url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://jp.mathworks.com/matlabcentral/answers/376934-for-loop-with-multiple-variables | 1,669,480,728,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00598.warc.gz | 374,473,522 | 29,765 | # For loop with multiple variables
492 ビュー (過去 30 日間)
Vaishali Nagpure 2018 年 1 月 13 日
As in java we have "for (int i = 0, j = 1, k = 2; i < 5; i++)"
I would like to do as
for st = 1:numel(pointst) , ed = 1:numel(pointed)
What is the way i can do this kind of operation
##### 3 件のコメント表示非表示 2 件の古いコメント
Vaishali Nagpure 2018 年 1 月 13 日
Yes here those are constants what i am trying is for (int i = 1, j = 0; i <= limit; i++, j++)
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### 採用された回答
Jan 2018 年 1 月 13 日
If you want this:
Java: for (int i = 1, j = 0; i <= limit; i++, j++)
in Matlab:
j = 0;
for i = 1:limit
j = j + 1;
...
end
Less nice alternatives:
index = [1:limit; ...
1:limit];
for k = 1:size(index, 2)
i = index(1, k);
j = index(2, k);
...
end
or
index = [1:limit; ...
1:limit];
for k = index
i = k(1);
j = k(2);
...
end
The last one might look strange. The FOR loop evaluates the given index vector along the 2nd dimension, such that the column vectors are assigned to k.
Maybe this is useful for you:
indexI = 1:limit;
indexJ = 1:limit; % Of course this is a test data only
for k = 1:numel(index1)
i = indexI(k);
j = indexJ(k);
...
end
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### その他の回答 (1 件)
Rik 2018 年 1 月 13 日
If you mean a nested loop:
for st = 1:numel(pointst)
for ed = 1:numel(pointed)
end
end
##### 1 件のコメント表示非表示 なし
Vaishali Nagpure 2018 年 1 月 13 日
I want pointst first field and pointed first field at the same time in one iteration. Then in second iteration i want second field of pointst and pointed and so on ....
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# Description
Zadanie1. From the city A to city B are 5 roads in and out of the city in the city with - three roads. How many paths that pass through in the lead from A to C? Target 2. There are six pairs of gloves of different sizes. In how many ways you can choose the one glove on his left hand, and one - to the right so that the gloves were selected in different sizes? Task 3. Five girls and three boys playing in the towns. How many ways-mi they can divide into two teams of 4 people, if each team must have at least one boy? Quest 9. outing went to 92 people. Sandwiches with bulb-soi took 47 people with cheese - 38 people, with ham - 42 people, with cheese and sausage - 28 people, and sausage and ham - 31 people, and with cheese and | 260 | 923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.973629 |
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#1 2017-01-11 04:34:09
Zeeshan 01
Member
Registered: 2016-07-22
Posts: 726
Complex Loop
Any one can print this
******* *******
****** * ******
***** * *****
**** * * * ****
*** * ***
** * **
* *
Last edited by Zeeshan 01 (2018-06-22 15:30:53)
M.M.Zeeshan.K
Offline
#2 2017-01-11 23:36:32
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 812
Re: Complex Loop
In which programming language do you want the loop program to be written?
Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam
Offline
#3 2017-01-12 04:40:02
Zeeshan 01
Member
Registered: 2016-07-22
Posts: 726
Re: Complex Loop
In which programming language do you want the loop program to be written?
I prefer Java.
Last edited by Zeeshan 01 (2018-06-22 15:34:03)
M.M.Zeeshan.K
Offline
#4 2017-01-12 18:28:23
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,616
Re: Complex Loop
Hi;
Zeeshan 01 wrote:
Any one can print this
******* *******
****** ******
***** *****
**** ****
*** ***
** **
* *
Help me
By using loops
In which programming language do you want the loop program to be written?
Zeeshan 01 wrote:
I prefer c or cpp....
For ,,while or do while !!!
I don't know c or cpp, but here's a code in BASIC that will do it...
`````` a\$="*******":b\$=" ":c=7:d=1
WHILE c>0
PRINT RIGHT\$(a\$,c)+RIGHT\$(b\$,d*3)+RIGHT\$(a\$,c)
c=c-1:d=d+1
WEND``````
It runs in LibertyBASIC (paid software) and JustBASIC (freeware), but in most other BASIC programs the code will need some tweaking to work.
EDIT: Here's a slightly more streamlined version:
`````` a\$="*******":b\$=" ":c=7
WHILE c>0
PRINT RIGHT\$(a\$,c)+b\$+RIGHT\$(a\$,c)
c=c-1:b\$=b\$+" "
WEND``````
...and another variation:
`````` a\$="*******":b\$=" "
WHILE LEN(a\$)>0
PRINT a\$+b\$+a\$
a\$=RIGHT\$(a\$,LEN(a\$)-1)
b\$=b\$+" "
WEND``````
All three programs give the following output:
******* *******
****** ******
***** *****
**** ****
*** ***
** **
* *
Last edited by phrontister (2018-11-22 23:01:47)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
#5 2017-01-12 20:50:49
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,616
Re: Complex Loop
Zeeshan 01 wrote:
b\$=" ":c=7
B is empty???
No, b\$ isn't empty: it contains 21 blank spaces.
Each of the seven lines of printing in your first post contains two equal groups of asterisks that are separated by a group of blank spaces. Line 1 has 3 blank spaces, increasing by 3 in each subsequent line until reaching 21 blank spaces in the last (7th) line.
EDIT: The following code explanation is for the first program in my previous post.
RIGHT\$(b\$,d*3) for line 1 selects the right-most 3 blank spaces from b\$, and, using the increasing value of d, for each subsequent loop selects 3 more spaces from b\$ than for the previous line.
RIGHT\$(a\$,c) for line 1 selects the right-most 7 asterisks from a\$, and, using the reducing value of c, for each subsequent loop selects 1 less asterisk from a\$ than for the previous line.
Last edited by phrontister (2018-11-24 02:22:43)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
#6 2017-01-12 20:57:05
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,616
Re: Complex Loop
Here's a different method...
`````` a=8:b=0
WHILE a>1
a=a-1:b=b+3
FOR aa=1 TO a
PRINT "*";
NEXT aa
FOR bb=1 TO b
PRINT " ";
NEXT bb
FOR aa=1 TO a
PRINT "*";
NEXT aa
PRINT
WEND``````
And the output:
******* *******
****** ******
***** *****
**** ****
*** ***
** **
* *
Last edited by phrontister (2018-11-22 23:04:30)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
Offline
#7 2018-11-21 00:52:03
phrontister
Real Member
From: The Land of Tomorrow
Registered: 2009-07-12
Posts: 4,616
Re: Complex Loop
Hi Zeeshan 01;
Zeeshan 01 wrote:
Any one can print this
******* *******
****** * ******
***** * *****
**** * * * ****
*** * ***
** * **
* *
I thought I'd give your new puzzle a go, but unlike with your original puzzle, a simple solution has me stumped this time because it seems to lack symmetry now that you've added that internal cross. It's trickier now...hence your change of the thread title from "Help in simple prob of loop" to "Complex Loop", I suppose.
Zeeshan 01 wrote:
Thanks for help but I need c language code
I still don't know c, nor the other two programming languages you mentioned (cpp and Java); but anyway, in case it helps someone, here's my effort in BASIC, based on my first solution to your original puzzle:
EDIT: It's closer to the second solution, actually.
`````` a\$="*******":c=7
WHILE c>0
IF c=7 THEN b\$=" "
IF c=6 THEN b\$=" * "
IF c=5 THEN b\$=" * "
IF c=4 THEN b\$=" * * * "
IF c=3 THEN b\$=" * "
IF c=2 THEN b\$=" * "
IF c=1 THEN b\$=" "
PRINT RIGHT\$(a\$,c)+b\$+RIGHT\$(a\$,c)
c=c-1
WEND``````
My program's output:
******* *******
****** * ******
***** * *****
**** * * * ****
*** * ***
** * **
* *
Last edited by phrontister (2018-11-25 00:06:37)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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• paltry sum (of money)
腐れ金
くされがね
## 関連用語
paltry sum of money: はした金、わずかな金額 "Did you hear that George bought Ferrari in cash!!?" "Oh yeah, that's a paltry sum (of money) to him. You know how much he makes a year?" 「ジョージがフェラーリを現金で買ったって聞いたか？」「ああもちろん。そんなのやつにとっては、はし
small (paltry) sum (of money): small (paltry) sum (of money) 端金 端た金 はしたがね
paltry sum: はした金、わずかな金額 "Did you hear that George bought Ferrari in cash!!?" "Oh yeah, that's a paltry sum (of money) to him. You know how much he makes a year?" 「ジョージがフェラーリを現金で買ったって聞いたか？」「ああもちろん。そんなのやつにとっては、はし
a sum of money: a sum of money 也 なり
sum of money: 金額、貨幣総額
paltry: paltry 零細 れいさい いじましい
a certain sum of money: a certain sum of money 某かの金 なにがしかのかね
amount or sum (of money): amount or sum (of money) 額 がく
appropriate a sum of money for education: 金を教育｛きょういく｝に充てる
big sum of money: 大金｛たいきん｝
colossal sum of money: 莫大｛ばくだい｝な金額｛きんがく｝
considerable sum of money: かなりの大金｛たいきん｝
determined sum of money: 一定｛いってい｝の金額｛きんがく｝
enormous sum of money: 大金｛たいきん｝
equal sum of money: 同額｛どうがく｝のお金 All three parties lose an equal sum of money. 三方一両損 | 492 | 1,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.463058 |
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# Calculate ?cell for the following concentration cell ? Cu(s) | Cu2+ (aq, 0.025M) ? Cu2+ (aq,...
Calculate ?cell for the following concentration cell ? Cu(s) | Cu2+ (aq, 0.025M) ? Cu2+ (aq, 0.30M) | Cu(s)
Cu2+ (aq) + 2e- ? Cu(s) ? ?° = 0.340V
A)0.0638 V
B)-0.0734 V
C)0.0734 V
D)-0.0319 V
E)0.0139 V
cell contain same electrodes with
different electrolyte concentration .Such ells are known
as concentration cell.Current flow will happen by
flow of electrons from lower concentration half cell to
high concentration half cell.
we have
Ecell = E0cell-0.0591/n log Q
This type of cell E0cell= 0 [since both are same electrods]
So Ecell = -RT/nFlog Q** is the formula
Q is the reaction quotient = [product]/[reactant]
Cell voltage for concentration cell
Ecell = - 0.0592/n *log[Cu2+(dil)/Cu2+(conc)]
=- 0.0592/2*log[0.025/0.3] = -0.0319 V
**
option D)-0.0319 V
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Posted on
# Sorting Algorithms
Sorting involves arranging data in a collection based on a comparison algorithm.
There are two general families of sorting algorithms;
1.Linear Sorting - treat the problem of sorting as a single large operation
2.Divide and Conquer - partition data to be sorted into smaller sets that can
be independently sorted.
The performance of sorting algorithms can be measured in terms of:
1.Comparisons - number of times two values of an input array are compared for relative equality.
2.Swaps - number of times two values stored in the input are swapped.
I am going to show you the implementation of 5 sorting algorithms in JavaScript:
• Bubble sort
• Selection Sort
• Insertion Sort
• Merge Sort
• Quick Sort
I found this site really helpful in visualizing these algorithms.
## Bubble Sort
This is the simplest.
It works by repeatedly swapping values if they are the wrong position. Higher values are generally to the right and lower values are to the left.
Pseudocode
``````set swap counter to a truthy value
Repeat until the swap counter is a falsy value
Reset swap counter to a falsy value
If two adjacent elements are not in order
Swap them and set swap counter to truthy value
``````
Code
``````function bubbleSort(arr) {
let swapCounter = 1;
while (swapCounter) {
swapCounter = 0;
for (let i = 0; i < arr.length - 1; i++){
if (arr[i] > arr[i + 1]) {
const swapElement = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = swapElement;
swapCounter = 1;
}
}
}
return arr;
}
let arr = [64, 34, 25, 12, 22, 11, 90];
console.log(bubbleSort(arr))
// >> [11, 12, 22, 25,34, 64, 90]
``````
Performance
Worst case - O(n^2)
Best case - O(n^2)
## Selection Sort
It works by finding the smallest unsorted element and add it to the array in the first unsorted location
Pseudocode
``````Repeat until no sorted element remains:
Search the unsorted part of the data to find the smallest value
Swap the smallest value with the first element of unsorted part
``````
Code
``````function selectionSort(arr){
for (let i = 0; i < arr.length; i++){
for (let j = i + 1; j < arr.length; j++){
if (arr[j] < arr[i]) {
const swapElement = arr[i];
arr[i] = arr[j];
arr[j] = swapElement;
}
}
}
return arr;
}
let arr = [4, 2, 5, 1, 3];
console.log(selectionSort(arr))
// >> [1, 2, 3, 4, 5]
``````
Performance
Worst case - O(n^2)
Best case - O(n)
## Insertion Sort
This algorithm sorts items are they are encoumtered
Pseudocode
``````Call the first element of the array 'sorted'
Repeat until all the elements are sorted :
Look at the next unsorted element . Insert into the 'sorted' position by
shifting the required number of elements
``````
Code
``````function insertionSort(arr) {
for (let i = 1; i < arr.length; i++){
let unsorted = arr[i];
let idx = i - 1;
while (idx >= 0 && unsorted < arr[idx]) {
arr[idx + 1] = arr[idx];
idx -= 1;
}
arr[idx + 1] = unsorted;
}
return arr;
}
let arr = [4, 2, 5, 1, 3];
console.log(insertionSort(arr))
// >> [1, 2, 3, 4, 5]
``````
Perfomance
Worst case - O(n^2)
Best case - O(n)
## Merge Sort
Works by recursively splitting an array into two, sorting them and then combining these arrays in a sorted order
Pseudocode
``````Sort the left half of the array (Assuming n > 1)
Sort the right half of the array (Assuming n > 1)
Merge the two halves together
``````
Code
``````function mergeSort(arr) {
let length = arr.length
// if n is not > 1
// list is considered sorted
if (length === 1) {
return arr;
}
let midIdx = Math.ceil(length / 2);
let leftHalf = arr.slice(0, midIdx);
let rightHalf = arr.slice(midIdx, length);
leftHalf = mergeSort(leftHalf);
rightHalf = mergeSort(rightHalf);
return merge(leftHalf, rightHalf)
}
// merge both halfs
function merge(leftHalf, rightHalf) {
const sorted = []
while (leftHalf.length > 0 && rightHalf.length > 0) {
const leftItem = leftHalf[0]
const rightItem = rightHalf[0]
if (leftItem > rightItem) {
sorted.push(rightItem)
rightHalf.shift()
} else {
sorted.push(leftItem);
leftHalf.shift()
}
}
// if left half is not empty
while (leftHalf.length !== 0) {
sorted.push(leftHalf[0])
leftHalf.shift()
}
// if right half is not empty
while (rightHalf.length !== 0) {
sorted.push(rightHalf[0])
rightHalf.shift()
}
return sorted;
}
let arr = [4, 2, 5, 1, 3];
console.log(mergeSort(arr))
// >> [1, 2, 3, 4, 5]
``````
Performance
Worst case - O(nlogn)
Best case - O(nlogn)
## Quick Sort
Pseudocode
``````Repeat until sorted
Pick a pivot value and partition array
Put all value smaller than pivot to the left and larger values to the right
Perform pivot and partition on the left and the right partition
``````
Code
``````function swap(arr, leftIndex, rightIndex) {
const temp = arr[leftIndex];
arr[leftIndex] = arr[rightIndex];
arr[rightIndex] = temp;
}
function partition(arr, left, right) {
let pivot = arr[Math.floor((right + left) / 2)], //middle element
i = left, //left pointer
j = right; //right pointer
while (i <= j) {
// while left pointer is less than pivot
// move pointer to the right
while (arr[i] < pivot) {
i++;
}
// while righ pointer is greater than pivot
// move pointer to the left
while (arr[j] > pivot) {
j--;
}
// if left pointer is less than or equal to right pointe
// swap elements
// increment left pointer n decrement right pointer
if (i <= j) {
swap(arr, i, j); //sawpping two elements
i++;
j--;
}
}
return i; // index of left pointer
}
function quickSort(arr, left, right) {
let index;
if (arr.length > 1) {
index = partition(arr, left, right); //index returned from partition
if (left < index - 1) { //more elements on the left side of the pivot
quickSort(arr, left, index - 1);
}
if (index < right) { //more elements on the right side of the pivot
quickSort(arr, index, right);
}
}
return arr;
}
let arr = [4, 2, 5, 1, 3];
console.log(quickSort(arr, 0, arr.length - 1));
// >> [1, 2, 3, 4, 5]
``````
Perfomance
Worst Case - O(n^2)
Best Case - O(nlogn)
Note: Bubble sort, insertion sort and selection sort are linear sorting algorithms while merge sort and quick sort are divide and conquer algorithms.
Happy coding đ | 1,751 | 6,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-40 | latest | en | 0.786071 |
https://www.unitsconverters.com/en/Hectare-To-Squarenanometre/Unittounit-307-317 | 1,627,265,914,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00227.warc.gz | 1,118,115,158 | 36,651 | Formula Used
1 Hectare = 10000 Square Meter
1 Square Meter = 1E+18 Square Nanometer
1 Hectare = 1E+22 Square Nanometer
Hectares to Square Nanometers Conversion
ha stands for hectares and nm² stands for square nanometers. The formula used in hectares to square nanometers conversion is 1 Hectare = 1E+22 Square Nanometer. In other words, 1 hectare is 1E+22 times bigger than a square nanometer. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
Convert Hectare to Square Nanometer
How to convert hectare to square nanometer? In the area measurement, first choose hectare from the left dropdown and square nanometer from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from square nanometer to hectare? You can check our square nanometer to hectare converter.
How many Square Meter is 1 Hectare?
1 Hectare is equal to 10000 Square Meter. 1 Hectare is 10000 times Bigger than 1 Square Meter.
How many Square Kilometer is 1 Hectare?
1 Hectare is equal to 0.01 Square Kilometer. 1 Hectare is 100 times Smaller than 1 Square Kilometer.
How many Square Centimeter is 1 Hectare?
1 Hectare is equal to 100000000 Square Centimeter. 1 Hectare is 100000000 times Bigger than 1 Square Centimeter.
How many Square Millimeter is 1 Hectare?
1 Hectare is equal to 10000000000 Square Millimeter. 1 Hectare is 10000000000 times Bigger than 1 Square Millimeter.
Hectares to Square Nanometers Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like area finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like ha to nm² through multiplicative conversion factors. When you are converting area, you need a Hectares to Square Nanometers converter that is elaborate and still easy to use. Converting Hectare to Square Nanometer is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Hectare to Square Nanometer conversion along with a table representing the entire conversion.
Let Others Know | 603 | 2,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-31 | latest | en | 0.8652 |
http://www.java-forums.org/java-2d/17082-developing-hex-board-strategy-game.html | 1,369,511,390,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706153698/warc/CC-MAIN-20130516120913-00099-ip-10-60-113-184.ec2.internal.warc.gz | 535,064,013 | 18,929 | # Thread: Developing an Hex-board strategy game
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## Developing an Hex-board strategy game
Hello,
I am pretty new in programming java games, i tried some easy stuff and i got used to Applets/Applications, drawing methods etc. Now to the point.
I want to develop a java version of the board game Battlemasters, since i've noticed that no one else did it and it would be great for such a beautiful and funny game.
The basic idea, is to have every single Hexagon as an object, which can understand who are his neighbours and have the proper listeners and stuff.
It would be really appreciated some help, especially with the board, because i can't find a good way yet to implement it :confused: :eek:
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I've got it! :)
Here's the code for drawing an hexagonal board, assuming that you have a 61*55 transparent gif image of the hexagon.
public void paint(Graphics g) {
int a = 61; //dimensions
int b = 55; //
for(int i=0; i<13; i++) {
for(int j=0; j<10; j++) {
if (i%2 == 0)
g.drawImage(image, i*a - (a/4 + 1) * i, j*b, this);
else
g.drawImage(image, i*a - (a/4 + 1) * i, j*b - b/2, this);
}
}
}
3. abuse reported
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wtf are u talking about mister?
5. I'm talking about the post with blatant advertising for porn that was in the post just above my post and has since been removed by the moderators at my request. Any more questions?
6. Java Code:
```import java.awt.*;
import java.awt.event.*;
import java.awt.font.*;
import java.awt.geom.*;
import java.util.*;
import java.util.List;
import javax.swing.*;
public class HexTest extends JPanel {
final static int SIDES = 6;
double MIN_DIST = 0;
boolean showCenters = false;
boolean tipHigh = false;
boolean scaleDown = false;
static boolean debug = false; // main method switch -d
boolean firstTime = true;
HexCell[] cells;
double scale = 3/4.0;
protected void paintComponent(Graphics g) {
super.paintComponent(g);
Graphics2D g2 = (Graphics2D)g;
g2.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
double w = getWidth();
double h = getHeight();
double R = Math.min(w,h)/8;
MIN_DIST = R/4;
if(debug && firstTime)
System.out.printf("w = %.1f h = %.1f R = %.1f%n", w, h, R);
double delta = (1.0 - scale)/2;
if(scaleDown) {
// Draw everything smaller to see all the cells.
g2.translate(delta*w, delta*h);
g2.scale(scale, scale);
}
Rectangle r = getBounds();
r.grow((int)(R*3/4), (int)(R*3/4));
if(cells == null) {
initHexCells(w, h, R, r);
}
if(scaleDown) {
g2.setPaint(Color.green.darker());
g2.draw(r);
g2.setPaint(Color.blue);
g2.draw(getBounds());
}
g2.setPaint(Color.black);
for(int i = 0; i < cells.length; i++) {
cells[i].draw(g2);
}
if(scaleDown) {
g2.translate(-delta*w, -delta*h);
g2.scale(1.0/scale, 1.0/scale);
}
firstTime = false;
}
public Dimension getPreferredSize() {
return new Dimension(400,400);
}
private void initHexCells(double w, double h, double R, Rectangle range) {
Path2D.Double path = getPath(w/2, h/2, R);
Rectangle2D bounds = path.getBounds2D();
if(debug && firstTime)
System.out.printf("bounds = [%.1f, %.1f, %.1f, %.1f]%n",
bounds.getX(), bounds.getY(),
bounds.getWidth(), bounds.getHeight());
if(debug && firstTime)
List<Point2D.Double> list =
getAllPoints(bounds.getCenterX(),
if(debug && firstTime)
System.out.printf("list size = %d%n", list.size());
cells = new HexCell[list.size()];
// For HexCell to find the side that starts at zero degrees.
double theta = tipHigh ? -Math.PI/6 : 0;
for(int i = 0; i < list.size(); i++) {
String id = String.valueOf(i);
Point2D.Double p = list.get(i);
double x = p.x - w/2;
double y = p.y - h/2;
AffineTransform at =
AffineTransform.getTranslateInstance(x, y);
Shape s = at.createTransformedShape(path);
cells[i] = new HexCell(id, p, s, showCenters, theta, adjacentIds);
}
}
private String[] getNeighbors(int index, double radius,
List<Point2D.Double> list) {
// Collect neighbors clockwise starting at zero degrees.
String[] ids = new String[SIDES];
double thetaInc = tipHigh ? Math.PI/3 : Math.PI/6;
Point2D.Double center = list.get(index);
// Make ellipse larger to include the points we're
// looking for so we can use the intersects method.
for(int i = 0; i < list.size(); i++) {
if(i == index) continue;
Point2D.Double p = list.get(i);
if(e.contains(p)) {
// Get bearing to p.
double phi = Math.atan2(p.y-center.y, p.x-center.x);
// Avoid case of -0.0 for negative phi.
if(phi < 0.0 && phi < -0.0001) phi += 2*Math.PI;
// Index into array found with thetaInc.
int j = (int)Math.round(phi/thetaInc);
if(!tipHigh) j /= 2;
if(j < 0) j += 5;
if(j < ids.length) {
ids[j] = String.valueOf(i);
}
}
}
return ids;
}
private List<Point2D.Double> getAllPoints(double cx, double cy,
Rectangle range) {
Point2D.Double center = new Point2D.Double(cx, cy);
List<Point2D.Double> list = new ArrayList<Point2D.Double>();
Point2D.Double[] points = { new Point2D.Double(cx, cy) };
List<Point2D.Double> subList = null;
do {
List<Point2D.Double> nextPoints = new ArrayList<Point2D.Double>();
for(int i = 0; i < points.length; i++) {
subList = getPoints(points[i].x, points[i].y,
for(int j = 0; j < subList.size(); j++) {
Point2D.Double p = subList.get(j);
if(!haveCloseEnoughPoint(p, list)) {
}
}
}
points = nextPoints.toArray(new Point2D.Double[nextPoints.size()]);
} while(points.length > 0);
return list;
}
private List<Point2D.Double> getPoints(double cx, double cy,
Point2D.Double center) {
List<Point2D.Double> list = new ArrayList<Point2D.Double>();
double minDist = center.distance(cx, cy);
for(int i = 0; i < SIDES; i++) {
double theta = i*Math.PI/3;
if(!tipHigh) theta += Math.PI/6;
double x = cx + radius*Math.cos(theta);
double y = cy + radius*Math.sin(theta);
double distance = center.distance(x, y);
if(r.contains(x, y) && distance > minDist) {
}
}
return list;
}
private boolean haveCloseEnoughPoint(Point2D.Double p,
List<Point2D.Double> list) {
for(int i = 0; i < list.size(); i++) {
Point2D.Double next = list.get(i);
if(next.distance(p) < MIN_DIST) {
return true;
}
}
return false;
}
private Path2D.Double getPath(double cx, double cy, double R) {
Path2D.Double path = new Path2D.Double();
double thetaInc = 2*Math.PI/SIDES;
double theta = tipHigh ? -Math.PI/2 : thetaInc;
double x = cx + R*Math.cos(theta);
double y = cy + R*Math.sin(theta);
path.moveTo(x, y);
for(int i = 0; i < SIDES; i++) {
theta += thetaInc;
x = cx + R*Math.cos(theta);
y = cy + R*Math.sin(theta);
path.lineTo(x, y);
}
return path;
}
private JPanel getControls() {
String[] ids = { "show centers", "tip high", "scale down" };
boolean[] state = { showCenters, tipHigh, scaleDown };
ActionListener al = new ActionListener() {
public void actionPerformed(ActionEvent e) {
String id = e.getActionCommand();
boolean selected = ((JCheckBox)e.getSource()).isSelected();
if(id.equals("show centers")) {
showCenters = selected;
for(int i = 0; i < cells.length; i++) {
cells[i].setShowCenter(selected);
}
repaint();
} else {
if(id.equals("tip high")) {
tipHigh = selected;
}
if(id.equals("scale down")) {
scaleDown = selected;
}
reset();
}
}
};
JPanel panel = new JPanel();
for(int i = 0; i < ids.length; i++) {
JCheckBox cb = new JCheckBox(ids[i], state[i]);
cb.setActionCommand(ids[i]);
}
return panel;
}
public static void main(String[] args) {
if(args.length > 0)
debug = args[0].equals("-d");
HexTest test = new HexTest();
JFrame f = new JFrame("click me");
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.pack();
f.setLocation(100,100);
f.setVisible(true);
}
private void reset() {
cells = null;
firstTime = true;
repaint();
}
private MouseListener switcher = new MouseAdapter() {
public void mousePressed(MouseEvent e) {
Point p = e.getPoint();
if(scaleDown) {
double cx = getWidth()/2.0;
double cy = getHeight()/2.0;
double x = cx + (p.x - cx)/scale;
double y = cy + (p.y - cy)/scale;
p.setLocation((int)x, (int)y);
}
for(int i = 0; i < cells.length; i++) {
if(cells[i].contains(p)) {
cells[i].toggleSelection();
break;
}
}
repaint();
}
};
private ComponentListener resizeMonitor = new ComponentAdapter() {
public void componentResized(ComponentEvent e) {
reset();
}
};
}
class HexCell {
String id;
Point2D.Double center;
Shape shape;
boolean showCenter;
double start;
String[] neighbors;
boolean isSelected = false;
public HexCell(String id, Point2D.Double center, Shape shape,
boolean showCenter, double start, String[] neighbors) {
this.id = id;
this.center = center;
this.shape = shape;
this.showCenter = showCenter;
this.start = start;
this.neighbors = neighbors;
}
public void draw(Graphics2D g2) {
if(showCenter) {
g2.setPaint(Color.red);
g2.fill(new Ellipse2D.Double(center.x-1.5, center.y-1.5, 4, 4));
g2.drawString(id, (float)(center.x+3), (float)(center.y+3));
g2.setPaint(Color.black);
}
g2.draw(shape);
if(isSelected) {
// Show that we know who our neighbors are and where they live.
Font font = g2.getFont();
FontRenderContext frc = g2.getFontRenderContext();
LineMetrics lm = font.getLineMetrics("0", frc);
float sh = lm.getAscent() + lm.getDescent();
Rectangle r = shape.getBounds();
int R = Math.max(r.width, r.height)/2;
double thetaInc = 2*Math.PI/HexTest.SIDES;
double theta = start;
double lastX = 0, lastY = 0;
for(int i = 0; i <= neighbors.length; i++) {
double x = center.x + R*Math.cos(theta);
double y = center.y + R*Math.sin(theta);
if(i > 0 && neighbors[i-1] != null) {
float midx = (float)(x - (x - lastX)/2);
float midy = (float)(y - (y - lastY)/2);
double phi = Math.atan2(midy-center.y, midx-center.x);
String s = neighbors[i-1];
double sw = font.getStringBounds(s, frc).getWidth();
double diag = Math.sqrt(sw*sw + sh*sh)/2;
float sx = (float)(midx - diag*Math.cos(phi) - sw/2);
float sy = (float)(midy - diag*Math.sin(phi)) +
lm.getDescent();
g2.drawString(s, sx, sy);
}
lastX = x;
lastY = y;
theta += thetaInc;
}
}
}
public void setShowCenter(boolean show) { showCenter = show; }
public void toggleSelection() { isSelected = !isSelected; }
public boolean contains(Point p) {
return shape.contains(p);
}
public String toString() {
return "HexCell[id:" + id + ", neighbors:" +
Arrays.toString(neighbors) + "]";
}
}```
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Thank you very much! I will definitely make a good use of this code ;)
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Thanks a lot for the code!
9. @Hardwired
Me and all other beekeepers thank you for this!
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• | 2,995 | 10,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2013-20 | longest | en | 0.866826 |
https://math.stackexchange.com/questions/295029/two-definitions-of-bounded-variation-function | 1,726,122,241,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651422.16/warc/CC-MAIN-20240912043139-20240912073139-00449.warc.gz | 356,167,176 | 38,715 | # Two definitions of "Bounded Variation Function"
As far as I know, a function $f$ defined on an interval $[a, b]$ is said to be of bounded variation if $$\tag{1}V_a^b(f)=\sup\left\{\sum_{P} \lvert f(x_{j+1})-f(x_j)\rvert \ :\ P\ \text{partition of }[a, b]\right\}<\infty.$$ Today I discovered that another definition is in use for a function defined in an open subset $\Omega$ of $\mathbb{R}^n$, namely (cfr. Wikipedia) we say that $f$ is of bounded variation if $$\tag{2}V(f; \Omega)=\sup\left\{ \int_{\Omega}f(x)\,\text{div}\,\phi (x)\, dx\ :\ \phi\in C^1_c(\Omega),\ \lVert \phi\rVert_{\infty}\le 1 \right\}<\infty.$$
Even if the cited Wikipedia article treats the two definitions as if they were equivalent when $\Omega=(a, b)$, this does not seem to me to be the case. The Dirichlet function $\chi_{\mathbb{Q}\cap [0, 1]}$ is not of bounded variation in $(0, 1)$ in the sense of definition (1) but it is in the sense of definition (2).
Question. What is the precise relationship between the two definitions?
Thank you for reading.
• The 2nd definition operates with equivalence classes up to equality a.e. I think in one dimension it corresponds to having a representative that is BV in the first sense.
– user53153
Commented Feb 5, 2013 at 4:12
• @5PM: You are probably right. If you take as $\phi$ a step function then the integral in $(2)$ reduces to a sum similar to the one in $(1)$ (but without the absolute value). Maybe playing with this observation one can show the equivalence of (1) and (2) up to a.e. equality. It would be nice to prove this thoroughly and then add it to the Wikipedia page which is a bit confusing as it is now. Commented Feb 5, 2013 at 19:12
The book A First Course in Sobolev Spaces by Giovanni Leoni is a very helpful resource for calculus aspects of function spaces. The first seven chapters deal with functions of one real variable. Chapter 2 is called "Functions of Bounded Pointwise Variation" which are defined by (1) and this class is denoted $BPV(I)$, rather than $BV(I)$. Here $I$ is the interval of definition. The author comments in a footnote on pp.39-40:
Although we do not like changing standard notation, unfortunately in the literature the notation $BV(I)$ is also used for a quite different (although strictly related) function space. This book studies both spaces, so we really had to change the notation for one of them.
Section 7.1 is titled "$BV(\Omega)$ Versus $BPV(\Omega)$". Here, $\Omega$ is an open subset of $\mathbb R$ (we are still in one dimension), and $BV(\Omega)$ is defined as the class of integrable functions $u\in L^1(\Omega)$ for which there is a finite signed Radon measure $\lambda$ such that $\int u\varphi'=-\int \varphi\,d\lambda$ for all $\phi\in C_c^1(\Omega)$. This is somewhat different from, but equivalent to (2): one direction is trivial and the other is a form of Riesz representation.
Theorem 7.2. Let $\Omega\subset\mathbb R$ be an open set. If $u:\Omega\to\mathbb R$ is integrable and if it belongs to $BPV(\Omega)$, then $u\in BV(\Omega)$ and $$|Du|(\Omega)\le \operatorname{Var}u$$ Conversely, if $u\in BV(\Omega)$, then $u$ admits a right continuous representative $\bar u$ in $BPV(\Omega)$ such that $$\operatorname{Var}\bar u = |Du|(\Omega)$$
This is proved thoroughly indeed; the proof takes three pages (pp. 216-218).
• Great! You nailed it completely. Thank you very much! Commented Feb 6, 2013 at 18:50
• This answer deserves a lot more upvotes than just one. Commented Feb 7, 2013 at 1:54
• @GiuseppeNegro Welcome to Math.SE! I see you are new here. :)
– user53153
Commented Feb 7, 2013 at 5:09 | 1,038 | 3,607 | {"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} | 3.46875 | 3 | CC-MAIN-2024-38 | latest | en | 0.889034 |
https://edurev.in/course/quiz/attempt/-1_Test-Solid-Mechanics-1/4d44848f-672c-45ab-a6ba-f6653fbdc613 | 1,628,199,698,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046157039.99/warc/CC-MAIN-20210805193327-20210805223327-00462.warc.gz | 232,152,429 | 42,910 | Courses
# Test: Solid Mechanics- 1
## 10 Questions MCQ Test GATE Civil Engineering (CE) 2022 Mock Test Series | Test: Solid Mechanics- 1
Description
This mock test of Test: Solid Mechanics- 1 for Civil Engineering (CE) helps you for every Civil Engineering (CE) entrance exam. This contains 10 Multiple Choice Questions for Civil Engineering (CE) Test: Solid Mechanics- 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Solid Mechanics- 1 quiz give you a good mix of easy questions and tough questions. Civil Engineering (CE) students definitely take this Test: Solid Mechanics- 1 exercise for a better result in the exam. You can find other Test: Solid Mechanics- 1 extra questions, long questions & short questions for Civil Engineering (CE) on EduRev as well by searching above.
QUESTION: 1
Solution:
QUESTION: 2
### In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to
Solution:
The engineering stress-strain curve for mild steel is
Ultimate tensile strength represents the maximum stress that a material can withstand without Fracture.
QUESTION: 3
### _______ is the capacity of material to absorb energy when it is elastically deformed and then upon unloading, to have this energy recovered.
Solution:
Resilience: It is energy absorbed by a member in elastic region. It denotes the capacity of material to absorb energy when it is elastically deformed and then upon unloading, to release this energy.
Toughness: It is energy absorbed by member just before its fracture.
QUESTION: 4
The shear modulus of a material is half of its Young’s modulus. What is the value of its Poisson’s ratio?
Solution:
QUESTION: 5
A mild steel wire is 10 mm in diameter and 1 m long. If the wire is subjected to an axial tensile load 10 kN, find an extension of the road (Take E = 200 × 10Pa) :
Solution:
QUESTION: 6
The ratio of strengths of solid to hollow shafts, both having outside diameter D and hollow having inside diameter D/2, in torsion, is
Solution:
QUESTION: 7
A simply supported beam is subjected to a uniformly distributed load. Which one of the following statements is true?
Solution:
For section AB
For equilibrium, ∑M0 = 0
For M to be maximum or minimum ⇒ dM/dx = 0
Hence dM/dx = V = 0
QUESTION: 8
A steel plate of side ‘a’ (2d) is enclosed in a closely fitting rigid support as shown. Assume friction between plate and support is zero. Determine the free expansion in x - direction of the plate due to heating by ΔT – (Poison ratio is μ & thermal co - eff. α)
Solution:
σx = 0
σy = −αTE
εy = −αT
εx = −με= μαT = δa/a
ϵx = Free expansion due to temperature change + Expansion due to lateral strain
= a α Δ T + μ a α Δ T (due to y direction)
= a α Δ T (1 + μ)
*Answer can only contain numeric values
QUESTION: 9
For a cylindrical bar of 30 mm dia& 900 mm length, during a tension test, it is found that longitudinal strain is 4 times of lateral strain. The bulk modulus for the bar is (X) × 105 N/mm2 (if E = 3 × 105 N/mm2) The value if X is ___________
Solution:
E = 3 × 105 N/mm2 & E = 3k (1 – 2μ).....(1)
From (1) we get K = E/3(1−2μ) [∴μ=0.25 given]
put μ=0.25 in equation 1,we get
k=2*105N/mm2.
QUESTION: 10
A cantilever beam has the cross-section of an isosceles triangle and is loaded as shown in figure. If the moment of inertia of the cross-section Izz = 1/36m4, then the maximum bending stress is
Solution: | 932 | 3,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-31 | longest | en | 0.891109 |
https://birost.com/a?ID=00000-04f9d86c-6a86-4fd1-a16c-50566d073289 | 1,657,070,348,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00412.warc.gz | 164,529,576 | 17,880 | # Introduction to Probability Theory and Mathematical Statistics
We have seen that probability theory actually reduces common sense to calculations. It enables us to use our rational minds to accurately evaluate intuitions and insights that are often unable to explain clearly. Attractive Yes, probability theory, a science that originated from thinking about games of chance, should have long been the most important part of human knowledge... What are the most important problems in life are mostly just problems of probability theory. Marquis of Laplace (Pierre-Simon Laplace) French mathematician and astronomer
"Probability Theory and Mathematical Statistics" is a commonly studied course in science and engineering, but after studying, students seem to have only learned some mathematical formulas, and they have not produced an intuitive understanding of statistical concepts and an understanding of the theoretical basis of the method. (It is related to the application conditions and limitations of the method), it is not clear to distinguish between subjective probability and objective probability. Mathematical statistics and probability theory are two closely related sister disciplines. Generally speaking: Probability theory is the basis of combing statistics, and mathematical statistics is an important application of probability theory.
Mathematical statistics is a highly applied subject with its methods, applications and theoretical foundations. In the West, the term "mathematical statistics" refers specifically to the part of the mathematical basic theory of statistical methods. In my country, it has a broader meaning, including methods, applications, and theoretical foundations, which is called in the West. For: "Statistics". In our country, because there is still a statistical subject that is recognized as a social science, it is sometimes necessary to use these two terms differently. "Probability Theory and Statistics"
## Preface
I want to understand some things, such as why the load factor of HashMap is 0.75. I am not satisfied that this is just a good explanation. My question is why 0.75 is just right. Then I saw on Zhihu that there are actually high-level languages similar to HashMap. The data structure of, such as 0.65 in Go, 0.8 in Dart, and 0.762 in python.
So why is HashMap in Java 0.75? I found the answer in the comments in HashMap:
Because TreeNodes are about twice the size of regular nodes, we use them only when bins contain enough nodes to warrant use (see TREEIFY_THRESHOLD). And when they become too small (due to removal or resizing) they are converted back to plain bins. In usages with well-distributed user hashCodes, tree bins are rarely used. Ideally, under random hashCodes, the frequency of nodes in bins follows a Poisson distribution ( en.wikipedia.org/wiki/Poisso ) with a parameter of about 0.5 on average for the default resizing threshold of 0.75, although with a large variance because of resizing granularity. Ignoring variance, the expected occurrences of list size k are (exp(-0.5) * pow(0.5, k)/factorial(k)).
(entry ) TREEIFY_THRESHOLD hash 0.75 , ( ) 0.5 k : $P(X=k)=\frac{0.5^k}{k!}e^{-0.5},k=0,1,$
0 8 0.75 8 StackOverFlow :
6001
?
:
:
It doesn t feel like a medical student, as if more bacteria are equal to dirty. In the toilet, most of the bacteria belong to the Actinobacteria, Bacteroides and other categories. These bacteria are very harmful to the human body, and some can even enter the air through the airflow of the toilet. Although there are many bacteria in human saliva, most of them are harmless. What I want to say is that whether it is dirty or not is not just a matter of the number of bacteria, but also the quality of bacteria. It s just that the teacher said that people should surrender to the light of reason in their heart, and not as if what the teacher said is all right. Everyone understands this truth, but everyone is superstitious of authority. I dislike this kind of attitude. The number of bacteria defines the idea of not being dirty. Salmonella has two lethal forms, including Salmonella enterica and Salmonella typhi. Salmonella typhi causes typhoid fever, which kills 216,000 people every year. It is spread through infections of the face and urine. There are about 500-1000 different kinds of bacteria in each person's body, and they can multiply about 100 trillion individual cells in an adult's body-about 10 times the total body cells of a person. Even if you take a bath every day, these bacteria can't be washed away. According to your statement, a person who pays attention to cleanliness is more dirty than a petri dish containing Salmonella. (If you don t believe me, you can check the number of bacteria on the human body. Each of us has more bacteria than the toilet. Every one of us is so dirty. It s the kind that can t be washed off. It s the kind of dirt. Even if it becomes a corpse, it is more dirty than the toilet.) Dirty is a perceptual realization. Your teacher is at least a graduate degree. Don t you understand? I always think this is a rumor, not knowledge. And are you from a science and engineering background? Seriously doubt it! If you offend, I hope to forgive me
For another example, in HashMap, the length of the hash bucket array table must be 2 to the nth power (it must be a composite number). This is an unconventional design. The conventional design is to design the size of the bucket as a prime number. Relatively speaking, the probability of a conflict caused by a prime number is less than a composite number. You see, here is the probability. So what is probability?
## What is probability?
### Subjective probability
The word probability appears very frequently in the real world. People often use probability to express a measure of whether a certain event has occurred. For example, when Xiao Yan swallowed the fire in the heart of Qinglian in "Fights Break Sphere" Description:
According to rough calculations, if there are no auxiliary items such as blood lotus pills, the success rate of swallowing abnormal fire is basically less than 1%, but with them, this success rate may be increased to about 10%.
We can find a lot of judgments like this in the real world. For example, when the weather is gloomy, people will roughly say that I am 90% sure that it will rain today, and there is a 90% chance that Shakespeare is true. Writing "Hamlet", the probability that Oswald assassinated President Kennedy alone is 80%.
But is this probability in mathematics? No. Simply put, it is a measure of people s confidence in what they say. In other words, less than 1% is because Xiao Yan is not so confident about whether he can successfully swallow the abnormal fire. Ninety% To be confident that it will rain today means that we are very sure that it will rain today. Probability as a measure of the degree of individual confidence is often called subjective probability. There is no such thing as the frequency in a coin toss experiment as a support for probability. It is just a measure of the degree of confidence. The difference in numbers given by different people does not have much meaning. Like today s gloomy weather, you think there will be an 85 percent chance of rain, but I think it s 90%. The only explanation for this is that I am more convinced that it will rain today.
The objective probability, the probability in mathematics is linked to the frequency. We always give the sample space first, that is, assuming that all possible results are known, we cannot know the results before the experiment, all possible results The formed set is called the sample space, and any subset of the sample space is called an event. An event is a set composed of some results of an experiment. If the result of the experiment is contained in E, then we say that event E has occurred.
Compared with the well-known coin tossing and dice tossing experiments, we can easily get the sample space. The sample space for coin tossing is as follows:
• Face up (digital side)
• Back side up (non-digital side)
If I ask what is the probability of being upright at this time? Maybe some students will answer one-half, which is 50%, without hesitation, but the assumption of the answer of 50% is that the coins are even. That is to say, the probability of occurrence of positive and negative sides is equal. After exhausting the sample space of the experiment, people often divide the number of events in the sample space by the total number of sample spaces to calculate the probability of an event. At this point, I remembered the discussion between me and my college classmates. When I started taking the course "Probability Theory and Mathematical Statistics", my classmates came up with the following conclusions:
• My probability of failing in this course is 50% (because there are only two events in the sample space: failing, not failing)
• The probability of being a monitor is also 50% (because there are only two sample spaces: being a leader, not being a leader)
The reason for this inference is that the reason is that there are too many questions. Many of the problems of probability theory are given equal possibilities. There are too many questions, whether it is a mathematical problem or the real world. The default and the possibility of presetting, etc., the above wrong conclusion is that the two coins of "being a class" and "being a monitor" are uniform in texture from the beginning. The long and improper heads are both one-half, so it can be concluded that the probability of a single event in a sample space with only two cases will be 50%. There is such a coincidence in the exercises in probability theory that the number of subsets in the sample space divided by the total number of sample spaces is exactly equal to the probability of this event (equal possible sample space). For example, if you roll a dice, there are six In this case, our general assumption is that the texture of the dice is uniform. This implies that the appearance of each face is equal to the probability, and then it is deduced that the probability of each face is one-sixth. This model We call it equal possible sample space in probability theory.
The wrong conclusion drawn above lies in the assumption of equal possibilities at the beginning. Taking the second thesis alone, we have no way of knowing what is the probability of an event happening in the sample space of being a monitor? Personally The inference is that the probability of being elected is 0.0000001%. Strictly speaking, I also gave the possibility of this model again, but here I just think that it is impossible for my college classmate to be elected as the monitor.
Subjective probability can be understood as a mentality or tendency. The root causes are basically two: 1. based on its experience and knowledge, for example, when we were young, our family was drying wheat on the road. The sky was gloomy that evening. It was decided that it would rain today, so I told my father that the wheat would be harvested in the evening. But a grandfather from the same village told my father that there is no need to collect it. The weather today is very good, the wind is very good, and it will not rain. My father finally listened to the opinion of the grandfather. Although the sky was gloomy that day, it did not rain. From this point of view, subjective probability also has an objective background, which is different from rhetoric after all. The second is based on a strong relationship. Take today s weather in Shanghai. It rained yesterday, but when I went out for breakfast, I still didn t bring an umbrella. The reason is that even if it rains, I don t care. I find it a bit troublesome.
The characteristic of subjective probability is that it is not established on a solid objective basis and is recognized by people, so it seems that it should be denied by science (the task of science is to explore the truth). It's not clear how to understand the subjective probability, but don't negate it completely. One of the reasons is that this concept has a broad foundation in life. We estimate the possibility of various situations almost all the time, and different people seldom reach agreement on an "objective" basis. The second reason is that this may also reflect a tendency of the cognitive subject, which has its social significance. For example, if you ask what is the probability of "the economic situation will be better in three years?", people of different economic conditions, social status, and even political inclinations will make different estimates. In terms of individual estimates, there may not be many reasons; but in general, it reflects the confidence of the general public in the long-term development of the society. This is very useful information for sociologists and even decision makers. The third reason is that in designing (economic and other) profit and loss decision-making issues, people in different positions and with different levels of information should weigh the possibility of an event with reference to these circumstances and consequences, and it is suitable for a certain event. A person's decision, such as a decision with less risk, may not be suitable for another person, and for another person, the risk of this decision may be somewhat greater. Therefore, the concept of subjective probability also has its practical basis. In fact, many decisions inevitably involve personal judgment, and this is subjective probability
### Objective probability
#### Experiments and events
The subjective probability we mentioned earlier is a kind of mentality or tendency to whether an event occurs, so let's examine the term "event" below. What is an event? In the usual sense, it often refers to a situation that has occurred, such as the XX air crash, the Japanese attack on Pearl Harbor in 1941, and so on. In probability theory, this is not the case. An event does not refer to a situation that has occurred, but a "statement" of a certain (or some) situation. It may or may not happen. Whether it happens or not depends on what is relevant." "Experiment" will only be known after the results have been obtained.
