You don't have to agree with these inputs, I recommend redoing the calculations yourself and playing with them. The DCF also does not consider the possible cyclicality of an industry, or a company's future capital requirements, so it does not give a full picture of a company's potential performance. Given that we are looking at Yue Yuen Industrial (Holdings) as potential shareholders, the cost of equity is used as the discount rate, rather than the cost of capital (or weighted average cost of capital, WACC) which accounts for debt. In this calculation we've used 7.1%, which is based on a levered beta of 0.915. Beta is a measure of a stock's volatility, compared to the market as a whole. We get our beta from the industry average beta of globally comparable companies, with an imposed limit between 0.8 and 2.0, which is a reasonable range for a stable business.
### Next Steps:
Valuation is only one side of the coin in terms of building your investment thesis, and it shouldn’t be the only metric you look at when researching a company. The DCF model is not a perfect stock valuation tool. Rather it should be seen as a guide to "what assumptions need to be true for this stock to be under/overvalued?" If a company grows at a different rate, or if its cost of equity or risk free rate changes sharply, the output can look very different. What is the reason for the share price to differ from the intrinsic value? For Yue Yuen Industrial (Holdings), I've compiled three further aspects you should further examine:
1. Financial Health: Does 551 have a healthy balance sheet? Take a look at our free balance sheet analysis with six simple checks on key factors like leverage and risk.
2. Future Earnings: How does 551's growth rate compare to its peers and the wider market? Dig deeper into the analyst consensus number for the upcoming years by interacting with our free analyst growth expectation chart.
3. Other High Quality Alternatives: Are there other high quality stocks you could be holding instead of 551? Explore our interactive list of high quality stocks to get an idea of what else is out there you may be missing!
PS. Simply Wall St updates its DCF calculation for every HK stock every day, so if you want to find the intrinsic value of any other stock just search here.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material. Thank you for reading. | 1,069 | 4,938 | {"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} | 2.71875 | 3 | CC-MAIN-2020-16 | latest | en | 0.926243 |
https://vnemart.com.vn/how-do-you-spell-11/ | 1,695,562,090,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00846.warc.gz | 662,667,687 | 12,759 | General 8 how do you spell 11 best, you should know
Nội dung chính
1 11 in Words – Write 11 in Words | 11 Spelling
• Author: cuemath.com
• Published Date: 03/16/2022
• Review: 4.96 (767 vote)
• Summary: 11 in Words = Eleven · Eleven in Numbers = 11
• Matching search results: Eleven in numerals is written as 11. Seven in numerals is written as 7, Now Eleven Minus Seven means subtracting 7 from 11, i.e. 11 – 7 = 4 which is read as Four.
2 How to write spell say 11 number in english words or Spelling?
• Author: numwrite.com
• Published Date: 01/11/2022
• Review: 4.89 (784 vote)
• Summary: spelling for 11 in english, number to words for 11 number. Spell Say Write 11 in english. 11 in english: ( eleven )… Search Number: <>
• Matching search results: INR -> Indian rupees -> eleven Indian rupees
3 11 in Words – How to Spell 11 | numbersinwords.net
• Author: numbersinwords.net
• Published Date: 10/17/2021
• Review: 4.91 (951 vote)
• Summary: 11th is the abbreviation of eleventh. To put it another way, 11th is the short form of the ordinal number for 11. 11th spelled out = eleventh. Below, we wrap
• Matching search results: What is 11 in words? As with any number, 11 is the idea, and the numeral is how we write it, either pronominal, or attributive.
4 How do you spell 11 in English?
• Author: valeur.com
• Published Date: 12/13/2021
• Review: 4.93 (713 vote)
• Summary: When communicating in English, it is sometimes useful to spell out the number 11 with words instead of simply writing 11. Here we show you how to spell 11
• Matching search results: When communicating in English, it is sometimes useful to spell out the number 11 with words instead of simply writing 11.
5 Spell 11 Number In English, Words And Spelling- Mathauditor
• Author: mathauditor.com
• Published Date: 05/31/2022
• Review: 4.83 (730 vote)
• Summary: 11 in words :eleven,11 in english :eleven
• Matching search results: Question : How would you write 11 in binary?
6 How Do You Spell 11 In Words Perfectly Without Mistakes
• Author: numbertowordsconverter.com
• Published Date: 09/13/2022
• Review: 4.93 (942 vote)
• Summary: To write the number 11 in dollar amount, the currency symbol is placed before the number, with no spaces : \$11 . If spelled out in words, the currency symbol
• Matching search results: – If you want to know how to spell another number in English, type any number in the number to words converter above to see how it is spelled in English.
7 11 in Words – Write 11 in Words | 11 Spelling
• Author: translateen.com
• Published Date: 03/25/2022
• Review: 4.84 (984 vote)
• Summary: What is the correct spelling of 11?
• Matching search results: If you have typed 11 English or something similar like eleven number in the search engine of your choosing, then you have come to the right site, too.
8 Spelling Number Words 1 to 20 (songs, videos, games, activities)
• Author: onlinemathlearning.com
• Published Date: 07/26/2022
• Review: 4.9 (809 vote)
• Summary: This children’s educational video teaches kids how to spell the numbers from 11 to 20: eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen,
• Matching search results: Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 922 | 3,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.813369 |
http://sciforums.com/Conversion-between-base-10-and-base-2-t-42243.html | 1,386,778,261,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164038825/warc/CC-MAIN-20131204133358-00028-ip-10-33-133-15.ec2.internal.warc.gz | 159,882,748 | 3,507 | SciForums.com > Technology > Computer Science & Culture > Conversion between base 10 and base 2 PDA View Full Version : Conversion between base 10 and base 2 Post ReplyCreate New Thread cfso195211-06-04, 09:03 PMHow to convert from base 2 to 10 or 10 to 2???????? Including conversion of integers, decimals How to carry out mathematical operations in base 2???????????????? geodesic11-06-04, 10:22 PMTo convert from base 10 to base 2: find out how many digits you need by taking log base 2 of your number, which we shall call x, rounded down plus 1. Take n equal to the number of digits. Set each digit (starting from the left) by doing integer division of number x, and then set x as the remainder. decrement n and repeat the previous step until n=0. eg. 168 has 8 digits, 10101000 To go from 2 to 10: Multiply each digit by 2 to the power of of its distance from the right minus 1. eg. 1001101=1x1+1x4+1x8+1x64=77. Why do you ask? Voodoo Child11-07-04, 04:21 AMdecimals depends upon the format of the number. As I recall the number is normalised in the form of .101110101(there is an implicit 1 at the start that need not be recorded). The exponent is used to tell how far to move the "decimal" point. There is also a sign bit. The IEEE convention is: 1 bit sign /8 bit exponent/23 bit mantissa: 0 11111111 0101010 10101010 10101010 the exponent(unsigned) is converted to decimal, and then has 127 subtracted from it. if -ve, the dp goes <---. If +ve the dp goes --->. So the mantissa, 11111111. (or 27 after the bias is removed) means the dp goes 7 to the right. That is your floating point. To convert a binary no to base human: eg. 101010.101 is merely 25 * 1 + 24 *0+ 23 *1 + 22 *0 + 21 *1 + 20 *0 + 2-1 *1 + 2-2 *0 + 2-3 *1 does that make any sense at all? geodesic11-07-04, 12:14 PMSurely the exponent is the power of two to which the number is raised? Your exponent (which you have mis-named the mantissa) is 128 in decimal, which means the dp. moves 128 places right. You're multiplying by the exponent. And isn't the exponent in twos complement form, where the msb is negative, and all the others are positive? That would make your exponent -1. Voodoo Child11-07-04, 11:38 PMYour exponent (which you have mis-named the mantissa) is 128 in decimal, which means the dp. moves 128 places right. True. Should read: "(or 27 after the bias is removed) means the dp goes 27 to the right. ". Which is also wrong(should be 27-1) And isn't the exponent in twos complement form, where the msb is negative, and all the others are positive? No, it's biased. The reason is that 2's comp would make comparisons too hard. from http://www.math.grin.edu/~stone/courses/fundamentals/IEEE-reals.html The exponent, which is a signed integer in the range from -126 to 127, is represented neither as a signed magnitude nor as a twos-complement number, but as a biased value. The idea here is that the integers in the desired range of exponents are first adjusted by adding a fixed ``bias'' to each one. The bias is chosen to be large enough to convert every integer in the range into a positive integer, which is then stored as a binary numeral. The IEEE single-precision representation uses a bias of 127. edit: was also incorrect in that the bias is added and not subtracted. Post ReplyCreate New Thread | 907 | 3,293 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2013-48 | latest | en | 0.857693 |
https://econometrics.io/yield-to-maturity-calculation-using-a-python-script/ | 1,607,066,607,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735395.99/warc/CC-MAIN-20201204071014-20201204101014-00434.warc.gz | 275,941,255 | 10,848 | When interested in buying bonds a term that often comes up is the definition of Yield To Maturity (YTM).
Yield to maturity (YTM) is the total return anticipated on a bond if the bond is held until the end of its lifetime. Yield to maturity is considered a long-term bond yield, but is expressed as an annual rate. In other words, it is the internal rate of return of an investment in a bond if the investor holds the bond until maturity and if all payments are made as scheduled. – Investopedia
Let us consider a bond with a par value of $\100$ and a coupon rate of $3\%$ with the bond maturing in $29$ years time. The current bond price is $\89$. We are interested in the bond’s YTM.
Knowing that we get $\3$ annual cashflow ($coupon\ rate \times par\ value = 0.03 \times \100 = \3$) we will have the following cashflow for the next $29$ years. Notice in the last year we will receive an additional $\100$ which is the bond’s principle:
Let us assume that we want to value these future cashflows and determine how much they are worth today. What we need to do, is to present value all of these future cashflows back to today and add up those present values $pv_i$. This sum would be the price of the bond.
A present value is the amount of money $pv$ one would need to invest today in order to have $c$ amount of money in $n$ years if he is earning $r%$ annual interest on his investment:
$pv = \frac{c}{(1 + r)^n}$
For example, if Bob wants to have $\100$ one year from now at a rate of $5\%$, he needs to invest $\95.24$ today ($\frac{100}{(1 + 0.05)} = 95.24$), since $\95.24 \times (1 + 0.05) \approx \100$.
Back to our bond’s present value of future cashflows the resulting formula would look as follows:
$bond\ price = \frac{future\ cashflow\ 1}{(1 + ytm)^1} + \frac{future\ cashflow\ 2}{(1 + ytm)^2} + \frac{future\ cashflow\ 3}{(1 + ytm)^3} + \dots + \frac{future\ cashflow\ 29}{(1 + ytm)^{29}}$
with $ytm$ being the yield to maturity we are looking for.
For our particular bond the formula will look like this:
$bond\ price = \frac{3}{(1 + ytm)^1} + \frac{3}{(1 + ytm)^2} + \frac{3}{(1 + ytm)^3} + \dots + \frac{3 + 100}{(1 + ytm)^{29}}$
So how can we go about finding $ytm$? Well, above we have learned that sum of present values of all future cash flows is our bond price. Since we already know the today’s bond price which is $\89$, the formula becomes like this:
$\frac{3}{(1 + ytm)^1} + \frac{3}{(1 + ytm)^2} + \frac{3}{(1 + ytm)^3} + \dots + \frac{3 + 100}{(1 + ytm)^{29}} = 89$
All we need to do now is to solve the above equation for $ytm$ so that it holds true. We will do this through trial-and-error.
Solving the equation by hand requires an understanding of the relationship between a bond’s price and its yield, as well as of the different types of bond pricings. Bonds can be priced at a discount, at par and at a premium. When the bond is priced at par, the bond’s interest rate is equal to its coupon rate. A bond priced above par (called a premium bond) has a coupon rate higher than the interest rate, and a bond priced below par (called a discount bond) has a coupon rate lower than the interest rate. So if an investor were calculating YTM on a bond priced below par, he or she would solve the equation by plugging in various annual interest rates that were higher than the coupon rate until finding a bond price close to the price of the bond in question (_quote from:_ Investopedia).
You can verify this quickly with a simple example. Assume a bond with current price $b$ and a par value of $100$ that matures in $1$ year. The coupon rate is $c$ and interest rate is $ytm$:
$b = \frac{100 + c \times 100}{(1 + ytm)}$
==>
$b \times (1 + ytm) = 100 + c \times 100 = 100 \times (1 + c)$
==>
$\frac{b}{100} = \frac{(1 + c)}{(1 + ytm)}$
If $\frac{b}{100} \textless 1$ then $(1 + ytm) \textgreater (1 + c)$. Similarly for the other cases.
Back to our bond. We had the following equation:
$\frac{3}{(1 + ytm)^1} + \frac{3}{(1 + ytm)^2} + \frac{3}{(1 + ytm)^3} + \dots + \frac{3 + 100}{(1 + ytm)^{29}} = 89$
Now we must solve for the interest rate $ytm$ which is where things start to get difficult. Yet, we do not have to start simply guessing random numbers if we stop for a moment to consider the relationship between bond price and yield. As was mentioned above, when a bond is priced at a discount from par, its interest rate will be greater than the coupon rate. In this example, the par value of the bond is $\100$, but it is priced below the par value at $\89$, meaning that the bond is priced at a discount. As such, the annual interest rate we are seeking must necessarily be greater than the coupon rate of $3\%$.
With this information we can now calculate and test a number of bond prices by plugging various annual interest rates that are higher than $3\%$ into the formula above. Using a few different interest rates above $3\%$, one would come up with the following bond prices:
I have used $0.00001$ increments and as we can see with an interest rate of $3.62\%$ we can approximate our current price of $\89$ quite good enough. So $3.62\%$ would be our Yield To Maturity.
Full code
Result: | 1,448 | 5,150 | {"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": 47, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.925431 |
http://stackoverflow.com/questions/3796266/change-the-class-of-many-columns-in-a-data-frame/20978400 | 1,469,613,969,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257826759.85/warc/CC-MAIN-20160723071026-00215-ip-10-185-27-174.ec2.internal.warc.gz | 245,593,111 | 23,269 | Dismiss
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# Change the class of many columns in a data frame
What is the quickest/best way to change a large number of columns to numeric from factor?
I used the following code but it appears to have re-ordered my data.
``````> head(stats[,1:2])
rk team
1 1 Washington Capitals*
2 2 San Jose Sharks*
3 3 Chicago Blackhawks*
4 4 Phoenix Coyotes*
5 5 New Jersey Devils*
6 6 Vancouver Canucks*
for(i in c(1,3:ncol(stats))) {
stats[,i] <- as.numeric(stats[,i])
}
rk team
1 2 Washington Capitals*
2 13 San Jose Sharks*
3 24 Chicago Blackhawks*
4 26 Phoenix Coyotes*
5 27 New Jersey Devils*
6 28 Vancouver Canucks*
``````
What is the best way, short of naming every column as in:
``````df\$colname <- as.numeric(ds\$colname)
``````
-
Isn't there any generic solution?. Some of the solutions proposed here only work with factors, other work always except with factors, and so on... – skan May 6 at 10:32
Further to Ramnath's answer, the behaviour you are experiencing is that due to `as.numeric(x)` returning the internal, numeric representation of the factor `x` at the R level. If you want to preserve the numbers that are the levels of the factor (rather than their internal representation), you need to convert to character via `as.character()` first as per Ramnath's example.
Your `for` loop is just as reasonable as an `apply` call and might be slightly more readable as to what the intention of the code is. Just change this line:
``````stats[,i] <- as.numeric(stats[,i])
``````
``````stats[,i] <- as.numeric(as.character(stats[,i]))
``````
This is FAQ 7.10 in the R FAQ.
HTH
-
No need for any kind of loop. Just use the indices and unlist(). Edit : I added an answer illustrating this. – Joris Meys Sep 26 '10 at 22:19
you have to be careful while changing factors to numeric. here is a line of code that would change a set of columns from factor to numeric. i am assuming here that the columns to be changed to numeric are 1, 3, 4 and 5 respectively and you could change it accordingly
``````cols = c(1, 3, 4, 5);
df[,cols] = apply(df[,cols], 2, function(x) as.numeric(as.character(x));
``````
-
This won't work correctly. Example: `x<-as.factor(1:3); df<-data.frame(a=x,y=runif(3),b=x,c=x,d=x)`. I don't think that `apply` is appropriate to this kind of problems. – Marek Sep 26 '10 at 15:07
apply works perfectly in these situations. the error in my code was using margin = 1, instead of 2 as the function needs to be applied column wise. i have edited my answer accordingly. – Ramnath Sep 26 '10 at 19:46
Now it works. But I think it could be done without `apply`. Check my edit. – Marek Sep 27 '10 at 10:00
... or Joris answer with `unlist`. And `as.character` conversion in your solution is not needed cause `apply` converts `df[,cols]` to `character` so `apply(df[,cols], 2, function(x) as.numeric(x))` will work too. – Marek Sep 27 '10 at 10:12
This can be done in one line, there's no need for a loop, be it a for-loop or an apply. Use unlist() instead :
``````# testdata
Df <- data.frame(
x = as.factor(sample(1:5,30,r=T)),
y = as.factor(sample(1:5,30,r=T)),
z = as.factor(sample(1:5,30,r=T)),
w = as.factor(sample(1:5,30,r=T))
)
##
Df[,c("y","w")] <- as.numeric(as.character(unlist(Df[,c("y","w")])))
str(Df)
``````
Edit : for your code, this becomes :
``````id <- c(1,3:ncol(stats)))
stats[,id] <- as.numeric(as.character(unlist(stats[,id])))
``````
Obviously, if you have a one-column data frame and you don't want the automatic dimension reduction of R to convert it to a vector, you'll have to add the `drop=FALSE` argument.
-
Small improvement could be setting `recursive` and `use.names` parameters of `unlist` both to `FALSE`. – Marek Sep 27 '10 at 10:10
@Marek : true. I love this game :-) – Joris Meys Sep 27 '10 at 11:49
I am just going to add for those looking for answers in the future, this is not equivalent to op + gavin's method if the dataframe is of only one column. It will convert to a vector in that case, whereas op's will still be a dataframe. – themartinmcfly Feb 27 '13 at 3:15
@themartinmcfly the good ol' `drop=FALSE`... But thx for pointing that out, I added it to the answer. – Joris Meys Feb 27 '13 at 13:37
@JorisMeys +1, very good. – themartinmcfly Feb 27 '13 at 22:31
I think that ucfagls found why your loop is not working.
In case you still don't want use a loop here is solution with `lapply`:
``````factorToNumeric <- function(f) as.numeric(levels(f))[as.integer(f)]
cols <- c(1, 3:ncol(stats))
stats[cols] <- lapply(stats[cols], factorToNumeric)
``````
Edit. I found simpler solution. It seems that `as.matrix` convert to character. So
``````stats[cols] <- as.numeric(as.matrix(stats[cols]))
``````
should do what you want.
-
lapply is pretty much designed for this
``````unfactorize<-c("colA","colB")
df[,unfactorize]<-lapply(unfactorize, function(x) as.numeric(as.character(df[,x])))
``````
-
Hi @transcom, and welcome to stackoverflow. Note that this question is about converting to numeric representation from a factor, not the other way around. See Marek's solution. – Aaron Feb 10 '14 at 16:54
@Aaron, understood. I posted this answer due to the ambiguity of the OP's title, operating under the assumption that others may land here looking for a way to convert multiple columns easily, regardless of class. Anyway, I've edited my answer to more appropriately address the question :) – transcom Feb 18 '14 at 16:56
I know this question is long resolved, but I recently had a similar issue and think I've found a little more elegant and functional solution, although it requires the magrittr package.
``````library(magrittr)
cols = c(1, 3, 4, 5)
df[,cols] %<>% lapply(function(x) as.numeric(as.character(x))
``````
The `%<>%` operator pipes and reassigns, which is very useful for keeping data cleaning and transformation simple. Now the list apply function is much easier to read, by only specifying the function you wish to apply.
-
I had problems converting all columns to numeric with an `apply()` call:
``````apply(data, 2, as.numeric)
``````
The problem turns out to be because some of the strings had a comma in them -- e.g. "1,024.63" instead of "1024.63" -- and R does not like this way of formatting numbers. So I removed them and then ran `as.numeric()`:
``````data = as.data.frame(apply(data, 2, function(x) {
y = str_replace_all(x, ",", "") #remove commas
return(as.numeric(y)) #then convert
}))
``````
Note that this requires the stringr package to be loaded.
-
I would like to point out that if you have NAs in any column, simply using subscripts will not work. If there are NAs in the factor, you must use the apply script provided by Ramnath.
E.g.
``````Df <- data.frame(
x = c(NA,as.factor(sample(1:5,30,r=T))),
y = c(NA,as.factor(sample(1:5,30,r=T))),
z = c(NA,as.factor(sample(1:5,30,r=T))),
w = c(NA,as.factor(sample(1:5,30,r=T)))
)
Df[,c(1:4)] <- as.numeric(as.character(Df[,c(1:4)]))
``````
Returns the following:
``````Warning message:
NAs introduced by coercion
x y z w
1 NA NA NA NA
2 NA NA NA NA
3 NA NA NA NA
4 NA NA NA NA
5 NA NA NA NA
6 NA NA NA NA
``````
But:
``````Df[,c(1:4)]= apply(Df[,c(1:4)], 2, function(x) as.numeric(as.character(x)))
``````
Returns:
``````> head(Df)
x y z w
1 NA NA NA NA
2 2 3 4 1
3 1 5 3 4
4 2 3 4 1
5 5 3 5 5
6 4 2 4 4
``````
- | 2,259 | 7,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-30 | latest | en | 0.851997 |
https://www.technocrazed.com/12-6-temperature-coefficient-of-resistance | 1,726,562,952,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00478.warc.gz | 929,708,421 | 16,520 | # 12.6 Temperature Coefficient of Resistance
You might have noticed on the table for specific resistances that all figures were specified at a temperature of 20° Celsius. If you suspected that this meant specific resistance of a material may change with temperature, you were right!
Resistance values for conductors at any temperature other than the standard temperature (usually specified at 20 Celsius) on the specific resistance table must be determined through yet another formula:
The “alpha” (α) constant is known as the temperature coefficient of resistance and symbolizes the resistance change factor per degree of temperature change. Just as all materials have a certain specific resistance (at 20° C), they also change resistance according to temperature by certain amounts. For pure metals, this coefficient is a positive number, meaning that resistance increases with increasing temperature. For the elements carbon, silicon, and germanium, this coefficient is a negative number, meaning that resistance decreases with increasing temperature. For some metal alloys, the temperature coefficient of resistance is very close to zero, meaning that the resistance hardly changes at all with variations in temperature (a good property if you want to build a precision resistor out of metal wire!). The following table gives the temperature coefficients of resistance for several common metals, both pure and alloy:
#### Temperature Coefficients of Resistance at 20 Degrees Celsius
* = Steel alloy at 99.5 percent iron, 0.5 percent carbon tys
Let’s take a look at an example circuit to see how temperature can affect wire resistance, and consequently circuit performance:
This circuit has a total wire resistance (wire 1 + wire 2) of 30 Ω at standard temperature. Setting up a table of voltage, current, and resistance values we get:
At 20° Celsius, we get 12.5 volts across the load and a total of 1.5 volts (0.75 + 0.75) dropped across the wire resistance. If the temperature were to rise to 35° Celsius, we could easily determine the change of resistance for each piece of wire. Assuming the use of copper wire (α = 0.004041) we get:
Recalculating our circuit values, we see what changes this increase in temperature will bring:
As you can see, voltage across the load went down (from 12.5 volts to 12.42 volts) and voltage drop across the wires went up (from 0.75 volts to 0.79 volts) as a result of the temperature increasing. Though the changes may seem small, they can be significant for power lines stretching miles between power plants and substations, substations and loads. In fact, power utility companies often have to take line resistance changes resulting from seasonal temperature variations into account when calculating allowable system loading.
REVIEW:
• Most conductive materials change specific resistance with changes in temperature. This is why figures of specific resistance are always specified at a standard temperature (usually 20° or 25° Celsius).
• The resistance-change factor per degree Celsius of temperature change is called the temperature coefficient of resistance. This factor is represented by the Greek lower-case letter “alpha” (α).
• A positive coefficient for a material means that its resistance increases with an increase in temperature. Pure metals typically have positive temperature coefficients of resistance. Coefficients approaching zero can be obtained by alloying certain metals.
• A negative coefficient for a material means that its resistance decreases with an increase in temperature. Semiconductor materials (carbon, silicon, germanium) typically have negative temperature coefficients of resistance.
• The formula used to determine the resistance of a conductor at some temperature other than what is specified in a resistance table is as follows:
Back to Book’s Main Index | 768 | 3,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.881058 |
https://discourse.hibernate.org/t/search-and-sort-other-than-asc-or-desc/7635 | 1,685,266,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00597.warc.gz | 250,280,671 | 5,868 | # Search and sort other than asc or desc
Hi
I have a photo site where I sort photos on aspect ratios and I have and indexed field with sortable on aspectRatio(derivedFrom width / height), and sorting on panorama or tall photos works fine, i.e aspectRatio desc or asc. But now I want to sort on photos close to a square as well. As such the aspect ratio should be as close to one as possible.
Example:
Photo 1 with aspect ratio: 3 (3000 / 1000)
Photo 2 with aspect ratio: 0.4 (400 / 1000)
Photo 3 with aspect ratio: 1 (500 / 500)
I would then like a list returned as {Photo 3, Photo 2, Photo 1}.
I guess a way to do it could be to make a new field with “@IndexingDependency” and having some calculations on that field, for example absolutte(aspectRatio -1) and then sort on that field. The above would give
Photo 1: 2 abs(3-1)
Photo 2 : 0.6 abs(0.4 -1)
Photo 3: 0.1 abs(1.1-1)
And then sort would give {Photo 3, Photo 2, Photo 1}
Unfortunately this require a fairly complex query since I have to change which field I sort on depending upon the sort {Panorama, tall, square}. I know the algorithm with abs(aspectRatio -1) does not really works since an photo with 200 / 1000 would give 0.8 meaning its close to a square which it is not. But I guess I could come up with something that would make it work but are there another better hibernate way to do it.
Anyone here have any tips or ideas
Hey,
Thanks for reaching out. Depending on the particular backend you are using, you might be able to leverage some backed-specific features. For example, with an Elasticsearch backend, you may try to use something like a script-based sorting. But this would just move the “query complexity” as you named it into the script and will slow things down.
Since you have three options (panorama, tall, square), I can’t think of a way to use just asc/desc sort order to achieve what you are asking.
What I can suggest for making the query construction cleaner is to have dedicated fields in the index and pass the sort option (panorama, tall, square) as an enum and store the field/order in the enum. Something along the next lines:
``````enum PhotoTypeSort {
SQUARE( "squareSort", SortOrder.DESC ),
TALL( "wideSort", SortOrder.ASC ),
PANORAMA( "wideSort", SortOrder.DESC ); // using the same field for wide photos, but can also have a dedicated field and then use same DESC order for all cases.
final String field;
final SortOrder order;
PhotoTypeSort(String field, SortOrder order) {
this.field = field;
this.order = order;
}
}
PhotoTypeSort sort = ... ; // get this from the search request
searchSession.search( Photo.class )
.where( /* any predicates you need or a match all */ )
.sort( f -> f.field( sort.field ).order( sort.order ) )
.fetchHits( 20 );
``````
Also, another idea – depending on your needs, maybe you’d be able to get away without sorting but instead have a derived multivalue field with “classifiers”. Instead of sorting the results at the query time, determine to which categories (panorama, tall, square) the photo is likely to belong and store those at an index time. Then during the search, have a filter for a type you need. Most photos would have one category, except maybe those that are close to the criteria of being wide/tall but still might be considered square ones.
1 Like
Thanks for good ideas. The classifiers idea sounds nice but then all panorama would be similar, i.e 2000 / 200 is more panorama than 2000 / 400. But I like the idea of having the PhotoTypeSort enum with the field, since then it is clear how the sorting is done. I will try that one. | 904 | 3,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.943361 |
https://www.math.unl.edu/~bdeng1/Teaching/s14m423.html | 1,516,766,418,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893300.96/warc/CC-MAIN-20180124030651-20180124050651-00270.warc.gz | 940,465,790 | 12,925 | Spring 2014 Math 423
Intro. Complex Variables
09:30-10:45, TTh.
Avery 19
Office Hours: 1:00-2:30 TTh
Office: 318 Avery
Email: bdeng1@math.unl.edu
Phone: 472-7219 (No voice messages, please)
Turn Off All Wireless Electronic Devices Except For Sanctioned Calculators During Lectures, Quizzes, And Exams
Final Exam:
10:00 to 12:00 noon
Thursday, May 8
(Picture IDs required)
· Test & Solution
· Solution Key to Selected Problems
· Sample Test
· Calendar
· Supplemental Notes
Syllabus
Textbook: Fundamentals of Complex Analysis with Applications to Engineering and Science, 3rd Ed., by E.B. Saff and A.D. Snider Week 1 Section 1.1 1.2 1.3 1.4 1.5 1.6 Exercise 1,4,9,15,24,26,27 1,10,14,15 10,13,14,15,16,22,23 10,12,13,20,21 5(a,b),7(a),10,13,15,17* 9,10,13,16,17,20,22*,24* Comment Problems with asterisk * are more challenging, and more theoretical. Problems in red are collected. 2 1.7 2.1 2,5,8,9 3,4,5,11 3 2.2 2.3 2.4 4,5,8-11,14,15,18,22,24 2,4,8,9,13-15 1,2,3,6-12,13*,14*,10 4 2.5 3.1 1,3,4,5,10,22 1,2,4,7,11,12,13(a,b), 15,(a,d),17 5 3.2 3.3 5,6,9,10,11,13(a,b),18,19, 20,23 (Due Thur. 02/13) 1,2,3,5,8,10,11,12 6 3.4 3.5 1,2,3,4 1(a,b),15(a,b) 7 4.1 4.2 4.3 1,5,8 2,6(a),8,10,14*,15 1(c,e,i),2,5,6,7,10,12 (Due 13/03) 8 Midterm Exam 9 4.4 2,4,5,9,10,,14,16 (Tue. After Spring Break) 10 4.5 1,2,4,5,8,9,10,13,15 (Thur. After Spring Break) 11 Spring Break 12 4.6 5.2 5.5 5.6 6.1 2,3,4,5,6,10,11,13,14 1(a,c,f),3(a,d),5(c,e),6 1(a,c),3,13 Due 5 pm Friday, 04/11. 1(a,b,c),2,4,6,8 1(a,b),3(a,b,c,g),6,7 13 6.2 6.3 6.7 1,4,6 1,3,6,11 Due 5 pm Thur, 04/17. 2,3,6,7,8 14 7.3 1,2,3,4,5,7,9,15,16,17 Due 5 pm Friday, 04/25 15 Review HWK Review | 853 | 1,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-05 | latest | en | 0.570243 |
https://math.answers.com/Q/How_do_you_convert_a_meter_to_a_kilometer | 1,675,429,979,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500056.55/warc/CC-MAIN-20230203122526-20230203152526-00560.warc.gz | 396,439,531 | 49,424 | 0
# How do you convert a meter to a kilometer?
Wiki User
2008-11-07 02:32:40
There are 1000 metres in 1 kilometre,so if you convert 3metres = 3/1000 km, and so on.
Wiki User
2012-09-14 02:11:50
Study guides
14 cards
## Which scientist used his knowledge of astronomy to publish a popular almanac
➡️
See all cards
3.81
555 Reviews
Wiki User
2009-01-16 01:20:29
1km = 1000m Divide the Kilometer value by 1000.
Wiki User
2008-11-07 02:32:40
Well 1000 meters is 1 kilometer | 175 | 484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-06 | latest | en | 0.789117 |
http://code.antonio081014.com/2011/04/uva10810ultra-quicksortcpp.html | 1,503,420,676,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112533.84/warc/CC-MAIN-20170822162608-20170822182608-00456.warc.gz | 85,009,965 | 18,706 | ## Thursday, April 14, 2011
### UVa_10810_Ultra-QuickSort.cpp
poj2299, uva10810,
## Problem B: Ultra-QuickSort
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence
9 1 0 5 4 ,
Ultra-QuickSort produces the output
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.
For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.
```5
9
1
0
5
4
3
1
2
3
0```
```6
0```
Stefan B�ttcher
## Source Code:
//Thu Apr 14 14:19:42 CDT 2011
#include<stdio.h>
#include <iostream>
using namespace std;
#define MAX 500001
long long MergeSort(long a[], long c[], int left, int right) {
int i, j, mid, temp, count;
long long total = 0;
if (right > left) {
mid = (left + right) / 2;
total += MergeSort(a, c, left, mid);
total += MergeSort(a, c, mid + 1, right);
count = 0;
for (i = left, j = mid + 1; i <= mid && j <= right;) {
if (a[i] > a[j]) {
total += mid - i + 1;
c[++count] = a[j++];
} else
c[++count] = a[i++];
}
while (i <= mid)
c[++count] = a[i++];
while (j <= right)
c[++count] = a[j++];
for (i = left; i <= right; i++)
a[i] = c[i - left + 1];
}
return total;
}
long a[MAX], c[MAX];
int main(void) {
//freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
long n;
int i, j;
long long total;
scanf("%ld", &n);
while (n != 0) {
for (i = 1; i <= n; i++)
scanf("%ld", &a[i]);
total = MergeSort(a, c, 1, n);
cout << total << endl;
scanf("%ld", &n);
}
//fclose(stdin);
//fclose(stdout);
return 0;
} | 681 | 2,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | longest | en | 0.532981 |
https://www.albert.io/ie/single-variable-calculus/fun-with-area-and-integral-notation | 1,490,235,054,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218186608.9/warc/CC-MAIN-20170322212946-00163-ip-10-233-31-227.ec2.internal.warc.gz | 856,229,747 | 14,649 | Free Version
Easy
Fun With Area and Integral Notation
SVCALC-OON@RK
Given $\int _{ 0 }^{ 2 }{ f(x)dx=4 }$ and $\int _{ 0 }^{ 2 }{ g(x)dx=-3 }$, and assuming $f(x)>g(x)$ for all $x$, find the area bounded by $y=f(x)$, $y=g(x)$, $x=0$, and $x=2$.
A
$1$
B
$-3$
C
$4$
D
$7$ | 127 | 280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-13 | latest | en | 0.451716 |
https://www.theyeshivaworld.com/coffeeroom/topic/question-for-democrats-and-i-guess-anyone-else-that-wants-to-chime-in | 1,708,938,045,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00414.warc.gz | 1,051,878,310 | 26,125 | # question for democrats (and i guess anyone else that wants to chime in)
Home Forums Decaffeinated Coffee question for democrats (and i guess anyone else that wants to chime in)
Viewing 50 posts - 1 through 50 (of 96 total)
• Author
Posts
• #615687
Participant
I don’t understand the rationale behind raising the minimum wage, doesn’t that make things more expensive?
it’s basic economics
the cost of a product is based on raw materials + direct labor (how much the worker works X wage) + manufacturing overhead, which is computed based off of labor costs (like .25 of labor costs)
for example if to make a salad in a restaurant, it costs .50 for the lettuce (im throwing random numbers here) \$2 for the labor (15 minutes @ 8 dollars an hour) and manufacturing overhead is .25 of labor costs or .50 the cost of the salad (not the price) is \$3 and the owner wants to make 10% profit the price would be \$3.30 but if you raise the cost of labor to 15 dollars an hour the cost would be .50 for the salad \$3.75 for the labor and .94 for the overhead it would cost \$5.20 (2 dollars more for the same salad) whereby the owner sells it for \$5.72 and actually the owner makes more money however food costs more therefore. it doesn’t help anyone
so someone please explain the rationale
#1145014
Joseph
Participant
Yes, the minimum wage raises the cost of products. No one, even Democrats, thinks otherwise. The minimum wage is designed to help the employee not the employer.
#1145015
👑RebYidd23
Participant
There are some particular circumstances where another factor upsets the balance and raising the minimum wage solves that. Rare, but possible.
#1145016
mosheemes2
Member
The price of a product isn’t the price of its inputs. For example, Starbucks sells coffee, which costs them almost nothing to make, at a couple of dollars, but gives away milk and (in most locations) usage of their real estate, which are relatively expensive, for free. Price is a function of where the meeting point is at which a product will be supplied and consumed.
In your example, if the owner was rational and could sell the salad for 5.72, that would be the price regardless of what wage the employees were getting and the surplus over the 10% he’d have been happy with would go to the owner. If you’re arguing for an increase in the minimum wage, you’re most likely saying that for some reason the market isn’t adequately allowing the worker’s to negotiate to get a share of that surplus. It’s more complicated than that, but that’s a short version of it.
#1145017
Participant
but joe, my example proves otherwise, unless you’re taking supply and demand into account, and less people would pay 2 dollars more for the salad, but either way the employee would have to raise the price of the salad
#1145018
👑RebYidd23
Participant
Joseph, that makes no sense. If the employer can’t afford to pay the employee, both suffer. And the employee also has to pay twice as much for his salad.
#1145019
Joseph
Participant
I said the minimum wage is designed to help the employee; I didn’t opine whether it is successful or not.
#1145020
homer
Member
Its called populism.
Similar example: TAX THE RICH! It doesnt really help you (or anyone for that matter) at the end of the day if the millionaire pays more, but it sounds good to the poor masses.
#1145021
BarryLS1
Participant
homer is right on target. It’s all about votes by an ignorant uneducated public that thinks they’re actually getting something. It’s not the wages that count, it’s what those wages can buy.
Increasing the minimum wage does two things; 1. It’s inflationary so the net effect is the buying power is diminished; 2. it causes layoff, since most people getting minimum wage are teens or entry level jobs whose purpose is to train people for the workforce.
In the case of a small businesses like a pizza shop, where they have a lot of employees and the business owner makes a decent living, but is certainly not getting rich, raising the minimum wage erodes his income, since he can only raise prices just so much, he will lay people off to compensate.
The same applies to most mom and pop businesses. Even larger businesses will make adjustments to compensate.
The bottom line is that raising the minimum wage harms the very people that it was designed for. If an employee produces more and is worth more money, they will get more than minimum wage anyway, but when it’s raised, a business owner is forced to raise most hourly wages in some proportion. It could destroy some small businesses.
#1145022
Joseph
Participant
Should the minimum wage be eliminated entirely?
#1145023
akuperma
Participant
If the minimum wage is set at or below what would normally be negotiated for an unskilled, untrained workers, it has no impact other than for n individual is not very good at negotiating, in which case the minimum wage helps him/her. If on the other hand, the minimum wage is set at higher than what would be worked out by the law of supply and demand, it helps those individuals who get the jobs, but results in employers cutting back employment (using automation, outsourcing, etc.). For example if the minimum wage is \$10/hour, an employer will choose not to hire an employee who is not producing \$10 of extra profits every hour – the higher the wage, the harder it is to justify the employee.
The Democrats believe that most employers are making a huge profit on each employee, and therefore the employers will not reduce employment if forced to raise wages. The Republicans tend to feel that the law of supply and demand works so that wages stay in sync with profits, and therefore raising the minimum wage about what would be produced by a free market results in lower employment.
#1145024
Yserbius123
Participant
Because many large companies can afford to pay a lot more to their employees so that raising the minimum wage won’t affect them. The affects only smaller and mid-sized companies who will be forced to lay off workers and/or hire illegals for under minimum wage. But the vast majority of minimum wage workers work for places that can well afford an extra 50k a year to pay their workers.
#1145025
Participant
yserbeius,
if mcdonalds (big company) sells a burger for a dollar, if it costs more than a dollar to make they’re going to raise the price
im specifically bringing mcdonalds in because deblasio wants to raise the minimum wage for restaurants to 15 dollars an hour
#1145026
ubiquitin
Participant
Coffee
Economics is not quite as simple as you lay out. Mcdonalds doesn’t just add up the cost of a burger add some profit and charge that amount. There is also an element of them charging what they think they can get for it. If they think they would sell just as many burgers charging \$1.50 vs \$1.00 of course they would charge the \$1.50! So increasing the minimum wage wouldn’t automatically increase the price. It’s not like the reason they dont charge the extra 50 cents is out of the goodness of their hearts to provide cheap burgers.
In other words the reason they charge a dollar for a burger instead of 1.50 is not just based on how much a burger cost them. It is far more complicated than that.
What might happen though is increasing the minum wage would put more money in the hands of the “burger buyers” thus letting them spend more for burgers thus increasing the price of the burger.
#1145027
Participant
ubiquitin,
see my comment to joe, i understand the laws of supply and demand apply here (as with most other places) however if all restaurants have to pay more money for the labor (\$15 dollars an hour!) they would have to charge more to the customer and if they can’t find a medium they go kaput and welcome to higher unemployment
additionally, if the price of burgers go up (because people have more money to spend on them as you say) then there is inflation where the burger is anyway the same percentage of a persons hourly wage
#1145028
Participant
what i don’t think is understood over here is a corporations drive for “profits”
it’s what gives bonuses to ceos
it raises the stock price of a company
it hires more employees
even if the company is going to make the same amount of sales (revenues) their expenses go up do to labor costs and viola less “profit” which kills the company at the quarterly report and drives down the stock price
#1145029
Rebbe Yid
Participant
Less profit also means less capital investment in new locations meaning fewer new jobs created. Less profit means less paid in taxes, because the company pays taxes at a higher rate than the employees it would be giving the money to. And that’s just for major corporations.
For a smaller business, raising the minimum wage could force it to close. So stop being so cavalier.
#1145030
☕ DaasYochid ☕
Participant
Ubiquitin, of course cost is a factor in price.
#1145031
ubiquitin
Participant
DY
I never said otherwise
“…is not just based on how much a burger cost them. It is far more complicated than that.”
CA
“additionally, if the price of burgers go up (because people have more money to spend on them as you say) then there is inflation where the burger is anyway the same percentage of a persons hourly wage”
That is essentially what I said, Barry said it above too.
When gas prices fluctuate obviously the cost of a burger (to the manufacturer) fluctuates as well since they need to truck the cows, meat etc. Do you think their burger prices fluctuate alongside increases/decreases in gas prices?