For example, it will not rain before 6 o'clock today. Of course, we didn't say that this has happened. Whether it happened or not, we have to wait for the results of the experiment. This experiment is to observe the weather conditions before 6 o'clock in the afternoon.
By extension, it is not difficult for us to understand: In probability theory, the general meaning of the term "event" is as follows: (1) There is a clearly defined experiment. The term "experiment" has artificial and active meaning As in the previous example, people are only in a passive position, only recording without interfering with the meteorological process. This type of situation is generally called "observation". In statistics, this one sometimes has practical meaning, but it is not important to the current discussion. You can include the word "experiment" as an observation. (2) All possible results of this experiment were clarified before the experiment. Taking the example above, there are only two possible results of the experiment: one is that it will not rain before 6 o clock today, which is represented by A, and the other is that it will rain before 6 o clock today.$\overline{A}$ , So this experiment can be written as {$A$ ,$\overline{A}$}. You don t have to complete the experiment, that is, you can know it until 6 pm, it s not${A}$ , just$\overline{A}$.
But in many cases, we cannot know exactly all possible results of an experiment, but we can know that it does not exceed a certain range. At this time, this range can also be used as all possible results of the experiment. For example, in the previous example, if we are not only interested in whether it rains before 6 pm, but need to record the rainfall before 6 pm (in millimeters), the experimental result will be a non-negative real number x, and we cannot determine the value of x Possible value range, but this range can be taken as [0,$\infty$ ], it can always contain all experimental results, although we know that some results, such as x> 10000, are impossible to appear. We can even take this range as (-$\infty$ ,$\infty$ ), there is a certain mathematical abstraction in it, which can bring great convenience, which will become clearer in the future.
(3) We have a clear statement that defines a certain part of all possible results of the experiment, or a certain part, which is called an event. For example, in the following example, A is all possible results ($A$ ,$\overline{A}$). In the example of rolling the dice, we can define many events, such as:
• E1 = {Throw an even number} = {2,4,6}
• E2 = {Throw prime points} = {2,3,5}
• E3 = {roll a multiple of 3} = {3,6}
And so on, they clearly defined a corresponding part of the set of all test results (1,2,...,6).
If we do the experiment now, that is, throw the dice once, when the result of the throw is 2 or 4, or 6, we say that event E1 "has happened", otherwise we say that event E1 "does not happen", so , We can also say: An event is a proposition related to the results of an experiment, and its correctness depends on the results of the experiment.
In probability theory, there are cases where a single experimental result is called a "fundamental event". In this way, one or some basic events are combined to form an event, and the basic event itself is also an event. In the example of rolling the dice, there are 6 basic events such as 1, 2,..., 6. Event E2 is composed of three basic events 2, 3, and 5.
Imagine you are in a situation where you throw a dice. If there are prime points, you will win the prize. Before the dice are thrown, you will think this way: Whether I can win depends on chance. Therefore, in probability theory, events are often referred to as "random events" or "incident events". The meaning of "random" is nothing more than to say that whether an event occurs in a certain experiment depends on chance. The extreme case is an "inevitable event" (things that must happen in an experiment, such as throwing a dice, whose number of points does not exceed 6) and "impossible events" (times that cannot happen in the experiment). There are no opportunities for these two situations, but for the convenience of calculation, we might as well regard them as special cases in random events.
#### Probability definition of classical probability
Continuing from the previous paragraph, suppose a certain experiment has a limited number of possible results e1, e2,..., en, assuming that from the analysis of the conditions and implementation methods of the experiment, we cannot find any reason to believe that one of the results, such as ei, It is more advantageous than any other result, such as ek (that is, more prone to happen), so we have to think that all the results e1,...,en have the same chance of appearing in the experiment, that is, 1/N When opportunities arise, such experimental results are often referred to as "equally possible".
Take the example of rolling a dice, if: The material of the dice is absolutely uniform; The dice is an absolute regular hexahedron; When the dice is thrown there is sufficient height from the ground, then most people will agree that the chances of each side appearing should be equal may. Of course, this can only be an approximation in real life, not to mention the number of points engraved on the dice will also affect its symmetry.
On the basis of the concept of "equal possibility", it is natural to introduce the definition of classical probability: Suppose an experiment has N equal possible outcomes, and event E contains exactly M results, then the probability of event E is recorded as P(E), defined as:
P(E) = M/N
The limitation of classical probability is obvious: it can only be used when all experimental results are finite and equal possibilities are true, but in some cases, this concept can be slightly extended to the case where there are infinitely many experimental results. This is the so-called "geometric probability".
#### Probability definition from statistics
From a practical point of view, the statistical definition of probability is nothing more than a method of estimating event probability through experiments. Take the example of "throwing a dice". If the dice is not a cube with a uniform texture, the probability of each side appearing at the time of throwing does not have to be the same. At this time, the probability of "6" E1 at this time cannot be determined by a theoretical consideration alone. But we can do an experiment: repeatedly throw this dice a large number of times, for example, n times, if there are 6 in these n throws and a total of m1 times, then m1/n is called E1. Throwing is counted as the "frequency" in an experiment). The main point of the statistical definition of frequency is to take this frequency m1/n as an estimate of the probability P(E1) of the event E1. The intuitive background of this probability is very simple: the probability of an event s occurrence should be characterized by the frequency of its occurrence in multiple repeated experiments.
The general situation is no different from this. Just in the above description, change "rolling a dice" to a general experiment, and change the event E1 of "6" to a general experiment. The main point is: The experiment must be repeated in large numbers under the same conditions so that we can observe the frequency of the event.
The shortcoming of the above definition is that frequency is only an estimate of probability rather than probability itself. Formally, the following statement can be used to solve this difficulty:
The probability of event E is defined as a number p with the following properties: When the experiment is repeated, the frequency of E swings around p, and when the number of repetitions increases, the swing becomes smaller and smaller, or simply: the probability is when the experiment The limit of frequency when the number of times increases indefinitely. To do so, the following questions must be answered: How do you prove the existence of p with the above properties, or that the existence of p is a hypothesis? Proponents who use frequency to define probability often answer the above questions like this: They think n(E)/n tends to the limit of a certain constant is a hypothesis or an axiom of the entire system. But this assumption seems extremely complicated, because, although the limit of frequency needs to be assumed in fact, it is not the most basic and simple assumption. At the same time, this assumption is not accepted by everyone. We are more inclined to define some simpler, more obvious axioms, and then prove that the frequency tends to the limit of a constant in a certain sense, isn't it more reasonable? This is also the axiomatic method of modern probability theory (implemented by the former Soviet mathematician Kolmogorov)
The importance of the statistical definition of probability is not that it provides a method of defining probability, it actually does not provide such a method, because you can never accurately define the probability of any event based on this definition. Its importance lies in two points: One is to provide a method of estimating probability. This has been mentioned above, there are many such applications. For example, in a population sample survey, the illiteracy rate of the entire population is estimated based on a small sample of people; in industrial production, the rejection rate of the product is estimated based on some products sampled. The second is that it provides a criterion for testing the correctness of the theory. It is assumed that the probability p of a certain event A is calculated based on a certain theory, assumption, etc. We are not sure whether this theory or assumption is consistent with reality, so we Recourse to experiments, that is, to conduct a large number of repeated experiments to observe the frequency of event A m/n. If m/n is similar to p, the experimental results are considered to support the relevant theory; if you want to go farther, you think the theory may have error. This type of problem belongs to an important branch of mathematical statistics-hypothesis testing.
## Mathematical expectations and average time complexity
500 1000 , 2/3, 1/3 , :
" " ( ) 1000 1000 3: 1. 750 250 :
$1000\times/frac{3}{4}$ + $0\times/frac{1}{4}$ = 750 $1000\times/frac{1}{4}$ + $0\times/frac{3}{4}$ = 250
X X ( ) X : 1000 0 $\frac{3}{4}$ $\frac{1}{4}$ X " "
X
This is the origin of the term "mathematics expectation" ("expectation" for short). This term originated from gambling. It does not sound very popular or easy to understand. It is not a very appropriate name, but it has a long history in probability theory. Recognized by everyone, it also has a firm foothold. Its another name is Mean, which is easy to understand and very commonly used. It will be explained below. Is the above explanation a bit abstract? We not only need to be abstract, but also need to be intuitive. If we carefully examine the above formula, is there a feeling of weighted average? The weight is the probability, and the weight is the possible value of X.
Another expected explanation comes from the frequency interpretation of probability. This explanation (we will talk about the strong theorem of large numbers later) believes that if an infinite number of independent repeated experiments are carried out, then the ratio of the number of occurrences of any event E to E is P(E). Assume that the possible values of the random variable X are x1, x2,..., xn. And the corresponding probabilities are P(x1), P(X2),...,P(Xn), and we interpret random variables as the winning unit of a game of chance. That is, in each game we use xi units of probability p(xi), i = 1,2...,n. Now we use frequency to explain, if we continue this game, then the proportion of our winning xi is p(xi). Since this is true for all i(i=1,2,...,n), our average game unit per game is:
$/sum_{i=1}^{n} x_ip(x_i)$ = E(X) E is the abbreviation of Expectation
Roughly speaking, we can think of the expectation, which we will call the average below, and understand it as the average win. If the average win is lower than your bet, then there is a high probability of losing money, because the game itself may be unfair. . Then we return to the average time complexity in the algorithm, we simply introduce a problem:
For a set of ordered numbers, what is the average time complexity of searching in order? This is actually an expected application of mathematics, that is, finding the average. We simply give an example to discuss: {1,2,3,4,5}. Find 1, you can find it immediately, and find 2 to find twice, in the second position. When searching, we cannot determine which number to find, which also involves the mean. Assuming that the probability of finding any number in an ordered sequence is equal, the mathematical expectation is $\sum_{i=1}^{n}/frac{1}{n}/times/frac{(n+1)/times n}{2} =/frac{(n+1) }{2}$
$\frac{(n+1)/times n}{2}$It is the number of digits compared when looking up the number. The mathematical expectation at this time is the average number of comparisons. This is the predecessor of the average complexity of the algorithm. Roughly speaking, the time complexity is how the time spent in the operation increases when n increases. The above is the sequential search, and we can say that the complexity of the sequential search is Linear growth because$\frac{(n+1) }{2}$ Is a linear function.
## Conclusion
This article is the beginning of probability theory and mathematical statistics, vaccine science, data structure and algorithm analysis. I did not introduce many mathematical formulas. I try to introduce probability and combing statistics in an intuitive way. We must not only be abstract, but also To restore abstraction to intuition. Roughly speaking, probability is the result of analysis, or prediction, when the cause is known. And mathematical statistics is to analyze the cause under the condition of known results. This article was originally intended to be called study notes, but it doesn't feel like study notes while writing. The "Basic Course of Probability Theory" and "Probability Theory and Mathematical Statistics" are mixed together. Because the learning articles of "Probability Theory and Mathematical Statistics" will be published later, this article will link the past and the future, so let's call it an introduction. | 6,102 | 28,605 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 21, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.960429 |
http://spotidoc.com/doc/292247/year-9--13---entrance-examination--sample-paper-mathematics | 1,585,661,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00090.warc.gz | 160,700,481 | 8,730 | # YEAR 9 (13+) ENTRANCE EXAMINATION SAMPLE PAPER MATHEMATICS
```YEAR 9 (13+) ENTRANCE EXAMINATION
SAMPLE PAPER
MATHEMATICS
Time allowed:
1 hour
Equipment needed:
Pen, pencil, lined paper, eraser, calculator
Information for candidates:
1.
2.
3.
4.
5.
Try to answer all questions, but don’t worry if you cannot complete all of
them. If you are stuck on a question just go on to the next one and, if you have
time left at the end, come back to any that you left.
There are 100 marks in total available for this paper.
The marks for each question or part question are shown in square brackets [ ].
Show all your working. You may be awarded marks for correct working eve if
3
following:
(a)
( 15) − ( 13) + ( 12)
(b)
0.45 ÷ 0.09
(c)
1
12 % of
2
(d)
83× 217
(e)
4
2
×3
5
3
(f)
75% of
−
−
−
£360
0.83 km .
[2, 2, 2, 3, 3, 3 ]
2.
Solve:
5 x + 2 = 37
[2]
(ii) 2 w + 18 = 5w − 3
[3]
(i)
(iii)
3y
+4=5
7
[3]
(iv)
6( x − 5) − 4( x − 6 ) = 0
[3]
3. Write down the next two terms of these sequences:
(a) 5, 12, 19, 26, ..., ....
[2]
(b) 32, 16, 8, ..., ....
[2]
(c)
3 5 7
, , , ....
4 6 8
[2]
4
4.
The number 325 is equal to a square number multiplied by a prime number.
a) What are these two numbers?
b) Find a number between 100 and 200 which is also equal to a square number
multiplied by a prime number.
[ 4]
5.
Simplify the following algebraic expressions:
(a)
12 x × 4 y
(b)
(8 y )2 − 4 y 2
[2]
[2]
(c)
8 x − 3 x + 15 − 2 x + 2
[2]
2
(d)
15 y + 5 y
5y
2
[2]
(e) 2( x + 1) − ( x + 3)
6.
[ 2]
Estimate the following, showing the values used to arrive at your estimate.
(a)
33.1 × 18.2
5.8
[2]
(b) 0.027 × 274
[2]
5
7.
Find the area of this shape, giving the correct unit, and name it.
12 cm
8 cm
7 cm
8.
9.
[4]
At Christmas, Ben, Sam and Tom received cards in the ration 2 : 3 : 12.
(a)
What fraction of the cards did Ben receive?
[2]
(b)
What fraction did Ben and Sam receive between them?
[2]
(c)
How many cards did Sam receive?
[3]
(d)
How many cards did they receive altogether?
[3, 1]
If a = 2.8 × 10 5 and b = 1.6 × 10 6
the value of
(i) a + b and (ii) a × b ,
[3, 3]
6
10. On graph paper draw a set of axes with values of x from -6 to +6 and y from -8 to +8.
i) draw the graph of y = 3x – 2, use a table with these 3 values.
x
y
0
1
3
ii) draw the graph of y = 2 – x use a table with these values
x
y
0
1
3
[8]
iii) What are the coordinates of the intersection of these two lines?
[2]
iv) draw the line x = 4 and hence find the area of the triangle between the lines x = 4,
y = 3x – 2 and y = 2 - x
[3]
v) What is the gradient of the line y = 5x – 4?
[1]
Where does this line cross the y axis ?
[1]
7
11. The shape opposite is a semicircle on top
of a rectangle.
Find its area using the π button on you
correct to the nearest cm 2 :
12.
12cm
6cm
The lines marked with arrows are parallel.
[ 4]
(i) Calculate the size of the angles lettered a, b c and d:
[2, 2, 2, 2]
z°
95 °
x°
(ii) Calculate the size of the angles
x, y and z:
[2, 2, 2]
15 °
y°
8
13. Here are the rules for an algebra grid.
Use these rules to copy and complete the algebra grids below.
Write your expressions as simply as possible.
9
(7)
14. A spinner had the numbers 1 to 4 on it.
Calculate the probability of spinning a number 3
15
[3]
The diagram below shows an isosceles triangle (A) and a semicircle (B). The
triangle has a base of length 12cm and the semicircle has diameter 12cm.
Find the area of A if A and B have the same perimeter.
END OF PAPER: IF YOU HAVE TIME, GO BACK AND CHECK YOUR WORK
10
[6]
``` | 1,262 | 3,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.863566 |
http://www.chegg.com/homework-help/mathematics-for-machine-technology-6th-edition-chapter-47-problem-24a-solution-9781111803964 | 1,455,094,348,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701159031.19/warc/CC-MAIN-20160205193919-00249-ip-10-236-182-209.ec2.internal.warc.gz | 337,060,983 | 15,760 | View more editions
# TEXTBOOK SOLUTIONS FOR Mathematics for Machine Technology 6th Edition
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PROBLEM
Chapter: Problem:
Determine the answers to the following problems, which are based on corresponding parts.
Identify the angle that corresponds with each angle listed. All dimensions are in inches.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 3
Consider the two triangles.
a) Here, angle is corresponds to angle since both the angles lies opposite to the smallest side.
Write the corresponding relation.
Hence, the corresponding angle of is .
• Chapter , Problem is solved.
Corresponding Textbook
Mathematics for Machine Technology | 6th Edition
9781111803964ISBN-13: 111180396XISBN: Authors: | 213 | 906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.821326 |
https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Signal_Processing_and_Modeling/Book%3A_Signals_and_Systems_(Baraniuk_et_al.)/10%3A_Sampling_and_Reconstruction/10.07%3A_Discrete_Time_Processing_of_Continuous_Time_Signals | 1,619,040,291,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039550330.88/warc/CC-MAIN-20210421191857-20210421221857-00307.warc.gz | 341,420,251 | 26,329 | # 10.7: Discrete Time Processing of Continuous Time Signals
## Introduction
Digital computers can process discrete time signals using extremely flexible and powerful algorithms. However, most signals of interest are continuous time signals, which is how data almost always appears in nature. Now that the theory supporting methods for generating a discrete time signal from a continuous time signal through sampling and then perfectly reconstructing the original signal from its samples without error has been discussed, it will be shown how this can be applied to implement continuous time, linear time invariant systems using discrete time, linear time invariant systems. This is of key importance to many modern technologies as it allows the power of digital computing to be leveraged for processing of analog signals.
## Discrete Time Processing of Continuous Time Signals
### Process Structure
With the aim of processing continuous time signals using a discrete time system, we will now examine one of the most common structures of digital signal processing technologies. As an overview of the approach taken, the original continuous time signal $$x$$ is sampled to a discrete time signal $$x_s$$ in such a way that the periods of the samples spectrum $$X_s$$ is as close as possible in shape to the spectrum of $$X$$. Then a discrete time, linear time invariant filter $$H_2$$ is applied, which modifies the shape of the samples spectrum $$X_s$$ but cannot increase the bandlimit of $$X_s$$, to produce another signal $$y_s$$. This is reconstructed with a suitable reconstruction filter to produce a continuous time output signal $$y$$, thus effectively implementing some continuous time system $$H_1$$. This process is illustrated in Figure $$\PageIndex{1}$$, and the spectra are shown for a specific case in Figure $$\PageIndex{2}$$.
Further discussion about each of these steps is necessary, and we will begin by discussing the analog to digital converter, often denoted by ADC or A/D. It is clear that in order to process a continuous time signal using discrete time techniques, we must sample the signal as an initial step. This is essentially the purpose of the ADC, although there are practical issues that which will be discussed later. An ADC takes a continuous time analog signal as input and produces a discrete time digital signal as output, with the ideal infinite precision case corresponding to sampling. As stated by the Nyquist-Shannon Sampling theorem, in order to retain all information about the original signal, we usually wish sample above the Nyquist frequency $$\omega_s≥2B$$ where the original signal is bandlimited to $$(−B,B)$$. When it is not possible to guarantee this condition, an anti-aliasing filter should be used.
The discrete time filter is where the intentional modifications to the signal information occur. This is commonly done in digital computer software after the signal has been sampled by a hardware ADC and before it is used by a hardware DAC to construct the output. This allows the above setup to be quite flexible in the filter that it implements. If sampling above the Nyquist frequency the. Any modifications that the discrete filter makes to this shape can be passed on to a continuous time signal assuming perfect reconstruction. Consequently, the process described will implement a continuous time, linear time invariant filter. This will be explained in more mathematical detail in the subsequent section. As usual, there are, of course, practical limitations that will be discussed later.
Finally, we will discuss the digital to analog converter, often denoted by DAC or D/A. Since continuous time filters have continuous time inputs and continuous time outputs, we must construct a continuous time signal from our filtered discrete time signal. Assuming that we have sampled a bandlimited at a sufficiently high rate, in the ideal case this would be done using perfect reconstruction through the Whittaker-Shannon interpolation formula. However, there are, once again, practical issues that prevent this from happening that will be discussed later.
### Discrete Time Filter
With some initial discussion of the process illustrated in Figure $$\PageIndex{1}$$ complete, the relationship between the continuous time, linear time invariant filter $$H_1$$ and the discrete time, linear time invariant filter $$H_2$$ can be explored. We will assume the use of ideal, infinite precision ADCs and DACs that perform sampling and perfect reconstruction, respectively, using a sampling rate $$\omega_s=2 \pi /T_s≥2B$$ where the input signal $$x$$ is bandlimited to $$(−B,B)$$. Note that these arguments fail if this condition is not met and aliasing occurs. In that case, pre-application of an anti-aliasing filter is necessary for these arguments to hold.
Recall that we have already calculated the spectrum $$X_s$$ of the samples $$x_s$$ given an input $$x$$ with spectrum $$X$$ as
$X_{s}(\omega)=\frac{1}{T_{s}} \sum_{k=-\infty}^{\infty} X\left(\frac{\omega-2 \pi k}{T_{s}}\right) .$
Likewise, the spectrum $$Y_s$$ of the samples $$y_s$$ given an output $$y$$ with spectrum $$Y$$ is
$Y_{s}(\omega)=\frac{1}{T_{s}} \sum_{k=-\infty}^{\infty} Y\left(\frac{\omega-2 \pi k}{T_{s}}\right) .$
From the knowledge that $$y_s=(H_1x)_s=H_2(x_s)$$, it follows that
$\sum_{k=-\infty}^{\infty} H_{1}\left(\frac{\omega-2 \pi k}{T_{s}}\right) X\left(\frac{\omega-2 \pi k}{T_{s}}\right)=H_{2}(\omega) \sum_{k=-\infty}^{\infty} X\left(\frac{\omega-2 \pi k}{T_{s}}\right).$
Because $$X$$ is bandlimited to $$(−\pi /T_s, \pi /T_s)$$, we may conclude that
$H_{2}(\omega)=\sum_{k=-\infty}^{\infty} H_{1}\left(\frac{\omega-2 \pi k}{T_{s}}\right)(u(\omega-(2 k-1) \pi)-u(\omega-(2 k+1) \pi)).$
More simply stated, $$H_2$$ is $$2 \pi$$ periodic and $$H_2(\omega)=H_1( \omega /T_s)$$ for $$\omega \in[-\pi, \pi)$$.
Given a specific continuous time, linear time invariant filter $$H_1$$, the above equation solves the system design problem provided we know how to implement $$H_2$$. The filter $$H_2$$ must be chosen such that it has a frequency response where each period has the same shape as the frequency response of $$H_1$$ on $$\left(-\pi / T_{s}, \pi / T_{s}\right)$$. This is illustrated in the frequency responses shown in Figure $$\PageIndex{2}$$.
We might also want to consider the system analysis problem in which a specific discrete time, linear time invariant filter $$H_2$$ is given, and we wish to describe the filter $$H_1$$. There are many such filters, but we can describe their frequency responses on $$\left(-\pi / T_{s}, \pi / T_{s}\right)$$ using the above equation. Isolating one period of $$H_2(\omega)$$ yields the conclusion that $$H_{1}(\omega)=H_{2}\left(\omega T_{s}\right)$$ for $$\omega \in\left(-\pi / T_{s}, \pi / T_{s}\right)$$. Because $$x$$ was assumed to be bandlimited to $$\left(-\pi / T_{s}, \pi / T_{s}\right)$$, the value of the frequency response elsewhere is irrelevant.
## Practical Considerations
As mentioned before, there are several practical considerations that need to be addressed at each stage of the process shown in Figure $$\PageIndex{1}$$. Some of these will be briefly addressed here, and a more complete model of how discrete time processing of continuous time signals appears in Figure $$\PageIndex{3}$$.
### Anti-Aliasing Filter
In reality, we cannot typically guarantee that the input signal will have a specific bandlimit, and sufficiently high sampling rates cannot necessarily be produced. Since it is imperative that the higher frequency components not be allowed to masquerade as lower frequency components through aliasing, anti-aliasing filters with cutoff frequency less than or equal to $$\omega_s/2$$ must be used before the signal is fed into the ADC. The block diagram in Figure $$\PageIndex{3}$$ reflects this addition.
As described in the previous section, an ideal lowpass filter removing all energy at frequencies above $$\omega_s/2$$ would be optimal. Of course, this is not achievable, so approximations of the ideal lowpass filter with low gain above $$\omega_s/2$$ must be accepted. This means that some aliasing is inevitable, but it can be reduced to a mostly insignificant level.
### Signal Quantization
In our preceding discussion of discrete time processing of continuous time signals, we had assumed an ideal case in which the ADC performs sampling exactly. However, while an ADC does convert a continuous time signal to a discrete time signal, it also must convert analog values to digital values for use in a digital logic device, a phenomenon called quantization. The ADC subsystem of the block diagram in Figure $$\PageIndex{3}$$ reflects this addition.
The data obtained by the ADC must be stored in finitely many bits inside a digital logic device. Thus, there are only finitely many values that a digital sample can take, specifically $$2N$$ where $$N$$ is the number of bits, while there are uncountably many values an analog sample can take. Hence something must be lost in the quantization process. The result is that quantization limits both the range and precision of the output of the ADC. Both are finite, and improving one at constant number of bits requires sacrificing quality in the other.
### Filter Implementability
In real world circumstances, if the input signal is a function of time, the future values of the signal cannot be used to calculate the output. Thus, the digital filter $$H_2$$ and the overall system $$H_1$$ must be causal. The filter annotation in Figure $$\PageIndex{3}$$ reflects this addition. If the desired system is not causal but has impulse response equal to zero before some time $$t_0$$, a delay can be introduced to make it causal. However, if this delay is excessive or the impulse response has infinite length, a windowing scheme becomes necessary in order to practically solve the problem. Multiplying by a window to decrease the length of the impulse response can reduce the necessary delay and decrease computational requirements.
Take, for instance the case of the ideal lowpass filter. It is acausal and infinite in length in both directions. Thus, we must satisfy ourselves with an approximation. One might suggest that these approximations could be achieved by truncating the sinc impulse response of the lowpass filter at one of its zeros, effectively windowing it with a rectangular pulse. However, doing so would produce poor results in the frequency domain as the resulting convolution would significantly spread the signal energy. Other windowing functions, of which there are many, spread the signal less in the frequency domain and are thus much more useful for producing these approximations.
### Anti-Imaging Filter
In our preceding discussion of discrete time processing of continuous time signals, we had assumed an ideal case in which the DAC performs perfect reconstruction. However, when considering practical matters, it is important to remember that the sinc function, which is used for Whittaker-Shannon interpolation, is infinite in length and acausal. Hence, it would be impossible for an DAC to implement perfect reconstruction.
Instead, the DAC implements a causal zero order hold or other simple reconstruction scheme with respect to the sampling rate $$\omega_s$$ used by the ADC. However, doing so will result in a function that is not bandlimited to $$(−\omega_s/2,\omega_s/2)$$. Therefore, an additional lowpass filter, called an anti-imaging filter, must be applied to the output. The process illustrated in Figure $$\PageIndex{3}$$ reflects these additions. The anti-imaging filter attempts to bandlimit the signal to $$(−\omega_s/2,\omega_s/2)$$, so an ideal lowpass filter would be optimal. However, as has already been stated, this is not possible. Therefore, approximations of the ideal lowpass filter with low gain above $$\omega_s/2$$ must be accepted. The anti-imaging filter typically has the same characteristics as the anti-aliasing filter.
## Discrete Time Processing of Continuous Time Signals Summary
As has been show, the sampling and reconstruction can be used to implement continuous time systems using discrete time systems, which is very powerful due to the versatility, flexibility, and speed of digital computers. However, there are a large number of practical considerations that must be taken into account when attempting to accomplish this, including quantization noise and anti-aliasing in the analog to digital converter, filter implementability in the discrete time filter, and reconstruction windowing and associated issues in the digital to analog converter. Many modern technologies address these issues and make use of this process. | 2,839 | 12,701 | {"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} | 2.671875 | 3 | CC-MAIN-2021-17 | latest | en | 0.914404 |
http://clay6.com/qa/48917/a-coin-is-tossed-if-it-shows-a-tail-we-draw-a-ball-from-a-box-which-contain | 1,575,720,829,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540499389.15/warc/CC-MAIN-20191207105754-20191207133754-00376.warc.gz | 32,852,479 | 9,318 | Comment
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# A coin is tossed .If it shows a tail,we draw a ball from a box which contains 2 red and 3 black balls.If it shows head,we throw a die.Find the sample space for this experiment.
$S=\left\{\begin{array}{1 1}H1,H2,H3,H4,H5,H6,\\TR_1,TR_2,TB_1,TB_2,TB_3\end{array}\right\}$ | 108 | 290 | {"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} | 2.671875 | 3 | CC-MAIN-2019-51 | latest | en | 0.767569 |
https://teachsimple.com/product/dividing-fractions-lesson-dividing-fractions-math-unit-mini-flip-book | 1,716,379,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00467.warc.gz | 491,591,209 | 119,245 | # Dividing Fractions Lesson - Dividing Fractions Math Unit Mini Flip Book
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### Introducing the Dividing Fractions Lesson - Dividing Fractions Math Unit Mini Flip Book
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## What's Included
1 PDF file with 10 pages.
## Resource Tags
fraction division arithmetic hands-on learning math unit interactive
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Check out these other great products | 435 | 2,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.88563 |
https://www.mathworks.com/matlabcentral/cody/problems/582-function-composition-harder/solutions/421108 | 1,580,030,820,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251687958.71/warc/CC-MAIN-20200126074227-20200126104227-00453.warc.gz | 984,186,233 | 15,491 | Cody
Problem 582. Function composition - harder
Solution 421108
Submitted on 21 Mar 2014 by Guillaume
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Test Suite
Test Status Code Input and Output
1 Pass
%% f1 = @(x)x+1; f2 = @(x)3*x; f3 = @sqrt; h = compose(f1,f2,f3); assert(isequal(h(9),10));
2 Pass
%% f = repmat({@(x)x+1},1,100); h = compose(f{:}); assert(isequal(h(0),100));
3 Pass
%% f = @(x)x; h = compose(f); assert(isequal(h(1234),1234)); | 175 | 519 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.684937 |
https://goprep.co/consider-a-box-with-three-terminals-on-top-of-it-as-shown-in-i-1nm4c5 | 1,611,433,405,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00569.warc.gz | 365,971,733 | 37,407 | Q. 38
# Consider a box with three terminals on top of it as shown in Fig.14.18 (a):Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.A student performs an experiment in which any two of these three terminals are connected in the circuit shown in Fig. 14.18 (b).The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit.The graphs are(i) when A is positive and B is negative(ii) when A is negative and B is positive(iii) When B is negative and C is positive(iv) When B is positive and C is negative(v) When A is positive and C is negative(vi) When A is negative and C is positiveFrom these graphs of current – voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B and C.
The first diagram is of the reverse bias of diode.
The 2nd diagram is simple ohm’s law but having a turn on voltage 0.7 volts.
Also slop =1000Ω.
Hence, R=1kΩ
That means a diode in a series connection with a 1kΩ resistance.
Furthermore, the 3rd and 4th diagram represents a diode in reverse and forward bias.
The last two diagrams shows a turn on potential of 1.4v. This is quite certain for having two 0.7v batteries in series or in other words both the diode is in same orientation.
The final structure will be as shown.
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view all courses | 424 | 1,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.92237 |
https://electronics.stackexchange.com/questions/392075/multiple-parallel-led-drivers-with-single-thermal-derating-ntc-thermistor | 1,725,753,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650926.21/warc/CC-MAIN-20240907225010-20240908015010-00704.warc.gz | 217,180,755 | 41,785 | # Multiple Parallel LED Drivers with Single Thermal Derating NTC Thermistor
I am looking at powering Some multi color High power LEDs e.g LED engin LZ4 or LZ7 (probably a string of five) using a tps92691 based Circuit
these being high power leds it seems wise to include thermal derating circuitry
the tps92691 suggests using a voltage divider with a NTC thermistor from its on voltage regulator e.g
simulate this circuit – Schematic created using CircuitLab
The LEDs however are either 4 or 7 LEDs in one die so using multiple PTC seems redundant and complicates track layout near the LEDs . I am therefore looking for the best way to connect multiple drivers to a single NTC thermistor.
The simplest circuit would be
simulate this circuit
but that makes assumptions about the TPS92691's 7.5V LDO Regulator and that they wont "fight" with each other.
and external regulator is possible but is that really necessary?
is this a sensible way of doing this or is there a better way?
• I don't think I could justify the LZ7. It does have a strong (1W) blacklight LED. The Luxeon Color C line have a very small package (2.0 x 2.0), are reasonably priced, and have good color mixing. You could fit 16 Luexon C in the same area of a LZ7. Luexon C provides flexibility in layout and color selection. Commented Aug 21, 2018 at 19:14
• Those look interesting but does bring up question of best way of getting the equivalent flux output(s), as they appear to operate at "base" current of 350mA and give roughly half the flux, of a LZ4/7 at that current, and while can operate at 700mA ("base" for LZ4/7) seem to give ~1.75 times there base flux at that current. I also cant find thermal derating information on there spreadsheet. Commented Aug 22, 2018 at 15:39
• Not only 350 mA vs. 700 mA but 85° vs 25°. Looking at min flux for red Luxeon C = 35 lm adjusted for temp => 58 lm, then adjust for current => 104 lm vs. LZ7's min flux of 60 lm which equates to Luxeon C red having 58% better performance than LZ7. Then adjust for efficacy Luxeon C red min Vf = 1.75 and LZ7 min Vf = 2.1 so add another 20% for Luxeon C when comparing lm/watt. Commented Aug 22, 2018 at 18:04 | 567 | 2,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.937946 |
http://docplayer.net/27330472-Math-311-test-iii-spring-2013-with-solutions.html | 1,531,684,925,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676588961.14/warc/CC-MAIN-20180715183800-20180715203800-00584.warc.gz | 108,549,965 | 26,890 | # Math 311 Test III, Spring 2013 (with solutions)
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1 Math 311 Test III, Spring 2013 (with solutions) Dr Holmes April 25, 2013 It is extremely likely that there are mistakes in the solutions given! Please call them to my attention if you find them. This exam will begin at 1030 am and end officially at 1145 am (at which time you will get a five minute warning to hand your papers in). There is an ID number on the first inside page of your paper by which your grade will be posted. You are allowed your test paper, drawing instruments, and your writing instrument. Cell phones must be turned off and inaccessible to you. One of the nine questions (the one on which you do worst or which you choose to omit) will be dropped. There may be other adjustments for length as seems appropriate. For the last two proof questions, there are alternative proof questions in brackets that you may do instead if you prefer. 1
2 1. Triangle congruence and notation (a) Write down the six congruence facts which make up the statement ABC = DEF. AB = DE BC = EF AC = DF BAC = EDF ABC = DEF BCA = EF D (b) List three of these which imply the full congruence by the SAS axiom. for example. AB = DE BC = EF ABC = DEF three possible answers (c) Explain why ABC = DEF is a different statement from ABC = DF E. For example the first statement requires AB to be congruent to DE and the second requires AB to be congruent to DF. Similar statements about angles work. I did not give credit for statements about points corresponding to each other: these do not express any information about the geometry. 2
3 (d) Describe a situation in which both of the two congruence statements in the previous part would be true. Both would hold if ABC = DEF and in addition DF were congruent to DE (and also AB congruent to BC) so both triangles are isosceles. Saying the triangles could be equilateral works. Saying that the triangles have to be equilateral for this to be true earns a deduction, as they do not have to be. 3
4 2. Describe a counterexample to the ASA Theorem in taxicab geometry, with angle measure defined as in the usual geometry of the plane (hint: use triangles with different orientations. Give all relevant details and calculations). The typically given counterexample involves two triangles differently oriented. For example A = (0, 0), B = (1, 1), C = (2, 0); D = (0, 0), E = (0, 1), F = (1, 1). The information you should specify is that both BAC and EDF have measure 45, and so do BCA and DF E, and both AC and DF have measure 2. This gives the conditions for ASA, but the triangles are not congruent (as for example AB has length 2 and DE has length 1. You lose a lot of points for giving an SAS counterexample, and a lot of points for using the square metric instead of the taxicab metric. 4
5 3. We give an interpretation of part of our geometrical language on a circle. A point is a point on the circumference of the circle. For two points A and B, d(a, B) is the length of the shorter of the two arcs between A and B (or of either of them if they are the same length). Define A B C as d(a, B) + d(b, C) = d(a, C). There is only one line in this interpretation, containing all the points on the circle. For each of the points A, B, A (at the opposite end of a diameter of the circle from A), X, Y, and Z, determine using the usual definition of ray with the new definition of betweenness whether it is on the ray AB. Explain your answer briefly for each one. You may judge betweenness facts visually. Sketch the ray AB on the diagram given. Lots of you know the facts (the ray is the semicircle from A to A inclusive including B and including its endpoints). Few of you gave the reasons, which were a huge part of the value. A and B are of course on ray AB by definition. X is on the ray because A X B. Y is on the ray because A B Y (nothing to do with A, which has no role in the definition of ray AB). A is on the ray because A B A. Z is not on the ray because neither A Z B nor A B Z is true. (nothing to do with half-planes, which have no role in the definition of rays). 5 points for the bare facts about what the ray is; one more for many of you who did write A X B or say X was between A and B. More points if you gave more of the above justifications. 5
6 4. Prove that if line L is perpendicular to line M, then all four angles formed by a ray from each line are right angles (hint: this is all about the Linear Pair Theorem). If L is perpendicular to M, then (A being their point of intersection) there are points B on L and C on M such that BAC is right. Choose B such that B A B. Angles BAC and B AC form a linear pair, so their measures add to 180 degrees, so the measure of B AC is also 90 degrees. Choose C so that C A C. Angles BAC and BAC form a linear pair, so their measures add to 180 degrees, so the measure of BAC is also 90 degrees. Similarly angles B AC and B AC form a linear pair, so their measures add to 180 degrees, so the measure of B AC is also 90 degrees. You lose two points for starting by saying something like Let L and M be perpendicular lines intersecting at A; let B be on L and C be on M; by definition of perpendicular lines the measure of BAC is 90 degrees : in fact, this is the result we are trying to prove. All you know is that one of the four angles is 90 degrees, and you have to start by specifying one of them as the 90 degree angle provided by the definition (without implying that any such angle is 90 degrees by definition, which is not in the definition). 6
7 5. Exterior angle theorem (a) Draw a triangle. Draw one of its exterior angles (and name it, which requires labelling of appropriate points). State what the remote interior angles for this exterior angle are. Not set up to draw here. (b) Prove that if one of the angles of a triangle is obtuse, the other two angles are acute. Suppose that ABC is a triangle. Suppose without loss of generality that ABC is obtuse (has measure greater than 90). Choose D so that B C D. Angles DBC and ABC form a linear pair, so their measures add to 180, so the measure of DBC is less than 90 degrees. Angles BAC and BCA have measure less than the measure of DBC by the exterior angle theorem, so they have measure less than 90, so they are acute. 7
8 6. Prove that if P is a point not on line L, and M is a line incident on P and perpendicular to L, then M is the only line incident on P and perpendicular to L. (uniqueness of dropped perpendiculars). You are not being asked to construct a perpendicular. Start by supposing that there is another one. If you use the diagram that I provide you need to label it appropriately and possibly justify some of the facts implied by the way it is drawn. You needed to say everything in words that was indicated on the diagram. Let Q be the point at which M intersects L. Let N be another line through P perpendicular to L (for the sake of a contradiction); let R be the intersection of N and L. Choose X so that Q R X. Then angle P RX is external to triangle P QR and so greater in measure than angle P QX by the exterior angle theorem. But both P RX and P QX have measure 90 by facts about perpendicular lines proved in another problem on this test, which is a contradiction. 8
9 7. Prove the existence of angle bisectors: for each angle BAC there is a ray AD which is an angle bisector of BAC. You are not being asked to prove uniqueness. You do need to prove that it has all the defining properties of an angle bisector: just constructing the right ray and saying that it is an angle bisector will not carry that much credit. You may use either method of proof. Since there are two methods that you have seen, I give both proofs. There is an error in the statement of the problem, caught by just one student: it is necessary to provide that the angle is not a zero angle. First proof: Let ABC be an angle which is not a zero angle. ABC has measure r > 0. There is a unique ray BD such that D is on the same side of line BC as A and DBC has measure r. The ray 2 BD is between BA and BC by the betweenness theorem for rays (not by the Protractor Postulate). Thus the sum of the measures of DBC and DBA is r, the measure of ABC, so the measure of DBA is also r. We have shown that DBA and DBC are 2 congruent and BD is between BA and BC, that is, that BD is an angle bisector for ABC. Second proof: Let ABC be an angle which is not a zero angle. Choose a point A on BA with BA = BC (point construction postulate). Let D be the midpoint of A C. Note that A BC is isosceles, so BCA = BA C (Isosceles Triangle Theorem); further A D = DC (since D is chosen as the midpoint of A C), and A B = CB by choice of A : this is the evidence for A BD = CBD (SAS). Now by this congruence A BD = ABD is congruent to CBD. Further, since A D C we have BD between BA = BA and BC (3.3.10), so BD is an angle bisector for ABC. It is really important to show that the bisector is between the two rays that make up the original angle as well as showing that it makes angles of the same measure with the two sides of the original angle. 9
10 8. Carefully state and prove the Angle-Side-Angle Theorem. [You may state and prove Angle-Angle-Side if you prefer.] A number of students proved AAS. There is a proof of AAS on the class announcements page. You cannot prove AAS without invoking the exterior angle theorem (no, the uniqueness part of the Angle Construction Postulate does not work), so proofs of AAS with the same gap that I call attention to in the proof of ASA are even more deficient than proofs of ASA with the same gap. Theorem (ASA): Let ABC and DEF be triangles with BC = EF, ABC = DEF and ACB = DF E: it follows that ABC = DEF. Proof of Theorem: Let A be chosen on BA such that BA = ED. It follows that A BC = DEF by SAS. Thus A CB = DF E. But also ACB is congruent to DF E by our initial assumptions, so ACB = A CB. A and A are on the same side of line BC by the ray theorem. By the uniqueness portion of the Angle Construction Postulate, it follows that µ( ACB) = µ( A CB) forces CA and CA to be the same ray. This is the main move of the proof. This does not give you A = A in one step trying to make this leap loses you a lot of points. Giving me an equation between segments instead of rays also loses you a lot of points, for the same reason. Now this shows that A and A are both on line CA as well as both being on line BA by construction, so they are the same point by the incidence axiom. Since A = A and we have already shown A BC = DEF, we get ABC = DEF by substitution of equals for equals. 10
11 9. Prove the Scalene Inequality in one direction: if ABC is a triangle and d(a, B) > d(b, C) then µ( ACB) > µ( (BAC). [You may prove existence of a perpendicular to a line L through a point P not on L if you prefer.] Let ABC be a triangle. Suppose d(a, B) > d(b, C). Choose D on BA such that BD = BC (point construction postulate). B D A by , so CD is between CA and CB by , so µ( BCD) + µ( DCA) = µ( ACB) by the Angle Addition Postulate, so BCD is smaller in measure than ACB. BCD is isosceles, so BCD has the same measure as BDC by the isosceles triangle theorem. Angle BDC is exterior to ADC so greater in measure than DAC = BAC by the exterior angle theorem. So we have shown µ( ACB) > µ( BCD) = µ( BDC) > µ( DAC) = µ( BAC) which completes the proof. No one came close to proving the bracketed theorem; there is a proof in the book. 11
12 1 Selected axioms, theorems and definitions: There is no intention here that every axiom theorem or definition will be given, or even every one that will be useful to you will be given. If you think I have omitted something that you think ought to be here, you may ask me during the exam, but I am most likely to tell you that I left it out on purpose. The form in which I state an axiom theorem or definition will not necessarily be literally identical to what is in the book: the meaning should be the same! taxicab distance: d((a, b), (c, d)) = a c + b d definition of a ray: Omitted on purpose. You should know it or be able to figure it out. It is defined in terms of betweenness : If B lies on AC then A B C iff d(a, B) < d(a, C). Point Construction Postulate: For any segment AB and ray CD, there is a unique point E on the ray CD such that AB = CE. definition of a linear pair: Angles ABD and DBC make up a linear pair iff A B C. Linear Pair Theorem: The sums of the measures of the two angles in a linear pair is exactly 180 degrees. interior of an angle, betweenness of rays: A point D is in the interior of an angle BAC iff D is on the same side of line AB as C and D is on the same side of line AC as B. A ray AD is between AB and AC iff D is in the interior of BAC : If BAC is an angle and D is on line BC, AD is between AB and AC iff B D C. Betweenness Theorem for Rays: Let A, B, C, D be distinct, with C, D on the same side of line AB. Then µ( BAD) < µ( BAC) iff AD is between AB and AC. Crossbar Theorem: If BAC is an angle and AD is between AB and AC then AD intersects BC at some point G. Protractor Postulate: There is a function µ from the set of all angles to the set of all real numbers with the following properties: 12
13 1. 0 µ( BAC) < 180 for any angle BAC. 2. µ( BAC) = 0 iff AB = AC. 3. Angle Construction Postulate: for each pair of distinct points A, B and each half plane H determined by line AB and each real number r such that 0 r < 180, there is a unique point C such that C is in H and µ( BAC) = r. 4. Angle Addition Postulate: If AD is between AB and AC then µ( BAD) + µ( DAC) = µ( BAC). 13
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### Geometry Chapter 5 Relationships Within Triangles
Objectives: Section 5.1 Section 5.2 Section 5.3 Section 5.4 Section 5.5 To use properties of midsegments to solve problems. To use properties of perpendicular bisectors and angle bisectors. To identify | 10,078 | 40,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-30 | longest | en | 0.94624 |
http://www.numbersaplenty.com/21386 | 1,542,317,378,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742937.37/warc/CC-MAIN-20181115203132-20181115225132-00479.warc.gz | 471,458,716 | 3,521 | Search a number
21386 = 217237
BaseRepresentation
bin101001110001010
31002100002
411032022
51141021
6243002
7116231
oct51612
932302
1021386
1115082
1210462
139971
147b18
15650b
hex538a
21386 has 12 divisors (see below), whose sum is σ = 34998. Its totient is φ = 9792.
The previous prime is 21383. The next prime is 21391. The reversal of 21386 is 68312.
Adding to 21386 its reverse (68312), we get a palindrome (89698).
It can be divided in two parts, 2 and 1386, that multiplied together give a palindrome (2772).
It can be written as a sum of positive squares in 3 ways, for example, as 21025 + 361 = 145^2 + 19^2 .
It is a hoax number, since the sum of its digits (20) coincides with the sum of the digits of its distinct prime factors.
It is an Ulam number.
It is a Curzon number.
It is a nialpdrome in base 13.
It is not an unprimeable number, because it can be changed into a prime (21383) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 560 + ... + 596.
221386 is an apocalyptic number.
21386 is a deficient number, since it is larger than the sum of its proper divisors (13612).
21386 is a wasteful number, since it uses less digits than its factorization.
With its predecessor (21385) it forms an eRAP, since the sums of their prime factors are consecutive (72 and 73).
21386 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 73 (or 56 counting only the distinct ones).
The product of its digits is 288, while the sum is 20.
The square root of 21386 is about 146.2395295397. The cubic root of 21386 is about 27.7572553601.
The spelling of 21386 in words is "twenty-one thousand, three hundred eighty-six".
Divisors: 1 2 17 34 37 74 289 578 629 1258 10693 21386 | 564 | 1,915 | {"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} | 3.59375 | 4 | CC-MAIN-2018-47 | latest | en | 0.901516 |
https://math.stackexchange.com/questions/2712815/not-too-slow-computation-of-euler-products-singular-series | 1,558,653,665,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00353.warc.gz | 550,688,119 | 38,634 | # Not-too-slow computation of Euler products / singular series
I'd like to compute, to at least a few digits of accuracy, the constants that arise in Hardy-Littlewood conjecture F / Bateman-Horn conjecture, in particular for just a single quadratic polynomial.
Specifically, consider (for instance) the following number, which arises when studying primes in the polynomial $n^2 + n + 1$:
\begin{align} C_{1} &= \frac{2}{1} \frac{2}{2} \frac{5}{4} \frac{5}{6} \frac{11}{10} \frac{11}{12} \frac{17}{16} \frac{17}{18} \frac{23}{22} \frac{29}{28} \frac{29}{30} \frac{35}{36} \frac{41}{40} \cdots \\ &= \prod_{p \equiv 1 \pmod3}\left(\frac{p - 2}{p - 1}\right) \prod_{p \equiv 2 \pmod3}\left(\frac{p}{p - 1}\right) \\ &= \prod_{p \equiv 1 \pmod3}\left(\frac{1 - 2/p}{1 - 1/p}\right) \prod_{p \equiv 2 \pmod3}\left(\frac{1}{1 - 1/p}\right) \\ \end{align} where the products run over all primes that leave remainder $1$ and $2$ when divided by $3$, respectively. (Note: we cannot really write it as two separate infinite products; the order matters, so excuse the notation.)