#1145032
☕ DaasYochid ☕
Participant
Ubiquitin, “So increasing the minimum wage wouldn’t automatically increase the price.”
It pretty much would.
#1145033
popa_bar_abba
Participant
I like burgers.
#1145034
☕ DaasYochid ☕
Participant
With Russian dressing.
#1145035
screwdriverdelight
Participant
Raising the minimum wage isn’t an end of itself. Next, they plan to put caps on all prices. For example, McDonald’s will be banned from selling burgers that cost more than a dollar. Pretty smart, actually.
#1145036
ubiquitin
Participant
DY
i’ll pose the same question to you as to CA:
“When gas prices fluctuate obviously the cost of a burger (to the manufacturer) fluctuates as well since they need to truck the cows, meat etc. Do you think their burger prices fluctuate alongside increases/decreases in gas prices?”
Ill also pose a second question
why do you think McDonalds doesnt charge 1.50 a burger instead of \$1. Do you think its because they dont think that 1.50 is a fair price since given the cost of meat, labor etc say it comes out to 50 cents a burger and they only deserve 50 cents profit?
As an aside I’m not sure how much labor factors in to the price per burger at all. For example, you may or may not now this but Dunkin donuts spends more on the cup then they do on the coffee. If the price of a coffee bean would double it would barely (if at all) effect the price of coffee at all since the coffee is so cheap to the provider in the first place.
#1145037
☕ DaasYochid ☕
Participant
When gas prices fluctuate obviously the cost of a burger (to the manufacturer) fluctuates as well since they need to truck the cows, meat etc. Do you think their burger prices fluctuate alongside increases/decreases in gas prices?
Of course, although the effect on prices will be less immediate and less drastic.
why do you think McDonalds doesnt charge 1.50 a burger instead of \$1. Do you think its because they dont think that 1.50 is a fair price since given the cost of meat, labor etc say it comes out to 50 cents a burger and they only deserve 50 cents profit?
Of course not, it’s because they’ll sell fewer burgers and therefore have lower profits. If costs were higher, though, they would profit more at higher prices.
Obviously, this is an oversimplification; for one thing, certain overhead costs may not fluctuate directly with volume. I am working with the assumption that the more burgers you sell, the more people you need to take deliveries, cook them, and sell them.
As an aside I’m not sure how much labor factors in to the price per burger at all. For example, you may or may not now this but Dunkin donuts spends more on the cup then they do on the coffee. If the price of a coffee bean would double it would barely (if at all) effect the price of coffee at all since the coffee is so cheap to the provider in the first place.
I’m not sure what that has to do with labor, which I assume to be a much higher percentage of the cost of a burger than the price of a coffee bean to a cup of coffee.
#1145038
☕ DaasYochid ☕
Participant
BTW, that doesn’t mean that I agree with the OP. I do agree that prices will go up and sales will go down, but the net result will still be wealth redistribution.
#1145039
ubiquitin
Participant
DY the thrust of my comment was correcting the OP’s oversimplification
“the cost of a product is based on raw materials + direct labor (how much the worker works X wage) + manufacturing overhead, which is computed based off of labor costs (like .25 of labor costs)”
This is not all there is, what is probably more important is what the public is welling (or has available) to spend on the item. I can have a cheap item that costs me 0.50 that only I know how to make but that is very popular and charge a lot of money. On the other hand if there is an expensive item that is in little demand I cant charge the cost and might have to sell it a loss.
#1145040
Participant
ubequitin,
“the cost of a product is based on raw materials + direct labor (how much the worker works X wage) + manufacturing overhead, which is computed based off of labor costs (like .25 of labor costs)”
means COST (like cost of goods sold) whereas
This is not all there is, what is probably more important is what the public is welling (or has available) to spend on the item
is talking about price, which is based off of supply and demand, 2 different things
however everyone agrees the price cannot be less than the cost of the product
#1145041
☕ DaasYochid ☕
Participant
I can have a cheap item that costs me 0.50 that only I know how to make
A lot of people besides Ronald McDonald know how to make hamburgers.
Bottom line: the increase in minimum wage would increase the price of fast food, which would lead to fewer sales and some layoffs. The remaining workers would get paid more, and that increase would very possibly be more than the lost wages of the laid off workers.
#1145042
ubiquitin
Participant
DY
I’m not sure about your bottom line. there are leading economists and many studies on both sides of the issue. Thus I dont claim to know either way.
The OP said “the cost of a product is based on raw materials + direct labor (how much the worker works X wage) + manufacturing overhead, which is computed based off of labor costs (like .25 of labor costs)”
This is simply not true. The determinants of cost are (mainly) supply and demand if those don’t change the price of an item wont change.
Now as I and Barry laid out earlier. Increasing minum wage might increase the affordibality for burgers and thus increase the demand and thus increase the price. But not for the reason the OP said
#1145043
ubiquitin
Participant
CA
“however everyone agrees the price cannot be less than the cost of the product”
This isnt remotely true. You can develop something with little demand say a mud-flavored lollipop. You spend a lot of money developing the exact taste of mud. Say it cost you \$5 a pop to develop.
Trust me with little demand you arent selling it for more than \$5.
You might be able to create some demand and sell it as a novelty item or a trick lollipop or something but you will have to sell it at less than cost. There is absolutely no doubt if you want to sell it your price will have to be less than cost (obviously you wont be profitable, but its better than not selling them at all)
As I said in the last comment. The main determinants of price (i.e what the consumer is willing to pay) are supply and demand.
#1145044
ubiquitin
Participant
CA
“however everyone agrees the price cannot be less than the cost of the product”
This isnt remotely true. You can develop something with little demand say a bad-tasting lollipop. You spend a lot of money developing a particular bad taste. Say it cost you \$5 a pop to develop.
Trust me with little demand you arent selling it for more than \$5.
You might be able to create some demand and sell it as a novelty item or a trick lollipop or something but you will have to sell it at less than cost. There is absolutely no doubt if you want to sell it your price will have to be less than cost (obviously you wont be profitable, but its better than not selling them at all)
As I said in the last comment. The main determinants of price (i.e what the consumer is willing to pay) are supply and demand.
#1145045
Participant
that’s correct ubequitin,
however i’m talking about a business, not something that flopped and the person wants to sell, albeit at a loss
a lot of businesses try out how good a product is (free sample) before making a gazillion of that product so they don’t have to lose money on the product
#1145046
Participant
yet again ubequitin,
you confuse COST and PRICE
PRICE is based off of supply and demand
COST isn’t
#1145047
ubiquitin
Participant
ca
and besides in your OP you said
“I don’t understand the rationale behind raising the minimum wage, doesn’t that make things more EXPENSIVE?…it’s basic economics…the cost of a product is based on raw materials + direct labor (how much the worker works X wage) + manufacturing overhead, which is computed based off of labor costs (like .25 of labor costs)”
Expensive for whom? In context it seems clear you meant the consumer, which is not necessarily correct.
#1145048
☕ DaasYochid ☕
Participant
He’s arguing that because the cost will go up, the price will go up. It’s hard to think he’s not right when it comes to fast food.
#1145049
Participant
ubequitin,
thats right that the consumer doesnt care about cost only price however the seller does, whereby if the seller doesn’t “break even” he won’t sell the product
yes your right about your second point because i presume that the PRICE factors in the COST one really has to be an idiot to sell under COST unless he was liquidating his inventory
#1145050
ubiquitin
Participant
CA
There is a third possibility: that Mcdonalds will still sell burgers at more than cost though make less profit than they do now.
alternativly they will recoup their losses in other ways by cutting down the size of the burger (slightly) less pickles lettuce etc… OR hire fewer workers
#1145051
☕ DaasYochid ☕
Participant
alternativly they will recoup their losses in other ways by cutting down the size of the burger (slightly) less pickles lettuce etc…
Because currently the portion size is determined by their generosity?
#1145052
ubiquitin
Participant
DY
Lol
No.
but if you dont like that option I gave 2 others
#1145053
☕ DaasYochid ☕
Participant
Option #1 is remote. If the cost goes up, the price point at which profits are maximized will most likely be higher.
Option #3 is just as fallacious as #2. Do you think the number of employees is a function of goodheartedness? If prices go up (or portion sizes reduced) and sales decrease, then yes, they will let go of some employees because there will be fewer sales and less work to do.
#1145054
Participant
ubequitin,
that’s great (option 2) now a dollar would get a smaller burger
3 is also good, they wouldn’t hire more people thereby keeping the unemployment numbers high for the next president
#1145055
ubiquitin
Participant
DY
Not neccesarily good heartedness, I’ts not like Mcdonalds is reevaluating their profit margins daily and checking all branches to make sure all minimum wage empoies are working as hard as they can. If the minumum wage goes up and their profits go down then they might be pushed to reevalute and come up with a solution.
I’m confused though. ARe you really saying the price of an item is not (primarly) determined by supply and demand?
#1145056
☕ DaasYochid ☕
Participant
Of course it’s determined by supply and demand, but as the price goes up (and it almost has to if cost goes up), demand goes down. Or, in other words raising the price doesn’t affect actual demand, but does affect the number of sales.
You said it yourself: why do you think McDonalds doesnt charge 1.50 a burger instead of \$1. Do you think its because they dont think that 1.50 is a fair price since given the cost of meat, labor etc say it comes out to 50 cents a burger and they only deserve 50 cents profit?
#1145057
Participant
wait a minute (i missed this on the first read)
“and they only deserve 50 cents profit?”
huh? who’s to decide how much profit a company desrves? the government
if a company could produce something for a penny and could sell it for 1,000 dollars the government has to take away 99% of it because he only “desrves 10 dollars (9.99 profit)
#1145058
☕ DaasYochid ☕
Participant
CA, he was being facetious, that a company sets its prices on factors besides profits. (Then he went on to make arguments that essentially assume that.)
#1145059
ubiquitin
Participant
CA
I was asking “Do you thi9nk…” Mcdonalds decided to make 50 cents a burger becasue they (Mcdonalds) feels thats all they deserve even though they can just as easily charge more money per burger and sell the same number of burgers.
I was asking that rhetoricly, since it obviously isnt true. Rather since price is detemrined by supply and demand. If Mcdonalds charged more per burger (artificaly increasing the price) demand would drop and they would not make more.
The reason they charge 1.00 a burger is NOT because that Is the cost + profit, like you said in the OP (and if they felt they deserved more profit they can just charge more). It is becuase based on current supply and demand models that is all they can charge (to make the money they make). If they could make the more money by charging 1.50 of course they would!
Granted this is an oversimplification. But your question in the OP is built on an incorrect premise
#1145060
Participant
dy,
hard to tell when it’s written instead of spoken
#1145061
ubiquitin
Participant
CA
Consider the reverse. Say instead of increasing the minumum wage. We approached Mcdonalds with a fresh batch of laborers who are willing to work at half the current minumum wage (PRetend that for whatever reason this is all legal) therby dropping th cost
Do you think Mcdonalds would say “no thank you, if we drop the cost that will just drop the price so we dont gain anything” Or would they jump at the oppurtunity to produce the same burger charge the same amount and make more profit?
#1145062
☕ DaasYochid ☕
Participant
If McDonald’s is the only one getting this offer, they’ll decide whether they can profit more from higher profit margin or more sales, and act accordingly
Even if all burger places got the offer, they might all lower prices, but would profit from more sales.
Either way, they gain. | 5,383 | 23,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-10 | latest | en | 0.944617 |
https://quizizz.com/en-us/perimeter-of-a-rectangle-worksheets-grade-6?page=1 | 1,721,867,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00535.warc.gz | 415,993,973 | 27,511 | Perimeter of Shapes
15 Q
6th - 8th
Area and Perimeter
12 Q
6th
Area and Perimeter Practice
20 Q
6th - 7th
Area and Perimeter
20 Q
6th - 8th
Area and Perimeter
10 Q
5th - 6th
Polygons on a Coordinate Plane-Practice 1
13 Q
6th
Perimeter, Area, & Volume Test
20 Q
4th - 6th
Area & Perimeter of Rectangle on Coordinate Plane
9 Q
6th
22 Q
5th - 6th
Rectangles and Squares Perimeter and Area
10 Q
6th
10 Q
5th - 6th
6th Grade: Perimeter and Area of Polygons
10 Q
6th - 8th
Area/Perimeter/Volume
12 Q
4th - 6th
Perimeter
10 Q
4th - 6th
Area
15 Q
6th
Chapter 9: Lessons 1-6
12 Q
6th
Perimeter of a Rectangle
10 Q
5th - 6th
Area Formulas +
15 Q
6th
Area of Squares and Rectangles
15 Q
4th - 6th
Coordinate Plane & Geometry Review
12 Q
6th
Area and Perimeter of Rectangles and Triangles
8 Q
6th
CPM/CC1 Chapter 5 Vocabulary: Multiplying Fractions and Area
13 Q
6th
Mid-Module 4 Review
16 Q
6th
Area and Perimeter of a Triangle and a Rectangle
20 Q
6th - 8th
Explore printable Perimeter of a Rectangle worksheets for 6th Grade
Perimeter of a Rectangle worksheets for Grade 6 are essential tools for teachers to help their students develop a strong foundation in math and geometry concepts. These worksheets focus on teaching students how to calculate the perimeter of a rectangle by using the formula P = 2(l+w), where P represents the perimeter, l represents the length, and w represents the width of the rectangle. By incorporating these worksheets into their lesson plans, teachers can provide ample practice opportunities for their students to master this important skill. Additionally, these worksheets can be easily adapted to cater to different learning styles, making them a versatile resource for any Grade 6 classroom. With the use of Perimeter of a Rectangle worksheets for Grade 6, teachers can ensure their students gain a thorough understanding of perimeter and its applications in math and geometry.
Quizizz is an excellent platform that offers a wide variety of resources, including Perimeter of a Rectangle worksheets for Grade 6, to help teachers create engaging and interactive learning experiences for their students. In addition to worksheets, Quizizz also provides teachers with access to a vast library of quizzes, games, and other educational materials that cover various topics in math, geometry, and other subjects. By utilizing Quizizz in their classrooms, teachers can easily monitor their students' progress and identify areas where additional support may be needed. Furthermore, Quizizz allows teachers to customize their content to suit the specific needs of their Grade 6 students, ensuring that each lesson is tailored to their unique learning requirements. With Quizizz, teachers can enhance their instruction and provide their students with a comprehensive and enjoyable learning experience in math, geometry, and beyond. | 705 | 2,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-30 | latest | en | 0.928593 |
https://analystprep.com/blog/timelines-your-best-friends-calculations-for-cfa-and-frm-exams/ | 1,720,795,398,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514404.71/warc/CC-MAIN-20240712125648-20240712155648-00473.warc.gz | 64,254,213 | 31,087 | # Timelines – Your Best Friends (Calculations for CFA® and FRM® Exams)
## What is a timeline?
A timeline is an physical illustration of the amount and the timing of cash flows associated with an investment project.
Some of the applications of a timeline include:
• Quantitative Methods: Time Value of Money
• Capital Budgeting
• Equity Valuation
• Fixed Income Valuation
• Derivatives Pricing & Valuation
## Quantitative Methods
The time value of money (TVM) is used in calculation of Future Values and Present values of investments, annuities, perpetuities, retirement planning, and mortgage payments. The time value of money (TVM) is based on the concept that money you have now is worth more than the identical sum in the future due to its potential earning capacity.
An illustration of a timeline of future values and present values is as follows:
## Capital Budgeting
Capital budgeting in corporate finance involves analyzing the concepts of net present value (NPV) and internal rate of return (IRR). We use NPV and IRR in determining the right project (in case of a single project) or the correct sequence of projects (in case of multiple projects) using NPV and IRR.
NPV and IRR help us in selecting the best possible project. For instance, when we have multiple projects, we will sort the various NPVs in descending order, such that we select the project with the highest positive NPV.
Note: we reject a project with a negative NPV.
We calculate the Net Present Value (NPV) using the formulae given below:
\begin{align*}NPV=&\sum_{t=1}^n\frac{CF_t}{(1-r)^t}-Outlay\end{align*}
Where,
$$CF_t$$= after-tax cash flow at time t
$$r$$= required rate of return for the investment
$$Outlay$$= Initial Investment Cashflow
### Example: Calculating NPV and IRR of a project
Assuming Company XYZ is considering investing in a four-year project characterized with an initial investment of $100, an annual after-tax operating cash flows of$20 and a terminal after-tax operating cash flow of 33 at a required rate of return on 8%. Determine whether the project is worth investing in. #### Solution We know that, \begin{align*}NPV=&\sum_{t=1}^n\frac{CF_t}{(1-r)^t}-Outlay\end{align*} Thus, \begin{align*}{NPV}=&\frac{20}{1.08^1}+\frac{20}{1.08^2}+\frac{20}{1.08^3}+\frac{33}{1.08^4}-100\\=&18.519+17.147+15.1877+24.256-100\\=&-24.201 million\end{align*} We realize that the NPV of this project is negative; therefore, we reject it. A timeline can be used to illustrate this calculation, as shown below: The timeline implies that at year zero, we have a cash outflow of100 (the initial investment), while at year one, two, three, and four, we have a cash inflow of $20,$20, $20,$33, respectively.
Also, we can calculate the IRR using excel, where;
$$-100+\frac{20}{(1+IRR)^1}+\frac{20}{(1+IRR)^2}+\frac{20}{(1+IRR)^3}+\frac{33}{(1+IRR)^4}=0:IRR=-2.62\%$$
Unlimited Package (for FRM Part I & Part II)
## Equity Valuation
An equity valuation involves using the dividend discount model, the free cash flow model, and the residual income model to value a stock. These models rely on discounting the cash flows. The method of computing the equity valuation remains the same despite determining the different cash flows that change in the different models. After determining what these cash flows are, you will put them on a timeline and then discounted them back to time zero. The application of the timeline is the same across all three models.
### Example: Equity Valuation
• For the next three years, the annual dividends of a stock are expected to be €1.00, €1.50, and €2.0.
• The stock price is expected to be €20.00 at the end of three years.
• If the required rate of return on the shares is 10 percent, what is the estimated value of a share?
#### Solution
By applying the timeline we see that;
Where;
\begin{align*}V_0&=\frac{1.0}{1.1}+\frac{1.5}{1.1^2}+\frac{(2.0+20)}{1.1^3}\\&= 18.67\end{align*}
According to this calculation, the stock should be valued at €18.67; thus, if in the market, the value is higher than €18.67, then we should not buy that stock because it is overvalued. However, if the stock market value is less than €18.67, it means that it is undervalued, and we should go ahead and buy that stock.
## Bond Valuation
Bond valuation is a method used to determine the theoretical fair value of a particular bond. It involves calculating the present value of a bond’s expected future coupon payments, or cash flow, and the bond’s value upon maturity, or face value.
#### Types of bond valuation
• Plain vanilla bond
• Zero coupon bonds
• Deferred coupon bonds
• Step-up bonds
Regardless of the type of bond we are dealing with, the valuation occurs in the timeline only. The bond valuation is similar to equity valuation, only that we take the coupons as the cashflow.
The timeline below illustrate bond valuation:
Where:
$$C$$= Coupon on the bond, i.e., what the issuer is giving
$$Y$$= Yield to Maturity (YTM), i.e., minimum rate that we require, given the risk
The value of the bond depends on the relationship between the Coupon of the bond (C) and Yield to Maturity (Y), where:
• If $$C>Y$$, then the bond trades at a premium
• If $$C<Y$$, then the bond trades at a discount
• If $$C=Y$$, then the bond trades at the par value
## Derivatives Pricing & Valuation
The different types of derivative pricing and valuation include:
• Forwards
• Futures
• Forward Rate Agreements (FRAs)
• Options
• Swaps
All of these types are valued using the timeline. However, the complexity could increase when dealing with options and swaps, but the fundamental mathematical concept is the timeline of discounted cashflows.
### Example: Forward Derivative Pricing and Valuation
Assuming that:
• Spot Price of a stock is \$90
• The risk-free rate is 4% p.a.
• Forward contract expires in one year
What is the correct forward price?
#### Solution
Applying the timeline we have:
Therefore, Forward Pricing Equation is given as:
\begin{align*}{F_0(T)}=&{S_0(1+r)^t}\end{align*}
Where:
$$F_0(T)$$= Forward pricing
$$s_0$$= spot price
$$t$$= time
$$r$$= risk-free rate
Thus,
\begin{align*}=&90(1+0.04)^1\\=&93.6 \text{dollars}\end{align*}
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Featured Shop FRM® Exam Prep Shop Actuarial Exams Prep Shop Graduation Admission Exam Prep | 1,666 | 6,329 | {"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": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.865626 |
https://www.folkstalk.com/2022/09/loop-array-js-with-code-examples.html | 1,696,402,356,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511361.38/warc/CC-MAIN-20231004052258-20231004082258-00103.warc.gz | 825,207,458 | 15,350 | # Loop Array Js With Code Examples
Loop Array Js With Code Examples
With this article, we’ll look at some examples of Loop Array Js problems in programming.
```function delay(time) {
return new Promise(resolve => setTimeout(resolve, time));
}
delay(1000).then(() => console.log('ran after 1 second1 passed'));```
Below is a list of different approaches that can be taken to solve the Loop Array Js problem.
```var colors = ["red","blue","green"];
for (var i = 0; i < colors.length; i++) {
console.log(colors[i]);
}```
```var colors = ['red', 'green', 'blue'];
colors.forEach((color, colorIndex) => {
console.log(colorIndex + ". " + color);
});```
```var arr = ["f", "o", "o", "b", "a", "r"];
for(var i in arr){
console.log(arr[i]);
}```
```const array = ["one", "two", "three"]
array.forEach(function (item, index) {
console.log(item, index);
});```
```var ar = [1, 2, 3, 4, 5, 6];
ar.length = 4; // set length to remove elements
console.log( ar ); // [1, 2, 3, 4]```
As we’ve seen, a lot of examples were used to address the Loop Array Js problem.
## Can you loop through an array in JavaScript?
If we want to loop through an array, we can use the length property to specify that the loop should continue until we reach the last element of our array.23-Jun-2022
## How do you iterate through an array?
Iterating over an array You can iterate over an array using for loop or forEach loop. Using the for loop − Instead on printing element by element, you can iterate the index using for loop starting from 0 to length of the array (ArrayName. length) and access elements at each index.02-Jul-2020
## How do you iterate in JavaScript?
• There are multiple ways one can iterate over an array in Javascript. The most useful ones are mentioned below.
• Using while loop. This is again similar to other languages.
• using forEach method.
• Using every method.
• Using map.
• Using Filter.
• Using Reduce.
• Using Some.
## Can we use loops in arrays?
We can use iteration with a for loop to visit each element of an array. This is called traversing the array. Just start the index at 0 and loop while the index is less than the length of the array.
## How do you iterate through an array in HTML?
There are 3 methods that can be used to properly loop through an HTMLCollection.
• Method 1: Using the for/of loop: The for/of loop is used to loop over values of an iterable object.
• Method 2: Using the Array.from() method to convert the HTMLCollection to an Array.
• Method 3: Using a normal for loop.
## How does for loop start in JavaScript?
JavaScript for loop start with first statement. The loop starts in position 0 ( let i = 0 ) . The loop automatically increments i for each run. The loop runs as long as i < array.02-Feb-2022
## What is traversing an array?
To traverse an array means to access each element (item) stored in the array so that the data can be checked or used as part of a process. In most high-level languages, it is necessary to create a variable that will track the position of the element currently being accessed.
## What is difference between Forin and for OF in JavaScript?
The main difference between them is in what they iterate over. The forin statement iterates over the enumerable string properties of an object, while the forof statement iterates over values that the iterable object defines to be iterated over.13-Sept-2022
## How do you create an array in JavaScript?
Creating an Array Using an array literal is the easiest way to create a JavaScript Array. Syntax: const array_name = [item1, item2, ]; It is a common practice to declare arrays with the const keyword.
## What is for loop in JS?
for – loops through a block of code a number of times. for/in – loops through the properties of an object. for/of – loops through the values of an iterable object. while – loops through a block of code while a specified condition is true. | 932 | 3,896 | {"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} | 2.640625 | 3 | CC-MAIN-2023-40 | latest | en | 0.711296 |
https://serc.carleton.edu/introgeo/studentresearch/examples/WatershedCommonTroubleSpots.html | 1,708,759,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474523.8/warc/CC-MAIN-20240224044749-20240224074749-00766.warc.gz | 523,080,465 | 17,045 | |
# Watershed Analysis - Teaching Notes and Tips
This material is replicated on a number of sites as part of the SERC Pedagogic Service Project
### Common Trouble Spots
• A student doesn't have a good grasp of geography (and place names) of their chosen area.
• Solution: have them work with maps from the area(e.g. Google Maps: http://maps.google.com/).
• No precipitation stations are located within the watershed of choice.
• Solution: have them use a precipitation station within a nearby watershed (but on same side of mountain range to ensure conditions similar to their watershed).
• The chosen precipitation station is at the same (or lower) elevation as the discharge station. This results in more water reported exiting than entering the catchment.
• Solution: Either have them redo the part of the assignment, or discuss the influence of the orographic effect on their results.
• Unit conversion. The most common error seems to occur in the conversion from daily discharge measurements (reported in cubic feet per second, cfs) to annual discharge (cubic feet per year, cfy).
• Solution: Students are usually comfortable converting the daily measurements in cfs to cfd (cubic feet per day). However, going from cfd to cfy presents the problem. Summing the daily numbers yields the total discharge for the year (cfy). The problem: students will often sum the data then convert days to years, resulting in a discharge value more than 300X too large. | 306 | 1,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.936499 |
https://www.jscamp.app/hi/docs/javascript30 | 1,721,549,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00639.warc.gz | 717,962,880 | 11,066 | मुख्य कंटेंट तक स्किप करें
Theory is good, but without practice how to fly with one wing, so in this section we will solve problems in JavaScript. We took the tasks from the Codewars - site this is a site where you can improve your skills not only in JavaScript, but also in other programming languages.
## Find numbers that are divisible by a given number
Complete a function that takes two arguments and returns all numbers that are divisible by the given divisor. The first argument is an array of numbers and the second is a divisor.
### Example
``divisibleBy([1, 2, 3, 4, 5, 6], 2) == [2, 4, 6]``
### Practice
लाइव एडिटर
परिणाम
See the answer only if you could not solve the problem yourself.
## Twice as old
• father's current age (years)
• the current age of the son (years)
Calculate how many years ago a father was twice as old as his son, or in how many years he will be twice as old.
लाइव एडिटर
परिणाम
## Count odd numbers below `n`
If number `n` is given, return the number of positive odd numbers less than `n`, EASY!
### Example
``oddCount(7) //=> 3, i.e [1, 3, 5]oddCount(15) //=> 7, i.e [1, 3, 5, 7, 9, 11, 13]``
लाइव एडिटर
परिणाम
## Convert string to number!
We need a function that can convert a string to a number. What ways do you know to achieve this?
Note: Don't worry, all inputs will be strings, and each string is a perfectly valid integer representation.
### Example
``stringToNumber('1234') == 1234stringToNumber('605') == 605stringToNumber('1405') == 1405stringToNumber('-7') == -7``
लाइव एडिटर
परिणाम | 451 | 1,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.799867 |
http://www.mathworks.com/matlabcentral/cody/problems/1286-matcat-reconstruct-x-from-its-x-rays/solutions/207977 | 1,484,720,004,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00345-ip-10-171-10-70.ec2.internal.warc.gz | 552,807,141 | 11,815 | Cody
# Problem 1286. MatCAT - Reconstruct X from Its X-rays
Solution 207977
Submitted on 23 Feb 2013 by Richard Zapor
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
%% row_sums = [3 5 11]; col_sums = [4 7 8]; x = matcat(row_sums,col_sums); assert(all(x(:)>=0)) assert(isequal(floor(x),x)) assert(isequal(sum(x,2)',row_sums)) assert(isequal(sum(x,1),col_sums))
2 Pass
%% row_sums = [2 2 2 2 2 6]; col_sums = [2 3 3 3 3 2]; x = matcat(row_sums,col_sums); assert(all(x(:)>=0)) assert(isequal(floor(x),x)) assert(isequal(sum(x,2)',row_sums)) assert(isequal(sum(x,1),col_sums))
3 Pass
%% row_sums = [65 65 65 65 65]; col_sums = [65 65 65 65 65]; x = matcat(row_sums,col_sums); assert(all(x(:)>=0)) assert(isequal(floor(x),x)) assert(isequal(sum(x,2)',row_sums)) assert(isequal(sum(x,1),col_sums))
4 Pass
%% row_sums = [22 34 33]; col_sums = [15 23 18 21 12]; x = matcat(row_sums,col_sums); assert(all(x(:)>=0)) assert(isequal(floor(x),x)) assert(isequal(sum(x,2)',row_sums)) assert(isequal(sum(x,1),col_sums))
5 Pass
%% row_sums = 55; col_sums = [1 2 3 4 5 6 7 8 9 10]; x = matcat(row_sums,col_sums); assert(all(x(:)>=0)) assert(isequal(floor(x),x)) assert(isequal(sum(x,2)',row_sums)) assert(isequal(sum(x,1),col_sums)) | 507 | 1,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.441035 |
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Hi...
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Buying/selling active licenses of online courses/tests/question banks etc. is illegal and unethical. Pls do not make such offers. Thank You
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This is having very high page rank in google
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ron321 wrote:
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PM me or my email is ron39686[at]gmail[dot]com
Thanks
Ron
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You cannot ask such favors openly.
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Display posts from previous: Sort by | 838 | 2,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.880728 |
http://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.467436.html | 1,369,482,732,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705948348/warc/CC-MAIN-20130516120548-00066-ip-10-60-113-184.ec2.internal.warc.gz | 308,718,605 | 4,765 | # SOLUTION: Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 60mph and train B is traveling at 80mph. Train A passes a station at 10:25 pm, if tra
Algebra -> Algebra -> Customizable Word Problem Solvers -> Mixtures -> SOLUTION: Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 60mph and train B is traveling at 80mph. Train A passes a station at 10:25 pm, if tra Log On
Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Word Problems: Mixtures Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Mixture Word Problems Question 467436: Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 60mph and train B is traveling at 80mph. Train A passes a station at 10:25 pm, if train B passes the same station at 10:40 pm, at what time will train B catch up to train A?Answer by mananth(12270) (Show Source): You can put this solution on YOUR website!Train A 60 kmph Train B 80 kmph Train A 10:25 Train B 10:40 Difference in time= 00:15 0.25 Train A will have covered 15 miles beforeTrain B starts from station catch up distance= 15 miles catch up speed = 80 -60 mph catch up speed = 20 mph Catchup time = catchup distance/catch up speed catch up time= 15 / 20 catch up time= 0.75 hours catch up time= 45 minutes 10:40 + 0:45= 11:25 pm | 425 | 1,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2013-20 | latest | en | 0.839819 |
https://mattermodeling.stackexchange.com/questions/10701/how-to-plot-homo-lumo-diagram-using-python | 1,696,281,555,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00741.warc.gz | 428,936,122 | 27,284 | # How to plot HOMO-LUMO diagram using python?
I'm trying to create a HOMO-LUMO diagram for a presentation, similar to the below figure from ResearchGate. I would like to learn how to do this with Python if possible, since I know this language (though I'm a beginner). Is there any easy way to make such an image in Python?
• I suspect this was a composite image, where presumably just the the red/blue line markers and the Y-axis (along with maybe the energy labels next to the markers) were added programmatically, while the rest of the figure (the molecule images, the molecule labels underneath, the HOMO/LUMO/Eg labels) were added in manually with a simple image editing program like Paint or SmartArt on Windows. For the basic figure, you could probably adapt it from code such as github.com/giacomomarchioro/PyEnergyDiagrams. It would be useful to know where you are stuck currently with making your figure.
– Tyberius
Apr 10 at 14:58
• Ideally the "Show more" button in the image would be removed. Apr 10 at 15:16
• Here is an article which clearly explains how to calculate eLumo and eHomo using rdkit and python. andersle.no/posts/2022/mo/mo.html Apr 10 at 15:51
The e_Lumo and e_Homo can be calculated using pyscf and scipy. I will demonstrate using a simple molecule such as ethanol:
## Code:
#### Helper functions and init:
from pyscf import gto, scf, lo, tools
from scipy.constants import physical_constants
HARTREE_TO_EV_FACTOR = dict(physical_constants)['Hartree energy in eV'][0]
def find_homo_lumo(mf):
lumo = float("inf")
lumo_idx = None
homo = -float("inf")
homo_idx = None
for i, (energy, occ) in enumerate(zip(mf.mo_energy, mf.mo_occ)):
if occ > 0 and energy > homo:
homo = energy
homo_idx = i
if occ == 0 and energy < lumo:
lumo = energy
lumo_idx = i
return homo, lumo
basis="sto-3g"
xyz = '''
C -0.887089 0.175064 -0.012535
C 0.460489 -0.515516 -0.046535
O 1.442965 0.307267 0.565572
H -0.847478 1.127768 -0.550817
H -1.658782 -0.453327 -0.465842
H -1.176944 0.403676 1.018306
H 0.768712 -0.724328 -1.075460
H 0.419486 -1.462073 0.500177
H 1.478640 1.141468 0.067135
'''
#### Main Code:
mol = gto.M(atom=xyz,basis=basis)
mol.build()
mf = scf.RHF(mol).run()
homo_hartree, lumo_hartree = find_homo_lumo(mf)
homo_eV = homo_hartree*HARTREE_TO_EV_FACTOR
lumo_eV = lumo_hartree*HARTREE_TO_EV_FACTOR
Eg = abs(homo_eV - lumo_eV)
So once you have the e_homo, e_lumo and Eg, you can use any image editing software to construct images like the ones in the question.
• thak you so much sir May 17 at 11:20 | 818 | 2,620 | {"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} | 2.84375 | 3 | CC-MAIN-2023-40 | longest | en | 0.853177 |
https://www.dsprelated.com/showthread/comp.dsp/62275-1.php | 1,638,655,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00114.warc.gz | 813,601,973 | 14,414 | # estimation of autocorrelation matrix or eigenvalue spread.
Started by August 10, 2006
```Hi everyone,
I am experimenting with some equalization schemes and to judge the
performance, I need to estimate the eigenvalue spread of the input to
the equalizer. Currently I am working with a single input stream of
many thousands of samples, and I am assuming that my data is more or
less uniformly distributed. To get the eigenvalues, I generate
R = Sum( x[k]*x[k]T) where x[k] is a vector of length N at time k of
input stream, x[k]T is the transpose of said vector, and k changes
over the data stream. Then I generate the eigenvalues of R or mainly
the ratio of max eig(R) to min eig(R). Does this make sense ? Any
other way(s) to estimate the eigenvalue spread of the input data?
Thanks.
```
```mk wrote:
> Hi everyone,
> I am experimenting with some equalization schemes and to judge the
> performance, I need to estimate the eigenvalue spread of the input to
> the equalizer. Currently I am working with a single input stream of
> many thousands of samples, and I am assuming that my data is more or
> less uniformly distributed. To get the eigenvalues, I generate
> R = Sum( x[k]*x[k]T) where x[k] is a vector of length N at time k of
> input stream, x[k]T is the transpose of said vector, and k changes
> over the data stream.
What's the limits of the summation?
> Then I generate the eigenvalues of R or mainly
> the ratio of max eig(R) to min eig(R). Does this make sense ?
Eh... what do you mean? It is usually a good idea to actually compute
the parameters one wants to analyse. But maybe that's not what you
> Any
> other way(s) to estimate the eigenvalue spread of the input data?
You have to estimate eigenvalues to be able to analyze
eigenvalues. As you may have seen already, computing
eigenvalues is not the cheapest of tasks, computationally
speaking, so there are a couple of ways to proceed:
- Are there particularly fast ways of computing what one needs?
- Can the same information be obtained by other strategies?
If you use something like matlab to compute the eigenvalues,
there is a huge overhead in those routines, since they need
both to be able to handle the general case, and the compute
everything, like all the eigenvalues and all the corresponding
eigenvectors.
You only need the largest and smallest eigenvalues of a
symmetric(?) matrix.
As far as I remember, there are particular recipes for those
very tasks in the book by Golub and van Loan. The algorithm
only computes what is necessary to get the job done, thus
saving lots of flops. The downside is that you have to implement
it yourself.
The other question was whether you can achieve the same
goals with other means than eigenvalues. You could try to
compute the AR(P) model for some fixed P and estimate
the ration between the geometric and arithmetic means
of the magnitudes of the zeros.
If you use matlab, the downside is that matlab solves for
roots of polynomials by expressing the polynomial as a
matrix and search for the eigenvalues of that matrix.
Which means it's a more computationally demanding task
than the "expensive" scheme above. So you'll have to
implement your own root solver...
Rune
```
```Hello everybody,
I am currently intented to work on Seismic
signals. iam new to seismic signals though have some exposure to
Digital signal processing.
iam awaiting suggestion
Arul
```
```Arul wrote:
> Hello everybody,
>
> I am currently intented to work on Seismic
> signals. iam new to seismic signals though have some exposure to
> Digital signal processing.
> iam awaiting suggestion
I suggest you ask a question.
Regards,
Andor
```
```Rune Allnor wrote:
...
> The other question was whether you can achieve the same
> goals with other means than eigenvalues. You could try to
> compute the AR(P) model for some fixed P and estimate
> the ration between the geometric and arithmetic means
> of the magnitudes of the zeros.
Interesting. Why the ratio of the geometric and arithmetic means of the
magnitudes? I would have guessed that an estimate for the eigenspread
of the signal is the square of the ratio of the max and the min of the
magnitudes of the zeros of the prediction polynomial.
As the eigenspread of a signal is more or less equal to the condition
number of the autocorrelation matrix, any condition number estimate
will do.
>
> If you use matlab, the downside is that matlab solves for
> roots of polynomials by expressing the polynomial as a
> matrix and search for the eigenvalues of that matrix.
> Which means it's a more computationally demanding task
> than the "expensive" scheme above. So you'll have to
> implement your own root solver...
To avoid that, I once used the maximum norm condition number of the
autocorrelation matrix to estimate the 2-norm condition number. Worked
quite well, and was cheaper to compute.
Regards,
Andor
```
```Andor wrote:
> Rune Allnor wrote:
> ...
> > The other question was whether you can achieve the same
> > goals with other means than eigenvalues. You could try to
> > compute the AR(P) model for some fixed P and estimate
> > the ration between the geometric and arithmetic means
> > of the magnitudes of the zeros.
>
> Interesting. Why the ratio of the geometric and arithmetic means of the
> magnitudes? I would have guessed that an estimate for the eigenspread
> of the signal is the square of the ratio of the max and the min of the
> magnitudes of the zeros of the prediction polynomial.
First: This has nothing at all to do with the eigenvalues, at least not
in a way I am aware of. It has to do with spectrum flatness. The OP
wanted to evaluate how well the equalizer worked, this is an
alternative
approach that may or may not be a bit cheaper to compute.
If you have a flat spectrum the zeros of the AR predictor tend to
be distributed evenly around the unit circle at fairly similar
magnitudes, say, around 0.5. Both the geometric and arithmetic
means may be fairly low here.
If you have a spectrum with lots of peaks, sone zeros are close to
the unit circle in order to cancel those peaks. The ratio between
the arithmetic and geometric means is nonlinear, and capable of
picking up that difference, even if there is only one or two poles that
have moved.
Both the AIC and MDL order estimators for AR models use this
concept.
Rune
```
```Andor wrote:
> Arul wrote:
>> Hello everybody,
>>
>> I am currently intented to work on Seismic
>> signals. iam new to seismic signals though have some exposure to
>> Digital signal processing.
>> iam awaiting suggestion
>
> I suggest you ask a question.
In a new thread with an appropriate subject.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
```
```"Andor" <andor.bariska@gmail.com> writes:
> Arul wrote:
>> Hello everybody,
>>
>> I am currently intented to work on Seismic
>> signals. iam new to seismic signals though have some exposure to
>> Digital signal processing.
>> iam awaiting suggestion
>
> I suggest you ask a question.
Oh, come now, Andor - you're too intolerant and demanding. Don't you
know this is the new Information Age, where any anonymous, insolent
person must only hint at a question and the whole world will clammer
--
% Randy Yates % "So now it's getting late,
%% Fuquay-Varina, NC % and those who hesitate
%%% 919-577-9882 % got no one..."
%%%% <yates@ieee.org> % 'Waterfall', *Face The Music*, ELO
```
```On 10 Aug 2006 22:16:37 -0700, "Rune Allnor" <allnor@tele.ntnu.no>
wrote:
>
>mk wrote:
>> Hi everyone,
>> I am experimenting with some equalization schemes and to judge the
>> performance, I need to estimate the eigenvalue spread of the input to
>> the equalizer. Currently I am working with a single input stream of
>> many thousands of samples, and I am assuming that my data is more or
>> less uniformly distributed. To get the eigenvalues, I generate
>> R = Sum( x[k]*x[k]T) where x[k] is a vector of length N at time k of
>> input stream, x[k]T is the transpose of said vector, and k changes
>> over the data stream.
>
>What's the limits of the summation?
I see that I wasn't very clear. In equalization, one deals with a
regressor of some certain size (here it's N, which is the same length
as that of the equalizer taps). So the vector x is of size N and
effectively it's a moving window over the input data stream. At any
point in time I am generating a matrix by calculating x[k] * x[k]T (x
is a column vector so R is NxN in size). Normally R, the
autocorrelation matrix is the expectation of x*xT but I only have one
data stream so I am moving x over this data stream and averaging the
resulting matrix to get an estimated autocorrelation matrix. The
limits of the summation above, which is the first portion of the
averaging process, is all the input data. ie get a slice of input data
of size N, generate x*xT, add to R, increment k which moves the x one
step, repeat.
At the end of this summation, I get calculate the eigenvalues of the
resulting matrix R. As I am mainly interested in the ratio of the
maximum eigenvalue to the minimum eigenvalue, the normalization of the
R is not really necessary. This ratio gives me the eigenvalue spread
of the input stream which is mainly caused by the ISI in the channel
and the higher that value, the more difficult it's for some equalizers
to converge.
>> Any
>> other way(s) to estimate the eigenvalue spread of the input data?
>
>You have to estimate eigenvalues to be able to analyze
>eigenvalues. As you may have seen already, computing
>eigenvalues is not the cheapest of tasks, computationally
>speaking, so there are a couple of ways to proceed:
Actually because the number of taps in the equalizer is rather small
(5 to 15), the computational burden of calculating the eigenvalues of
the autocorrelation matrix is not that high. The main question I am
trying to answer is whether the averaging process I am using to
approximate the expectation is really meaningful or not.
```
```mk wrote:
> On 10 Aug 2006 22:16:37 -0700, "Rune Allnor" <allnor@tele.ntnu.no>
> wrote:
>
> >
> >mk wrote:
> >> Hi everyone,
> >> I am experimenting with some equalization schemes and to judge the
> >> performance, I need to estimate the eigenvalue spread of the input to
> >> the equalizer. Currently I am working with a single input stream of
> >> many thousands of samples, and I am assuming that my data is more or
> >> less uniformly distributed. To get the eigenvalues, I generate
> >> R = Sum( x[k]*x[k]T) where x[k] is a vector of length N at time k of
> >> input stream, x[k]T is the transpose of said vector, and k changes
> >> over the data stream.
> >
> >What's the limits of the summation?
>
> I see that I wasn't very clear. In equalization, one deals with a
> regressor of some certain size (here it's N, which is the same length
> as that of the equalizer taps). So the vector x is of size N and
> effectively it's a moving window over the input data stream. At any
> point in time I am generating a matrix by calculating x[k] * x[k]T (x
> is a column vector so R is NxN in size). Normally R, the
> autocorrelation matrix is the expectation of x*xT but I only have one
> data stream so I am moving x over this data stream and averaging the
> resulting matrix to get an estimated autocorrelation matrix. The
> limits of the summation above, which is the first portion of the
> averaging process, is all the input data. ie get a slice of input data
> of size N, generate x*xT, add to R, increment k which moves the x one
> step, repeat.
> At the end of this summation, I get calculate the eigenvalues of the
> resulting matrix R. As I am mainly interested in the ratio of the
> maximum eigenvalue to the minimum eigenvalue, the normalization of the
> R is not really necessary. This ratio gives me the eigenvalue spread
> of the input stream which is mainly caused by the ISI in the channel
> and the higher that value, the more difficult it's for some equalizers
> to converge.
>
> >> Any
> >> other way(s) to estimate the eigenvalue spread of the input data?
> >
> >You have to estimate eigenvalues to be able to analyze
> >eigenvalues. As you may have seen already, computing
> >eigenvalues is not the cheapest of tasks, computationally
> >speaking, so there are a couple of ways to proceed:
>
> Actually because the number of taps in the equalizer is rather small
> (5 to 15), the computational burden of calculating the eigenvalues of
> the autocorrelation matrix is not that high. The main question I am
> trying to answer is whether the averaging process I am using to
> approximate the expectation is really meaningful or not.
Ah. Much clearer. Thanks.
I would be very reluctant to use an covariance accumulator like that.
The first N-length frame totally dominates the structure of R, the
next
a little less so, and as the number of samples increases the relative
contribution to R from each new sample vanishes.
You can handle this in at least two ways. Each new frame can be
added with a forgetting factor a, 0<a<1, like
R_{n} = a*x[n]x'[n] + (1-a)R_{n-1}
where x[n] is the N-length subsequence of x that ends at sample n.
This way, the last frame x[n] always make a large contribution to
the covariance matrix.
Or you can use a sliding window covariance estimator like
R_{n} = R_{n-1} + x[n]x'[n] - x[n-M]x'[n-M]
i.e. you add a new frame and remove the contributions from the
M'th oldest frame.
Rune
``` | 3,685 | 14,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-49 | latest | en | 0.878236 |
https://inches.guru/1.5-feet-in-cm/ | 1,685,766,815,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649105.40/warc/CC-MAIN-20230603032950-20230603062950-00015.warc.gz | 363,350,381 | 4,661 | # Convert 1.5 feet in cm
1.5 feet = 45.72 cm
## How many cm is 1.5 feet?
What is 1.5 feet to centimeters? The answer is that 1.5 feet equals 45.72 centimeters.
How to convert 1.5 feet to cm? We know that one foot is equal to 30.48 centimeters, so to convert 1.5 feet to centimeters, we must multiply the amount of centimeters by 30.48, which means that 1.5 feet is equal to 1.5 feet × 30.48 = 45.72 cm. | 141 | 406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-23 | longest | en | 0.925569 |
http://stackoverflow.com/tags/matrix/new | 1,411,050,928,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657127503.54/warc/CC-MAIN-20140914011207-00129-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 255,759,901 | 24,720 | # Tag Info
0
MATLAB uses column-major order for historical reasons. Very early versions of MATLAB were implemented in FORTRAN and relied heavily on the LINPACK and EISPACK FORTRAN libraries which, unlike C, use column-major order. Even though it was (mostly) converted to a C implementation when it was initially commercialized, it retained the use of column-major order. ...
0
Also: nm1 <- unique(unlist(dimnames(x))) x1 <- matrix(NA, ncol=length(nm1), nrow=length(nm1), dimnames=list(nm1, nm1)) x1[match(rownames(x), nm1), match(colnames(x), nm1)] <- x library(Matrix) x2 <- forceSymmetric(x1) diag(x2) <- 0 x2 #6 x 6 Matrix of class "dsyMatrix" # 1 2 3 4 5 6 #1 0.000 0.043 0.080 0.106 ...
1
First, you'd have to add an additional row of NAs to your matrix (since right now it is 5 by 6): x <- rbind(x, "6"=NA) Then you can turn the entire lower half (including the diagonals) to 0s, then add it to its own transpose: x[!upper.tri(x)] <- 0 result <- x + t(x) (Note that in your case, x[is.na(x)] <- 0 would also work).
0
My exercise for this question, though some Linqs can still be shortcircuited, this may be more clear: [TestMethod] public void LinqArraySkipFirstColumnAndLine() { var inputString = @"SkipThisLine 1 4 2 1 3 5 2 3 1 2 4 5 3 4 5 3 2 1 ...
1
I'd advise to simply skip foreach and use a regular for loop. After trimming and splitting the line by space(s), simply skip the 1st element by initializing the index to 1 instead of 0. var cols = line.Trim().Split(' '); //use string split option to remove empty entries for robustness for (int j = 1; j < cols.Length; j++) //note the j is initialized ...
0
There are many variations on how to do this, but you want to start j at 1 instead of 0. that was you're missing element 0, which I believe if what you're after, so: foreach (var line in File.ReadAllLines(@"input_5_5.txt", Encoding.GetEncoding(1250))) { int j = 1; foreach (var col in line.Trim().Split(' ')) { result[i, j] = ...
0
You should instead use (if you need to know which line/col). Col is initialized with 1 since you don't need first column. string[] fileLines = System.IO.File.ReadAllLines(@"input_5_5.txt", Encoding.GetEncoding(1250)); for(int line = 0; line < fileLines.Length; line++) { string[] splittedLines = fileLines[line].Trim().Split(' '); for(int col = 1; ...
0
If you flatten the boolean mask like: m2[np.asarray(m2[:,1]>10).flatten()] you get the same result, but I would recommend using np.array instead of np.matrix for the reasons given in this answer.
2
Particularly if you're using the programmable pipeline, you have almost complete freedom about the coordinate system you work in, and how you transform your geometry. But once all your transformations are applied in the vertex shader (resulting in the vector assigned to gl_Position), there is still a fixed function block in the pipeline between the vertex ...
0
Seems correct to me. If you boil down your question, you are really saying: Given a matrix B(m*n), we can represent this in a vector V with length m*n such that V exactly recreates B, or B == V The number of possible values for B is the same as the number of possible values for V.
0
The issue you're experiencing comes down to the fact that operations on a matrix return always return a 2-dimensional array. When you build the mask on the first array, you get: In [24]: a2[:,1] > 10 Out[24]: array([False, False, True, True, True], dtype=bool) which, as you can see, is a 1-dimensional array. When you do the same thing with the ...
1
Here I tried to clean it up a bit and generalize the use of m: random_block_sample <- function(a_matrix, m = 2L) { N <- nrow(a_matrix) stopifnot(m <= N) n <- ceiling(N / m) s <- sample(N - m + 1L, n, TRUE) # start_point i <- unlist(lapply(s, seq, length.out = m)) b_matrix <- a_matrix[i, , drop = FALSE] ...
1
The problem is that you cannot assign the result of the product to one of the terms while still evaluating it, since that will destroy the original values you still need to calculate the other elements. Since you have a working binary *, the simple way to implement *= is to make it as *this = *this * other. There can be shortcuts, but requires matrices to ...
1
You are overwriting the LHS of the operator while still trying to compute the product. Take the matrices: 1 0 0 1 and 2 0 0 2 For i = 0 and j = 0, you end of making the first matrix; 0 0 0 1 before you start the multiplication. You know you are not going to get the right answer after that. I don't know if there is a technique you can use to ...
0
I have a few suggestions: Reshape: b = reshape(A.',1,[]) b = 1 0 1 0 1 1 0 1 0 1 Transpose then create a vector: c = A.'; d = c(:).' d = 1 0 1 0 1 1 0 1 0 1 Indexing: e = A([1:size(e,1):end, 2:size(e,1):end]) e = 1 0 1 0 1 1 0 1 0 ...
0
Here is a function: def f(r,v): def p(x): return vector([0]+list(x[:-1r])) + x[-1r]*v m = [r] for _ in xrange(len(r)-1r): m.append(p(m[-1r])) return matrix(m) Example: sage: r = vector(ZZ,(0, 1, 2, 3,)) sage: v = vector(ZZ,(3, 5, 7, 9,)) sage: f(r,v) [ 0 1 2 3] [ 9 15 22 29] [ 87 154 218 283] [ 849 ...
0
Create this function: function transpose(arr) { return Object.keys(arr[0]).map(function (c) { return arr.map(function (r) { return r[c]; }); }); } and then: var transposedArray = transpose(originalArray);
3
If you just want to change all the 4s to 5s then: x(x==4)=5 basically x==4 will result in a logical matrix with 1s everywhere there was a 4 in x: [0 0 0 1 0 1 1 0 0 1 0 1] We then use logical index to only affect the values of x where those 1s are and change them all to 5s. If you wanted to do this using a loop (which I highly recommend against) ...
2
Short answer to achieve what you want: x(x==4) = 5 Answer to why your code doesn't do what you expected: You are changing the item to a 5. But that item is a new variable, it does not point to the same item in your matrix x. Hence the original matrix x remains unchanged.
3
An alternative using ndgrid: [I, J] = ndgrid(1:m, 1:n); A = I.^2 + J.^2;
5
bsxfun based solution and as such has to be pretty efficient with runtime performance, as it is well-suited for these expansion related problems - A = bsxfun(@plus,[1:m]'.^2,[1:n].^2)
0
I've wrote a modified version of matlab's function rref, called g2rref that works great! https://gist.github.com/esromneb/652fed46ae328b17e104
0
Overview Consider a sample matrix could look like this: ABCD EFGH IJKL MNOP For the purposes of my explanation, ABCD is considered as row 0, EFGH is row 1, and so on. The first pixel of row 0 is A. Also, when I talk about the outer shell, I am referring to: ABCD E H I L MNOP First let's look at the code that moves the values. int top = ...
1
First, a streaming loader. Provide it with a function that, given a max, returns a vector of data (aka loader<T>). It can store internal state, but it will be copied, so store that internal state in a std::shared_ptr. I guarantee that only one copy of it will be invoked. You are not responsible for returning all max data from your loader, but as ...
0
This is in the vein of Michal's answer (from which I will steal the nice graphic). Matrix: min ..... b ..... c . . . . II . I . . . . d .... mid .... f . . . . III . IV . . . . g ..... h ..... max Min and max are the smallest and largest values, respectively. ...
3
This is just a minor improvement over Divakar's answer. It is a little faster because it replaces a 3D-array permute with a 2D-array permute: B = bsxfun(@times, permute(A, [1 3 2]), permute(A, [3 1 2]));
4
To state the obvious, have you tried a simple for-loop: [m,n] = size(A); B = zeros(m,m,n); for i=1:n B(:,:,i) = A(:,i) * A(:,i).'; end You'll be surprised how competitively fast it is.
3
This - B = permute(bsxfun(@times,A,permute(A,[3 2 1])),[1 3 2])
0
The matrix firstMultiplyMat itself is fine memory-wise; if you push the values into it that you had put into a, it'll invert correctly. Your problem lies in the values that are inside firstMultiplyMat, which are not exactly the same values as what get printed, because what is printed is truncated. Essentially, your matrix may not be invertible. You can work ...
1
the easy way would be to use only the output Allocation and use getElementAt_<type> to read from the input Allocations of a different size. I'm curious--if we were to add this as an intrinsic, do you specifically need only GEMM, or are there other BLAS operations you'd like supported?
0
Given a non-singular matrix A. Construct an algorithm using Gaussian elimination to find \$A^{-1}\$ I had the same problem and did it In python: #Helper functions: def check_zeros(A,I,row, col=0): """ returns recursively the next non zero matrix row A[i] """ if A[row, col] != 0: return row else: if row+1 == len(A): return "The Determinant is ...
1
The Mat objects are complex objects that have internal allocations of memory. When you clear the vector, one will need to iterate through every instance of Mat contained and run its destructor which is itself a non-trivial operation. Also remember that free-store memory perations are non-trivial, so depending on your heap implementation, the heap may decide ...
5
Java does not support returning multiple things at once. However, you could create a small class that does this: public class TwoArrays { public final int[][] A; public final int[][] B; public TwoArrays(int[][] A, int[][] B) { this.A = A; this.B = B; } } Then make your method like this: public static TwoArrays ...
2
You can't return multiple values from a method. You can return a single Object that contains two arrays as members though. You could also return a multi-dimensional array that contains both arrays, but that's not a very OOP solution.
5
Make one array which contains both arrays. In your case int[2][][] = { {{1,2},{2,3},{4,5}}, {{1,2},{2,3},{4,5}} }; Or better, make an object which contains both arrays.
1
NOTE: getComputedStyle method only work's for elements which are appended to your document let's start with a function that make's a hidden element and append's it into your document : var make_test_el = function () { var el = document.createElement("div"); // some browsers doesn't add transform styles if display is inline ...
-1
Be careful using optimization, it can drive the debugger crazy. If you were to do this in a function and simply let the vector go out of scope?? Since the elements are not pointers I think this will work.
0
Here is the final code that prints different parts of a matrix as shown in the picture below: Here is the code for the output given above: package com.codopedia.java7.sep2014; /** * * @author www.codopedia.com */ public class TwoDArrayExp1 { public static void main(String args[]) { int row = 15, column = 15, k = 10; int my2dArray1[][] = new ...
0
The easiest thing to do here would be to encapsulate this into a single for loop, and change the way you're accessing the core variables so that you're using the loop index instead. Looking at your code, I'm assuming that sat_look_tcs_pass1 is a 3 x 79 matrix. I'm also going to assume that the output height h_test is a single value because when you're ...
0
>>> xs = np.arange(16) >>> xs array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]) >>> xs.reshape(4, 4) array([[ 0, 1, 2, 3], [ 4, 5, 6, 7], [ 8, 9, 10, 11], [12, 13, 14, 15]]) >>> xs = xs.reshape(4, 4) >>> a, b, c, d = xs[:2, :2], xs[2:, :2], xs[:2, 2:], xs[2:, 2:] ...
2
It looks like you are reusing your distance variable in an unintentional way, as well as your velocity variable at the end of your for loop. You're mutating the distance variable, and then trying to reshape it as a matrix. You'll need to change this variable, and then call where you want to store your matrix as something else. Also, distance and velocity ...
0
Your data: mydata<-c(1,2,3,4,5,2,3,4,6,7,3,3,5,7,0,4,9,2,4,5) mat<-matrix(mydata,ncol=5, byrow=T) mat # [,1] [,2] [,3] [,4] [,5] #[1,] 1 2 3 4 5 #[2,] 2 3 4 6 7 #[3,] 3 3 5 7 0 #[4,] 4 9 2 4 5 use mat[mat[,2]==3,] result being: [,1] [,2] [,3] [,4] [,5] [1,] 2 3 4 6 ...
2
This fine question is the best I have seen in a week on StackOverflow. Answer: Round it! A major point of the QR decomposition is precisely that the factors in question be well-conditioned, that there be no tiny/huge eigenvalues in Q. When there are no tiny/huge eigenvalues, rounding doesn't hurt. What point would there be in the QR if it did not let ...
2
For the specific case of QR decomposition, you usually don’t even bother computing the matrix entries that you know are going to be zeroed out by a Householder reflection; you simply treat them as zero from that point on (in fact, you generally don’t even bother storing the zeros, and instead use that space to store the reflection itself).
1
This line is not right: scanf("%lf\t", *(*(m1+i)+j)); //Scanning input The type of *(*(m1+i)+j) is double, not double*, which is what you need for scanf. You need to use scanf("%lf\t", *(m1+i)+j); //Scanning input or, a simpler form: scanf("%lf\t", &m1[i][j]); //Scanning input You have similar errors in the other loops.
1
Here is an alternative solution using datasample in which , similarly to @RTL's solution, you first create a matrix of zeros and then add a given number of ones: clc; clear all; A = 5; B = zeros(5,5); y = datasample(1:size(B,1)*size(B,2),A) % Randomly select 5 (i.e. A) linear indices which will be replaced by 1. B(y) = 1 y = 12 17 1 22 14 ...
7
Here a matrix of zeros, B, is created with the required size and then A random locations of B are changed to 1 A = 5; B = zeros(5,5); B(randperm(numel(B),A)) = 1 output B = 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0 0
0
Here's a link to a blog which has a pretty good explanation of the issue you're having. Judging from other people having this error it seems to be caused by the content on the page. The content of the web page may be the cause of this error, not your iOS app programming. I found the error occurred in my app only while on Yahoo. Google.com, no problem. ...
1
This should do it: from sklearn.feature_extraction.text import CountVectorizer cv = CountVectorizer() words_csm = cv.fit_transform(records) words_csm.todense() matrix([[1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 1, 0, 0, 0, 0], [0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1], [0, 1, 1, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0]], dtype=int64)
1
Yes you can do this by writing a procedure like this one: procedure assign_row(signal slm : out T_SLM; slv : STD_LOGIC_VECTOR; constant RowIndex : NATURAL) is variable temp : STD_LOGIC_VECTOR(slm'high(2) downto slm'low(2)); -- Xilinx iSIM work-around, because 'range(2) evaluates to 'range(1); tested with ISE XST/iSim 14.2 begin temp := slv; for i in ...
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www.convertunits.com/from/mg/(25+(mg/ml))/to/ml
Error: We couldn't find a conversion between mg/(25 (mg/ml)) and ml. Having trouble with a unit conversion? Try doing a search, or posting to the forum. You may also be looking for a date difference or molecular weight. You can also contact us directly if you find any missing units or errors. ›› Want other units?
www.easycalculation.com/unit-conversion/Milligram-mg-Milliliter-mL.html
Conversion of mg to ml is very simple. Since 1 Milligram is equal to 0.001 milliliters, multiply the entered milligram with 0.001 to get the result. For example, when the given number of milligrams is 15, then the conversion of milligrams to milliLitres is 15 x 0.001 is 0.015 mL.
www.thecalculatorsite.com/articles/units/convert-mg-to-ml.php
Q: I would like to know how to convert mg to ml. Specifically, what is 15mg in liquid (IE: ml)? Let's start with some explanation on why this type of conversion isn't as easy as it sounds. After that, we'll try an example conversion and then, at the bottom of the article, discuss how to convert an order when given a dosage label.
www.endmemo.com/sconvert/mg_mlper.php
Concentration solution unit conversion between milligram/mL and percentage, percentage to milligram/mL conversion in batch, mg/mL per conversion chart. ENDMEMO.
www.endmemo.com/sconvert/mg_lmg_ml.php
Concentration solution unit conversion between milligram/liter and milligram/mL, milligram/mL to milligram/liter conversion in batch, mg/L mg/mL conversion chart. ENDMEMO.
www.convertunits.com/from/ml/to/mg
The millilitre (ml or mL, also spelled milliliter) is a metric unit of volume that is equal to one thousandth of a litre. It is a non-SI unit accepted for use with the International Systems of Units (SI). It is exactly equivalent to 1 cubic centimetre (cm³, or, non-standard, cc). ›› Definition: Milligram
It depends on the substance, since mg is a measure of mass and ml is a measure of volume. If you are talking about water, then a gram, or 1000 milligrams (mg) makes a millilitre (ml); so 25 mg ...
www.easycalculation.com/unit-conversion/mgml.php
Mg to Ml Converter. This online Mg to Ml Converter is used to convert Milligram to Milliliter based on the liquid density. To convert Mg to Ml, you can multiply the milligram by its liquid density and then divided by 1,000. The liquid density of a substance is its mass per unit volume.
www.thecalculatorsite.com/conversions/common/mcg-to-mg.php
Convert between micrograms (mcg) and milligrams (mg) using this simple tool. To begin the conversion, select a unit to convert from and to, enter your figure and click the 'Convert!' button. Default rounding is set to a maximum of 14 decimal places.
calculator-converter.com/milliliters-to-grams.htm
1 ml (milliliter) = 1/1000 L (Liter, the official SI unit of volume).Weight of 1 milliliter (ml) of pure water at temperature 4 °C = 1 gram (g) .The gram (British spelling: gramme, abbreviation: g) is a unit of mass in the SI system (metric system). One gram is equal to one thousandth of the kilogram (kg), which is the current SI (Metric system) base unit of mass.
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# Flamer Area of Effect
## 16 posts in this topic
So the CRB is a little ambiguous on the subject of exactly what sort of area is effected when I flamer is used to make an attack. My GM and I are trying to come up with a good compromise between us, but I'm curious if there's a more official answer. Assuming a standard flamer with a range of 20m, does that 20m cover just a line bewteen you and the target, or do you sweep it around in a cone that covers more area? That flamers do hella damage against hordes suggests that the latter might be correct, but I don't know.
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It's explicitly stated: a 30-degree cone.
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If you've played the tabletop minis game, you've probably seen the flamethrower template. I think that gives me the best sense of what it should look like. I'm sure I could Google image it.
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I am using grid paper in which one square = 2 meters. And since man and I get along about as well as a dolphin and a Japanese man, I just ruled that at 20m, the spread would be 6m at its widest point.
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It's explicitly stated: a 30-degree cone.
So it is, page 142. I read right over that. Thanks, mate.
On a similar topic, I found this while poking around on the web. It's from the FFG Dark Heresy forum (and there's a nifty flamethrower video included).
http://community.fantasyflightgames.com/index.php?/topic/14215-flame-weapons/
So doing the math out on a sheet of graph paper, I've determined that if 1 square = 1 meter, an Astartes flamer can bathe a total of 91 squares (45 squares on your paper, pearldrum1) in cleansing fire and at its widest point, it will be just under 11 meters, meaning that it can hit two targets that are 10 meters apart at its widest point, in addition to whoever might be foolish enough to be standing inside its blast radius. Which is about what I'd expect.
pearldrum1 likes this
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A 30-degree cone 20 meters long is freakin' huge. One of my players (an unemployed graphic designer) created a template to scale with our miniatures- it's pretty jaw-dropping when you see it for the first time. I'll try to remember to take a picture of it at our next session and post it.
pearldrum1 and bogi_khaosa like this
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Using mathses.
Edge A=20
Edge B=20
Edge C=X
angle A=30
C2=A2+B2-2AB cos(A)
X2=202+202-2(20)(20) cos(30)
X2=400+400-2(400)(0.866)
X2=800-692.8
X2=107.2
X=10.35
The width of the far edge of a 30 degree cone extending 20 meters is 10.35 meters.
Honestly though, real flame throwers do not simply throw fire. That is a blowtorch or bunsen burner. Flame throwers shoot a preasurised stream of liquid jellied fuel (napalm is jellied gasoline). Realistically a flame thrower would still require a too hit roll as you still have to aim the stream. As the stream has the propensity to splatter on impact they should have either a blast or the ability to "saturate" a limited area by moving the stream.
Also, flame throwers do not "light you on fire" they shoot burning fuel at you which then sticks to you. You aren't on fire, the lit gas which is sticking to you (thanks jelly) is. Amounts to the same thing.
Alas, the game isn't really realistic when it comes to flamethrowers, put it down to popular knowledge.
Edited by herichimo
DJSunhammer and pearldrum1 like this
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TOO MUCH MATH.
10m wide at 20m (range) is what the math says, we will go by this.
Good with you, FIx?
You're the DM.
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Then stop being a little ***** and play.
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You're the DM.
GM! What do you think this is, Dungeons & Dragons ? The Emperor will be displeased.
pearldrum1 likes this
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It is like he never learns, DDMW.
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As promised, here are some pics of my friend's to-scale flamer template. The picture quality isn't very good, but you get an idea of how big it is.
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Nice! Except if you're not using a 1:1 map
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Ugh. Flamer range in the 40k RPGs borders on insane. IIRC, flamethrowers in world war 2 could reach, at most, about 20 feet. And that is with a full two handed unit and backpack fuel supply. Someone here is going to think, "but the technology is better in 40k", which is pretty obvious, but a person with a pint sized, hand held flame thrower should not be capable of shooting a 30 foot by 10 foot flame. It just doesn't make sense.
Range increments should be something like 3m, 6m and 10m for the whole group.
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Someone here is going to think, "but the technology is better in 40k"
Not exactly, more like it's different. It's fiction!
Kshatriya likes this
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# LEP
1.3.03
Newtons Law / Air track
R
PHYWE series of publications Laboratory Experiments Physics PHYWE SYSTEME GMBH 37070 Gttingen, Germany 21303
1
Related topics
Velocity, acceleration, force, acceleration due to gravity.
The distance-time law, the velocity-time law, and the relation-
ship between mass, acceleration and force are determined
with the aid of the air track rail for uniformly accelerated
motion in a straight line.
Equipment
Air track rail 11202.17 1
Blower 13770.93 1
Pressure tube, l = 1.5 m 11205.01 1
Glider f. air track 11202.02 1
Screen with plug, l = 100 mm 11202.03 1
Hook with plug 11202.07 1
Starter system 11202.13 1
Magnet w. plug f. starter system 11202.14 1
Precision pulley 11201.02 1
Stop, adjustable 11202.19 1
Fork with plug 11202.08 1
Endholder for air track rail 11202.15 1
Light barrier, compact 11207.20 4
Timer 4-4 13605.99 1
Slotted weight, 10 g, black 02205.01 8
Slotted weight, 50 g, black 02206.01 4
Weight holder 1 g 02407.00 1
Silk thread, 200 m 02412.00 1
Slotted weight, 1 g, natur.colour 03916.00 20
Portable balance Mod. LS2000 46002.93 1
Barrel base 02006.10 4
Support rod -PASS-, square, l = 400 mm 02026.55 4
Right angle clamp -PASS- 02040.55 4
Connecting cord, 1000 mm, red 07363.01 4
Connecting cord, 1000 mm, yellow 07363.02 4
Connecting cord, 1000 mm, blue 07363.04 4
Connecting cord, 2000 mm, yellow 07365.02 4
Connecting cord, 2000 mm, black 07365.05 4
Problems
Determine the following for uniformly accelerated motion in a
straight line:
1. Distance travelled as function of time
2. Velocity as a function of time
3. Acceleration as a function of the accelerated mass
4. Acceleration as a function of force.
Set-up and procedure
The experimental set-up is shown in Fig. 1.
The starting device is mounted in such a manner that trigger-
ing unit releases the glider without giving it an initial impulse
when triggered. It is connected with the two Start jacks on
the timer; when connecting it, ensure that the polarity is cor-
rect. The red jack on the starting device is connected with the
yellow jack of the timer. The four light barriers are connected
in sequence from left to right with the control input jacks 1
to 4 on the timer. Connect jacks having the same colour
when doing so.
The mass of the glider can be altered by adding slotted
weights. Always place weights having the same mass on the
gliders weight-bearing pins, as optimum gliding properties are
provided only with symmetrically loading. The accelerating
force acting on the glider can be varied by changing the num-
Fig. 1. Experimental set-up for investigation of uniformly accelerated motion.
LEP
1.3.03
Newtons Law / Air track
R
21303 PHYWE series of publications Laboratory Experiments Physics PHYWE SYSTEME GMBH 37070 Gttingen, Germany
2
ber of weights (on the weight holder) acting via the silk thread
and the precision pulley. Determine the mass of the glider
without the supplementary slotted weights by weighing it.
Position the four light barriers in a manner such that they
divide the measuring distance into approximately equal seg-
ments. Place the last light barrier such that the glider with
screen passes through it before the accelerating weight
touches the floor. Position the adjustable stop with the fork
and plug on the track in such a manner that the glider is gent-
ly braked by the rubber band just before the accelerating
weight touches the floor. Measure the distances travelled
s
1
s
4
between the front edge of the screen from the starting
position to the respective light barriers exactly for the evalua-
tion. Perform all subsequent measurements without changing
the light barriers positions.
After measuring the times t
1
t
4
required for the four distanc-
es travelled s
1
s
4
with the timer in the s(t) operating mode
(see operating instructions), determine the corresponding
velocities with the v(t) operating mode. While doing so, the
1
t
4
of the four light barriers are measured;
from them the mean values of the velocity for the correspond-
ing distance travelled are determined with reference to the
screens length. These mean velocities correspond to the
instantaneous velocities represented by the times t
1
t
4
in
accordance with the following:
t
n
'
= t
n
+
To determine the acceleration as a function of the mass,
increase the mass of the glider progressively by 20 g incre-
ments (10 g on each side), and measure the instantaneous
velocity at a predetermined position.
In determining the acceleration as a function of force, the total
mass remains constant. Successively transfer 2 g (1 g from
each side) from the glider to the weight holder and measure
the instantaneous velocity at a fixed position. The accelerated
mass must not exceed 20 g.
Before beginning with the measurements, it is advisable to
check the tracks adjustment.
Theory and evaluation
Newtons equation of motion for a mass point of mass m to
which a force
## F is applied is given by the following:
m
a =
F ,
where
a =
is the acceleration.
The velocity v obtained by application of a constant force is
given as a function of the time t by the expression
v (t) = t
for
v (0) = 0 .
Assuming that
v (0) = 0 ;
r (0) = 0
the position of
## r of the mass point is
r (t) = t
2
. (0)
In the present case the motion is unidimensional and the force
produce by a weight of m
1
is
|
F
|
= m
1
|
g
|
m
1
g
where g is the acceleration of gravity. If the total mass of the
glider is m
2
the equation of motion is given by
(m
2
+ m
1
)
|
a
|
= m
1
g ; (1)
The velocity is
|
v (t)
|
v = t (2)
and the position is
|
r (t)
|
s (t) = t
2
. (3)
The evaluation is now illustrated by the following sample
measurements. For all the measurements the distances of the
four light barriers from the starting point were 22 cm, 44 cm,
66 cm and 88 cm, respectively. The mass of the glider includ-
ing the screen, hooks and support magnet was 201 g.
Fig. 2 shows the distance travelled s as a function of the time
t (m
1
= 10 g, m
2
= 201 g). In Fig. 3 the distance travelled is
illustrated as a function of t
2
for the same measured values. A
linear correlation results, as was expected from the theory.
The slope is 0.246 m/s
2
and the following is thus obtained
from Equation (0):
F = 2 (m
1
+ m
2
) 0.246 ms
-2
= 0.104 N.
m
1
g
m
1
+ m
2
1
2
m
1
g
m
1
+ m
2
F
m
1
2
F
m
d
2
r
dt
2
t
n
2
Fig. 2. The distance travelled s plotted as a function of the
time t ; m
1
= 10 g, m
2
= 201 g.
As a good approximation, this corresponds to the weight force
of the mass m1 (0.010 kg); F = m
1
g = 0.0981 N.
Under the same experimental conditions, the correlation v(t)
presented in Fig. 4 is obtained by measuring the shading time
of the four light barriers due to the screen, which has a length
of 10 cm. The slope of the compensation line drawn through
the origin correspond to the acceleration a in this case. For
the presented sample measurement a = 0.473 ms
-2
. We
expect a to be equal to
a = = 0.465 ,
This value agrees with the acceleration determined using Fig.
4. very well.
In the same manner as shown in the example in Fig. 4, the
accelerations are measured in two measuring series as a func-
tion of the inert mass m
1
+ m
2
(F = constant) and as a function
of the force (m
1
+ m
2
= constant). In the process, one can
make the evaluation work much easier by using a computer
with a spreadsheet program (e. g. Microsoft Excel
).
Fig. 5 shows the acceleration due to the mass m
1
= 10 g as a
function of the inert mass. If the acceleration is plotted against
the reciprocal of the inert mass using the same measured val-
ues, a linear correlation results, as expected (Fig. 6). The slope
of the straight lines should be equal to the accelerating force,
m
s
2
m
1
g
m
1
+ m
2
LEP
1.3.03
Newtons Law / Air track
R
PHYWE series of publications Laboratory Experiments Physics PHYWE SYSTEME GMBH 37070 Gttingen, Germany 21303
3
Fig. 5. The acceleration a as a function of the inert mass
m
1
+ m
2
measured at a constant accelerating (weight)
force due to the mass m
1
= 10 g.
Fig. 3. The same measurement as in Fig. 2 plotted against t
2
.
Fig. 4. The velocity v plotted as a function of the time t ;
m
1
= 10 g, m
2
= 201 g.
Fig. 6. The same measurement as in Fig. 5 plotted against the
reciprocal inert mass.
LEP
1.3.03
Newtons Law / Air track
R
21303 PHYWE series of publications Laboratory Experiments Physics PHYWE SYSTEME GMBH 37070 Gttingen, Germany
4
m
1
g = 0.981 N. The evaluation of the present example in
Fig. 6. results in a slope of 0.999 kgms
2
= 0.999 N.
In conclusion, Fig. 7 shows the dependence of the accelera-
tion on the accelerating force F. One sees the linear propor-
tionality between the two parameters. The reciprocal slope is
0.213 kg and corresponds well to the inert mass m
1
+ m
2
=
0.217 kg.
Fig. 7. The acceleration a as a function of the force F for
constant inert mass m
1
+ m
1
= 217 g. | 2,580 | 8,890 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.752303 |
https://www.scribd.com/document/329720655/kindin-lec1 | 1,566,165,968,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314130.7/warc/CC-MAIN-20190818205919-20190818231919-00047.warc.gz | 968,497,304 | 60,212 | You are on page 1of 25
# Kinematika dan dinamika
## 2nd half of semester
Lecture note #1: Newton 2nd Law for particle
Dept. of Mechanical Engineering
Lecturer/instructor
Dr. Eng. Radon Dhelika (S3 2015, Tokyo
Institute of Technology
Departemen Teknik Mesin, Fakultas Teknik UI
Ruangan: Gedung MRC lantai 5, Lab
perancangan mekanikal dan biomekanik
Syllabus
Kinematics:
NO force
## To study the interaction between those properties
Involving force, work, energy, impulse, etc
Dynamics
## 1. Particle: force and acceleration
2. Particle: work and energy
3. Particle: impulse and momentum
4. Particle: collision and relative motion
5. Systems of particles
6. Rigid body: force and acceleration
7. Rigid body: work and energy
8. Rigid body: impulse and momentum
Schedule (tentative)
Lecture No
Topic
1 (25 Oct)
2 (27 Oct)
3 (1 Nov)
4 (3 Nov)
5 (8 Nov)
6 (10 Nov)
7 (15 Nov)
8 (17 Nov)
9 (22 Nov)
Torque (1)
10 (24 Nov)
11 (29 Nov)
## Quiz 3 (7.5%), Torque (2)
12 (1 Dec)
Angular momentum
13 (6 Dec)
Quiz 4 (7.5%)
14 (8 Dec)
14 (13 Dec)
UAS
Quiz #1: 5%
Quiz #2: 5%
Quiz #3: 7.5%
Quiz #4: 7.5%
UAS: 25%
TOTAL: 50%
References
Young, D. H., and A. R. Freedman. "University Physics
with Modern Physics, 13th."
Kraige, L. G., and J. L. Meriam. "Engineering Mechanics:
Dynamics."
Kleppner, Daniel, and Robert Kolenkow. An introduction
to mechanics. Cambridge University Press
## Approaches to dynamics problems
General approaches to the
solution of kinetics problems
(1)
Application of
Newtons 2nd Law
(Force-MassAcceleration
method)
(2).
Work and
energy
principles
(3)
Impulse and
momentum methods
## Newton's Law of Motion
First Law: an object either remains
at rest or continues to move at a
constant velocity, unless acted
upon by an external force
Second Law: The vector sum of the
external forces F on an object is
equal to the mass m of that object
multiplied by the acceleration
vector a of the object: F = ma
Third Law: When one body exerts
a force on a second body, the
second body simultaneously exerts
a force equal in magnitude and
opposite in direction on the first
body
8
Equation of motion
a
F1
ma
Inertia Force
F2
m
Fn
F-ma=0
F= ma
## First simplification: point particle assumption
Second simplification: motion in one dimension
only
10
Forces
Four common types of forces:
Normal force
Tension force
## How do you confirm the
existance of normal force?
Friction force
Weight or
gravitational force
11
Superposition of forces
The force
## axis, may be replaced by its rectangular component
vectors
and
12
Solved problem 1
A gymnast with mass 50 kg suspends herself from the lower end of a
hanging rope of negligible mass. The upper end of the rope is attached
to the gymnasium ceiling.
(a) What is the gymnasts weight?
(b) What force (magnitude and direction) does the rope exert on her?
(c) What is the tension at the top of the rope?
13
Solved problem 1
(a)
(b)
(c)
14
Solved problem 2
Find the tension at each end of the rope in Solved Problem 1 if the
weight of the rope is 120 N
15
Solved problem 2
## At each end, the tension of the rope is -610 N and 610 N
16
Solved problem 3
a car engine with weight w hangs from a chain
that is linked at ring O to two other chains,
one fastened to the ceiling and the other to the
wall. Find expressions for the tension in each
of the three chains in terms of w. The weights
of the ring and chains are negligible compared
with the weight of the engine.
17
Solved problem 3
Engine
O ring
18
Solved problem 4
An iceboat is at rest on a frictionless horizontal surface. A wind is
blowing along the direction of the runners so that 4.0 s after the iceboat
is released, it is moving at 6 m/s. What constant horizontal force Fw
does the wind exert on the iceboat? The combined mass of iceboat and
rider is 200 kg
Ans: Fw = 300 N
19
Solved problem 4
20
## Useful kinematics formulas
21
Solved problem 5
An elevator and its load have a combined mass of 800 kg. The
elevator is initially moving downward at 10 m/s; it slows to a stop
with constant acceleration in a distance of 25.0 m. What is the
tension T in the supporting cable while the elevator is being
brought to rest?
Ans: T = 9440 N
22
Solved problem 5
23
Solved problem 6
## Dynamics with constant force
A particle weighing 4 N is pulled up a smooth plane by a
force F. Determine the force of the plane on the particle
(N1) and the acceleration along the plane
Ans: N1 = 2.94 N, ax = 2.34 m/s2
24
Solved problem 1
N1 = 2.94 N
ax = 2.34 m/s2
25 | 1,324 | 4,543 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-35 | latest | en | 0.729915 |
https://cn.maplesoft.com/support/help/maple/view.aspx?path=GraphTheory/SeidelSpectrum | 1,708,995,851,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474669.36/warc/CC-MAIN-20240226225941-20240227015941-00013.warc.gz | 179,806,664 | 21,556 | SeidelSpectrum - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
GraphTheory
SeidelSpectrum
compute Seidel spectrum
Calling Sequence SeidelSpectrum(G)
Parameters
G - graph
Description
• The SeidelSpectrum command returns the Seidel spectrum of the eigenvalues of a specified graph. That is the set of eigenvalues of the matrix $J-I-2A$ where $J$ is the all-one matrix, $I$ is the identity matrix and $A$ is the adjacency matrix of the graph. The output is a list $L$. Each element of $L$ is a list of size 2, where the first element is an eigenvalue and the second element is its multiplicity.
Examples
> $\mathrm{with}\left(\mathrm{GraphTheory}\right):$
> $\mathrm{with}\left(\mathrm{SpecialGraphs}\right):$
> $G≔\mathrm{ClebschGraph}\left(\right)$
${G}{≔}{\mathrm{Graph 1: an undirected graph with 16 vertices and 40 edge\left(s\right)}}$ (1)
> $\mathrm{SeidelSpectrum}\left(G\right)$
$\left[\left[{-3}{,}{10}\right]{,}\left[{5}{,}{6}\right]\right]$ (2)
> $P≔\mathrm{PetersenGraph}\left(\right)$
${P}{≔}{\mathrm{Graph 2: an undirected graph with 10 vertices and 15 edge\left(s\right)}}$ (3)
> $\mathrm{SeidelSpectrum}\left(P\right)$
$\left[\left[{-3}{,}{5}\right]{,}\left[{3}{,}{5}\right]\right]$ (4) | 401 | 1,286 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.491799 |
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Topic: ODE treatment of PDE
Replies: 6 Last Post: Mar 7, 2012 8:52 PM
Messages: [ Previous | Next ]
mecej4 Posts: 3 Registered: 3/6/12
Re: ODE treatment of PDE
Posted: Mar 6, 2012 12:16 PM
On 2/12/2012 3:28 PM, Gib Bogle wrote:
> I'm showing my ignorance by asking this question, but here goes...
>
> It's possible to represent a discretised PDE by a system of ODEs. For
> example, the simple 1D diffusion equation
>
> dC/dt = k.d2C/dx2 (where derivatives are partial)
>
> can be represented on a grid of points with spacing h by
>
> dC(i)/dt = K.(C(i-1) -2.C(i) + C(i+1))/h^2
>
> With suitable treatment of initial and boundary conditions, this ODE
> system could be solved with one of several methods.