Now, if I try to compute this the naive way:
#include <cstdio>
#include "FJ64_16k.h" // For a fast is_prime function: http://ceur-ws.org/Vol-1326/020-Forisek.pdf
int main() {
double ans = 0.5;
for (uint64_t p = 0; ; ++p) {
if (p % 1000000 == 0) {
printf("%lld %.9f\n", p, ans);
}
if (!is_prime(p)) continue;
if (p % 3 == 1) ans *= (p - 2.0) / (p - 1.0);
if (p % 3 == 2) ans *= p / (p - 1.0);
}
}
Then even after computing with primes up to 100 million, we seem to have only have about four decimal digits of accuracy:
99000000 1.120724721
100000000 1.120725012
101000000 1.120727310
For another example, if we start with the polynomial $n^2 + 5n + 1$ (discriminant $21$), then we need to compute (something like) the constant $$C_2 = \prod_{p \in P_1 \cap Q_1 \cup P_2 \cap Q_2} \frac{(p-2)}{(p-1)} \prod_{p \in P_1 \cap Q_2 \cup P_2 \cap Q_1} \frac{p}{(p-1)}$$ where $P_1 = \{p \equiv 1 \pmod 3\}$, $P_2 = \{p \equiv 2 \pmod 3\}$, $Q_1 = \{p \equiv \text{$1$,$2$, or$4$}\pmod 7\}$, $Q_2 = \{p \equiv \text{$3$,$5$or$6$}\pmod 7\}$ which is messier: we do one thing for $p \equiv 1, 4, 5, 16, 17, 20\pmod 21$ and another for $p \equiv 2, 8, 10, 11, 13, 19$.
Question: How can we compute these constants reasonably quickly?
I have seen the related questions on MathOverflow, and tried to read their answers and some of the mentioned references:
But the papers take quite a lot of effort for me to follow (I've never encountered Dirichlet L-series before, for example), and I keep suspecting there should be something simpler, if we don't want so much. (The older papers are simpler, but Western's paper simply says “by the use of a transformation suggested by Mr. Littlewood” without a reference or explaining anything.) In particular, I don't care about getting thousands of digits, I'll be happy if I can get, say six (ten would be fantastic). (Or even 4 digits, if they can be computed a lot faster.) And I don't care about computing such series that arise out of really large numbers; like the $3$ and $7$ in the second example above I might have at most two-digit factors.
With these looser constraints, is there something simpler that suffices, to compute these constants? What I'm looking for is either:
• ideally, a method of computation with sufficient detail, and which can be translated into a few lines of code that compute it from scratch (as in the program above) (after some reasonable amount of work on paper if necessary),
• or, the same, but it's ok to rely on certain constants (like $\zeta(2)$ say) that I can just look up and hard-code into the program, or on some existing functions (such as the logarithmic integral say), if they are well-known functions commonly available (in software libraries that are easy to install or have an online interface).
I'm looking for an elementary exposition at undergraduate level, say. For instance, the answer by KConrad makes sense (though it's not clear how good the convergence is), but I don't know how to compute $L(1,\chi_D)$.
• Fortunately, evaluating $L(1, \chi_D)$ is actually very straightforward. If you want to do this in a very sophisticated way, you could use the class number formula. But you could also evaluate it explicitly and simply through the use of Hurwitz zeta functions. $L(s, \chi)$ is a finite sum of Hurwitz zeta functions $\zeta(s, a)$, and $\zeta(0, a)$ is given explicitly by a Bernouilli number (like the regular zeta function). This will get you $L(0, \chi)$, and the functional equation will then get you $L(1, \chi)$. – davidlowryduda Mar 30 '18 at 0:35
• If you think this line of reasoning is interesting, I could return and flesh out an answer distilling the relevant parts. But you might think it is not elementary enough, perhaps? – davidlowryduda Mar 30 '18 at 0:36
• @mixedmath Thank you, that's great to know! I'm willing to accept on faith, any theorem, identity, or formula, no matter how “deep” (at least, if I can understand the terms and what is being claimed!), but ultimately I want to be able to turn what I learn into a computer program, so I'd hope for sufficient detail… assuming that's not too much work, I'd definitely love to hear it! – ShreevatsaR Mar 30 '18 at 0:44
• If you need to list all the primes up to a number using a sieve algorithm is faster than primality checking one by one. – Sophie Apr 4 '18 at 23:10
The intent here is to build off of Keith's answer, and thus discuss how to evaluate the $L$-function at $1$ and to discuss the rate of convergence.
## Evaluating $L(1, \chi)$
We will use the following theorem.
Theorem
Let $\chi$ be a primitive Dirichlet character of modulus $q$, and let $j$ be any positive integer. Then $$L(1-j,\chi) = - \frac{q^{j-1}}{j} \sum_{1 \leq a \leq q} \chi(a)B_{j}(\tfrac{a}{q}),$$ where $B_j(x)$ is the $j$th Bernoulli polynomial, defined by $$\frac{t e^{Xt}}{e^t-1} = \sum_{n=0}^\infty B_n(X) \frac{t^n}{n!}.$$
As a corollary, choosing $j = 1$ and evaluating that $B_1(x) = x - \tfrac{1}{2}$, we have that $$L(0, \chi) = - \sum_{1 \leq a \leq q} \chi(a) \left(\frac{a}{q} - \frac{1}{2}\right).$$ If $\chi \neq 1$, then $\sum_a \chi(a) = 0$, so that this simplifies to $$L(0, \chi) = - \sum_{1 \leq a \leq q} \chi(a) \frac{a}{q}.$$ This is tremendously straightforward to actually compute.
Recall the functional equation for a Dirichlet $L$-function, which states that $$\Lambda(s, \chi) := \left( \frac{\pi}{q} \right)^{-(s + A)/2} \Gamma(\tfrac{s + A}{2})L(s, \chi),$$ where $$A = \begin{cases} 0 & \text{if } \chi(-1) = 1 \\ 1 & \text{if } \chi(-1) = -1 \end{cases},$$ satisfies the functional equation $$\Lambda(1-s, \overline{\chi}) = \frac{i^A \sqrt q}{\tau(\chi)} \Lambda(s, \overline{\chi}).$$
There are several pieces here to disentangle. Note that $\overline{\chi}$ is the character obtained by (complex) conjugating $\chi$. For the example in your post, $\chi = \overline{\chi}$ since $\chi$ is real-valued. The function $\tau(\chi)$ is a Gauss sum, which are a bit annoying in general but ultimately they are a short finite sum. Note that when $\chi$ is quadratic, as in your example problem in the post, the Gauss sum is already known and $\tau(\chi) = \sqrt q$ if $q \equiv 1 \bmod 4$ and $\tau(\chi) = i \sqrt q$ if $q \equiv 3 \bmod 4$.
### An example application: your question
Let's specialize to your post now. The underlying character $\chi$ is the primitive character mod $3$, given by $$\chi(a) = \begin{cases} 0 & \text{if } a \equiv 0 \bmod 3 \\ 1 & \text{if } a \equiv 1 \bmod 3 \\ -1 & \text{if } a \equiv 2 \bmod 3 \end{cases}.$$ For concreteness, from the Euler product representation, we see that $$L(1, \chi) = \prod_p (1 - \chi(p)/p)^{-1} = \prod_{p \equiv 1 \bmod 3} \left(1 - \frac{1}{p}\right) \prod_{p \equiv 2 \bmod 3} \left(1 + \frac{1}{p}\right).$$
From the corollary to the theorem above, we see that $$L(0, \chi) = -\frac{1}{3} \left(1 - 2\right) = \frac{1}{3}.$$ Since $3 \equiv 3 \bmod 4$, we have that $\tau(\chi) = i \sqrt 3$, and $A = 1$ (as appears in the functional equation above). Note again that $\overline{\chi} = \chi$.
Thus $$\Lambda(1, \chi) = \frac{i \sqrt 3}{i \sqrt 3} \Lambda(0, \chi),$$ or rather $$L(1, \chi) \left( \frac{\pi}{3} \right)^{-1} \Gamma(1) = L(0, \chi) \left( \frac{\pi}{3} \right)^{-1/2} \Gamma(1/2).$$ Rearranging, we find that $$L(1, \chi) = L(0, \chi) \left( \frac{\pi}{3} \right)^{1/2} \Gamma(1/2) / \Gamma(1).$$ I note that $\Gamma(1/2) = \pi$, $\Gamma(1) = 0! = 1$, and (from above) $L(0, \chi) = 1/3$, so ultimately $$L(1, \chi) = \frac{1}{3} \frac{\pi}{\sqrt 3} = \frac{\pi\sqrt 3}{9}.$$
(Aside: this can be quickly verified in sage, and this is the output of quadratic_L_function_exact(1, -3). The - in -3 comes from a choice of numerator and denominator in specifying Dirichlet characters).
## Convergence of Keith's answer
Keith Conrad showed that one reduces this computation to computing $$\frac{1}{L(1,\chi_D)}\prod_{p} \frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} = \frac{1}{L(1,\chi_D)}\prod_{p} \left(1 - \left(\frac{\chi(p)}{(p-1)(p-\chi(p))} \right) \right)$$ (where I've kept the simplified RHS explicit, instead of replacing it with a big Oh term). Generically, the convergence of an infinite product, written in this form, is the same speed as the convergence of the corresponding infinite series $$\sum_{p} \left( \frac{\chi(p)}{(p-1)(p-\chi(p))} \right),$$ which converges at a rate at worst the same as $$\sum_n \frac{1}{n^2}.$$ But in fact the series is approximately alternating, and the rate of convergence will be much faster.
From heuristics coming from the alternating series test, you might expect that to compute the first $d$ digits, you should need to use the first $\sqrt d$ terms or so.
## Proof of Theorem
I will later edit in some exposition of this theorem, but this follows from representing the Dirichlet $L$-function as a sum of Hurwitz zeta functions and then evaluating each of these Hurwitz zeta functions at negative integer arguments. This is roughly done in Apostol's Intro to Analytic Number Theory, chapter 12.
But the point is that almost each step is actually very elementary, with the possible exception of proving the functional equations. Apostol actually uses the functional equations of Hurwitz zeta functions to prove the functional equation for Dirichlet $L$-functions, but some complex analysis gets involved.
• Thank you! Are there any additional complexities when $3$ in the example is replaced by a higher number: that is, when working, instead of with $\left(\frac{3}{p}\right)$, with say $\left(\frac{437}{p}\right)$ (here $437 = 19 \times 23$, the discriminant of a quadratic polynomial that I had happened to start this exploration with and am still interested in)? (I think in my example I was working with $-3$ instead of $3$ btw.) – ShreevatsaR Mar 30 '18 at 22:28
• No, there will be no significant differences for higher numbers (unless you happen to have a discriminant which is divisible by a nontrivial square number. Then the Dirichlet character isn't primitive, and a few minor changes are necessary). – davidlowryduda Mar 31 '18 at 0:57
• Could you clarify what exactly $\chi$ is (for the case of $\left(\frac{D}{p}\right)$), and what's the $q$, and how to find them? Is $\chi(n)=\left(\frac{D}{n}\right)$ even if $n$ may not be prime? For example, for $D=-3$, indeed I can see that we care about the character of modulus $3$, given in the answer. But for $D=3$, we have $\left(\frac{D}{p}\right)=1$ when $p \equiv1\pmod{12}$ and $\left(\frac{D}{p}\right)=-1$ when $p \equiv7\pmod{12}$, so I think in fact $q=12$. This does not fit either $q\equiv1\pmod4$ or $q\equiv3\pmod4$. (I think OEIS A037449 is related.) – ShreevatsaR Apr 8 '18 at 6:13 | 3,731 | 11,756 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-22 | latest | en | 0.608086 |
https://www.unitsconverters.com/en/Radianpersquaresecond-To-Cyclepersquaresecond/Unittounit-4771-7458 | 1,701,610,139,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00220.warc.gz | 1,185,967,444 | 37,534 | Formula Used
1 Radian per Square Second = 0.1592 Cycle per Square Second
1 Radian per Square Second = 0.1592 Cycle per Square Second
## Radian per Square Seconds to Cycle per Square Seconds Conversion
rad/s² stands for radian per square seconds and cyc/s² stands for cycle per square seconds. The formula used in radian per square seconds to cycle per square seconds conversion is 1 Radian per Square Second = 0.1592 Cycle per Square Second. In other words, 1 radian per square second is 7 times smaller than a cycle per square second. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Radian per Square Second to Cycle per Square Second
How to convert radian per square second to cycle per square second? In the angular acceleration measurement, first choose radian per square second from the left dropdown and cycle per square second from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from cycle per square second to radian per square second? You can check our cycle per square second to radian per square second converter.
How to convert Radian per Square Second to Cycle per Square Second?
The formula to convert Radian per Square Second to Cycle per Square Second is 1 Radian per Square Second = 0.1592 Cycle per Square Second. Radian per Square Second is 6.28140703517588 times Smaller than Cycle per Square Second. Enter the value of Radian per Square Second and hit Convert to get value in Cycle per Square Second. Check our Radian per Square Second to Cycle per Square Second converter. Need a reverse calculation from Cycle per Square Second to Radian per Square Second? You can check our Cycle per Square Second to Radian per Square Second Converter.
How many Radian per Square Minute is 1 Radian per Square Second?
1 Radian per Square Second is equal to 0.1592 Radian per Square Minute. 1 Radian per Square Second is 6.28140703517588 times Smaller than 1 Radian per Square Minute.
How many Revolution per Square Second is 1 Radian per Square Second?
1 Radian per Square Second is equal to 0.1592 Revolution per Square Second. 1 Radian per Square Second is 6.28140703517588 times Smaller than 1 Revolution per Square Second.
How many Revolution per Square Minute is 1 Radian per Square Second?
1 Radian per Square Second is equal to 0.1592 Revolution per Square Minute. 1 Radian per Square Second is 6.28140703517588 times Smaller than 1 Revolution per Square Minute.
## Radian per Square Seconds to Cycle per Square Seconds Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like angular acceleration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like rad/s² to cyc/s² through multiplicative conversion factors. When you are converting angular acceleration, you need a Radian per Square Seconds to Cycle per Square Seconds converter that is elaborate and still easy to use. Converting Radian per Square Second to Cycle per Square Second is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Radian per Square Second to Cycle per Square Second conversion along with a table representing the entire conversion.
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https://publicaawaz.com/learn-how-to-mannequin-a-magnet-falling-via-a-conducting-pipe/ | 1,674,772,601,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00720.warc.gz | 496,030,897 | 54,964 | HomePhysicsLearn how to Mannequin a Magnet Falling Via a Conducting Pipe
Learn how to Mannequin a Magnet Falling Via a Conducting Pipe
Introduction
In an earlier article, we examined a magnet falling by way of a solenoid. We argued that the purpose dipole mannequin can account for the fundamental options of the induced emf throughout the solenoid ends. Right here, we lengthen the mannequin to a magnet falling by way of a conducting pipe alongside its axis.
With the falling dipole second oriented alongside the vertical ##z##-axis, the electrical discipline ##E(rho,z)## is tangent to circles centered on the axis. The induced emf round a closed loop of radius ##rho## is ##textual content{emf}(z)=2pi rho~E(rho,z)##. There’s a similarity and a distinction between a solenoid and a conducting pipe positioned in that house. The similarity is of their modeling as a stack of conducting rings. The distinction is within the conceptual connection of the constituent rings. Within the solenoid, the rings are in sequence and the general emf throughout them is of curiosity. Within the pipe, the rings are type of in parallel and the whole present by way of them is of curiosity. “Kind of” signifies that the emf, being position-dependent, is just not the identical from one ring to the subsequent.
Nonetheless, Lenz’s regulation dictates that the induced currents will oppose the movement of the falling magnet. Within the section beneath the dipole, they are going to be in repulsion whereas within the section above the dipole they are going to be in attraction with the magnet. We are going to add these forces to seek out the online electromagnetic drive on the magnet. We are going to then clear up Newton’s second regulation equation to acquire an expression appropriate for elementary knowledge evaluation.
A single ring
We have now seen that the motional emf in a coaxial single ring of radius ##a## positioned at ##z’## is \$\$start{align}textual content{emf}_{textual content{ring}}= -frac{3mu_0ma^2}{2} frac{ (z-z’)~v} {left[a^2+(z-z’)^2right]^{5/2}}.finish{align}\$\$Right here, the constructive ##z## axis is “down”. The dipole with magnetic second ##mathbf{m}=m~hat{z}## is on the axis at ##z##.
Discovering the present and its discipline
The present within the ring is \$\$start{align}I=frac{textual content{emf}}{R}= -frac{3mu_0ma^2}{2R} frac{ (z-z’)~v} {left[a^2+(z-z’)^2right]^{5/2}}finish{align}\$\$the place ##R## is the ring’s resistance. The on-axis magnetic discipline generated by this present on the location of the dipole is \$\$mathbf{B}_{textual content{ring}}=frac{mu_0 a^2 I~hat{z}}{left[(a^2+(z-z’)^2right]^{3/2}}.\$\$ We assume that the ring is beneath the dipole (##z-z'<0##) and that the dipole factors within the constructive ##z## route. By Lenz’s regulation, the sphere and the dipole second have to be antiparallel.
Discovering the drive
We are going to receive the drive on the dipole utilizing power concerns. The potential power is \$\$U=-mathbf{m}cdotmathbf{B}_{textual content{ring}}=+m{B}_{textual content{ring}}=frac{mu_0 m a^2 I}{left[(a^2+(z-z’)^2right]^{3/2}}\$\$and the drive on the dipole is \$\$mathbf{F}=-mathbf{nabla}U=-mu_0 m a^2 Ifrac{partial}{partial z}left{frac{1}{left[(a^2+(z-z’)^2right]^{3/2}}proper}~hat z=frac{3 mu _0 a^2 I (z-z’)}{2 left[a^2+(z-z’)^2right]^{5/2}}~hat z.\$\$With the present from equation (2) the magnetic drive on the dipole is \$\$start{align}mathbf{F}=-frac{ 9 mu _0^2 m^2 a^4 (z-z’) ^2 ~v ~hat{z}}{4 R left[a^2+(z-z’) ^2right]^5}.finish{align}\$\$The drive exerted by the ring on the dipole is detrimental (up) for all values of the dipole-loop separation ##z-z’.## A separate calculation (not proven) of the integral ##oint I dmathbf{l} occasions mathbf{B}_{dip}## confirms that the drive exerted on the ring by the dipole is in step with Newton’s third regulation and equation (3).
Determine 1 reveals the dependence of the magnitude of the drive on the dipole as a operate of the decreased dipole-loop separation ##zeta=(z-z’)/a.## The drive is zero when the dipole passes by way of the middle of the ring. That’s as a result of the present goes by way of zero earlier than it reverses route. Nevertheless, the drive additionally goes by way of zero on the heart, but it doesn’t reverse route and shows a attribute two hump profile. Additionally, the drive is important solely inside a area of about ##pm## 2 ring radii from the middle.
Determine 1. The spatial dependence of the drive on the dipole exerted by a present loop as a operate of the decreased distance between the 2.
From ring to conducting pipe
We place the origin on the midpoint of a pipe of size ##L##. The web drive on the dipole is the vector sum of all ring contributions over the size of the pipe. First, we convert equation (3) to a drive factor ##dmathbf{F}##. We think about a hoop of radius ##rho’## and rectangular cross-sectional space ##dA’ =dz’occasions drho’##. The resistance of this ring is \$\$dR=frac{2pi rho’}{sigma~drho’~dz’}\$\$the place ##sigma## is the conductivity. Then \$\$dmathbf{F}=-frac{ 9sigma mu _0^2 m^2 {rho’}^3 (z-z’) ^2 ~v }{8pi left[{rho’}^2+(z-z’) ^2right]^5}~drho’~ dz’~hat{z}\$\$
We outline a drive issue ##mathcal{F}(z)## because the double integral with out the constants, the rate and the detrimental signal:
\$\$start{align}mathcal{F}(z)equivint_{-L/2}^{L/2}(z-z’)^2~dz’int_{a}^{b}frac{{rho’}^3}{left[{rho’}^2+(z-z’) ^2right]^5}~drho’.finish{align}\$\$ Inbuilt it’s the geometric info of the pipe: size ##L##, interior diameter ##a## and outer diameter ##b##.
The integrals
Assisted by Mathematica, we discover the radial integral,
\$\$int_a^b frac{ {rho’}^3 ~drho’}{left[{rho’}^2+(z-z’) ^2right]^5}=frac{4 a^2+(z-z’) ^2}{24 left[a^2+(z-z’) ^2right]^4}-frac{4 b^2+(z-z’) ^2}{24 left[b^2+(z-z’) ^2right]^4}.\$\$
We are going to break up and compact the outcomes of the axial integration over ##z’## as a way to match them on the display. We have now,
\$\$ start{align}& int_{-L/2}^{L/2}frac{left(z-z’)^2[4 a^2+(z-z’) ^2right]}{24 left[a^2+(z-z’) ^2right]^4}dz’=nonumber & left .frac{1}{384a^3}left{frac{a (z-z’) left[5 a^4-8 a^2 left(z-z’right)^2-5 (z-z’)^4right]}{left[a^2+left(z-z’right)^2right]^3}-5 tan ^{-1}left(frac{z-z’}{a}proper)proper}
proper |_{z’=-L/2}^{z’=L/2}nonumber finish{align}\$\$
and
\$\$ start{align}&- int_{-L/2}^{L/2}frac{left(z-z’)^2[4 b^2+(z-z’) ^2right]}{24 left[b^2+(z-z’) ^2right]^4}dz’=nonumber & left . -frac{1}{384b^3}left{frac{b (z-z’) left[5 b^4-8 b^2 left(z-z’right)^2-5 (z-z’)^4right]}{left[b^2+left(z-z’right)^2right]^3}-5 tan ^{-1}left(frac{z-z’}{b}proper)proper} proper |_{z’=-L/2}^{z’=L/2}nonumber finish{align}\$\$
The sum of the evaluated phrases above is the drive issue ##mathcal{F}(z).##
The drive issue
Determine 2 reveals two plots for the drive issue ##mathcal{F}(z)## for 2 outer radii, ##b=1.25a## (blue line) and ##b=2a## (orange line). The pipe size was set at ##L=100a##. The plots point out that the drive issue rises quick, is flat, and rises increased for the thicker pipe. We are going to take care of these options and what all of it means in FAQ format.
Determine 2. The drive issue as a operate of place for 2 values of the outer diameter, 1.25 a (blue) and a couple of.0 a (orange). The size of the pipe is 100 a.
How briskly does it rise?
For the reason that efficient vary of the dipole is 4 interior radii, we might anticipate ##mathcal{F}(z)## to alter appreciably inside that distance from the ends. Determine 3 under reveals a element of Determine 2 and confirms this. The primary vertical line is on the entry level at ##z/a=-L/(2a)=-50.## The second vertical line is at at ##z/a=-L/(2a)+2=-48.## It’s the place we think about the dipole to be “simply totally inside” the pipe. Clearly, the drive issue is kind of flat there. That’s fairly quick.
Determine 3. Element of Determine 2 close to the purpose of entry at -50.
The plateau on the level of entry could be understood by referring to the double-humped drive proven in Determine 1. Very close to the purpose of entry the drive because of the first infinitesimal ring and its close to neighbors is small and reducing, goes by way of zero, after which will increase. The whole drive stays flat till the dipole is much sufficient inside for the second hump to kick in. As soon as totally inside there’s a steady-state. New rings forward of the dipole come into vary whereas previous ones behind the dipole fade out of vary.
Sure, however how flat is “flat”?
We take as a measure of flatness is the % ratio ##mathcal{F}(-48)/mathcal{F}(0)##. It’s the ratio of the worth of the drive issue at 2 ring radii in to the worth within the center. For the pipe with ##b=1.25a## (blue line), the ratio is 99.3%; for the pipe with ##b=2~a## it’s 98.2%. That’s fairly flat.
How does the wall thickness have an effect on it?
We calculated the worth on the midpoint ##mathcal{F}(0)## as a operate of the decreased outer radius ##b/a##. Determine 4 under reveals the ensuing plot. Evidently past ##bapprox 2a##, making the wall thicker produces diminishing returns. This is smart. On one hand, growing the wall thickness introduces extra cost carriers within the house. Then again, the magnetic discipline that induces these currents turns into weaker with distance at a quicker charge.
Determine 4. The drive issue at z=0 as a operate of the outer radius for ##lambda##=100. See equation (8) within the Appendix.
What’s the backside line?
To an excellent approximation, ##mathcal{F}(z) approx mathcal{F}(0)## for ##-L/2 leq z leq L/2##. This validates the assertion that the magnetic drive is proportional to the rate contained in the pipe. Then we will write the magnitude of the drive on the dipole as \$\$start{align}F(z) =frac{ 9sigma mu _0^2 m^2 v }{8pi}mathcal{F}(0).finish{align}\$\$ The analytic expression for ##mathcal{F}(0)## seems within the Appendix as equation (8) and is plotted in Determine 4 for ##lambda=L/a=100.##
The kinematics
We write the magnetic drive on the dipole as ##F=-kappa~v## and begin with the differential equation \$\$Mfrac{dv}{dt}=-kappa~v+Mg~~~~left(kappa equiv frac{ 9sigma mu _0^2 m^2}{8pi}mathcal{F}(0)proper)\$\$ the place ##M## is the mass of the dipole.
We solid the well-known resolution by way of the terminal velocity, ##v_{textual content{ter}}=dfrac{Mg}{kappa}##: \$\$v=v_{textual content{ter}}+(v_0-v_{textual content{ter}})e^{-g~t/v_{textual content{ter}}}implies e^{-g~t/v_{textual content{ter}}}=frac{(v-v_{textual content{ter}})}{(v_0-v_{textual content{ter}})}.\$\$
Additionally, \$\$z=int_0^television~dt=v_{textual content{ter}}t+frac{v_{textual content{ter}}(v_0-v_{textual content{ter}})}{g}left[1-e^{-g~t/v_{text{ter}}}right].\$\$ By eliminating the exponential, the 2 equations could be mixed to yield \$\$start{align}z=v_{textual content{ter}}t-frac{v_{textual content{ter}}left(v-v_0right)}{g}.finish{align}\$\$
A prompt experiment
An apparent experiment to carry out and analyze can be to drop a magnet right into a pipe and use photogates and knowledge logging software program to measure the entry velocity ##v_0##, exit velocity ##v_{!f}##, and transit time ##T##. Equation (6) turns into \$\$start{align}L=v_{textual content{ter}}T-frac{v_{textual content{ter}}left(v_{!f}-v_0right)}{g}.finish{align}\$\$It may be solved to seek out the terminal velocity. Repeating for different decisions of entry velocity ought to confirm the fidelity of the terminal velocity. The worth of the terminal velocity would then decide the worth of ##kappa.## Notice that this dedication relies on kinematics and is impartial of the magnet mannequin.
Furthermore, equation (5) reveals that the drive issue (decided by the pipe geometry), and the conductivity (decided by the selection of pipe materials), could be mathematically separated out of ##kappa.## One can check the validity of this separation by performing a sequence of experiments wherein the identical magnet is dropped by way of pipes of varied wall thicknesses, lengths, and conductivities. When ##mathcal{F}(0)## and the conductivity are factored out of ##kappa##, a continuing shared by all experimental runs ought to be left. One can then clear up equation (5) to seek out an efficient magnetic second acceptable to a given magnet.
As an apart and since there isn’t a different appropriate place within the article, the creator can not resist mentioning the next commentary. With the usual definitions of common velocity ##bar v=L/T##, common acceleration ##bar a=(v_{!f}-v_0)/T## and exponential time fixed ##tau=v_{textual content{ter}}/g##, equation (7) turns into the curiously suggestive relation \$\$v_{textual content{ter}}=bar v+bar atau.\$\$It brings collectively two express constants, ##v_{textual content{ter}}## and ##kappa##, an implicit impartial variable, ##v_0## and two implicit dependent variables, ##v_{!f}## and ##T##. Thus, regardless of its SUVAToid look, it’s a constraint relation on the dependent variables.
Abstract and afterthoughts
The magnetic dipole mannequin of a magnet falling by way of a pipe is enough to point out mathematically that the magnetic drive opposing the movement of the falling dipole contained in the solenoid is fast-rising and could be written as a continuing ##kappa## multiplied by the instantaneous velocity. The mannequin additionally affords the calculation of a drive issue that enables the separation of the pipe geometry and conductivity from the fixed ##kappa.## As soon as that separation is finished, one might extract a worth for an efficient dipole second that might function enter to a extra real looking mannequin than the purpose dipole.
Moreover, if a sensible mannequin is a objective, evidently place to focus one’s experimental efforts are the entry and exit factors of the magnet. As Determine 3 reveals, at these factors, the drive profile is most delicate to the form of the magnet and its efficient vary. No matter one’s objective, the purpose dipole mannequin has fairly a bit to say; its simplicity makes it place to start out the mathematical evaluation of a magnet falling by way of a conducting pipe.
Acknowledgment
The creator needs to thank person @Delta2 for useful feedback and remarks on an earlier model of this text.
Appendix
With decreased lengths ##lambda= L/a## and ##beta = b/a##, the worth of the drive issue is
\$\$start{align}mathcal{F}(0)= & frac{lambda}{96a^3}left[frac{80 beta ^4-32 beta ^2 lambda ^2-5 lambda ^4}{beta ^2 left(4 beta ^2+lambda ^2right)^3}-frac{80-32 lambda ^2-5 lambda ^4}{left(lambda ^2+4right)^3}right] nonumber & + frac{5}{192a^3}left[tan ^{-1}left(frac{lambda }{2}right)-frac{1}{beta ^3}tan ^{-1}left(frac{lambda }{2 beta }right)right]
.finish{align}\$\$
References
For readers who want to discover different approaches. The checklist is just not meant to be full.
C.S. MacLatchy, P. Backman and L. Bogan, Am. J. Phys. 61, 1096 (1993); doi: 10.1119/1.17356
Ok.D. Hahn, E.M. Johnson, A. Brokken and Steven Baldwinc, Am. J. Phys. 66, 1066 (1998); doi: 10.1119/1.19060
Y. Levin, F.L. da Silveira and F.B. Rizzato, Am. J. Phys. 74, 815 (2006); doi: 10.1119/1.2203645
G. Donoso, C.L. Ladera and P. Martin, Eur. J. Phys. 30 (2009), 855–869 doi:10.1088/0143-0807/30/4/018
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1. ## real numbers of an equation
Hi, I need to find all the real numbers to satisfy x & y for the following equation 2x-1+2yi=4x-5i HELP
2. Originally Posted by babs21
Hi, I need to find all the real numbers to satisfy x & y for the following equation 2x-1+2yi=4x-5i HELP
The real and imaginary parts of this equation both need to be satisfied. So we have that:
$\displaystyle 2x - 1 = 4x$
and
$\displaystyle 2y = -5$
Can you take it from here?
-Dan
3. so x = -1/2 and y = -5/2 but what about i I am lost
4. Originally Posted by topsquark
The real and imaginary parts of this equation both need to be satisfied. So we have that:
$\displaystyle 2x - 1 = 4x$
and
$\displaystyle 2y = -5$
Can you take it from here?
-Dan
Originally Posted by babs21
so x = -1/2 and y = -5/2 but what about i I am lost
I was factored out of the equation
$\displaystyle 2yi=-5i$
to make
$\displaystyle 2y=-5$ | 301 | 935 | {"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} | 3.90625 | 4 | CC-MAIN-2018-22 | latest | en | 0.918388 |
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### 遍历字典
```testMap := map[string]int{
"one": 1,
"two": 2,
"three": 3,
}
for key, value := range testMap {
fmt.Println(key, value)
}```
```three 3
one 1
two 2```
```for _, value := range testMap {
fmt.Println(value)
}```
```for key := range testMap {
fmt.Println(key)
}```
### 键值对调
```invMap := make(map[int] string, 3)
for k, v := range testMap {
invMap[v] = k
}
for k, v := range invMap {
fmt.Println(k, v)
}```
```3 three
1 one
2 two```
### 字典排序
```keys := make([]string, 0)
for k, _ := range testMap {
keys = append(keys, k)
}
sort.Strings(keys) // 对键进行排序
fmt.Println("Sorted map by key:")
for _, k := range keys {
fmt.Println(k, testMap[k])
}```
```Sorted map by key:
one 1
three 3
two 2```
```values := make([]int, 0)
for _, v := range testMap {
values = append(values, v)
}
sort.Ints(values) // 对值进行排序
fmt.Println("Sorted map by value:")
for _, v := range values {
fmt.Println(invMap[v], v)
}```
```Sorted map by value:
one 1
two 2
three 3```
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# The author seems well qualified to write the book: she is
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The author seems well qualified to write the book: she is [#permalink]
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11 Jun 2013, 01:33
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The author seems well qualified to write the book: she is fluent in Chinese and English, has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does.
1 has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does.
2 has reported from Beijing – where most of the story takes place – and, like the main character in the story, emigrated to the New York area when she was a teenager.
3 has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just like the main character in the story does.
4 has reported from Beijing – where most of the story takes place – and, she, like the main character in the story does, emigrated to the New York area when she was a teenager.
5 has reported from Beijing – where most of the story takes place – and, she, like the main character in the story, emigrated to the New York area when she was a teenager.
Please explain! Thanks
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Re: The author seems well qualified to write the book: she is fl [#permalink]
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The author seems well qualified to write the book: she is fluent in Chinese and English, has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does.
Focus on the parallelism:
she is fluent, has reported from Beijing ...
1 and she emigrated
2 and, like the main character in the story, emigrated to the New York area when she was a teenager.
3 and she emigrated
4 and, she, like the main character in the story does, emigrated
5 and, she, like the main character in the story, emigrated
There are three actions that the subject does. All those have to be parallel.
In a parallel structure of this kind there is no need to repet the subject: since all actions are parallel the subject is the same.
"My brother lives (...), plays (...) and hetravels (...)
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Re: The author seems well qualified to write the book: she is fl [#permalink]
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11 Jun 2013, 01:57
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The author seems well qualified to write the book: she is fluent in Chinese and English, has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does.
1 has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does. - This option changes the intended meaning of the sentence. This option conveys that the both the author & character migrated when they were teenager rather than both just migrated.
2 has reported from Beijing – where most of the story takes place – and, like the main character in the story, emigrated to the New York area when she was a teenager. - CORRECT
3 has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just like the main character in the story does - LIKE must be followed by a Phrase & not by a clause.
4 has reported from Beijing – where most of the story takes place – and, she, like the main character in the story does, emigrated to the New York area when she was a teenager. - LIKE must be followed by a Phrase & not by a clause. Moreover this option introduces modifier error
5 has reported from Beijing – where most of the story takes place – and, she, like the main character in the story, emigrated to the New York area when she was a teenager. - Changes the intended meaning of the sentence. This option means that the Author is similar to the main character of the story
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Re: The author seems well qualified to write the book: she is fl [#permalink]
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11 Jun 2013, 05:28
The author seems well qualified to write the book: she is fluent in Chinese and English, has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does.
1 has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just as the main character in the story does.
there should be a "comma" after the and, as the list contains more than two items
2 has reported from Beijing – where most of the story takes place – and, like the main character in the story, emigrated to the New York area when she was a teenager.
Correct
3 has reported from Beijing – where most of the story takes place – and she emigrated to the New York area when she was a teenager, just like the main character in the story does.
there should be a "comma" after the and, as the list contains more than two items
4 has reported from Beijing – where most of the story takes place – and, she, like the main character in the story does, emigrated to the New York area when she was a teenager.
like compares nouns, hence like the main character in the story does is wrong as it is comparing action, as should be used instead
5 has reported from Beijing – where most of the story takes place – and, she, like the main character in the story, emigrated to the New York area when she was a teenager.
and, she, like the main character in the story - the subject is not required again as it breaks the parallelism
Hope it helps!
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Re: The author seems well qualified to write the book: she is [#permalink]
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Re: The author seems well qualified to write the book: she is [#permalink] 08 Sep 2015, 22:28
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<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">John,</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">You know how to ask the right questions.<span style> </span>Your last question about my statement: “Glutamate
behaves the way it does BECAUSE of its redness quality.” Helped <span style> </span>me to understand that this is not the right
way to say it.<span style> </span>“BECAUSE” implies a cause
and effect relationship, which is not right.<span style>
</span>The two are simply two different ways to look at the same physically detectable
stuff:</span></p><p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"><br></span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">1: The Strawberry is causally red,
and this physical property can be detected.</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">2: The strawberry reflects 650 nm
light</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">Now you are asking another good
question: “if X fully represents Y how can X not have all the qualities of Y?”</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">Let me see if it helps to put it
this way:</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">Would you agree that something,
detectable, is responsible for the elemental redness quality you can experience?</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">If so, what do you imagine it to
be?<span style> </span>Certainly, it is not infinite in
complexity, whatever it is?</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">And would you agree that the word
red (nor any other zombie knowledge) has or is the same as whatever that is?</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">Yet, you can interpret enough
abstract symbols, (which by definition do not have redness) to represent anything you want, but only if you know how to
qualitatively interpret it back to the real thing that does have, or is
responsible for redness?</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif">Brent</span></p>
<p class="MsoNormal" style="text-align:left" align="left"><span style="font-size:12pt;font-family:"Times New Roman",serif"> </span></p>
</div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Jan 29, 2015 at 2:07 PM, John Clark <span dir="ltr"><<a href="mailto:johnkclark@gmail.com" target="_blank">johnkclark@gmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">Brent, you say "Zombie Red: Any representation of red that can fully represent its referent, but which does not have the quality it represents", but if X fully represents Y how can X not have all the qualities of Y?<div><br></div><div> John K Clark<br><div><br></div><div><br></div><div><br></div><div><br></div><div><br><div><div class="gmail_extra"><div class="gmail_quote"><div><div><div class="gmail_extra"><div class="gmail_quote"><br></div></div></div></div></div><br></div></div></div></div></div>
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<br></blockquote></div><br></div> | 1,836 | 5,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-39 | latest | en | 0.094762 |
https://physics.stackexchange.com/questions/365012/square-of-annihilation-operator | 1,721,783,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00499.warc.gz | 399,333,052 | 43,472 | # Square of annihilation operator
Sorry about this question, but due to my limited background in quantum physics I wasn't able to figure out why in ALL the refs I have searched they state it as iffor sure: Why is the square of a creation/annihilation operator zero? Given that this is for a Fock space system.
$$\hat{a}^\dagger~\propto~(\omega\hat{q} - i\hat{p}) \tag{2.14}$$ $$\hat{a}^{\dagger 2} = 0$$
The 2 means that the operator is in power two. The same holds true for the annihilation operator. I found some phoney proof (based on the fact that this describes a harmonic oscillator) which I don't like and wish to get feedback from experienced people.
References:
1. C.C. Gerry & P.L. Knight, Introductory Quantum Optics, 2004; eq. (2.45).
• Fermionic operators satisfy $(A^\dagger)^2=0$ since you cannot put two fermions in the same state. Commented Oct 25, 2017 at 17:59
• Hi @Mazem: Is this from a textbook? Which page? Which formula? Commented Oct 25, 2017 at 18:03
• @Qmechanic Yeah. “Introductory Quantum Optics ”, Jerry & Knight. ~p. 15 There is a load of other handouts here and there from the net. They all state the same thing. Commented Oct 25, 2017 at 18:34
• @Qmechanic It has been the case to find the variance of the electric field where the creation + annihilation operators are raised to power 2. I can completely understand that this is not a commutative operation (I.e. a.b =/= b.a) but prefer to approach the creation square(and annihilation square) from a mathematical view point of possible) If it is not & better thought of as an axiom as ZeroTheHero refers, I’ll be quite grateful to cite the ref for this. Commented Oct 25, 2017 at 18:45
• Are you talking about eq. (2.45)? Commented Oct 25, 2017 at 18:51
1. Without context, it first sounded like you were considering fermionic creation & annihilation operators, which square to zero.
2. However, Ref. 1 is considering bosonic creation & annihilation operators, so they definitely do not square to zero.
3. If the $\hat{a}^{2}$ or $\hat{a}^{\dagger 2}$ are sandwiched between a bra vector $\langle m|$ and a ket vector $|n\rangle$, then the result is zero if $m-n\neq \pm 2$. Comparison with Ref. 1 reveals that this is exactly what's going on.
References:
1. C.C. Gerry & P.L. Knight, Introductory Quantum Optics, 2004; eq. (2.45).
• In equation (2.45) as shown ${\hat{a}}^{\dagger 2}$ disappears and the same holds true for ${\hat{a}}^2$. Probably, it is their addition what goes to zero. BTW, their addition gives you $\frac{1}{\hbar \omega}(\omega^2 \hat{q}^2 - \hat{p}^2)$. How could this value be zero? Does it have to do eigen propoerty of vectors? Commented Oct 26, 2017 at 10:37
• They disappear independently! Commented Oct 26, 2017 at 10:42
• is there a way to prove that mathematically? Commented Oct 26, 2017 at 10:45
• Yes. You need: 1. $\hat{a}|0\rangle=0$; 2. $\langle 0|\hat{a}^{\dagger }=0$; 3. the CCR; 4. and a couple of definitions. Commented Oct 26, 2017 at 10:49
There are two different types of creation and annihilation operators:
• One is the bosonic set, which obeys the commutation relation $[a,a^\dagger]=1$ (and trivially $[a,a]=0=[a^\dagger,a^\dagger]$), and for which you can define quadratures via $q=\frac12(a+a^\dagger)$ and $p=\frac{1}{2i}(a-a^\dagger)$ which obey $[q,p]=i$, i.e. $a=q+ip$ and $a^\dagger=q-ip$ can be seen as the ladder operators for a harmonic oscillator with hamiltonian $H=\frac12(p^2+q^2)$.
• The other is the fermionic set, which obeys the anticommutation relations $\{c,c^\dagger\}=1$, $\{c,c\}=0=\{c^\dagger,c^\dagger\}$, from which follows that $c^2=0=(c^\dagger)^2$.
The question incorrectly conflates properties of the two sets, i.e. the question as written asserts that 'the' creation operator obeys both $(a^\dagger)^2=0$ and $a^\dagger = q-ip$ where $q$ and $p$ are reasonably-behaved quadratures. There is no such operator: it's either fermionic, with the first property, or bosonic, with the second property.
However, OP's reference for that first property makes a much weaker claim: Gerry & Knight's eq. (2.45) does not require that $(a^\dagger)^2=0$ or that $a^2=0$; instead, it only requires that their expectation values over Fock states vanish, i.e. $$⟨n|a^2|n⟩=0=⟨n|(a^\dagger)^2|n⟩.$$ These two properties are complex conjugates of each other, and they can be rigorously proved from the facts that $$a|n⟩=\sqrt{n}|n-1⟩$$ so therefore $$a^2|n⟩=\sqrt{n(n-1)}|n-2⟩$$ and thus $$⟨n|a^2|n⟩=\sqrt{n(n-1)}⟨n|n-2⟩=0.$$ However, just because the expectation values of an operator vanish in the Fock basis (like they do for $a$ and $a^\dagger$ themselves) does not mean that the operator is identically zero, so e.g. $⟨0|a^2|2⟩ = \sqrt{2}\neq 0$ as a simple example.
• Many thanks for your clarification. Now that is a complete answer! Honestly, I managed to prove that $<n|a^{\dagger 2}|n> = 0$ just before your post and was just about to answer my own question. But this is more comprehensive and covers the subtlty between fermionic and bosonic operators. Commented Oct 29, 2017 at 13:34 | 1,553 | 5,054 | {"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} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.920384 |
https://www.expertsmind.com/library/what-is-the-frequency-of-the-a-allele-in-the-population-51134718.aspx | 1,721,085,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00467.warc.gz | 662,456,772 | 14,337 | What is the frequency of the a allele in the population
Assignment Help Biology
Reference no: EM131134718
In a population of 2,454 moths, you observe 532 moths that have large spots on their wings. You identify that having large spots is dominant over small spots (aa), but the trait displays incomplete dominance (Aa individuals have medium spots).
What is the frequency of the A allele in the population?
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How many phage were in the original stock
You serially dilute a stock of bacteriophage by sequentiallytaking: 10 ul into 1 ml, 10 ul into 1 ml, 100 ul into 0.9 ml andplating 0.1 ml on a sensitive strain of bacteria. You later count96 plaques on that plate. How many phage per ml were in th.. | 988 | 4,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-30 | latest | en | 0.950696 |
https://brilliant.org/problems/this-is-jee-2006/ | 1,495,627,056,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607813.12/warc/CC-MAIN-20170524112717-20170524132717-00150.warc.gz | 722,592,222 | 18,710 | # Area of a Triangle Just From a Circle?
Geometry Level 4
The given triangle is isosceles with vertex angle $$120^\circ$$ and inradius of length $$\sqrt{3}.$$ In simplest form, the area of the triangle is $$a + b\sqrt{c}.$$ Find $$a+b+c.$$
× | 72 | 244 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-22 | longest | en | 0.711157 |
https://www.crackmnc.com/2014/05/data-structure-aptitude.html | 1,721,066,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00251.warc.gz | 650,559,658 | 36,749 | # Data Structure Aptitude
1. What is data structure?
A data structure is a way of organizing data that considers not only the items stored, but also their relationship to each other. Advance knowledge about the relationship between data items allows designing of efficient algorithms for the manipulation of data.
2. List out the areas in which data structures are applied extensively?
Compiler Design,
Operating System,
Database Management System,
Statistical analysis package,
Numerical Analysis,
Graphics,
Artificial Intelligence,
Simulation
3. What are the major data structures used in the following areas : RDBMS, Network data model & Hierarchical data model.
RDBMS – Array (i.e. Array of structures)
Network data model – Graph
Hierarchical data model – Trees
4. If you are using C language to implement the heterogeneous linked list, what pointer type will you use?
The heterogeneous linked list contains different data types in its nodes and we need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So we go for void pointer. Void pointer is capable of storing pointer to any type as it is a generic pointer type.
5. Minimum number of queues needed to implement the priority queue?
Two. One queue is used for actual storing of data and another for storing priorities.
6. What is the data structures used to perform recursion?
Stack. Because of its LIFO (Last In First Out) property it remembers its ‘caller’ so knows whom to return when the function has to return. Recursion makes use of system stack for storing the return addresses of the function calls.
Every recursive function has its equivalent iterative (non-recursive) function. Even when such equivalent iterative procedures are written, explicit stack is to be used.
7. What are the notations used in Evaluation of Arithmetic Expressions using prefix and postfix forms?
Polish and Reverse Polish notations.
8. Convert the expression ((A + B) * C – (D – E) ^ (F + G)) to equivalent Prefix and Postfix notations.
Prefix Notation:
^ - * +ABC - DE + FG
Postfix Notation:
AB + C * DE - - FG + ^
9. Sorting is not possible by using which of the following methods?
(a) Insertion
(b) Selection
(c) Exchange
(d) Deletion
(d) Deletion.
Using insertion we can perform insertion sort, using selection we can perform selection sort, using exchange we can perform the bubble sort (and other similar sorting methods). But no sorting method can be done just using deletion.
10. A binary tree with 20 nodes has_______null branches?
21
Let us take a tree with 5 nodes (n=5)
It will have only 6 (i.e., 5+1) null branches. In general, A binary tree with n nodes has exactly n+1null nodes.
11. What are the methods available in storing sequential files?
Straight merging,
Natural merging,
Polyphase sort,
Distribution of Initial runs.
12. How many different trees are possible with 10 nodes ?
1014
For example, consider a tree with 3 nodes (n=3), it will have the maximum combination of 5 different (i.e., 23 - 3 = 5) trees.
In general:
If there are n nodes, there exist 2n-n different trees.
13. List out few of the Application of tree data-structure?
The manipulation of Arithmetic expression,
Symbol Table construction,
Syntax analysis.
14. List out few of the applications that make use of Multilinked Structures?
Sparse matrix,
Index generation.
15. In tree construction which is the suitable efficient data structure?
(a) Array
(c) Stack
(d) Queue
(e) none
16. What is the type of the algorithm used in solving the 8 Queens problem?
Backtracking
17. In an AVL tree, at what condition the balancing is to be done?
If the ‘pivotal value’ (or the ‘Height factor’) is greater than 1 or less than –1.
18. What is the bucket size, when the overlapping and collision occur at same time?
One. If there is only one entry possible in the bucket, when the collision occurs, there is no way to accommodate the colliding value. This results in the overlapping of values.
19. Traverse the given tree using Inorder, Preorder and Postorder traversals.
Inorder : D H B E A F C I G J
Preorder: A B D H E C F G I J
Postorder: H D E B F I J G C A
20. There are 8, 15, 13, 14 nodes were there in 4 different trees. Which of them could have formed a full binary tree?
15.
In general:
There are 2n-1 nodes in a full binary tree.