>
> Is there a significant difficulty with this approach?
It is more a question of efficiency rather than difficulty. The spatial
discretization has error O(h^2); if mixed boundary conditions apply and
are not discretized properly, the error can degenerate to O(h). In such
circumstances, using a fourth order ODE solver such as RK4 is overkill.
If the equation being solved has a source term that depends on C, the
problem is nonlinear and iterations become necessary. In such cases,
efficiency becomes important.
For these reason, it is more common to use first order integration
schemes (Forward and Backward Euler) or second order (Midpoint)
time-differencing than higher order schemes such as RK4.
-- mecej4 | 440 | 1,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | longest | en | 0.930865 |
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```
While exploring a cave, a spelunker starts at the entrance
and moves the following distances: She goes
75.0 m north, 250 m east, 125 m at an angle 30.0° north
of east, and 150 m south. Find the resultant displacement
from the cave entrance.
```
7 years ago
Sanjana Gupta
16 Points
``` resultant displacement=358.25 i - 12.5 j
magnitude of displacemant will be = 358.46m
direction , theta= -(tan -1^ 0.03)
```
7 years ago
510 Points
``` total displacement is 250.3 metres
```
7 years ago
Think You Can Provide A Better Answer ?
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Anonymous asked in Science & MathematicsMathematics · 4 weeks ago
# Calculating chance with repeated action and multiple changing variables?
Let's say I have a deck of 60 cards, with 6 jokers. If I draw 6 cards, what's the chance of getting all 6 if I repeat the action 6 times? It is my understanding that since we have to assume each draw will net a joker, meaning one less joker the next and one less card in the deck, the math goes like this:
(6/60) (5/59) (4/58) (3/57) (2/56) (1/55) = .0000000199744886 = 00000199744886% | 155 | 539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-34 | latest | en | 0.909394 |
https://activegaliano.org/how-much-was-5-worth-in-1940-update/ | 1,674,809,732,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494974.98/warc/CC-MAIN-20230127065356-20230127095356-00875.warc.gz | 105,870,378 | 21,302 | Home » How Much Was \$5 Worth In 1940? Update
# How Much Was \$5 Worth In 1940? Update
Let’s discuss the question: how much was \$5 worth in 1940. We summarize all relevant answers in section Q&A of website Activegaliano.org in category: Blog Marketing. See more related questions in the comments below.
## How much was \$1 in 1940 worth now?
Value of \$1 from 1940 to 2022
\$1 in 1940 is equivalent in purchasing power to about \$20.54 today, an increase of \$19.54 over 82 years. The dollar had an average inflation rate of 3.75% per year between 1940 and today, producing a cumulative price increase of 1,953.60%.
## How much was \$5 worth in 1942?
\$5 in 1942 is equivalent in purchasing power to about \$88.68 today, an increase of \$83.68 over 80 years. The dollar had an average inflation rate of 3.66% per year between 1942 and today, producing a cumulative price increase of 1,673.67%.
### 1940 NICKEL VALUES – RARE NICKELS WORTH MONEY!!
1940 NICKEL VALUES – RARE NICKELS WORTH MONEY!!
1940 NICKEL VALUES – RARE NICKELS WORTH MONEY!!
## What was the value of \$5 in 1945?
Value of \$5 from 1945 to 2022
\$5 in 1945 is equivalent in purchasing power to about \$79.86 today, an increase of \$74.86 over 77 years. The dollar had an average inflation rate of 3.66% per year between 1945 and today, producing a cumulative price increase of 1,497.24%.
## What would \$5 in 1950 be worth now?
Value of \$5 from 1950 to 2022
\$5 in 1950 is equivalent in purchasing power to about \$59.65 today, an increase of \$54.65 over 72 years. The dollar had an average inflation rate of 3.50% per year between 1950 and today, producing a cumulative price increase of 1,092.96%.
## How much was a loaf of bread in 1940?
YEAR Cost of 1 lb. of Bread
1930 \$0.09
1940 \$0.10
1950 \$0.12
1960 \$0.23
## How much was 25 cents in the 1940s?
Value of \$0.25 from 1940 to 2022
\$0.25 in 1940 is equivalent in purchasing power to about \$5.16 today, an increase of \$4.91 over 82 years.
## What was \$5 worth 1920?
\$5 in 1920 is equivalent in purchasing power to about \$72.28 today, an increase of \$67.28 over 102 years. The dollar had an average inflation rate of 2.65% per year between 1920 and today, producing a cumulative price increase of 1,345.54%.
## What would \$5 be worth today?
This means that today’s prices are 22.82 times higher than average prices since 1800, according to the Bureau of Labor Statistics consumer price index.
Value of \$5 from 1800 to 2022.
Cumulative price change 2,181.78%
Average inflation rate 1.42%
Converted amount (\$5 base) \$114.09
Price difference (\$5 base) \$109.09
## How much was 5 cents 1900?
\$5 in 1900 is equivalent in purchasing power to about \$172.09 today, an increase of \$167.09 over 122 years. The dollar had an average inflation rate of 2.94% per year between 1900 and today, producing a cumulative price increase of 3,341.77%.
### RARE FIVE DOLLAR BILLS WORTH MONEY – MISPRINTED MONEY IN YOUR POCKETS!!
RARE FIVE DOLLAR BILLS WORTH MONEY – MISPRINTED MONEY IN YOUR POCKETS!!
RARE FIVE DOLLAR BILLS WORTH MONEY – MISPRINTED MONEY IN YOUR POCKETS!!
## How much was \$1000 1950?
\$1,000 in 1950 is equivalent in purchasing power to about \$11,929.63 today, an increase of \$10,929.63 over 72 years. The dollar had an average inflation rate of 3.50% per year between 1950 and today, producing a cumulative price increase of 1,092.96%.
## How much was 5 dollars during the Great Depression?
\$5 in 1930 is equivalent in purchasing power to about \$86.56 today, an increase of \$81.56 over 92 years. The dollar had an average inflation rate of 3.15% per year between 1930 and today, producing a cumulative price increase of 1,631.19%.
## How much was 5\$ 1850?
\$5 in 1850 is equivalent in purchasing power to about \$185.33 today, an increase of \$180.33 over 172 years. The dollar had an average inflation rate of 2.12% per year between 1850 and today, producing a cumulative price increase of 3,606.53%.
## How much was a car in 1940?
A new car back in 1940s was about 800 dollars and a gallon of gas was whooping 18 cents. On average most cars got about 15 to 20 miles per gallon.
## How much did a beer cost in 1940?
The 1940s: \$3.55 per Keg.
## How much was minimum wage in 1940?
Actual minimum wage (2016\$)
1940 \$ 4.43
1941 \$ 4.22
1942 \$ 3.81
1943 \$ 3.59
Jul 25, 2016
## What would \$25 in 1940 be today?
Value of \$25 from 1940 to 2022
\$25 in 1940 is equivalent in purchasing power to about \$513.40 today, an increase of \$488.40 over 82 years. The dollar had an average inflation rate of 3.75% per year between 1940 and today, producing a cumulative price increase of 1,953.60%.
## How much was 5 cents 1940?
\$0.05 in 1940 is equivalent in purchasing power to about \$1.03 today, an increase of \$0.98 over 82 years.
### RARE Nickels from 1940 YOU Should Know About!!
RARE Nickels from 1940 YOU Should Know About!!
RARE Nickels from 1940 YOU Should Know About!!
See also How To Install A Trailer Hitch On A Minivan? New
## What was \$2 worth 1920?
\$2 in 1920 is equivalent in purchasing power to about \$28.91 today, an increase of \$26.91 over 102 years. The dollar had an average inflation rate of 2.65% per year between 1920 and today, producing a cumulative price increase of 1,345.54%.
## What was \$1 worth 1920?
Buying power of \$1 in 1920
Initial value Equivalent value
\$1 dollar in 1920 \$12.94 dollars in 2020
\$5 dollars in 1920 \$64.70 dollars in 2020
\$10 dollars in 1920 \$129.41 dollars in 2020
\$50 dollars in 1920 \$647.03 dollars in 2020
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You have just come across an article on the topic how much was \$5 worth in 1940. If you found this article useful, please share it. Thank you very much. | 2,020 | 6,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.927613 |
http://slideplayer.com/slide/4168329/ | 1,526,890,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863967.46/warc/CC-MAIN-20180521063331-20180521083331-00365.warc.gz | 265,765,786 | 20,568 | # CS138A 1999 1 Single Source Shortest Paths Peter Schröder.
## Presentation on theme: "CS138A 1999 1 Single Source Shortest Paths Peter Schröder."— Presentation transcript:
CS138A 1999 1 Single Source Shortest Paths Peter Schröder
CS138A 1999 2 Definitions Shortest path problems weighted directed graph G=(V,E) with real valued weight function w weight of a path v 0,v 1,…,v k is sum of edge costs shortest path negative weight edges negative weight cycles representation of shortest path
CS138A 1999 3 Shortest Paths Predecessor subgraph G p =(V p,E p ) contains all vertices reachable from source and edges induced by predecessor relationship eventually G p will be shortest paths tree shortest paths are not necessarily unique nor are the shortest path trees
CS138A 1999 4 Relaxation Finding shortest paths by relaxation Lemma 1: optimal substructure property; any subpath of a shortest path is itself a shortest path Corollary 2: suppose a shortest path from s to v can decomposed into a path from s to u and the edge (u,v) then the cost of the shortest path from s to v is the sum of the cost of the shortest path from s to u plus w(u,v)
CS138A 1999 5 Relaxation Shortest path properties Lemma 3: Let G=(V,E) be a weighted directed graph with weight function w and source s. Then for all edges (u,v) in E we have all algorithms will use relaxation; each vertex contains a field d[] containing an upper bound for the shortest path length; this will repeatedly be reduced
CS138A 1999 6 Shortest Paths Algorithm 1. Initialize-Single-Source(G,s) 2. for( v in V[G] ) 3. d[v] = infinity; p[v] = nil; 4. d[s] = 0 1. Relax(u,v,w) 2. if( d[v] > d[u] + w(u,v) ) 3. d[v] = d[u] + w(u,v) 4. p[v] = u;
CS138A 1999 7 Algorithms Main differences order in which edges are relaxed number of times each edge is relaxed Properties Lemma 4: Let G=(V,E) be weighted, directed graph with weight function w; then immediately after relaxing (u,v) in E we have
CS138A 1999 8 Relaxation Properties Lemma 5: Let G=(V,E) be a weighted, directed graph with weight function w and source s; after intialization for all v in V and this invariant is maintained during relaxation. Moreover, once d[v] achieves its lower bound it never changes again Corollary 6: suppose a vertex is not reachable; then the above still holds
CS138A 1999 9 Relaxation Properties Lemma 7: Let G=(V,E) bawdgwwfw and source s; let s to u to v be a shortest path in G for some vertices u,v; after initialization suppose a sequence of relaxation steps including Relax(u,v,w) is run; if any time prior to the call then anytime thereafter
CS138A 1999 10 Dijkstra’s Algorithm Single source shortest paths algorithm for a directed, weighted graph G=(V,E) with all edge weights non-negative 1. Dijkstra(G,w,s) 2. Initialize-Single-Source(G,s) 3. S = {}; Q = V[G]; 4. while( !Q.empty() ) 5. u = Q.extractMin(); S = S + {u} 6. for( v in Adj[u] ) 7. Relax(u,v,w)
CS138A 1999 11 Analysis Complexity Correctness greedy strategy… optimal substructure property already established Theorem 10: when running Dijkstra on a weighted directed graph with non-negative weight function w from s then at termination for all u in V
CS138A 1999 12 Bellman-Ford Single source shortest paths algorithm allows for negative weight edges but of course no negative weight cycles 1. Bellmann-Ford(G,s) 2. Initialize-Single-Source(G,s) 3. for( i = 1 to |V[G]|-1 ) 4. for( (u,v) in E[G] ) 5. Relax(u,v,w) 6. for( (u,v) in E[G] ) 7. if( d[v] > d[u] + w(u,v) ) 8. return false 9. return true
CS138A 1999 13 Analysis Complexity Correctness Lemma 12: Let G=(V,E) be a weighted directed graph with source s and weight function w and assume it contains no negative weight cycles reachable from s. Then upon termination of Bellman-Ford we have for all vertices u reachable from s
CS138A 1999 14 Analysis Correctness Theorem 14: Let Bellman-Ford be run on a wdgwssawfw and no negative weight cycles reachable from s then the algorithm returns true and for all u in V and the predecessor subgraph G p is a shortest paths tree rooted at s. If G contains a negative weight cycles reachable from s it will return false.
CS138A 1999 15 Predecessor Graphs Still have to prove things about predecessor trees… Lemma 8: Let G=(V,E) be awdgwssawfw with no negative weight cycles reachable from s. Then after the graph is initialized the predecessor subgraph G p forms a rooted (at s) tree and any sequence of relaxation steps maintains this invariant.
CS138A 1999 16 Predecessor Subgraphs Correctness Lemma 9: Let G=(V,E) be awdgwssawfw with no negative weight cycles reachable from s. After initialization and any relaxation sequence which produces for all u in V, then the predecessor subgraph G p is a shortest paths tree rooted at s.
CS138A 1999 17 All Pairs Shortest paths between all pairs of vertices could run single source algorithms |V| times O(VE ln V) for Dijkstra O(V 2 E) for Bellman-Ford which can be very high… mostly use adjacency matrix representation yield predecessor matrices Example: dynamic programming solution
CS138A 1999 18 Approach Recall dynamic programming characterize the structure of an optimal solution recursively define the value of an optimal solution compute the solution bottom up Structure optimal substructure property (Lemma 1)
CS138A 1999 19 All Pairs 1. Extend-Shortest-Paths(D,W) 2. n = rows[D] 3. for( i = 1 to n ) 4. for( j = 1 to n ) 5. d’ ij = infinity 6. for( k = 1 to n ) 7. d’ = min( d’ ij, d ik + w kj ) 8. return D’ | 1,484 | 5,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.830305 |
https://getrevising.co.uk/revision-cards/redox_reactions_5 | 1,719,003,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00335.warc.gz | 244,271,323 | 12,244 | # Redox Reactions
?
• Created by: Zoe
• Created on: 14-05-13 15:21
## OIL- RIG
• Redox reactions are all around and even inside us, for example, repsiration and photosynthesis.
• Redox Reactions are reactions that involve both oxidation and reduction
• Oxidation is the loss of electrons from an atom or ion
• Reduction is the gain of electrons from an atom or ion.
Oxidation
Is
Loss
Reduction
Is
Gain
1 of 5
## OIL- RIG
• In a redox reaction one atom, ion or molecule will gain electrons from the one that lost it
• Oxidising agent oxidise other chemicals so they themselves are reduced, they are electron acceptors
• Reducing agent reduces other chemicals so they themselves are oxidised as they lose electrons, they are electron donors
2 of 5
## Half Equations
• Half equations show the loss or gain of electrons by one chemical
• Half equations show what chemical is oxidised and which is reduced.
• The electron will appear on the left of the equation when a chemical is reduced and on the right of the equation if oxidised. An easy way to remember this is to think there are a reduced number of left handers in the world.
• Half equations only show half the story, they show how each chemical is effected within the Redox Reaction
• Spectator Ions- Ions that take no part in the reaction, they are unchanged through the reaction they simple "watch" the reaction take place.
3 of 5
## Combining Half equations
• When you are given two half-equations you will need to join them together.
• The aim is to make the same number of electrons equal the same as each other.
How to combine the half equations
• Step 1: Write the two half equations
• Step 2: Take a note of how many electrons are lost or gained in each half equations and multiply either the one or both equations to make them both have the same number of electrons in their equations
• Step 3: Now multiply the number of products and reactants in each equation by the number that you multiplied the electrons with.
• Step 4: Now combine the equations together and cancel out the electrons as the number of electrons should be the same in each equation.
4 of 5
## Oxidation States
• An oxidation state is the number of electrons needed to be gained or lost to make a neutral atom.
• Oxidation states help to work out how oxidized or how reduced something is. It is similar to the charge on the ions, except that it is used for covalent compounds.
Working out oxidation states
Rules:
• Elements always have an oxidation state of 0
• In a compound the sum of the oxidation numbers equals zero
• In an ion, the sum of the oxidation numbers equals the charge
5 of 5 | 619 | 2,647 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.931883 |
https://www.computerhope.com/jargon/a/asterisk.htm | 1,660,850,515,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00377.warc.gz | 623,771,534 | 5,479 | # Asterisk
Updated: 06/02/2020 by Computer Hope
Sometimes called a star, big dot, and multiplication symbol, the asterisk is a symbol ( * ) found above the "8" key on standard US keyboards and on the number pad.
## Where is the asterisk key on the keyboard?
Below is an overview of a computer keyboard with the asterisk key highlighted in blue on the main keyboard and the numeric keypad.
## How to create the * symbol
### Creating the * symbol on a U.S. keyboard
To create an asterisk using a U.S. keyboard, hold down the Shift and press 8 on the keyboard.
Tip
The asterisk is easier to use on the numeric keypad with math formulas.
### Creating the * symbol on a smartphone or tablet
To create an asterisk on a smartphone or tablet, open the keyboard, go into the numbers (123) section, and then the (#+=) or symbols (sym) section. Press your finger on the * symbol.
## What is an asterisk used for on a computer?
Below are examples of how the asterisk is used on a computer.
### An asterisk in a math formula
In a math formula, an asterisk is used to represent a multiplication (times). For example, in the below formula you can see that 5 <times> 2 is equal to 10.
`5 * 2 = 10`
### An asterisk in a wildcard
The asterisk represents a wildcard or wild character with computers. For example, when typing *.TXT in a search or find field, the computer would look for any file ending with .TXT.
### An asterisk in computer programming
Another example of an asterisk is in the nonexecutable statement used with some programming languages. For example, in the following PHP code, the comment "hidden text" would not show when executed because it is surrounded by a forward slash and an asterisk.
```<?php
/* hidden text */
echo "hello world";
?>```
In Python and other programming languages, a ** (double asterisk) can be used as an operator to calculate the power of a number. For example, i**j would be "i" to the power of "j" in the script.
### A star symbol in a spreadsheet program
In spreadsheet programs like Microsoft Excel, a star symbol (asterisk) is used in formulas for multiplication. For example, in the following formula cell A1 is multiplied by cell B1.
`=sum(A1*B1)`
### An asterisk as a reference
Finally, an asterisk is also commonly used as a reference mark in a document. | 546 | 2,316 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-33 | longest | en | 0.845553 |
https://physics.stackexchange.com/questions/372765/is-angular-momentum-conserved-in-this-system | 1,560,955,594,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999000.76/warc/CC-MAIN-20190619143832-20190619165832-00112.warc.gz | 546,938,559 | 36,152 | # Is angular momentum conserved in this system?
Quick question, I'm a bit confused by this problem:
A mass $m$ attached to a string is moving horizontally with constant velocity: Clearly angular momentum with respect to $M$ is conserved. Is angular momentum with respect to point A conserved as well in this setup? I'd say it isn't because due to the momentum pointing tangentially along the circular path of the mass and $l \times p$ changes direction whilst the mass $m$ is moving. So what's the torque acting on the mass? Is it the constraint force (cross $r$) of the string?
You have correctly pointed out that angular momentum about A (red arrow) will change. Forces on mass m are Tension and Gravity. Now if you consider torque about point A then torque due to tension becomes zero because tension force is parallel to l but torque due to gravity is non zero and that torque provides the change in angular momentum.
$\vec{\tau}= \vec{l} \times m\vec{g}$
Torque due to gravity about point A is along the tangent of the circle (green arrow). So, torque will only induce a change of angular momentum only along the tangential direction, so vertical component of angular momentum doesn't change, only horizontal component changes. The left side of the figure shows this clearly.
Let's call $\vec h=AM$, such that $\vec l=\vec h+\vec r$. Then the torque $$\vec \tau=\frac{d}{dt}(\vec l\times \vec p)=\frac{d}{dt}((\vec h+\vec r)\times \vec p)=\vec h\times \frac{d}{dt}\vec p+ \frac{d}{dt}(\vec r\times \vec p)$$ The last term is zero, since the angular momentum with respect to $M$ is conserved. I've also used the fact that $\vec h$ is a constant. Then, since $\vec p$ is always in the horizontal plane, $d\vec p/dt$ is also in the horizontal plane, so the force you are interested in is the resultant of the tension in the string and gravity.
• The angular velocity with respect to A changes in direction, but the magnitude is constant – Andrei Dec 5 '17 at 18:28
It is clear that the motion of the mass is in horizontal plane & not having a single effect in vertical plane (i.e. no acceleration in vertical plane) If we resolve the tension (T) on the mass due to string.. We get, T sin (alpha) as the centrifugal force ( direction :- radially inwards, in plane of the circle). T cos (alpha) in vertical plane (direction:- upwards ). And we have mg as the gravitational force acting in opposite directions to Tcos(alpha) . Now, As there is no acceleration in vertical plane, the magnitude of Tcos(alpha) will be equal to mag. of mg..
Considering torques of all the present forces about point A on the axis AM, Torque due to centripetal force Tsin(alpha) will be ZERO. In vertical plane, magnitudes of torques due to Tcos(alpha) & mg will be equal to Tcos(alpha)r & mgr respectively.. Tcos(alpha) = mg.. so both torques will be of equal magnitude and will be in opposite directions cancelling each other...
So, No torque acts due to forces in vertical plane.
Hence, No torque is acting about point A on axis AM..... .. angular momentum is conserved...
• If we consider an axis passing through point A & parallel to plane of horizontal circle, the only torque acting will be due to centripetal force (Tsin(alpha)) will be present & will vary in it's directions.. So will the angular momentum be conserved about this axis???? – user177698 Dec 5 '17 at 19:24 | 816 | 3,369 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-26 | longest | en | 0.896816 |
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Converting hours over 24 into decimal format
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## Converting hours over 24 into decimal format
#1
July 4th 07, 11:42 AM posted to microsoft.public.excel.worksheet.functions
Ary external usenet poster Posts: 2
Converting hours over 24 into decimal format
How would I convert hours such as 128:24:12 to 128.403?
In previous discussions, I saw someone suggest formatting the cell as
[hh]:mm, but I have no idea what this means nor how to do it.
--
-Ary
#2
July 4th 07, 11:48 AM posted to microsoft.public.excel.worksheet.functions
Pete_UK external usenet poster Posts: 8,856
Converting hours over 24 into decimal format
If you want them in decimal format then multiply by 24:
=A1*24
and format the cell as number with appropriate decimal places.
If you want to see them in time format, then click Format | Cells |
Custom and enter [hh]:mm:ss in the panel then click OK.
Hope this helps.
Pete
On Jul 4, 11:42 am, Ary > wrote:
> How would I convert hours such as 128:24:12 to 128.403?
>
> In previous discussions, I saw someone suggest formatting the cell as
> [hh]:mm, but I have no idea what this means nor how to do it.
> --
> -Ary
#3
July 4th 07, 11:56 AM posted to microsoft.public.excel.worksheet.functions
Mike H external usenet poster Posts: 11,501
Converting hours over 24 into decimal format
Multiply time * 24
select your cell and then format:cells:custom and enter a format of [hh]:mm:ss
Enter your time in that cell 18:22:13 (say cell A1)
In b1 typr =A1*24
Change thje format to general and your done.
Mike
"Ary" wrote:
> How would I convert hours such as 128:24:12 to 128.403?
>
> In previous discussions, I saw someone suggest formatting the cell as
> [hh]:mm, but I have no idea what this means nor how to do it.
> --
> -Ary
#4
July 4th 07, 11:56 AM posted to microsoft.public.excel.worksheet.functions
Toppers external usenet poster Posts: 4,340
Converting hours over 24 into decimal format
Multiply by 24 and format cell as General.
=A1*24
a1=128:24:12 (formatted as [h]:mm:ss)
Format of [h]:mm:ss allows hours of > 24 to be displayed. 128:24:12 with
hh:mm:ss would be display as 8:24:12 [removing 120 (5*24 hours)].
"Ary" wrote:
> How would I convert hours such as 128:24:12 to 128.403?
>
> In previous discussions, I saw someone suggest formatting the cell as
> [hh]:mm, but I have no idea what this means nor how to do it.
> --
> -Ary
#5
July 4th 07, 12:10 PM posted to microsoft.public.excel.worksheet.functions
Ary external usenet poster Posts: 2
Converting hours over 24 into decimal format
I don't think I understand. When I do this, I get a completely different time:
A1 A2 =A1*24
126:47:59 19:11:36
I need it to read 126:47:59 and give me 126.7997
--
-Ary
"Pete_UK" wrote:
> If you want them in decimal format then multiply by 24:
>
> =A1*24
>
> and format the cell as number with appropriate decimal places.
>
> If you want to see them in time format, then click Format | Cells |
> Custom and enter [hh]:mm:ss in the panel then click OK.
>
> Hope this helps.
>
> Pete
>
> On Jul 4, 11:42 am, Ary > wrote:
> > How would I convert hours such as 128:24:12 to 128.403?
> >
> > In previous discussions, I saw someone suggest formatting the cell as
> > [hh]:mm, but I have no idea what this means nor how to do it.
> > --
> > -Ary
>
>
>
#6
July 4th 07, 12:26 PM posted to microsoft.public.excel.worksheet.functions
Toppers external usenet poster Posts: 4,340
Converting hours over 24 into decimal format
format A2 as GENERAL
"Ary" wrote:
> I don't think I understand. When I do this, I get a completely different time:
>
> A1 A2 =A1*24
> 126:47:59 19:11:36
>
> I need it to read 126:47:59 and give me 126.7997
> --
> -Ary
>
>
> "Pete_UK" wrote:
>
> > If you want them in decimal format then multiply by 24:
> >
> > =A1*24
> >
> > and format the cell as number with appropriate decimal places.
> >
> > If you want to see them in time format, then click Format | Cells |
> > Custom and enter [hh]:mm:ss in the panel then click OK.
> >
> > Hope this helps.
> >
> > Pete
> >
> > On Jul 4, 11:42 am, Ary > wrote:
> > > How would I convert hours such as 128:24:12 to 128.403?
> > >
> > > In previous discussions, I saw someone suggest formatting the cell as
> > > [hh]:mm, but I have no idea what this means nor how to do it.
> > > --
> > > -Ary
> >
> >
> >
#7
July 4th 07, 12:34 PM posted to microsoft.public.excel.worksheet.functions
Toppers external usenet poster Posts: 4,340
Converting hours over 24 into decimal format
"Toppers" wrote:
> format A2 as GENERAL
>
> "Ary" wrote:
>
> > I don't think I understand. When I do this, I get a completely different time:
> >
> > A1 A2 =A1*24
> > 126:47:59 19:11:36
> >
> > I need it to read 126:47:59 and give me 126.7997
> > --
> > -Ary
> >
> >
> > "Pete_UK" wrote:
> >
> > > If you want them in decimal format then multiply by 24:
> > >
> > > =A1*24
> > >
> > > and format the cell as number with appropriate decimal places.
> > >
> > > If you want to see them in time format, then click Format | Cells |
> > > Custom and enter [hh]:mm:ss in the panel then click OK.
> > >
> > > Hope this helps.
> > >
> > > Pete
> > >
> > > On Jul 4, 11:42 am, Ary > wrote:
> > > > How would I convert hours such as 128:24:12 to 128.403?
> > > >
> > > > In previous discussions, I saw someone suggest formatting the cell as
> > > > [hh]:mm, but I have no idea what this means nor how to do it.
> > > > --
> > > > -Ary
> > >
> > >
> > >
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All times are GMT +1. The time now is 03:21 PM. | 2,051 | 7,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | longest | en | 0.889002 |
http://www.nytimes.com/2000/11/10/arts/a-rendezvous-with-2-giants.html?pagewanted=3&src=pm | 1,472,633,172,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471983580563.99/warc/CC-MAIN-20160823201940-00114-ip-10-153-172-175.ec2.internal.warc.gz | 514,700,057 | 12,860 | (Page 3 of 3)
According to the ''Great Trees'' guide, the Staten Island tree is 146 feet high. If true, it would easily be the victor over the Queens Giant. But Mr. Leverett is an expert at busting overblown claims.
''We're trying to bring truth into the big-tree numbers,'' he said. The big-tree-hunting world, it turned out, is rife with inaccurate measurements. But no arboreal claimant can hide from Mr. Leverett's laser range finder. For example, he and his colleagues at the Eastern Native Tree Society discovered that a red oak in Michigan, which was listed as the state champion, was overestimated by 90 feet. ''Ninety feet, that's a whole tree,'' he said.
The Staten Island tree, which we dubbed the Clove Lakes colossus, was clearly younger than its Queens rival, and it had had the benefit of little competition. Whereas the Queens giant was losing its crown, struggling to get enough sun, the colossus was lord of the lawn, spreading out in every direction with abandon. The only hassle it appeared to face was children, running about on its massive buttressed trunk.
Mr. Leverett measured the circumference of the trunk. He hooked the tape to the bark and vanished for what seemed to be a long time as he made his way around. At 20 feet, the tape was not long enough this time, and we had to put a finger on the spot so he could measure the remainder. It was a whopping 21.4 feet around, bigger in circumference than the Queens tree.
Walking backward across the lawn, trying to get a bead on the tree's leader, Mr. Leverett commented on how easy it was to measure a tree in an open field. ''It's almost like shooting fish in a barrel,'' he said.
He lasered the tree with his range finder and worked his mathematical magic with the clinometer and calculator. ''The height,'' he announced, as we waited anxiously, ''is 119 feet.'' That's 27 feet shorter than the height advertised in the ''Great Trees'' guide, but, more importantly, 12.2 feet shorter than the Queens' giant.
However, Mr. Kershner pointed out that the colossus had more limbs and a more massive trunk. And we had to admit that the trunk was overwhelming. Mr. Leverett, who's no wood sprite, looked like a finger puppet standing next to it.
But it was all going to come down to calculations he would make later. Height is not the be-all and end-all when it comes to determining a tree's bigness. With more measurements (height to the first bough, crown spread), Mr. Leverett planned to use a mathematical model to estimate the tree's overall volume. ''I have to sit down with a pencil and calculator for an hour or so,'' he said. ''But I can tell you it's going to be a close one.''
We headed our separate ways and waited nervously for the results. The next evening we received word via e-mail. Both trees had an estimated volume of 1,750 cubic feet and weighed in the neighborhood of 50,000 pounds. The Queens tree was probably a bit more voluminous, but the Staten Island tree was slightly heavier.
So what Mr. Leverett was saying was that it was a dead heat. Until further review, we had two trees worthy of being called the New York Giant.
''At this point,'' wrote Mr. Leverett, ''I would call them co-champions. Should you want to take the contest further, we would need to have both trees climbed with periodic girth measurements taken for at least the first 75 feet. Until that is done, I'm willing to call it a draw.''
And so, until some hardy spirit clambers to the top of both these behemoths, bragging rights in this heavyweight-tree contest can be shared by both boroughs. As for the other counties, Manhattan and the Bronx seem to be out of the running and, while trees may grow in Brooklyn, they grow taller in Queens and Staten Island.
Finding the Trees
To reach the Queens giant, a tulip tree measuring 133.8 feet tall and 18.6 feet in circumference, head for a section of Alley Pond Park where the Long Island Expressway and the Cross Island Parkway intersect. At East Hampton Boulevard and the Horace Harding Expressway (a service road of the Long Island Expressway) look for a nearby trail into the woodlands. The tree, which is surrounded by a small fence, is a five-minute walk from the trailhead.
The Clove Lakes colossus, a tulip tree measuring 119 feet tall and 21.4 feet in circumference, is situated in the northernmost part of Staten Island's Clove Lakes Park near the intersection of Forest Avenue and Clove Road.
From Forest Avenue, walk south across the park's lawn for about 200 feet to reach the giant tree.
''Great Trees of New York City'' is a 48-page guide that describes more than 100 city trees of impressive size, age, species, form and historic association. For detailed instructions on measuring big trees, visit the Eastern Native Tree Society's Web site at www.uark.edu/misc/ents.
Photos: The Staten Island tulip tree in Clove Lakes Park. (Bruce Kershner/''Secret Places of Staten Island'')(pg. E31); The king of Queens: the giant tulip tree in Alley Pond Park, a 635-acre swath in the borough. At 133.8 feet, it is as tall as a 13-story building. (Bruce Kershner)(pg. E44) Map of Staten Island highlighting location of Clove Lakes colossus: Map of Queens highlighting location of Queens giant. (pg. E44) | 1,208 | 5,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2016-36 | latest | en | 0.97988 |
https://stats.stackexchange.com/questions/339790/what-are-the-implicit-assumptions-that-justify-the-usage-of-the-relative-error | 1,695,364,630,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506329.15/warc/CC-MAIN-20230922034112-20230922064112-00801.warc.gz | 614,114,195 | 39,103 | # What are the implicit assumptions that justify the usage of the “relative error”?
In applied statistics, for example, analyzing data from a science experiment, we sometimes use "absolute error" while at the most of time, we calculate the "relative error". Generally, under what circumstances should we use the "relative error" instead of "the absolute error"?
If we use relative error, are we basically implicitly assuming that the variance of the sample distribution is positively correlated with the value of the mean? Is there a fundamental axiom in mathematics or statistics called "scale independence"?
It is similar to the financial math literature, the variance of the random variable $X_t$ is usually correlated with the exact value of $X_{t−1}$. If $X_{t−1}$ is greater, the variance of $X_t$ is greater. This often leads to something like a Lognormal distribution.
I think you are getting a bit ahead of yourself. Both absolute error and relative error are measures of error that are useful for different purposes. There are no assumptions required to use either of these measures$^\dagger$, so long as they are correctly interpreted. Of course, the statistical properties of these measures will depend on the underlying distribution of the values being predicted and the predictions being made, so if you want to assert a particular distributional property of one of these measures, you may need to make some assumptions. So, if you are asking what assumptions are needed, then naturally you first need to ask, for what? What statistical property are you seeking?
If you have a prediction $\hat{x}_t$ of an unknown value $x_t$ then the absolute error $\varepsilon_t = |\hat{x}_t - x_t|$ gives a measure of the amount by which you are wrong in your prediction. Often it is the case that this is not particularly useful on its own, since this amount might be large or small in comparison to the size of the thing you are predicting. For this reason, analysts often prefer to measure their error using the relative error $\eta_t = |\hat{x}_t - x_t| / |x_t|$.
$^\dagger$ Strictly speaking, you need to have $x_t \neq 0$ to calculate the relative error, so this could possibly be taken as a required assumption. Alternatively you could merely adopt the convention that $\eta_t = \infty$ when $x_t = 0$. | 499 | 2,314 | {"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} | 3.578125 | 4 | CC-MAIN-2023-40 | latest | en | 0.917374 |
https://www.exactlywhatistime.com/time-ago/74-years-ago-from-today | 1,718,263,589,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00119.warc.gz | 692,634,173 | 7,597 | # What was the date 74 years ago?
## Tuesday June 13, 1950
0
74 years in the past was 13 Jun 1950, a Tuesday. Subtracting 74 years in the past is usually simple. Anything under a decade can usually be counted on one hand. The biggest challenge will be skipping decades behind or even centuries. Additionally, we’re 13 days from the end of June, so being in the middle of of the month, you'll need to consider monthly changes as well. Weekly and daily changes most likely won't impact 74 years ago.
## How we calculated 74 years before today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how 74 years ago gets calculated on each visit:
1. We started with date inputs: used current day of 13 Jun, then set the calculation - 74 years, and factored in the year 2024
2. Noted your current time of year: 74 years in June will bring us back to May or further.
3. Counted backwards years from current day: date - 13 Jun, factoring in the 13 days left in June to calculate Tuesday June 13, 1950
4. Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the fiscal calendar
### Tips when solving for June 13
• Current date: 13 Jun
• Day of the week: Tuesday
• New Date: Tuesday June 13, 1950
• New Date Day of the week: Tuesday
• June is the beginning of summer so be careful, counting backwards could put you in the chilly months of spring and winter.
• The solution crosses into a different year.
## Ways to calculate 74 years ago
1. Calculate it: Start with a time ago calculator. 74 years is easiest solved on a calculator. For ours, we've already factored in the 13 days in June + all number of days in each month and the number of days in . Simply add your years and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use June's calendar: Begin by identifying 13 Jun on a calendar, note that it’s Tuesday, and the total days in May (trust me, you’ll need this for smaller calculations) and days until last year (double trust me, you'll need this for larger calculations). From there, count backwards 74 times years by years, subtracting years from until your remainder of years is 0.
3. Use excel: For more complex years calculations or if you h8 our site (kidding), I use Excel functions like =TODAY()-74 to get or =WORKDAY(TODAY()), -74, cell:cell) for working years.
## Working years in 74 calendar years
74 years is Tuesday June 13, 1950 or could be if you only want workdays. This calculation takes 74 years and only subtracts by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Tuesday June 13, 1950 date.
Work years Solution
Monday
74 years back
Tuesday
June 13
Wednesday
Thursday
Friday
Saturday
Sunday
## The past 74 years is equivalent to:
Counting back from today is Tuesday June 13, 1950 using a full calendar, and is also 648240 hours ago and 44.93% of the year.
Did you know?
Tuesday Tuesday June 13, 1950 was the 164 day of the year. At that time, it was 44.93% through 1950.
## In 74 years, the average person Spent...
• 5801748.0 hours Sleeping
• 771405.6 hours Eating and drinking
• 1264068.0 hours Household activities
• 375979.2 hours Housework
• 414873.6 hours Food preparation and cleanup
• 129648.0 hours Lawn and garden care
• 2268840.0 hours Working and work-related activities
• 2087332.8 hours Working
• 3416224.8 hours Leisure and sports
• 1853966.4 hours Watching television
## What happened on June 13 (74 years ago) over the years?
### On June 13:
• 1850 Poet Alfred Tennyson (40) weds Emily Sellwood (36) | 956 | 3,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-26 | latest | en | 0.950435 |
https://jeremykun.com/tag/categories/ | 1,696,458,898,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00064.warc.gz | 360,357,267 | 45,893 | # The Universal Properties of Map, Fold, and Filter
A lot of people who like functional programming often give the reason that the functional style is simply more elegant than the imperative style. When compelled or inspired to explain (as I did in my old post, How I Learned to Love Functional Programming), they often point to the three “higher-order” functions map, fold, and filter, as providing a unifying framework for writing and reasoning about programs. But how unifying are they, really? In this post we’ll give characterizations of these functions in terms of universal properties, and we’ll see that in fact fold is the “most” universal of the three, and its natural generalization gives a characterization of transformations of standard compound data types.
By being universal or having a universal property, we don’t mean that map, fold, and filter are somehow mystically connected to all programming paradigms. That might be true, but it’s not the point of this article. Rather, by saying something has a universal property we are making a precise mathematical statement that it is either an initial or final object (or the unique morphism determined by such) in some category.
That means that, as a fair warning to the reader, this post assumes some basic knowledge of category theory, but no more than what we’ve covered on this blog in the past. Of particular importance for the first half of this post is the concept of a universal property, and in the followup post we’ll need some basic knowledge of functors.
## Map, Filter, and Fold
Recalling their basic definitions, map is a function which accepts as input a list $L$ whose elements all have the same type (call it $A$), and a function $f$ which maps $A$ to another type $B$. Map then applies $f$ to every entry of $L$ to produce a new list whose entries all have type $B$.
In most languages, implementing the map function is an elementary exercise. Here is one possible definition in ML.
fun map(_, nil) = nil
| map(f, (head::tail)) = f(head) :: map(f, tail)
Next, filter takes as input a boolean-valued predicate $p : A \to \textup{bool}$ on types $A$ and a list $L$ whose entries have type $A$, and produces a list of those entries of $L$ which satisfy the predicate. Again, it’s implementation in ML might be:
fun filter(_, nil) = nil
| filter(p, (head::tail)) = if p(head)
then (head :: filter(p, tail))
else filter(p, tail)
Finally, fold is a function which “reduces” a list $L$ of entries with type $A$ down to a single value of type $B$. It accepts as input a function $f : A \times B \to B$, and an initial value $v \in B$, and produces a value of type $B$ by recursively applying $f$ as follows:
fun fold(_, v, nil) = v
| fold(f, v, (head::tail)) = f(head, fold(f, v, tail))
(If this definition is mysterious, you’re probably not ready for the rest of this post.)
One thing that’s nice about fold is that we can define map and filter in terms of it:
fun map(f, L) = fold((fn x, xs => (f(x):xs)), [], L)
fun filter(p, L) = fold((fn x, xs => if p(x) then x:xs else xs end), [], L)
We’ll see that this is no accident: fold happens to have the “most general” universality of the three functions, but as a warm-up and a reminder we first investigate the universality of map and filter.
## Free Monoids and Map
Map is the easiest function of the three to analyze, but to understand it we need to understand something about monoids. A monoid is simple to describe, it’s just a set $M$ with an associative binary operation denoted by multiplication and an identity element for that operation.
monoid homomoprhism between two monoids $M, N$ is a function $f: M \to N$ such that $f(xy) = f(x)f(y)$. Here the multiplication on the left hand side of the equality is the operation from $M$ and on the right it’s the one from $N$ (this is the same definition as a group homomorphism). As is to be expected, monoids with monoid homomorphisms form a category.
We encounter simple monoids every day in programming which elucidate this definition. Strings with the operation of string concatenation form a monoid, and the empty string acts as an identity because concatenating a string to the empty string has no effect. Similarly, lists with list concatenation form a monoid, where the identity is the empty list. A nice example of a monoid homomorphism is the length function. We know it’s a homomorphism because the length of a concatenation of two lists is just the sum of the lengths of the two lists.
Integers also form a monoid, with addition as the operation and zero as the identity element. However, the list and string monoids have an extra special property that integers do not. For a number $n$ you can always find $-n$ so that $n + (-n) = 0$ is the identity element. But for lists, the only way to concatenate two lists and get the empty list is if both of the lists were empty to begin with. A monoid with this property is called free, and to fully understand it we should make a definition which won’t seem to mean anything at first.
Definition: Let $\mathbf{C}$ be a category. Given a set $A$, the free object over $A$, denoted $F(A)$, is an object of $\mathbf{C}$ which is universal with respect to set-maps $A \to B$ for any object $B$ in $\mathbf{C}$.
As usual with a definition by universal property, we should elaborate as to exactly what’s going on. Let $\mathbf{C}$ be a category whose objects are sets and let $A$ a set, possibly not in this category. We can construct a new category, $\mathbf{C}_{A \to X}$ whose objects are set-maps
$\displaystyle f: A \to X$
and whose morphisms are commutative diagrams of the form
where $\varphi$ is a morphism in $\mathbf{C}$. In other words, we simply require that $\varphi(f_1(a)) = f_2(a)$ for every $a \in A$.
By saying that the free monoid on $A$ satisfies this universal property for the category of monoids, we really mean that it is initial in this category of set-maps and commutative diagrams. That is, there is a monoid $F(A)$ and a set-map $i_A: A \to F(A)$ so that for every monoid $Y$ and set-map $f: A \to Y$ there is a unique monoid homomorphism from $i_A$ to $f$. In other words, there is a unique monoid homomorphism $\varphi$ making the following diagram commute:
For example, if $A$ is the set of all unicode characters, then $F(A)$ is precisely the monoid of all strings on those characters. The map $i_A$ is then the map taking a character $a$ to the single-character string “$a$”. More generally, if $T$ is any type in the category of types and computable functions, then $F(T)$ is the type of homogeneous lists whose elements have type $T$.
We will now restrict our attention to lists and types, and we will denote the free (list) monoid on a type $A$ as $[A]$. The universal property of map is then easy to describe, it’s just a specific instance of this more general universal property for the free monoids. Specifically, we work in the sub-category of homogeneous list monoids (we only allow objects which are $[T]$ for some type $T$). The morphism $i_A: A \to [A]$ just takes a value $a$ of type $A$ and spits out a single-item list $[a]$. We might hope that for any mapping $A \to [B]$, there is a unique monoid homomorphism $[A] \to [B]$ making the following diagram commute.
And while this is true, the diagram lies because “map” does not achieve what $\varphi$ does. The map $f$ might send all of $A$ to the empty list, and this would cause $\varphi$ to be the trivial map. But if we decomposed $f$ further to require it to send $a$ to a single $b$, such as
then things work out nicely. In particular, specifying a function $f: A \to B$ uniquely defines how a list homomorphism operates on lists. In particular, the diagram forces the behavior for single-element lists: $[a] \mapsto [f(a)]$. And the property of $\varphi$ being a monoid homomorphism forces $\varphi$ to preserve order.
Indeed, we’ve learned from this analysis that the structure of the list involved is crucial in forming the universal property. Map is far from the only computable function making the diagram commute, but it is clearly the only monoid homomorphism. As we’ll see, specifying the order of operations is more generally described by fold, and we’ll be able to restate the universal property of map without relying on monoid homomorphisms.
## Filter
The filter function is a small step up in complexity from map, but its universal property is almost exactly the same. Again filter can be viewed through the lens of a universal property for list monoids, because filter itself takes a predicate and produces a list monoid. We know this because filtering two lists by the same predicate and concatenating them is the same as concatenating them first and then filtering them. Indeed, the only difference here is that the diagram now looks like this
where $p$ is our predicate, and $j_A$ is defined by $(a, T) \mapsto [a]$ and $(a,F) \mapsto []$. The composition $j_A \circ p$ gives the map $A \to [A]$ that we can extend uniquely to a monoid homomorphism $[A] \to [A]$. We won’t say any more on it, except to mention that again maintaining the order of the list is better described via fold.
## Fold, and its Universal Property
Fold is different from map and filter in a very concrete way. Even though map and filter do specify that the order of the list should be preserved, it’s not an important part of their definition: filter should filter, and map should map, but no interaction occurs between different entries during the computation. In a sense, we got lucky that having everything be monoid homomorphisms resulted in preserving order without our specific intention.
Because it doesn’t necessarily produce lists and the operation can be quite weird, fold cannot exist without an explicit description of the order of operations. Let’s recall the definition of fold, and stick to the “right associative” order of operations sometimes specified by calling the function “foldr.” We will use foldr and fold interchangeably.
fun foldr(_, v, nil) = v
| foldr(f, v, (head::tail)) = f(head, foldr(f, v, tail))
Even more, we can’t talk about foldr in terms of monoids. Even after fixing $f$ and $v$, foldr need not produce a monoid homomorphism. So if we’re going to find a universal property of foldr, we’ll need a more general categorical picture.
One first try would be to view foldr as a morphism of sets
$\displaystyle B \times \textup{Hom}(A \times B, B) \to \textup{Hom}([A], B)$
The mapping is just $f,b \mapsto \textup{foldr } f \textup{ } b$, and this is just the code definition we gave above. One might hope that this mapping defines an isomorphism in the category of types and programs, but it’s easy to see that it does not. For example, let
$A = \left \{ 1 \right \}, B = \left \{ 1,2 \right \}$
Then the left hand side of the mapping above is a set of size 8 (there are eight ways to combine a choice of element in $B$ with a map from $A \times B \to B$). But the right hand size is clearly infinite. In fact, it’s uncountably infinite, though not all possible mappings are realizable in programs of a reasonable length (in fact, very few are). So the morphism can’t possibly be a surjective and hence is not an isomorphism.
So what can we say about fold? The answer is a bit abstract, but it works out nicely.
Say we fix a type for our lists, $A$. We can define a category $\mathbf{C}_A$ which has as objects the following morphisms
By $1$ we mean the type with one value (null, if you wish), and $f$ is a morphism from a coproduct (i.e. there are implicit parentheses around $A \times B$). As we saw in our post on universal properties, a morphism from a coproduct is the same thing as a pair of functions which operates on each piece. Here one operates on $1$ and the other on $A \times B$. Since morphisms from $1$ are specified by the image of the sole value $f(1) = b$, we will often write such a function as $b \amalg h$, where $h: A \times B \to B$.
Now the morphisms in $\mathbf{C}_A$ are the pairs of morphisms $\varphi, \overline{\varphi}$ which make the following diagram commute:
Here we write $\overline{\varphi}$ because it is determined by $\varphi$ in a canonical way. It maps $1 \to 1$ in the only way one can, and it maps $(a,b) \mapsto (a, \varphi(b))$. So we’re really specifying $\varphi$ alone, but as we’ll see it’s necessary to include $\overline{\varphi}$ as well; it will provide the “inductive step” in the definition of fold.
Now it turns out that fold satisfies the universal property of being initial in this category. Well, not quite. We’re saying first that the following object $\mathscr{A}$ is initial,
Where cons is the list constructor $A \times [A] \to [A]$ which sends $(a_0, [a_1, \dots, a_n]) \mapsto [a_0, a_1, \dots, a_n]$. By being initial, we mean that for any object $\mathscr{B}$ given by the morphism $b \amalg g: 1 \coprod A \times B \to B$, there is a unique morphism from $\mathscr{A} \to \mathscr{B}$. The “fold” is precisely this unique morphism, which we denote by $(\textup{fold g, b})$. We implicitly know its “barred” counterpart making the following diagram commute.
This diagram has a lot going on in it, so let’s go ahead and recap. The left column represents an object $\mathscr{A}$ we’re claiming is initial in this crazy category we’ve made. The right hand side is an arbitrary object $\mathscr{B}$, which is equivalently the data of an element $b \in B$ and a mapping $g: A \times B \to B$. This is precisely the data needed to define fold. The dashed lines represent the unique morphisms making the diagram commute, whose existence and uniqueness is the defining property of what it means for an object to be initial in this category. Finally, we’re claiming that foldr, when we fix $g$ and $b$, makes this diagram commute, and is hence the very same unique morphism we seek.
To prove all of this, we need to first show that the object $\mathscr{A}$ is initial. That is, that any two morphisms we pick from $\mathscr{A} \to \mathscr{B}$ must be equal. The first thing to notice is that the two objects $1 \coprod A \times [A]$ and $[A]$ are really the same thing. That is, $\textup{nil} \amalg \textup{cons}$ has a two-sided inverse which makes it an isomorphism in the category of types. Specifically, the inverse is the map $\textup{split}$ sending
$\textup{nil} \mapsto \textup{nil}$
$[a_1, \dots, a_n] \mapsto (a_1, [a_2, \dots, a_n])$
So if we have a morphism $\varphi, \overline{\varphi}$ from $\mathscr{A} \to \mathscr{B}$, and the diagram commutes, we can see that $\varphi = (b \amalg g) \circ \overline{\varphi} \circ \textup{split}$. We’re just going the long way around the diagram.
Supposing that we have two such morphisms $\varphi_1, \varphi_2$, we can prove they are equal by induction on the length of the input list. It is trivial to see that they both must send the empty list to $b$. Now suppose that for lists of length $n-1$ the two are equal. Given a list $[a_1, \dots, a_n]$ we see that
$\displaystyle \varphi_1([a_1, \dots, a_n]) = \varphi_1 \circ \textup{cons} (a_1, [a_2, \dots, a_n]) = g \circ \overline{\varphi_1} (a_1, [a_2, \dots, a_n])$
where the last equality holds by the commutativity of the diagram. Continuing,
$\displaystyle g \circ \overline{\varphi_1} (a_1, [a_2, \dots, a_n]) = g (a_1, \varphi_1([a_2, \dots, a_n])) = g(a_1, \varphi_2(a_2, \dots, a_n))$
where the last equality holds by the inductive hypothesis. From here, we can reverse the equalities using $\varphi_2$ and it’s “barred” version to get back to $\varphi_2([a_1, \dots, a_n])$, proving the equality.
To show that fold actually makes the diagram commute is even simpler. In order to make the diagram commute we need to send the empty list to $b$, and we need to inductively send $[a_1, \dots, a_n]$ to $g(a_1, (\textup{fold g b})([a_2, \dots, a_n]))$, but this is the very definition of foldr!
So we’ve established that fold has this universal property. It’s easy now to see how map and filter fit into the picture. For mapping types $A$ to $C$ via $f$, just use the type $[C]$ in place of $B$ above, and have $g(a, L) = \textup{cons}(f(a), L)$, and have $b$ be the empty list. Filter similarly just needs a special definition for $g$.
A skeptical reader might ask: what does all of this give us? It’s a good question, and one that shouldn’t be taken lightly because I don’t have an immediate answer. I do believe that with some extra work we could use universal properties to give a trivial proof of the third homomorphism theorem for lists, which says that any function expressible as both a foldr and a foldl can be expressed as a list homomorphism. The proof would involve formulating a universal property for foldl, which is very similar to the one for foldr, and attaching the diagrams in a clever way to give the universal property of a monoid homomorphism for lists. Caveat emptor: this author has not written such a proof, but it seems likely that it would work.
More generally, any time we can represent the requirements of a list computation by an object like $\mathscr{B}$, we can represent the computation as a foldr. What good does this do us? Well, it might already be obvious when you can and can’t use fold. But in addition to proving when it’s possible to use fold, this new perspective generalizes very nicely to give us a characterization of arbitrary computations on compound data types. One might want to know when to perform fold-like operations on trees, or other sufficiently complicated beasts, and the universal property gives us a way to see when such a computation is possible.
That’s right, I said it: there’s more to the world than lists. Shun me if you must, but I will continue dream of great things.
In an effort to make the egregiously long posts on this blog slightly shorter, we’ll postpone our generalization of the universal property of fold until next time. There we’ll define “initial algebras” and show how to characterize “fold-like” computations any compound data type.
Until then!
# Functoriality
Last time we worked through some basic examples of universal properties, specifically singling out quotients, products, and coproducts. There are many many more universal properties that we will mention as we encounter them, but there is one crucial topic in category theory that we have only hinted at: functoriality.
As we’ve repeatedly stressed, the meat of category theory is in the morphisms. One natural question one might ask is, what notion of morphism is there between categories themselves? Indeed, the most straightforward way to see category theoretic concepts in classical mathematics is in a clever choice of functor. For example (and this example isn’t necessary for the rest of the article) one can “associate” to each topological space a group, called the homology group, in such a way that continuous functions on topological spaces translate to group homomorphisms. Moreover, this translation is functorial in the following sense: the group homomorphism associated to a composition is the composition of the associated group homomorphisms. If we denote the association by a subscripted asterisk, then we get the following common formula.
$\displaystyle (fg)_* = f_* g_*$
This is the crucial property that maintains the structure of morphisms. Again, this should reinforce the idea that the crucial ingredient of every definition in category theory is its effect on morphisms.
## Functors: a Definition
In complete generality, a functor is a mapping between two categories which preserves the structure of morphisms. Formally,
Definition: Let $\mathbf{C}, \mathbf{D}$ be categories. A functor $\mathscr{F}$ consists of two parts:
• For each object $C \in \mathbf{C}$ an associated object $\mathscr{F}(C) \in \mathbf{D}$.
• For each morphism $f \in \textup{Hom}_{\mathbf{C}}(A,B)$ a corresponding morphism $\mathscr{F}(f) \in \textup{Hom}_{\mathbf{D}}(\mathscr{F}(A), \mathscr{F}(B))$. Specifically, for each $A,B$ we have a set-function $\textup{Hom}_{\mathbf{C}}(A,B) \to \textup{Hom}_{\mathbf{C}}(\mathscr{F}(A), \mathscr{F}(B))$.
There are two properties that a functor needs to satisfy to “preserve structure.” The first is that the identity morphisms are preserved for every object; that is, $\mathscr{F}(1_A) = 1_{\mathscr{F}(A)}$ for every object $A$. Second, composition must be preserved. That is, if $f \in \textup{Hom}_{\mathbf{C}}(A,B)$ and $g \in \textup{Hom}_{\mathbf{C}}(B,C)$, we have
$\displaystyle \mathscr{F}(gf) = \mathscr{F}(g) \mathscr{F}(f)$
We often denote a functor as we would a function $\mathscr{F}: \mathbf{C} \to \mathbf{D}$, and use the function application notation as if everything were with sets.
Let’s look at a few simple examples.
Let $\mathbf{FiniteSet}$ be the poset category of finite sets with subsets as morphisms, and let $\mathbf{Int}$ be the category whose objects are integers where there is a unique morphism from $x \to y$ if $x \leq y$. Then the size function is a functor $\mathbf{Set} \to \mathbf{Int}$. Continuing with $\mathbf{Int}$, remember that $\mathbf{Int}$ forms a group under addition (also known as $\mathbb{Z}$). And so by its very definition any group homomorphism $\mathbb{Z} \to \mathbb{Z}$ is a functor from $\mathbf{Int} \to \mathbf{Int}$. A functor from a category to itself is often called an endofunctor.
There are many examples of functors from the category $\mathbf{Top}$ of topological spaces to the category $\mathbf{Grp}$ of groups. These include some examples we’ve seen on this blog, such as the fundamental group and homology groups.
One trivial example of a functor is called the forgetful functor. Let $\mathbf{C}$ be a category whose objects are sets and whose morphisms are set-maps with additional structure, for example the category of groups. Define a functor $\mathbf{C} \to \mathbf{Set}$ which acts as the identity on both sets and functions. This functor simply “forgets” the structure in $\mathbf{C}$. In the realm of programs and types (this author is thinking Java), one can imagine this as a ‘type-cast’ from String to Object. In the same vein, one could define an “identity” endofunctor $\mathbf{C} \to \mathbf{C}$ which does absolutely nothing.
One interesting way to think of a functor is as a “realization” of one category inside another. In particular, because the composition structure of $\mathbf{C}$ is preserved by a functor $\mathbf{C} \to \mathbf{D}$, it must be the case that all commutative diagrams are “sent” to commutative diagrams. In addition, isomorphisms are sent to isomorphisms because if $f, g$ are inverses of each other, $1_{\mathscr{F}(A)} = \mathscr{F}(1_A) = \mathscr{F}(gf) = \mathscr{F}(g)\mathscr{F}(f)$ and likewise for the reverse composition. And so if we have a functor $\mathscr{F}$ from a poset category (say, the real numbers with the usual inequality) to some category $\mathbf{C}$, then we can realize the structure of the poset sitting inside of $\mathbf{C}$ (perhaps involving only some of the objects of $\mathbf{C}$). This view comes in handy in a few places we’ll see later in our series on computational topology.
## The Hom Functor
There is a very important and nontrivial example called the “hom functor” which is motivated by the category of vector spaces. We’ll stick to the concrete example of vector spaces, but the generalization to arbitrary categories is straightforward. If the reader knows absolutely nothing about vector spaces, replace “vector space” with “object” and “linear map” with “morphism.” It won’t quite be correct, but it will get the idea across.
To each vector space $V$ one can define a dual vector space of functions $V \to \mathbb{C}$ (or whatever the field of scalars for $V$ is). Following the lead of hom sets, the dual vector space is denoted $\textup{Hom}(V,\mathbb{C})$. Here the morphisms in the set are those from the category of vector spaces (that is, linear maps $V \to \mathbb{C}$). Indeed, this is a vector space: one can add two functions pointwise ($(f+g)(v) = f(v) + g(v)$) and scale them ($(\lambda f)(v) = \lambda f(v)$), and the properties for a vector spaces are trivial to check.
Now the mapping $\textup{Hom}(-, \mathbb{C})$ which takes $V$ and produces $\textup{Hom}(V, \mathbb{C})$ is a functor called the hom functor. But let’s inspect this one more closely. The source category is obviously the category of vector spaces, but what is the target category? The objects are clear: the hom sets $\textup{Hom}(V,\mathbb{C})$ where $V$ is a vector space. The morphisms of the category are particularly awkward. Officially, they are written as
$\displaystyle \textup{Hom}(\textup{Hom}(V, \mathbb{C}), \textup{Hom}(W, \mathbb{C}))$
so a morphism in this category takes as input a linear map $V \to \mathbb{C}$ and produces as output one $W \to \mathbb{C}$. But what are the morphisms in words we can understand? And how can we compose them? Before reading on, think about what a morphism of morphisms should look like.
Okay, ready?
The morphisms in this category can be thought of as linear maps $W \to V$. More specifically, given a morphism $\varphi: V \to \mathbb{C}$ and a linear map $g: W \to V$, we can construct a linear map $W \to \mathbb{C}$ by composing $\varphi g$.
And so if we apply the $\textup{Hom}$ functor to a morphism $f: W \to V$, we get a morphism in $\textup{Hom}(\textup{Hom}(V, \mathbb{C}), \textup{Hom}(W, \mathbb{C}))$. Let’s denote the application of the hom functor using an asterisk so that $f \mapsto f_*$.
But wait a minute! The mapping here is going in the wrong direction: we took a map in one category going from the $V$ side to the $W$ side, and after applying the functor we got a map going from the $W$ side ($\textup{Hom}(W, \mathbb{C})$) to the $V$ side ($\textup{Hom}(V, \mathbb{C})$). It seems there is no reasonable way to take a map $V \to \mathbb{C}$ and get a map in $W \to \mathbb{C}$ using just $f$, but the other way is obvious. The hom functor “goes backward” in a sense. In other words, the composition property for our “functor” makes the composite $(g f)_*$ to the map taking $\varphi$ to $\varphi g f$. On the other hand, there is no way to compose $g_* f_*$, as they operate on the wrong domains! It must be the other way around:
$\displaystyle (gf)_* = f_* g_*$
We advise the reader to write down the commutative diagram and trace out the compositions to make sure everything works out. But this is a problem, because it makes the hom functor fail the most important requirement. In order to fix this reversal “problem,” we make the following definition:
Definition: A functor $\mathscr{F} : \mathbf{C} \to \mathbf{D}$ is called covariant if it preserves the order of morphism composition, so that $\mathscr{F}(gf) = \mathscr{F}(g) \mathscr{F}(f)$. If it reverses the order, we call it contravariant.
And so the hom functor on vector spaces is a contravariant functor, while all of the other functors we’ve defined in this post are covariant.
There is another way to describe a contravariant functor as a covariant functor which is often used. It involves the idea of an “opposite” category. For any category $\mathbf{C}$ we can define the opposite category $\mathbf{C}^{\textup{op}}$ to be a category with the same objects as $\mathbf{C}$, but with all morphisms reversed. That is, we define
$\displaystyle \textup{Hom}_{\mathbf{C}^{\textup{op}}}(A,B) = \textup{Hom}_{\mathbf{C}}(B,A)$
We leave it to the reader to verify that this is indeed a category. It is also not hard to see that $(\mathbf{C}^{\textup{op}})^{\textup{op}} = \mathbf{C}$. Opposite categories give us a nice recharacterization of a contrvariant functor. Indeed, because composition in opposite categories is reversed, a contravariant functor $\mathbf{C} \to \mathbf{D}$ is just a covariant functor on the opposite category $\mathbf{C}^{\textup{op}} \to \mathbf{D}$. Or equivalently, one $\mathbf{C} \to \mathbf{D}^{\textup{op}}$. More than anything, opposite categories are syntactical sugar. Composition is only reversed artificially to make domains and codomains line up, but the actual composition is the same as in the original category.
## Functors as Types
Before we move on to some code, let’s take a step back and look at the big picture (we’ve certainly plowed through enough details up to this point). The main thesis is that functoriality is a valuable property for an operation to have, but it’s not entirely clear why. Even the brightest of readers can only assume such properties are useful for mathematical analysis. It seems that the question we started this series out with, “what does category theory allow us to do that we couldn’t do before?” still has the answer, “nothing.” More relevantly, the question of what functoriality allows us to do is unclear. Indeed, once again the answer is “nothing.” Rather, functoriality in a computation allows one to analyze the behavior of a program. It gives the programmer a common abstraction in which to frame operations, and ease in proving the correctness of one’s algorithms.
In this light, the best we can do in implementing functors in programs is to give a type definition and examples. And in this author’s opinion this series is quickly becoming boring (all of the computational examples are relatively lame), so we will skip the examples in favor of the next post which will analyze more meaty programming constructs from a categorical viewpoint.
So recall the ML type definition of a category, a tuple of operations for source, target, identity, and composition:
datatype ('object, 'arrow)Category =
category of ('arrow -> 'object) *
('arrow -> 'object) *
('object -> 'arrow) *
('arrow * 'arrow -> 'arrow)
And so a functor consists of the two categories involved (as types), and the mapping on objects, and the mapping on morphisms.
datatype ('cObject, 'cArrow, 'dObject, 'dArrow)Functor =
aFunctor of ('cObject, 'cArrow)Category *
('cObject -> 'dObject) *
('cArrow -> 'dArrow) *
('dObject -> 'dArrow)Category
We encourage the reader who is uncomfortable with these type definitions to experiment with them by implementing some of our simpler examples (say, the size functor from sets to integers). Insofar as the basic definitions go, functors are not all that interesting. They become much more interesting when additional structure is imposed on them, and in the distant future we will see a glimpse of this in the form of adjointness. We hope to get around to analyzing statements like “syntax and semantics are adjoint functors.” For the next post in this series, we will take the three beloved functions of functional programming (map, foldl(r), and filter), and see what their categorical properties are.
Until then!
# Universal Properties
Previously in this series we’ve seen the definition of a category and a bunch of examples, basic properties of morphisms, and a first look at how to represent categories as types in ML. In this post we’ll expand these ideas and introduce the notion of a universal property. We’ll see examples from mathematics and write some programs which simultaneously prove certain objects have universal properties and construct the morphisms involved.
## A Grand Simple Thing
One might go so far as to call universal properties the most important concept in category theory. This should initially strike the reader as odd, because at first glance universal properties are so succinctly described that they don’t seem to be very interesting. In fact, there are only two universal properties and they are that of being initial and final.
Definition: An object $A$ in a category $\mathbf{C}$ is called initial if for every object $B$ there is a unique morphism $A \to B$. An object $Z$ is called final if for every object $B$ there is a unique morphism $B \to Z$. If an object satisfies either of these properties, it is called universal. If an object satisfies both, it is called a zero object.
In both cases, the existence of a unique morphism is the same as saying the relevant Hom set is a singleton (i.e., for initial objects $A$, the Hom set $\textup{Hom}_{\mathbf{C}}(A,B)$ consists of a single element). There is one and only one morphism between the two objects. In the particular case of $\textup{Hom}(A,A)$ when $A$ is initial (or final), the definition of a category says there must be at least one morphism, the identity, and the universal property says there is no other.
There’s only one way such a simple definition could find fruitful applications, and that is by cleverly picking categories. Before we get to constructing interesting categories with useful universal objects, let’s recognize some universal objects in categories we already know.
In $\mathbf{Set}$ the single element set is final, but not initial; there is only one set-function to a single-element set. It is important to note that the single-element set is far from unique. There are infinitely many (uncountably many!) singleton sets, but as we have already seen all one-element sets are isomorphic in $\mathbf{Set}$ (they all have the same cardinality). On the other hand, the empty set is initial, since the “empty function” is the only set-mapping from the empty set to any set. Here the initial object truly is unique, and not just up to isomorphism.
It turns out universal objects are always unique up to isomorphism, when they exist. Here is the official statement.
Proposition: If $A, A’$ are both initial in $\mathbf{C}$, then $A \cong A’$ are isomorphic. If $Z, Z’$ are both final, then $Z \cong Z’$.
Proof. Recall that a mophism $f: A \to B$ is an isomorphism if it has a two sided inverse, a $g:B \to A$ so that $gf = 1_A$ and $fg=1_B$ are the identities. Now if $A,A’$ are two initial objects there are unique morphisms $f : A \to A’$ and $g: A’ \to A$. Moreover, these compose to be morphisms $gf: A \to A$. But since $A$ is initial, the only morphism $A \to A$ is the identity. The situation for $fg : A’ \to A’$ is analogous, and so these morphisms are actually inverses of each other, and $A, A’$ are isomorphic. The proof for final objects is identical. $\square$
Let’s continue with examples. In the category of groups, the trivial group $\left \{ 1 \right \}$ is both initial and final, because group homomorphisms must preserve the identity element. Hence the trivial group is a zero object. Again, “the” trivial group is not unique, but unique up to isomorphism.
In the category of types with computable (halting) functions as morphisms, the null type is final. To be honest, this depends on how we determine whether two computable functions are “equal.” In this case, we only care about the set of inputs and outputs, and for the null type all computable functions have the same output: null.
Partial order categories are examples of categories which need not have universal objects. If the partial order is constructed from subsets of a set $X$, then the initial object is the empty set (by virtue of being a subset of every set), and $X$ as a subset of itself is obviously final. But there are other partial orders, such as inequality of integers, which have no “smallest” or “largest” objects. Partial order categories which have particularly nice properties (such as initial and final objects, but not quite exactly) are closely related to the concept of a domain in denotational semantics, and the language of universal properties is relevant to that discussion as well.
The place where universal properties really shine is in defining new constructions. For instance, the direct product of sets is defined by the fact that it satisfies a universal property. Such constructions abound in category theory, and they work via the ‘diagram categories’ we defined in our introductory post. Let’s investigate them now.
## Quotients
Let’s recall the classical definition from set theory of a quotient. We described special versions of quotients in the categories of groups and topological spaces, and we’ll see them all unified via the universal property of a quotient in a moment.
Definition: An equivalence relation on a set $X$ is a subset of the set product $\sim \subset X \times X$ which is reflexive, symmetric, and transitive. The quotient set $X / \sim$ is the set of equivalence classes on $X$. The canonical projection $\pi : X \to X/\sim$ is the map sending $x$ to its equivalence class under $\sim$.
The quotient set $X / \sim$ can also be described in terms of a special property: it is the “largest” set which agrees with the equivalence relation $\sim$. On one hand, it is the case that whenever $a \sim b$ in $X$ then $\pi(a) = \pi(b)$. Moreover, for any set $Y$ and any map $g: X \to Y$ which equates equivalent things ($g(a) = g(b)$ for all $a \sim b$), then there is a unique map $f : X/\sim \to Y$ such that $f \pi = g$. This word salad is best translated into a diagram.
Here we use a dashed line to assert the existence of a morphism (once we’ve proven such a morphism exists, we use a solid line instead), and the symbol $\exists !$ signifies existence ($\exists$) and uniqueness (!).
We can prove this explicitly in the category $\mathbf{Set}$. Indeed, if $g$ is any map such that $g(a) = g(b)$ for all equivalent $a,b \in X$, then we can define $f$ as follows: for any $a \in X$ whose equivalence class is denoted by $[a]$ in $X / \sim$, and define $f([a]) = g(a)$. This map is well defined because if $a \sim b$, then $f([a]) = g(a) = g(b) = f([b])$. It is unique because if $f \pi = g = h \pi$ for some other $h: X / \sim \to Y$, then $h([x]) = g(x) = f([x])$; this is the only possible definition.
Now the “official” way to state this universal property is as follows:
The quotient set $X / \sim$ is universal with respect to the property of mapping $X$ to a set so that equivalent elements have the same image.
But as we said earlier, there are only two kinds of universal properties: initial and final. Now this $X / \sim$ looks suspiciously like an initial object ($f$ is going from $X / \sim$, after all), but what exactly is the category we’re considering? The trick to dissecting this sentence is to notice that this is not a statement about just $X / \sim$, but of the morphism $\pi$.
That is, we’re considering a diagram category. In more detail: fix an object $X$ in $\mathbf{Set}$ and an equivalence relation $\sim$ on $X$. We define a category $\mathbf{Set}_{X,\sim}$ as follows. The objects in the category are morphisms $f:X \to Y$ such that $a \sim b$ in $X$ implies $f(a) = f(b)$ in $Y$. The morphisms in the category are commutative diagrams
Here $f_1, f_2$ need to be such that they send equivalent things to equal things (or they wouldn’t be objects in the category!), and by the commutativity of the diagram $f_2 = \varphi f_1$. Indeed, the statement about quotients is that $\pi : X \to X / \sim$ is an initial object in this category. In fact, we have already proved it! But note the abuse of language in our offset statement above: it’s not really $X / \sim$ that is the universal object, but $\pi$. Moreover, the statement itself doesn’t tell us what category to inspect, nor whether we care about initial or final objects in that category. Unfortunately this abuse of language is widespread in the mathematical world, and for arguably good reason. Once one gets acquainted with these kinds of constructions, reading between the limes becomes much easier and it would be a waste of time to spell it out. After all, once we understand $X / \sim$ there is no “obvious” choice for a map $X \to X / \sim$ except for the projection $\pi$. This is how $\pi$ got its name, the canonical projection.
Two last bits of terminology: if $\mathbf{C}$ is any category whose objects are sets (and hence, where equivalence relations make sense), we say that $\mathbf{C}$ has quotients if for every object $X$ there is a morphism $\pi$ satisfying the universal property of a quotient. Another way to state the universal property is to say that all maps respecting the equivalence structure factor through the quotient, in the sense that we get a diagram like the one above.
What is the benefit of viewing $X / \sim$ by its universal property? For one, the set $X / \sim$ is unique up to isomorphism. That is, if any other pair $(Z, g)$ satisfies the same property, we automatically get an isomorphism $X / \sim \to Z$. For instance, if $\sim$ is defined via a function $f : X \to Y$ (that is, $a \sim b$ if $f(a) = f(b)$), then the pair $(\textup{im}(f), f)$ satisfies the universal property of a quotient. This means that we can “decompose” any function into three pieces:
$\displaystyle X \to X / \sim \to \textup{im}(f) \to Y$
The first map is the canonical projection, the second is the isomorphism given by the universal property of the quotient, and the last is the inclusion map into $Y$. In a sense, all three of these maps are “canonical.” This isn’t so magical for set-maps, but the same statement (and essentially the same proof) holds for groups and topological spaces, and are revered as theorems. For groups, this is called The First Isomorphism Theorem, but it’s essentially the claim that the category of groups has quotients.
This is getting a bit abstract, so let’s see how the idea manifests itself as a program. In fact, it’s embarrassingly simple. Using our “simpler” ML definition of a category from last time, the constructive proof that quotient sets satisfy the universal property is simply a concrete version of the definition of $f$ we gave above. In code,
fun inducedMapFromQuotient(setMap(x, pi, q), setMap(x, g, y)) =
setMap(q, (fn a => g(representative(a))), y)
That is, once we have $\pi$ and $X / \sim$ defined for our given equivalence relation, this function accepts as input any morphism $g$ and produces the uniquely defined $f$ in the diagram above. Here the “representative” function just returns an arbitrary element of the given set, which we added to the abstract datatype for sets. If the set $X$ is empty, then all functions involved will raise an “empty” exception upon being called, which is perfectly fine. We leave the functions which explicitly construct the quotient set given $X, \sim$ as an exercise to the reader.
## Products and Coproducts
Just as the concept of a quotient set or quotient group can be generalized to a universal property, so can the notion of a product. Again we take our intuition from $\mathbf{Set}$. There the product of two sets $X,Y$ is the set of ordered pairs
$\displaystyle X \times Y = \left \{ (x,y) : x \in X, y \in Y \right \}$
But as with quotients, there’s much more going on and the key is in the morphisms. Specifically, there are two obvious choices for morphisms $X \times Y \to X$ and $X \times Y \to Y$. These are the two projections onto the components, namely $\pi_1(x,y) = x$ and $\pi_2(x,y) = y$. These projections are also called “canonical projections,” and they satisfy their own universal property.
The product of sets is universal with respect to the property of having two morphisms to its factors.
Indeed, this idea is so general that it can be formulated in any category, not just categories whose objects are sets. Let $X,Y$ be two fixed objects in a category $\mathbf{C}$. Should it exist, the product $X \times Y$ is defined to be a final object in the following diagram category. This category has as objects pairs of morphisms
and as morphisms it has commutative diagrams
In words, to say products are final is to say that for any object in this category, there is a unique map $\varphi$ that factors through the product, so that $\pi_1 \varphi = f$ and $\pi_2 \varphi = g$. In a diagram, it is to claim the following commutes:
If the product $X \times Y$ exists for any pair of objects, we declare that the category $\mathbf{C}$ has products.
Indeed, many familiar product constructions exist in pure mathematics: sets, groups, topological spaces, vector spaces, and rings all have products. In fact, so does the category of ML types. Given two types ‘a and ‘b, we can form the (aptly named) product type ‘a * ‘b. The canonical projections exist because ML supports parametric polymorphism. They are
fun leftProjection(x,y) = x
fun rightProjection(x,y) = y
And to construct the unique morphism to the product,
fun inducedMapToProduct(f,g) = fn a => (f(a), g(a))
We leave the uniqueness proof to the reader as a brief exercise.
There is a similar notion called a coproduct, denoted $X \coprod Y$, in which everything is reversed: the arrows in the diagram category go to $X \coprod Y$ and the object is initial in the diagram category. Explicitly, for a fixed $X, Y$ the objects in the diagram category are again pairs of morphisms, but this time with arrows going to the central object
The morphisms are again commutative diagrams, but with the connecting morphism going away from the central object
And a coproduct is defined to be an initial object in this category. That is, a pair of morphisms $i_1, i_2$ such that there is a unique connecting morphism in the following diagram.
Coproducts are far less intuitive than products in their concrete realizations, but the universal property is no more complicated. For the category of sets, the coproduct is a disjoint union (in which shared elements of two sets $X, Y$ are forcibly considered different), and the canonical morphisms are “inclusion” maps (hence the switch from $\pi$ to $i$ in the diagram above). Specifically, if we define the coproduct
$\displaystyle X \coprod Y = (X \times \left \{ 1 \right \}) \cup (Y \times \left \{ 2 \right \})$
as the set of “tagged” elements (the right entry in a tuple signifies which piece of the coproduct the left entry came from), then the inclusion maps $i_1(x) = (x,1)$ and $i_2(y) = (y,2)$ are the canonical morphisms.
There are similar notions of disjoint unions for topological spaces and graphs, which are their categories’ respective coproducts. However, in most categories the coproducts are dramatically different from “unions.” In group theory, it is a somewhat complicated object known as the free product. We mentioned free products in our hasty discussion of the fundamental group, but understanding why and where free groups naturally occur is quite technical. Similarly, coproducts of vector spaces (or $R$-modules) are more like a product, with the stipulation that at most finitely many of the entries of a tuple are nonzero (e.g., formal linear combinations of things from the pieces). Even without understanding these examples, the reader should begin to believe that relatively simple universal properties can yield very useful objects with potentially difficult constructions in specific categories. The ubiquity of the concepts across drastically different fields of mathematics is part of why universal properties are called “universal.”
Luckily, the category of ML types has a nice coproduct which feels like a union, but it is not supported as a “native” language feature like products types are. Specifically, given two types ‘a, ‘b we can define the “coproduct type”
datatype ('a, 'b)Coproduct = left of 'a | right of 'b
Let’s prove this is actually a coproduct: fix two types ‘a and ‘b, and let $i_1, i_2$ be the functions
fun leftInclusion(x) = left(x)
fun rightInclusion(y) = right(y)
Then given any other pair of functions $f,g$ which accept as input types ‘a and ‘b, respectively, there is a unique function $\varphi$ operating on the coproduct type. We construct it as follows.
fun inducedCoproductMap(f,g) =
let
theMap(left(a)) = f(a)
theMap(right(b)) = g(b)
in
theMap
end
The uniqueness of this construction is self-evident. This author finds it fascinating that these definitions are so deep and profound (indeed, category theory is heralded as the queen of abstraction), but their realizations are trivially obvious to the working programmer. Perhaps this is a statement about how well-behaved the category of ML types is.
## Continuing On
So far we have seen three relatively simple examples of universal properties, and explored how they are realized in some categories. We should note before closing two things. The first is that not every category has objects with these universal properties. Unfortunately poset categories don’t serve as a good counterexample for this (they have both products and coproducts; what are they?), but it may be the case that in some categories only some pairs of objects have products or coproducts, while others do not.
Second, there are many more universal properties that we haven’t covered here. For instance, the notion of a limit is a universal property, as is the notion of a “free” object. There are kernels, pull-backs, equalizers, and many other ad-hoc universal properties without names. And for every universal property there is a corresponding “dual” property that results from reversing the arrows in every diagram, as we did with coproducts. We will visit the relevant ones as they come up in our explorations.
In the next few posts we’ll encounter functors and the concept of functoriality, and start asking some poignant questions about familiar programmatic constructions.
Until then!
# Properties of Morphisms
This post is mainly mathematical. We left it out of our introduction to categories for brevity, but we should lay these definitions down and some examples before continuing on to universal properties and doing more computation. The reader should feel free to skip this post and return to it later when the words “isomorphism,” “monomorphism,” and “epimorphism” come up again. Perhaps the most important part of this post is the description of an isomorphism.
## Isomorphisms, Monomorphisms, and Epimorphisms
Perhaps the most important paradigm shift in category theory is the focus on morphisms as the main object of study. In particular, category theory stipulates that the only knowledge one can gain about an object is in how it relates to other objects. Indeed, this is true in nearly all fields of mathematics: in groups we consider all isomorphic groups to be the same. In topology, homeomorphic spaces are not distinguished. The list goes on. The only way to do determine if two objects are “the same” is by finding a morphism with special properties. Barry Mazur gives a more colorful explanation by considering the meaning of the number 5 in his essay, “When is one thing equal to some other thing?” The point is that categories, more than existing to be a “foundation” for all mathematics as a formal system (though people are working to make such a formal system), exist primarily to “capture the essence” of mathematical discourse, as Mazur puts it. A category defines objects and morphisms, but literally all of the structure of a category lies in its morphisms. And so we study them.
The strongest kind of morphism we can consider is an isomorphism. An isomorphism is the way we say two objects in a category are “the same.” We don’t usually care whether two objects are equal, but rather whether some construction is unique up to isomorphism (more on that when we talk of universal properties). The choices made in defining morphisms in a particular category allow us to strengthen or weaken this idea of “sameness.”
Definition: A morphism $f : A \to B$ in a category $\mathbf{C}$ is an isomorphism if there exists a morphism $g: B \to A$ so that both ways to compose $f$ and $g$ give the identity morphisms on the respective objects. That is,
$gf = 1_A$ and $fg = 1_B$.
The most basic (usually obvious, but sometimes very deep) question in approaching a new category is to ask what the isomorphisms are. Let us do this now.
In $\mathbf{Set}$ the morphisms are set-functions, and it is not hard to see that any two sets of equal cardinality have a bijection between them. As all bijections have two-sided inverses, two objects in $\mathbf{Set}$ are isomorphic if and only if they have the same cardinality. For example, all sets of size 10 are isomorphic. This is quite a weak notion of “sameness.” In contrast, there is a wealth of examples of groups of equal cardinality which are not isomorphic (the smallest example has cardinality 4). On the other end of the spectrum, a poset category $\mathbf{Pos}_X$ has no isomorphisms except for the identity morphisms. The poset categories still have useful structure, but (as with objects within a category) the interesting structure is in how a poset category relates to other categories. This will become clearer later when we look at functors, but we just want to dissuade the reader from ruling out poset categories as uninteresting due to a lack of interesting morphisms.
Consider the category $\mathbf{C}$ of ML types with ML functions as morphisms. An isomorphism in this category would be a function which has a two-sided inverse. Can the reader think of such a function?
Let us now move on to other, weaker properties of morphisms.
Definition: A morphism $f: A \to B$ is a monomorphism if for every object $C$ and every pair of morphisms $g,h: C \to A$ the condition $fg = fh$ implies $g = h$.
The reader should parse this notation carefully, but truly think of it in terms of the following commutative diagram:
Whenever this diagram commutes and $f$ is a monomorphism, then we conclude (by definition) that $g=h$. Remember that a diagram commuting just means that all ways to compose morphisms (and arrive at morphisms with matching sources and targets) result in an identical morphism. In this diagram, commuting is the equivalent of claiming that $fg = fh$, since there are only two nontrivial ways to compose.
The idea is that monomorphisms allow one to “cancel” $f$ from the left side of a composition (although, confusingly, this means the cancelled part is on the right hand side of the diagram).
The corresponding property for cancelling on the right is defined identically.
Definition: A morphism $f: A \to B$ is an epimorphism if for every object $C$ and every pair of morphism $g,h: B \to C$ the condition $gf = hf$ implies $g = h$.
Again, the relevant diagram.