By the method of elimination:
Full binary trees contain odd number of nodes. So there cannot be full binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete binary tree but not a full binary tree. So the correct answer is 15.
*Note:
Full and Complete binary trees are different. All full binary trees are complete binary trees but not vice versa.
21. In the given binary tree, using array you can store the node 4 at which location?
At location 6
22. Sort the given values using Quick Sort?
65 70 75 80 85 60 55 50 45
Sorting takes place from the pivot value, which is the first value of the given elements, this is marked bold. The values at the left pointer and right pointer are indicated using L and R respectively.
65 70L 75 80 85 60 55 50 45R
Since pivot is not yet changed the same process is continued after interchanging the values at L and R positions
65 45 75L 80 85 60 55 50R 70
65 45 50 80L 85 60 55R 75 70
65 45 50 55 85L 60R 80 75 70
65 45 50 55 60R 85L 80 75 70
When the L and R pointers cross each other the pivot value is interchanged with the value at right pointer. If the pivot is changed it means that the pivot has occupied its original position in the sorted order (shown in bold italics) and hence two different arrays are formed, one from start of the original array to the pivot position-1 and the other from pivot position+1 to end.
60L 45 50 55R 65 85L 80 75 70R
55L 45 50R 60 65 70R 80L 75 85
50L 45R 55 60 65 70 80L 75R 85
In the next pass we get the sorted form of the array.
45 50 55 60 65 70 75 80 85
23. For the given graph, draw the DFS and BFS?
BFS: A X G H P E M Y J
DFS: A X H P E Y M J G
24. Classify the Hashing Functions based on the various methods by which the key value is found.
Direct method,
Subtraction method,
Modulo-Division method,
Digit-Extraction method,
Mid-Square method,
Folding method,
Pseudo-random method.
25. What are the types of Collision Resolution Techniques and the methods used in each of the type?
The methods used include:
Overflow block,
The methods used include:
Binary tree…
26. In RDBMS, what is the efficient data structure used in the internal storage representation?
B+ tree. Because in B+ tree, all the data is stored only in leaf nodes, that makes searching easier. This corresponds to the records that shall be stored in leaf nodes.
27. Draw the B-tree of order 3 created by inserting the following data arriving in sequence –
92 24 6 7 11 8 22 4 5 16 19 20 78
28. Of the following tree structure, which is, efficient considering space and time complexities?
(a) Incomplete Binary Tree
(b) Complete Binary Tree
(c) Full Binary Tree
(b) Complete Binary Tree.
By the method of elimination:
Full binary tree loses its nature when operations of insertions and deletions are done. For incomplete binary trees, extra storage is required and overhead of NULL node checking takes place. So complete binary tree is the better one since the property of complete binary tree is maintained even after operations like additions and deletions are done on it.
29. What is a spanning Tree?
A spanning tree is a tree associated with a network. All the nodes of the graph appear on the tree once. A minimum spanning tree is a spanning tree organized so that the total edge weight between nodes is minimized.
30. Does the minimum spanning tree of a graph give the shortest distance between any 2 specified nodes?
No.
Minimal spanning tree assures that the total weight of the tree is kept at its minimum. But it doesn’t mean that the distance between any two nodes involved in the minimum-spanning tree is minimum.
31. Convert the given graph with weighted edges to minimal spanning tree.
the equivalent minimal spanning tree is:
32. Which is the simplest file structure?
(a) Sequential
(b) Indexed
(c) Random
(a) Sequential
33. Whether Linked List is linear or Non-linear data structure?
• According to Access strategies Linked list is a linear one.
• According to Storage Linked List is a Non-linear one.
34. Draw a binary Tree for the expression :
A * B - (C + D) * (P / Q)
35. For the following COBOL code, draw the Binary tree?
01 STUDENT_REC.
02 NAME.
03 FIRST_NAME PIC X(10).
03 LAST_NAME PIC X(10).
02 YEAR_OF_STUDY.
03 FIRST_SEM PIC XX.
03 SECOND_SEM PIC XX. | 2,129 | 8,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.855285 |
kaspar.website | 1,708,824,457,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00848.warc.gz | 336,731,053 | 12,119 | # Neural Processes as distributions over functions
In this year’s ICML, some interesting work was presented on Neural Processes. See the paper conditional Neural Processes and the follow-up work by the same authors on Neural Processes which was presented in the workshop.
Neural Processes (NPs) caught my attention as they essentially are a neural network (NN) based probabilistic model which can represent a distribution over stochastic processes. So NPs combine elements from two worlds:
• Deep Learning – neural networks are flexible non-linear functions which are straightforward to train
• Gaussian Processes – GPs offer a probabilistic framework for learning a distribution over a wide class of non-linear functions
Both have their advantages and drawbacks. In the limited data regime, GPs are preferable due to their probabilistic nature and ability to capture uncertainty. This differs from (non-Bayesian) neural networks which represent a single function rather than a distribution over functions. However the latter might be preferable in the presence of large amounts of data as training NNs is computationally much more scalable than inference for GPs. Neural Processes aim to combine the best of these two worlds.
I found the idea behind NPs interesting, but I felt I was lacking intuition and a deeper understanding how NPs behave as a prior over functions. I believe, often the best way towards understanding something is implementing it, empirically trying it out on simple problems, and finally explaining this to someone else. So here is my attempt at reviewing and discussing NPs.
Before reading my post, I recommend the reader to take a look at both original papers. Even though here I discuss [NPs], you might find it easier to start with [conditional NPs] which are essentially a non-probabilistic version of NPs.
### What is a Neural Process?
The NP is a neural network based approach to represent a distribution over functions. The broad idea behind how the NP model is set up and how it is trained is illustrated in this schema:
Given a set of observations $(x_i, y_i)$, they are split into two sets: “context points” and “target points”. Given the pairs $(x_c, y_c)$ for $c = 1, …, C$ in the context set and given unseen inputs $x_t^{\ast}$ for $t = 1, …, T$ in the target set, our goal is to predict the corresponding function values $y_t^{\ast}$. We can think of NPs as if they were modelling the target set conditional on the context. Information flows from the context set (on the left) to making new predictions on the target set (on the right) via the latent space $z$. The latter is essentially a finite-dimensional embedding of mappings from x to y. The fact that $z$ is a random variable makes NP a probabilistic model and lets us capture uncertainty over functions. Once we have trained the model, we can use our (approximate) posterior distribution of $z$ as a prior to make predictions at test time.
At first sight, the split into context and target sets may look like the standard train and test split of the data, but this is not the case, as the target set is directly used in training the NP model – our (probabilistic) loss function is explicitly defined over the target set. This will allow the model to avoid overfitting and achieve better out-of-sample generalisation. In practice, we would repeatedly split our training data into randomly chosen context and target sets to obtain good generalisation.
Let us consider two scenarios:
1. Inferring a distribution over functions, based on a single data set
2. Inferring a distribution over functions, when we have access to multiple data sets which we believe to be related in some way
The first scenario corresponds to a standard (probabilistic) supervised learning setup: Given a data set of $N$ samples, i.e. given $(x_i, y_i)$ for $i = 1, … ,N$, and assuming there is an underlying true $f$ which has generated the $y_i = f(x_i)$ values, our goal is to learn the posterior distribution over $f$ and use it to get predictive densities at test points $f(x^{\ast})$.
The second scenario can be seen from the meta-learning viewpoint. Given $D$ data sets $d = 1, …, D$, each consisting of $N_d$ pairs $(x_i^{(d)}, y_i^{(d)})$, if we assume that every data set $d = 1, …, D$ has its own underlying function $f_d$ which has generated the values $y_i = f_d(x_i)$, we might want to learn the posterior of every $f_d$ as well as generalise to a new data set $d^{\ast}$. The latter is especially useful when every data set has only a small number of observations. This information sharing is achieved by specifying that there exists a shared process which underlies all functions $f_d$. For example, in the context of GPs, one can assume that $f_d \sim \mathcal{GP}$ share kernel hyperparameters. Having learned the shared process, when given a new data set $d^{\ast}$, one can use the posterior over functions as a prior and carry out few-shot function regression.
The reason I wanted to highlight these two scenarios is the following. Usually, in the most standard setting, GP-regression is carried out under the first scenario. This tends to work well even when $N$ is small. However, the motivation behind NPs seems to be mainly coming from the meta-learning setup – in this setting the latent $z$ can be thought of as a mechanism to share information across different data sets. Nevertheless, having elements of a probabilistic model, NPs should be applicable in both scenarios. Below we will investigate how NPs behave when trained only on a single data set, as well as the second setup where we have access to a large number of function draws in order to train the NP.
### How are NPs implemented?
Here is a more detailed schema of the NP generative model:
Going through this generative mechanism step-by-step:
• First, the context points $(x_c, y_c)$ are mapped through a NN $h$ to obtain a latent representation $r_c$.
• Then, the vectors $r_c$ are aggregated (in practice: averaged) to obtain a single value $r$ (which has the same dimensionality as every $r_c$).
• This $r$ is used to parametrise the distribution of $z$, i.e. $p(z | x_{1:C}, y_{1:C}) = \mathcal{N}(\mu_z( r ), \sigma_z^2( r ))$
• Finally, to obtain a prediction at a target $x_t^{\ast}$, we sample $z$ and concatenate this with $x_t^{\ast}$, and map $(z, x_t^{\ast})$ through a NN $g$ to obtain a sample from the predictive distribution of $y_t^{\ast}$.
Inference for the NP is carried out in the variational inference framework. Specifically, we introduce two approximate distributions:
• $q(z | \text {context})$ to approximate the conditional prior $p(z | \text {context})$
• $q(z | \text {context}, \text {target})$ to approximate the respective $p(z | \text {context}, \text {target})$ where we have denoted $\text {context} := (x_{1:C}, y_{1:C})$ and $\text {target} := (x_{1:T}^{\ast}, y_{1:T}^{\ast})$.
The approximate posterior $q(z | \cdot)$ is chosen to have the specific form as illustrated in the inference model diagram below. That is, we use the same $h$ to map both the context set as well as the target set to obtain the aggregated $r$, which in turn is mapped to $\mu_z$ and $\sigma_z$. These parametrise the approximate posterior $q(z | \cdot) = \mathcal{N}(\mu_z, \sigma_z)$.
The variational lower bound
$$\text {ELBO} = \mathbb{E}_{q(z | \text {context}, \text {target})} \left[ \sum_{t=1}^T \log p(y_t^{\ast} | z, x_t^{\ast}) + \log \frac{q(z | \text {context})}{q(z | \text {context}, \text {target})} \right]$$
contains two terms. The first is the expected log-likelihood over the target set. This is evaluated by first sampling $z \sim q(z | \text {context}, \text {target})$, as indicated on the left part of the inference diagram, and then using these $z$ values for predictions on the target set, as on the right part of the diagram. The second term in ELBO has a regularising effect – it is the negative KL divergence between $q(z | \text {context}, \text {target})$ and $q(z | \text {context})$. Note that this differs slightly from the most commonly encountered variational inference setup with $\text{KL}(q || p)$, where $p$ would be the prior $p(z)$. This is because in our generative model, we have specified a conditional prior $p(z | \text {context})$ instead of directly specifying $p(z)$. As this conditional prior depends on $h$, we do not have access to its exact value and instead need to use an approximate $q(z | \text {context})$.
### Experiments
#### NP as a prior over functions
Let’s start by exploring the behaviour of NPs as a prior over functions, i.e. in the setting where we haven’t observed any data and haven’t yet trained the model. Having initialised the weights (here I initialised them independently from a standard normal), we can sample $z \sim \mathcal{N}(0, I)$ and generate from the prior predictive distribution over a grid of $x^{\ast}$ values to plot the functions.
As opposed to GPs which have interpretable kernel hyperparameters, the NP prior is much less explicit. There are various architectural choices involved (such as how many hidden layers to use, what activation functions to use etc) which all implicitly affect our prior distribution over the function space.
For example, when using sigmoid activations and varying the dimensionality of $z$ in $\{1, 2, 4, 8\}$, typical draws from the (randomly initialised) NP prior may look as follows:
But when deciding to use the ReLU activations instead, we have placed the prior probability mass over a different set of functions:
#### Training NP on a small data set
Suppose all we have is the following five data points:
The training procedure for NPs will involve separating the context set and target set. One option is to use a fixed size context set, another is to cover a wider range of scenarios by training using varying context set sizes (e.g. at every iteration we could randomly draw the number of context points from the set $\{1, 2, 3, 4\}$). Once we have trained the model on these random subsets, we use the trained model as our prior and now condition on all of our data (i.e. we take all these five points to be the context points) and plot draws from the posterior. The animation below illustrates how the predictive distribution of the NP changes as it is being trained:
So the NP seems to have successfully learned a distribution over mappings which go through all of our five points. Now let’s explore how well it generalises to other mappings, i.e. what happens if we use this trained NP for prediction on a different context set. Here is the posterior when conditioning on the red points instead:
Not very surprisingly, the flexible NP model which was trained only on subsets of the five blue points, doesn’t generalise to a different set of context points. To get a model which would generalise better, we could consider (pre)training the NP on a larger set of functions.
#### Training NPs on a small class of functions
So far, we have explored the training scenario using a single (fixed) data set. To have an NP which would generalise similarly to GPs, it seems that we should train it on a much larger class of functions. But before that, let’s explore how NPs behave in a simpler setting.
That is, let’s consider a toy scenario, where instead of a single function we observe a small class of functions. Specifically, let’s consider all functions of the form $a \cdot \sin(x)$ where $a \in [-1, 1]$.
It would be interesting to see:
1. Is the NP able to capture this class of functions
2. Will it generalise beyond this class of functions
Let’s use the following training procedure:
• Draw $a$ uniformly $a \sim U(-2, 2)$
• Draw $x_i \sim U(-3, 3)$
• Define $y_i := f(x_i)$, where $f(x) = a \sin(x)$
• Divide pairs $(x_i, y_i)$ randomly into context and target sets and perform an optimisation step
• Repeat
Here we used a two-dimensional $z$ so that we could visualise what the model has learned. Having trained the NP, we visualised the function draws corresponding to various $(z_1, z_2)$ values on a grid, as shown below:
It seems that the direction from left-to-right essentially encodes our parameter $a$. Here is another visualisation of the same effect, where we vary one of the latent dimensions (either $z_1$ or $z_2$):
Note that above we did not use any context set at prediction time, but simply pre-specified various $(z_1, z_2)$ values. Now let’s look into using this trained model for predicting.
Taking the context set to be the point $(0, 0)$, shown on the left, will result in quite a broad posterior which looks quite nice, covering functions which resemble $a \sin(x)$ for a certain range of values of $a$ (but note that not for all $a \in [-2, 2]$ it was trained on).
Adding a second context point $(1, \sin(1))$ will result in the posterior shown in the middle. The posterior has changed compared to the previous plot, e.g. functions with a negative values of $a$ are not included any more, but none of the functions goes through the given point. When increasing the number of context points which follow $f(x) = 1.0 \sin(x)$ then the NP posterior will become reasonably close to the true underlying function, as shown on the right.
Now let’s explore how well the trained NP will generalise beyond the class of functions it was trained on. Specifically, let’s explore how it will generalise to the following functions $2.5 \sin(x)$ and $| \sin(x) |$. The first requires some extrapolation from the training data. The second one has a similar shape to the functions in training set but unlike the rest its values are non-negative.
As seen from the plots, the NP has not been able to generalise beyond what it had seen during training. In both cases, the model behaviour is somewhat expected (e.g. on the left, $a=2$ corresponds to the best fit within the class of functions the model has seen). However, note that there is not much uncertainty in the NP predictive distributions. Of course over-confident predictions are not specific to NPs, however their black-box nature may make them more difficult to diagnose, compared to more interpretable models.
#### Training NPs on functions drawn from GPs
Based on experiments so far, it seems that NPs are not out-of-the-box replacements for GPs, as the latter have more desirable properties regarding posterior uncertainty. In order to achieve a similar behaviour with an NP, we could train it using a large number of draws from a GP prior. We could do it as follows:
• Draw $f \sim \mathcal{GP}(0, k_{\theta}(\cdot))$
• Draw $x_i \sim U(-3, 3)$
• Define $y_i := f(x_i)$
• Divide pairs $(x_i, y_i)$ randomly into context and target sets and perform an optimisation step
• Repeat
The above procedure can be carried out with fixed kernel hyperparameters $\theta$ or a mixture of different values. The RBF (or squared exponential) kernel has two parameters: one controls the variance (essentially the range of function values) and the other “wigglyness”. The latter is called the lengthscale parameter and its effect is illustrated here, by drawing functions from the GP prior with lengthscale values in $\{1, 2, 3\}$:
To cover a variety of functions in the NP training, we could specify a prior $p(\theta)$ where to draw samples from. In this toy experiment, I varied lengthscale, as above, uniformly in $\{1, 2, 3\}$. As previously, choosing the latent $z$-space to be 2D, we can visualise what the NP has learned:
This looks pretty cool! The NP has learned the two-dimensional $z$ space where we can smoothly interpolate between different functions.
Now let’s explore the predictions we get using this NP, and let’s see how its behaviour compares to a Gaussian Process posterior. Using an increasing number of context points $\{3, 5, 11\}$, let’s consider two functions:
First, using a relatively smooth function $f(x) = \sin(0.5x)$, the predictions look as follows:
Second, let’s consider $f(x) = \sin(1.5x)$:
In the first case, the NP predictions follow the observations quite closely, whereas in the second case, with three observations it looks good, but when given more points it hasn’t been able to capture the pattern. This effect is quite likely due to our architectural choices to use quite small NNs and a low-dimensional $z$. In order to improve the model behaviour so that it would resemble GPs more closely, we could consider applying the following changes:
• Using only a 2D $z$-space might be quite restrictive in what we are able to learn, we could consider using a higher-dimensional $z$. And similarly for $r$.
• We could consider using a larger number of hidden units in NNs $h$ and $g$, and consider making them deep.
• Observing a larger number of function draws as well as a larger variety of functions (i.e. more variability in GP kernel hyperparameters) during the training phase could lead to better generalisation.
Such changes can indeed lead to more desirable results. For example, having increased $\text {dim}( r )$ to 32 and $\text {dim}(z)$ to 4 together with a larger number of hidden units in $h$ and $g$, we observe much nicer behaviour:
### Conclusions
Even though Neural Processes combine elements from both NNs and GP-like Bayesian models to capture distributions over functions, on this spectrum NPs lie closer to neural models. By making careful choices regarding the neural architectures as well as the training procedure for NPs, it is possible to achieve desirable model behaviour, e.g. GP-like predictive uncertainties. However, these effects are mostly implicit which make NPs more challenging to interpret as a prior.
### Implementation
My implementation using TensorFlow in R (together with code for the experiments above) is available in github.com/kasparmartens/NeuralProcesses.
### Acknowledgements
I would like to thank Hyunjik Kim for clarifying my understanding of the NP papers, Jin Xu for sharing his results on NPs (which stimulated me to think about more complex NNs), Tanel Pärnamaa and Leon Law for their comments, and Chris Yau for being an awesome supervisor. | 4,222 | 18,065 | {"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} | 2.765625 | 3 | CC-MAIN-2024-10 | latest | en | 0.962396 |
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bijna 2 jaar ago | 1,526 | 5,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.404872 |
https://www.geeksforgeeks.org/tcs-coding-practice-question-average-of-2-numbers/ | 1,701,722,669,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100534.18/warc/CC-MAIN-20231204182901-20231204212901-00130.warc.gz | 883,691,267 | 50,154 | # TCS Coding Practice Question | Average of 2 Numbers
Given two numbers, the task is to find the average of two numbers using Command Line Arguments. Examples:
```Input: n1 = 10, n2 = 20
Output: 15
Input: n1 = 100, n2 = 200
Output: 150```
Approach:
• Since the numbers are entered as Command line Arguments, there is no need for a dedicated input line
• Extract the input numbers from the command line argument
• This extracted numbers will be in String type.
• Convert these numbers into integer type and store it in variables, say num1 and num2
• Find the average of the numbers using the formula ((num1 + num2)/2)
• Print or return the average
Program:
## C
`// C program to compute the average of two numbers ` `// using command line arguments ` ` ` `#include ` `#include /* atoi */ ` ` ` `// Function to compute the average of two numbers ` `int` `average(``int` `a, ``int` `b) ` `{ ` ` ``return` `(a + b) / 2; ` `} ` ` ` `// Driver code ` `int` `main(``int` `argc, ``char``* argv[]) ` `{ ` ` ` ` ``int` `num1, num2; ` ` ` ` ``// Check if the length of args array is 1 ` ` ``if` `(argc == 1) ` ` ``printf``(``"No command line arguments found.\n"``); ` ` ` ` ``else` `{ ` ` ` ` ``// Get the command line argument and ` ` ``// Convert it from string type to integer type ` ` ``// using function "atoi( argument)" ` ` ``num1 = ``atoi``(argv[1]); ` ` ``num2 = ``atoi``(argv[2]); ` ` ` ` ``// Find the average and print it ` ` ``printf``(``"%d\n"``, average(num1, num2)); ` ` ``} ` ` ``return` `0; ` `} `
## Java
`// Java program to compute the average of two numbers ` `// using command line arguments ` ` ` `class` `GFG { ` ` ` ` ``// Function to compute the average of two numbers ` ` ``static` `int` `average(``int` `a, ``int` `b) ` ` ``{ ` ` ``return` `(a + b) / ``2``; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ` ` ``// Check if length of args array is ` ` ``// greater than 0 ` ` ``if` `(args.length > ``0``) { ` ` ` ` ``// Get the command line argument and ` ` ``// Convert it from string type to integer type ` ` ``int` `num1 = Integer.parseInt(args[``0``]); ` ` ``int` `num2 = Integer.parseInt(args[``1``]); ` ` ` ` ``// Find the average ` ` ``int` `res = average(num1, num2); ` ` ` ` ``// Print the average ` ` ``System.out.println(res); ` ` ``} ` ` ``else` ` ``System.out.println(``"No command line "` ` ``+ ``"arguments found."``); ` ` ``} ` `} `
## Python
`# python program to find average of two numbers ` `def` `average(n1,n2): ` ` ``return``(``int``((n1``+``n2)``/``2``)) ` ` ` `# taking user input ` `n1,n2``=``input``().split() ` `# Convert it from string type to integer type ` `n1``=``int``(n1) ` `n2``=``int``(n2) ` `#find the average ` `result``=``(average((n1),(n2))) ` `# print average ` `print``(result)`
Output:
• In C:
• In Java:
• in Python
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https://m.wikihow.com/Do-a-Tempo-Run | 1,579,699,924,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607118.51/warc/CC-MAIN-20200122131612-20200122160612-00552.warc.gz | 548,493,192 | 57,246 | # How to Do a Tempo Run
Co-authored by wikiHow Staff
Updated: December 16, 2019
Tempo runs are great exercises for runners to build up endurance. Before starting a tempo run, calculate your tempo pace using an online calculator or doing the math yourself with your PR times as your base. Warm up to get your muscles stretched before running. There are several different tempo runs to try, with the basic one being to run for 20 minutes at your tempo pace. If you find yourself breathing very heavily or needing to rest, it's time to slow down. Cool down once you finish your run to help prevent injuries.
### Method 1 of 3: Calculating Your Tempo Pace
1. 1
Use an online calculator to help you figure out your tempo pace. Calculating your perfect tempo pace can be tricky, so people have designed an easy way for you to figure yours out without having to personally do the math. Plug your information, such as a distance and the time it takes you to run that distance, into an online running calculator to discover your pace.[1]
• To find a running calculator, type "tempo pace calculator" into your online search engine.
2. 2
Calculate your tempo pace using your 10k pace. Your average 10k pace is often the pace you can maintain comfortably for at least a full hour, which is perfect for figuring out your tempo pace. Take your average 10k time and divide it by 6.2 miles (the number of miles in a 10k) to get your average pace per mile. This is your own suggested tempo pace.[2]
• For example, if you run a 10k in 50 minutes, divide 50 by 6.2, meaning your pace is 8 minutes per mile.
• To find your average pace per kilometer, divide your 10k time by 10.
• If you don't know your 10k time, make an educated guess as to how long it would take you to run a 10k and divide this time by 6.2 to get your average pace per mile (which you can always adjust if needed).
3. 3
Keep your heart rate within 85-90% of your maximum heart rate. By staying aware of your heart rate during tempo runs, you'll help ensure you're going at the right pace. If you have a heart rate monitor in the form of a watch or wristband, this is a great tool to use to tell whether you’re going at a good pace. Otherwise, calculate your heart rate yourself using a regular watch.[3]
• Your maximum heart rate is the average maximum number of times your heart should beat during exercise per minute. To calculate your maximum heart rate, subtract your age from 220.
• To calculate your heart rate yourself, place your fingers on the inside of your wrist and count how many pulses (heart beats) you feel per minute.
### Method 2 of 3: Undertaking a Tempo Run
1. 1
Take the weather into account when planning your tempo run. Outdoor conditions like the temperature, wind, and elevation can impact your tempo pace. Be prepared to make small adjustments to your speed or the time of day that you run so that you're tempo run goes smoothly.[4]
• Try running early in the morning when the temperature is cooler and the sun isn't as strong during warmer weather.
2. 2
Warm up for 15-20 minutes. Instead of jumping right into your tempo run, give your body a chance to get stretched and ready. Touch your toes, do some jumping jacks, or practice walking lunges before starting your run.[5]
• Even taking a 10-minute brisk walk can help you warm up.
3. 3
Practice good running form to prevent injuries and strengthen your pace. While you're running, try to look roughly 10 ft (3.0 m) in front of you to help keep your neck in a neutral position and your feet from tripping. Relax your arms and lean forward just slightly to keep you balanced and in a straight line as you run.[6]
• If you're feeling pain during or after your run, examine your form and consider consulting a doctor to see what the problem is.
• Good running form can vary from person to person, but the main goal is for it to not be painful.
4. 4
Run at a slower pace whenever you’re in doubt. The worst thing you can do in a tempo run is run faster than you should. A good tempo run should feel challenging but doable. If you’re pushing yourself to go too fast, it’s not going to help your tempo training.[7]
• If you’re out of breath when you’re running, you’re going too fast and need to slow down.
• If you’re doing a tempo run and it feels easy, avoid speeding up. Calculate your tempo again the next time you do a race to see if you can train faster.
5. 5
Cool down for 10-15 minutes before stopping. This allows your heartbeat to return to a normal pace slowly. After you’ve finished your run, take a few walking laps while stretching to let your body and muscles cool down.[8]
### Method 3 of 3: Choosing Your Workout
1. 1
Run at your tempo pace for 20 minutes if you’re training for a 5k or 10k. When you start your run, try to stay at your tempo pace for at least 20 minutes. Look out for physical signs that you might be going too fast such as difficulty breathing or needing to rest.[9]
• If you’re running at your tempo pace for 20 minutes with no problem, try running for a total of 30 minutes on your next run.
2. 2
Train for a marathon or half marathon by running for roughly 40 minutes. If you’re training for a half marathon or full marathon, you’re likely a more experienced runner (or soon will be!). Try to run at your tempo pace for 35-40 minutes, making sure not to go too fast so you’re able to run without stopping for the full amount of time.[10]
• Some runners run as long as 50-70 minutes at their tempo speed when training for a half or full marathon.
3. 3
Run for 10 minutes with 3-minute recoveries to try out a different exercise. Instead of running for 20-40 minutes at your tempo pace and then finishing, run for 10 minutes at your tempo pace and jog for 3 minutes. Repeat this 3 or 4 times for your workout.[11]
• Giving your body a break every 10 minutes with an easy jog helps take your mind off of the long running time and keeps your pace on track.
4. 4
Do a new tempo workout by giving yourself 30-60 second recoveries while running. Run at your tempo pace for 3 minutes, and then jog for 30-60 seconds, depending on your skill level. Repeat this 6-10 times for a different approach to tempo training.[12]
• This tempo workout helps you run for a longer amount of time if desired due to the frequent recovery periods.
## Community Q&A
200 characters left
## Tips
• If you notice that you're running too fast, start running 10-20 seconds slower than your tempo pace to let your heart rate go down.
Thanks!
• Different running coaches and resources will have different methods for calculating your perfect tempo pace. The most important thing is to listen to your body's cues for how fast or slow you should be going.
Thanks!
## Warnings
• Avoid speeding up your pace until your racing PR shows that you're able to train faster.
Thanks!
Co-Authored By:
wikiHow Staff Editor
This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 12 references. | 1,648 | 7,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-05 | latest | en | 0.937547 |
https://mathoverflow.net/questions/248897/a-question-about-julia-set-for-quadratic-family | 1,558,547,520,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256887.36/warc/CC-MAIN-20190522163302-20190522185302-00368.warc.gz | 566,686,889 | 31,823 | Let $P_{c}(z)=z^2+c$. It seems from the software that the map between the parameter $c$ and the Julia set $J(P_c)$ is an injective map. Is there some reference about it? Any comments and reference will be appreciated.
I expect that you will find the answer (and a lot more) in the following paper, since it's easy to figure out which quadratic polynomials (if any) commute with one another:
Commuting polynomials and polynomials with same Julia set, Pau Atela, Jun Hu, Int. J. Bifurcation Chaos 06, 2427 (1996). DOI: http://dx.doi.org/10.1142/S0218127496001570
Abstract: It has been known since Julia that polynomials commuting under composition have the same Julia set. More recently in the works of Baker and Eremenko, Fern\'andez, and Beardon, results were given on the converse question: When do two polynomials have the same Julia set? We give a complete answer to this question and show the exact relation between the two problems of polynomials with the same Julia set and commuting pairs.
The paper is also available on the ArXiv
https://arxiv.org/abs/math/9504210
• Thank you, professor. Very nice paper. This is first time I know Fernandez's result, which claims that if f and g has the same degree and the same leading coefficient, then $J(f)=J(g)$ implies $f=g$. – yaoxiao Sep 2 '16 at 15:05
Joe's answer and the cited paper are, indeed, quite nice. I think the basic idea is fairly self-contained and probably easier than the reference.
Two complex polynomials $f$ and $g$ of the same degree have geometrically similar Julia sets precisely when there is an affine function $\varphi(z)=mz+b$ with $m\neq0$ such that $$\varphi(f(z)) = g(\varphi(z)).$$ The function $\varphi$ is called a conjugacy and it's easy to see why this should work. By induction, $$\varphi(f^n(z)) = g^n(\varphi(z)).$$ As a result, $\varphi$ maps an orbit of $f$ containing the point $z_0$ onto an orbit of $g$ containing the point $\varphi(z_0)$ so we expect the dynamics of $f$ and $g$ to be closely related; the nicer the function $\varphi$. The nicer the relationship and our $\varphi$ is about as nice as it could be.
Now the question is, when is there an affine conjugacy between $$f(z)=z^2+c_1\: \text{ and } \: g(z)=z^2+c_2?$$ If you simply expand the conjugation equation or $$m(z^2+c_1)+b = (mz+b)^2+c_2$$ and collect terms you find the system \begin{align} m^2-m &= 0 \\ 2bm &= 0 \\ mc_1-c_2-b^2+b &= 0. \end{align} Recalling that $m$ can't be zero for $\varphi$ to be an affine function, we see that the first equation implies that $m=1$ so that the second equation implies $b=0$. Plugging those into the last equation we see that $c_1=c_2$.
• But isn't the hard part proving that if $J(f)=J(g)$, then there is an affine conjugacy? Once you know that, then rest, as you've indicated, is straightforward. – Joe Silverman Sep 2 '16 at 19:28
For $z^2+c$ it is really easy: the answer is yes. In general, if Julia sets of polynomials coincide, these polynomials are equal unless their common Julia set has a rotation symmetry, or polynomials are permutable. (Baker, Eremenko, Ann. Acad. Sci. Fenn., 12 (1987) 229-236.)
For $z^2+c$ we never have rotational symmetry and never have permutable polynomials. The result cited above was reproved and generalized by many other people. For rational functions, the question in full generality is still open.
• Thanks for your comment and reference, professor. Would you think it may be too optimistic to guess that $\lambda\rightarrow J(\lambda*\exp)$ be injective map for the paramters $\lambda$ for which $J(\lambda*exp)\neq \mathbb{C}$. – yaoxiao Sep 7 '16 at 14:33 | 998 | 3,614 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-22 | longest | en | 0.906808 |
http://krysstal.com/maths.html | 1,485,197,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560282937.55/warc/CC-MAIN-20170116095122-00288-ip-10-171-10-70.ec2.internal.warc.gz | 157,713,166 | 8,795 | A great way to test your math knowledge online is though by playing games of chance such as blackjack and baccarat at. There are several great options for players like these found here on OnlineCasinoGambling.co.za a leading South African guide.
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# Essays on Mathematical Topics
Real, rational, irrational, imaginary. An explanation of numbers of different kinds, a little about infinite series and an surprising relationship. Types of integers: odd, even, squares; perfect, amicable and prime numbers.
How letters are used to solve equations. Notation and rules. Simple, simultaneous and quadratic equations covered for the beginner. Biquadratics, simple cubics and complex roots.
A number triangle known by the Chinese and named after a French mathematician. This triangle is used for algebra, probabilities, and calculations. The Binomial Theorem and series.
Right-angled triangles. Introducing the trigonometric functions: Sine, Cosine and Tangent. Trigonometrical calculations, formulas and series. Cosine and Sine rules. Secants, Cosecants and Cotangents.
An introductory look at the strange world of logarithms, how they are used for calculations, how they are evaluated with series, and using logarithms to solve algebraic equations. Definitions of index and base.
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The trigonometric identities and graphs of Sine and Cosine. Solving equations containing Sines, Cosines and Tangents.
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Spherical Trigonometry is the trigonometry of triangles drawn on a sphere. It is used for many areas in geography and astronomy including navigation, mapping and sundials.
More algebra. Determinants and their use in solving simultaneous equations. Partial fractions. An introduction to Limits, including L'Hôpital's Rule.
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A brief History of Mathematics
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# KryssTal Glossary
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Encyclopedia.Com
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Worksheet Library
School resource site for various ages.
HappyChild
is an excellent educational resource for children, full of interesting information. | 861 | 4,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-04 | longest | en | 0.928194 |
https://mersenneforum.org/showthread.php?s=af6d2df48673b0f97d8bc7dd0f24203c&t=20417 | 1,606,693,791,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00563.warc.gz | 387,424,812 | 10,114 | mersenneforum.org x.265 half the size, double the computation; so if you double again? 1/4th?
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2015-08-14, 05:21 #1 jasong "Jason Goatcher" Mar 2005 1101101100012 Posts x.265 half the size, double the computation; so if you double again? 1/4th? In case my title wasn't clear, I'm wondering about the connection between doubling the computation time for x.265 and the reduction of the size of the video. Basically, the wiki says that if you give twice the computation time you would give to an x.264 video to an x.265 video, the result will be about half the size of the resulting x.264 video. So, then, my question is... What happens if you keep doubling the processing time for the x.265 video? Obviously, there are diminishing returns, but does it continue to be about half the size for every doubling of processing time? Are the numbers given simply a coincidence? (the coincidence being half the time is 1/2, double the processing time is 2, and they're reciprocals)
2015-08-14, 06:38 #2 VBCurtis "Curtis" Feb 2005 Riverside, CA 448910 Posts If the answer to your question were yes, what would happen if you gave it enormous processing time? A 1-byte video? It's nice that trying twice as hard results in half the file size, but that specific ratio is coincidence.
2015-08-14, 07:01 #3
jasong
"Jason Goatcher"
Mar 2005
DB116 Posts
Quote:
Originally Posted by VBCurtis If the answer to your question were yes, what would happen if you gave it enormous processing time? A 1-byte video? It's nice that trying twice as hard results in half the file size, but that specific ratio is coincidence.
I know taking it to the extreme creates irrational conclusions, but I'm wondering if it's approximately true in the short term.
2015-08-14, 07:15 #4
retina
Undefined
"The unspeakable one"
Jun 2006
My evil lair
10110111101112 Posts
Quote:
Originally Posted by jasong I know taking it to the extreme creates irrational conclusions, but I'm wondering if it's approximately true in the short term.
Perhaps you could so some empirical tests and plot the results on a graph. If you are not confident in analysing the results then post the graph here if you like and see how others would analyse the curve.
2015-08-14, 14:17 #5 Mark Rose "/X\(β-β)/X\" Jan 2013 2×31×47 Posts Personally I'm more excited about the Daala codec. https://xiph.org/daala/ https://www.youtube.com/watch?v=Dmho4gcRvQ4
2015-08-15, 11:23 #6 VictordeHolland "Victor de Hollander" Aug 2011 the Netherlands 22268 Posts The same 1080p film encoded with x264 and x265 (half the size) I find the x264 has more visual sharpnes/details. With the same bitrate, x265 looks better (obviously). So I would say its closer to 0.7-0.8 the size of x264 for the same visual sharpness on 1080p content. With 4k, UHD, 2160p (or however you want to call) it might be closer to 0.5 as there are more pixels to encode and the algorithms of x265 work more efficient. I still mainly use x264, because my smartphone/TV/tablet have hardware accelleration for it and storage is very cheap.
2015-08-17, 08:58 #7
jasong
"Jason Goatcher"
Mar 2005
5·701 Posts
Quote:
Originally Posted by VictordeHolland The same 1080p film encoded with x264 and x265 (half the size) I find the x264 has more visual sharpnes/details. With the same bitrate, x265 looks better (obviously). So I would say its closer to 0.7-0.8 the size of x264 for the same visual sharpness on 1080p content. With 4k, UHD, 2160p (or however you want to call) it might be closer to 0.5 as there are more pixels to encode and the algorithms of x265 work more efficient. I still mainly use x264, because my smartphone/TV/tablet have hardware accelleration for it and storage is very cheap.
I read that x.265 was created mainly because of concerns about mobile bandwidth in the near future.
Personally, I'd love to see Wifi routers installed at the top of light poles and along highways, with maybe some insurance offered by the Federal Government so investment losses aren't huge. Although that type of thing might already exist, I'm no expert on investment losses as they relate to American taxation.
I know a big problem with my idea is crime, so I'm thinking cheap cameras on or near the poles would be a good idea as well. Any motion not on the road itself would trigger taking a picture and then sending it back to the main server, and there'd be no requirement to view any pictures unless there was peculiar internet behavior at that point.
2015-08-17, 10:56 #8
xilman
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
32×11×103 Posts
Quote:
Originally Posted by jasong I read that x.265 was created mainly because of concerns about mobile bandwidth in the near future. Personally, I'd love to see Wifi routers installed at the top of light poles and along highways, with maybe some insurance offered by the Federal Government so investment losses aren't huge. Although that type of thing might already exist, I'm no expert on investment losses as they relate to American taxation. I know a big problem with my idea is crime, so I'm thinking cheap cameras on or near the poles would be a good idea as well. Any motion not on the road itself would trigger taking a picture and then sending it back to the main server, and there'd be no requirement to view any pictures unless there was peculiar internet behavior at that point.
That approach might work better in a society where people aren't likely to take potshots at street fittings.
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Sun Nov 29 23:49:51 UTC 2020 up 80 days, 21 hrs, 3 users, load averages: 0.99, 1.27, 1.34 | 1,579 | 5,956 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.936316 |
http://www.reddit.com/r/AskReddit/comments/dc8qi/hey_reddit_can_you_help_me_with_this_math/ | 1,386,744,396,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164032288/warc/CC-MAIN-20131204133352-00047-ip-10-33-133-15.ec2.internal.warc.gz | 501,480,985 | 15,266 | [–] 1 point2 points ago
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It's 30. cm2. The cm3 in the first term cancels out with the cm in the denominator, then it turns into cm2 minus cm2. Here's the math on google http://www.google.com/search?client=safari&rls=en&q=(151.0/4.6)-(12.6801/4.3)&ie=UTF-8&oe=UTF-8 .
[–] 0 points1 point ago
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Excellent!
[–] 0 points1 point ago
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I got 29.9. Anyway, CellarDorre needs to be able to understand this: http://www.chem.sc.edu/faculty/morgan/resources/sigfigs/index.html
[–] 1 point2 points ago
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Don't you need need two sigfigs, since the answer must have the same number of sigfigs as the number with the lowest number of digits?
[–] 0 points1 point ago
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Haha, oops. I hate doing math in Google!
You're right. I just wanted to point out that in order to solve this problem, you need to know to combine units and use sigfigs.
[–] 0 points1 point ago
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I understand how to use significant figures (or digits, if you will), it was mostly the cm business that I was unsure about as today we were learning about converting units and stuff and at one point we had to square a set of numbers and I wasn't sure if the same idea applied to this. I'm glad to see it didn't!
Thank you for the link though! I'm sure it will come in handy.
[–] 0 points1 point ago
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As far as units goes, it's always helpful to think of the concept the unit represents. A square with 2 cm by 2 cm has an area of 4 cm 2, aka it's made up of 4 squares of 1 cm by 1 cm. Sounds basic, but it helps to know if you're dividing a volume by an area, or an area into a number of parts, like in this case. Another example is speed aka distance/time. Always write the units for distance on top, and the units for time on the bottom -- do it like they do in the book, and always cross things out carefully like they show
[–] 0 points1 point ago
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Thanks! I'll keep all of that in mind!
[–] 0 points1 point ago
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x3/x = x2... Same goes for cm? The question is numerically just as difficult as if there were no units at all.
[–] 1 point2 points ago
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Or that, since it's way easier way to explain and solve this way. It must be a long time since I dealt with a math problem...
[–] 0 points1 point ago
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Good! That's what I was hoping!
[–] -1 points0 points ago
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You need a common denominator, so the simplest way to do this would be multiply (151.0cm3/4.60cm) by (4.3/4.3) and (12.6801cm2/4.3) by (4.60/4.60). I would then also reduce the units of your first fraction to cm2/#, to divide it more easily later on. Then subtract the numerator and divide that by the denominator. Hope that helps. | 891 | 3,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2013-48 | latest | en | 0.974233 |
https://www.csdn.net/tags/MtTakg2sNTI4NjMtYmxvZwO0O0OO0O0O.html | 1,628,189,863,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046156141.29/warc/CC-MAIN-20210805161906-20210805191906-00638.warc.gz | 722,363,151 | 46,962 | • 随机森林随机回归预测Why do we try to predict happiness? Being able to predict happiness means that we are able to manipulate or try to improve certain components in order to increase our own happiness...
随机森林随机回归预测
Why do we try to predict happiness? Being able to predict happiness means that we are able to manipulate or try to improve certain components in order to increase our own happiness, and possibly national happiness for governments. I found Random Forest (RF) to be the simplest and most efficient package, so let’s get started!
为什么我们要尝试预测幸福？ 能够预测幸福意味着我们能够操纵或尝试改善某些组成部分，以增加我们自己的幸福，甚至可能增加政府的国民幸福。 我发现随机森林(RF)是最简单，最有效的软件包，所以让我们开始吧！
内容： (Contents:)
1. The Data
数据
2. Random Forest Model
随机森林模型
3. Data Cleaning
数据清理
4. Training and Testing
培训与测试
5. Feature Importances
功能重要性
6. Modifying number of variables
修改变量数
7. Evaluating the Model
评估模型
数据：(The Data:)
The data obtained from the #WorldValuesSurvey contains >290 questions & consist of ~69k responses after removing missing data for happiness levels. It is a cross-national survey across the years, and the questionnaire can be found on the website. In particular, we will be looking at the 2017–2020 data set. The size of the data set makes it optimal for machine learning.
#WorldValuesSurvey获得的数据包含超过290个问题，并在删除幸福水平缺失的数据后包含约69k响应。 这是多年来的跨国调查，其问卷可以在网站上找到。 特别是，我们将研究2017-2020年的数据集。 数据集的大小使其最适合机器学习。
随机森林模型：(Random Forest Model:)
To start with, we will be using the RF classifier* since we would like the machine to predict the level of happiness in groups (Very happy, Quite happy, Not very happy, Not at all happy).*side-note, a RF regressor is used when looking for a number that can take a range of values e.g any value between 0 and 1.
首先，我们将使用RF分类器*，因为我们希望机器预测小组中的幸福程度(非常高兴，非常高兴，不太高兴，根本不高兴)。 *旁注，当寻找一个可以取值范围(例如0到1之间的任何值)的数字时，将使用RF回归器。
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
from sklearn.ensemble import RandomForestClassifier
from sklearn.model_selection import train_test_split
from sklearn import metrics
数据清理：选择数据(Data cleaning: Selecting the data)
Let’s start by getting the columns of only the questions and removing negative values* in the responses to Q46 that asks about happiness levels.
让我们从仅获得问题的列开始，并在对Q46的询问中询问幸福水平的答案中删除负值*。
var="Q46"
df=df[df.columns[32:349]]
df=df[df[var]>0]
*Negative values are either respondents saying they don’t know, have no answer, were not asked or the response was missing. These values would make it harder for the machine to classify them, since it increases the number of categories and are not what we are looking for.
*负值是指受访者表示他们不知道，没有答案，没有被要求或答案丢失。 这些值会使机器更难以对其进行分类，因为这会增加类别的数量，而这并不是我们要寻找的。
The data set remaining is shown below:
剩余的数据集如下所示：
进一步的数据清理：(Further data cleaning:)
The next concern is that we would have to deal with missing values in other columns. There are 3 options to consider:
下一个需要考虑的问题是，我们将不得不处理其他列中缺失的值。 有3个选项可供考虑：
1. Replace the missing values with 0
将缺失的值替换为0
2. Replace the missing values with the mean
用均值替换缺失值
3. Drop the rows with missing values (data set becomes empty).
删除缺少值的行(数据集为空)。
Since the third option is not viable, we will have to check which option, 1 or 2, would give the highest accuracy. In this case, I found that replacing with 0 makes it more accurate.
由于第三个选项不可行，我们将必须检查哪个选项1或2将提供最高的准确性。 在这种情况下，我发现用0代替会使它更准确。
df.fillna(0, inplace=True)
准备火车标签：(Prepare train labels:)
Now we set the ‘label’ for the machine to recognize the feature that I want it to predict and split the data into train and test sets.
现在，我们为机器设置“标签”，以识别我希望它预测和将数据分为训练和测试集的功能。
train_labels = pd.DataFrame(df[var])
train_labels = np.array(df[var])
train_features= df.drop(var, axis = 1)
feature_list = list(train_features.columns)
train_features = np.array(train_features)
train_features, test_features, train_labels, test_labels = train_test_split(train_features, train_labels, test_size = 0.25, random_state = 42)
训练和测试模型：(Train and Test the Model:)
The process of training and testing is simple. To improve the predictive power and/or model speed, we can simply modify the parameters within the RF classifier.
培训和测试的过程很简单。 为了提高预测能力和/或模型速度，我们可以简单地在RF分类器中修改参数。
精度提高：(Increasing accuracy:)
n_estimators — number of trees the algorithm builds before majority voting
n_estimators-算法在多数表决之前构建的树数
max_features — maximum number of features random forest considers to split a node
max_features —随机森林考虑拆分节点的最大特征数
min_sample_leaf — the minimum number of leafs required to split an internal node.
min_sample_leaf —拆分内部节点所需的最小叶子数。
提高速度：(Increasing speed:)
n_jobs — number of processors it is allowed to use. If = 1, only use one processor. If =-1, no limit
n_jobs-允许使用的处理器数量。 如果= 1，则仅使用一个处理器。 如果= -1，则没有限制
random_state — makes the model’s output replicable i.e always produce the same results given the same hyperparameters and training data
random_state —使模型的输出可复制，即在给定相同的超参数和训练数据的情况下始终产生相同的结果
oob_score: random forest cross-validation method
oob_score：随机森林交叉验证方法
rf=RandomForestClassifier(n_estimators = 1000, oob_score = True, n_jobs = -1,random_state =42,max_features = “auto”, min_samples_leaf = 12)
rf.fit(train_features, train_labels)
predictions = rf.predict(test_features)
print(metrics.accuracy_score(test_labels, predictions))
The model takes 1.3 minutes to train ~52k training rows and >290 columns, and 1 second to test. The accuracy was 63.70%. If we had chosen to fill the missing values with the mean, the accuracy would be 63.55%. But what’s important is finding out what influences the machine’s prediction, since those would be the variables that we want to look at. We certainly cannot expect everyone to answer 290+ questions, or try to work on all 290 aspects to improve happiness (that’s going to cost a lot). So we’ll be looking at the feature importances.
该模型需要1.3分钟来训练约52k训练行和> 290列，并且需要1秒进行测试。 准确度是63.70％ 。 如果我们选择用平均值填充缺失值，则准确度将为63.55％ 。 但是重要的是找出影响机器预测的因素，因为这就是我们要查看的变量。 当然，我们当然不能期望每个人都能回答290多个问题，或者尝试在290个方面进行工作以提高幸福感(这将花费很多)。 因此，我们将研究功能的重要性。
功能重要性：(Feature Importances:)
If you recall, feature_list contains the columns of all other variables except Q46. The goal is to understand which are the variables that influence the prediction.