Whenever $f$ is an epimorphism and this diagram commutes, we can conclude $g=h$.
Now one of the simplest things one can do when considering a category is to identify the monomorphisms and epimorphisms. Let’s do this for a few important examples.
## Monos and Epis in Various Categories
In the category $\mathbf{Set}$, monomorphisms and epimorphisms correspond to injective and surjective functions, respectively. Lets see why this is the case for monomorphisms. Recall that an injective function $f$ has the property that $f(x) = f(y)$ implies $x=y$. With this property we can show $f$ is a monomorphism because if $f(g(x)) = f(h(x))$ then the injective property gives us immediately that $g(x) = h(x)$. Conversely, if $f$ is a monomorphism and $f(x) = f(y)$, we will construct a set $C$ and two convenient functions $g, h: C \to A$ to help us show that $x=y$. In particular, pick $C$ to be the one point set $\left \{ c \right \}$, and define $g(c) = x, h(c) = y$. Then as functions $fg = fh$. Indeed, there is only one value in the domain, so saying this amounts to saying $f(x) = fg(c) = fh(c) = f(y)$, which we know is true by assumption. By the monomorphism property $g = h$, so $x = g(c) = h(c) = y$.
Now consider epimorphisms. It is clear that a surjective map is an epimorphism, but the converse is a bit trickier. We prove by contraposition. Instead of now picking the “one-point set,” for our $C$, we must choose something which is one element bigger than $B$. In particular, define $g, h : B \to B’$, where $B’$ is $B$ with one additional point $x$ added (which we declare to not already be in $B$). Then if $f$ is not surjective, and there is some $b_0 \in B$ which is missed by $f$, we define $g(b_0) = x$ and $g(b) = b$ otherwise. We can also define $h$ to be the identity on $B$, so that $gf = hf$, but $g \neq h$. So epimorphisms are exactly the surjective set-maps.
There is one additional fact that makes the category of sets well-behaved: a morphism in $\mathbf{Set}$ is an isomorphism if and only if it is both a monomorphism and an epimorphism. Indeed, isomorphisms are set-functions with two-sided inverses (bijections) and we know from classical set theory that bijections are exactly the simultaneous injections and surjections. A warning to the reader: not all categories are like this! We will see in a moment an example of a nontrivial category in which isomorphisms are not the same thing as simultaneous monomorphisms and epimorphisms.
The category $\mathbf{Grp}$ is very similar to $\mathbf{Set}$ in regards to monomorphisms and epimorphisms. The former are simply injective group homomorphisms, while the latter are surjective group homomorphisms. And again, a morphisms is an isomorphism if and only if it is both a monomorphism and an epimorphism. We invite the reader to peruse the details of the argument above and adapt it to the case of groups. In both cases, the hard decision is in choosing $C$ when necessary. For monomorphisms, the “one-point group” does not work because we are constrained to send the identity to the identity in any group homomorphism. The fortuitous reader will avert their eyes and think about which group would work, and otherwise we suggest trying $C = \mathbb{Z}$. After completing the proof, the reader will see that the trick is to find a $C$ for which only one “choice” can be made. For epimorphisms, the required $C$ is a bit more complex, but we invite the reader to attempt a proof to see the difficulties involved.
Why do these categories have the same properties but they are acquired in such different ways? It turns out that although these proofs seem different in execution, they are the same in nature, and they follow from properties of the category as a whole. In particular, the “one-point object” (a singleton set for $\mathbf{Set}$ and $\mathbb{Z}$ for $\mathbf{Grp}$) is more categorically defined as the “free object on one generator.” We will discuss this more when we get to universal properties, but a “free object on $n$ generators” is roughly speaking an object $A$ for which any morphism with source $A$ must make exactly $n$ “choices” in its definition. With sets that means $n$ choices for the images of elements, for groups that means $n$ consistent choices for images of group elements. On the epimorphism side, the construction is a sort of “disjoint union object” which is correctly termed a “coproduct.” But momentarily putting aside all of this new and confusing terminology, let us see some more examples of morphisms in various categories.
Our recent primer on rings was well-timed, because the category $\mathbf{Ring}$ of rings (with identity) is an example of a not-so-well-behaved category. As with sets and groups, we do have that monomorphisms are equivalent to injective ring homomorphisms, but the argument is trickier than it was for groups. It is not obvious which “convenient” object $C$ to choose here, since maps $\mathbb{Z} \to R$ yield no choices: 1 maps to 1, 0 maps to 0, and the properties of a ring homomorphism determine everything else (in fact, the abelian group structure and the fact that units are preserved is enough). This makes $\mathbb{Z}$ into what’s called an “initial object” in $\mathbf{Ring}$; more on that when we study universal properties. In fact, we invite the reader to return to this post after we talk about the universal property of polynomial rings. It turns out that $\mathbb{Z}[x]$ is a suitable choice for $C$, and the “choice” made is where to send the indeterminate $x$.
On the other hand, things go awry when trying to apply analogous arguments to epimorphisms. While it is true that every surjective ring homomorphism is an epimorphism (it is already an epimorphism in $\mathbf{Set}$, and the argument there applies), there are ring epimorphisms which are not surjections! Consider the inclusion map of rings $i : \mathbb{Z} \to \mathbb{Q}$. The map $i$ is not surjective, but it is an epimorphism. Suppose $g, h : \mathbb{Q} \to R$ are two parallel ring morphisms, and they agree on $\mathbb{Z}$ (they will always do so, since there is only one ring homomorphism $\mathbb{Z} \to R$). Then $g,h$ must also agree on $\mathbb{Q}$, because if $p,q \in \mathbb{Z}$ with $q \neq 0$, then
$\displaystyle g(p/q) = g(p)g(q^{-1}) = g(p)g(q)^{-1} = h(p)h(q)^{-1} = h(p/q)$
Because the map above is also an injection, the category of rings is a very naturally occurring example of a category which has morphisms that are both epimorphisms and monomorphisms, but not isomorphisms.
There are instances in which monomorphisms and epimorphisms are trivial. Take, for instance any poset category. There is at most one morphism between any two objects, and so the conditions for an epimorphism and monomorphism vacuously hold. This is an extreme example of a time when simultaneous monomorphisms and epimorphisms are not the same thing as isomorphisms! The only isomorphisms in a poset category are the identity morphisms.
## Morals about Good and Bad Categories
The inspection of epimorphisms and monomorphisms is an important step in the analysis of a category. It gives one insight into how “well-behaved” the category is, and picks out the objects which are special either for their useful properties or confounding trickery.
This reminds us of a quote of Alexander Grothendieck, one of the immortal gods of mathematics who popularized the use of categories in mainstream mathematics.
A good category containing some bad objects is preferable to a bad category containing only good objects.
I suppose the thesis here is that the having only “good” objects yields less interesting and useful structure, and that one should be willing to put up with the bad objects in order to have access to that structure.
Next time we’ll jump into a discussion of universal properties, and start constructing some programs to prove that various objects satisfy them.
Until then! | 15,639 | 61,943 | {"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} | 3.765625 | 4 | CC-MAIN-2023-40 | latest | en | 0.921638 |
https://numberworld.info/36696 | 1,596,580,129,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735885.72/warc/CC-MAIN-20200804220455-20200805010455-00572.warc.gz | 393,618,189 | 3,992 | Number 36696
Properties of number 36696
Cross Sum:
Factorization:
2 * 2 * 2 * 3 * 11 * 139
Divisors:
1, 2, 3, 4, 6, 8, 11, 12, 22, 24, 33, 44, 66, 88, 132, 139, 264, 278, 417, 556, 834, 1112, 1529, 1668, 3058, 3336, 4587, 6116, 9174, 12232, 18348, 36696
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
8f58
Base 32:
13qo
sin(36696)
0.80978558654839
cos(36696)
-0.58672591882282
tan(36696)
-1.3801769456054
ln(36696)
10.510423036277
lg(36696)
4.5646187271239
sqrt(36696)
191.56200040718
Square(36696)
Number Look Up
Look Up
36696 which is pronounced (thirty-six thousand six hundred ninety-six) is a special number. The cross sum of 36696 is 30. If you factorisate the number 36696 you will get these result 2 * 2 * 2 * 3 * 11 * 139. The figure 36696 has 32 divisors ( 1, 2, 3, 4, 6, 8, 11, 12, 22, 24, 33, 44, 66, 88, 132, 139, 264, 278, 417, 556, 834, 1112, 1529, 1668, 3058, 3336, 4587, 6116, 9174, 12232, 18348, 36696 ) whith a sum of 100800. The figure 36696 is not a prime number. The number 36696 is not a fibonacci number. The figure 36696 is not a Bell Number. 36696 is not a Catalan Number. The convertion of 36696 to base 2 (Binary) is 1000111101011000. The convertion of 36696 to base 3 (Ternary) is 1212100010. The convertion of 36696 to base 4 (Quaternary) is 20331120. The convertion of 36696 to base 5 (Quintal) is 2133241. The convertion of 36696 to base 8 (Octal) is 107530. The convertion of 36696 to base 16 (Hexadecimal) is 8f58. The convertion of 36696 to base 32 is 13qo. The sine of the figure 36696 is 0.80978558654839. The cosine of the number 36696 is -0.58672591882282. The tangent of the figure 36696 is -1.3801769456054. The root of 36696 is 191.56200040718.
If you square 36696 you will get the following result 1346596416. The natural logarithm of 36696 is 10.510423036277 and the decimal logarithm is 4.5646187271239. I hope that you now know that 36696 is special number! | 831 | 2,060 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-34 | latest | en | 0.747687 |
https://socratic.org/questions/56573a52581e2a53a3d36cc5 | 1,581,935,082,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141806.26/warc/CC-MAIN-20200217085334-20200217115334-00405.warc.gz | 574,051,148 | 7,217 | # What is the coordination number of cobalt in the [Co(en)_2Br_2]^+ complex?
Nov 26, 2015
$6$
#### Explanation:
First thing first, if you're dealing with the dibromobis(ethylenediamine) cobalt(III) complex ion, ${\text{Co"("en")_2"Br}}_{2}^{+}$, then I'm not sure about the formula you have for the complex ion.
The charge of the chloride anion is $\left(1 -\right)$, which means that you would either need two of these complex ions to balance it, or you have cobalt in the $+ 4$ oxidation state, which is not very common at all.
I think you could get cobalt in its $+ 4$ oxidation state by using fluoride anions, since they have a much better chance at being able to stabilize the charge on the cobalt cation.
So my guess is that you're either dealing with ["Co"("en")_2"Br"_2]"Cl", or with ["Co"("en")_2"Br"_2]_2"Cl"_2, maybe?
Anyway, this doesn't actually matter for finding the coordination number of cobalt.
As you know, the coordination number of an atom is determined by counting the number of other atoms to which it is bonded. Remember, single, double, or triple bonds have the same significance here.
So, take a look at the dibromobis(ethylenediamine) cobalt(III) ion. The central cobalt atom is bonded to
• two ethylenediamine molecules ,$\text{en}$
• two bromide anions, ${\text{Br}}^{-}$
Now, it's important to realize that ethylenediamine is a bidentate ligand, which means that it uses two donor atoms to bind to the central cobalt atom.
More specifically, these donor atoms are the two nitrogen atoms
This means that the central cobalt atom will actually be bonded to a total of six atoms
• two nitrogen atoms from the first ethylenediamine molecule
• two nitrogen atoms from the second ethylenediamine molecule
• two bromine atoms
Therefore, the coordination number of cobalt in the ${\text{Co"("en")_2"Br}}_{2}^{+}$ complex ion is equal to $6$.
Nov 26, 2015
The coordination number of $C o$ in ${\left[C o {\left(e n\right)}_{2} B {r}_{2}\right]}^{+}$ is 6.
#### Explanation:
Ethylenediamine is a bidentate ligand; it has 2 donor atoms to coordinate to the metal centre in a 5-membered chelate ring. Note that this (I presume) is a $C {o}^{3 +}$ complex, i.e. ${d}^{6}$ configuration, whose formula is ${\left[C o {\left(e n\right)}_{2} B {r}_{2}\right]}^{+}$, not as quoted in your question. | 675 | 2,330 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-10 | longest | en | 0.900725 |
https://stats.stackexchange.com/questions/106356/interpretation-of-spearmans-rank-correlation-coefficient-beyond-its-significa | 1,722,802,756,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00042.warc.gz | 434,167,763 | 40,907 | # Interpretation of Spearman's rank correlation coefficient - beyond its significance
I calculated the Spearman's rank correlation coefficient interpretation for a given 2D dataset. I then tested its significance by doing a permutation test and obtained a p-value.
I have a problem with the interpretation of the coefficient value. While I understand that a Spearman's rank coef. value should not be mistaken/be interpreted as giving information about its significance, I still do not have a simple interpretation for the coefficient value. The significance test shows us basically how likely is the coefficient to be larger that the observed one when the Null Hypothesis is respected, but says nothing about the observed value that uses as starting point. Can for instance a value of 0.60 mean that there are 60% more ranked pairs in my data set following a monotonically crescent discrete function than otherwise?
• Wouldn't that be $0.60^{2}=0.36$%? Commented Jul 9, 2014 at 16:34
• I do not know, it was just my intuition. Could you please elaborate on how that is the case or point a reference? Commented Jul 9, 2014 at 16:35
• Well, the coefficient of determination ($R^{2}$) is often interpreted in the manner that you propose. And for simple OLS (meeting its assumptions) that $R^{2}=r^{2}$, as in Pearson's $r$ squared. So I wonder if you are proposing a similar interpretation with Spearman's rank correlation coefficient? Hence, I wonder if you mean to square it. Commented Jul 9, 2014 at 16:42
• I see your point. But does this make sense even if the linear function assumed/fitted is not a good representation of reality? After all, we are only assuming a monotonically crescent function. Commented Jul 9, 2014 at 17:42
• Are you asking whether it is possible to arrive at erroneous conclusions when misapplying statistical tests and measures? Let me think on that... hmmm.... HMMMM... mm-hmmm... mmm-hmmm. YES. Commented Jul 9, 2014 at 17:44
The Spearman's rank c. c. is the Pearson' c.c. of the ranked variables; in its turn the Pearson's c.c. is defined as the mean of the product of the paired standardized scores $z(X_i)$, $z(Y_i)$.
$$r(X,Y) = \Sigma_i[z(X_i) z(Y_i)]/(n-1)$$
in which $n$ is the sample size and the standard scores
$$z(X_i) = [X_i - \bar{X}]/std(X)$$
$$z(Y_i) = [Y_i - \bar{Y}]/std(Y)$$
are relative to the ranked variables ($X_i$, $Y_i$). Squaring $r(X_i, Y_i)$ we obtain the coefficient of determination $r²$, which we can equate to the fraction of explained variance. So if my Spearman's rank c.c. is of 0.6, I can deduce that the variance of the ranked variables is shared at 36%.
From the first equation and attempting at a simpler way of explaining $r(X,Y)$, I would say is the average value of concordance of z-score variations. For instance, let us say I repeat an experiment by increasing the sample size $n$ and calculate $r(X,Y)$ for both the small sample and the larger one. Let us say that associated to an increase in n of $~3$ I get a decrease in $r(X,Y)$ of roughly 50%; this corresponds to a decrease in standard scores concordance of 50%. My interpretation should be then that the latest dataset provides weaker evidence for the presence of a correlation in the data. | 817 | 3,230 | {"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": 3, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-33 | latest | en | 0.921265 |
https://ryanwingate.com/other-interests/business/total-interest-over-life-of-loan/ | 1,642,952,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304287.0/warc/CC-MAIN-20220123141754-20220123171754-00168.warc.gz | 561,403,676 | 5,140 | # Total Interest over Life of Loan
import pandas as pd
import numpy as np
import numpy_financial as npf
Calculate the total interest and payment for a $10K loan over 60 months at a 3% annual interest rate. annual_interest_rate = 0.03 monthly_interest_rate = annual_interest_rate / 12 n_periods = 60 periods = np.arange(n_periods) + 1 loan_amount = 10000. monthly_payment = (npf.pmt(monthly_interest_rate, n_periods, loan_amount) .round(2) * -1 ) interest_per_month = (npf.ipmt(monthly_interest_rate, periods, n_periods, loan_amount) .round(2) * -1 ) total_interest = interest_per_month.sum() loan_amount, annual_interest_rate, n_periods, total_interest, monthly_payment (10000.0, 0.03, 60, 781.23, 179.69) Now, parametrize and calculate the same for loan amounts from \$10K to \\$45K.
loans = []
annual_interest_rate = 0.03
monthly_interest_rate = annual_interest_rate / 12
n_periods = 60
periods = np.arange(n_periods) + 1
for loan_amount in np.arange(10000, 46000, 1000):
monthly_payment = (npf.pmt(monthly_interest_rate,
n_periods,
loan_amount)
.round(2) * -1 )
interest_per_month = (npf.ipmt(monthly_interest_rate,
periods,
n_periods,
loan_amount)
.round(2) * -1 )
total_interest = interest_per_month.sum()
loans.append((loan_amount, total_interest, monthly_payment, annual_interest_rate, n_periods))
loans = (pd.DataFrame(loans,
columns=['loan_amount', 'total_interest', 'monthly_payment',
'annual_interest_rate', 'n_periods'])
.set_index('loan_amount'))
loans
total_interest monthly_payment annual_interest_rate n_periods
loan_amount
10000 781.23 179.69 0.03 60
11000 859.30 197.66 0.03 60
12000 937.47 215.62 0.03 60
13000 1015.58 233.59 0.03 60
14000 1093.72 251.56 0.03 60
15000 1171.85 269.53 0.03 60
16000 1249.99 287.50 0.03 60
17000 1328.02 305.47 0.03 60
18000 1406.16 323.44 0.03 60
19000 1484.33 341.41 0.03 60
20000 1562.46 359.37 0.03 60
21000 1640.57 377.34 0.03 60
22000 1718.64 395.31 0.03 60
23000 1796.80 413.28 0.03 60
24000 1874.88 431.25 0.03 60
25000 1952.99 449.22 0.03 60
26000 2031.19 467.19 0.03 60
27000 2109.30 485.15 0.03 60
28000 2187.37 503.12 0.03 60
29000 2265.57 521.09 0.03 60
30000 2343.67 539.06 0.03 60
31000 2421.80 557.03 0.03 60
32000 2499.85 575.00 0.03 60
33000 2577.97 592.97 0.03 60
34000 2656.12 610.94 0.03 60
35000 2734.26 628.90 0.03 60
36000 2812.30 646.87 0.03 60
37000 2890.46 664.84 0.03 60
38000 2968.65 682.81 0.03 60
39000 3046.72 700.78 0.03 60
40000 3124.87 718.75 0.03 60
41000 3202.99 736.72 0.03 60
42000 3281.13 754.69 0.03 60
43000 3359.23 772.65 0.03 60
44000 3437.38 790.62 0.03 60
45000 3515.49 808.59 0.03 60 | 1,064 | 2,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-05 | longest | en | 0.534226 |
https://forum.solar-electric.com/discussion/354407/wire-sizing | 1,561,616,566,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000894.72/warc/CC-MAIN-20190627055431-20190627081431-00401.warc.gz | 425,362,115 | 11,490 | # Wire sizing
Registered Users Posts: 4
I'm sure this question has been asked and answered many times but I'm still confused.
I have 2 Canadian Solar panels ready to go onto my roof. They are 235 watt with Voc 39.6V, Isc 8.46A. My intention is to hook them up in series to give about 75 volts Voc at 8.46A, wire it to a 40Amp MPPT controller that's about 40 feet distance from the panels and feed the MPPT 12 volt output into 4 x 6 volt FLA batteries arranged in a series/parallel configuration..
My question is, do I calculate the safe amperage capacity of the wire from the solar panels going to the MPPT using the 8.46 amps?
I have used many on line calculators and they often come up with contradictory results, ranging from 12AWG to 02AWG.
Assuming your MPPT solar charge controller can manage the Voc-cold from the solar panels (need >100 Vpanel input rating--details matter here), then two calculations... First based on the voltage drop from the Array to the charge controller. We use 3% to 1% as typical values, but it can be more than 3% (MPPT controllers "don't care" as long as the minimum input voltage for your battery bank is met).
Assuming these are the specs:
Nominal Maximum Power (Pmax):235W Optimum Operating Voltage (Vmp):30.1V Optimum Operating Current (Imp):7.82A Open Circuit Voltage (Voc):37.2V Short Circuit Current (Isc):8.34A
First, the voltage drop:
• 2*Vmp= 2 * 30.1 volts Vmp = 60.2 volts Vmp
• 1x*Imp= 7.82 Amps Imp
Using 60.2 volts and 7.82 amps and 40 feet (one way run) for this calculator. For 3% maximum drop:
https://www.calculator.net/voltage-drop-calculator.html?material=copper&wiresize=8.286&voltage=60.1&phase=dc&noofconductor=1&distance=40&distanceunit=feet&eres=7.82&x=58&y=27
14 AWG:
Voltage drop: 1.58
Voltage drop percentage: 2.63%
Voltage at the end: 58.52
And for 1% maximum drop:
https://www.calculator.net/voltage-drop-calculator.html?material=copper&wiresize=3.277&voltage=60.1&phase=dc&noofconductor=1&distance=40&distanceunit=feet&eres=7.82&x=58&y=27
10 AWG:
Voltage drop: 0.62
Voltage drop percentage: 1.04%
Voltage at the end: 59.48
And from the panel specification, your maximum series fuse/breaker would be 15 Amps... So, a minimum of 14 AWG cable would be required (following US NEC code).
For single and 2x arrays, you do not need a series fuse/breaker(s). However, it is nice to use a breaker because you can use it as a switch when debugging/working on your system (turn off array without having to pull a power lead from the controller).
-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 4
Thank you so much.
Looking at various solar company sites I see many offering 50 foot extension cables with MC4 connectors already attached. So 2 of those in 10 AWG will be ordered.
Again, many thanks
My pleasure.
-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 3,621 ✭✭✭✭✭
Unless you have a need for mc4 connector on both ends of the extension, it may make more sense to get a 100' extention. The extension is then cut in half (or whatever is needed for each run), leaving a male and a female mc4 connector for the pv end, and a cut end for combiner or controller end.
Getting a single extension may be a bit cheaper, and can work better if the +ive and -ive ends of the array are a bit different wire distance from combiner/controller.
Off-grid.
Main daytime system ~4kw panels into 2xMNClassic150 370ah 48v bank 2xOutback 3548 inverter 120v + 240v autotransformer
Night system ~1kw panels into 1xMNClassic150 700ah 12v bank morningstar 300w inverter
As always, make sure you color wrap (colored electrical tape) your wiring (positive, negative, ground)...
Solar panels, if connected backwards, are simply "diodes", and can conduct large amounts of short circuit current (which will ruin the panel(s) if connected to other panels in parallel and/or connected directly to a battery bank).
-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Solar Expert Posts: 1,296 ✭✭✭✭
edited March 19 #7
So, the Female MC-4 connector of the extension cable plugs into the Male, positive output of the panels. Some people make the mistake of assuming the male connector is always the positive. Yes as Bill said Get some red electrical tape, if you don't already have some and carefully label the +ive with red tape.
2.1 Kw Suntech 175 mono, Classic 200, Trace SW 4024 ( 15 years old but brand new out of sealed factory box Jan. 2015), Bogart Tri-metric, 700 ah @24 volt AGM battery bank. Plenty of Baja Sea of Cortez sunshine. | 1,308 | 4,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-26 | latest | en | 0.855576 |
http://www.tutorialspoint.com/quantitative_aptitude/aptitude_progression.htm | 1,660,450,125,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571993.68/warc/CC-MAIN-20220814022847-20220814052847-00559.warc.gz | 88,488,090 | 8,343 | # Aptitude - Progression
## Sequence
A succession of numbers formed and arranged in a definite order according to a certain definite rule is called a sequence.
## Arithmetic Progression (A.P.)
It is a sequence in which each term, except the first one differs the preceding term by a constant. This constant is called the common difference. We denote the first term by a, common difference by d, nth term by Tṇ and the sum of first n terms by Sṇ.
### Examples
```5, 8,11,14,17...is an A.P. in which a=5 and d = (8-5) =3.
8, 5, 2,-1,-4,-7.... is an A.P. in which a = 8 and d = (5-8) = -3.
```
## General Term of an A.P.
In a given A.P., let first term =a, common difference =d. Then,
```Tn= a + (n-1) d.
Sum of n terms of an A.P.
Sn = n/2[2a+ (n-1) d]
Sn = n/2 (a + L), where L is the last term.
```
## Geometrical Progression (G.P.)
A sequence in which each term, except the first one bears a constant ratio with its preceding term, is called a geometrical progression, written as G.P. The constant ratio is called the common ratio of the G.P. We denote its first term by a and common ratio by r.
### Example
```2, 6, 18, 54, is a G.P.in which a=2 and r=6/2=3.
24, 12, 6, 3... Is a G.P. in which a = 24 and r = 12/24=1/2.
```
General Term of a G.P.: In a G.P. we have
```Tn= arn-1
Sum of n terms of a G.P.
Sn = a (1-rn)/ (1-r), When r < 1
a (r - 1n)/(r-1), When r > 1
```
## Arithmetic Mean
```A.M. of a and b = 1/2(a+b).
```
## Geometric Mean
```G.M. of a and b =√ab
```
## Some General Series
```(i) 1+2+3+4+…….+n=1/2n (n+1).
(ii) 12+22+32+42+……+n2 = n(n+1)(2n+1)/6
(iii) 13+23+33+43+…..+n3= {1/2 n(n+1)}2
```
## Solved Examples
Solved Examples
aptitude_progression.htm | 613 | 1,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-33 | latest | en | 0.924089 |
http://www.universetoday.com/14437/how-far-is-earth-from-the-sun/ | 1,490,335,876,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187717.19/warc/CC-MAIN-20170322212947-00351-ip-10-233-31-227.ec2.internal.warc.gz | 732,794,189 | 13,411 | ## How Far is Earth from the Sun?
Article Updated: 16 Mar , 2017
by
It’s amazing to think that for the majority of human history, we had almost no understanding about the Sun. We didn’t know what it was made of, how it formed, or how it produced energy. We didn’t know how big it was, and we didn’t know how far away it was.
We orbit the Sun at a distance of about 150 million kilometers. This number is actually an average, since we follow an elliptical path. At its closest point, the Earth gets to 147 million km, and at its most distant point, it’s 152 million km.
Distances in the Solar System are so vast that astronomers use this distance as a standard for measurement, and so the average distance from the Earth to the Sun is called an astronomical unit. Instead of saying that Pluto is 5.87 billion kilometers away from the Sun, astronomers say that it’s 39 astronomical units, or AUs.
You might be surprised to know that the distance from the Sun to the Earth was only determined within the last few hundred years. There were just too many variables. If astronomers knew how big it was, they could figure out how far away it was, or vice versa, but both of these numbers were mysteries.
Ancient astronomers, especially the Greeks, tried estimating the distance to the Sun in several different ways: measuring the length of shadows on Earth, or comparing the size of the Moon and its orbit to the Sun. Unfortunately, their estimates were off at least by a factor of 10.
The key to figuring out the distance to the Sun came from observing Venus as it passed directly in front of the Sun. This rare event, known as a Transit of Venus, happens only twice every 108 years. Once devised, the best opportunities for taking this precise measurement came during the Venus transits of 1761 and 1769. Astronomers were dispatched to remote corners of the globe to observe the precise moment when Venus began to move in front of the Sun, and when it had moved completely across the surface.
By comparing these measurements, astronomers could use geometry to calculate exactly how far away the Sun is. Their initial calculations put the distance at 24,000 times the radius of the Earth. Not bad considering our modern measurement of 23,455 times the radius of the Earth.
Modern astronomers can use radar and laser pulses to calculate the distance to objects in the Solar System. For example, they fire an intense beam of radio waves at a distant object, like Mercury, and then calculate how long it takes for the waves to bounce off the planet and return to Earth. Since the speed of light is well known, the return travel time tells you how far away the planet is.
Astronomy has truly helped us find our place in the Universe. It nice to be living in a time when many of these big mysteries have been solved. I don’t know about you, but I can’t wait to see what’s around the corner of the next discovery.
### 4 Responses
1. Shaynydra Lee says:
I’m happy that its a website on science because you can learn more about different planets and different areas!
2. cortez says:
this is a very cool website
3. b says:
nice program
4. jimmy says:
this website rox | 709 | 3,171 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-13 | latest | en | 0.957624 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/385/C100.C4.html | 1,713,219,656,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00546.warc.gz | 423,096,779 | 7,144 | Copied to
clipboard
## G = C100.C4order 400 = 24·52
### 1st non-split extension by C100 of C4 acting faithfully
Aliases: C20.1F5, C100.1C4, D50.3C4, C251M4(2), Dic25.6C22, C25⋊C81C2, C4.(C25⋊C4), C5.(C4.F5), C50.2(C2×C4), C10.7(C2×F5), (C4×D25).3C2, C2.4(C2×C25⋊C4), SmallGroup(400,29)
Series: Derived Chief Lower central Upper central
Derived series C1 — C50 — C100.C4
Chief series C1 — C5 — C25 — C50 — Dic25 — C25⋊C8 — C100.C4
Lower central C25 — C50 — C100.C4
Upper central C1 — C2 — C4
Generators and relations for C100.C4
G = < a,b | a100=1, b4=a50, bab-1=a43 >
Smallest permutation representation of C100.C4
On 200 points
Generators in S200
```(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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100)(101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200)
(1 103 76 128 51 153 26 178)(2 110 25 171 52 160 75 121)(3 117 74 114 53 167 24 164)(4 124 23 157 54 174 73 107)(5 131 72 200 55 181 22 150)(6 138 21 143 56 188 71 193)(7 145 70 186 57 195 20 136)(8 152 19 129 58 102 69 179)(9 159 68 172 59 109 18 122)(10 166 17 115 60 116 67 165)(11 173 66 158 61 123 16 108)(12 180 15 101 62 130 65 151)(13 187 64 144 63 137 14 194)(27 185 50 146 77 135 100 196)(28 192 99 189 78 142 49 139)(29 199 48 132 79 149 98 182)(30 106 97 175 80 156 47 125)(31 113 46 118 81 163 96 168)(32 120 95 161 82 170 45 111)(33 127 44 104 83 177 94 154)(34 134 93 147 84 184 43 197)(35 141 42 190 85 191 92 140)(36 148 91 133 86 198 41 183)(37 155 40 176 87 105 90 126)(38 162 89 119 88 112 39 169)```
`G:=sub<Sym(200)| (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200), (1,103,76,128,51,153,26,178)(2,110,25,171,52,160,75,121)(3,117,74,114,53,167,24,164)(4,124,23,157,54,174,73,107)(5,131,72,200,55,181,22,150)(6,138,21,143,56,188,71,193)(7,145,70,186,57,195,20,136)(8,152,19,129,58,102,69,179)(9,159,68,172,59,109,18,122)(10,166,17,115,60,116,67,165)(11,173,66,158,61,123,16,108)(12,180,15,101,62,130,65,151)(13,187,64,144,63,137,14,194)(27,185,50,146,77,135,100,196)(28,192,99,189,78,142,49,139)(29,199,48,132,79,149,98,182)(30,106,97,175,80,156,47,125)(31,113,46,118,81,163,96,168)(32,120,95,161,82,170,45,111)(33,127,44,104,83,177,94,154)(34,134,93,147,84,184,43,197)(35,141,42,190,85,191,92,140)(36,148,91,133,86,198,41,183)(37,155,40,176,87,105,90,126)(38,162,89,119,88,112,39,169)>;`
`G:=Group( (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100)(101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200), (1,103,76,128,51,153,26,178)(2,110,25,171,52,160,75,121)(3,117,74,114,53,167,24,164)(4,124,23,157,54,174,73,107)(5,131,72,200,55,181,22,150)(6,138,21,143,56,188,71,193)(7,145,70,186,57,195,20,136)(8,152,19,129,58,102,69,179)(9,159,68,172,59,109,18,122)(10,166,17,115,60,116,67,165)(11,173,66,158,61,123,16,108)(12,180,15,101,62,130,65,151)(13,187,64,144,63,137,14,194)(27,185,50,146,77,135,100,196)(28,192,99,189,78,142,49,139)(29,199,48,132,79,149,98,182)(30,106,97,175,80,156,47,125)(31,113,46,118,81,163,96,168)(32,120,95,161,82,170,45,111)(33,127,44,104,83,177,94,154)(34,134,93,147,84,184,43,197)(35,141,42,190,85,191,92,140)(36,148,91,133,86,198,41,183)(37,155,40,176,87,105,90,126)(38,162,89,119,88,112,39,169) );`
`G=PermutationGroup([[(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100),(101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144,145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176,177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192,193,194,195,196,197,198,199,200)], [(1,103,76,128,51,153,26,178),(2,110,25,171,52,160,75,121),(3,117,74,114,53,167,24,164),(4,124,23,157,54,174,73,107),(5,131,72,200,55,181,22,150),(6,138,21,143,56,188,71,193),(7,145,70,186,57,195,20,136),(8,152,19,129,58,102,69,179),(9,159,68,172,59,109,18,122),(10,166,17,115,60,116,67,165),(11,173,66,158,61,123,16,108),(12,180,15,101,62,130,65,151),(13,187,64,144,63,137,14,194),(27,185,50,146,77,135,100,196),(28,192,99,189,78,142,49,139),(29,199,48,132,79,149,98,182),(30,106,97,175,80,156,47,125),(31,113,46,118,81,163,96,168),(32,120,95,161,82,170,45,111),(33,127,44,104,83,177,94,154),(34,134,93,147,84,184,43,197),(35,141,42,190,85,191,92,140),(36,148,91,133,86,198,41,183),(37,155,40,176,87,105,90,126),(38,162,89,119,88,112,39,169)]])`
34 conjugacy classes
class 1 2A 2B 4A 4B 4C 5 8A 8B 8C 8D 10 20A 20B 25A ··· 25E 50A ··· 50E 100A ··· 100J order 1 2 2 4 4 4 5 8 8 8 8 10 20 20 25 ··· 25 50 ··· 50 100 ··· 100 size 1 1 50 2 25 25 4 50 50 50 50 4 4 4 4 ··· 4 4 ··· 4 4 ··· 4
34 irreducible representations
dim 1 1 1 1 1 2 4 4 4 4 4 4 type + + + + + + + image C1 C2 C2 C4 C4 M4(2) F5 C2×F5 C4.F5 C25⋊C4 C2×C25⋊C4 C100.C4 kernel C100.C4 C25⋊C8 C4×D25 C100 D50 C25 C20 C10 C5 C4 C2 C1 # reps 1 2 1 2 2 2 1 1 2 5 5 10
Matrix representation of C100.C4 in GL4(𝔽7) generated by
4 6 5 6 2 1 4 6 1 4 0 5 2 1 3 4
,
5 0 1 2 1 5 6 0 3 4 1 5 3 3 0 3
`G:=sub<GL(4,GF(7))| [4,2,1,2,6,1,4,1,5,4,0,3,6,6,5,4],[5,1,3,3,0,5,4,3,1,6,1,0,2,0,5,3] >;`
C100.C4 in GAP, Magma, Sage, TeX
`C_{100}.C_4`
`% in TeX`
`G:=Group("C100.C4");`
`// GroupNames label`
`G:=SmallGroup(400,29);`
`// by ID`
`G=gap.SmallGroup(400,29);`
`# by ID`
`G:=PCGroup([6,-2,-2,-2,-2,-5,-5,24,121,55,50,3364,2896,178,5765,2897]);`
`// Polycyclic`
`G:=Group<a,b|a^100=1,b^4=a^50,b*a*b^-1=a^43>;`
`// generators/relations`
Export
×
𝔽 | 4,181 | 7,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.175471 |
https://prepfortech.in/leetcode-solutions/parallel-courses | 1,709,015,746,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474671.63/warc/CC-MAIN-20240227053544-20240227083544-00097.warc.gz | 470,289,501 | 9,768 | ## Similar Problems
Similar Problems not available
# Parallel Courses
## Companies:
LeetCode: Parallel Courses Leetcode Solution
Difficulty: Unknown
Topics:
Problem Statement:
You are given an integer n which denotes the number of courses offered by a university labeled from 1 to n. Some courses may have prerequisites, which is shown in a pair of integers [a, b] where a is the prerequisite course for b.
For example, if [1, 2] is a pair of integers showing that course 1 is a prerequisite for course 2, it means that to take course 2, you have to first take course 1.
Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.
If there are multiple correct answers, return any of them. If it is impossible to finish all courses, return an empty array.
Solution:
The problem can be represented as a graph problem. We will represent the prerequisites as edges between the graph nodes and then we can use a topological sort to find the valid path that satisfies all the prerequisites.
The basic idea for this algorithm is to first create an adjacency list of all the courses, where the prerequisite course points towards the course it is a prerequisite for (reversed order from the input array). We also create an array to store all the incoming edges for each course, which is a simple count of the number of times a prerequisite is required.
We create a queue and add the courses which don’t require prerequisites to be completed (i.e., those courses which have no incoming edges). This is the initial set of courses that can be completed.
We then follow the following steps:
1. Dequeue a course from the queue.
2. Check if this course leads to any other courses by checking its adjacency list.
3. For each course that the current course leads to, decrement the incoming edges count by one.
4. If this course leads to any other courses which have no incoming edges, add those courses to the queue as well.
5. Repeat steps 1-4 until the queue is empty.
After this algorithm finishes, we need to check if all courses have been completed or not.
Implementation:
We can use a hashmap to create the adjacency list. The key represents the course and the value is a list of all the courses that this course leads to.
To count incoming edges, simply loop over all of the prerequisites, where each prerequisite (b) adds an incoming edge count to the course (a), i.e., incomingEdges[b]++.
Code:
Here's the code that implements the above algorithm.
``````class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] incomingEdges = new int[numCourses];
Map<Integer, List<Integer>> adjList = new HashMap<>();
int[] result = new int[numCourses];
for (int i = 0; i < numCourses; i++) {
}
for (int[] prereq : prerequisites) {
incomingEdges[prereq[0]]++;
}
for (int i = 0; i < incomingEdges.length; i++) {
if (incomingEdges[i] == 0) {
queue.offer(i);
}
}
int index = 0;
while (!queue.isEmpty()) {
int course = queue.poll();
result[index++] = course;
for (int c : adjList.get(course)) {
incomingEdges[c]--;
if(incomingEdges[c] == 0) {
queue.offer(c);
}
}
}
if (index != numCourses) {
return new int[0];
}
return result;
}
}
``````
## Solution Implementation
``1`` | 746 | 3,287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-10 | latest | en | 0.918543 |
https://www.javahelps.com/2015/03/ | 1,547,932,843,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583681597.51/warc/CC-MAIN-20190119201117-20190119223117-00626.warc.gz | 835,008,510 | 48,358 | ## Bridge Design Pattern
Name: Bridge Design Pattern
Type: Structural Design Pattern
Purpose: Decouple abstraction from implementation
(In other words, using composite over inheritance)
Sample Problem and Solution:
Assume that you are required to create some Shapes with two different patterns. Some sample shapes are SolidColorCircle, GradientCircle, SolidColorSquare, and GradientSquare. A possible design is provided below.
public interface Shape {
public void draw();
}
public abstract class Circle implements Shape {
}
@Override
public void draw() {
}
}
public class SolidColorCircle extends Circle {
private String color;
public SolidColorCircle(double radius, String color) {
this.color = color;
}
@Override
public void draw() {
super.draw();
System.out.println("Fill with: " + color);
}
}
public class GradientCircle extends Circle {
private String colorX;
private String colorY;
public SolidColorCircle(double radius, String colorX, String colorY) {
this.colorX = colorX;
this.colorY = colorY;
}
@Override
public void draw() {
super.draw();
System.out.println("Fill with: " + colorX + " & " + colorY);
}
}
Square, SolidColorSquare, and GradientSquare are provided in GitHub.
Even though this tutorial is named as Advanced Android Database, there is nothing advanced in Android database. The last Basic Android Database tutorial explains the quick and dirty way of using SQLite database in your application. This tutorial uses the same problem discussed in the last article, but to explain the recommended way of database access in Android.
The Create, Read, Update and Delete (CRUD) related methods are not explained in this article since those are already explained in the Basic Android tutorial. If you are new to Android Database, please read the Basic Android Database article first and then continue this article in order to learn the best practices.
## Install Apache Tomcat on Ubuntu
Step 1:
Step 2:
Open the Terminal (Ctrl + Alt + T) and enter the following command to change the directory.
cd /opt/
Step 3:
Enter the command given below to extract the Tomcat from the ~/Downloads directory. If your downloaded file is in any other directory, replace the last parameter by the actual file path.
Step 4: (Optional)
Rename the folder name to apache-tomcat.
sudo mv apache-tomcat-8.0.26/ apache-tomcat/
Step 5:
Tomcat creates some files at the runtime inside this folder (Log files and some other configuration files). If you want to start Tomcat without root privilege, it is required to change the permission of this directory. Enter the following command to change the permission of apache-tomcat folder.
sudo chmod -R 777 apache-tomcat/
Step 6:
An environment variable has to be added to the system. Enter the following command in the terminal to open the /etc/environment in gedit.
sudo gedit /etc/environment
Step 7:
Add the following line at the end of the file. Save and close the gedit.
CATALINA_HOME="/opt/apache-tomcat"
The /etc/environment before the modification.
The /etc/environment after the modification.
Step 8:
Reload the environment variables to the current terminal using this command.
source /etc/environment
Step 9:
Enter the following command to start the Tomcat server.
\$CATALINA_HOME/bin/startup.sh
Step 10:
Visit to the following URL.
http://localhost:8080/
If you get the Apache Tomcat home page, you have successfully installed the Apache Tomcat in your system.
Step 11:
To stop the Tomcat, enter the following command.
\$CATALINA_HOME/bin/shutdown.sh
##### Integrate Tomcat with Eclipse
If you do not have the Eclipse, follow this link and install the Eclipse first.
Step 1:
Open the Eclipse.
Step 2:
Goto Windows → Preferences in the menu bar and select the Runtime Environments under the Server.
Step 3:
Step 4:
Select your Apache Tomcat version under the 'Apache' root and click 'Next'.
Step 5:
Browse and select the Tomcat installation directory.
Step 6:
Click on the 'Finish' button
Step 7:
Click on the 'OK' button to close the Preferences dialog.
Now you are ready to use Apache Tomcat from Eclipse.
## Install the latest Eclipse in Ubuntu
Step 1:
Step 2:
Open the Terminal (Ctrl + Alt + T) and enter the following command to change the directory.
cd /opt
Step 3:
Enter the command given below to extract the Eclipse from ~/Downloads directory. If your downloaded file is in any other directory, replace the last parameter by the actual file path.
Step 4:
Open another Terminal (Ctrl + Alt + T) and enter the following command to create a shortcut file for eclipse.
gedit eclipse.desktop
Step 5:
In the opened gedit, copy and paste the following text.
[Desktop Entry]
Name=Eclipse
Type=Application
Exec=/opt/eclipse/eclipse
Terminal=false
Icon=/opt/eclipse/icon.xpm
Comment=Integrated Development Environment
NoDisplay=false
Categories=Development;IDE;
Name[en]=Eclipse
Name[en_US]=Eclipse
Step 6:
Save and close the gedit.
Step 7:
Enter the following command in the terminal to install the shortcut.
sudo desktop-file-install eclipse.desktop
Now search for Eclipse in the dashboard and open it.
In Ubuntu, the default tooltip background is black, which makes it unreadable in Eclipse. To fix this problem, follow these steps.
Step 1:
Enter the following commands one by one, in a Terminal
sudo gedit /usr/share/themes/Ambiance/gtk-2.0/gtkrc
sudo gedit /usr/share/themes/Ambiance/gtk-3.0/settings.ini
sudo gedit /usr/share/themes/Ambiance/gtk-3.0/gtk-main.css
Change the values of tooltip_fg_color and tooltip_bg_color in all the files, to the following values and save the changes.
tooltip_fg_color:#000000
tooltip_bg_color:#f5f5c5
Step 2:
Restart the Eclipse.
## Factory Design Pattern
Name: Factory Design Pattern
Type: Creational Design Pattern
Purpose:
• Hide object creation logic
• Decouple classes from the clients
Sample Problem and Solution:
Consider a problem where you are required to create a library which provides some 2D Shapes. Other developers will use your library to create some shapes with random measurements. A possible design is provided here.
package com.javahelps.shapes;
public interface Shape {
public void draw();
}
package com.javahelps.shapes;
public class Circle implements Shape {
}
@Override
public void draw() {
System.out.println("Circle");
}
}
package com.javahelps.shapes;
public class Square implements Shape {
private int width;
public Square(int width) {
this.width = width;
}
@Override
public void draw() {
System.out.println("Square");
}
}
package com.javahelps.shapes;
public class Rectangle implements Shape {
private int width;
private int height;
public Rectangle(int width, int height) {
this.width = width;
this.height = height;
}
@Override
public void draw() {
System.out.println("Rectangle");
}
}
## Android Spinner
In Android combo box is called as Spinner. This tutorial shows you, how to use Spinner in your application.
Step 1:
Create an Android project with a project name “Spinner Sample” and a package name “com.javahelps.sample.spinner”.
Step 2:
Modify the activity_main.xml as shown below. Notice that there is a UI component Spinner which is the combo box used to display some names of animals.
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
tools:context=".MainActivity">
<TextView
android:id="@+id/txtInfo"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_alignParentTop="true"
android:text="Select an animal"
android:textAppearance="?android:attr/textAppearanceMedium" />
<Spinner
android:id="@+id/cmbAnimals"
android:layout_width="match_parent"
android:layout_height="wrap_content"
android:layout_alignParentLeft="true"
android:layout_alignParentStart="true"
android:layout_below="@+id/txtInfo" />
<RelativeLayout
android:layout_width="match_parent"
android:layout_height="match_parent"
android:layout_below="@id/cmbAnimals">
<ImageView
android:id="@+id/imgAnimal"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_centerHorizontal="true"
android:layout_centerVertical="true"/>
</RelativeLayout>
</RelativeLayout>
Step 3:
Right click on the drawable folder and select “Paste”.
## Custom Font in Android
This tutorial shows you, how to develop an application in your local language. Since some Android phones have problems with Unicode, you can use custom true-type fonts for your application. The same technique can be used to decorate your application using stylish English fonts as well. This tutorial develops an application to display a welcome message in Tamil. You can use your own font to display whatever you want.
Step 1:
Create an Android application “Custom Font” with a package name “com.javahelps.customfont”.
Step 2:
Add a TextView to the activity_main.xml as shown below.
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
tools:context=".MainActivity">
<TextView
android:id="@+id/txtMessage"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_centerHorizontal="true"
android:layout_centerVertical="true"
android:textAppearance="?android:attr/textAppearanceLarge" />
</RelativeLayout>
Step 3:
Right click on the “res” folder and select NewFolderAssets Folder.
## Immutability Of String
Java defines some objects as immutable objects; developers cannot modify those objects once they are created. String objects and wrapper class objects are the best examples for immutable objects. This article covers the immutable objects and the reason for having immutable objects using String as an example.
Before getting into the actual content, it is required to have a basic knowledge about the equality operator (==). Usually equality operator is used to compare two primitive data types, but it has a different meaning when it is used with references. To compare the content of two objects, equals method is used and the equality operator is used to check whether two references are referring same object.
public class EqualityOperator {
public static void main(String[] args) {
Student stu1 = new Student(100);
Student stu2 = new Student(100);
Student stu3 = stu1;
if (stu1 == stu2) { // false
System.out.println("stu1 == stu2");
}
if (stu1.equals(stu2)) { // true
System.out.println("stu1 equals stu2");
}
if (stu1 == stu3) { // true
System.out.println("stu1 == stu3");
}
}
}
For the complete project click on this link.
In the above example stu1 == stu2 returns false, because they refer two different objects. stu1 == stu3 returns true because both references are referring the same object.
## Android: Simple Calculator
This tutorial guide you to develop a basic Calculator Android application in Android Studio.
NOTE: There is a new article written on How to develop an Android calculator application using Kotlin: Android: Simple Calculator in Kotlin. The new article is the Kotlin version of this article. If you like to start with Kotlin, the recent article is recommended for you.
Step 1:
Create a new Android application project with an application name: “Calculator” and package name: “com.javahelps.calculator”.
Step 2:
By default, Android uses a green Android robot icon. In this project, we are going to use a custom application icon. Therefore, delete the default ic_launcher icon set from the “mipmap” folder.
Step 3:
Get any PNG image file for the application icon. (It is recommended to have a minimum size 256x256 pixels). This icon is used to display in the Google Play as well as in the applications menu of Android devices.
Step 4:
Right click on the “mipmap” folder and select New → Image Asset
Step 5:
Browse and select your icon as the image file and click on Next → Finish buttons. (Make sure that the resource name is: ic_launcher)
Step 6:
Replace the content of the activity_main.xml file by the following code. This code creates a TextView as the calculator number screen and some necessary buttons. TextView is used instead of EditText, in order to prevent manual user input using the default keypad of Android. In this code, some common properties of Buttons are not provided to reduce the length of this tutorial. In your code make sure that you have included these four attributes for all the Buttons.
android:layout_width="0dp"
android:layout_height="match_parent"
android:layout_weight="1"
android:textSize="30sp"
Name
Email *
Message * | 2,780 | 12,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-04 | longest | en | 0.75527 |
http://slideplayer.com/slide/3223100/ | 1,534,498,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211935.42/warc/CC-MAIN-20180817084620-20180817104620-00279.warc.gz | 358,570,854 | 24,996 | # Multiplication – Microprocessor
## Presentation on theme: "Multiplication – Microprocessor"— Presentation transcript:
Multiplication 353156 – Microprocessor
Asst. Prof. Dr. Choopan Rattanapoka and Asst. Prof. Dr. Suphot Chunwiphat
Review : ARM Instruction so far
ADD : Addition SUB : Subtraction MOV : Move AND : And BIC : And not ORR : Or EOR : Xor LSL : Shift left LSR : Shift right ASR : Arithmetic Shift Right ROR : Rotate right
Decimal Multiplication
In decimal Multiplying by 10 is same as shifting left by 1 54 * 10 = 540 120 * 10 = 1200 Multiplying by 100 is same as shifting left by 2 54 * 100 = 5400 120 * 100 = 12000 Multiplying by 10n is same as shifting left by n
Binary Multiplication
In Binary Multiplying by 2 (102) is same as shifting left by 1 112 * 102 = 1102 1012 * 102 = 10102 Multiplying by 4 (1002) is same as shifting left by 2 112 * 1002 = 11002 1012 * 1002 = Multiplying by 2n is same as shifting left by n
Multiply by Power of 2 via Shift Left
Since shifting is so much faster than the multiplication A good compiler usually notice for example C code multiplies by power of 2 and compiles it to a shift instruction Example In C: A = A * 8 In Assembly : MOV R0, R0, LSL #3
Shift, Add and Subtract for Multiplication
In Depth techniques : Example 1 : In C : R0 = 5 * R1 In ARM Assembly: R0 = (4 + 1) * R1 = (4 * R1) + R1 = R1 + (4 * R1) ADD R0, R1, R1, LSL #2 Example 2 : In C : R0 = 105 * R1 In ARM Assembly : R0 = (15 * 7) * R1 = (16 – 1) * (8 - 1) * R1 RSB R0, R1, R1, LSL # ; (16 – 1) * R1 RSB R0, R0, R0, LSL # ; (8 – 1) * R0
Multiply (unsigned) : Example
Multiplicand (810) Multiplier (910) X 1 1 1 1 =7210 m bits x n bits = m + n bit product Thus, 32 bits X 32 bits = 64 bits value
Multiply (signed) : Example
Multiplicand (-810) Multiplier (-710) X 1 1 1 1 = 7210
2’s Complement Multiplication
A sequence of 2’s complement addition except for last step, we do the subtraction. Multiplicand (-810) Multiplier (-710) X 1 1 1 1 1 1 1 1 2’s complement.. =5610 1 1 1 1
2’s Complement Multiplication (more example)
Multiplicand (-810) Multiplier ( 310) X 1 1 1 1 1 1 1 1 1 2’s complement.. 1 1 1 1 1 2’s complement.. 1 1 =2410
Multiplication Instructions
BASIC ARM Multiplication MUL Rd, Rm, Rs ; Rd = Rm * Rs ;signed [31:0] Warning : Rd can’t be the same as Rm Multiply-Long UMULL RdHi, RdLo, Rm, Rs ;unsigned SMULL RdHi, RdLo, Rm, Rs ;signed RdHi = Rm * Rs [63:32] RdLo = Rm * Rs [31:0] ** No Division Instruction in ARM ** Division has to be done in software through a sequence of shift/subtract/add instruction (too much for this class)
Exercise 1 Given After execute the instruction
R2 stores value -10 R3 stores value 2 After execute the instruction SMULL R0, R1, R2, R3 What is the value stores in R0 and R1 ?
353156 – Microprocessor Asst. Prof. Dr. Choopan Rattanapoka
Memory Access – Microprocessor Asst. Prof. Dr. Choopan Rattanapoka
ARM and Memory Up until now, we learnt the ARM instructions
ADD, SUB, LSL, LSR, ORR, XOR, MOV, ..etc. Remember ?, ARM arithmetic instructions only operate on registers, never directly on memory. Data Transfer Instructions transfer data between registers and memory: Memory to Register Register to Memory
Review : MOV with Immediate
MOV R0, #12 Immediate value must be fit in 8 bit Immediate value can get by rotate even number of bit MOV R0, #0xFF ; 0xFF rotate right by 24 bit MOV R0, #0x doesn’t work !! LDR R0, =0x123456 Assembler will allocate memory and store value 0x for you, then LDR instruction will load value from that address to the register Why not always use LDR instruction for immediate value ? Loading value from memory is much SLOWER than generating it by ALU.
Data Transfer : Memory to Register (1)
To transfer a word of data, we need to specify two things : Register Memory Address Think of memory as a single one- dimension array, so we can address it simply by supplying a pointer to a memory address Other times, we want to be able to offset from this pointer pointer 0x 0x 0x 0x C 0x offset
Data Transfer : Memory to Register (2)
To specify a memory address to copy from, specify two things : A register which contains a pointer to memory A numerical offset (in bytes), or a register which contain an offset The desired memory address is the sum of these two values Example 1: [ v1 , #8 ] Specifies the memory address pointed by the value in v bytes Example 2: [ v1, v2 ] Specifies the memory address pointed by the value in v1 + v2
Data Transfer : Memory to Register (3)
Load Instruction Syntax : LDR 1, [ 2, 3 ] 1 : register that will receive value 2 : register containing pointer to memory 3 : numerical offset in bytes or another shifted index register Example LDR R0, [ R1, #8] LDR R0, [ R1, R2] Offset Base register Index register
Example 1 Given a table of memory address and its stored data, What the value in R0, R1, and R2 after executing the programs AREA ex1, CODE, READONLY ENTRY start LDR R2, =0x MOV R3, #4 LDR R0, [R2, R3] LDR R1, [R2, #16] END Value Address 0x10 0x 0x20 0x 0x30 0x 0x50 0x C 0x 0x45 0x 0xFF 0x
Pre Indexed Load Instruction Syntax : LDR 1, [ 2, 3]! 1 : register that will receive value 2 : register containing pointer to memory 3 : numerical offset in bytes or another shifted index register This instruction will take 2 step 1) Register “1” loads the value from the memory pointed by the pointer in “2”, add “3” bytes 2) “2” is updated by computed sum of “2” and “3” Example : Let R1 = 0x LDR R0, [R1, #0x12] ! Load R0 mem[ R1 + 0x12 ] ; R0 mem [0x ] Update base register : R1 R1 + 0x ; now R1 = 0x
Example 2 Given a table of memory address and its stored data, What the value in R0, R1, R2 after execution the programs AREA ex2, CODE, READONLY ENTRY start LDR R2, =0x MOV R3, #4 LDR R0, [R2, R3] ! LDR R1, [R2, #16] ! END Value Address 0x10 0x 0x20 0x 0x30 0x 0x50 0x C 0x 0x45 0x 0xFF 0x
Post Indexed Load Instruction Syntax : LDR 1, [ 2 ] , 3 1 : register that will receive value 2 : register containing pointer to memory 3 : numerical offset in bytes or another shifted index register This instruction will take 2 step 1) Register “1” loads the value from the memory pointed by the pointer in “2” 2) “2” is updated by computed sum of “2” and “3” Example : Let R1 = 0x LDR R0, [ R1 ], #0x12 Load R0 mem[ R1] ; R0 mem [0x ] Update base register : R1 R1 + 0x ; now R1 = 0x
Example 3 Given a table of memory address and its stored data, What the value in R0, R1 and R2 after execution the programs AREA ex3, CODE, READONLY ENTRY start LDR R2, =0x MOV R3, #4 LDR R0, [R2], R3 LDR R1, [R2], #16 END Value Address 0x10 0x 0x20 0x 0x30 0x 0x50 0x C 0x 0x45 0x 0xFF 0x
Data Transfer : Register to Memory
For storing data in register to memory, we will use STR instruction STR instruction syntax is identical to LDR instruction syntax STR : store from register, so 32 bits or one word are stored from register to memory at a time Example : Let R1 = 0x , R2 = 0x8 STR R0, [R1, #0x12] Store R0 mem[ R1 + 0x12 ] ; R0 mem [0x ] STR R0, [R1, R2] Store R0 mem[ R1 + R2] ; R0 mem [0x ]
Pre Indexed Store Instruction
Pre Indexed Store Instruction Syntax : STR 1, [ 2, 3]! 1 : register that will send value 2 : register containing pointer to memory 3 : numerical offset in bytes or another shifted index register This instruction will take 2 step 1) Register “1” stores the value to the memory pointed by the pointer in “2”, add “3” bytes 2) “2” is updated by computed sum of “2” and “3” Example : Let R1 = 0x STR R0, [R1, #0x12] ! Store R0 mem[ R1 + 0x12 ] ; R0 mem [0x ] Update base register : R1 R1 + 0x12 ; now R1 = 0x
Post Indexed Store Instruction
Post Indexed Store Instruction Syntax : STR 1, [ 2 ] , 3 1 : register that will send value 2 : register containing pointer to memory 3 : numerical offset in bytes or another shifted index register This instruction will take 2 step 1) Register “1” stores the value to the memory pointed by the pointer in “2” 2) “2” is updated by computed sum of “2” and “3” Example : Let R1 = 0x STR R0, [ R1 ], #0x12 Store R0 mem[ R1] ; R0 mem [0x ] Update base register : R1 R1 + 0x ; now R1 = 0x
Example 4 Complete the values in the given table of memory address and its stored data after execution the programs AREA ex4, CODE, READONLY ENTRY start MOV R0, #0xFF LDR R2, =0x MOV R3, #8 STR R0, [R2, R3] STR R0, [R2], R3 STR R0, [R2, R3] ! END Value Address 0x 0x 0x 0x C 0x 0x
We can work with Local on-chip memory from address 0x4000 0000
ARM : Memory Map ON-CHIP RAM CODE (ROM) We can work with Local on-chip memory from address 0x to address 0x4000 FFFFF
Memory Alignment Don’t forget that sequential word addresses in ARM with byte addressing do not differ by 1 ARM requires that all words start at addresses that are multiples of 4 bytes 3 2 1 Aligned Not Aligned
Assignment 6 AREA hw6, CODE, READONLY
ENTRY start MOV R1, #1 LDR R2, =0x MOV R3, #4 STR R1, [R2, R3]! LSL R1, #2 LSL R1, #2 STR R1, [R2, R3]! LDR R2, =0x LDR R0, [R2, #8 ]! LDR R1, [R2] , #4 ADD R0, R0, R1, LSL #4 STR R0, [R2, #4] END Solve the following expression in 4- bits register step-by-step -5 x -2 5 x -2 Complete the value in memory table after program is executed Value Address ? 0x 0x 0x 0x C 0x | 3,033 | 9,195 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-34 | longest | en | 0.78991 |
http://schoolbag.info/mathematics/gmat/32.html | 1,485,123,966,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281649.59/warc/CC-MAIN-20170116095121-00321-ip-10-171-10-70.ec2.internal.warc.gz | 256,427,568 | 3,410 | Rectangular Solids and Cylinders - Geometry - Math Review - GMAT Quantitative Review
## 3.0Math Review
### 9. Rectangular Solids and Cylinders
rectangular solid is a three-dimensional figure formed by 6 rectangular surfaces, as shown below. Each rectangular surface is a face. Each solid or dotted line segment is an edge, and each point at which the edges meet is a vertex. A rectangular solid has 6 faces, 12 edges, and 8 vertices. Opposite faces are parallel rectangles that have the same dimensions. A rectangular solid in which all edges are of equal length is a cube.
The surface area of a rectangular solid is equal to the sum of the areas of all the faces. The volume is equal to
.
In the rectangular solid above, the dimensions are 3, 4, and 8. The surface area is equal to . The volume is equal to .
The figure above is a right circular cylinder. The two bases are circles of the same size with centers O and P, respectively, and altitude (height) is perpendicular to the bases. The surface area of a right circular cylinder with a base of radius r and height h is equal to (the sum of the areas of the two bases plus the area of the curved surface).
The volume of a cylinder is equal to , that is,
.
In the cylinder above, the surface area is equal to
,
and the volume is equal to
.
| 326 | 1,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-04 | longest | en | 0.855449 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-test-for-convergence-for-1-2n-1 | 1,726,802,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00894.warc.gz | 84,283,474 | 6,518 | # How do you test for convergence for 1/((2n+1)!) ?
Jul 30, 2015
In the case you meant "test the convergence of the series : sum_(n=1)^(oo)1/((2n+1)!)"
the Answer is : it $\textcolor{b l u e}{\text{converges}}$
#### Explanation:
To find out, we can use the ratio test.
That is, if $\text{U"_"n}$ is the ${n}^{\text{th}}$ term of this series
Then if, we show that ${\lim}_{n \rightarrow + \infty} \left\mid {\text{U"_("n"+1)/"U}}_{n} \right\mid < 1$
it means that the series converges
On the other if lim_(nrarr+oo)abs(("U"_("n"+1))/"U"_n)>1
it means that the series diverges
In our case
"U"_n=1/((2n+1)!)
$\text{ }$ and
"U"_("n"+1)=1/([2(n+1)+1]!)=1/([2n+3]!)
Hence, "U"_("n"+1)/"U"_n=1/((2n+3)!)÷1/((2n+1)!)=((2n+1)!)/((2n+3)!)
$\text{Notice that} :$
(2n+3)! =(2n+3)xx(2n+2)xx(2n+1)!
Just like : 10! =10xx9xx8!
We subtract $1$ each time to get the next
So we have,
"U"_("n"+1)/"U"_n=((2n+1)!)/((2n+3)(2n+2)(2n+1)!)=1/((2n+3)(2n+2))
Next we test,
${\lim}_{n \rightarrow + \infty} \left\mid {\text{U"_("n"+1)/"U}}_{n} \right\mid$
$= {\lim}_{n \rightarrow + \infty} \left\mid \frac{1}{\left(2 n + 3\right) \left(2 n + 2\right)} \right\mid = {\lim}_{n \rightarrow + \infty} \frac{1}{\left(4 {n}^{2} + 10 n + 6\right)} = \frac{1}{+ \infty} = 0 \text{ }$ and $0$ is less than $1$
Hence, it's quite safe to conclude that the series color(blue)"converges" ! | 606 | 1,369 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 20, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.682887 |
https://www.physicsforums.com/threads/how-can-i-calculate-the-force-that-is-applied-on-a-tube-by-an-another.767904/ | 1,527,116,604,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00038.warc.gz | 810,435,698 | 15,632 | How can I calculate the force that is applied on a tube by an another?
1. Aug 28, 2014
Transstars
Let's say there is two tubes(cylinders with no tops or bottoms) with charges $q_1$ and $q_2$, radii $b_1$ and $b_2$, lengths $l_1$ and $l_2$. These tubes are located along the axis of each other's surfaces like in this figure:
I know I have to use Coulomb's law rather than Gauss law since the tubes are along the axis of each other's surfaces.The question is how can I calculate the force between these two tubes? If I have to integrate something please show me how to do it since I have never taken a course on calculus.
P.S: I asked this question on another platform and I got that answer but I don't know how to use integral so I would be grateful if someone could make the equations that is mentioned in the answer below.
" 1. Start with the force of two point charges: you know this equation
$$F=\frac{Q_1Q_2}{4\pi\epsilon_0 r^2}$$
2. Integrate this force over an infinitesimally thin ring of charge: now you have the force of a ring on an off-axis point (hint: you only need the axial component - the radial components will cancel due to symmetry in the next step). The distance $R$ will be a function of position along the ring (since the point charge is off-axis)
3. Integrate over all possible points that constitute a second ring: now you have the force of one ring on another; with the two rings on the same axis, the force will be along that axis. This is easy since the axial force is the same everywhere (so no difficult integration needed - just multiply by $2\pi$ and take account of the "charge per unit length").
4. Integrate over the length of the first cylinder: now you have the force of a cylinder on a ring. This is a bit harder - you are in essence integrating the force over a series of rings of variable (axial) distance
5. Integrate over the second cylinder: this is the sum of the force between a cylinder and a series of rings of different distance to the cylinder.
Note that your expression for the on-axis force of the cylinder is not terribly helpful since the charges of the second cylinder are off-axis."
Last edited by a moderator: Aug 28, 2014
2. Aug 28, 2014
Andrew Mason
Welcome to PF Transstars!
Are you asking us to give you a course in calculus so you can work out the integral?
AM
3. Aug 28, 2014
Transstars
Thank you. Yes, that would be great if you could give me enough knowledge to make the equations and solve them. I can solve basic integrals but just them. I can't solve the complex ones. Also I would be such pleased if someone check my equations and answer after I learned and worked out the integrals. I hope you have understood me.
That would be also great that somebody could make the equations and solve them for me. | 669 | 2,782 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-22 | latest | en | 0.940888 |
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## #1 2019-01-11 04:04:18
xxStellaxx
Guest
### I don't how to solve this problem!
I don't know how to solve this problem! I just don't understand it. I know they want me to find the radius but I just don't how. Here's the word problem and the question I'm stuck on.
My mom made a round cake with a 10 inch radius. She cut the cake into 16 equal slices. When 1 slice was left, my sister and I both wanted it, so we agreed to cut it in half, but I like the icing more than she does, so we decided to cut it the "other way." In other words, the two pieces would not be symmetrical. The inside piece would contain less icing, and the outer piece would contain the icing.
5. What is the radius of a half-piece when it is cut as shown in the picture? (ie, where do I need to cut to make two equal halves out of a piece?)
## #2 2019-01-11 20:51:46
bob bundy
Registered: 2010-06-20
Posts: 8,442
### Re: I don't how to solve this problem!
hi xxStellaxx
Welcome to the forum.
This question has been asked many times. Post 18 of this thread:
http://www.mathisfunforum.com/viewtopic.php?id=18391
has a diagram that may help.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy
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Q: Psychology ( Answered , 1 Comment )
Question
Subject: Psychology Category: Miscellaneous Asked by: mbroom-ga List Price: \$25.00 Posted: 23 Sep 2003 06:50 PDT Expires: 23 Oct 2003 06:50 PDT Question ID: 259362
```A boy is walking down the street to get to his friends house, he doesn't have much time. He approaches the corner and he can make a left or a right. There is a boy standing on the corner (knowing that 50% of people are good and 50% are bad - (meaning the boy can lie), what is one question he can ask the boy to find out which way to go?```
```Hi mbroom and thanks for the question! I will assume that if a person is bad then he or she always lies and if a person is good then he or she always tells the truth. Furthermore, I will assume that the boy standing at the corner knows the correct direction. The traveler should ask the boy on the corner the following question: Question: "If I were to ask you if going to the left is the correct direction, would you answer 'yes'?" Let us first consider the case in which the boy on the corner is a liar: If the left is the correct direction, then the boy will answer 'yes' since answering 'yes' would be a lie: if the traveler were to ask him if going to the left is the correct direction he would *not* say 'yes'. If the left is the incorrect direction, then the boy will answer 'no' since answering 'no' would be a lie: if the traveler were to ask him if going to the left is the correct direction he would say 'yes'. Now let us consider the case in which the boy on the corner tells the truth: If the left is the correct direction, then the boy will answer 'yes' since answering 'yes' would be the truth: if the traveler were to ask him if going to the left is the correct direction, he would say 'yes' If the left is the incorrect direction, then the boy will answer 'no' since answering 'no' would be the truth: if the traveler were to ask him if going to the left is the correct direction, he would *not* say 'yes' OK, so the interesting result is that whether the boy is a liar or a truthteller, the answer he gives to the question will be the same. If he answers 'yes' then the traveler should go to the left. If he answers 'no' then the traveler should go to the right. Additional links: This puzzle is a version of a puzzle given by Raymond Smullyan. More such puzzles may be found in his book entitled: What is the Name of this Book? which can be found at: http://www.amazon.com/exec/obidos/tg/detail/-/0139550887/qid=1064327727/sr=1-13/ref=sr_1_13/104-1019769-4771157?v=glance&s=books More info on Smullyan can be found at: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Smullyan.html Search strategy: smullyan smullyan fork If you are unsatisfied with any portion of this answer please request clarification prior to submitting a rating. Best of luck! -blinkwilliams-ga```
mbroom-ga rated this answer: ```Blinkwilliams answer was not only thorough it was quick as well! Thanks, Blinkwilliams!```
```A more straightforward question is: "If I were to ask you which way to go, what would your answer be?" Following the same logic as above, the boy would then have to give you the correct directions. alex_p``` | 803 | 3,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-43 | latest | en | 0.963857 |
https://forums.eveonline.com/t/capital-ore-calculator/32191 | 1,544,474,289,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00041.warc.gz | 621,319,237 | 6,064 | # Capital Ore Calculator
(le mediateur) #1
Hey Hey
Im trying to work out the most efficient way to mine the roids that i need minerals for
Is there such a programme that tells you the precise amounts of ore required if the minerals are
inputted for e.g.
If i entered 54m trit, 6.6m pyerite, 1.2m mex, 500k isogen etc etc.
The programme would tell me that the optimal ores i need to mine would be for e.g. 1000 spod, 500 bistot, 199, arkonor etc etc ??
Thanks
(ISD Sakimura) #2
It is quite simple, each batch of 100 units of the ore (or 1 Compressed) yield Specific amounts of minerals, for example a Batch of Spodumain (as per āShow Infoā) will yield:
Tritanium (56000 Units)
Pyerite (12050 Units)
Mexallon (2100 Units)
Isogen (450 Units)
So with this information you just need to factor in your Repocessing Yield, lets say it is 85%, then you just multiply the yield by 0.85 and you will know how much you will receive per batch:
Tritanium (47600 Units)
Pyerite (10242 Units)
Mexallon (1785 Units)
Isogen (382 Units)
So now that you know how much you get per batch of Spodumain you can easily calculate how much of it you need to mine by: āMinerals neededā / āMinerals from reprocessingā = ānumber of batchesā * 100 = units of Spodumain you need to mine to achieve your required minerals.
As a source to figure out which ores to mine to get specific minerals, you can use this 3rd party website here
(PJHustle) #3
I donāt think that the response above is hitting what the OP was asking for. I have a spreadsheet that utilizes a google sheet add-on called Linear Optimization. You tell the config sheet what you need and what each ore gives you it then can calculate the best way to get to the required materials. There is some tweaking you need to do in order to get the minimal waste product but I have been working on this sheet for a few days and I am happy with the outcome.
I will help you create your own sheet if this is what you are referring to in your post. I have it so you can choose a ship (Capital or T1) and that references my database on build materials, sums it all up after calculating the ME level on the BPās and dumps the required totals of minerals. You then go to the calculations page (Linear Optimization) and it will then tell you how many units of compressed ores you need to reach your goal.
If this is something you are after I will help create it with you from the information I have forged together on my sheet. I would just give you mine but it is not the most user friendly and may layouts are often disliked
(le mediateur) #4
Hey. Yes that is exactly what iām looking for and a bonus that it covers Capital construction and differing ME levels.
I hope you can get in touch
Tnx
(Sgt Ocker) #5
Try this https://www.fuzzwork.co.uk/compression/ from the same site mentioned by ISD Sakimura but it actually does what you were asking.
(le mediateur) #6
I tested the fuzzwork site by asking for 100000 Trit and it came back suggesting i mine crokite??? Not sure that right LOL and the google spreadsheet wants me to request permissionsā¦
(le mediateur) #7
PJHustle knows what i need by the looks of his answer
(Betty Blap) #8
(Sgt Ocker) #9
you can adjust the fuzzwork calculator by changing how much extra you want. It will then default to ābetter valueā ores.
As for the spreadsheet. If you want to contact me in game I will give you a direct link to the one i use. No permissions needed.
(von Susla) #10
wrote for myself
https://forest-tales.tk
allows not only set minerals but also choose items to build and select allowed for using ores
(Astra Roux) #11
does https://forest-tales.tk calculate the best ore to buy? Eg: compressed Veldspar vs Compressed Arkonor because itās cheaper to buy the velspar?
(von Susla) #12
It optimizes result by the lowest cost (At least I think it so)
(KenFlorian) #13
@betty_blap
Thanks for a great tool.
(Ag3nt Jita) #14
There was a program called mining_optimizer that i used in the past. It does what you need dude
(KenFlorian) #15
@Betty_Blap
Thanks for the pointer to this tool. Very powerful. Did you right it?
Please confirm one thingā¦on the āOutput M3ā tab is the value in Column A the amount of Column B required to satisfy the requested amount of minerals?
Eg. I ran for 350M Tritaniumā¦Column A is 1,018,593 and Column B is Compressed Dense Veldspar.
(system) #16
This topic was automatically closed 90 days after the last reply. New replies are no longer allowed. | 1,183 | 4,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.915916 |
https://exponentiations.com/2-to-the-70th-power | 1,686,406,414,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00708.warc.gz | 286,619,473 | 23,595 | Home » Powers of 2 » 2 to the 70th Power
# 2 to the 70th Power
Welcome to 2 to the 70th power, our post about the mathematical operation exponentiation of 2 to the power of 70.
If you have been looking for 2 to the seventieth power, or if you have been wondering about 2 exponent 70, then you also have come to the right place. 🙂
The number 2 is called the base, and the number 70 is called the exponent. In this post we are going to answer the question what is 2 to the 70th power.
=
Reset
## What is 2 to the 70th Power?
2 to the 70th power is conventionally written as 270, with superscript for the exponent, but the notation using the caret symbol ^ can also be seen frequently: 2^70.
270 stands for the mathematical operation exponentiation of two by the power of seventy.
As the exponent is a positive integer, exponentiation means a repeated multiplication:
2 to the 70th power =
The exponent of the number 2, 70, also called index or power, denotes how many times to multiply the base (2).
Thus, we can answer what is 2 to the 70th power as
2 to the power of 70 = 270 = 11 8059162071 7411303424.
If you have come here in search of an exponentiation different to 2 to the seventieth power, or if you like to experiment with bases and indices, then use our calculator above.
To stick with 2 to the power of 70 as an example, insert 2 for the base and enter 70 as the index, also known as exponent or power.
2 to the 70th power is an exponentiation which belongs to the category powers of 2.
Similar exponentiations on our site in this category include, but are not limited, to:
Ahead is more info related to 2 to the 70 power, along with instructions how to use the search form, located in the sidebar or at the bottom, to obtain a number like 2^70.
## 2 to the Power of 70
Reading all of the above, you already know most about 2 to the power of 70, except for its inverse which is discussed a bit further below in this section.
Using the aforementioned search form you can look up many numbers, including, for instance, 2 to the power 70, and you will be taken to a result page with relevant posts.
Now, we would like to show you what the inverse operation of 2 to the 70th power, (270)−1, is. The inverse is the 70th root of 270, and the math goes as follows:
(270)−1
=
=
=
=
2
Because the index of 70 is a multiple of 2, which is even, in contrast to odd numbers, the operation produces two results: (270)−1
; the positive value is the principal root.
Make sure to understand that exponentiation is not commutative, which means that 270 ≠ 702, and also note that (270)-1 ≠ 2-70, the inverse and reciprocal of 270, respectively.
You already know what 2 to the power of 70 equals, but you may also be interested in learning what 2 to the negative 70th power stands for.
Next is the summary of our content.
## Two to the Seventieth Power
You have reached the concluding section of two to the seventieth power = 270. Two to the seventieth power is, for example, the same as 2 to the power 70 or 2 to the 70 power.
Exponentiations like 270 make it easier to write multiplications and to conduct math operations as numbers get either big or small, such as in case of decimal fractions with lots of trailing zeroes.
If you have been looking for 2 power 70, what is 2 to the 70 power, 2 exponent 70 or 70 power of 2, then it’s safe to assume that you have found your answer as well.
If our explanations have been useful to you, then please hit the like button to let your friends know about our site and this post 2 to the 70th power.
And don’t forget to bookmark us.
## Conclusion
In summary,
If you like to learn more about exponentiation, the mathematical operation conducted in 270, then check out the articles which you can locate in the header menu of our site.
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Posted: September 8, 2014 in Advance, Isolation Drills
6.RP.A.1
Representing ratios
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Finding Percents
Percentage Word Problems 1
Solving Ratios Problems withProportions 1
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Analyzing an Identifying Proportional Relationships
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Discount, Tax and Tip Word Problems
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Using MUL & SHRD instead of DIV
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```Hi all,
I read somewhere someone saying that these days
it is common to use multiply & shift in lieu of
division because doing so is faster.
I did a quick test and found that indeed it is
faster with a Core 2 DUo. What took 102 seconds
for DIV required only 10.6 seconds using MUL with SHRD.
; Using DIV
mov eax, ecx
mov esi, 123
xor edx, edx
div esi
; Using MUL & SHRD
mov eax, ecx
mov esi, (1<<25)/123
mul esi
shrd eax, edx, 25
However there is a problem: The multiply & shift
approach is giving me results that are not
precise. They are usually off by 1.
Can anyone suggest a way to improve
the precision of this much-faster approach?
Thanks.
```
0
See related articles to this posting
```On Jun 29, 8:48=A0pm, Noop <questione...@nospicedham.yahoo.com> wrote:
> Hi all,
>
> I read somewhere someone saying that these days
> it is common to use multiply & shift in lieu of
> division because doing so is faster.
>
> I did a quick test and found that indeed it is
> faster with a Core 2 DUo. What took 102 seconds
> for DIV required only 10.6 seconds using MUL with SHRD.
>
> ; Using DIV
> mov eax, ecx
> mov esi, 123
> xor edx, edx
> div esi
>
> ; Using MUL & SHRD
> mov eax, ecx
> mov esi, (1<<25)/123
> mul esi
> shrd eax, edx, 25
>
> However there is a problem: The multiply & shift
> approach is giving me results that are not
> precise. They are usually off by 1.
>
> Can anyone suggest a way to improve
> the precision of this much-faster approach?
Picking the right constant to multiply by is a bit more complex than
what you're doing, plus many divisions require an adjustment step
after the multiplication by the reciprocal. For an excellent
discussion (and programs to generate the sequences needed), see the
"Software Optimization Guide for AMD64 Processors," particularly 8.1
and 8.8:
http://support.amd.com/us/Embedded_TechDocs/25112.PDF
For example, an unsigned division by 123, generates:
G:\wessel>udiv
Unsigned division by constant
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D
enter divisor: 123
; dividend: register other than EAX or memory location
MOV EAX, 085340853h
MUL dividend
SHR EDX, 6
; quotient now in EDX
While a division by 1000 generates the rather simpler:
G:\wessel>udiv
Unsigned division by constant
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D
enter divisor: 1000
; dividend: register other than EAX or memory location
MOV EAX, 010624DD3h
MUL dividend
SHR EDX, 6
; quotient now in EDX
```
0
```"Noop" wrote:
> I read somewhere someone saying that these days
> it is common to use multiply & shift in lieu of
> division because doing so is faster.
>
> I did a quick test and found that indeed it is
> faster with a Core 2 DUo. What took 102 seconds
> for DIV required only 10.6 seconds using MUL with SHRD.
>
> ; Using DIV
> mov eax, ecx
> mov esi, 123
> xor edx, edx
> div esi
>
> ; Using MUL & SHRD
> mov eax, ecx
> mov esi, (1<<25)/123
> mul esi
> shrd eax, edx, 25
>
> However there is a problem: The multiply & shift
> approach is giving me results that are not
> precise. They are usually off by 1.
>
> Can anyone suggest a way to improve
> the precision of this much-faster approach?
quick and dirty:
_____
2**32/123 = 34918433.3
lets now truncute the fraction and show it in hex: 0214D021h
MOV ecx,00214D021h ;reciprocal divisor
MOV eax,dividend ;
MUL ecx
;no shift required ;edx holds the result,
precise enough ?
If not then you need to imply some fraction-bits ie:
_____
2**34/123 = 139673733.2
this become 08534085h but needs an additional shr edx 2
and the valid input range is limited.
still not enough precise ?
then you may end up with Robert's example from AMD:
_____
2**38/123 = 2234779731 (fraction disappeared now!) == 85340853h
__
wolfgang
```
0
```Noop wrote:
> Hi all,
>
> I read somewhere someone saying that these days
> it is common to use multiply& shift in lieu of
> division because doing so is faster.
[snip]
> However there is a problem: The multiply& shift
> approach is giving me results that are not
> precise. They are usually off by 1.
>
> Can anyone suggest a way to improve
> the precision of this much-faster approach?
I reinvented this approach many years ago, and so did Agner Fog. He has
written a very nice guide in one of his optimization guides (google is
The key is that if you want N-bit accurate results, you need an N+1 bit
accurate scaled reciprocal value. Said reciprocal must also be rounded
UP, so that after the truncation which happens after the multiplication,
the slightly too large result becomes exact.
The easiest method is to first calculate the largest power of two, such
that when you divide by the divisor the result is still less than 2^32.
If the fraction is greater than 0.5, you can simply round up the result
and you have a 32-bit reciprocal which is 33-bit accurate.
Another poster used x / 1000 as an example:
512 is the largest power of two that is less than 1000, so we calculate
512 * 65536 * 65536 / 1000 = 2199023255.552
Rounding up we get 2199023256, so to divide by 1000 we can do:
mov eax,2199023256
mul ebx ; Number to be divided
shr edx,9 ; 512 = 2^9
and we get the result in about 5 cycles.
If otoh the fraction is less than 0.5, you need one more bit in the
reciprocal, or you must accept that the reciprocal only works for a
smaller range of input numbers.
Let's try x / 123:
C:\>perl -e "print(64*65536*65536/123)"
2234779731.25203
Since the fraction is less than 0.5 we need to multiply by 2234779731.5
and shift right 6 bits:
mov eax,2234779731
mul ebx
shr ebx,1
shr edx,6
We can also multiply by 4469559462.50407 (i.e. 4469559463 when rounded
up) and shift down 7 bits.
4469559463 is 2^32 + 174592167, i.e. we multiply the original value by
174592167 and then add the same value to EDX:
mov eax,174592167
mul ebx
shr edx,7
It is ~15 years since I originally worked out this, afair there might by
some corner cases for that last simplified approach.
I do remember that you can calculate this during runtime when you know
that you are going to make a lot of divisions by the same value and use
constant code to do the divide:
mov eax,[reciprocal]
mul ebx
and eax,ebx
mov ecx,[shift]
shr edx,cl
I.e. the code above takes 6-8 cycles, vs 40+ for a DIV.
Terje
--
- <Terje.Mathisen at tmsw.no>
"almost all programming can be viewed as an exercise in caching"
```
0
```On Jun 29, 11:36=A0pm, "robertwess...@yahoo.com"
<robertwess...@nospicedham.yahoo.com> wrote:
> MOV EAX, 085340853h
> MUL dividend
> SHR EDX, 6
Interesting, thanks.