回想一下，feature_list包含除Q46之外的所有其他变量的列。 目的是了解哪些因素会影响预测。
importances = list(rf.feature_importances_)
feature_importances = [(feature, round(importance, 2)) for feature, importance in zip(feature_list, importances)]
feature_importances = sorted(feature_importances, key = lambda x: x[1], reverse = True)
[print('Variable: {:20} Importance: {}'.format(*pair)) for pair in feature_importances]
x_values = list(range(len(importances)))# Make a bar chart
plt.bar(x_values, importances, orientation = 'vertical', color = 'r', edgecolor = 'k', linewidth = 1.2)
# Tick labels for x axis
plt.xticks(x_values, feature_list, rotation='vertical')
# Axis labels and title
plt.ylabel('Importance'); plt.xlabel('Variable'); plt.title('Variable Importances');
Feature importances sum to 1 and what we notice is that certain variables have a greater influence over the prediction compared to the others, and almost every variable has some form of influence, albeit extremely small because there are just too many variables. The next thing is to continue improving our model to allow us to better understand happiness.
特征重要性的总和为1，我们注意到，与其他变量相比，某些变量对预测的影响更大，并且几乎每个变量都有某种形式的影响，尽管由于变量太多而影响很小。 接下来的事情是继续改进我们的模型，以使我们更好地了解幸福。
修改变量数：(Modifying the number of variables:)
Let’s take the top 20 features, and set up a new model using just these 20 variables (+ var itself). We’ll repeat the data cleaning and same RF model. I got an accuracy of 64.47%. If we had chosen to replace missing values with the mean, the accuracy would be 64.41%. What is surprising here is that with smaller number of variables, the model becomes more accurate (from 63.70% to 64.47%). This is likely because the other variables were generating noise in the model and causing it to be less accurate.
让我们采用前20个功能，并仅使用这20个变量(+ var本身)来建立新模型。 我们将重复数据清理和相同的RF模型。 我的准确度是64.47％。 如果我们选择用均值代替缺失值，则准确度将为64.41％ 。 令人惊讶的是，变量数量越少，模型变得越准确(从63.70％64.47％ )。 这可能是因为其他变量在模型中产生了噪音，并导致其准确性降低。
让我们再次看一下功能重要性：(Let’s look at the Feature Importances again:)
This time, it is clearer to tell which variables were more important. You may refer to the questionnaire found on WVS for more detailed information. I will give a summary of the topics that the questions covered.
这次，更清楚地指出哪些变量更重要。 您可以参考WVS上的调查表以获取更多详细信息。 我将总结这些问题涉及的主题。
评估模型：(Evaluating the model:)
Let’s look at the graph of actual vs predicted values for the first 200 test values. For greater visibility of the whole test set, let’s also do a simple count for the difference in values of predicted and actual (predicted minus actual).
让我们看一下前200个测试值的实际值与预测值的关系图。 为了更好地了解整个测试集，我们还对预期值和实际值(预期值减去实际值)之间的差进行简单计数。
The model appears to be slightly more negative than positive in predicting the happiness levels, but would still be considered otherwise balanced!
在预测幸福水平时，该模型似乎比肯定模型更具负面性，但在其他方面仍然可以认为是平衡的！
见解：(Insights:)
What I have done is to examine the key questions out of >290 in the WVS that is more relevant to happiness levels. This would mean that we can try to focus specifically on these aspects when examining happiness.
我要做的是研究WVS中超过290个与幸福感水平更相关的关键问题。 这意味着我们在检查幸福时可以尝试着重于这些方面。
Looking at the questionnaire, we would also notice that Q261 and Q262 are the same thing (age and year born), so we could remove 1 of them to include another feature. For Q266,267,268 (country of birth of the respondent and parents) they appear to be repeats, but are not exactly the same thing since immigration/cross-cultural marriage may occur. Nonetheless, we could consider removing 2 of them since the occurrence is minimal.
通过问卷调查，我们还会注意到Q261和Q262是相同的东西(年龄和出生年份)，因此我们可以删除其中的一个以包括另一个功能。 对于Q266,267,268(受访者和父母的出生国家)，它们似乎是重复的，但由于可能发生移民/跨文化婚姻，因此并非完全相同。 尽管如此，由于发生的可能性很小，我们可以考虑删除其中的2个。
常规主题是：(The general topics are:)
Individual level:Life satisfaction, health, finances, freedom, age, safety, religion, marriage, and family.National level:country, perception of corruption, democracy/political influence, national pride
个人层面：生活满意度，健康，财务，自由，年龄，安全，宗教，婚姻和家庭。 国家层面：国家，对腐败的看法，民主/政治影响，民族自豪感
In particular, health, finances and age were the top features that were deemed as important by the machine. In this sense, the individual level factors has a greater influence on one’s happiness level compared to the national level factors.
特别是，健康，财务和年龄是机器认为重要的重要功能。 从这个意义上说，个人水平因素比国家水平因素对一个人的幸福水平影响更大。
However, I noticed that the WVS did not have data on sleep hours, which was a key element that was observed in my earlier post. Nonetheless, it is still very much useful as we can consider those aspects for further analysis! I’ll be back with more insights into the correlation between those aspects and happiness, to determine how we can improve our happiness levels. Until then, remember to stay happy!
但是，我注意到WVS没有睡眠时间的数据，这是我早先文章中观察到的关键因素。 但是，它仍然非常有用，因为我们可以考虑对这些方面进行进一步分析！ 我将在这些方面与幸福之间的相关性方面提供更多见解，以确定我们如何提高幸福水平。 在此之前，请记住保持快乐！
随机森林随机回归预测
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• 随机森林气温预测数据+代码 随机森林回归模型 万事俱备，我们可以来建立随机森林模型啦，首先导入工具包，先建立1000个树试试吧，其他参数先用默认值，之后我们会再深入到调参任务中： # 导入算法 from sklearn....
随机森林气温预测数据+代码
随机森林回归模型
万事俱备，我们可以来建立随机森林模型啦，首先导入工具包，先建立1000个树试试吧，其他参数先用默认值，之后我们会再深入到调参任务中：
# 导入算法
from sklearn.ensemble import RandomForestRegressor
# 建模
rf = RandomForestRegressor(n_estimators= 1000, random_state=42)
# 训练
rf.fit(train_features, train_labels)
由于数据样本量还是非常小的，所以很快就可以得到结果了，这里我们先用MAPE指标来进行评估，也就是平均绝对百分误差，其实对于回归任务，评估方法还是比较多，给大家列出来几种，很简单就可以实现出来，也可以选择其他指标来进行评估：
测试
# 预测结果
predictions = rf.predict(test_features)
# 计算误差
errors = abs(predictions - test_labels)
# mean absolute percentage error (MAPE)
mape = 100 * (errors / test_labels)
print ('MAPE:',np.mean(mape))
MAPE: 6.011244187972058
可视化展示树
# 导入所需工具包
from sklearn.tree import export_graphviz
import pydot #pip install pydot
# 拿到其中的一棵树
tree = rf.estimators_[5]
# 导出成dot文件
export_graphviz(tree, out_file = 'tree.dot', feature_names = feature_list, rounded = True, precision = 1)
# 绘图
(graph, ) = pydot.graph_from_dot_file('tree.dot')
# 展示
graph.write_png("somefile.png")
print('The depth of this tree is:', tree.tree_.max_depth)
The depth of this tree is: 15
执行完上述代码，会在指定的目录下（如果只指定其名字，会在代码所在路径下）生成一个tree.png文件，这就是绘制好的一棵树的模型，如图 所示。树模型看起来有点太大，观察起来不太方便，可以使用参数限制决策树的规模， 预剪枝方案在这里可以派上用场。
预剪枝
# 限制一下树模型 n_estimators棵树采用的数据集都是独立自助采样的，最大深度和随机数种子
rf_small = RandomForestRegressor(n_estimators=10, max_depth = 3, random_state=42)
rf_small.fit(train_features, train_labels)
# 提取一颗树
tree_small = rf_small.estimators_[8]
# 保存
export_graphviz(tree_small, out_file = 'small_tree.dot', feature_names = feature_list, rounded = True, precision = 1)
(graph, ) = pydot.graph_from_dot_file('small_tree.dot')
graph.write_png('small_tree.png')
对生成的树模型中各项指标的含义进行了标识，看起来还是比较好理解，其中非叶子节点中包括 4项指标∶ 所选特征与切分点、评估结果、此节点样本数量、节点预测结果（回归中就是平均）。
特征重要性
讲解随机森林算法的时候，曾提到使用集成算法很容易得到其特征重要性，在sklearn工具包中也有现成的函数，调用起来非常容易∶
# 得到特征重要性
importances = list(rf.feature_importances_)
# 转换格式
feature_importances = [(feature, round(importance, 2)) for feature, importance in zip(feature_list, importances)]
# 排序
feature_importances = sorted(feature_importances, key = lambda x: x[1], reverse = True)
# 对应进行打印
[print('Variable: {:20} Importance: {}'.format(*pair)) for pair in feature_importances]
上述输出结果分别打印了当前特征及其所对应的特征重要性，绘制成图表分析起来更容易∶
上述代码可以可以明显发现，temp_1和average这两个特征的重要性占据总体的绝大部分，其他特征的重要性看起来微乎其微。
# 转换成list格式
x_values = list(range(len(importances)))
# 绘图
plt.bar(x_values, importances, orientation = 'vertical')
# x轴名字
plt.xticks(x_values, feature_list, rotation='vertical')
# 图名
plt.ylabel('Importance'); plt.xlabel('Variable'); plt.title('Variable Importances');
那么，只用最厉害的特征来建模，其效果会不会更好呢?其实并不能保证效果一定更好，但是速度肯定更快，先来看一下结果∶
# 选择最重要的那两个特征来试一试
rf_most_important = RandomForestRegressor(n_estimators= 1000, random_state=42)
# 拿到这俩特征
important_indices = [feature_list.index('temp_1'), feature_list.index('average')]
train_important = train_features[:, important_indices]
test_important = test_features[:, important_indices]
# 重新训练模型
rf_most_important.fit(train_important, train_labels)
# 预测结果
predictions = rf_most_important.predict(test_important)
errors = abs(predictions - test_labels)
# 评估结果
mape = np.mean(100 * (errors / test_labels))
print('mape:', mape)
6.229055723613811
从损失值上观察，并没有下降，反而上升了，说明其他特征还是有价值的，不能只凭特征重要性就否定部分特征数据，一切还要通过实验进行判断。
但是，当考虑时间效率的时候，就要好好斟酌一下是否应该剔除掉那些用处不大的特征以加快构建模型的速度。到目前为止，已经得到基本的随机森林模型，并可以进行预测，下面来看看模型的预测值与真实值之间的差异∶
# 日期数据
months = features[:, feature_list.index('month')]
days = features[:, feature_list.index('day')]
years = features[:, feature_list.index('year')]
# 转换日期格式
dates = [str(int(year)) + '-' + str(int(month)) + '-' + str(int(day)) for year, month, day in zip(years, months, days)]
dates = [datetime.datetime.strptime(date, '%Y-%m-%d') for date in dates]
# 创建一个表格来存日期和其对应的标签数值
true_data = pd.DataFrame(data = {'date': dates, 'actual': labels})
# 同理，再创建一个来存日期和其对应的模型预测值
months = test_features[:, feature_list.index('month')]
days = test_features[:, feature_list.index('day')]
years = test_features[:, feature_list.index('year')]
test_dates = [str(int(year)) + '-' + str(int(month)) + '-' + str(int(day)) for year, month, day in zip(years, months, days)]
test_dates = [datetime.datetime.strptime(date, '%Y-%m-%d') for date in test_dates]
predictions_data = pd.DataFrame(data = {'date': test_dates, 'prediction': predictions})
# 真实值
plt.plot(true_data['date'], true_data['actual'], 'b-', label = 'actual')
# 预测值
plt.plot(predictions_data['date'], predictions_data['prediction'], 'ro', label = 'prediction')
plt.xticks(rotation = '60');
plt.legend()
# 图名
plt.xlabel('Date'); plt.ylabel('Maximum Temperature (F)'); plt.title('Actual and Predicted Values');
通过上述输出结果的走势可以看出，模型已经基本能够掌握天气变化情况，接下来还需要深入数据，考虑以下几个问题。
1. 如果可利用的数据量增大，会对结果产生什么影响呢?
2. 加入新的特征会改进模型效果吗?此时的时间效率又会怎样?
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• - 《RandomForest：随机森林预测生物标记biomarker——分类》， 大家可以学习此文，实现分组挖掘两组或多组的特异Features，也可以展示特征的贡献度，获得分类评估的准确度，以及使用新数据进行预测，无监督的随机...
关于随机森林的简介和应用理论，请阅读之前分享的文章：
关于随机森林进行分类的入门实战，请阅读
之前分享的
- 《RandomForest：随机森林预测生物标记biomarker——分类》
大家可以学习此文，实现分组挖掘两组或多组的特异Features，也可以展示特征的贡献度，获得分类评估的准确度，以及使用新数据进行预测，无监督的随机森林等基础技能。
今天我们讲使用randomForest实现回归分析的实战代码。回归的应用主要包括时间序列预测模式，如预测股票、尸体死亡时间等。
本节不需要先难知识可也直接学习使用。
RandomForest安装与加载
# 安装
install.packages("randomForest")
# 加载
library(randomForest)
回归Classification
先了解一下输入数据格式，方便准备
使用R内置按天记录的空气质量数据
data(airquality)
数据包括157天中，臭氧、太阳强度、风和温度，部分有缺失。前4列属性数据，后2列时间月和日为分组数据。
Ozone Solar.R Wind Temp Month Day
1 41 190 7.4 67 5 1
2 36 118 8.0 72 5 2
3 12 149 12.6 74 5 3
4 18 313 11.5 62 5 4
5 NA NA 14.3 56 5 5
6 28 NA 14.9 66 5 6
设置随机数种子保证结果可重复
set.seed(315)
随机森林回归臭氧与其它所有属性
ozone.rf= randomForest(Ozone ~ ., data=airquality, mtry=3,
importance=TRUE, na.action=na.omit)
print(ozone.rf)
结果如下：包括分析的命令，分析类型，树数量，重要的变量(Feature)个数，平均残差平方，解析率。
Call:
randomForest(formula = Ozone ~ ., data = airquality, mtry = 3, importance = TRUE, na.action = na.omit)
Type of random forest: regression
Number of trees: 500
No. of variables tried at each split: 3
Mean of squared residuals: 304.4269
% Var explained: 72.26
查看每个变量的分类贡献度，显示仅保留两位小数可读性更好
round(importance(ozone.rf), 2)
%IncMSE IncNodePurity
Solar.R 10.40 10833.08
Wind 23.55 43838.66
Temp 47.13 53731.95
Month 2.04 1504.72
Day 0.91 6306.42
结果为每个相关变量对应两列值。%IncMSE是Increased in mean squared error (%)，直译为增长的错误率平方均值，即去除该变量后，对目标预测的准确度下降的低，可理解为对目标变量预测准确的贡献度。IncNodePurity是Increased node purity，是另一种评估的方法。这里我们只关注%IncMSE就够了。
varImpPlot(ozone.rf)
交叉验证cross-validation
# 先清空NA的样本，验证不允许有NA
airquality = na.omit(airquality)
myairquality= cbind(airquality[1:6], matrix(runif(96 * nrow(airquality)), nrow(airquality), 96))
# 交驻验证添加了随机数的训练集，分组，交叉验证的次数
result= rfcv(myairquality, airquality$Ozone, cv.fold=3) # 绘制错误率曲线，观查错误率与使用Markers数量的变化 with(result, plot(n.var, error.cv, log="x", type="o", lwd=2)) 我们看到一个现象，不是feature越多越好，无关的feature如果多了，反而错误率上升，会影响预测的准确度。 # 使用replicate进行多次交叉验证，可选 result= replicate(5, rfcv(myairquality, airquality$Ozone), simplify=FALSE)
error.cv= sapply(result, "[[", "error.cv")
matplot(result[[1]]\$n.var, cbind(rowMeans(error.cv), error.cv), type="l",
lwd=c(2, rep(1, ncol(error.cv))), col=1, lty=1, log="x",
xlab="Number of variables", ylab="CV Error")
多次验证结果类型，更能说明结果的可信度。
严谨总没有坏处，好的结果都是多角度证明的。
大家学习随机森林的分类、和回归。将来有时间，将带大家上手重复一些高水平文章中的分析，结合具体生物学问题会更有意思。
Reference
写在后面
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• 本文是Python商业数据挖掘实战的第3篇1 - 基于不平衡数据的反欺诈模型实战2 - Apriori算法实现智能推荐3- 随机森林预测宽带客户离网前言 组合算法也叫集成学习，在金融行...
本文是Python商业数据挖掘实战的第3篇
前言
组合算法也叫集成学习，在金融行业或非图像识别领域，效果有时甚至比深度学习还要好。能够理解基本原理并将代码用于实际的业务案例是本文的目标，本文将详细介绍如何利用Python实现集成学习中随机森林这个经典的方法来预测宽带客户的流失，主要将分为两个部分：
• 详细原理介绍
• Python代码实战
集成学习
本文的主角是随机森林，所以我们将以随机森林所属的分支 —— 装袋法 入手，深入浅出该集成学习方法的原理步骤。装袋法流程如下
乍一看图中的步骤可能有些复杂，现在来逐步拆解。装袋法中的装袋二字是精髓，顾名思义即将多个模型装入同一个袋子后，让这个袋子作为一个新的模型来实现预测需求，仅此而已。换句话说，即把多个模型组合起来形成一个新的大模型，这个大模型最终给出的预测结果是由这多个小模型综合决定的，决定方式为少数服从多数。
假设有10万条原始数据，用这些数据来做十棵决策树（当然也可以是其他模型），最后这10棵树将被装进了同一个袋子中。这时候取其中一条数据放入这个袋子，便会得出10个预测值(每棵树各一个)，假如其中三棵树给出的预测值为0，剩余的七棵给出的为1，那我们便可知道这个袋子对这个数据的预测结果为 0 的概率是 3/10。
为了更深入的理解装袋法，下面将回答三个与装袋法有关的常见问题
问：袋子中的每个模型使用的样本量范围应为多少合适？
答：如果是上面的例子，袋子里面有十棵树，源数据总量为 10万 条，则每棵树取用的样本量的最小值为最少是1w个(10w/10棵 = 1w/棵)，因为至少要保证不能浪费样本，但每棵树最多可取用多少样本呢？其实在样本量已知，同一袋子中模型个数为n的情况下，样本的选择比例为1/n ~ 0.8最好。每个小模型取用 100% 的样本是绝对没有意义的，那就跟没抽是一样的，这样也就没有体现出装袋，只有每个模型用到的数据都有一定的不同，组合起来后每个的投票（预测结果）也才有意义。
问：袋中模型们之间的相关性会影响最后的决策结果吗？
答：装袋法思路最重要的一点：袋子中每个模型之间不能相关，越不相关越好，这里的不相关主要体现在用于训练每个模型的样本不一样。其次，每个模型的精度越高越好，这样它的投票才更有价值。
PS：训练模型的样本不一样这一点可以理解为总统选举，抽 10 波选民来投票，这 10 波选民的差异性越大越好，这样一来，只有在选民千差万别的情况下你依然脱颖而出，才足以说明你的实力，如果这10波选民中每一波之间的差异性都很小，比如都是本来就偏袒于总统候选人，那投票结果的说服力就会大减。
问：上面所说的模型精度高是不是哪怕模型很复杂也可以，如果每个模型的精度高但都过度拟合怎么办？
答：在装袋法中，模型是越精确越好，哪怕是过度拟合的也没有关系。因为一个模型要想在训练集上做到越精确越好，而精确程度与模型的复杂度大多是成正比的，所以出现过拟合的情况也是正常且情有可原的。复杂和过度拟合只是对袋子中每个模型而言，因为最后都会被加权，所以整个袋子（整体）并不会出现过度拟合的情况。
随机森林
随机森林的实现步骤如下：
有关随机森林算法，本文说明以下几个问题
问：为什么在列上也要随机抽样？
答：在引入笔者最最喜欢的一个比喻之前，先来看一个实际的业务场景，来自某城市商业银行。我们有一大个电子表格存着大量的历史数据，大概50多个变量(50多列)，变量们来自几个不同的公司如人行，电信等(同一个客户在不同公司)，最后希望预测的是该客户是否会违约。电子表格组成如下：
而根据基础的业务知识可知，与银行有关的数据中往往会存在许多缺失值，以上图为例，通常情况下只有待预测的变量这一列的数据是齐全的，毕竟客户们是否违约这个行为的历史数据很容易查找，但蓝框和绿框这两部分的缺失值往往较多，而且较随意，具体随意程度参见下图：
红框表示数据缺失，这里只展示了部分行和部分列数据，如果这份数据表的规模为 4万行 * 50列，那这数据缺失的分布得有多随意啊 ????？？所以，到底该如何充分利用这残次不齐的数据就成了呈待解决的关键问题。这时候就可以祭出超级生动形象的 “岛屿 - 湖泊 - 椰子树”比喻了：
• 整个表格看成一座巨大的岛屿，岛屿的长和宽分别对应电子表格横轴长和纵轴的长度
• 表中缺失的数据段看成一个个分布随意的小湖泊，有数据的地方看成陆地
• 整个小岛地底埋藏着巨大的价值（数据价值），通过在随意的种树（用装袋法在行列上进行随机抽样）来吸取地底的养分，毕竟湖泊上种不了树，所以只要足够随机，就总能充分的利用陆地。
正因为是行列都随机，才能够做到真正的把整个数据表随机切分成多份，每个模型使用一份，只要模型的数量足够，总有模型能够在最大程度上获取数据集的价值。而且因变量的分类往往又是极不平衡的，可以参考原理+代码｜手把手教你使用Python实战反欺诈模型至于如何将这些种好的树的信息又再收集，便可以将陆地上比较近的几棵树上面再弄一个收集器，把这几棵树从陆地上收集到的养分再递进一层汇总，最终实现陆地养分汇总于树木，树木养分汇总于收集器，收集器养分汇总于更上层的另一个收集器，最终实现整片数据海洋中多个岛屿的信息汇总，这便是周志华团队和蚂蚁金服的合作的用分布式深度随机森林算法检测套现欺诈
随机森林第一步之后的操作完全可以参照集成学习——装袋法中提及的步骤。
问：既然每个模型给出的预测结果最后都会被加权，所以随机森林中每棵决策树的权重是多少？
问：装袋法中袋子中的模型越多越好吗？袋中用来训练每个模型的源数据比例也是越多越好吗？
答：袋子中模型多一点好，袋中用来训练每个模型的源数据比例小一点好，但这并不代表越多越好与越小越好，还得结合数据集特性和一些深层次的模型算法知识。
装袋法的优势如下：
• 准确率明显高于组合中任何单个分类器
• 对于较大的噪音，表现不至于很差，并且具有鲁棒性
• 不容易过度拟合
随机森林算法的优点
• 准确率有时可以和神经网络媳美，比逻辑回归高
• 对错误和离群点更加鲁棒性
• 决策树容易过度拟合的问题会随着森林的规模而削弱
• 大数据情况下速度快(分布式)，性能好
数据探索
import pandas as pd
import numpy as np
参数说明
df.rename(str.lower, axis='columns', inplace=True)
from collections import Counter
## Broadband: Counter({0: 908, 1: 206}) 比较不平衡。
## 根据原理部分，可知随机森林是处理数据不平衡问题的利器
接着拆分测试集与训练集，客户id没有用，故丢弃cust_id,
X = df.iloc[:, 1:-1]
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y,
test_size=0.4, random_state=12345)
决策树建模
我们先进行完整的决策树建模来和随机森林进行对比
import sklearn.tree as tree
# 直接使用交叉网格搜索来优化决策树模型，边训练边优化
from sklearn.model_selection import GridSearchCV
# 网格搜索的参数：正常决策树建模中的参数 - 评估指标，树的深度，
## 最小拆分的叶子样本数与树的深度
param_grid = {'criterion': ['entropy', 'gini'],
'max_depth': [2, 3, 4, 5, 6, 7, 8],
'min_samples_split': [4, 8, 12, 16, 20, 24, 28]}
# 通常来说，十几层的树已经是比较深了
clf = tree.DecisionTreeClassifier() # 定义一棵树
clfcv = GridSearchCV(estimator=clf, param_grid=param_grid,
scoring='roc_auc', cv=4)
# 传入模型，网格搜索的参数，评估指标，cv交叉验证的次数
## 这里也只是定义，还没有开始训练模型
clfcv.fit(X=X_train, y=y_train)
# 使用模型来对测试集进行预测
test_est = clfcv.predict(X_test)
# 模型评估
import sklearn.metrics as metrics
print("决策树准确度:")
print(metrics.classification_report(y_test,test_est))
# 该矩阵表格其实作用不大
print("决策树 AUC:")
fpr_test, tpr_test, th_test = metrics.roc_curve(y_test, test_est)
print('AUC = %.4f' %metrics.auc(fpr_test, tpr_test))
AUC 大于0.5是最基本的要求，可见模型精度还是比较糟糕的，决策树的调优技巧就不再过多展开，我们将在随机森林调优部分展示
随机森林建模
随机森林建模一样是使用网格搜索，有关Python实现随机森林建模的详细参数解释可以看代码的注释
param_grid = {
'criterion':['entropy','gini'],
'max_depth':[5, 6, 7, 8], # 深度：这里是森林中每棵决策树的深度
'n_estimators':[11,13,15], # 决策树个数-随机森林特有参数
'max_features':[0.3,0.4,0.5],
# 每棵决策树使用的变量占比-随机森林特有参数（结合原理）
'min_samples_split':[4,8,12,16] # 叶子的最小拆分样本量
}
import sklearn.ensemble as ensemble # ensemble learning: 集成学习
rfc = ensemble.RandomForestClassifier()
rfc_cv = GridSearchCV(estimator=rfc, param_grid=param_grid,
scoring='roc_auc', cv=4)
rfc_cv.fit(X_train, y_train)
# 使用随机森林对测试集进行预测
test_est = rfc_cv.predict(X_test)
print('随机森林精确度...')
print(metrics.classification_report(test_est, y_test))
print('随机森林 AUC...')
fpr_test, tpr_test, th_test = metrics.roc_curve(test_est, y_test)
# 构造 roc 曲线
print('AUC = %.4f' %metrics.auc(fpr_test, tpr_test))
可以看到，模型的精度大大提升
为什么要打印梯度优化给出的最佳参数？打印梯度优化结果的最佳参数的目的是为了判断这个分类模型的各种参数是否在决策边界上，简言之，我们不希望决策边界限制了这个模型的效果。（通常这时候会先把复杂度放一边）
不难发现，参数max_depth, min_samples_split, 和n_estimators 这三个参数的范围设置可能有限制模型精度的可能，所以需要适当调整
"""
{'criterion': 'gini',
'max_depth': 8, 在最大值边界上，所以这个参数的最大值范围应该再调大
'max_features': 0.5, 也在最大值边界上，说明这个参数的最小值范围应该再调大
'min_samples_split': 4, 同理，在最小边界上，可考虑把范围调小
'n_estimators': 15 同理，在最大边界上，可以适当调大范围
"""
# 调整结果
param_grid = {
'criterion':['entropy','gini'],
'max_depth':[7, 8, 10, 12],
# 前面的 5，6 也可以适当的去掉，反正已经没有用了
'n_estimators':[11, 13, 15, 17, 19], #决策树个数-随机森林特有参数
'max_features':[0.4, 0.5, 0.6, 0.7],
#每棵决策树使用的变量占比-随机森林特有参数
'min_samples_split':[2, 3, 4, 8, 12, 16] # 叶子的最小拆分样本量
现在来查看再次建模的结果
此时都在决策边界内了，但其实调整参数是门技术活，并不只是通过决策边界这一单一指标来调整，后续推文会陆续更新。
小结
最后总结一下：随机森林是集成学习中非常经典的一种方法，基础原理简单，实现优雅，可即学即用。而且随机森林应用十分广泛，并不只是局限于常见的金融领域，只要数据不平衡或者随机缺失严重，都值得尝试。如果你也对本文使用的数据和代码感兴趣，可以在后台回复0726获取，我们下个案例见。
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• 1. 2. 参考链接： https://www.jianshu.com/p/d0d7809007a1 http://blog.csdn.net/lulumi2016/article/details/52662069
• 随机森林分类器中的树的个数，最小样本数及最小子叶数都是可以修改的。我们可以通过循环尝试找出最佳参数。这里以树的个数为例。 from sklearn.ensemble import RandomForestClassifier a = [] b = [] for i in ... | 11,419 | 26,055 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-31 | latest | en | 0.506033 |
https://jakubmarian.com/integral-of-logx-or-lnx/ | 1,719,024,476,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00718.warc.gz | 298,181,116 | 16,011 | # Integral of $\log(x)$ (or $\ln(x)$)
by Jakub Marian
Tip: See my list of the Most Common Mistakes in English. It will teach you how to avoid mistakes with commas, prepositions, irregular verbs, and much more.
The natural (base $e$) logarithm is usually denoted $\log(x)$ in mathematics, but for the sake of students of other disciplines, we will denote it $\ln(x)$ in this article.
How to go about the integration? The task is actually very simple with the help of integration by parts, but it requires a little trick. As you can see, there is only one function in
$$∫ \ln(x)\,dx\,,$$
but integration by parts requires two. The trick is to write $\ln(x)$ as $1⋅\ln(x)$ and then apply integration by parts by integrating the $1$ and differentiating the logarithm:
$$∫ \ln(x)\,dx = ∫ \overbrace{1\vphantom{\ln(x)}}^{u’}\overbrace{\ln(x)}^v\,dx = \overbrace{x\vphantom{\ln(x)}}^u\overbrace{\ln(x)}^v - ∫ \overbrace{x\vphantom{\frac1x}}^u\overbrace{\frac1x}^{v’}\,dx = x\ln(x)-x+c\,.$$
Finally, if you found this article because you are wondering what the logarithm of $\log_{10}(x)$ is, then you can use the equality $\log_{10}(x) = \ln(x)/\ln(10)$, so
$$∫ \log_{10}(x)\,dx = ∫ \frac{\ln(x)}{\ln(10)}\,dx = \frac{1}{\ln(10)} ∫ \ln(x)\,dx\,.$$
To get the integral, simply multiply the result above by $\frac{1}{\ln(10)}$.
By the way, I have written several educational ebooks. If you get a copy, you can learn new things and support this website at the same time—why don’t you check them out? | 480 | 1,508 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-26 | latest | en | 0.850073 |
http://www.geom.uiuc.edu/~demo5337/s97a/trans-wsgspans.html | 1,544,630,710,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824059.7/warc/CC-MAIN-20181212155747-20181212181247-00062.warc.gz | 376,701,512 | 1,572 | # A Teacher's Guide to Investigations of Translational Symmetry Using Geometer's Sketchpad
Students will start by being reminded about the definitions of the types of symmetry.
There is a link to the Sketchpad activity on the Translation page, but not to the questions below.
Students may wish to just explore the Geometry Sketchpad site first before getting a list of questions. There are a few suggestions on the SKETCH, but the list of questions below is more leading for those students who need more guidance.
1a) What happens to the green polygon when you move point S? The green polygon move the same distance and direction as point S is moved.
b) How are the black and green polygons related? The green polygon is a translation of the black polygon by vector RS.
2a) What happens when the Blue Angle gets bigger or smaller? The blue polygon moves.
b)How are the black and the blue polygons related? The black polygon is rotated through the Blue Angle to the blue polygon.
3a) What happens when the Red Angle gets bigger or smaller? The red polygon moves.
b)How are the blue and the red polygons related? The blue polygon is rotated through the Red Angle to the red polygon.
4) What happens when the red and blue lines are parallel? The red polygon (the second reflection) looks like a translation.
5) How are translation and reflection related? A translation is like a double reflection through parallel mirrors.
6) What happens when the red and blue lines intersect? The red polygon (the second reflection) looks like a rotation.
7) How are reflection and rotation related? Reflections through intersection mirrors is like a rotation about the intersection point. | 342 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.892104 |
https://www.learnamic.com/learning_resources/money-and-taxes-in-a-micro-business | 1,632,039,136,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00692.warc.gz | 876,638,830 | 15,438 | # Money and Taxes in a Micro Business
## Description
After you have started a micro business and run it for a while, you may wonder…
How can I make more money? Have I made a profit? Do I owe any taxes on my profits? How can I reduce my taxes? How can I measure my success? How can I plan to stay in business? How can I avoid failure? This third book in the Micro Business for Teens series, Money and Taxes in a Micro Business, will help you answer all those questions.
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We receive a small share of sales from those links to help us keep this service free for all learners. Thanks for your support! | 721 | 3,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.918787 |
https://www.physicsforums.com/threads/asymptotes-and-holes.535576/ | 1,542,416,244,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743247.22/warc/CC-MAIN-20181116235534-20181117021534-00497.warc.gz | 978,689,380 | 12,184 | # Homework Help: Asymptotes and holes
1. Oct 1, 2011
### Johnyi
1. The problem statement, all variables and given/known data
Find the oblique asymptote and sketch the graph
2. Relevant equations
f(x)=x^3+3x^2-x-5/x^2-1
3. The attempt at a solution
My question is that shouldnt there be a hole in the graph when x=-1 or 1? The denominator would end up being zero.
When should i be using an hole or when do i know it is an asymptote?
2. Oct 1, 2011
### Mentallic
If you have a rational function of the form $$f(x)=\frac{x-b}{x-a}$$ then there doesn't exist a hole in the function at x=a, instead there is a vertical asymptote at that point. You will have a hole in the function when it is of the form $$f(x)=\frac{(x-a)(x-b)}{x-a}$$ since this simplifies into the linear equation $$f(x)=x-b$$ but at x=a there is a hole.
So basically, if the fraction is of the form 0/0 at some point x=a, then it could be a hole (since x=a is a zero of both the numerator and denominator, you can factor out x-a from both and then cancel). If it is of the form a/0 for some non-zero a, then it is a vertical asymptote. | 323 | 1,111 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2018-47 | latest | en | 0.881286 |
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# What is the area of the circle above with center O.
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What is the area of the circle above with center O. [#permalink]
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01 May 2013, 05:32
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What is the area of the circle above with center O?
(1) The length of major arc CA is 27pi
(2) The perimeter of OABC is 72
[Reveal] Spoiler:
[Reveal] Spoiler: OA
Last edited by Bunuel on 01 May 2013, 05:36, edited 1 time in total.
Edited the question.
Kudos [?]: 37 [0], given: 94
Current Student
Joined: 04 Mar 2013
Posts: 68
Kudos [?]: 58 [0], given: 27
Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
GMAT 1: 770 Q50 V44
GPA: 3.66
WE: Operations (Manufacturing)
Re: What is the area of the circle above with center O. [#permalink]
### Show Tags
01 May 2013, 05:40
The Quadrilateral OABC is evidently a square as
1. OA = OC
2. /_AOC is 90deg.
3. /_OAB is 90deg. as tangent is perpendicular to radius (as percieved from figure)
So statement A gives the value of pi*(360-90)/360*r, from which r can be calculated
Statement B gives the perimieter of the square which is 4r so r can be calculated
Both are independently sufficient
_________________
When you feel like giving up, remember why you held on for so long in the first place.
Kudos [?]: 58 [0], given: 27
Intern
Joined: 22 Jul 2010
Posts: 32
Kudos [?]: 37 [0], given: 94
Re: What is the area of the circle above with center O. [#permalink]
### Show Tags
01 May 2013, 07:21
aceacharya wrote:
The Quadrilateral OABC is evidently a square as
1. OA = OC
2. /_AOC is 90deg.
3. /_OAB is 90deg. as tangent is perpendicular to radius (as percieved from figure)
So statement A gives the value of pi*(360-90)/360*r, from which r can be calculated
Statement B gives the perimieter of the square which is 4r so r can be calculated
Both are independently sufficient
You have assumed that quadrilateral is a square. My question and doubt is precisely that, to which level we can assume things from the figure? It is not mentioned in question that AB and CB is tangent to the circle.
Kudos [?]: 37 [0], given: 94
Re: What is the area of the circle above with center O. [#permalink] 01 May 2013, 07:21
Display posts from previous: Sort by | 896 | 3,095 | {"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} | 4.03125 | 4 | CC-MAIN-2017-47 | latest | en | 0.871273 |
http://archive.ambermd.org/201102/0349.html | 1,575,906,510,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540519149.79/warc/CC-MAIN-20191209145254-20191209173254-00366.warc.gz | 11,408,175 | 4,660 | # [AMBER] Scale sander.MPI to 64 ncpus - Amber11
From: Leong Wye Kit Damien <damienl.acrc.a-star.edu.sg>
Date: Fri, 18 Feb 2011 01:05:09 +0800
Good Morning Amber user group
I hope someone can give me some clues or point me to the correct direction.
I'm helping an amber ( Amber11 ) user to scale his sander.MPI computation from 32 ncpus to 64 ncpus. Everything seems to work fine when we use 32 ncpus, once we try to scale up to 64 ncpus ( eventually, if possible 128 npcus ), the results did not compute. I believe there is a limit to 256 ncpus in sander.MPI ( please correct me if I'm wrong ).
If it did compute correctly ( with 32 cpus ), I should be able to see:
Example
------------------------------
| Local SIZE OF NONBOND LIST = 139474
| TOTAL SIZE OF NONBOND LIST = 4808242
NSTEP ENERGY RMS GMAX NAME NUMBER
1 -4.0791E+04 3.1384E+01 2.8014E+03 O 1754
BOND = 77.0003 ANGLE = 215.9942 DIHED = 295.2149
VDWAALS = 6162.9204 EEL = -48661.2899 HBOND = 0.0000
1-4 VDW = 81.5555 1-4 EEL = 1037.1123 RESTRAINT = 0.0000
NSTEP ENERGY RMS GMAX NAME NUMBER
50 -5.0768E+04 1.9930E+00 1.1115E+01 H1 12978
BOND = 2690.2212 ANGLE = 57.6311 DIHED = 266.7564
VDWAALS = 4728.9004 EEL = -59638.3394 HBOND = 0.0000
1-4 VDW = 73.5857 1-4 EEL = 1053.2269 RESTRAINT = 0.0000
NSTEP ENERGY RMS GMAX NAME NUMBER
100 -5.2988E+04 9.6958E-01 7.1804E+00 H2 14272
---------------------------------------
OUR PROBLEM :
We have try various methods on 64 cpus, we end up at ( incorrect ):
---------------------------------------
4. RESULTS
APPROXIMATING switch and d/dx switch using CUBIC SPLINE INTERPOLATION
using 5000.0 points per unit in tabled values
TESTING RELATIVE ERROR over r ranging from 0.0 to cutoff
| CHECK switch(x): max rel err = 0.2738E-14 at 2.422500
| CHECK d/dx switch(x): max rel err = 0.8314E-11 at 2.736960
---------------------------------------------------
| Local SIZE OF NONBOND LIST = 75749
| TOTAL SIZE OF NONBOND LIST = 4808242
---------------------------------------
Nothing comes up further, it just did not continue, we have try to give it a longer CPU walltime, but it still does not continue.
We are using a GNU/Linux X86 cluster, each compute node has 8 cpu cores, there is 200 compute nodes using LSF Job Scheduler.
We are using Intel MPI (4.00 ).
Our LSF-sander.MPI script:
----------------------
#!/bin/bash
#BSUB -q normal
#BSUB -R "span[ptile=8]"
#BSUB -W 1:00
#BSUB -J "min64"
#BSUB -o lsf%J.o
#BSUB -e lsf%J.e
#BSUB -R "rusage[mem=1000]"
#BSUB -n 64
TOTAL_CPUS=64
NODES=8
export AMBERHOME=/apps/Amber11/amber11
export PATH=\$PATH:\$AMBERHOME/bin
MACHINEFILE=mymacs.\$LSB_JOBID
for i in `echo \$LSB_HOSTS`; do echo \$i; done > \$MACHINEFILE
/usr/local/intel/impi/4.0.0.025/intel64/bin/mpdboot -n \$NODES -f \$MACHINEFILE
if [ \$? -ne 0 ] ; then
exit 1
fi
RESULT=`/usr/local/intel/impi/4.0.0.025/intel64/bin/mpdtrace | wc -l`
if [ "\$RESULT" != "\$NODES" ] ; then
/usr/local/intel/impi/4.0.0.025/intel64/bin/mpdallexit
exit 1
fi
mpiexec -machinefile \$MACHINEFILE -n \$TOTAL_CPUS /apps/Amber11/amber11/bin/sander.MPI -O -i min.in -o min_ARNO.out -p ARNO.parmtop -c ARNO.inpcrd -r ARNO.rst
/usr/local/intel/impi/4.0.0.025/intel64/bin/mpdallexit
---------------------------------------------
More info:
-------------------
cat min.in
ringheterodim : initial minimisation prior to MD, whole system
&cntrl
imin = 1,
maxcyc = 1000,
ncyc = 500,
ntb = 1,
ntr = 0,
cut = 10,
/
---------------------
We believe the amber11 compiling or binaries are in order, else we should not be able to compute ( sander.MPI ) successfully with 32 ncpus, What we
do not understand is why we cannot scale to 64 ncpus and beyond.
Some helpful folks, please advise or at least diverts us to the correct direction. Many Thanks.
If you have any clarification, please let us know. Thanks.
Cheers
Damien Leong
(Computing Systems Group )
A*STAR Compute Resource Centre (A*CRC). Biopolis
_______________________________________________
AMBER mailing list
AMBER.ambermd.org
http://lists.ambermd.org/mailman/listinfo/amber
Received on Thu Feb 17 2011 - 09:30:04 PST
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# Two accounting professors decided to compare the
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Question 6.Question :Two accounting professors decided to compare the variance of theirgrading procedures. To accomplish this, they each graded the same 10 exams, with the following results:At the 1% level of significance, what is the decision? Student Answer:Reject the null hypothesis and conclude the variances are different. Fail to reject the null hypothesis and conclude the variances are different. Reject the null hypothesis and conclude the variances are the same. Fail to reject the null hypothesis and conclude the variances are the same. Instructor Explanation:The null hypothesis is . The computed value of F is found by taking the ratio of the two sample variances: . For the F ratio, , the degrees of freedom in both the numerator and the denominator are 9, found by n1- 1 and n2- 1, or 10 - 1. Using the "Critical values of the F distribution at a 1 percent level
of significance" table, the critical value of F with 9 degrees of freedom in the numerator, 9 degrees of freedom in the denominator, and the .01 significance level is 5.35. The computed value of 3.484 is less than the critical value of 5.35. Therefore, fail to reject the null hypothesis. Points Received: 1 of 1 Comments: Question 7.Question :Two accounting professors decided to compare the variance of theirgrading procedures. To accomplish this, they each graded the same 10 exams, with the following results:What is the null hypothesis? Student Answer: Instructor Explanation: The null hypothesis is that the variance in the two populations is the same, . Points Received: 0 of 1 Comments: Question 8.Question :A large department store examined a sample of the 18 credit card sales and recorded the amounts charged for each of three types of credit cards: MasterCard, Visa, and Discover. Six MasterCard sales,seven Visa, and five Discover sales were recorded. The store used an ANOVA to test if the mean sales for each credit card were equal.
What are the degrees of freedom for the Fstatistic? Points Received: 1 of 1 Comments: | 507 | 2,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-17 | latest | en | 0.922509 |
https://metanumbers.com/29994 | 1,600,821,223,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00230.warc.gz | 511,403,130 | 7,456 | ## 29994
29,994 (twenty-nine thousand nine hundred ninety-four) is an even five-digits composite number following 29993 and preceding 29995. In scientific notation, it is written as 2.9994 × 104. The sum of its digits is 33. It has a total of 3 prime factors and 8 positive divisors. There are 9,996 positive integers (up to 29994) that are relatively prime to 29994.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 33
• Digital Root 6
## Name
Short name 29 thousand 994 twenty-nine thousand nine hundred ninety-four
## Notation
Scientific notation 2.9994 × 104 29.994 × 103
## Prime Factorization of 29994
Prime Factorization 2 × 3 × 4999
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 29994 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 29,994 is 2 × 3 × 4999. Since it has a total of 3 prime factors, 29,994 is a composite number.
## Divisors of 29994
1, 2, 3, 6, 4999, 9998, 14997, 29994
8 divisors
Even divisors 4 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 60000 Sum of all the positive divisors of n s(n) 30006 Sum of the proper positive divisors of n A(n) 7500 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 173.188 Returns the nth root of the product of n divisors H(n) 3.9992 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 29,994 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 29,994) is 60,000, the average is 7,500.
## Other Arithmetic Functions (n = 29994)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 9996 Total number of positive integers not greater than n that are coprime to n λ(n) 4998 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3248 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 9,996 positive integers (less than 29,994) that are coprime with 29,994. And there are approximately 3,248 prime numbers less than or equal to 29,994.
## Divisibility of 29994
m n mod m 2 3 4 5 6 7 8 9 0 0 2 4 0 6 2 6
The number 29,994 is divisible by 2, 3 and 6.
## Classification of 29994
• Arithmetic
• Abundant
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
• Sphenic
## Base conversion (29994)
Base System Value
2 Binary 111010100101010
3 Ternary 1112010220
4 Quaternary 13110222
5 Quinary 1424434
6 Senary 350510
8 Octal 72452
10 Decimal 29994
12 Duodecimal 15436
20 Vigesimal 3eje
36 Base36 n56
## Basic calculations (n = 29994)
### Multiplication
n×i
n×2 59988 89982 119976 149970
### Division
ni
n⁄2 14997 9998 7498.5 5998.8
### Exponentiation
ni
n2 899640036 26983803239784 809352194374081296 24275709718056194392224
### Nth Root
i√n
2√n 173.188 31.0703 13.1601 7.85972
## 29994 as geometric shapes
### Circle
Diameter 59988 188458 2.8263e+09
### Sphere
Volume 1.13029e+14 1.13052e+10 188458
### Square
Length = n
Perimeter 119976 8.9964e+08 42417.9
### Cube
Length = n
Surface area 5.39784e+09 2.69838e+13 51951.1
### Equilateral Triangle
Length = n
Perimeter 89982 3.89556e+08 25975.6
### Triangular Pyramid
Length = n
Surface area 1.55822e+09 3.18007e+12 24490
## Cryptographic Hash Functions
md5 8875c52a7bcbd3e60c07ce65b152fefc 90b1062944fd07bcf5ab7781db98feb8a69253f9 c8e37435cfe35c0ef814df831095cf8d19ffc753c68b57781a1e9304c587bba0 f443e24897f2bd25aa37d409e5a08f26549bb236fa796382b4ba5e9c52b7aeacaf76cd3d6e345105bcc7fac99ac053dd20a5f8f3506ffe55fd6cb1d7bb846b39 6dd6c7bc8cceafa7826c073ee0ec510473a15990 | 1,469 | 4,167 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-40 | latest | en | 0.814237 |
https://zsf24.pl/ball-mill/174/calculate.html | 1,660,603,881,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572212.96/warc/CC-MAIN-20220815205848-20220815235848-00189.warc.gz | 945,609,865 | 5,757 | # Calculate Actual Velocity Of A Ball Mill
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### Calculete Actual Velocity Of A Ball Mill
Derivation of critical velocity of ball mill. critical speed of ball mill 766. Velocity Calculation In Ball Mills , equation formula critical speed of the ball mill Calculate Ball Mill Critical Speed, formula to ball mill.
Nov 18, 2013 formula for calculating the critical speed of a ball mill, ball mill speed newbie questions 2 most if the actual speed of a 6 ft diameter ball mill how to calculate critical speed of a ball mill - YouTube.
### Critical Speed Calculation Formula Of Ball Mill
Derive the equation of critical speed of ball mill. formula for critical speed of ball mill If the actual speed of a 6 ft diameter ball mill is 25 rpm, calculate lifter face inclination angle φ, steel ball.
The theoretical formula to calculate formula for finding critical velocity . Critical Speed Formula For Ball Mill - How Much Crusher defination of ball mill critical speed critical speed of ball mill 76.6 calculator for ball mill critical speed critical speed formula for ball mill in meters.
### Ball Mill Parameter Selection & Calculation
• Empty heights of both the chambers measured to calculate the ball − The velocity of gases calculated through mill is 1 29 m Formulas kiln Upload amp Share PowerPoint presentations and cement mill formulas mill critical velocity 76 d 1 2 mill actual velocity 32 d 1 2 d theta’ mill .
Mill Cyclone Design Calculations Xls. ball mill design calculation xls and pdf file ball mill design power calculation the basic cyclone design equations formulas calculator air filtration quality patent crusher the acknowledged leader in fluid bed roller mill design plant hpc cone crusher xsd sand washer hp 500 cone crushers in fujairah thailand ,atox raw mill calculation,smidth's atox.