I decided to put this ADD/ADC extra step to the test, and
wrote assembly code that tries to comprehensively test every
possible value -- a brute force approach.
I have found that very large dividends, e.g.
0x80000000, don't work. Smaller ones like 0x10000000
down to 1 are fine. But in any program where
dividends could be any value, this faster approach won't work.
For example, my code shows that dividends starting
at 0x4000000 and decreasing, I get:
Success with divisor 4 using const=3D0x40000000
Success with divisor 6 using const=3D0x2aaaaaaa
Success with divisor 7 using const=3D0x24924924
Success with divisor 10 using const=3D0x19999999
Success with divisor 20 using const=3D0x0ccccccc
Success with divisor 26 using const=3D0x09d89d89
....
I get far fewer successes:
Success with divisor 641 using const=3D0x00663d80
```
0
```Terje Mathisen wrote:
> I reinvented this approach many years ago, and so did Agner Fog. He has
> written a very nice guide in one of his optimization guides (google is
And so am I :-)
http://www.agner.org/optimize/
http://www.agner.org/optimize/optimizing_assembly.pdf
16.9 Division (all processors)
Integer division by a constant (all processors)
Dividing by a constant can be done by multiplying with the reciprocal.
A useful algorithm for integer division by a constant is as follows:
[snip]
```
0
```On Jun 30, 6:36=A0am, "robertwess...@yahoo.com"
<robertwess...@nospicedham.yahoo.com> > > I read somewhere someone
saying that these days
> > it is common to use multiply & shift in lieu of
> > division because doing so is faster.
>
> > I did a quick test and found that indeed it is
> > faster with a Core 2 DUo. What took 102 seconds
> > for DIV required only 10.6 seconds using MUL with SHRD.
If your benchmark is large amount of divisions and not much else then
results might be accurate, but not useful. It's hard to find use-case
where the workload is a lot of divisions and not much else. ;)
It would be interesting to see how much of the processor you are
testing with is idle. You only found a bottleneck, but it won't be one
when you don't need the result immediately. The (i)div isn't a
blocking instruction. Of course, if this is used in a really tight
loop and the result is needed in the current iteration then it might
again stall the pipe and slow your code down.
Just saying that things aren't as simple as counting clock cycles for
individual instructions.
```
0
```hanukas wrote:
> On Jun 30, 6:36 am, "robertwess...@yahoo.com"
> <robertwess...@nospicedham.yahoo.com> > > I read somewhere someone
> saying that these days
>>> it is common to use multiply& shift in lieu of
>>> division because doing so is faster.
>>
>>> I did a quick test and found that indeed it is
>>> faster with a Core 2 DUo. What took 102 seconds
>>> for DIV required only 10.6 seconds using MUL with SHRD.
>
>
> If your benchmark is large amount of divisions and not much else then
> results might be accurate, but not useful. It's hard to find use-case
> where the workload is a lot of divisions and not much else. ;)
>
> It would be interesting to see how much of the processor you are
> testing with is idle. You only found a bottleneck, but it won't be one
> when you don't need the result immediately. The (i)div isn't a
> blocking instruction. Of course, if this is used in a really tight
> loop and the result is needed in the current iteration then it might
> again stall the pipe and slow your code down.
>
> Just saying that things aren't as simple as counting clock cycles for
> individual instructions.
For AMD:
DIV reg/mem _____________ VectorPath
IDIV reg/mem _____________ VectorPath
The fastest DIV_ is 16 VP, the slowest 72 VP
The fastest IDIV is 18 VP, the slowest 79 VP
IMUL reg8 ________________ DirectPath Single 3
IMUL reg16 _______________ VectorPath ______ 4
IMUL reg16, imm16 ________ VectorPath ______ 4
IMUL reg16, mem16 ________ DirectPath Single 6
IMUL reg16, mem16, imm ___ VectorPath ______ 7
IMUL reg16, reg16 ________ DirectPath Single 3
IMUL reg16, reg16, imm ___ VectorPath ______ 4
IMUL reg32 DirectPath ____ Double __________ 3
IMUL reg32, imm32 ________ DirectPath Single 3
IMUL reg32, mem32 ________ DirectPath Single 6
IMUL reg32, mem32, imm ___ VectorPath ______ 7
IMUL reg32, reg32 ________ DirectPath Single 3
IMUL reg32, reg32, imm ___ DirectPath Single 3
IMUL reg64 _______________ DirectPath Double 5
IMUL reg64, imm32 ________ DirectPath Single 4
IMUL reg64, mem64 ________ DirectPath Single 7
IMUL reg64, mem64, imm ___ VectorPath ______ 8
IMUL reg64, reg64 ________ DirectPath Single 4
IMUL reg64, reg64, imm32 _ DirectPath Single 4
IMUL mem8 ________________ DirectPath Single 6
IMUL mem16 _______________ VectorPath ______ 7
IMUL mem32 _______________ DirectPath Double 6
IMUL mem64 _______________ DirectPath Double 8
MUL_ reg8 ________________ DirectPath Single 3
MUL_ reg16 _______________ VectorPath ______ 4
MUL_ reg32 _______________ DirectPath Double 3
MUL_ reg64 _______________ DirectPath Double 5
MUL_ mem8 ________________ DirectPath Single 6
MUL_ mem16 _______________ VectorPath ______ 7
MUL_ mem32 _______________ DirectPath Double 6
MUL_ mem64 _______________ DirectPath Double 8
VP blocks all three pipes for the given amount
of cycles. MUL and IMUL execute in pipe 0 (ex-
clusive). Subsequential MULs/IMULs are limited
to one operation per given amount of cycles.
AMD surely did not suggest to replace division
by multiplication if both were on par - 64 bit
IMUL reg or MUL reg need 5 cycles, the corres-
ponding DIV 72 and IDIV 79 cycles. Calculating
the speed gain: 1440 % for MUL vs. DIV, 1580 %
for IMUL vs. IDIV. Isn't that amazing? ;)
Greetings from Augsburg
Bernhard Schornak
```
0
```hanukas wrote:
> On Jun 30, 6:36 am, "robertwess...@yahoo.com"
> <robertwess...@nospicedham.yahoo.com> > > I read somewhere someone
> saying that these days
>>> it is common to use multiply& shift in lieu of
>>> division because doing so is faster.
>>
>>> I did a quick test and found that indeed it is
>>> faster with a Core 2 DUo. What took 102 seconds
>>> for DIV required only 10.6 seconds using MUL with SHRD.
>
>
> If your benchmark is large amount of divisions and not much else then
> results might be accurate, but not useful. It's hard to find use-case
> where the workload is a lot of divisions and not much else. ;)
Please take a look at the Ogg Vorbis audio codec:
A decoder needs to do a bit-perfect recreation of an encoded frequency
envelope, this basically requires a DIV for every sample (think pixel)
on the curve.
When I wrote the world's fastest decoder a few years ago, replacing
those divisions with on-the-fly (during header decoding) calculated
reciprocals was one of the optimizations that worked well.
(At the end I had to try about 10 different optimizations before I found
one that actually helped at least for most of my sample sound files.)
AFAIR, I also found one other location where reciprocals made sense, but
I'd have to re-read my code to verify this.
Terje
--
- <Terje.Mathisen at tmsw.no>
"almost all programming can be viewed as an exercise in caching"
```
0 | 4,051 | 13,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.887019 |
https://playtaptales.com/numbers/novendecinongentillinovendeciducentillion/ | 1,675,760,206,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00113.warc.gz | 465,637,869 | 4,445 | Novendecinongentillinovendeciducentillion
A Novendecinongentillinovendeciducentillion (1 Novendecinongentillinovendeciducentillion) is 10 to the power of 2757660 (10^2757660). This is a truly astronomical number!
How many zeros in a Novendecinongentillinovendeciducentillion?
There are 2,757,660 zeros in a Novendecinongentillinovendeciducentillion.
What's before Novendecinongentillinovendeciducentillion?
A Novendecinongentillioctodeciducentillion is smaller than a Novendecinongentillinovendeciducentillion.
What's after Novendecinongentillinovendeciducentillion?
A Novendecinongentillivigintiducentillion is larger than a Novendecinongentillinovendeciducentillion.
Novendecinongentillinovendeciducentillionaire
A Novendecinongentillinovendeciducentillionaire is someone whos assets, net worth or wealth is 1 or more Novendecinongentillinovendeciducentillion. It is unlikely anyone will ever be a true Novendecinongentillinovendeciducentillionaire. If you want to be a Novendecinongentillinovendeciducentillionaire, play Tap Tales!
Is Novendecinongentillinovendeciducentillion the largest number?
Novendecinongentillinovendeciducentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
Novendecinongentillinovendeciducentillion written out
Novendecinongentillinovendeciducentillion is written out as:
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| 520,947 | 1,042,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-06 | latest | en | 0.555724 |
https://www.physicsforums.com/threads/closed-timelike-curves-question.801321/ | 1,511,098,012,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805578.23/warc/CC-MAIN-20171119115102-20171119135102-00319.warc.gz | 830,748,126 | 15,040 | Closed timelike curves question
1. Mar 4, 2015
georgir
[Mentor's note: split from https://www.physicsforums.com/threads/gravity-instantaneous.801033/ ]
If masses can not just appear and disappear, how do solutions with closed timelike curves work? If you have some mass happily looping around such curve, from some other point of view you have some "moments" where that mass exists (twice, even, for both branches of the CTC) and then some later moments where it does not...
I'm having trouble imagining what could be the boundary between those and how it does not form a "discontinuity"...
I guess this is not entirely theoretical question too, as I've read that virtual particle-antiparticle pairs can actually be viewed as a single particle on a CTC.
Last edited by a moderator: Mar 4, 2015
2. Mar 4, 2015
Staff: Mentor
The conservation law for the stress-energy tensor doesn't work with "moments"; it doesn't depend on how you split up spacetime into space and time, and it doesn't say that the total amount of mass in a spacelike slice must be the same for every spacelike slice. It just says that, for any small region of spacetime, the amount of stress-energy "coming in" must equal the amount of stress-energy "going out". If you pick out any small region of spacetime that contains a short segment of the CTC followed by the mass, then the mass "comes in" to that small region and "goes out" of that small region, and it's the same amount of mass both times, so the conservation law is satisfied.
3. Mar 5, 2015
georgir
This is interesting... but what defines if the mass is "coming in" or is "going out" ?
If you take a small region of spacetime around the formation of the particle pair, you could say that it is just one particle, coming in and going out... or you could say that it is two particles going out, and require having enough energy coming in to that region from elsewhere to compensate.
4. Mar 5, 2015
Staff: Mentor
The mass is described by a 4-momentum vector; you just look at where the vector is pointing. On one side of a given small region of spacetime, the vector will be pointing in; on the other side, it will be pointing out. (The direction of the vector corresponds to the direction in which a clock carried by the mass is increasing its elapsed time.) | 551 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-47 | longest | en | 0.958866 |
http://mathhelpforum.com/pre-calculus/62126-completing-square.html | 1,529,344,796,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860684.14/warc/CC-MAIN-20180618164208-20180618184208-00127.warc.gz | 207,443,669 | 12,321 | 1. ## Completing the Square
What is the minimum point of the graph of the equation
y = 2x^2 + 8x + 9?
I understand the minimum point is the vertex (h, k).
To complete the square, I set the function to zero first.
2x^2 + 8x + 9 = 0
I then subtracted 9 from both sides.
2x^2 + 8x = -9
I then took half of the second term and squared it.
2x^2 + 8x = 16 - 9
Where do I go from here?
2. Hello !
y = 2x² + 8x + 9
y = 2 (x² + 4x + 9/2)
y = 2 ((x+2)²+1/2)
Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1
3. Originally Posted by magentarita
What is the minimum point of the graph of the equation
y = 2x^2 + 8x + 9?
I understand the minimum point is the vertex (h, k).
To complete the square, I set the function to zero first.
2x^2 + 8x + 9 = 0
I then subtracted 9 from both sides.
2x^2 + 8x = -9
I then took half of the second term and squared it.
2x^2 + 8x = 16 - 9
Where do I go from here?
===============================================
magentarita,
I see where your problem is.
2x^2 + 8x = -9 ------ ok so far.
Before you take half of the secod term and square,
Make sure the coefficient of suared term 2x^2 to be 1.
Therefore, 2x^2 + 8x = -9
becomes (x^2) + 4x = -9/2 -- simply didived by 2.
(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get
(x^2) + 4x + 2^2 = (-9/2 ) + 2^2
(x + 2)^2 = -1/2 now complete the square part is done at this point.
Rewrite original question to y = [(x + 2)^2 ]+ 1/2
Can you see y = a ( x-h )^2 + k --- standard form.
Vertex is at ( h, k ) = ( -2, 1/2) and a =1
Opens up vertically, therefor, mini. point is at vertex.
4. ## ok............
Originally Posted by 2976math
===============================================
magentarita,
I see where your problem is.
2x^2 + 8x = -9 ------ ok so far.
Before you take half of the secod term and square,
Make sure the coefficient of suared term 2x^2 to be 1.
Therefore, 2x^2 + 8x = -9
becomes (x^2) + 4x = -9/2 -- simply didived by 2.
(x^2) + 4x = -9/2 -- now you are ready to take half of the secod term and square, you get
(x^2) + 4x + 2^2 = (-9/2 ) + 2^2
(x + 2)^2 = -1/2 now complete the square part is done at this point.
Rewrite original question to y = [(x + 2)^2 ]+ 1/2
Can you see y = a ( x-h )^2 + k --- standard form.
Vertex is at ( h, k ) = ( -2, 1/2) and a =1
Opens up vertically, therefor, mini. point is at vertex.
ok but none the choices given for the minimum point is (-2, 1/2).
(2,33)
(2,17)
(-2,-15)
(-2,1)
5. ## ok....
Originally Posted by running-gag
Hello !
y = 2x² + 8x + 9
y = 2 (x² + 4x + 9/2)
y = 2 ((x+2)²+1/2)
Therefore the minimum point of the graph is such as x+2=0 => x=-2 and y=1
You got the right answer. But how did you get y = 1?
I got y = 1/2
6. Originally Posted by magentarita
You got the right answer. But how did you get y = 1?
I got y = 1/2
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1
7. Originally Posted by magentarita
ok but none the choices given for the minimum point is (-2, 1/2).
(2,33)
(2,17)
(-2,-15)
(-2,1)
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.
Reason: I forgot to divid y by 2.
Remember orginal question: y =( 2x^2 ) + 8x +9
(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side
(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side
y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side
y = a [ ( x-h)^2 ] + k
where a=2, h=-2, k=1 sorry
8. I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.
From $\displaystyle y=ax^2+bx+c$, you want to convert to $\displaystyle y=a(x-h)^2+k$, where $\displaystyle (h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.
$\displaystyle y=2x^2+8x+9$ Factor 2 out of first 2 terms
$\displaystyle y=2(x^2+4x)+9$ Next, complete the square in parentheses.
$\displaystyle y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side
$\displaystyle y=2(x+2)^2+1$ Now, we're in vertex form.
$\displaystyle V(h, k)=V(-2, 1) = \ \ minimum \ \ point$
9. ## ok.........
Originally Posted by Moo
Substitute x= -2 into the expression :
2 [ (-2+2)² + 1/2 ] = 2 [ 0+1/2 ] = 1
By subbing -2 for x, I get y = 1.
I got it.
10. ## ok..........
Originally Posted by 2976math
I found the problem!
y should be 1 not 1/2, so answer (-2,1) is correct.
Reason: I forgot to divid y by 2.
Remember orginal question: y =( 2x^2 ) + 8x +9
(1/2) y = (x^2) + 4x + ( 9/2) -- divid by 2 both side
(1/2) y = [( x +2)^2 ] + (1/2) -- complete square right side
y = 2 [( x +2)^2 ] + 1 ----- multiply 2 both side
y = a [ ( x-h)^2 ] + k
where a=2, h=-2, k=1 sorry
Be be sorry, be right. We all make mistakes, right?
11. ## ok..........
Originally Posted by masters
I think you were originally trying to solve for the minimum point by finding the vertex through the method of 'completing the square'.
From $\displaystyle y=ax^2+bx+c$, you want to convert to $\displaystyle y=a(x-h)^2+k$, where $\displaystyle (h, k)$ represent the vertex of your parabola and thus the maximum or minimum point depending on a<0 or a>0.
$\displaystyle y=2x^2+8x+9$ Factor 2 out of first 2 terms
$\displaystyle y=2(x^2+4x)+9$ Next, complete the square in parentheses.
$\displaystyle y=2(x^2+4x+4)+9-8$ Notice that we added 2 times 4, so we subtract 8 to balance the side
$\displaystyle y=2(x+2)^2+1$ Now, we're in vertex form.
$\displaystyle V(h, k)=V(-2, 1) = \ \ minimum \ \ point$
I enjoyed the steps as a guide. Well-done! | 2,018 | 5,592 | {"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} | 4.34375 | 4 | CC-MAIN-2018-26 | latest | en | 0.900637 |
https://www.jiskha.com/display.cgi?id=1265987445 | 1,503,380,784,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110471.85/warc/CC-MAIN-20170822050407-20170822070407-00372.warc.gz | 925,358,829 | 3,911 | # algebra
posted by .
Investment. Part of \$25,000 is invested at 10 percent, another part is invested at 12 percent, and a third part is invested at 16 percent. The total yearly income from these three investments is \$3,200. Furthermore, the income from the 16 percent investment yields the same amount as the sum of the incomes from the other two investments. How much is invested at each rate?
• algebra -
Let A, B, C be the amounts invested at 10, 12, and 16 percent.
A*.10+B*.12+C*.16=3200
C*.16=A*.10+B*.12
A+B+C=25,000
Can you solve it now?
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https://handwiki.org/wiki/Branch_and_bound | 1,721,075,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00370.warc.gz | 256,275,559 | 22,922 | # Branch and bound
Short description: Optimization by eliminating non optimal solutions to sub-problems
Branch and bound (BB, B&B, or BnB) is a method for solving optimization problems by breaking them down into smaller sub-problems and using a bounding function to eliminate sub-problems that cannot contain the optimal solution. It is an algorithm design paradigm for discrete and combinatorial optimization problems, as well as mathematical optimization. A branch-and-bound algorithm consists of a systematic enumeration of candidate solutions by means of state space search: the set of candidate solutions is thought of as forming a rooted tree with the full set at the root. The algorithm explores branches of this tree, which represent subsets of the solution set. Before enumerating the candidate solutions of a branch, the branch is checked against upper and lower estimated bounds on the optimal solution, and is discarded if it cannot produce a better solution than the best one found so far by the algorithm.
The algorithm depends on efficient estimation of the lower and upper bounds of regions/branches of the search space. If no bounds are available, the algorithm degenerates to an exhaustive search.
The method was first proposed by Ailsa Land and Alison Doig whilst carrying out research at the London School of Economics sponsored by British Petroleum in 1960 for discrete programming,[1][2] and has become the most commonly used tool for solving NP-hard optimization problems.[3] The name "branch and bound" first occurred in the work of Little et al. on the traveling salesman problem.[4][5]
## Overview
The goal of a branch-and-bound algorithm is to find a value x that maximizes or minimizes the value of a real-valued function f(x), called an objective function, among some set S of admissible, or candidate solutions. The set S is called the search space, or feasible region. The rest of this section assumes that minimization of f(x) is desired; this assumption comes without loss of generality, since one can find the maximum value of f(x) by finding the minimum of g(x) = −f(x). A B&B algorithm operates according to two principles:
• It recursively splits the search space into smaller spaces, then minimizing f(x) on these smaller spaces; the splitting is called branching.
• Branching alone would amount to brute-force enumeration of candidate solutions and testing them all. To improve on the performance of brute-force search, a B&B algorithm keeps track of bounds on the minimum that it is trying to find, and uses these bounds to "prune" the search space, eliminating candidate solutions that it can prove will not contain an optimal solution.
Turning these principles into a concrete algorithm for a specific optimization problem requires some kind of data structure that represents sets of candidate solutions. Such a representation is called an instance of the problem. Denote the set of candidate solutions of an instance I by SI. The instance representation has to come with three operations:
• branch(I) produces two or more instances that each represent a subset of SI. (Typically, the subsets are disjoint to prevent the algorithm from visiting the same candidate solution twice, but this is not required. However, an optimal solution among SI must be contained in at least one of the subsets.[6])
• bound(I) computes a lower bound on the value of any candidate solution in the space represented by I, that is, bound(I) ≤ f(x) for all x in SI.
• solution(I) determines whether I represents a single candidate solution. (Optionally, if it does not, the operation may choose to return some feasible solution from among SI.[6]) If solution(I) returns a solution then f(solution(I)) provides an upper bound for the optimal objective value over the whole space of feasible solutions.
Using these operations, a B&B algorithm performs a top-down recursive search through the tree of instances formed by the branch operation. Upon visiting an instance I, it checks whether bound(I) is equal or greater than the current upper bound; if so, I may be safely discarded from the search and the recursion stops. This pruning step is usually implemented by maintaining a global variable that records the minimum upper bound seen among all instances examined so far.
### Generic version
The following is the skeleton of a generic branch and bound algorithm for minimizing an arbitrary objective function f.[3] To obtain an actual algorithm from this, one requires a bounding function bound, that computes lower bounds of f on nodes of the search tree, as well as a problem-specific branching rule. As such, the generic algorithm presented here is a higher-order function.
1. Using a heuristic, find a solution xh to the optimization problem. Store its value, B = f(xh). (If no heuristic is available, set B to infinity.) B will denote the best solution found so far, and will be used as an upper bound on candidate solutions.
2. Initialize a queue to hold a partial solution with none of the variables of the problem assigned.
3. Loop until the queue is empty:
1. Take a node N off the queue.
2. If N represents a single candidate solution x and f(x) < B, then x is the best solution so far. Record it and set Bf(x).
3. Else, branch on N to produce new nodes Ni. For each of these:
1. If bound(Ni) > B, do nothing; since the lower bound on this node is greater than the upper bound of the problem, it will never lead to the optimal solution, and can be discarded.
2. Else, store Ni on the queue.
Several different queue data structures can be used. This FIFO queue-based implementation yields a breadth-first search. A stack (LIFO queue) will yield a depth-first algorithm. A best-first branch and bound algorithm can be obtained by using a priority queue that sorts nodes on their lower bound.[3] Examples of best-first search algorithms with this premise are Dijkstra's algorithm and its descendant A* search. The depth-first variant is recommended when no good heuristic is available for producing an initial solution, because it quickly produces full solutions, and therefore upper bounds.[7]
#### Pseudocode
A C++-like pseudocode implementation of the above is:
// C++-like implementation of branch and bound,
// assuming the objective function f is to be minimized
CombinatorialSolution branch_and_bound_solve(
CombinatorialProblem problem,
ObjectiveFunction objective_function /*f*/,
BoundingFunction lower_bound_function /*bound*/)
{
// Step 1 above
double problem_upper_bound = std::numeric_limits<double>::infinity; // = B
CombinatorialSolution heuristic_solution = heuristic_solve(problem); // x_h
problem_upper_bound = objective_function(heuristic_solution); // B = f(x_h)
CombinatorialSolution current_optimum = heuristic_solution;
// Step 2 above
queue<CandidateSolutionTree> candidate_queue;
// problem-specific queue initialization
candidate_queue = populate_candidates(problem);
while (!candidate_queue.empty()) { // Step 3 above
// Step 3.1
CandidateSolutionTree node = candidate_queue.pop();
// "node" represents N above
if (node.represents_single_candidate()) { // Step 3.2
if (objective_function(node.candidate()) < problem_upper_bound) {
current_optimum = node.candidate();
problem_upper_bound = objective_function(current_optimum);
}
// else, node is a single candidate which is not optimum
}
else { // Step 3.3: node represents a branch of candidate solutions
// "child_branch" represents N_i above
for (auto&& child_branch : node.candidate_nodes) {
if (lower_bound_function(child_branch) <= problem_upper_bound) {
candidate_queue.enqueue(child_branch); // Step 3.3.2
}
// otherwise, bound(N_i) > B so we prune the branch; step 3.3.1
}
}
}
return current_optimum;
}
In the above pseudocode, the functions heuristic_solve and populate_candidates called as subroutines must be provided as applicable to the problem. The functions f (objective_function) and bound (lower_bound_function) are treated as function objects as written, and could correspond to lambda expressions, function pointers and other types of callable objects in the C++ programming language.
### Improvements
When $\displaystyle{ \mathbf{x} }$ is a vector of $\displaystyle{ \mathbb{R}^n }$, branch and bound algorithms can be combined with interval analysis[8] and contractor techniques in order to provide guaranteed enclosures of the global minimum.[9][10]
## Applications
This approach is used for a number of NP-hard problems:
Branch-and-bound may also be a base of various heuristics. For example, one may wish to stop branching when the gap between the upper and lower bounds becomes smaller than a certain threshold. This is used when the solution is "good enough for practical purposes" and can greatly reduce the computations required. This type of solution is particularly applicable when the cost function used is noisy or is the result of statistical estimates and so is not known precisely but rather only known to lie within a range of values with a specific probability.[citation needed]
## Relation to other algorithms
Nau et al. present a generalization of branch and bound that also subsumes the A*, B* and alpha-beta search algorithms.[16]
## Optimization Example
Branch and bound can be used to solve this problem
Maximize $\displaystyle{ Z=5x_1+6x_2 }$ with these constraints
$\displaystyle{ x_1+x_2\leq 50 }$
$\displaystyle{ 4x_1+7x_2\leq280 }$
$\displaystyle{ x_1, x_2\geq0 }$
$\displaystyle{ x_1 }$ and $\displaystyle{ x_2 }$ are integers.
The first step is to relax the integer constraint. We have two extreme points for the first equation that form a line: $\displaystyle{ \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=\begin{bmatrix}50 \\0\end{bmatrix} }$ and $\displaystyle{ \begin{bmatrix}0 \\50\end{bmatrix} }$. We can form the second line with the vector points $\displaystyle{ \begin{bmatrix}0\\40\end{bmatrix} }$ and $\displaystyle{ \begin{bmatrix} 70\\0\end{bmatrix} }$.
the two lines.
The third point is $\displaystyle{ \begin{bmatrix}0\\0\end{bmatrix} }$. This is a convex hull region so the solution lies on one of the vertices of the region. We can find the intersection using row reduction, which is $\displaystyle{ \begin{bmatrix}70/3\\80/3\end{bmatrix} }$, or $\displaystyle{ \begin{bmatrix} 23.333\\26.667\end{bmatrix} }$ with a value of 276.667. We test the other endpoints by sweeping the line over the region and find this is the maximum over the reals.
We choose the variable with the maximum fractional part, in this case $\displaystyle{ x_2 }$ becomes the parameter for the branch and bound method. We branch to $\displaystyle{ x_2\leq26 }$ and obtain 276 @ $\displaystyle{ \langle 24,26\rangle }$. We have reached an integer solution so we move to the other branch $\displaystyle{ x_2\geq27 }$. We obtain 275.75 @$\displaystyle{ \langle 22.75, 27\rangle }$. We have a decimal so we branch $\displaystyle{ x_1 }$ to $\displaystyle{ x_1\leq22 }$ and we find 274.571 @$\displaystyle{ \langle 22,27.4286\rangle }$. We try the other branch $\displaystyle{ x_1\geq23 }$ and there are no feasible solutions. Therefore, the maximum is 276 with $\displaystyle{ x_1\longmapsto 24 }$ and $\displaystyle{ x_2\longmapsto 26 }$.
• Backtracking
• Branch-and-cut, a hybrid between branch-and-bound and the cutting plane methods that is used extensively for solving integer linear programs.
• Evolutionary algorithms
• Alpha-beta pruning
## References
1. A. H. Land and A. G. Doig (1960). "An automatic method of solving discrete programming problems". Econometrica 28 (3): 497–520. doi:10.2307/1910129.
2. Template:Cite tech report
3. Little, John D. C.; Murty, Katta G.; Sweeney, Dura W.; Karel, Caroline (1963). "An algorithm for the traveling salesman problem". Operations Research 11 (6): 972–989. doi:10.1287/opre.11.6.972.
4. Balas, Egon; Toth, Paolo (1983). Branch and bound methods for the traveling salesman problem (Report). Carnegie Mellon University Graduate School of Industrial Administration.
5. Bader, David A.; Hart, William E.; Phillips, Cynthia A. (2004). "Parallel Algorithm Design for Branch and Bound". in Greenberg, H. J.. Tutorials on Emerging Methodologies and Applications in Operations Research. Kluwer Academic Press. Retrieved 2015-09-16.
6. Mehlhorn, Kurt; Sanders, Peter (2008). Algorithms and Data Structures: The Basic Toolbox. Springer. p. 249.
7. Moore, R. E. (1966). Interval Analysis. Englewood Cliff, New Jersey: Prentice-Hall. ISBN 0-13-476853-1.
8. Jaulin, L.; Kieffer, M.; Didrit, O.; Walter, E. (2001). Applied Interval Analysis. Berlin: Springer. ISBN 1-85233-219-0.
9. Hansen, E.R. (1992). Global Optimization using Interval Analysis. New York: Marcel Dekker.
10. Conway, Richard Walter; Maxwell, William L.; Miller, Louis W. (2003). Theory of Scheduling. Courier Dover Publications. pp. 56–61.
11. Fukunaga, Keinosuke; Narendra, Patrenahalli M. (1975). "A branch and bound algorithm for computing k-nearest neighbors". IEEE Transactions on Computers (7): 750–753. doi:10.1109/t-c.1975.224297.
12. Narendra, Patrenahalli M.; Fukunaga, K. (1977). "A branch and bound algorithm for feature subset selection". IEEE Transactions on Computers C-26 (9): 917–922. doi:10.1109/TC.1977.1674939.
13. Hazimeh, Hussein; Mazumder, Rahul; Saab, Ali (2020). "Sparse Regression at Scale: Branch-and-Bound rooted in First-Order Optimization". arXiv:2004.06152 [stat.CO].
14. Nowozin, Sebastian; Lampert, Christoph H. (2011). "Structured Learning and Prediction in Computer Vision". Foundations and Trends in Computer Graphics and Vision 6 (3–4): 185–365. doi:10.1561/0600000033. ISBN 978-1-60198-457-9.
15. Nau, Dana S.; Kumar, Vipin; Kanal, Laveen (1984). "General branch and bound, and its relation to A∗ and AO∗". Artificial Intelligence 23 (1): 29–58. doi:10.1016/0004-3702(84)90004-3. | 3,318 | 13,803 | {"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} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.929099 |
https://nl.mathworks.com/matlabcentral/answers/1647580-defining-sequence-of-functions-in-matlab?s_tid=prof_contriblnk | 1,670,066,808,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00700.warc.gz | 448,527,526 | 27,763 | # Defining sequence of functions in MATLAB
20 views (last 30 days)
Saurabh Madankar on 11 Feb 2022
Commented: Saurabh Madankar on 11 Feb 2022
Hello,
The Haar mother wavelet is defined as-
Now we can define a sequence of functions -
where
So for i=1, j=0 and k=0, i=2, j=1,k=0, i=3, j=1, k=1 and so on and we will define the functions accordingly. Now I could define these functions individually. But if I need to define till then how do I define them using for loop or any other way so I don't have to define each function individually?
Saurabh Madankar on 11 Feb 2022
Edited: Saurabh Madankar on 11 Feb 2022
Can we define a function inside for loop? So if we can define a function whose input is i and outputs are j and k, then for i=1, the outputs will be j=0, k=0, for i=2 the outputs will be j=1, k=0, for i=3, j=1,k=1, for i=4, j=2,k=0,...,for i=7, j=2,k=3 and so on. So in each loop we will get j and k for the corresponding i before defining the main sequence function.
Walter Roberson on 11 Feb 2022
Well, you can do
syms h(t) [1 5]
whos
Name Size Bytes Class Attributes h 1x1 8 symfun h1 1x1 8 symfun h2 1x1 8 symfun h3 1x1 8 symfun h4 1x1 8 symfun h5 1x1 8 symfun t 1x1 8 sym
but although that creates abstract functions with those names, it does not in itself give you a way to define the functions.
syms() does give a way to define symbols as existing without assigning to them, but when you use syms() that way, you cannot give them expressions. sym() (without an s) cannot be used to define variables without assignining to them, and symfun() cannot be used to define variables without assigning to them either. And the "assigning to them" is the problem, as the name to be assigned to is changing.
It isn't that it can't be done... but it is seldom a good idea to use the dirty parts of MATLAB that allow defining variables with constructed names.
I would recommend that you instead use cell arrays.
h{i}(t) = expression involving h{i-1}
Note that even if you did use one of the techniques for creating dynamic variable names, each of the functions is going to end up fully expanding all earlier functions, rather than just referring to the functions. For example,
f(t) = t + 1
f(t) =
g(t) = f^2 + 2
g(t) =
Notice that g(t) does not refer to f(t) -- it evaluated f(t) to get t+1, and squared that result. This is inherent in the way that Symbolic Toolbox works.
Saurabh Madankar on 11 Feb 2022
Thanks, I used cell arrays to define the sequence of functions and using function inside for loop like I said in the comment I was able to solve the whole problem. This was a huge help. Thanks for your time. | 755 | 2,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.841818 |
https://www.physicsforums.com/threads/decimal-places.772737/ | 1,508,295,343,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822668.19/warc/CC-MAIN-20171018013719-20171018033719-00072.warc.gz | 968,146,592 | 21,969 | # Decimal places?
1. Sep 24, 2014
### Beer w/Straw
In 1 km races, runner 1 on track 1 (with time 2 min, 27.95
s) appears to be faster than runner 2 on track 2 (2 min, 28.15 s).
However, length L2 of track 2 might be slightly greater than length
L1 of track 1. How large can L2 - L1 be for us still to conclude that
runner 1 is faster?
$$\frac{1000 m}{147.95 s}=\frac{L m}{148.15 s}$$
I get 1.35 the text book gives an answer 1.4
I would have posted this in the homework section, but things have changed here and I haven't posted for years.
2. Sep 24, 2014
### jbriggs444
The lap times are given to a precision of .01 seconds. The delta in lap time is .30 seconds. Two significant figures. One way to approach the problem (not the approach that you took) would have used the delta time and made this more obviously important. Two sig figs in the inputs, two sig figs in the outputs.
The distance that is computed by taking 1000m times 148.15 divided by 147.95 (the approach that you did take) results in an answer with 5 significant figures. To 5 significant figures, that result is 1001.4. Subtract 1000 and you get 1.4.
Remember that when adding or subtracting, it is the least precise decimal place that controls, not the number of significant figures.
[insert some handwaving here about the precision of the 1000 meter figure being unimportant as long as it is known to at least 2 significant figures]
3. Sep 24, 2014
### Beer w/Straw
The way I did it got me 16 significant figures 1 001.351 808 043 258 I dropped the one thousandth and since the information was given to a precision of two decimals, I kept the last two.
Why is that a wrong answer?
:EDIT:
I tried posting this in the other subforum, but was having problems with the tex and and it wasn't user friendly.
Last edited: Sep 24, 2014
4. Sep 24, 2014
### jbriggs444
That's not the way significant figures work.
When you multiply or divide a set of numbers, the rule of thumb is that the number of significant figures in the result is taken from the number of significant figures in the inputs (whichever input has the least).
12.3 * 0.045678 = .562 (three digits)
When you add or subtract a set of numbers, the rule of thumb is that the decimal position of the last significant figure is taken from the decimal position of the last significant figure in the inputs (whichever input has it leftmost).
123.45 + 0.123 = 123.57 (the one hundredth's place).
In the case at hand you have both multiplication/division and you have addition/subtraction. So you do the operations in sequence. First the multiplication/division part which leaves you with five signficant figures. Then the addition/subtraction part. The last significant figure in the result is in the tenth's place. So the result of the addition/subtraction part is precise only to that digit.
You should keep all the computed digits in your intermediate results (good job) and keep track of which ones are significant so that you can round off at the end.
5. Sep 24, 2014
### Beer w/Straw
...I guess the text book wants me to use a less precise formula and I'm wrong because I'm doing it a different way.
6. Sep 24, 2014
### Khashishi
No, that's not the right attitude to take here. jbriggs has given the right answer and it has nothing to do with the book being less precise. The book is as precise as the data allows.
7. Sep 24, 2014
### Beer w/Straw
jbriggs444 said the precision was to two decimals .01 seconds. The answer is to .1 seconds.
What am I missing?
I did change 1km to 1000m . Isn't it only 1 significant digit for the whole thing by the 1km?
Last edited: Sep 24, 2014
8. Sep 24, 2014
### Staff: Mentor
The answer does not give seconds, it gives meters.
1.35 is not wrong, but the last 5 is not a true significant figure any more - it could equally be 6 or 4 (check that!) without a change in the problem statement, so we cannot trust this digit.
Another way to estimate the accurary: how far do they go in .01 seconds? It does not make sense to give an answer that is more precise than that.
9. Sep 24, 2014
### jbriggs444
Following the rules of significant digits, it would be 1 significant digit, yes. But given the context, it is also a race track whose length is roughly the same as another race track to within something in the neighborhood of 1.4 meters!! So we can justifiably consider it to be accurate to four significant figures.
10. Sep 24, 2014
### Beer w/Straw
And the text answer, in more than one edition, is only two figures with one decimal. While it explicitly gives a precision to two decimals and ambiguously gives the the number of significant figures in the question, -but gives no hint that it is only two significant figures.
$${Solve}\left[\frac{1000 m}{147.95 s}=\frac{L m}{148.15 s},L\right]$$
$$L\to 1001.35$$ Is the first answer I can get. (I only checked further to see if I should round the thousandth $$1001.351808043258$$
I was thinking about faxing them (I can't find an email) and I think I will. People may think less of me, but it gives me something to do.
:EDIT:
My tex shows up in preview but not on post
11. Sep 24, 2014
### jbriggs444
How many significant digits do you count in 1001.35? How many can be justified based on the input?
12. Sep 24, 2014
### Beer w/Straw
Don't you keep more significant digits during the process and then round off at the end?
1.4 is two, 1.35 is three are they not?
How many significant digits is "2 min 27.95 s"?
13. Sep 24, 2014
### jbriggs444
How many significant digits do you count in 1001.35? How many can be justified based on the input?
14. Sep 25, 2014
### Beer w/Straw
Isn't the significant digits a Red Herring in this case and the decimal precision is what matters?
If I do it differently, and fudge the math by rounding immediately, I could end up with answers of 0 or 1.5. However, the same approach without fudging the math gives me 1.35
Last edited: Sep 25, 2014
15. Sep 25, 2014
### dean barry
Constant speed is assumed (no acceleration at the start)
You can calculate the track length that gives runner two the identical velocity.
Start by calculating runner 1 velocity :
= 1000 / 147.95 = 6.7590420 m/s
Now calculate the track length required by runner 2 travelling at an identical speed :
= 6.7590420 * 148.15
= 1001.3518 metres
16. Sep 25, 2014
### jbriggs444
It is not clear what you mean by "decimal precision". Possibly you mean that since there are two places to the right of the decimal point in the inputs, there should be two decimal places to the right of the decimal point in the outputs. If so, that's flatly wrong.
Take 1234.5 times 0.0000054321. The "decimal precision" is at the 0.1 level. Rounding based on this gives, a result of 0.0. Using significant digits, the correctly rounded result is 0.0067059.
A proper error analysis would examine how uncertainty in the inputs propagates to uncertainty in the computed result. For instance, one might write down the complete formula, take its partial derivative with respect to each of the inputs and multiply each partial by the corresponding uncertainty. Keeping track of significant digits is a short cut rule of thumb approach that can produce roughly the same computed uncertainty without as much work.
17. Sep 25, 2014
### Beer w/Straw
And the number of significant figures for the question are one, four, or five. To two decimal places.
The text gives an answer of only two figures to one decimal place.
Is the text telling me I shouldn't bother computation and round immediately to have only one or two decimal positions? Or should I compute a certain amount of figures and then round at the end to something that gives less precision than the input?
How many significant figures should the answer to this yield 1000/147.95?
Should I round to two decimal places?
Last edited: Sep 25, 2014
18. Sep 25, 2014
### jbriggs444
The number of significant figures in "1000" is ambiguous. It can reasonably be taken as having one significant figure. It can reasonably be taken as having four. Without any further context one would reasonably say that 1000/147.95 has one significant figure.
However... In post #16 above I mentioned an approach using partial derivatives. Write down the complete formula for the result you will report in terms of the inputs you have:
$r$ is the computed result -- the number of meters longer track 2 needs to be than track 1
$l$ is length of the first track
$t_1$ is the first runner's time
$t_2$ is the second runner's time
$r = \frac{l t_2}{t_1} - l$
The partial derivitive of r with respect to l is $\frac{t_2}{t_1} - 1$ which is about 0.0014. An error of 1 meter in l results in an error of 1 millimeter in r. In other words, the number of significant digits in 1000 is not that signficant. In post #2 above I mentioned a handwave. This is what makes that handwave rigorous.
No. You should not round to two decimal places. You should pay attention to significant figures. Treat $l$ as having unlimited precision. $\frac{l t_2}{t_1}$ has five significant figures because $t_1$ and $t_2$ have five significant figures and the formula consists entirely of multiplications and divisions. That result is 1001.3518... The rightmost significant digit is the in the 0.1's place.
Now again treat $l$ as having unlimited precision. As above, the $\frac{l t_2}{t_1}$ has its rightmost significant digit in the 0.1's place. $\frac{l t_2}{t_1} - l$ is a formula involving addition and or subtraction of two terms, one of which is only precise to the 0.1's place. So that's where the result has its rightmost significant digit.
So the result is 1001.3518 - 1000 = 1.3518 rounded in the 0.1's digit to 1.4
19. Sep 25, 2014
### Beer w/Straw
You mean I should round before completing the entire calculation. Concerning myself more with the method rather than an answer.
20. Sep 25, 2014
### jbriggs444
No. You should not round before completing the calculation. But you should keep track of the digit position that you would have rounded to. That way you know the precision of the intermediate results. | 2,630 | 10,111 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2017-43 | longest | en | 0.953007 |
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