### How To Calculate Mill Critcal Speed
The mill critical speed will be calculated based on the diameter (above) less twice this shell liner width. Mill Actual RPM Enter the measured mill rotation in revolutions per minute. Result 1 This mill would need to spin at RPM to be at 100 critical speed. ++. How To Calculate Critical Speed Of A Ball Mill.
Ball mill critical speed formula equation derivation. ball mill model equations – Grinding Mill China. ball mill design equations and modelling pdf Free Download (pdf,doc,xls,rtf,ppt,pps documents) ball mill model equations ball mill equation derivation.
### Velocity Calculation In Ball Mills Crusher Mills Cone
Ball mills are used primary for single stage fine grinding, regrinding, and as the second stage in two stage grinding circuits According to the need of customers , formula for calculate the ball mill velocity - BINQ , ball mill critical speed calculation - beltconveyers formula for critical speed of ball mill -.
Mar 08, 2013 calculation of ball mill grinding efficiency. dear experts . please tell me how to calculate the grinding efficiency of a closed ckt open ckt ball mill. in literatures it is written that the grinding efficiency of ball mill is very less [less than 10 ]. please expalin in a n excel sheet to calcualte the same. thanks. sidhant. reply.
### How To Calculate Cement Mill Velocity
Mill Critical Speed Determination. The Critical Speed for a grinding mill is defined as the rotational speed where centrifugal forces equal gravitational forces at the mill shell's inside surface. This is the rotational speed where balls will not fall away from the mill's shell. Result 1 This mill would need to spin at RPM to be at 100 critical speed.
The Lining Plate Of Ball Mill Is Gradually Replaced By Rubber And Polyurethane Lining Plate In China, But With The Continuous Application Of Rubber Polyurethane Lining Plate In The Lining Plate Of Ball Mill, It Has Gradually Replaced Manganese Steel And Other Lining Plates And Become The Mainstream Of Market Development.
### Calculation Of Residence Time In Ball Mill
Formula for critical velocity of ball mill pdf. The Ball Mill Driving Device Fault and the Main Bearing - IAES Mar 4 2013 consumption are rotational speed and friction status This paper also put forward to make sure kept the ball mill the formula 2 the mechanical efficiency only considers the friction between the transmission number is greater than the critical value the sliding will be in the.
Ball mill critical speed derivation . Ball Mills Mine EngineerCom ball mill critical speed derivation,If the peripheral speed of the mill is too great, it begins to act like a centrifuge and the balls do not fall back, but stay on the perimeter of the mill The point where the mill becomes a centrifuge is called the Critical Speed, and dragThe second is air drag, which.
### Calculation For Inclination Of Ball Mill
Jan 07, 2015 Raw mills usually operate at 72-74 critical speed and cement mills at 74-76 . 3.2 Calculation of the Critical Mill Speed G weight of a grinding ball in kg. w Angular velocity of the mill tube in radial second. w = 2 3.14 (n 60) Di inside mill diameter in meter (effective mill diameter). n Revolution per minute in rpm. 7.
• Empty heights of both the chambers measured to calculate the ball charge, filling and estimate the power − The velocity of gases calculated through mill is 1.29 m sec. This velocity is sufficient for OPC33 filter (180-200 mmWG). The actual flow measured is only 4 of the total separator air.
### Critical Speed Of Ball Mill Formula Derivation
Measurement of Velocity Fields in Ball Mills1.05 Мб. There are various techniques for measuring the velocity. In our case, it is difficult to put sensors in a ball mill or inside the balls.The y-z plane is used as an example to explain how to calculate the true position y from the position of a bright spot ~.
Produced by the ball mill, so the cyclone overflow allows a lot of these middle size particles to escape the grinding circuit. This results in a decrease of material that the ball mill would have to grind, which lowers its efficiency. Also, unwanted particle sizes are leaving the grinding circuit to next process stages.
### How To Calculate Residence Time In A Ball Mill
9.3.4 Mill Speed. During normal operation the mill speed tends to vary with mill charge. According to available literature, the operating speeds of AG mills are much higher than conventional tumbling mills and are in the range of 80–85 of the critical speed.
Nov 01, 1992 The code incorporates a scheme to calculate the applied torque, and hence power input to the mill. Therefore the power draft or torque of the actual mill is the corresponding value for the slice multiplied by the number of slices. The simulations were done in the fractional critical speed range of 0.6-1.0 at a constant ball filling of 38.5.
### Vertical Roller Mill Design Calculation Pdfsand Filter
Milling Speed and Feed Calculator Determine the spindle speed (RPM) and feed rate (IPM) for a milling operation, as well as the cut time for a given cut length. Milling operations remove material by feeding a workpiece into a rotating cutting tool with sharp teeth, such as an end mill or face mill.
Of the balls is mainly in the direction of the rotation of the mill and not in the direction of the main axis. 2. We consider a perfect layer structure. This means that the balls are lifted up in layers with distance of twice the radius of a ball and with the angular velocity of the mill.
### Ball Mill Critical Speed Mineral Processing & Metallurgy
Calculation Of Ball Mill Residence Time. Calculation Of Ball Mill Residence Time Our company has been devoted to mining machinery for 40 years. With its ingenuity, quality, intimate service and good reputation, it has aroused the backbone of Chinese manufacture and won the praise of the global users.
The velocity of a ball is given by the velocity vector,v, ergy Ball Mills using Discrete Events”, Rawmill – Wikipedia, the free encyclopedia A high velocity of hot gas flow is maintained close to the dish so that fine particles are swept away as soon as they are produced. | 1,688 | 8,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.687218 |
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# Word problem
Six dozen equally priced oranges cost a total of n dollars. In terms of n, what is the cost, in cents, of one orange.
a. 25n/18 b. 18n/25 c. n/72 d. 72/n
#### Solution Preview
Solution:
Six dozen oranges = 72 oranges
as Six dozen = 72
cost ...
#### Solution Summary
A step by step detailed solution is provided to solve the word problem.
\$2.19 | 110 | 393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-22 | latest | en | 0.856725 |
http://7167508.ru/complex-analysis-by-ahlfors-pdf.html | 1,563,402,594,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525414.52/warc/CC-MAIN-20190717221901-20190718003901-00310.warc.gz | 2,796,865 | 3,752 | # Complex Analysis By Ahlfors Pdf
In Ahlfors visited Harvard University. Ahlfors was born in Helsinki, Finland. Since each root are angle multiplies about the origin, they will be n equally spaced points. Prove that the closure of a connected set is connected. Let S be a discrete set in R or C.
However, S is not path connected. In order for both to be analytic, they both need to satisfy the Cauchy-Riemann equations. The latter space is bounded in the sense that all distances lie under a fixed bound. Use compactness to prove that a closed bounded set of real numbers has a maximum. Therefore, g f z satisfies the Cauchy- Riemann so it is analytic.
Since the union of two regions is not a region, the regions are separated. Every collection of closed sets with an empty intersection contains a finite subcollection with an empty intersection. Let P tn be the nth partial sum of bn. In he was one of the first two people to be awarded the Fields Medal. Thus, if E is a sequentially compact metric space, then E is compact.
We could also use the Cauchy-Riemann equations. From Wikipedia, the free encyclopedia. When the limit is one, we can draw no conclusion about convergence, but when the limit is greater than one, the sequence diverges.
## Andr s Vargas
For the finally implication, suppose they have a point in common and the union of two regions is not a region. Prove analytically that the midpoints of parallel chords to a circle lie on a diameter perpendicular to the chords. Book Category Mathematics portal. Then we can integrate with respect to y to find v x, y. Let A and B be these two nonempty regions.
Prove that a subset of the real line which is topologically equivalent to an open interval is an open interval. Show that the sum of an absolutely convergent series does not change if the terms are rearranged. Differential Equations I Course of Lectures.
He returned to Finland in to take up a professorship at the University of Helsinki. Try to rearrange the proof so economically that the absolute convergence of the second series is not needed. They are said to be equivalent if they determine the same open sets.
Express them through cos iz and sin iz. Let A and B be two nonempty separated regions. Let k be the degree of R z. If g w and f z are analytic functions, show that g f z is also analytic.
Thus, fn x is pointwise Pn convergent to f x. We have reached contradiction so if they have a point in common, then the union of two regions is are a region. We have reached a contradiction so if the union of two regions is a region, then they have a point in common. Express arctan w in terms of the logarithm.
His book Complex Analysis is the classic text on the subject and is almost certainly referenced in any more recent text which makes heavy use of complex analysis. Given an alternate proof of the fact that every bounded sequence of complex numbers has a convergent subsequence for instance by use of the limes inferior. Write the equation of an ellipse, hyperbola, dslr photography for dummies pdf free parabola in complex form.
Let a and b be the foci of a hyperbola. Verify that E is not locally connected.
In particular, how are the zeros and poles related to each other? Therefore, S is countable. Express the result in symmetric form. Consider the function tan z.
Laureates of the Wolf Prize in Mathematics. Show that a discrete set in a separable metric space is countable. Determine the conjugate harmonic function and the corresponding analytic function by integration and by the formal method. Therefore, an zn converges absolutely with a radius of convergence of R.
Since we are dealing with a set of real numbers, we are speaking of compact metric spaces. Now, zi are limit points of E as well. Show that the system of all matrices of the special form! What are the components of E? | 839 | 3,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.939371 |
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# Financial Analysis
Not what you're looking for? Search our solutions OR ask your own Custom question.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The MorTex Company assembles garments entirely by hand even though a textile machine exists that can assemble garments faster than a human can.
Workers cost \$50 per day, and each additional laborer can produce 200 more units per day (ie. marginal product is constant and equal to 200).
Installation of the first textile machine on the assembly line will increase output by 1800 units daily. Currently the firm assembles 5400 units per day.
a). The financial analysis department at MorTex estimates that the price of a textile machine is \$600 per day. Can management reduce the cost of assembling 5400 units per day by purchasing a textile machine and using less labor? why or why not?
b) The Textile Workers of America is planning to strike for higher wages. Management predicts that if the strike is successful, the cost of labor will increase to \$100 per day. If the strike is successful, how would this affect the decision in part a to purchase a textile machine"
https://brainmass.com/economics/finance/financial-analysis-274229
#### Solution Preview
a) If the MorTex Company assembles Garments entirely by hand, the total cost can be calculated as follows:
Workers cost \$50 per day and Marginal productivity of labor MPL = 200.
The firm assembles 5400 units per day.
Thus in order to produce 5400 units, 5400/200 = 27 labors is required.
Therefore the total cost of assembling 5400 units per day entirely by ...
#### Solution Summary
The solution discusses a financial analysis problem.
\$2.49 | 376 | 1,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.896741 |
http://perplexus.info/show.php?pid=7634&cid=47184 | 1,537,717,350,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159561.37/warc/CC-MAIN-20180923153915-20180923174315-00115.warc.gz | 189,266,166 | 4,109 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Maybe? (Posted on 2012-01-15)
Is there a function F(x), such that for every x, larger than -1, the following is true:
F(x)=1+ F(x/(x+1)) ?
No Solution Yet Submitted by Ady TZIDON Rating: 3.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Final answer (spoiler) | Comment 1 of 3
No, there is no such function.
Proof:
(a) Assume such a function exists.
(b) Let x = 0
(c) Then F(0) = 1+F(0), which is not true, not even if F(0) is undefined
Posted by Steve Herman on 2012-01-15 20:06:11
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Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information | 295 | 1,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | longest | en | 0.798778 |
https://leanplan.com/starting-costs/?amp | 1,701,741,865,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00854.warc.gz | 411,173,199 | 33,518 | # New Business Starting Costs
Startups should estimate the new business startup costs as part of their lean business plan. “
How much will it cost to start that business? Here’s how to create a useful educated guess. It’s essential two lists. You can do this.
## Two lists: expenses and assets
Knowing the starting costs before you start a business is a matter of two simple lists:
• Startup expenses: These are expenses that happen before the beginning of the plan, before the first month of operations. For example, many new companies incur expenses for legal work, logo design, brochures, site selection and improvements, and signage. If there is a business location, then normally the startup pays rent for a month or more before opening. And if employees start receiving compensation before the opening, then those disbursements are also startup expenses.
• Startup assets: Typical startup assets are cash (the money in the bank when the company starts), business or plant equipment, office furniture, vehicles, and starting inventory for stores or manufacturers.
“I knew that if I failed I wouldn’t regret that, but I knew the one thing I might regret is not trying.” — Jeff Bezos, founder of Amazon.com
## A Simple Starting Costs Example
I used an example in Starting Balances in Chapter 18, on the Balance Sheet. These are estimated starting costs for the sample bicycle store:
Notice that on the same worksheet Garrett used to estimate starting costs, he also estimated starting funding, on the right side of the illustration. Books have to balance, so the initial estimates need to include not just the money you spend, but also where it comes from. In the case above, Garrett had to find \$124,500, and you can see that he financed it with Accounts Payable, debt, and investment in various categories. That Starting Balances illustration in the Projected Balance Sheet chapter also shows how those initial balances go into the Projected Balance Sheet.
## Another Simple Starting Costs Example
Here is another simple example: the starting costs worksheet that Magda developed for the restaurant I used in Chapter 7 for basic numbers. Magda’s list includes rent and payroll, the same as in her monthly spending, but here they are included in starting costs because these expenses happen before the launch.
I included rent and payroll because they point out the importance in timing. The difference between these as startup expenses and running expenses is timing, and nothing else. Magda could have chosen to plan startup expenses as a running worksheet on expenses, starting a few months before launch, as in the illustration below. I prefer the separate lists, because I like the way the two lists create an estimate of starting costs. But that’s an option.
## How to Estimate Starting Costs
Obviously the goal with starting costs isn’t just to track them, but to estimate them ahead of time so you have a better idea, before you start a new business, of what the financial costs might be. Breaking the items down into a practical list makes the educated guess a lot easier. Ideally, you know the business you want to start, you are already familiar with the industry, so you can do a useful estimate for most of the startup costs from your own experience. If you don’t have enough firsthand knowledge, then you should be talking to people who do. For others, such as insurance, legal costs, or graphic design for logos, call some providers or brokers, and talk to partners; educate those guesses.
### Starting Cash is the Hardest and Most Important
How much cash do you need in the bank, as you launch? That’s usually the toughest starting cost question. It’s also prone to misinformation, such as those alleged rules of thumb you can find everywhere, saying you need to have a year’s worth of expenses, or six months’ worth, before you start. It’s not that simple. For most businesses, the startup cash isn’t a matter of what’s ideal, or what some expert says is the rule of thumb – it’s how much money you have, can get, and are willing to risk.
The best way is to do a Projected Cash Flow while leaving the supposed starting cash balance at zero, which shows how much (at least in theory, according to assumptions) the startup really needs in cash to support the business as it grows, before it reaches a monthly cash flow break-even point. Magda did that to determine the \$12,000 needed as starting cash for her restaurant. Note how, in the illustration here, the lowest point in cash is slightly less than \$12,000:
That low point comes, theoretically, in the third month of the business, March. The low point is \$11,609. Obviously that’s just an educated guess, but it’s based on assumptions for sales forecast, expense budget, and important cash flow factors including sales on account and purchasing inventory. So it’s better than a stab in the dark, or some rule of thumb.
Just as an example, the total spending with the estimates shown here, the theoretical “year’s worth of spending,” is \$182,000 (which you don’t see on the illustration, by the way, but take my word for it). The total for the first six months is \$93,000. If Magda sticks to those old formulas, she can’t start the business. She is able to raise enough money, between loans and her savings, to put \$12,000 into the starting cash balance. So that’s what she does. Then she launches and continues to have her monthly reviews, and watch the performance of all key indicators very carefully.
## Find Your Startup Costs Sweet Spot
There is no magic startup costs estimate for a given business. Every startup has its own natural level of startup costs. It’s built into the circumstances, like strategy, location, and resources. Call it the natural startup level; or maybe the sweet spot.
### 1. The Plan
For example, in the illustrations above, Magda’s restaurant deli in the office park needs about \$60,000, and Garrett’s bicycle store needs about \$125,000. The level is determined by factors like strategy, scope, founders’ objectives, location, and so forth. In both cases, the entrepreneurs have lists of assets they need and expenses they’ll incur. Let’s call these lists the natural startup level, which is built into the nature of the business, something like DNA.
Startup cost estimates have three parts: a list of expenses, a list of assets needed, and an initial cash number calculated to cover the company through the early months when most startups are still too young to generate sufficient revenue to cover their monthly costs.
It’s not just a matter of industry type or best practices; strategy, resources, and location make huge differences. The fact that it’s a Vietnamese restaurant or a graphic arts business or a retail shoe store doesn’t determine the natural startup level, by itself. A lot depends on where, by whom, with what strategy, and what resources.
While we don’t know it for sure ever — because even after we count the actual costs, we can always second-guess our actual spending — I do believe we can understand something like natural levels, related to the nature of the specific startup.
Marketing strategy, for example, might make a huge difference. The company planning to buy Web traffic will naturally spend much more in its early months than the company planning to depend on viral word of mouth. It’s in the plan.
So too with location, product development strategy, management team and compensation, lots of different factors. They’re all in the plan. They result in our natural startup level.
### 2. Funding or Not Funding
There’s an obvious relationship between the amount of money needed and whether or not there’s funding, and where and how you seek that funding. It’s not random, it’s related to the plan itself. Here again is the idea of a natural level, of a fit between the nature of the business startup, and its funding strategy.
It seems that you start with your own resources, and if that’s enough, you stop there too. You look at what you can borrow. And you deal with realities of friends and family (limited for most people), angel investment (for more money, but also limited by realities of investor needs, payoffs, etc.), and venture capital (available for only a few very high-end plans, with good teams, defensible markets, scalability, etc.).
### 3. Launch or Revise
Somewhere in this process is a sense of scale and reality. If the natural startup cost is \$2 million but you don’t have a proven team and a strong plan, then you don’t just raise less money, and you don’t just make do with less. No — and this is important — at that point, you have to revise your plan. You don’t just go blindly on spending money (and probably dumping it down the drain) if the money raised, or the money raisable, doesn’t match the amount the plan requires.
Revise the plan. Lower your sites. Narrow your market. Slow your projected growth rate.
Bring in a stronger team. New partners? More experienced people? Maybe a different ownership structure will help.
What’s really important is you have to jump out of a flawed assumption set and revise the plan. I’ve seen this too often: people do the plan, set the amounts, fail the funding, and then just keep going, but without the needed funding.
And that’s just not likely to work. More important, it is likely to cause you to fail and lose money.
Repetition for emphasis: you revise the plan to give it a different natural need level. You don’t just make do with less. You also do less. Otherwise, it’s not realistic. | 2,004 | 9,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.932269 |
https://www.sscexamtricks.com/2015/02/reasoning-short-cut-tricks-for-input-output-machine-questions.html | 1,718,330,047,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861520.44/warc/CC-MAIN-20240614012527-20240614042527-00030.warc.gz | 922,255,859 | 14,315 | ### REASONING SHORT CUT TRICKS for Input Output Machine Questions - sscexamtricks.com
REASONING SHORT CUT TRICKS TO QUALIFY / CRACK REASONING SECTION / SCORE MAXIMUM IN EXAM –FASTEST AND RELIABLE INPUT-OUTPUT MACHINE QUESTIONS-HOW TO SOLVE MACHINE INPUT-OUTPUT IN BANK EXAMS and Other competitive Exams
Reasoning Short Cut Tricks for Input Output questions are one of the fastest and accurate tricks used to solve such question within seconds in the exams. It is easy in understand and no more revision / practice is required once you understand the concept only one time. Input Output questions can be easily solved for SSC, Bank or other competitive exams. Reasoning Short cut Tricks for INPUT-OUTPUT is one of the mandatory part covers 5 questions in BANK or other competitive exam format which need to be solved with 2-3 minutes. These are used to check quick understanding of a candidate. INPUT-OUTPUT is very easy section to solve so must be candidate first choice to crack reasoning section and helps in getting cut-off score to crack exam.
READ ONCE- MEMORIZE FOREVER BASED TRICK
A trick named TABLE METHOD is used to solve Reasoning Input Output questions in a very short duration of time. Table Method to solve Reasoning Input Output questions is one of the best methods with 100% accuracy to solve such questions. It is quick and reliable method also.
Reasoning Short Cut Tricks for Input Output Questions- RULES FOR TABLE METHOD:
1. Check the complete arrangement step by step or all steps carefully.
2. Check the arrangement at the both ends of each step.
3. Read the provided table and check the ascending or descending order of numbers or letters.
4. Note down the reading as ascending or descending order.
5. Make the Table and Cross & Number it according to arrangement.
Question: Let input: flight 37 delay an 53 87 hour 19 by 46
Step 1: 87 flight 37 delay 53 hour 19 by 46 an
Step 2: 87 53 flight 37 delay hour 19 46 by an
Step 3: 87 53 46 flight 37 hour 19 delay by an
Step 4: 87 53 46 37 hour 19 flight delay by an
Step 5: 87 53 46 37 19 hour flight delay by an
This machine has step 5 as last step.
Applying Reasoning Short Cut Tricks Rules for Input-Output:
1. Checked the complete arrangement step by step or all steps carefully.
2. Checked the arrangement at the both ends of each step.
3. In above Input-output question, it is clearly seen that NUMBERS are DESCENDING order from Left to right and LETTERS are ASCENDING order from Right to left.
4. Mark the readings of arrangement as ascending or descending order ( as shown above in Bold ).
5. Make the Table and Cross & Number it according to arrangement shown as below Table.
How many steps will be of this input provided below?
√ I-R 2-L 4-R I-L √ 5-L 3-R 2-R 3-L 22 case 80 read 90 sir 12 jeans head 40 X X X X X X X X X X
Note: In above Table
Do not count already arranged number or letter while making the arrangement.
In 1st Row = No. of steps
I, II, III……so on = Step No.’s &
L=From Left, R=from Right
In 2nd Row = Input of machine
In 3rd Row = Elimination during output arrangement | 797 | 3,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.81739 |
http://www.mathmotivation.com/lessons/complex-numbers-bonus.html | 1,519,381,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814566.44/warc/CC-MAIN-20180223094934-20180223114934-00740.warc.gz | 487,558,420 | 1,824 | HomePage MathVideos MathLessons DeductiveReasoning SymbolicReasoning ScienceApplications Math andMoney All MathApplications How MathHelped Me
# Complex Numbers Bonus Question
Which of the following is false? Why?
1. √(-49) = 7i
2. i5 = -i
3. (1 + i)2 = 0
4. i(i + 1) + 1 = i | 89 | 281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.684293 |
http://exxamm.com/blog/Blog/14388/zxcfghfgvbnm4?Class%2012 | 1,558,554,227,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256948.48/warc/CC-MAIN-20190522183240-20190522205240-00083.warc.gz | 75,172,662 | 22,042 | Mathematics ATOMIC SPECTRA AND BOHR MODEL OF THE HYDROGEN ATOM
### Topic Covered
color{blue}{star} ATOMIC SPECTRA
color{blue}{star} BOHR MODEL OF THE HYDROGEN ATOM
### ATOMIC SPECTRA
color{blue} ✍️As we know, each element has a characteristic spectrum of radiation, which it emits. When an atomic gass or vapor is excited at low pressure, usually by passing an electric current through it, the emitted radiation has a spectrum which contains certain specific wavelengths only.
color{blue} ✍️A spectrum of this kind is termed as emission line spectrum and it consists of bright lines on a dark background. The spectrum emitted by atomic hydrogen is shown in Fig. 12.5.
color{blue} ✍️Study of emission line spectra of a material can therefore serve as a type of “fingerprint” for identification of the gas. When white light passes through a gas and we analyse the transmitted light using a spectrometer we find some dark lines in the spectrum.
color{blue} ✍️These dark lines correspond precisely to those wavelengths which were found in the emission line spectrum of the gas. This is called the absorption spectrum of the material of the gas.
color{brown}bbul("Spectral series")
color{blue} ✍️We might expect that the frequencies of the light emitted by a particular element would exhibit some regular pattern. Hydrogen is the simplest atom and therefore, has the simplest spectrum.
color{blue} ✍️In the observed spectrum, however, at first sight, there does not seem to be any resemblance of order or regularity in spectral lines. But the spacing between lines within certain sets of the hydrogen spectrum decreases in a regular way (Fig. 12.5).
color{blue} ✍️Each of these sets is called a spectral series. In 1885, the first such series was observed by a Swedish school teacher Johann Jakob Balmer in the visible region of the hydrogen spectrum. This series is called Balmer series (Fig. 12.6).
color{blue} ✍️The line with the longest wavelength, 656.3 nm in the red is called Hα; the next line with wavelength 486.1 nm in the bluegreen is called Hβ, the third line 434.1 nm in the violet is called Hγ; and so on.
color{blue} ✍️As the wavelength decreases, the lines appear closer together and are weaker in intensity. Balmer found a simple empirical formula for the observed wavelengths
color{navy}(1/lamda = R (1/(2^2) - 1/(n^2)))
......(12.5)
where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc.
The value of R is color{purple}(1.097 × 10^7 m^(–1)). This equation is also called Balmer formula.
"Paschen series:"
color{blue} ✍️Taking n = 3 in Eq. (12.5), one obtains the wavelength of the Hα line:
color{purple}(1/lamda = 1.097xx 10^7( 1/(3^2) - 1/(n^2))
color{purple}(=1.522 xx10^6m^(-1))
color{purple}(i.e., lamda= 656.3nm)
color{blue} ✍️For n = 4, one obtains the wavelength of H_β line, etc. For n = ∞, one obtains the limit of the series, at λ = 364.6 nm. This is the shortest wavelength in the Balmer series. Beyond this limit, no further distinct lines appear, instead only a faint continuous spectrum is seen. Other series of spectra for hydrogen were subsequently discovered. These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae:
color{brown}bbul("Lyman series:")
color{navy}(1/lamda = R (1/(1^2) - 1/(n^2)) \ \ \ \ n = 2,3,4)
......... (12.6)
color{brown}bbul("Paschen series:")
color{blue}(1/lamda = R (1/(3^2) - 1/(n^2)) \ \ \ \ n = 4,5,6)
......... (12.7)
color{brown}bbul("Brackett series:")
color{blue}(1/lamda = R(1/(4^2)-1/(n^2)) \ \ \ n = 5,6,7)
........... (12.8)
color{brown}bbul("Pfund series:")
color{navy}(1/lamda = R(1/(5^2)-1/(n^2)))
............ (12.9)
color{blue} ✍️The Lyman series is in the ultraviolet, and the Paschen and Brackett series are in the infrared region. The Balmer formula Eq. (12.5) may be written in terms of frequency of the light, recalling that
color{purple}(c = v lamda)
or color{purple}(1/lamda= v/c)
color{blue} ✍️Thus, Eq. (12.5) becomes
color{blue}(v = Rc (1/2^2-1/n^2))
.... (12.10)
color{blue} ✍️There are only a few elements (hydrogen, singly ionised helium, and doubly ionised lithium) whose spectra can be represented by simple formula like Eqs. (12.5) – (12.9).
color{blue} ✍️Equations (12.5) – (12.9) are useful as they give the wavelengths that hydrogen atoms radiate or absorb. However, these results are empirical and do not give any reasoning why only certain frequencies are observed in the hydrogen spectrum.
### BOHR MODEL OF THE HYDROGEN ATOM
color{blue} ✍️The model of the atom proposed by Rutherford assumes that the atom, consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates.
color{blue} ✍️However, there are some fundamental differences between the two situations. While the planetary system is held by gravitational force, the nucleus-electron system being charged objects, interact by Coulomb’s Law of force.
color{blue} ✍️We know that an object which moves in a circle is being constantly accelerated – the acceleration being centripetal in nature. According to classical electromagnetic theory, an accelerating charged particle emits radiation in the form of electromagnetic waves. The energy of an accelerating electron should therefore, continuously decrease. The electron would spiral inward and eventually fall into the nucleus (Fig. 12.7).
color{blue} ✍️Thus, such an atom can not be stable. Further, according to the classical electromagnetic theory, the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of revolution.
color{blue} ✍️As the electrons spiral inwards, their angular velocities and hence their frequencies would change continuously, and so will the frequency of the light emitted. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed. Clearly Rutherford model tells only a part of the story implying that the classical ideas are not sufficient to explain the atomic structure.
Q 3159267114
According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Class 12 Chapter 12 Example 4
Solution:
From Example 12.3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m is 2.2 × 10–6 m//s. Thus, the frequency of the electron moving around the proton is
v=v/(2pir) = (2.2xx10^6 ms^(-1))/(2pi(5.3xx10^(-11)m)
≈ 6.6 × 10^(15) Hz.
According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 10^(15) Hz.
color{blue} ✍️It was Niels Bohr who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model.
color{blue} ✍️Faced with the dilemma as discussed above, Bohr, in 1913, concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena, it could not be applied to the processes at the atomic scale. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra.
color{blue} ✍️Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are :
color{blue} {(i)} Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom.
color{blue} {(ii)} Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/(2π) where h is the Planck’s constant (= 6.6 × 10^(–34) J s).
color{blue} ✍️Thus the angular momentum (L) of the orbiting electron is quantised. That is
color{blue} {L = nh//2π}
........ (12.11)
color{blue} {(iii)} Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by
color{blue} {hν = E_i – E_f}
......(12.12)
color{blue} ✍️where E_i and E_f are the energies of the initial and final states and E_i > E_f . For a hydrogen atom, Eq. (12.4) gives the expression to determine the energies of different energy states. But then this equation requires the radius r of the electron orbit. To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The angular momentum L is given by color{purple} {L = mvr}
color{blue} ✍️Bohr’s second postulate of quantisation [Eq. (12.11)] says that the allowed values of angular momentum are integral multiples of h//2π.
color{blue}(L_n = mv_nr_n = (nh)/(2pi))
........(12.13)
color{blue} ✍️where n is an integer, r_n is the radius of nth possible orbit and n^th is the speed of moving electron in the nth orbit. The allowed orbits are numbered 1, 2, 3 ..., according to the values of n, which is called the principal quantum number of the orbit.
From Eq. (12.3), the relation between v_n and r_n is
color{purple}(v_n = e/sqrt(4piε_0mr_n))
color{blue} ✍️Combining it with Eq. (12.13), we get the following expressions for v_n and r_n,
color{blue}(v_n = 1/n (e^2)/(4piε_0) 1/(h/22pi))
..............(12.14)
and
color{blue}(r_n = ((n^2)/m) (h/(2pi))^2 (4piε_0)/(e^2))
.............. (12.15)
color{blue} ✍️Eq. (12.14) depicts that the orbital speed in the nth orbit falls by a factor of n. Using Eq. (12.15), the size of the innermost orbit (n = 1) can be obtained as
color{purple}(r_1 = (h^2ε_0)/(pime^2))
color{blue} ✍️This is called the Bohr radius, represented by the symbol a_0. Thus,
color{blue}(a_0 = (h^2ε_0)/(pime^2))
.....(12.16)
Substitution of values of h, m, ε0 and e gives color{purple}(a0 = 5.29 × 10^(–11)) m. From Eq. (12.15), it can also be seen that the radii of the orbits increase as n^2.
color{blue} ✍️The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as
color{purple}(E_n = - (e^2)/(8piε_0) (m/(n^2)) ((2pi)/h)^2 (e^2)/(4piε_0))
or
color{blue}(E_n = - (me^4)/(8n^2ε_(0)^(2)h^2))
............ (12.17)
color{blue} ✍️Substituting values, Eq. (12.17) yields
color{blue}(E_n = (2.18xx10^(-18))/((n^2) J)
........ (12.18)
color{blue} ✍️Atomic energies are often expressed in electron volts (eV) rather than joules. Since color{purple} {1 eV = 1.6 × 10^(–19) J}, Eq. (12.18) can be rewritten as
color{blue}(E_n = - (13.6)/(n^2) eV)
..... (12.19)
color{blue} ✍️The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. Energy will thus be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus (or proton in hydrogen atom).
color{blue} ✍️The derivation of Eqs. (12.17) – (12.19) involves the assumption that the electronic orbits are circular, though orbits under inverse square force are, in general elliptical. (Planets move in elliptical orbits under the inverse square gravitational force of the sun.) However, it was shown by the German physicist Arnold Sommerfeld (1868 – 1951) that, when the restriction of circular orbit is relaxed, these equations continue to hold even for elliptic orbits.
Q 3139567412
A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.
Class 12 Chapter 12 Example 5
Solution:
From Eq. (12.13), we have
m v_n rn = n+h//2π
Here m = 10 kg and r_n = 8 × 10^6 m. We have the time period T of the
circling satellite as 2 h. That is T = 7200 s.
Thus the velocity v_n = 2π rn//T.
The quantum number of the orbit of satellite
n = (2π r_n)^2 × m//(T × h).
Substituting the values,
n = (2π × 8 × 10^6 m)^2 × 10//(7200 s × 6.64 × 10^(–34) J s)
= 5.3 × 10^(45)
Note that the quantum number for the satellite motion is extremely large! In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.
### Energy levels
color{blue} ✍️The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits.
color{blue} ✍️The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a_0. The energy of this state (n = 1), E_1 is –13.6 eV.
color{blue} ✍️Therefore, the minimum energy required to free the electron from the ground state of the hydrogen atom is 13.6 eV. It is called the "ionisation energy" of the hydrogen atom.
color{blue} ✍️This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy. At room temperature, most of the hydrogen atoms are in ground state. When a hydrogen atom receives energy by processes such as electron collisions, the atom may acquire sufficient energy to raise the electron to higher energy states.
color{blue} ✍️The atom is then said to be in an excited state. From Eq. (12.19), for n = 2; the energy E_2 is –3.40 eV. It means that the energy required to excite an electron in hydrogen atom to its first excited state, is an energy equal to E_2 – E_1 = –3.40 eV – (–13.6) eV = 10.2 eV.
color{blue} ✍️Similarly, E_3 = –1.51 eV and E_3 – E_1 = 12.09 eV, or to excite the hydrogen atom from its ground state (n = 1) to second excited state (n = 3), 12.09 eV energy is required, and so on.
color{blue} ✍️From these excited states the electron can then fall back to a state of lower energy, emitting a photon in the process. Thus, as the excitation of hydrogen atom increases (that is as n increases) the value of minimum energy required to free the electron from the excited atom decreases.
color{blue} ✍️The energy level diagram for the stationary states of a hydrogen atom, computed from Eq. (12.19), is given in Fig. 12.8. The principal quantum number n labels the stationary states in the ascending order of energy.
color{blue} ✍️In this diagram, the highest energy state corresponds to n =∞ in Eq, (12.19) and as an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ∞) from the nucleus and is at rest. Observe how the energies of the excited states come closer and closer together as n increases. | 4,218 | 15,690 | {"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} | 3.390625 | 3 | CC-MAIN-2019-22 | latest | en | 0.917803 |
https://mtp.tools/converters/data-storage/kilobyte-to-megabit-calculator | 1,590,966,584,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413786.46/warc/CC-MAIN-20200531213917-20200601003917-00078.warc.gz | 464,780,155 | 5,620 | # Kilobyte to Megabit calculator (kB to Mb)
Convert kilobytes to megabits (kB to Mb) by typing the amount of kilobytes in the input field below and then clicking in the "Convert" button. If you want to convert from megabits to kilobytes, you can use our megabit to kilobyte converter.
## Formula
Formula used to convert kB to Mb:
F(x) = x / 125
For example, if you want to convert 1 kB to Mb, just replace x by 1 [kB]:
1 kB = 1 / 125 = 0.008 Mb
## Steps
1. Divide the amount of kilobytes by 125.
2. The result will be expressed in megabits.
## Kilobyte to Megabit Conversion Table
The following table will show the most common conversions for Kilobytes (kB) to Megabits (Mb):
Kilobytes (kB) Megabits (Mb)
0.001 kB 0.000008 Mb
0.01 kB 0.00008 Mb
0.1 kB 0.0008 Mb
1 kB 0.008 Mb
2 kB 0.016 Mb
3 kB 0.024 Mb
4 kB 0.032 Mb
5 kB 0.04 Mb
6 kB 0.048 Mb
7 kB 0.056 Mb
8 kB 0.064 Mb
9 kB 0.072 Mb
10 kB 0.08 Mb
20 kB 0.16 Mb
30 kB 0.24 Mb
40 kB 0.32 Mb
50 kB 0.4 Mb
60 kB 0.48 Mb
70 kB 0.56 Mb
80 kB 0.64 Mb
90 kB 0.72 Mb
100 kB 0.8 Mb
A kilobyte is a unit of measurement for digital information and computer storage. The prefix kilo (which is expressed with the letter k) is defined in the International System of Units (SI) as a multiplier of 10^3 (1 thousand). Therefore, 1 kilobyte is equal to 1,000 bytes. The symbol used to represent a kilobyte is kB.
Many people think that 1 kilobyte is equal to 1024 bytes. This is wrong because is contrary to international standards. The only unit used to represent 1024 bytes is a kibibyte.
A megabit is a unit of measurement for digital information and computer storage. The prefix mega (which is expressed with the letter M) is defined in the International System of Units (SI) as a multiplier of 10^6 (1 million). Therefore, 1 megabit is equal to 1,000,000 bits and equal to 1,000 kilobits. The symbol commonly used to represent a megabit is Mb (sometimes as Mbit).
## FAQs for Kilobyte to Megabit converter calculator
### What is Kilobyte to Megabit converter calculator?
Kilobyte to Megabit converter is a free and online calculator that converts Kilobytes to Megabits.
### How do I use Kilobyte to Megabit converter?
You just have to insert the amount of Kilobytes you want to convert and press the "Convert" button. The amount of Megabits will be outputed in the input field below the button.
### Which browsers are supported?
All mayor web browsers are supported, including Internet Explorer, Microsoft Edge, Firefox, Chrome, Safari and Opera.
### Which devices does Kilobyte to Megabit converter work on?
Kilobyte to Megabit converter calculator works in any device that supports any of the browsers mentioned before. It can be a smartphone, desktop computer, notebook, tablet, etc. | 794 | 2,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.746711 |
https://how-to-bike-2.netlify.app/how-to-size-a-road-bike-chain.html | 1,642,612,130,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00636.warc.gz | 354,155,523 | 7,493 | # How To Size A Road Bike Chain
### The type of chain you need will depend on the type of bike you are riding.
How to size a road bike chain. Consider the bike type. Methods to determine chain length method 1. Size to old chain count links if your old chain was sized correctly it can be used to determine the length of your new chain. A number of different chain widths are available to suit specific bike types such as bmx or to fit different road and mtb drivetrains e g.
It is only possible to join inner plates to outer plates. For single speed bicycles or bikes with internal gear hubs see. First consider what kind of bike you want sizing varies greatly between road and mountain bikes as well as between men s and women s models and even between. For a more thorough presentation watch how to size a bicycle chain https youtu be o0yibmdwbaw calvin runs through the two ways to size a chain in this.
This video will take you step by step through sizing a bike chain including determining whether your old chain was correctly sized sizing to an old chain and the largest cog chainring method. What is an optimal chain length for a bicycle with derailleurs. A new bicycle chain usually comes with. Chains cannot be cut to 52 1 4 53 1 8 or.
You can compare the original chain to the new chain and match the rivets. That is answered in this post. Because of this chains can only be connected at whole integral inch increments. It is often asked.
To check if it was the correct size shift the bike to the smallest sprockets and check the chain for any slack. Determining optimal single speed chain length. If you need to replace the chain on your bike determine how long the new chain needs to be. An 11 speed cassette requires a different chain to a 9 speed one the chain must be thinner in order to fit into the narrower spaces between the sprockets.
Bicycle chains consist of inner and outer plates. If you don t have the original chain or it s the wrong size use the largest cog and chainring to find the cutting point on the new chain. Chain sizing by equation.
### Fixing Stiff Bicycle Chain Links The Other Cause Of Skipping
Source : pinterest.com | 469 | 2,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-05 | latest | en | 0.921203 |
https://www.6witze.com/figures-tutoring-3-principles-steps-to-master-looking-at-arithmetic-formulas/ | 1,620,986,189,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990449.41/warc/CC-MAIN-20210514091252-20210514121252-00185.warc.gz | 629,622,187 | 9,211 | Others
Figures Tutoring – 3 Principles Steps to Master Looking at Arithmetic Formulas
Understanding how in order to read maths formulas demands a basic understanding of the formula vocabulary and how to understand formula reading patterns. Most of us will focus on the way to read Mathematical formulations and learn how this particular solution reading style can be used with formulations through different subjects (i. age. Algebra, Geometry, Hormones, Physics). Knowing how in order to read Mathematics remedies is essential for maximum understanding and easy memory call to mind.
This is my trust that you see a new pattern using reading formulations all over distinct subjects. So why will be discovering math best results across topics so critical? Students often feel like these people are learning something new each time they are usually introduced to some sort of Mathematics formula in a further type or course. Simple fact remains to be, the same procedures you make use of to read formulas inside Algebra are the just same strategies used in order to read formulas inside Trigonometry, Physics, Hormones, Economics, etc. So the key is excellence of reading treatments inside Algebra.
Step 1: Understand what a formula is. What is a new math solution? An situation (i. age. F sama dengan ma) which expresses a general truth, rule, as well as principle.
Step two: Identify and understand the basic Mathematics formula vocabulary and even use mainly because often as possible even though doing problems. A fine arithmetic educator (e. g. instructor, mentor, teacher,… ) will assist you to engage this vocabulary since you are working on your current problems. This kind of vocabulary is usually useful when reading Figures instructions, doing word issues, or solving Math troubles. Let’s define a fundamental set in place of basic Math concepts solution (equations) vocabulary words under:
Variable – a new correspondence or symbol used around numerical expressions to signify a quantity that will possess different values (i. at the. x or perhaps P)
Units – the parameters utilized to measure quantities ( i. age. length(cm, m, in, ft), mass (g, kg, lbs, etc))
Constant – the quantity having a set value that will not really change or fluctuate
Pourcentage – a range, mark, or even variable put in advance of a mysterious quantity figuring out the amount of occasions it will be increased
Operations – basic precise processes which include addition (+), subtraction (-), multiplication (*), and division (/)
Expressions-a combination a number of numbers, albhabets and precise symbols representing a amount. (i. electronic. 4, 6x, 2x+4, sin(O-90))
Equation : An equation is a new statement connected with equality involving two math expressions.
Answer – an answer to a problem (i. e. x sama dengan 5)
Step 3: Read through formulas as a complete concept as well as statement-do not SIMPLY read the letters together with emblems in a method. What am i saying? Most people help make the repeated miscalculation of reading the words inside a formula rather when compared with studying what the letters represent inside the formula. This may well appear simple, nevertheless this simple and easy step enables some sort of scholar to indulge the formula. By way of browsing the letters together with signs only, one can not online the formula along with specific words words and the reason of the method.
For instance , most people read this formula to get area connected with a circle (A = “pi”r2) just as it can be written – A equates to professional indemnity r squared. Rather than just reading the letters in addition to symbols in the formulation, most of us propose reading recipes such as a = “pi”r2 seeing that a finish thought making use of all the detailed thoughts for each letter: The spot (A) of a round is usually (=) pi increased by way of the radius (r) with the circle squared. Do you observe the solution is a comprehensive assertion or thought? For that reason, one should read recipes as a complete record (thought) as frequently as possible. It reinforces wht is the formula means found in the mind of this reader. Without a crystal clear association of Math recipes with their respective words, much more applications of those people formulas next to impossible.
Knowing the units for each volume represented in these remedies plays a good key function in solving problems, reading word problems, and remedy interpretations, nonetheless not simply reading the formulas.
Work with actions as a reference point and learn how to read through Mathematics formulas a lot more confidently. Once you master the fundamentals of formulas, you can be some sort of Learner4Life within different subjects apply Figures formulas! | 945 | 4,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-21 | latest | en | 0.89738 |
http://www.jiskha.com/members/profile/posts.cgi?name=sydney&page=8 | 1,386,836,883,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164572870/warc/CC-MAIN-20131204134252-00087-ip-10-33-133-15.ec2.internal.warc.gz | 400,561,340 | 3,407 | Thursday
December 12, 2013
# Posts by sydney
Total # Posts: 322
Geography
conduct research to examine the sphere of influence of an organisation that renders a service in a community .eg.clinic,hospital,a local high school
statistics
Is this correct: The probability of rolling a die once and getting a number greater than 3 or an odd number is 5/6.
College Chemistry
Which contains more sulfur, 35.2g CaSO4 or 35.7g NaSO4?
Chemistry
Calculate the heat released when 64.5g of steam at 119 C is converted to water at 36 C. Assume the specific heat is 4.184 J/g*C for water, the specific heat of steam is 1.99J/g*C, and the heat of vaporization is 40.79 kJ/mol for water.
FRENCH
Do Passer and Ranger have an accent at the end when they are in past tense? Or are they just Passe and Range. I added that accent on them in that activity.
FRENCH
Spot my own error numero 9 9. s'habiller bien pour sortir Sydney: Tu t'es habillée bien pour sortir? Brittany: Non, je ne suis pas habillée bien pour sortir. Is a "me" supposed to go between ne and suis?
FRENCH
From the same Activity--- Number 3. 3. ranger sa chambre. Sydney: Tu as rangée sa chambre? Sam: Non, je n'ai pas rangée sa chambre. Does sa chambre change the sentence at all? Is it supposed to be "ta" chambre? I'm asking--did you clean your roo...
FRENCH
Please check to see if I've done this activity correctly. There's just so much to cover. Reflexive verbs, vandertramp, and avoir are all in this activity. Interviewez des élèves de votre classe pour déterminer ce qu'ils ont fait pendant le week...
FRENCH
Qu'est-ce que tu fais quand tu es en retard? J'essaie de se dépêcher. Before I had it as.. Je me dépêche ---> Since it was reflexive this way. But I had to add the "try" in the sentence to make sense.
FRENCH
1. Tu te prépares pour l'école en combien de minutes? Je prépare pour l'ecole en trente minutes. 2. Est-ce que tu te rases chaque jour? How would I say.. No I do not shave ( Obviously not, since I'm a girl) Non, je ne me rase pas ? 3. Qu'es...
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Members | 627 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.557894 |
https://reverseengineering.stackexchange.com/questions/4368/experiences-from-reverse-engineers-in-detecting-recursive-calls?noredirect=1 | 1,627,964,993,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154420.77/warc/CC-MAIN-20210803030201-20210803060201-00311.warc.gz | 490,530,571 | 36,301 | Experiences from reverse engineers in detecting recursive calls
I have absolutely no experience in reversing real-world binary codes, so I wonder how the obfuscated codes prevent reversers. I doubt that the reverses always find some ways to understand what are hiden inside, even for heavy obfuscated codes, but I do not know how they think about them.
That partly comes from this question in detecting recursive calls where both two answers give a static approach: looking recursively in the functions called by the original function whether it is re-called.
In somehow theoretical manner, this approach can be bypassed if the programmer uses the continuation passing style, that is because there is no more explicit
``````call myself
``````
inside the code. The following program I have implemented to test out this idea:
``````template<typename T>
auto obf_if(bool p, T a, T b) -> T
{
T* pts[4] = { &a, &b, &a + 1, &b + 1 };
return *pts[int{ p }];
}
template<typename T>
auto obf_cmp(T a, T b) -> int
{
return obf_if<int>(a == b, 0, obf_if<int>(a < b, -1, 1));
}
using obf_strcmp_t = std::function < int(char*, char*) >;
auto h_strcmp(obf_strcmp_t func, char* str1, char* str2) -> int
{
return obf_if<int>((*str1 == *str2) && (*str1 != 0),
func(str1 + 1, str2 + 1), obf_cmp<int>(*str1, *str2));
}
using h_strcmp_t = decltype(h_strcmp);
obf_strcmp_t y_strcmp(h_strcmp_t func)
{
return std::bind(func, std::bind(y_strcmp, func),
std::placeholders::_1, std::placeholders::_2);
}
int main(int argc, char* argv[])
{
char str1[] = "ab";
char str2[] = "ac";
return y_strcmp(h_strcmp)(str1, str2);
}
``````
This is a trivial implementation of `strcmp` using the y combinator. But this piece of codes leads to the fact that there is no more direct call inside the implementation (even no conditional jump), except the first one
``````y_strcmp(h_strcmp)(str1, str2)
``````
As an amateur, I have even loaded the binary code (compiled by VS2013) in IDA and see a big mess where calls are replaced by
`````` call edx
``````
However because I write it I know how to detect this (e.g. the implicit recursive calls are detected by tracing the arguments passed into the function, the value of edx can only be one of passed arguments), and I think so do the reversers. So my question is:
Suppose that you do not know this trick, does it prevent you in understanding the binary code?
NB Because w-s has suggested that this question is only an opinion-based one, so it will be closed sooner or later, but I very appreciate if someone gives an idea.
• Strictly speaking it is opinion based question. Less strictly speaking there is dynamic analysis. Such kind of obfuscation (as almost any other obfuscation) will not prevent anyone to read this code because there is no obfuscation that can prevent reading the code at all, obfuscation just makes code reading slower. – w s May 18 '14 at 10:15
• Thanks a lot w-s, I am very agree with you. That leads me to this question because when I read some very nice approachs in code obfuscation, the question in my head is always: if the reversers know this trick, does it prevent them anymore. – Ta Thanh Dinh May 18 '14 at 10:20
• It is much more psychological question then question of reverse engineering. I know a lot of reversers that will stop reversing any code when they see any kind of obfucation but I also know some reverse engineers that will not recognize here any obfuscation and will define this code as "over-designed" and over C++ed :) – w s May 18 '14 at 10:40
• I got your joke w-s :D. I think that I should delete this psycho question before it is down-voted by many people and I lose all of my reputation points. – Ta Thanh Dinh May 18 '14 at 11:16
• This will only slow down a dedicated reverser, but because the code itself is not hiding its functionality in any way, it can be reversed given enough time. And like the answer below, there are ways to identify recursive functions, no matter how elusive the programmer may try to be. stack based traces are a good way to spot this. – gandolf May 27 '14 at 13:53 | 1,037 | 4,077 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.752948 |
https://boards.straightdope.com/t/a-totally-freaky-test/518895 | 1,600,603,351,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400197946.27/warc/CC-MAIN-20200920094130-20200920124130-00178.warc.gz | 308,282,442 | 5,932 | # A totally freaky test
Think of a country that starts with D, an animal that starts with K, and fruit that starts with O. Everything else in the text is distraction.
Me, I think of a Jackal in Djibouti eating a Lemon…
Dog in Deutschland eating a Guava?
a cat in the Dominican Republic licking a tomato?
Ice bear in Dubai ravishing raspberries.
Originally, I had a Koi in Denmark, but couldn’t think of any fruit with i…
The sum of the digits of any number produced by multiplying 1-10 by 9 always equals nine. I think I learned that in third grade.
Once you know that, it isn’t that “freaky”.
Not that freaky. The digits of any number where 9 is a factor always add up to 9, so the country always starts with D. Everything else is a small matter of psychology.
Err. Ravaging, even. :o
Hey, some of those raspberries can be pretty damn sexy.
I had a camel in the Dominican Republic eating a lime.
They shoulda specified a whole number from 1 to 10.
I couldn’t finish it. Does anyone know a fruit that starts with a Y.
You want creepy? Try the Mystic Ball. It took me quite some time to figure it out. (Of course, now some Doper is going to make me feel stupid by saying how obvious it is).
Yangmei.
Koala in Denmark eating an apple. <shrug>
You’re not the only one. I’m actually amazed.
I had an Ibex in Dubai eating a… What fruit starts with X?
Well, this might have worked out if I could add. Because 2 + 7 is so totally 10, you guys.
I ended up with an amoeba in Ethiopia eating an avocado.
Xigua. (It’s very much like a watermelon). | 394 | 1,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.9569 |
https://www.studypool.com/discuss/282239/how-do-you-multiply-monomials?free | 1,495,524,714,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607591.53/warc/CC-MAIN-20170523064440-20170523084440-00381.warc.gz | 946,055,759 | 14,314 | Time remaining:
##### how do you multiply monomials
Mathematics Tutor: None Selected Time limit: 1 Day
I dont understand how to do them.
Nov 17th, 2014
1y *1 y = 2y^2
y^2 * y^3 = y^5
2y^2 * Y^3 = 2Y^5
When you multiply a monomial, you add the exponents N^x
x is the exponent
When you multiply a monomial , the coefficients times just like
Nov 17th, 2014
...
Nov 17th, 2014
...
Nov 17th, 2014
May 23rd, 2017
check_circle | 156 | 432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-22 | longest | en | 0.80641 |
https://www.texasgateway.org/resource-index/?f%5B0%5D=im_field_resource_subject%3A1&f%5B1%5D=im_field_resource_subject%3A27551&f%5B2%5D=sm_field_resource_grade_range%3A7&f%5B3%5D=im_field_resource_subject%3A4&f%5B4%5D=im_field_resource_subject%3A2&%3Bf%5B1%5D=sm_field_resource_grade_range%3A4&%3Bamp%3Bf%5B1%5D=im_field_resource_subject%3A3&%3Bamp%3Bf%5B2%5D=bundle%3Aresource&%3Bamp%3Bf%5B3%5D=sm_field_resource_grade_range%3A9&%3Bamp%3Bf%5B4%5D=sm_field_resource_grade_range%3A3&%3Bamp%3Bf%5B5%5D=sm_field_resource_audience%3Astudents | 1,569,078,865,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00337.warc.gz | 1,029,342,784 | 15,830 | • Resource ID: S8M3L2
• Grade Range: 6–8
• Subject: Science
### Gravity
Using interactives, students will demonstrate that gravity is the force that governs the motion of our solar system.
• Resource ID: M7M2L17
• Grade Range: 7
• Subject: Math
### Using Theoretical and Experimental Probability to Make Predictions
Given an event to simulate, the student will use theoretical probabilities and experimental results to make predictions and decisions.
• Resource ID: R4SCI0001
• Grade Range: 6–12
• Subject: Science
### Potential and Kinetic Energy
This resource provides Tier I instruction ideas for Grade 6+ science teachers in the area of potential and kinetic energy.
• Resource ID: TEA001
• Grade Range: K–8
• Subject: Math
### TXRCFP: Texas Response to Curriculum Focal Points for K-8 Mathematics Revised 2013
The Texas Response to Curriculum Focal Points Revised 2013 was created from the 2012 revision of the TEKS as a guide for implementation of effective mathematics instruction by identifying critical areas of content at each grade level.
• Resource ID: R4SCI0011
• Grade Range: 6–12
• Subject: Science
### Introduction to Plate Tectonics
This resource is intended to use for Tier I classroom instruction.
• Resource ID: R4SCI0012
• Grade Range: 7–12
• Subject: Science
### Human Impact
This resource can be used, in conjunction with best practices, for Tier I classroom instruction.
• Resource ID: Revised_Math_TEKS_VA
• Grade Range: K–12
• Subject: Math
### Vertical Alignment Charts for Revised Mathematics TEKS
This resource provides vertical alignment charts for the revised mathematics TEKS.
• Resource ID: TEKS12_MATH_07_003
• Grade Range: 7
• Subject: Math
### Exploring the Ratio of Circumference to Diameter
This activity provides an opportunity for students to explore the ratio of the circumference of its circle to the length of its diameter in order to generalize the ratio for pi.
• Resource ID: R4SCI0028
• Grade Range: 7
• Subject: Science
### Food Chains, Food Webs, and Energy Pyramids
This resource provides flexible alternate or additional learning opportunities for students to diagram the flow of energy through living systems, Seventh Grade Science TEKS (7)(5)(C).
• Resource ID: GM3L2A
• Grade Range: 7
• Subject: Math
### Creating Nets for Three-Dimensional Figures
Given nets for three-dimensional figures, the student will apply the formulas for the total and lateral surface area of three-dimensional figures to solve problems using appropriate units of measure.
• Resource ID: MATH_IMG_001
• Grade Range: K–8
• Subject: Math
### Interactive Math Glossary
The Interactive Math Glossary is provided by the Texas Education Agency to help teachers explore and understand mathematics vocabulary.
• Resource ID: M7M2L16
• Grade Range: 7
• Subject: Math
### Finding the Probabilities of Dependent and Independent Events
Given problem situations, the student will find the probability of the dependent and independent events.
• Resource ID: M8M5L5
• Grade Range: 5–7
• Subject: Math
### Selecting and Using Representations for Collected Data
Given a variety of data (including line plots, line graphs, stem and leaf plots, circle graphs, bar graphs, box and whisker plots, histograms, and Venn diagrams), the student will select and use an appropriate representation for presenting and displaying relationships among the collected data with and without the use of technology
• Resource ID: M7M2L20
• Grade Range: 7
• Subject: Math
### Recognizing Misuses of Graphical or Numerical Information
Given a problem situation, the student will analyze data presented in graphical or tabular form by evaluating the predictions and conclusions based on the information given.
• Resource ID: R4SCI0038
• Grade Range: 7–12
• Subject: Science
### Dichotomous Keys Murder Mystery
A resource for the instruction of (7)(11)(A) involving the use of dichotomous keys for identification.
• Resource ID: M7M2L6
• Grade Range: 7
• Subject: Math
### Using Multiplication by a Constant Factor
Given problems involving proportional relationships, the student will use multiplication by a constant factor to solve the problems.
• Resource ID: M7M2L2
• Grade Range: 7
• Subject: Math
### Predicting, Finding, and Justifying Data from a Table
Given data in table form, the student will use the data table to interpret solutions to problems.
• Resource ID: M7M2L4b
• Grade Range: 7
• Subject: Math
### Predicting, Finding, and Justifying Data from Verbal Descriptions
Given data in a verbal description, the student will use equations and tables to solve and interpret solutions to problems.
• Resource ID: M8M2L6*
• Grade Range: 6–8
• Subject: Math
### Determining Slopes from Equations, Graphs, and Tables
Given algebraic, tabular, and graphical representations of linear functions, the student will determine the slope of the relationship from each of the representations.
• Resource ID: S8M1L2
• Grade Range: 6–8
• Subject: Science
### Matter and Energy - Elements versus Compounds
Given illustrations or descriptions, students will differentiate between elements and compounds. | 1,200 | 5,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2019-39 | latest | en | 0.767304 |
https://studylib.net/doc/7526711/ch231-wq2-2007 | 1,575,859,664,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00401.warc.gz | 564,284,872 | 41,715 | # CH231-WQ2-2007
```Chemistry 231
Weekly Quiz (?) #2
G. Marangoni
October 2, 2007.
Answer the following questions in the time allotted. There are no tricks on this quiz, so don’t
look for any. Marks are indicated in the [
]
Name:
ID# :
1. 1.00 mole of a diatomic ideal gas (Cv,m = 5/2 R) is expanded from 30.0 L and 300 K to 90.0
L. Calculate final temperature and pressure of the gas and the values of U, H and q and w
if the expansion is carried out [10 marks]
a) isothermally and reversibly;
A.
U = H = 0
Isothermal so T2 = 300 K
And P2 = 1/3 P1 = 0.831 bar /3 = 0.277 bar
w = -nRT ln (V2/V1)
= -1.00 mole x 8.314 J/ (K mole) x 300 K x ln (90.00 L / 30.00 L)
= -2740 J = -2.74 kJ
q = +2.74 kJ
B.
Q = 0 and U = w
Need to calculate the final temperature
V
T2 T1 x 1
V2
R/C v, m
30.00
T2 300 K x
90.00
= 193 K
2/5
U = 5/2 x 8.314 J/(K mol) (-107 K)
= -2224 J
= -2.22 kJ
w = -2.22 kJ
H = 7/2 x 8.314 J/(K mol) (-107 K)
= -3114 J
= -3.11 kJ
P2
Lbar
x193 K
nRT2 1.00 mol x 0.08314
K mol
V2
90.00 L
0.178 bar
``` | 544 | 1,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.731617 |
https://dev.to/eteimz/find-closest-value-in-binary-search-tree-30li | 1,642,616,013,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00568.warc.gz | 269,164,814 | 27,545 | ## DEV Community is a community of 785,073 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Youdiowei Eteimorde
Posted on
# Find closest value in Binary Search Tree
Binary Search Tree is a sorted binary tree meaning given a root node all the values to the left are less than it and all the values to the right are greater it. This property enables Binary Search to be performed on the tree.
Let exploit this capability, Given a target value let's find a number that is the closest to it in the tree. First we create our BST node which has a value and two pointers. One to the right and one to the left.
``````# A BST node
class BST:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
``````
Now we create our tree by attaching our nodes together
``````# 7 <-- Root Node
# 5 12 <-- Depth One
# 3 6 9 15 <-- Depth Two
# 1 4 8 10 13 17 <-- Depth Three
# Create the tree
tree = BST(7)
# depth one
tree.right = BST(5)
tree.left = BST(12)
# Right node --> depth two
tree.right.right = BST(15)
tree.right.left = BST(9)
# Right node --> depth three (right node)
tree.right.right.right = BST(17)
tree.right.right.left = BST(13)
# Right node --> depth three (left node)
tree.right.left.right = BST(10)
tree.right.right.left = BST(8)
# Left node --> depth two
tree.left.right = BST(6)
tree.left.left = BST(3)
# Left node --> depth three (left node)
tree.left.left.right = BST(4)
tree.right.right.left = BST(1)
``````
The Psuedo Code below is the solution to the problem.
Pseudo Code:
1. Assign the closest value to the root node.
2. Check if |target - closest| > |target - tree's value|.
• If yes assign the tree's value as the closest value
• If no the continue using the closest value
3. Check if the target's value > current tree's value
• If yes Go to the right sub tree.
• If no move on.
4. Check if the target's value < current tree's value
• If yes Go to the left sub tree.
• If no move on.
5. Check if there is are no sub trees left
• If yes return current closest value
• If no go back to step 2
The Pseudo Code above can be implemented recursively or iteratively
``````# Recursive solution
closest = tree.value # Initial closest value is the root's value
def findClosestValueInBstRecursive(tree, target, closest):
if tree is None:
return closest
if abs(target - closest) > abs(target - tree.value):
closest = tree.value
if target < tree.value:
return findClosestValueInBstRecursive(tree.left, target, closest)
elif target > tree.value:
return findClosestValueInBstRecursive(tree.right, target, closest)
else:
return closest
# Iterative solution
def findClosestValueInBstIterative(tree, target, closest):
currentNode = tree
while currentNode is not None:
if abs(target - closest) > abs(target - currentNode.value):
closest = currentNode.value
if target < currentNode.value:
currentNode = currentNode.left
elif target > currentNode.value:
currentNode = currentNode.right
else:
break
return closest
``````
Now time to test both solutions given the value of 14 find its closest value within the tree defined earlier.
``````tar = 14
findClosestValueInBstRecursive(tree, tar, closest)
# Output: 15
tar = 14
findClosestValueInBstIterative(tree, tar, closest)
#Output: 15
``````
Both solutions arrived at the same answer 15 which just happens to be the closest value in the tree. | 894 | 3,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | longest | en | 0.749952 |
http://openstudy.com/updates/4f3992d3e4b0fc0c1a0e362a | 1,394,591,683,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021097827/warc/CC-MAIN-20140305120457-00047-ip-10-183-142-35.ec2.internal.warc.gz | 140,187,531 | 8,344 | ## BLITZ506 What must be true about the slopes of two perpendicular lines, neither of which is vertical? 2 years ago 2 years ago
1. ChrisV
they must equal each other
2. ChrisV
i mean be inverts of each other
3. ChrisV
parallel equals and perpendicular inverts
4. ChrisV
so if one line has a slope of 3 the perpendicular line with have a slope of 1/3
5. Calcmathlete
It's the negative reciprocal, so if it's 4, it'd be -1/4.
6. ChrisV
yea i forgot the term :oP, been forever
7. ChrisV
what he said | 151 | 510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2014-10 | latest | en | 0.887072 |
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# Basic pointers question
P: n/a I'm trying to understand pointers in a little more detail, and have written a test program (reproduced below) to experiment with passing pointers between functions. Thinking only about the variable x in main, I would expect the following to happen: - &x is passed to func_one, so int* p1x is pointing to x (int* p1x = &x) - func_one then dereferences p1x to increment x; at this point x = 1 - func_one passes p1x to func_two, so int* p2x = p1x and pointers p2x and p1x both point to x - func_two then dereferences p2x to increment x; at this point x = 2 - control passes back to main() and the value of x is printed Therefore, I would expect this program to produce the output "x = 2; y = 2; c = 1", but it doesn't. Can anyone provide any insight into why this is the case? Thanks in advance Simon --code follows -- #include void func_two (int* p2x, int* p2y) { *px ++; *py ++; } void func_one (int* p1x, int* p1y, int* p1c) { *pc ++; *px ++; *py ++; func_two (px, py); } int main (void) { int x, y, c, d; x = y = c = 0; func_one (&x, &y, &c); printf("x = %d; y = %d; c = %d", x, y, c); d = getchar(); return 0; } Nov 17 '06 #1
6 Replies
P: n/a Simon said: I'm trying to understand pointers in a little more detail, and have written a test program (reproduced below) to experiment with passing pointers between functions. Thinking only about the variable x in main, I would expect the following to happen: - &x is passed to func_one, so int* p1x is pointing to x (int* p1x = &x) - func_one then dereferences p1x to increment x; at this point x = 1 Here is your mistake. *p1x++ does not increment x (the int pointed to by p1x)! This is what happens. Firstly, we need to evaluate p1x++ (because the grammar says so). The result of this evaluation is the prior value of p1x (because ++ is POST-increment), and the side effect is to increment p1x at some time before the next sequence point. The result of the evaluation, as I say, is the /prior/ value of p1x, which is of course still a pointer to your original int. The * is then applied to this pointer value, yielding the value of x, with which you do nothing. If you want to increment x through p1x, you can do this: ++*p1x; or this: (*p1x)++; -- Richard Heathfield "Usenet is a strange place" - dmr 29/7/1999 http://www.cpax.org.uk email: normal service will be restored as soon as possible. Please do not adjust your email clients. Nov 17 '06 #2
P: n/a Richard Heathfield wrote: Here is your mistake. *p1x++ does not increment x (the int pointed to by p1x)! Argh!!! - I new it was something simple! I forgot the relative binding strengths of ++ and *. Incidentally - there was another mistake in the code above which was due to me mis-typing it into Google Groups. All now works perfectly - thanks a lot for your help! Simon Nov 17 '06 #3
P: n/a Simon wrote: I'm trying to understand pointers in a little more detail, There is hardly any detail to know. Really. void func_two (int* p2x, int* p2y) { *px ++; Dereference `px` and throw the value away. Increment `px`. I suspect you meant `(*px)++`, which I would write as `*px += 1` because I think that makes what's going on clearer. -- Chris "hantwig efferko VOOM!" Dollin "- born in the lab under strict supervision -", - Magenta, /Genetesis/ Nov 17 '06 #4
P: n/a Richard Heathfield wrote: Simon said: This is what happens. Firstly, we need to evaluate p1x++ (because the grammar says so). (*p1x)++; I would request to kindly elaborate on this. I mean to say how from the C grammar we can determine this ? In some of the posts, that I read earlier in this newsgroup I came to know that C grammar does not specify any precedence. So, using the C grammar how post decrement operator is evaluated before the unary (pointer) operator. I was looking at the lex and yacc grammar of C, which I downloaded from the following links: http://www.lysator.liu.se/c/ANSI-C-grammar-l.html (lex file) http://www.lysator.liu.se/c/ANSI-C-g...ml#declaration (yacc file) Can you please explain how the expression (*p1x++) will be evaluated using the lex and yacc files at the above mentioned links ? thanks a lot for any help in advance .... Nov 17 '06 #5
P: n/a ju**********@yahoo.co.in wrote: > Richard Heathfield wrote: >Simon said:This is what happens. Firstly, we need to evaluate p1x++ (because thegrammar says so). (*p1x)++; I would request to kindly elaborate on this. I mean to say how from the C grammar we can determine this ? We look at the productions of the grammar and the meaning of the constructs. In some of the posts, that I read earlier in this newsgroup I came to know that C grammar does not specify any precedence. More accurately, the C grammar does not use the idea of "precedence" to disambiguate its expression grammar. It just uses an unambiguous grammar. In the expression `*p++`, there are only a few productions of interest. Here is a (strongly selected) view: primary-expression: identifier postfix-expression: postfix-expression ++ primary-expression unary-expression: unary-operator postfix-expression unary-operator: * expression: ... eventually, unary-expression ... We see from this that `*p++` /must/ be parsed as the unary-operator `*` applied to the postfix-expression `p++`. There's no other way to do it (the other productions of the grammar don't help). So, using the C grammar how post decrement operator is evaluated before the unary (pointer) operator. The evaluation rules are different from the grammar. The grammar says how expressions are /structured/. The evaluation rules tell you how those structures are /evaluated/. They're in the Standard, but I expect they're not in the gramamr files you downloaded, since they're not part of the grammar. To evaluate `*p++`, we evaluate the unary-expression with operator `*` and operand `p++`, because that's how it is parsed. To evaluate `*E`, we evaluate `E` and then dereference the result. Here `E` is `p++`. To evaluate `X++`, we get the lvalue `X`, deliver its current value, and note that we must increment `X` by the next sequence point. So we return the value of `p` (which will be dereferenced by the `*`) and will eventually increment in. In the original post, `*p++` is the full expression forming the body of the expression-statement `*p++;`, which supplies the sequence point; after the semicolon, `p` will have been incrememented. -- Chris "hantwig efferko VOOM!" Dollin "Our future looks secure, but it's all out of our hands" - Magenta, /Man and Machine/ Nov 17 '06 #6
P: n/a In the expression `*p++`, there are only a few productions of interest. Here is a (strongly selected) view: primary-expression: identifier postfix-expression: postfix-expression ++ primary-expression unary-expression: unary-operator postfix-expression unary-operator: * expression: ... eventually, unary-expression ... We see from this that `*p++` /must/ be parsed as the unary-operator `*` applied to the postfix-expression `p++`. There's no other way to do it (the other productions of the grammar don't help). So, using the C grammar how post decrement operator is evaluated before the unary (pointer) operator.The evaluation rules are different from the grammar. The grammar says how expressions are /structured/. The evaluation rules tell you how those structures are /evaluated/. They're in the Standard, but I expect they're not in the gramamr files you downloaded, since they're not part of the grammar. To evaluate `*p++`, we evaluate the unary-expression with operator `*` and operand `p++`, because that's how it is parsed. To evaluate `*E`, we evaluate `E` and then dereference the result. Here `E` is `p++`. To evaluate `X++`, we get the lvalue `X`, deliver its current value, and note that we must increment `X` by the next sequence point. So we return the value of `p` (which will be dereferenced by the `*`) and will eventually increment in. In the original post, `*p++` is the full expression forming the body of the expression-statement `*p++;`, which supplies the sequence point; after the semicolon, `p` will have been incrememented. Thanks Chris for the nice explanation. thanks a lot. Nov 17 '06 #7
### This discussion thread is closed
Replies have been disabled for this discussion. | 2,160 | 8,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.888124 |
http://math.stackexchange.com/questions/240819/whats-the-name-of-the-theoremmedian-of-right-triangle-hypotenuse-is-always-hal/240837 | 1,455,110,590,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701159376.39/warc/CC-MAIN-20160205193919-00318-ip-10-236-182-209.ec2.internal.warc.gz | 151,959,890 | 20,171 | # what's the name of the theorem:median of right-triangle hypotenuse is always half of it
This question is related to one of my previous questions.
The answer to that question included a theorem: "The median on the hypotenuse of a right triangle equals one-half the hypotenuse".
When I wrote the answer out and showed it a friend of mine, he basically asked me how I knew that the theorem was true, and if the theorem had a name.
So, my question:
-Does this theorem have a name?
-If not, what would be the best way to describe it during a math test? Or is it better to write out the full prove every time?
-
I independently proved it in fifth grade and my math teacher organized a column in the school newspaper which referred to the theorem using my surname. – user02138 Nov 19 '12 at 19:39
If you inscribe a right angled triangle in a circle, namely it's circumcircle, the median is the same as the radius of the circle and the hypotenuse is the diameter. – Gautam Shenoy Nov 19 '12 at 19:47
One could call it the converse of the Theorem of Thales. – André Nicolas Nov 19 '12 at 20:02
## 4 Answers
Here is a proof without words:
-
If you are interested in seeing proofs: see these proofs. You'll actually find two proofs (and illustrations) of the following theorems:
1. If a triangle is a right triangle, the median drawn to the hypotenuse has the measure half the hypotenuse (from which it follows that the median drawn to the hypotenuse divides the triangle in two isosceles triangles); and
2. If in a triangle a median has the measure half the length of the side it is drawn, then the triangle is a right triangle.
The proofs are not at all elaborate, and utilize properties you already know, so they are easy enough to reconstruct, if necessity dictates that you do so. Understanding "why" these theorems hold is the important point; correctly "naming" them is a less important.
-
Very nice guidance +1 – Amzoti May 11 '13 at 0:39
$\small{+1}^{\Large{\infty}}$ – S. Snape Jun 9 '13 at 12:11
@Babak: you're worth ${\small\infty}^{\Large\infty}$ – amWhy Jun 9 '13 at 14:21
You could call it a special case of Apollonius's theorem, or of the parallelogram law.
-
This is exactly how I remember it - as a special case of Appolonius theorem :-) – TenaliRaman Nov 19 '12 at 19:51
In a circle, a triangle which has for side a diameter is rectangle. This means the center of the circle is the mean point of the diameter, and the median is a radius of the circle.
You can prove this as a corollary to the fact that inscribed angles are half of the center ones. So I guess that is a corollary to this corollary.
- | 678 | 2,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2016-07 | longest | en | 0.941115 |
https://www.studypool.com/discuss/1282195/Need-calculus-help-to-complete-the-following-question-about-a-2x2-table-show-all-work-?free | 1,505,933,734,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687428.60/warc/CC-MAIN-20170920175850-20170920195850-00152.warc.gz | 858,695,764 | 14,389 | ##### Need calculus help to complete the following question about a 2x2 table (show all work.)
label Calculus
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Nov 20th, 2015
.
(d) Here Alice is fixed on 1 hat, meaning that we must restrict our choices to the top 2 squares. Of those, Bob's optimal choice lies on the upper right square, where he makes 2 hats.
(e) The process here is the same as before, but now since Alice is fixed on two hats, we focus on the bottom squares. The value where B is highest is on the bottom right square, and would be Bob's best response.
(f) Now we flip the scenarios. If Bob is fixed on making 1 hat, then we focus on the left column. Alex's optimal choice is located on the bottom left square.
For when Bob makes 2 hats, we look to the right column. Alice's optimal choice is now located on the bottom right sqaure.
(g) Regardless of who chooses first, the other is going to fall to the sqaure where they can make 2 hats.. What this means is that in the end they will decide upon the bottom right square,. Here they both will be maximizing their profit since both are able to sell two hats rather than one.
Nov 20th, 2015
j
Dec 19th, 2015
...
Nov 20th, 2015
...
Nov 20th, 2015
Sep 20th, 2017
check_circle | 330 | 1,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.922013 |
https://ngss.nsta.org/Resource.aspx?ResourceID=819 | 1,680,029,837,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00371.warc.gz | 484,376,851 | 17,684 | # Integrated STEM Through Tumblewing Gliders
Contributor
Scott R. Bartholomew
Type Category
Instructional Materials
Types
Activity , Article , Lesson/Lesson Plan
Note
This resource, vetted by NSTA curators, is provided to teachers along with suggested modifications to make it more in line with the vision of the NGSS. While not considered to be "fully aligned," the resources and expert recommendations provide teachers with concrete examples and expert guidance using the EQuIP rubric to adapted existing resources. Read more here.
### Reviews
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This is the correct link: https://polytechnic.purdue.edu/sites/default/files/files/STEMlessonplan-TumblewingChallenge1.pdf
## Description
Students observe and measure a tumblewing glider's motion, conducting a series of investigations to test the effect of changes to the glider's structural features on the flight path and duration of their gliders. The data collected provides evidence to support their own prototype's design solution and predicted future motion.
Intended Audience
Educator
Educational Level
• Upper Elementary
Language
English
Access Restrictions
Free access - The right to view and/or download material without financial, registration, or excessive advertising barriers.
#### Performance Expectations
3-5-ETS1-3 Plan and carry out fair tests in which variables are controlled and failure points are considered to identify aspects of a model or prototype that can be improved.
Clarification Statement: none
Assessment Boundary: none
This resource is explicitly designed to build towards this performance expectation.
Students will conduct multiple trials to test how the width of the glider, length of the glider, the side fold lengths and the end fold lengths affect the flight of the glider. The material used to construct the glider and the procedure for launching and powering the tumblewing are controlled and stated in the criteria and constraints. To best meet this performance expectation, students could identify, which variables could be tested once an initial prototype is built and tested.
3-PS2-2 Make observations and/or measurements of an object’s motion to provide evidence that a pattern can be used to predict future motion.
Clarification Statement: Examples of motion with a predictable pattern could include a child swinging in a swing, a ball rolling back and forth in a bowl, and two children on a see-saw.
Assessment Boundary: Assessment does not include technical terms such as period and frequency.
This resource appears to be designed to build towards this performance expectation, though the resource developer has not explicitly stated so.
Students test the following variables: width of the glider, length of the glider, the side fold lengths and the end fold lengths. Students make observations and measurements of the glider’s motion as they conduct their trials. Students will then analyze the data to make an argument for which combination of glider design will produce the longest flight for their tumblewing glider.
#### Science and Engineering Practices
This resource appears to be designed to build towards this science and engineering practice, though the resource developer has not explicitly stated so.
Students will work in teams to test the structural features of the glider. They will test one feature at a time, conducting four trials for each feature investigated. To more explicitly align to this practice it is recommended that students collaboratively identify the variables that could be tested based on the performance of the initial prototype, and plan how the data will be collected.
This resource is explicitly designed to build towards this science and engineering practice.
Once the testing of the glider's structural features are concluded students will make a claim as to which combination of variables will produce the longest flight and supporting their claim with evidence. They will then create and test a prototype based on this claim. To fully meet criteria, the glider must fly 9.1 meters. It is recommended that students then discuss the merits of their design solution, citing collected evidence based from the flight trials.
#### Disciplinary Core Ideas
This resource is explicitly designed to build towards this disciplinary core idea.
Investigations of how the measurement of four structural features could affect the flight of the glider will provide data that can improve its performance from that of the original prototype.
This resource is explicitly designed to build towards this disciplinary core idea.
Students will analyze the flight of their glider as they test the different variables that affect its motion to identify the measurements of each structural feature that will enable the longest flight.
#### Crosscutting Concepts
This resource appears to be designed to build towards this crosscutting concept, though the resource developer has not explicitly stated so. | 955 | 5,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.875801 |
https://web2.0calc.com/questions/did-i-solve-it-correctly | 1,585,676,169,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370502513.35/warc/CC-MAIN-20200331150854-20200331180854-00334.warc.gz | 783,826,778 | 6,722 | +0
# did i solve it correctly?
0
193
2
+1656
On January 22,2003 the distance between Sun and Pioneer 10 was 21,275,527,680,000m=2.12*10^13m
now on january 2009 the pioneer 10 has alreay covered 6 yrs of time
so , 6yrs= 1.892*10^8sec
and S=V.t
the relative velocity V=12.24Km/s and t=1.892*10^8sec
then in 6 years the pioneer 10 has covered distance d=12.24*1.892*10^8Km=23.1508080*10^8km=231508080000m
so overall distance=21275527680000m + 2315080800000m=.2150*10^14m
Aldebaran was in the distance 6.15*10^7m
then in 6 years=1.892*10^8sec the distance will be 6.15*10^7*1.892*10^8m=1.16358*10^16m
in 2009 the distance between aldebaran and pioneer 10 is=1.16358*10^16-.2150*10^14m=1.1614*10^16m
the speed of pioneer= 12.24km/s=12240m/s
so pioneer covers 12240m in one sec
pioneer covers 1m in 1/12240sec
then pioneer covers 1.1614*10^16m in 1.1614*10^16*1/12240sec=.0000947*10^16sec=3.029*10^-12yr
Sep 26, 2019
#1
+2833
+1
Does your teacher check for the exact answer? If you are a hundredth of a thousandth off, does your teacher consider it wrong?
Your answers and solution steps seem accurate, so you should be fine.
You can see what other people say though.
Sep 26, 2019
#2
+2
Here is how long it should take Pioneer 10 spacesraft to reach the star Aldebaran. According to Wikipedia, the distance to Aldebaran is ~65.3 light years from the Sun.
The speed of light is: 299,792.458 km/second.
The speed of Pioneer 10 is 12.24 km/second.
299,792.458 / 12.24 =24,492.85 times is the speed of light faster than the speed of the spacecraft.
Therefore, it will take Pioneer 10:
24,492.85 x 65.3 =1,599,383 Earth years to reach Aldebaran. Make sure that your final answer agrees roughly with this number.
Sep 27, 2019
edited by Guest Sep 27, 2019 | 650 | 1,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-16 | latest | en | 0.864676 |
https://hackage.haskell.org/package/lens-4.17/docs/Control-Lens-Internal-Level.html | 1,539,658,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509960.34/warc/CC-MAIN-20181016010149-20181016031649-00204.warc.gz | 692,823,354 | 5,401 | lens-4.17: Lenses, Folds and Traversals
Control.Lens.Internal.Level
Contents
Description
This module provides implementation details of the combinators in Control.Lens.Level, which provides for the breadth-first Traversal of an arbitrary Traversal.
Synopsis
# Levels
data Level i a Source #
This data type represents a path-compressed copy of one level of a source data structure. We can safely use path-compression because we know the depth of the tree.
Path compression is performed by viewing a Level as a PATRICIA trie of the paths into the structure to leaves at a given depth, similar in many ways to a IntMap, but unlike a regular PATRICIA trie we do not need to store the mask bits merely the depth of the fork.
One invariant of this structure is that underneath a Two node you will not find any Zero nodes, so Zero can only occur at the root.
Constructors
Two !Word !(Level i a) !(Level i a) One i a Zero
Instances
Source # Instance detailsDefined in Control.Lens.Indexed Methodsitraverse :: Applicative f => (i -> a -> f b) -> Level i a -> f (Level i b) Source #itraversed :: (Indexable i p, Applicative f) => p a (f b) -> Level i a -> f (Level i b) Source # Source # Instance detailsDefined in Control.Lens.Indexed MethodsifoldMap :: Monoid m => (i -> a -> m) -> Level i a -> m Source #ifolded :: (Indexable i p, Contravariant f, Applicative f) => p a (f a) -> Level i a -> f (Level i a) Source #ifoldr :: (i -> a -> b -> b) -> b -> Level i a -> b Source #ifoldl :: (i -> b -> a -> b) -> b -> Level i a -> b Source #ifoldr' :: (i -> a -> b -> b) -> b -> Level i a -> b Source #ifoldl' :: (i -> b -> a -> b) -> b -> Level i a -> b Source # Source # Instance detailsDefined in Control.Lens.Indexed Methodsimap :: (i -> a -> b) -> Level i a -> Level i b Source #imapped :: (Indexable i p, Settable f) => p a (f b) -> Level i a -> f (Level i b) Source # Functor (Level i) Source # Instance detailsDefined in Control.Lens.Internal.Level Methodsfmap :: (a -> b) -> Level i a -> Level i b #(<$) :: a -> Level i b -> Level i a # Source # Instance detailsDefined in Control.Lens.Internal.Level Methodsfold :: Monoid m => Level i m -> m #foldMap :: Monoid m => (a -> m) -> Level i a -> m #foldr :: (a -> b -> b) -> b -> Level i a -> b #foldr' :: (a -> b -> b) -> b -> Level i a -> b #foldl :: (b -> a -> b) -> b -> Level i a -> b #foldl' :: (b -> a -> b) -> b -> Level i a -> b #foldr1 :: (a -> a -> a) -> Level i a -> a #foldl1 :: (a -> a -> a) -> Level i a -> a #toList :: Level i a -> [a] #null :: Level i a -> Bool #length :: Level i a -> Int #elem :: Eq a => a -> Level i a -> Bool #maximum :: Ord a => Level i a -> a #minimum :: Ord a => Level i a -> a #sum :: Num a => Level i a -> a #product :: Num a => Level i a -> a # Source # Instance detailsDefined in Control.Lens.Internal.Level Methodstraverse :: Applicative f => (a -> f b) -> Level i a -> f (Level i b) #sequenceA :: Applicative f => Level i (f a) -> f (Level i a) #mapM :: Monad m => (a -> m b) -> Level i a -> m (Level i b) #sequence :: Monad m => Level i (m a) -> m (Level i a) # (Eq i, Eq a) => Eq (Level i a) Source # Instance detailsDefined in Control.Lens.Internal.Level Methods(==) :: Level i a -> Level i a -> Bool #(/=) :: Level i a -> Level i a -> Bool # (Ord i, Ord a) => Ord (Level i a) Source # Instance detailsDefined in Control.Lens.Internal.Level Methodscompare :: Level i a -> Level i a -> Ordering #(<) :: Level i a -> Level i a -> Bool #(<=) :: Level i a -> Level i a -> Bool #(>) :: Level i a -> Level i a -> Bool #(>=) :: Level i a -> Level i a -> Bool #max :: Level i a -> Level i a -> Level i a #min :: Level i a -> Level i a -> Level i a # (Read i, Read a) => Read (Level i a) Source # Instance detailsDefined in Control.Lens.Internal.Level MethodsreadsPrec :: Int -> ReadS (Level i a) #readList :: ReadS [Level i a] #readPrec :: ReadPrec (Level i a) #readListPrec :: ReadPrec [Level i a] # (Show i, Show a) => Show (Level i a) Source # Instance detailsDefined in Control.Lens.Internal.Level MethodsshowsPrec :: Int -> Level i a -> ShowS #show :: Level i a -> String #showList :: [Level i a] -> ShowS # newtype Deepening i a Source # This is an illegal Monoid used to construct a single Level. Constructors Deepening FieldsrunDeepening :: forall r. Int -> (Level i a -> Bool -> r) -> r Instances Semigroup (Deepening i a) Source # Instance detailsDefined in Control.Lens.Internal.Level Methods(<>) :: Deepening i a -> Deepening i a -> Deepening i a #sconcat :: NonEmpty (Deepening i a) -> Deepening i a #stimes :: Integral b => b -> Deepening i a -> Deepening i a # Monoid (Deepening i a) Source # This is an illegal Monoid. Instance detailsDefined in Control.Lens.Internal.Level Methodsmempty :: Deepening i a #mappend :: Deepening i a -> Deepening i a -> Deepening i a #mconcat :: [Deepening i a] -> Deepening i a # deepening :: i -> a -> Deepening i a Source # Generate the leaf of a given Deepening based on whether or not we're at the correct depth. newtype Flows i b a Source # This is an illegal Applicative used to replace the contents of a list of consecutive Level values representing each layer of a structure into the original shape that they were derived from. Attempting to Flow something back into a shape other than the one it was taken from will fail. Constructors Flows FieldsrunFlows :: [Level i b] -> a Instances Functor (Flows i b) Source # Instance detailsDefined in Control.Lens.Internal.Level Methodsfmap :: (a -> b0) -> Flows i b a -> Flows i b b0 #(<$) :: a -> Flows i b b0 -> Flows i b a # Applicative (Flows i b) Source # This is an illegal Applicative. Instance detailsDefined in Control.Lens.Internal.Level Methodspure :: a -> Flows i b a #(<*>) :: Flows i b (a -> b0) -> Flows i b a -> Flows i b b0 #liftA2 :: (a -> b0 -> c) -> Flows i b a -> Flows i b b0 -> Flows i b c #(*>) :: Flows i b a -> Flows i b b0 -> Flows i b b0 #(<*) :: Flows i b a -> Flows i b b0 -> Flows i b a # Apply (Flows i b) Source # Instance detailsDefined in Control.Lens.Internal.Level Methods(<.>) :: Flows i b (a -> b0) -> Flows i b a -> Flows i b b0 #(.>) :: Flows i b a -> Flows i b b0 -> Flows i b b0 #(<.) :: Flows i b a -> Flows i b b0 -> Flows i b a #liftF2 :: (a -> b0 -> c) -> Flows i b a -> Flows i b b0 -> Flows i b c # | 1,838 | 6,247 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-43 | latest | en | 0.698894 |
http://simulationresearch.lbl.gov/fmu/EnergyPlus/export/userGuide/mathematics.html | 1,544,875,801,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826856.55/warc/CC-MAIN-20181215105142-20181215131142-00292.warc.gz | 256,555,570 | 3,292 | 6. Mathematical DescriptionΒΆ
This section describes the algorithm for exchanging data between EnergyPlus, packaged as an FMU, and an external program.
Suppose we have a system with two simulation programs. Let simulation program 1 be EnergyPlus, the slave simulation program, which is packaged as an FMU for co-simulation; and let simulation program 2 be the master simulation program which supports the import of FMU for co-simulation. Suppose each program solves an initial-value ordinary differential equation that is coupled to the differential equations of the other program.
Let $$N$$ denote the number of time steps and let $$k$$ denote the time step. We will use the subscripts 1 and 2 to denote the state variable and the function that computes the next value of the state variable of the simulator 1 and 2, respectively. Then program 1 computes the sequence
$$x_{1}(k+1) = f_{1}(x_{1}(k), x_{2}(k))$$,
and, similarly, program 2 computes the sequence
$$x_{2}(k+1) = f_{2}(x_{2}(k), x_{1}(k))$$,
with initial conditions $$x_{1}(0) = x_{1,0}$$ and $$x_{2}(0) = x_{2,0}$$.
To advance from time $$k$$ to $$k+1$$, each program uses its own integration algorithm. At the end of the time step, program 1 sends its new state $$x_{1}(k+1)$$ to program 2, and receives the state $$x_{2}(k+1)$$ from program 2. The same procedure is done with the program 2. Program 2, which is the master simulation program, imports the FMU, and manages the data exchange between the two programs. In comparison to numerical methods of differential equations, this scheme is equal to an explicit Euler integration, which is an integration algorithm that solves an ordinary differential equation with specified initial values,
$$dx/dt = h(x)$$,
$$x(0) = x_{0}$$,
on the time interval $$t \in [0, 1]$$, using he following steps:
Step 0:
Initialize counter $$k=0$$ and number of steps, $$N > 0$$.
Set initial state $$x(k) = x_{0}$$ and set time step $$\Delta t = 1/N$$.
Step 1:
Compute new state $$x(k+1) = x(k) + h(x(k)) \Delta t$$.
Replace $$k$$ by $$k+1$$.
Step 2:
If $$k=N$$ stop, else go to Step 1.
However, this scheme does not require each simulation tool to use the explicit Euler method for its internal time integration algorithm; the analogy to explicit Euler applies only to the data exchange between the programs. In the situation where the differential equation is solved using co-simulation, the above algorithm becomes
Step 0:
Initialize counter $$k=0$$ and number of steps, $$N > 0$$.
Set initial state $$x_{1}(k) = x_{1,0}$$ and $$x_{2}(k) = x_{2,0}$$. Set the time step $$\Delta t = 1/N$$.
Step 1:
Compute new states
$$x_{1}(k+1) = x_{1}(k) + f_{1}(x_{1}(k), x_{2}(k)) \Delta t$$, and
$$x_{2}(k+1) = x_{2}(k) + f_{2}(x_{2}(k), x_{1}(k)) \Delta t$$.
Replace $$k$$ by $$k+1$$.
Step 2:
If $$k=N$$ stop, else go to Step 1. | 817 | 2,847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-51 | latest | en | 0.879446 |
https://hedberg.ccnysites.cuny.edu/viewers/slides.php?course=introphysics&topic=mathvectors&l=cal | 1,653,263,032,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662550298.31/warc/CC-MAIN-20220522220714-20220523010714-00225.warc.gz | 337,414,849 | 8,534 | # Math, Trigonometry and Vectors
## Geometry
What is the angle $\beta$ equal to?
### Trig Definitions
Here's a familiar image.
This one might be a little less familiar, but the same rules apply.
### Inverse functions
We can also use the inverse trigonmetric functions.
It's 450 meters from the corner of Hamilton Place and Broadway to 142nd St and Broadway. It's 489 meters from the same corner to 142nd if you walk along Hamilton Place. a) What is the angle between Hamilton and Broadway?
### Pythagorean Theorem
We'll use this relationship all the time.
$$a^2 + b^2 = c^2$$.
It's 450 meters from the corner of Hamilton Place and Broadway to 142nd St and Broadway. It's 489 meters from the same corner to 142nd if you walk along Hamilton Place. b) How far is it from Broadway to Hamilton Pl. walking along 142nd St?
## Scalars and Vectors
These are two different mathematical or physical entities.
5. 0 | 227 | 918 | {"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} | 3.546875 | 4 | CC-MAIN-2022-21 | latest | en | 0.90117 |
http://www.talkstats.com/threads/cohens-kappa-problem.18700/ | 1,590,487,356,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390755.1/warc/CC-MAIN-20200526081547-20200526111547-00440.warc.gz | 205,403,382 | 9,392 | # Cohen's Kappa Problem
#### bigsur
##### New Member
I am aiming to check the inter-rater reliability of a scale using Cohen's Kappa. I have input the 5 scores as their own variables for Rater A and the same again for Rater B. (1A, 2A, 3A, 4A, 5A and 1B, 2B, 3B etc).
In each variable, the scores range from 0-40 and are not categories - just scores. (It would be impossible to categorise them due to the wide variation in scores)
I run Cohen's Kappa on SPSS and for 2A*2B and 3A*3B i get an answer no problem. My problem is when i run it on 1A*1B, 4A*4B and 5a*5B. I get the error message "Kappa statistics cannot be computed. They require a symmetric 2-way table in which the values of the first variable match the values of the second variable."
Is this because the two scorers have not used exactly the same numbers? (For example, rater A gives a score of 8 somewhere in the results and rater B does not give that score at all?)
Any way to overcome this? I can't quite understand why it works for 2 of the variables and not the other 3. I've read somewhere about weighting kappa(?), but i'm not sure how to go about this or whether it would work.
Any help would be appreciated. | 328 | 1,188 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.924685 |
http://lessonshah123.s3-website-us-east-1.amazonaws.com/balancing-word-equations-worksheet-answers.html | 1,718,791,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00209.warc.gz | 15,577,813 | 6,860 | # Balancing Word Equations Worksheet Answers
Balancing Word Equations Worksheet Answers. There are 23 chemical equations and 4 types of equations: Chloride + hydrogen (2) + phosphoric acid chlorate water carbonate + oxygen(4) (4) acetate (4) word equations chapter 10.
Web given the following word equations, write the corresponding balanced chemical equations. 1) ___ nano3 + ___ pbo ( ___ pb(no3)2 + ___ na2o. Web if you are looking for word equations worksheet answers you’ve come to the right place.
### Web Balancing Chemical Equations Word Problems Worksheets:
Worksheets are balancing word equations chapter 9, work 1 writing and balancing formula equations, work word. Web practise questions on balanced chemical equations. Web word equations chapter 10 that is in parenthesis.
### Web Balancing Chemical Equations Vocabulary Match Up.
A) ___ al + ___ naoh ___ na 3 alo 3 + ___ h 2 b) ___ c 12 h 22 o 11 + ___ o 2 ___ co 2 + ___ h 2. Web here is data on worksheet. 1) 1 n2 + 3 h2 2 nh3 2) 2 kclo3 2 kcl + 3 o2 3) 2 nacl.
### Ca(Oh) 2 + H 3 Po 4 → Ca.
Web if you are looking for word equations worksheet answers you’ve come to the right place. There are 23 chemical equations and 4 types of equations: Web substitute symbols and formulas for words, then balance each equation.
### Unbalanced Chemical Equation An Unbalanced Equation Has The Mass And/Or The Charge Of Reactants Unequal To Those Of The Products.
Web balancing chemical equations with ease (100 free worksheets) angela october 2, 2019 worksheets in this article, you will learn about how to balance. Sodium chloride + lead (ii) nitrate lead (ii) chloride + sodium nitrate iron + chlorine iron (iii) chloride barium. We have 35 worksheets about word equations worksheet answers.
### Web Given The Following Word Equations, Write The Corresponding Balanced Chemical Equations.
Web displaying 8 worksheets for balancing word equations. (nh 4) 2 cr 2 o 7 (s) → cr 2 o 3 (s) + n 2 (g) + h 2 o(g) 2. Worksheets are balancing word equations chapter 9, work 1 writing and balancing formula equation. | 530 | 2,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.740971 |
http://stackoverflow.com/questions/14105590/how-to-find-a-number-x-if-we-move-the-last-digit-to-the-first-will-equal-doubl/14105732 | 1,417,109,322,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931008919.73/warc/CC-MAIN-20141125155648-00146-ip-10-235-23-156.ec2.internal.warc.gz | 283,191,555 | 19,918 | # how to find a number X, if we move the last digit to the first, will equal double X
"Move the last digit to the first" means: 12345 --> 51234
Someone tell me there must be an answer. I thought it should be very simple, but I just can not find any X even with coding. Anybody help?
``````public static void main(String[] args) {
for(int i=10; s<99999; i++){
if(2*i==convertInt(i)){
System.out.println(i);
}
}
}
private static int convertInt(int i) {
String s = i+"";
int index = s.length()-1;
String newS = s.charAt(index) + s.substring(0, index);
return Integer.parseInt(newS);
}
``````
-
Well maybe there isn't an `x` less than 99999 for which that works - that doesn't mean there is no answer. – arshajii Dec 31 '12 at 20:04
What is "s" in your for loop? – Majid L Dec 31 '12 at 20:11
yes, that could be the only reason why I can not find that number, I used to make i a long number, the performance become very bad and still find nothing. I hope someone can solve this problem with elegant way. – user1146450 Dec 31 '12 at 20:14
In fact the numbers for which this works are quite large (one is 105263157894736842). The brute force approach is unlikely to work. – arshajii Dec 31 '12 at 20:15
Googling the value given by @A.R.S. gives some interesting results. Namely en.wikipedia.org/wiki/Parasitic_number – Jan Dvorak Dec 31 '12 at 20:18
``````Let's start with the number without the two. Call it n.
If you multiply this by 10 and add 2 you would have your first number:
10n + 2
Doubling this you'd have:
20n + 4
This should be equivalent to taking n and putting a two in the front.
n + 2 * 10^d
Here d represents the number of digits.
20n + 4 = n + 2* 10^d
19n + 4 = 2 * 10^d
n = (2 * 10^d - 4) / 19
If you try values of d = 1, 2, 3, etc. you eventually get to n = 17 that gives you an integer result:
n = (2 * 10^17 - 4) / 19
n = (200000000000000000 - 4) / 19
n = 199999999999999996 / 19
n = 10526315789473684
Just put a 2 at the end and you have your answer:
105263157894736842
x 2
---------------------------------
210526315789473684
Edit: I just read Joseph's method and I like it better.
2 / 2 = 1 remainder 0 (carry down 1 to the next division)
1 / 2 = 0 remainder 1 (carry down 10 to the next division)
10 / 2 = 5 remainder 0 (carry down 5 to the next division)
5 / 2 = 2 remainder 1 (carry down 12 to the next division)
12 / 2 = 6 remainder 0 (carry down 6 to the next division)
6 / 2 = 3 remainder 0 (carry down 3 to the next division)
3 / 2 = 1 remainder 1 (carry down 11 to the next division)
11 / 2 = 5 remainder 1 (carry down 15 to the next division)
15 / 2 = 7 remainder 1 (carry down 17 to the next division)
17 / 2 = 8 remainder 1 (carry down 18 to the next division)
18 / 2 = 9 remainder 0 (carry down 9 to the next division)
9 / 2 = 4 remainder 1 (carry down 14 to the next division)
14 / 2 = 7 remainder 0 (carry down 7 to the next division)
7 / 2 = 3 remainder 1 (carry down 13 to the next division)
13 / 2 = 6 remainder 1 (carry down 16 to the next division)
16 / 2 = 8 remainder 0 (carry down 8 to the next division)
8 / 2 = 4 remainder 0 (carry down 4 to the next division)
4 / 2 = 2 remainder 0
You can stop here because you have gotten to a 2. Now read the digits going down:
105263157894736842
Note: technically you could repeat the process and get a larger number, but it would just be this sequence of digits repeated, e.g.:
105263157894736842 105263157894736842
One final note:
You could even do this the other direction and it might be easier:
The lowest digit is 2. If you double this, you get 4.
..42
Then double the 4:
..842
Then double the 8 (remember the carry):
..(1)6842
Then double the 6 and add the carry (13), that gets you another carry:
..(1)36842
Double the 3 and add the carry (7)
..736842
Double the 7 (remember the carry)
..(1)4736842
Double the 4 and add the carry (9)
..94736842
Double the 9 and remember the carry
..(1)894736842
Double the 8 and add the carry (17), remember the new carry:
..(1)7894736842
etc.
..(1)57894736842
..(1)157894736842
..3157894736842
..63157894736842
..(1)263157894736842
..5263157894736842
..(1)05263157894736842
Now when you double you get back to the digit 2 so you are done:
105263157894736842
``````
-
I used this function:
``````public static boolean isMultiple(int x) {
String g = String.valueOf(x);
g = g.substring(g.length()-1) + g.substring(0, g.length()-1);
int y = Integer.parseInt(g);
if(y == x * 2) return true;
return false;
}
``````
In this loop:
``````int x = 10;
while(!isMultiple(x)) {
System.out.println(x + ", " + isMultiple(x));
x++;
}
``````
And I have gotten no true value up to 650 million. I have a strong feeling this isn't actually possible. My guess is that no number will work for this, but I'm going to leave this running. I'll edit this if I find one.
Edit: It seems there are numbers like this. See above.
-
@A.R.S. Just found a number: 105263157894736842. I suggest you stop running this linear search. – irrelephant Dec 31 '12 at 20:18
en.wikipedia.org/wiki/Parasitic_number lists one: 105263157894736842 – Jan Dvorak Dec 31 '12 at 20:19
Oh, huh. That's an interesting number. Good to know, thanks! [edited] – Emrakul Dec 31 '12 at 20:19
In `isMultiple`, you can just `return Integer.parseInt(String.valueOf(x % 10) + (x / 10)) == x*2` to avoid the substring calls. This may let you get a little higher in the loop but still nowhere close to the actual value. – arshajii Dec 31 '12 at 20:27
That wikipedia link also gives an elegant way of finding such a number... – arcy Dec 31 '12 at 20:51
Somebody much smarter than me figured out the mathematical equation for this question:
Parasitic Number Equation
...which gives you a good way to code a non-brute force approach.
(where n = number of digits, and a = offset)
``````txtResult.Text = string.Empty;
for (int n = 2; n <= 100; n++)
{
if (((ulong)(Math.Pow(10, n)) - 2) % 19 == 0)
{
for (ulong a = 1; a <= 9; a++)
{
var omgwut = (ulong)((decimal)Math.Pow(10, n+1) / 19m) * a;
txtResult.Text = string.Format("{0}, {1}", omgwut.ToString(), txtResult.Text);
}
}
}
``````
Which ultimately gets you these results:
``````473684210526315789,
421052631578947368,
368421052631578947,
315789473684210526,
263157894736842105,
210526315789473684,
157894736842105263,
105263157894736842,
52631578947368421
``````
I've decided that the person who posited this question to you is a sadist. :)
-
You cab actually do this with a pencil and a piece of paper. you know the 18 digit number ends with 2. so the number twice as big must end with 4. therefore, the digit after 2 is 4. and the next is eight. you need to know how to multiply and retain the dozens (8X2 = 16, 6X2+1= 13, 3X2+1=7) and you go on until you get 2 without any 1 marks above it. this is how i found 210526315789473684 and 105263157894736842 at the same time
- | 2,152 | 6,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2014-49 | latest | en | 0.850327 |
https://mrmealor.wordpress.com/2008/12/18/twelve-days-of-christmas/ | 1,501,174,072,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549428325.70/warc/CC-MAIN-20170727162531-20170727182531-00500.warc.gz | 676,434,831 | 32,413 | # Twelve Days of Christmas
#1 How many total gifts are given just on day 10?
#2 How many total gifts are given during the twelve days?
#3 Suppose the song was changed to the “Nineteen Days of Christmas”? How many total gifts would my true love give to me?
#4 Generalize. How many gifts would be given on day n? How many total gifts would have been given on day n?
Here’s a discussion of the relationship of the song to Pascal’s triangle. The answers to #4 are C(n + 1, n – 1) and C(n + 2, 3). [Note: C(n+1, n-1) is translated as a “combination of n+1 things taken n-1 at a time”. Notation for combinations varies.] Can you prove that the two answers given will always work? | 191 | 682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.946965 |
https://books.google.com.jm/books?id=r4Q9Gyn1h6UC&dq=editions:UOM39015063895950&output=html_text&lr= | 1,725,811,429,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00475.warc.gz | 133,023,077 | 11,480 | # A New and Complete System of Arithmetick: Composed for the Use of the Citizens of the United States
Thomas & Andrews, 1808 - Algebra - 480 pages
### Popular passages
Page 181 - RULE. •f- 1 . Separate the given number into periods of three figures each, by putting a point over the unit figure, and every third figure beyond the place of units. 2. Find the greatest cube in the left hand period, and put its root in the quotient. 3.
Page 183 - Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring down the next period for a new dividend.
Page 245 - The present worth of any sum, due some time hence, is such a sum, as, if put to interest, would in that time, and at the rate per cent. for which the discount is to be made, amount to the sum or debt then due.
Page 175 - Distinguish the given number into periods of two figures each, by putting a point over the place of units, another over the place of hundreds, and so on, which points show the number of figures the root will consist of.
Page 61 - Divide by any number that will divide two or more of the given numbers without a remainder, and set the quotients, together with the undivided numbers, in a line beneath.
Page 134 - If 8 men can build a wall 20 feet long, 6 feet high and 4 feet thick, in 12 days ; in what time will 24 men build one 200 feet long, 8 feet high, and 6 feet thick ? 8 : 12 : : 20x6x4 24: 200x8x6 80 (lays, Jlns.
Page 175 - ... and subtract it therefrom, and to the remainder bring down the next period for a dividend. 3. " Place the double of the root, already found, on the lefl hand of the dividend for a divisor. 4. " Seek how often the divisor is contained...
Page 170 - Tret are practical rules for deducing certain allowances, •which are made by merchants and tradesmen in selling their goods by weight.
Page 79 - ... from the right hand of the quotient, point off so many places for decimals, as the decimal places in the dividend exceed those in the divisor.
Page 181 - Seek how often the divisor may be had in the dividend, and place the result in the quotient. | 539 | 2,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.905115 |
https://leetcode.ca/2019-10-23-1423-Maximum-Points-You-Can-Obtain-from-Cards/ | 1,719,082,877,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00719.warc.gz | 299,441,208 | 8,815 | 1423. Maximum Points You Can Obtain from Cards
Description
There are several cards arranged in a row, and each card has an associated number of points. The points are given in the integer array cardPoints.
In one step, you can take one card from the beginning or from the end of the row. You have to take exactly k cards.
Your score is the sum of the points of the cards you have taken.
Given the integer array cardPoints and the integer k, return the maximum score you can obtain.
Example 1:
Input: cardPoints = [1,2,3,4,5,6,1], k = 3
Output: 12
Explanation: After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.
Example 2:
Input: cardPoints = [2,2,2], k = 2
Output: 4
Explanation: Regardless of which two cards you take, your score will always be 4.
Example 3:
Input: cardPoints = [9,7,7,9,7,7,9], k = 7
Output: 55
Explanation: You have to take all the cards. Your score is the sum of points of all cards.
Constraints:
• 1 <= cardPoints.length <= 105
• 1 <= cardPoints[i] <= 104
• 1 <= k <= cardPoints.length
Solutions
Solution 1: Sliding Window
We can use a sliding window of length $k$ to simulate this process.
Initially, we place the window at the end of the array, i.e., the $k$ positions from index $n-k$ to index $n-1$. The sum of the points of the cards in the window is denoted as $s$, and the initial value of the answer $ans$ is also $s$.
Next, we consider the situation where we take $1, 2, …, k$ cards from the beginning of the array in turn. Suppose the card taken is $cardPoints[i]$. Then we add it to $s$. Due to the length limit of the window is $k$, we need to subtract $cardPoints[n-k+i]$ from $s$. In this way, we can calculate the sum of the points of the $k$ cards taken and update the answer $ans$.
The time complexity is $O(k)$, where $k$ is the integer given in the problem. The space complexity is $O(1)$.
• class Solution {
public int maxScore(int[] cardPoints, int k) {
int s = 0, n = cardPoints.length;
for (int i = n - k; i < n; ++i) {
s += cardPoints[i];
}
int ans = s;
for (int i = 0; i < k; ++i) {
s += cardPoints[i] - cardPoints[n - k + i];
ans = Math.max(ans, s);
}
return ans;
}
}
• class Solution {
public:
int maxScore(vector<int>& cardPoints, int k) {
int n = cardPoints.size();
int s = accumulate(cardPoints.end() - k, cardPoints.end(), 0);
int ans = s;
for (int i = 0; i < k; ++i) {
s += cardPoints[i] - cardPoints[n - k + i];
ans = max(ans, s);
}
return ans;
}
};
• class Solution:
def maxScore(self, cardPoints: List[int], k: int) -> int:
ans = s = sum(cardPoints[-k:])
for i, x in enumerate(cardPoints[:k]):
s += x - cardPoints[-k + i]
ans = max(ans, s)
return ans
• func maxScore(cardPoints []int, k int) int {
n := len(cardPoints)
s := 0
for _, x := range cardPoints[n-k:] {
s += x
}
ans := s
for i := 0; i < k; i++ {
s += cardPoints[i] - cardPoints[n-k+i]
ans = max(ans, s)
}
return ans
}
• function maxScore(cardPoints: number[], k: number): number {
const n = cardPoints.length;
let s = cardPoints.slice(-k).reduce((a, b) => a + b);
let ans = s;
for (let i = 0; i < k; ++i) {
s += cardPoints[i] - cardPoints[n - k + i];
ans = Math.max(ans, s);
}
return ans;
}
• /**
* @param {number[]} cardPoints
* @param {number} k
* @return {number}
*/
var maxScore = function (cardPoints, k) {
const n = cardPoints.length;
let s = cardPoints.slice(-k).reduce((a, b) => a + b);
let ans = s;
for (let i = 0; i < k; ++i) {
s += cardPoints[i] - cardPoints[n - k + i];
ans = Math.max(ans, s);
}
return ans;
};
• class Solution {
/**
* @param Integer[] $cardPoints * @param Integer$k
* @return Integer
*/
function maxScore($cardPoints,$k) {
$n = count($cardPoints);
$s = array_sum(array_slice($cardPoints, -$k));$ans = $s; for ($i = 0; $i <$k; ++$i) {$s += $cardPoints[$i] - $cardPoints[$n - $k +$i];
$ans = max($ans, $s); } return$ans;
}
}
• # @param {Integer[]} card_points
# @param {Integer} k
# @return {Integer}
def max_score(card_points, k)
n = card_points.length
s = card_points[-k..].sum
ans = s
k.times do |i|
s += card_points[i] - card_points[n - k + i]
ans = [ans, s].max
end
ans
end
• object Solution {
def maxScore(cardPoints: Array[Int], k: Int): Int = {
val n = cardPoints.length
var s = cardPoints.takeRight(k).sum
var ans = s
for (i <- 0 until k) {
s += cardPoints(i) - cardPoints(n - k + i)
ans = ans.max(s)
}
ans
}
}
• public class Solution {
public int MaxScore(int[] cardPoints, int k) {
int n = cardPoints.Length;
int s = cardPoints[^k..].Sum();
int ans = s;
for (int i = 0; i < k; ++i) {
s += cardPoints[i] - cardPoints[n - k + i];
ans = Math.Max(ans, s);
}
return ans;
}
}
• impl Solution {
pub fn max_score(card_points: Vec<i32>, k: i32) -> i32 {
let n = card_points.len();
let k = k as usize;
let mut s: i32 = card_points[n - k..].iter().sum();
let mut ans: i32 = s;
for i in 0..k {
s += card_points[i] - card_points[n - k + i];
ans = ans.max(s);
}
ans
}
}
• class Solution {
func maxScore(_ cardPoints: [Int], _ k: Int) -> Int {
let n = cardPoints.count
var s = cardPoints.suffix(k).reduce(0, +)
var ans = s
for i in 0..<k {
s += cardPoints[i] - cardPoints[n - k + i]
ans = max(ans, s)
}
return ans
}
} | 1,661 | 5,277 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.890472 |
https://www.jiskha.com/questions/1255003/a-medical-rescue-helicopter-is-flying-at-an-average-speed-of-172-miles-per-hour-toward-its | 1,603,155,472,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107867463.6/warc/CC-MAIN-20201019232613-20201020022613-00196.warc.gz | 781,938,332 | 5,232 | # algebra
A medical rescue helicopter is flying at an average speed of 172 miles per hour toward its base hospital. At 2:42 p.m., the helicopter is 80 miles from the hospital. Use this information for Items 4–6.
4. Which equation can be used to determine m, the number of minutes it will take the helicopter to
reach the hospital? A. 172(60m) = 80
60
C. 172(m)=80
B. 172 (60 )= 80 172
D. 60m=80
5. Solve the equation, and interpret the solution.
m
6. An emergency team needs to be on the roof of the hospital 3 minutes before the helicopter arrives. It takes the team 4 minutes to reach the roof. At what time should the team start moving to the roof to meet the helicopter? Explain your reasoning.
Please explain step by step how u did it I have a test tomorrow I need help
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2. 👎 1
3. 👁 553
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http://compgroups.net/comp.soft-sys.matlab/error-in-fsolve-cat-arguments-dimensions/395222 | 1,386,646,579,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164006951/warc/CC-MAIN-20131204133326-00097-ip-10-33-133-15.ec2.internal.warc.gz | 39,335,174 | 8,081 | COMPGROUPS.NET | Search | Post Question | Groups | Stream | About | Register
### Error in fsolve: CAT arguments dimensions are not consistent
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```I try to solve a couple of nonlinear equations with the following function
--------------------
function F = SolveNG(x,cx,cy,cz,ax,ay,az,P4,P5,CP)
n = cross(P4-CP',P4-P5);
n = n(:).'/norm(n)
D1 = -n(1,1)*P4(1,1) - n(1,2)*P4(1,2) - n(1,3)*P4(1,3)
P1=P4(1,1);
P2=P4(1,2);
P3=P4(1,3);
N1=n(1,1)
N2=n(1,2)
N3=n(1,3)
F = [N1*x(1) + N2*x(2) + N3*x(3) + D1;
((x(1)-cx)/ax)^2 + ((x(2)-cy)/ay)^2 + ((x(3)-cz)/az)^2 -1;
((2*(x(1)-cx))/(ax^2))- x(4);
((2*(x(2)-cy))/(ay^2))- x(5);
((2*(x(3)-cz))/(az^2))- x(6);
x(4)*x(1) + x(5)*x(2) + x(6)*x(3)+ x(7);
x(4)*P1 + x(5)*P2 + x(6)*P3+ x(7)];
end
-------------------
In the main function I use this code to call my function
-------------------
x0 = [-80; -40; -40;-30;-30;-30;-30]; % Make a starting guess at the solution
options=optimset('Display','iter'); % Option to display output
fsolve(@SolveNG, x0, options, cx,cy,cz,ax,ay,az,P4,P5,CP)
------------------
But I get an error which I do not understand:
??? Error using ==> vertcat
CAT arguments dimensions are not consistent.
Error in ==> SolveNG at 12
F = [N1*x(1) + N2*x(2) + N3*x(3) + D1;
Error in ==> fsolve at 253
fuser = feval(funfcn{3},x,varargin{:});
Error in ==> TestSimpeSphere at 726
fsolve(@SolveNG, x0, options, cx,cy,cz,ax,ay,az,P4,P5,CP)
Caused by:
Failure in initial user-supplied objective function evaluation.
FSOLVE cannot continue.
How can I solve this??
```
0
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Hi folks, I have the (for me) interesting case that I have an online backup where I can apply all the archived redo logs I have but the database will never be consistent. What have done: 1. alter tablespace ... begin backup; 2. cp datafiles to other machine 3. alter tablespace ... end backup; 4. ALTER DATABASE BACKUP CONTROLFILE TO ...; 5. cp backup controlfile, init.ora, archived redo logs etc. to other machine On the other machine: 1. startup mount; 2. recover database using backup controlfile; 3. applied all the archived redo logs When I now try to open the database I always get the info that system.dbf needs more recovery to be consistent. When I tried to create a new control file I get the following: SQL> RECOVER DATABASE ORA-00279: change 566146715 generated at 11/07/2003 04:03:48 needed for thread 1 ORA-00289: suggestion : /shop/oracle/arch/ISTOY2/arch_1_6424.arc ORA-00280: change 566146715 for thread 1 is in sequence #6424 Specify log: {<RET>=suggested | filename | AUTO | CANCEL} /shop/oracle/arch/ISTOY2/arch_1_6425.arc ORA-00310: archived log contains sequence 6425; sequence 6424 required ORA-00334: archived log: '/shop/oracle
How to calculate the number of arguments a function takes. (its arity).
Hello, Is there a portable way to calculate a function's arity in CL? I've skimmed the CLtL2 book without success. I suppose it can't be done because a function can take optional arguments... I'm I missing something obvious? If there is no way... How would you approach that need? I'm doing a genetic programming library and I want operators to be functions taking an unspecified number of arguments. I'd like to apply them this way: (let ((operator (select-operator-from-weighted-list operators-list)) (apply operator (select-n-individuals (arity-of operator) population))) That would be the clean, desirable way. Unfortunately there is no arity-of function. :( Right now I'm creating operators as subclasses of a CLOS class called operator which has an arity attribute. And the operator is implemented as a method taking an object of that class and the individuals as arguments. But it is ugly code. I've also thought about implementing it as a macro which would add an 'arity' property to the function's property list, but I've never used property lists. Would it be possible? How | 2,589 | 8,800 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2013-48 | latest | en | 0.621974 |
https://mathoverflow.net/questions/182021/asymptotic-eigenvalue-analysis-for-a-sparse-random-matrix | 1,702,167,453,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100989.75/warc/CC-MAIN-20231209233632-20231210023632-00292.warc.gz | 429,848,264 | 27,038 | # Asymptotic eigenvalue analysis for a sparse random matrix
We have an asymptotic analysis problem for the eigenvalue performance of the following random matrix: $H=\{h_{ij}\}_{N_r\times N_t}$,
where each entry $h_{ij}$ is with a probability $p$ to obey the Gaussian distribution $N(0,σ^2)$, and with a probability $1-p$ to be zero.
Then we have following questions
a)Can we use a specific distribution (pdf) to describe the entries of this $N_r \times N_t$ matrix as $h_{ij}\sim (1-p)\delta(0)+p\frac{1}{\sigma\sqrt{2\pi}}e^{-j\frac{x^2}{2\sigma^2}}\delta(x)$?
b)When $N_r,N_t\rightarrow \infty$, and $N_r/N_t\rightarrow \beta$ (determined value), does the eigenvalues’ PDF go to a deterministic distribution? If the answer is Yes, what is the explicit expression of this deterministic distribution?
For question b), one potentially useful tool maybe the Mar˘cenko-Pastur law in the random matrix theory, but I do not know if it can be used indeed.
• The matrices aren't square. Do you mean singular values instead of eigenvalues? Sep 29, 2014 at 13:06
• Yes, for a general case, singular values should be considered, and the Marchenko-Pastur law may can be used. But from my comparison between the asymptotic results and the numerical results, there is a mismatch, I cannot explain the reason. Sep 30, 2014 at 0:12 | 362 | 1,323 | {"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} | 2.8125 | 3 | CC-MAIN-2023-50 | latest | en | 0.759848 |
https://cstheory.stackexchange.com/questions/18203/multi-way-spectral-partitioning-and-higher-order-cheeger-inequalities | 1,610,946,222,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00647.warc.gz | 280,992,777 | 32,610 | # Multi-way spectral partitioning and higher-order Cheeger inequalities
I am reading the paper above by Lee, Gharan and Trevisan, and I am having trouble with lemma 4.8 on page 23. How do we form the set $T_1, \ldots, T_{r'}$ such that $r'>(1-\delta/2)k$ and each set $T_i$ satisfies both the upper bound and lower bound with respect to $\epsilon$ as in the paper ? I am struggling with this for a while, so any help is greatly appreciated.
• Why not ask the authors? – vb le Jul 1 '13 at 20:58
• $T_1$ is the union of $S_1, \ldots, S_{i_1}$ where $i_1$ is the first number s.t. the bounds on top of p. 23 are satisfied. $T_2$ is the union of $S_{i_1 + 1}, \ldots, S_{i_2}$ where $i_2$ is defined similarly. You get the idea... – Sasho Nikolov Jul 1 '13 at 22:43
• Thank you Sasho. You meant $diam(S_{i_1})\leq \Delta$ ? The partition guarantees $S_i$ always satisfies that for all i. The problem is that when we union a few $S_i$, that property may no longer hold. Since the $T_i$'s have to satisfy both the upper and the lower bounds at the same time, taking union of too few $S_i$ may not give us the lower bound while taking union of too many $S_i$ may violates $diam(T_i)\leq \Delta$ (thus, will not give us the upper bound). – Son P Nguyen Jul 5 '13 at 3:42 | 403 | 1,265 | {"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} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.884464 |
http://branajazyku.cz/site/d2s1hxr.php?page=55-miles-per-hour-to-kilometers-per-second-004bac | 1,652,702,802,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00040.warc.gz | 9,616,554 | 7,110 | If 10 years is a decade, then what is 4 years? You can sign in to vote the answer. So, 55 miles per hour times 0.00044704 is equal to 0.02459 km/s.
Among others. A mile per hour is a unit of speed commonly used in the United States. For quick reference purposes, below is a conversion table that you can use to convert from mph to km/s. Using the conversion formula above, you will get: Value in km/s = 55 × 0.00044704 = 0.02459 km/s. A foot per second is a unit of speed.
It is equal to exactly 1.609344 kilometers per hour. 1) AB > BC Hi, I'm really stuck on this math question: What is a value of k for, x^2 - kx + 7 = 0? In ΔABC, m∠A = 60, m∠B = 80, and m∠C = 40. Road speed limits are given in miles per hour which is abbreviated as mph or mi/h. If you write the units, you can see how the units cancel out: 55 mi/h * h/60min * min/60s = 55/3600 mi/s = 0.0152777778 mi/s. A mile per hour … 2) AC > BC Please help, Thank you!? This is a measurement of speed typically used in non-metric countries for transport such as the USA. 55 mph to fps conversion. 55 mph to m/s conversion. To convert any value in miles per hour to kilometers per second, just multiply the value in miles per hour by the conversion factor 0.00044704. 55 mph=55*1.6 kph=88 kph (1 mile=1600 metres is a close approximation), 55 mph = 88 km / 3600 sec = 0.024 km-per-sec, 55 miles/hour * 1.609 kilometers/mile * 1/3600 hours/second = 0.0246 km/sec.
0.0152777778 mi/s * 1.609344 km/mi = 0.0245872 km/s. Miles per Hour to Kilometers per Hour Conversions (some results rounded) Graph the image of the polygon after a reflection in the line y=-x ?
Please, choose a physical quantity, two units, then type a value in any of the boxes above. What are three numbers that have a sum of 35 if the greatest number is 14 more than the least number Volume to (Weight) Mass Converter for Recipes, Weight (Mass) to Volume to Converter for Recipes. Miles per hour. What is(are) the solution(s)? How do you think about the answers? What is 55 miles per hour in feet per second? Which inequality is true? All information in this site is provided “as is”, with no guarantee of completeness, accuracy, timeliness or of the results obtained from the use of this information. 55 miles/hour * 1.609 kilometers/mile * 1/3600 hours/second = 0.0246 km/sec. If you write the units, you can see how the units cancel out: mi/h * km/mi = km/h
What is the conversion factor to convert from mph to km/s? Still have questions? Conversion Table. Something traveling at one kilometer per hour is traveling about 0.278 meters per second, or about 0.621 miles per hour.
55 miles per hour = 0.02459 kilometer per second Formula: multiply the value in miles per hour by the conversion factor '0.00044704'. So, 55 miles per hour = 55 × 0.00044704 = 0.0245872 kilometer per second. There is 0.02459 kilometers per second in 55 miles per hour. The United Kingdom also uses this on the roads although officially the metric system has been adopted. Miles per hour to Kilometers per second Conversion Example Task: Convert 2,500 miles per hour to kilometers per second (show work) Formula: mph x 0.00044704 = km/s Calculations: 2,500 mph x 0.00044704 = 1.1176 km/s Result: 2,500 mph is equal to 1.1176 km/s. 1 Miles per hour = 1.6093 Kilometres per hour: 10 Miles per hour = 16.0934 Kilometres per hour: 2500 Miles per hour = 4023.36 Kilometres per hour: 2 Miles per hour = 3.2187 Kilometres per hour: 20 Miles per hour = 32.1869 Kilometres per hour: 5000 Miles per hour = 8046.72 Kilometres per hour: 3 Miles per hour = 4.828 Kilometres per hour: 30 Miles per hour = 48.2803 Kilometres per hour The graph of a quadratic equation is shown below.
To calculate a mile per hour value to the corresponding value in km/s, just multiply the quantity in mph by 0.00044704 (the conversion factor). | 1,103 | 3,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-21 | latest | en | 0.85103 |
http://mathhelpforum.com/discrete-math/146035-summation-print.html | 1,527,076,595,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00172.warc.gz | 191,403,466 | 3,097 | # Summation
• May 22nd 2010, 10:43 PM
helloying
Summation
Show that $\displaystyle \sum (-1)^{r+1} r^2$ = -n(2n+1)
summation is from r=1 to 2n . sry i dont know how to type
Thx:)
• May 23rd 2010, 04:25 AM
Quote:
Originally Posted by helloying
Show that $\displaystyle \sum (-1)^{r+1} r^2$ = -n(2n+1)
summation is from r=1 to 2n . sry i dont know how to type
Thx:)
hi
\sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate
$\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$
$\displaystyle =1-4+9-16+25-36+...$
$\displaystyle =1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2$
$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n-1)$
$\displaystyle =-3-7-...-(4n-1)$
which is an AP with first term -3 , last term -(4n-1)
$\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$
simplify this and you are done.
• May 23rd 2010, 07:10 AM
oldguynewstudent
Quote:
hi
\sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate
$\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$
$\displaystyle =1-4+9-16+25-36+...$
$\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$
$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$
$\displaystyle =-3-4-...-(4n-1)$
which is an AP with first term -3 , last term -(4n-1)
$\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$
simplify this and you are done.
How did you get $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$?
When I do this I come up with $\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $\displaystyle (-1)^{2n+1} {2n}^2$ = $\displaystyle -{2n}^2$.
• May 24th 2010, 01:06 AM
helloying
Quote:
Originally Posted by oldguynewstudent
How did you get $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$?
When I do this I come up with $\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $\displaystyle (-1)^{2n+1} {2n}^2$ = $\displaystyle -{2n}^2$.
yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$
it shd be -3-7 and not -3-4
• May 27th 2010, 11:38 PM
Quote:
Originally Posted by helloying
yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$
it shd be -3-7 and not -3-4
yup , some algebraic error here and there . Edited . | 1,040 | 2,249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-22 | latest | en | 0.677303 |
https://www.daniweb.com/programming/software-development/threads/444506/segmentation-fault-in-my-program | 1,669,849,326,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710777.20/warc/CC-MAIN-20221130225142-20221201015142-00554.warc.gz | 753,262,431 | 16,913 | Hi
I have written a program, it doesn't work and i can't figure out why.
It does compile, but it doesn't run.
The whole program:
`````` #include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main()
{
int a;
int j;
int c;
int d;
int dig;
c = 0;
int i;
a = 0;
d = 0;
for (i = 0;i == i;i++) {
a = a + i;
// from this
for (j = 1;j < a; j++){
dig = a / j;
if (isdigit(dig)){
d++;
}
//to this doesn't work
if (d == 500) {
printf("%d\n",a);
system("sleep 10s");
}
else {
d = 0;
}
}
}
return 0;
}
``````
The part which doesn't work should count how many divisors the number 'a' has.
What am i doing wrong?
## All 2 Replies
Your code is difficult to read due to the poor naming of variables and the complete lack of comments. Aside from that it just doesn't make much sense. Trying to make sense of it I added some comments:
``````#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(void)
{
int a = 0; // Will be sum(i).
int d = 0; // You probably try to let this contain the amount of divisors.
int j; // For-loop variable.
int i; // For-loop variable.
int dig; // Will contain how many times a 'j' fits into 'a'.
// Warning: "i == i" will result in an endless loop. Is this what you want?
for (i = 0; i == i; i++)
{
// 'a' will be sum(i).
a = a + i;
// Go past every number smaller than sum(i), starting at 1.
// You might want to change '<' to '<=' or let j start at 0.
for (j = 1; j < a; j++)
{
// dig will contain the amount of times 'j' fits into 'a'.
// This might not be what you want it to do, going by your next statement.
dig = a / j;
// 'isdigit' wants a char and you're providing it with ASCII values now. If 'a' isn't divisible by 'j' dig will contain the highest multitude of
// j smaller than a. (e.g. 5/2 = 2) Search for "integer division" for more information. If you want
// to check for divisibility you could use the module operator, '%'. (e.g. if "x % y" results in 0 then
// x is divisible by y)
if (isdigit(dig))
{
// I think your goal was to let 'd' contain the amount of divisors. This is not what it now contains.
// Instead it will end up being equal to 'a - 1' if a is bigger than 1. So not what you want.
d++;
}
// What are you even trying to do here? What is '500' supposed to represent?
if (d == 500)
{
printf("%d\n",a);
system("sleep 10s");
}
else
{
// Don't you want to reset 'd' for every 'i'?
d = 0;
}
}
}
return 0;
}
``````
I'm not even going to begin looking into what the problem could be before I know what it is you're trying to do.
-edit-
Modified.
isdigit expects a single character. Just one. It's meant to be used to check if a single character is a digit (so the input is expect to have a numerical value in the range zero to 127 for a char, because those are the values a char takes - see http://en.cppreference.com/w/c/string/byte/isdigit and look at that nice table at the bottom). I expect that isdigit is implemented in your system something like this:
``````#define isdigit(c) ((map[c] & FLAG_DIGIT)==FLAG_DIGIT)
``````
where map is an array of some fixed size. Your code manages to feed it many values larger than the array size, so it segFaults eventually when the OS spots you trying to read memory that isn't yours. What are you actually trying to test? Are you trying to test for dig being a single digit? That's not what isdigit does.
You could have identified that the segFault was being thrown by the function isdigit yourself if you used a debugger. The ten minutes it will take you to learn how to use a debugger will save you vast, vast amounts of time in the future.
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We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, learning, and sharing knowledge. | 1,053 | 3,775 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.811116 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/257/S3xC3xC15.html | 1,642,563,828,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00624.warc.gz | 463,999,364 | 5,629 | Copied to
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## G = S3×C3×C15order 270 = 2·33·5
### Direct product of C3×C15 and S3
Aliases: S3×C3×C15, C331C10, C323C30, C3⋊(C3×C30), (C3×C15)⋊8C6, C153(C3×C6), (C32×C15)⋊5C2, SmallGroup(270,24)
Series: Derived Chief Lower central Upper central
Derived series C1 — C3 — S3×C3×C15
Chief series C1 — C3 — C15 — C3×C15 — C32×C15 — S3×C3×C15
Lower central C3 — S3×C3×C15
Upper central C1 — C3×C15
Generators and relations for S3×C3×C15
G = < a,b,c,d | a3=b15=c3=d2=1, ab=ba, ac=ca, ad=da, bc=cb, bd=db, dcd=c-1 >
Subgroups: 104 in 64 conjugacy classes, 36 normal (12 characteristic)
C1, C2, C3, C3, C3, C5, S3, C6, C32, C32, C32, C10, C15, C15, C15, C3×S3, C3×C6, C33, C5×S3, C30, C3×C15, C3×C15, C3×C15, S3×C32, S3×C15, C3×C30, C32×C15, S3×C3×C15
Quotients: C1, C2, C3, C5, S3, C6, C32, C10, C15, C3×S3, C3×C6, C5×S3, C30, C3×C15, S3×C32, S3×C15, C3×C30, S3×C3×C15
Smallest permutation representation of S3×C3×C15
On 90 points
Generators in S90
(1 42 70)(2 43 71)(3 44 72)(4 45 73)(5 31 74)(6 32 75)(7 33 61)(8 34 62)(9 35 63)(10 36 64)(11 37 65)(12 38 66)(13 39 67)(14 40 68)(15 41 69)(16 56 78)(17 57 79)(18 58 80)(19 59 81)(20 60 82)(21 46 83)(22 47 84)(23 48 85)(24 49 86)(25 50 87)(26 51 88)(27 52 89)(28 53 90)(29 54 76)(30 55 77)
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15)(16 17 18 19 20 21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40 41 42 43 44 45)(46 47 48 49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72 73 74 75)(76 77 78 79 80 81 82 83 84 85 86 87 88 89 90)
(1 32 65)(2 33 66)(3 34 67)(4 35 68)(5 36 69)(6 37 70)(7 38 71)(8 39 72)(9 40 73)(10 41 74)(11 42 75)(12 43 61)(13 44 62)(14 45 63)(15 31 64)(16 88 46)(17 89 47)(18 90 48)(19 76 49)(20 77 50)(21 78 51)(22 79 52)(23 80 53)(24 81 54)(25 82 55)(26 83 56)(27 84 57)(28 85 58)(29 86 59)(30 87 60)
(1 82)(2 83)(3 84)(4 85)(5 86)(6 87)(7 88)(8 89)(9 90)(10 76)(11 77)(12 78)(13 79)(14 80)(15 81)(16 38)(17 39)(18 40)(19 41)(20 42)(21 43)(22 44)(23 45)(24 31)(25 32)(26 33)(27 34)(28 35)(29 36)(30 37)(46 71)(47 72)(48 73)(49 74)(50 75)(51 61)(52 62)(53 63)(54 64)(55 65)(56 66)(57 67)(58 68)(59 69)(60 70)
G:=sub<Sym(90)| (1,42,70)(2,43,71)(3,44,72)(4,45,73)(5,31,74)(6,32,75)(7,33,61)(8,34,62)(9,35,63)(10,36,64)(11,37,65)(12,38,66)(13,39,67)(14,40,68)(15,41,69)(16,56,78)(17,57,79)(18,58,80)(19,59,81)(20,60,82)(21,46,83)(22,47,84)(23,48,85)(24,49,86)(25,50,87)(26,51,88)(27,52,89)(28,53,90)(29,54,76)(30,55,77), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)(16,17,18,19,20,21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75)(76,77,78,79,80,81,82,83,84,85,86,87,88,89,90), (1,32,65)(2,33,66)(3,34,67)(4,35,68)(5,36,69)(6,37,70)(7,38,71)(8,39,72)(9,40,73)(10,41,74)(11,42,75)(12,43,61)(13,44,62)(14,45,63)(15,31,64)(16,88,46)(17,89,47)(18,90,48)(19,76,49)(20,77,50)(21,78,51)(22,79,52)(23,80,53)(24,81,54)(25,82,55)(26,83,56)(27,84,57)(28,85,58)(29,86,59)(30,87,60), (1,82)(2,83)(3,84)(4,85)(5,86)(6,87)(7,88)(8,89)(9,90)(10,76)(11,77)(12,78)(13,79)(14,80)(15,81)(16,38)(17,39)(18,40)(19,41)(20,42)(21,43)(22,44)(23,45)(24,31)(25,32)(26,33)(27,34)(28,35)(29,36)(30,37)(46,71)(47,72)(48,73)(49,74)(50,75)(51,61)(52,62)(53,63)(54,64)(55,65)(56,66)(57,67)(58,68)(59,69)(60,70)>;
G:=Group( (1,42,70)(2,43,71)(3,44,72)(4,45,73)(5,31,74)(6,32,75)(7,33,61)(8,34,62)(9,35,63)(10,36,64)(11,37,65)(12,38,66)(13,39,67)(14,40,68)(15,41,69)(16,56,78)(17,57,79)(18,58,80)(19,59,81)(20,60,82)(21,46,83)(22,47,84)(23,48,85)(24,49,86)(25,50,87)(26,51,88)(27,52,89)(28,53,90)(29,54,76)(30,55,77), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)(16,17,18,19,20,21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45)(46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75)(76,77,78,79,80,81,82,83,84,85,86,87,88,89,90), (1,32,65)(2,33,66)(3,34,67)(4,35,68)(5,36,69)(6,37,70)(7,38,71)(8,39,72)(9,40,73)(10,41,74)(11,42,75)(12,43,61)(13,44,62)(14,45,63)(15,31,64)(16,88,46)(17,89,47)(18,90,48)(19,76,49)(20,77,50)(21,78,51)(22,79,52)(23,80,53)(24,81,54)(25,82,55)(26,83,56)(27,84,57)(28,85,58)(29,86,59)(30,87,60), (1,82)(2,83)(3,84)(4,85)(5,86)(6,87)(7,88)(8,89)(9,90)(10,76)(11,77)(12,78)(13,79)(14,80)(15,81)(16,38)(17,39)(18,40)(19,41)(20,42)(21,43)(22,44)(23,45)(24,31)(25,32)(26,33)(27,34)(28,35)(29,36)(30,37)(46,71)(47,72)(48,73)(49,74)(50,75)(51,61)(52,62)(53,63)(54,64)(55,65)(56,66)(57,67)(58,68)(59,69)(60,70) );
G=PermutationGroup([[(1,42,70),(2,43,71),(3,44,72),(4,45,73),(5,31,74),(6,32,75),(7,33,61),(8,34,62),(9,35,63),(10,36,64),(11,37,65),(12,38,66),(13,39,67),(14,40,68),(15,41,69),(16,56,78),(17,57,79),(18,58,80),(19,59,81),(20,60,82),(21,46,83),(22,47,84),(23,48,85),(24,49,86),(25,50,87),(26,51,88),(27,52,89),(28,53,90),(29,54,76),(30,55,77)], [(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15),(16,17,18,19,20,21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45),(46,47,48,49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75),(76,77,78,79,80,81,82,83,84,85,86,87,88,89,90)], [(1,32,65),(2,33,66),(3,34,67),(4,35,68),(5,36,69),(6,37,70),(7,38,71),(8,39,72),(9,40,73),(10,41,74),(11,42,75),(12,43,61),(13,44,62),(14,45,63),(15,31,64),(16,88,46),(17,89,47),(18,90,48),(19,76,49),(20,77,50),(21,78,51),(22,79,52),(23,80,53),(24,81,54),(25,82,55),(26,83,56),(27,84,57),(28,85,58),(29,86,59),(30,87,60)], [(1,82),(2,83),(3,84),(4,85),(5,86),(6,87),(7,88),(8,89),(9,90),(10,76),(11,77),(12,78),(13,79),(14,80),(15,81),(16,38),(17,39),(18,40),(19,41),(20,42),(21,43),(22,44),(23,45),(24,31),(25,32),(26,33),(27,34),(28,35),(29,36),(30,37),(46,71),(47,72),(48,73),(49,74),(50,75),(51,61),(52,62),(53,63),(54,64),(55,65),(56,66),(57,67),(58,68),(59,69),(60,70)]])
135 conjugacy classes
class 1 2 3A ··· 3H 3I ··· 3Q 5A 5B 5C 5D 6A ··· 6H 10A 10B 10C 10D 15A ··· 15AF 15AG ··· 15BP 30A ··· 30AF order 1 2 3 ··· 3 3 ··· 3 5 5 5 5 6 ··· 6 10 10 10 10 15 ··· 15 15 ··· 15 30 ··· 30 size 1 3 1 ··· 1 2 ··· 2 1 1 1 1 3 ··· 3 3 3 3 3 1 ··· 1 2 ··· 2 3 ··· 3
135 irreducible representations
dim 1 1 1 1 1 1 1 1 2 2 2 2 type + + + image C1 C2 C3 C5 C6 C10 C15 C30 S3 C3×S3 C5×S3 S3×C15 kernel S3×C3×C15 C32×C15 S3×C15 S3×C32 C3×C15 C33 C3×S3 C32 C3×C15 C15 C32 C3 # reps 1 1 8 4 8 4 32 32 1 8 4 32
Matrix representation of S3×C3×C15 in GL3(𝔽31) generated by
5 0 0 0 5 0 0 0 5
,
14 0 0 0 1 0 0 0 1
,
1 0 0 0 5 5 0 0 25
,
1 0 0 0 1 0 0 4 30
G:=sub<GL(3,GF(31))| [5,0,0,0,5,0,0,0,5],[14,0,0,0,1,0,0,0,1],[1,0,0,0,5,0,0,5,25],[1,0,0,0,1,4,0,0,30] >;
S3×C3×C15 in GAP, Magma, Sage, TeX
S_3\times C_3\times C_{15}
% in TeX
G:=Group("S3xC3xC15");
// GroupNames label
G:=SmallGroup(270,24);
// by ID
G=gap.SmallGroup(270,24);
# by ID
G:=PCGroup([5,-2,-3,-3,-5,-3,4504]);
// Polycyclic
G:=Group<a,b,c,d|a^3=b^15=c^3=d^2=1,a*b=b*a,a*c=c*a,a*d=d*a,b*c=c*b,b*d=d*b,d*c*d=c^-1>;
// generators/relations
×
𝔽 | 4,125 | 6,920 | {"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} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.293068 |
https://www.askiitians.com/forums/Magical-Mathematics%5BInteresting-Approach%5D/29/32573/x-to-power-infinity.htm | 1,680,103,107,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00164.warc.gz | 737,575,055 | 32,640 | if x to power x to power x......upto infinity = 2 then find x...............I got my ans as (underroot 2)............but if I change the value from 2 to 4 or 6 or 8 ...............the ans is same ------(underroot 2).......the answer is both correct and wrong.........how is it possible???? see how i got my ans---- let x to power x to power x......upto infinity = y then x to power y = y y = 2 x squre = 2 x = uderoot 2 PLS CORRECT ME IF I AM WRONG!!!!!
Mir Noaman Ali IIT-Roorkee
57 Points
11 years ago
IF WE EQUATE X POWER X POWER X AND SO ON UPTO INFINITY TO = n then x = nth root of n .. it accidently happens to be that 4th root of 4 = 2nd root of 2 that is why you are getting ambiguity | 210 | 694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-14 | latest | en | 0.863191 |
https://headinside.blogspot.com/2009/09/magic-of-9.html | 1,653,424,973,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00386.warc.gz | 360,029,182 | 270,192 | 2
## The Magic of 9
Published on Thursday, September 10, 2009 in , , ,
Since yesterday was 9/9/09, it seems like the perfect time to look at the mysteries of 9.
BTW, the actor who was singing about 9s in yesterday's video, and who apparently learned about 9s from Fleet the Genii, is Reg E. Cathey, whom you may know from a surprising number of places.
We'll start off with something simple. The following video teaches an easy way to learn the 9 times table from 1 up to 10. It's so straightforward, there's no audio required to help you understand it:
Probably one of the most famous 9 tricks would have to be the classic 1089 trick. The explanation is surprisingly simple, despite the large numbers involved.
Why is 9 so surprising? Sol from Wild About Math took a look at this phenomenon last year, and explains it pretty well. The best part of this article is the mathematical magic tricks you learn while trying to understand the principles being taught. Yes, the 1089 trick shows up here, as well.
Sol's article briefly mentions “casting out nines”, which I remember learning about long ago. It's a great way to check your math on an arithmetic problem, but the teaching of it seems to have fallen out of favor decades ago. The usefulness and simplicity of it, once learned, is so elegant, that there are people today who, when they finally learn about casting out 9s, become enraged that they were never taught it!
Yes, 9s can simplify magic tricks, but math as well. Sophie B. shows us how to use 9's amazing properties to multiply by 9, 99, 999 and even 19! I not only like the methods taught here, but the amusing way the story is used to present it.
Even if you missed 9/9/09 itself, you still have the rest of the month to improve and show off your 9 prowess in 9/09!
### 2 Response to The Magic of 9
Jay
6:55 PM
Hey, Jay here, long time Grey Matters. I am crazy about mental calculation and noticed an interesting thing. You can cast out 99's just like 9's by adding up pairs of numbers. This has teh added bonus of catching place values errors. With casting out nine's it will not guard against errors such as getting 789 instead of 7890. The nines test doesn't guard against place value shifts. The nine's test catches 89% of errors whereas the 99's test catches 99% of errors! Just thought I'd share
Anonymous
2:21 AM
This post refers to a webpage at themathlab.com which teaches "casting out nines."
The internet is great for spreading misinformation. If you link to another webpage, you should be responsible for pointing out errors on that webpage.
And the folks at themathlab.com should held responsible for their careless reading of the Asimov book cited on their page. Asimov does not misuse the word "augend," but themathlab.com article does. (See link at end of third from last paragraph.)
If one is going to show off by using obscure words, one ought to make sure he/she is using the word correctly! | 697 | 2,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-21 | latest | en | 0.968217 |
https://www.teacherspayteachers.com/Product/Exponent-Rules-Card-Sort-Math-Centers-866108 | 1,493,438,855,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123270.78/warc/CC-MAIN-20170423031203-00100-ip-10-145-167-34.ec2.internal.warc.gz | 946,021,978 | 25,275 | Total:
\$0.00
Exponent Rules Card Sort - Math Centers
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
1.68 MB | 7 pages
PRODUCT DESCRIPTION
Common Core Standards: 8.EE.1
This activity is a fun card sort that will help students practice applying the exponent rules. Students will use the rules for multiplying and dividing powers, raising a power to a power, zero exponents, negative exponents, and evaluating exponents with order of operations. Half the cards have a problem or expression and half have an equivalent expression in the simplest form.
The activity has options for differentiation and self-checking built in. This is a great activity for math centers since students can check themselves and work with a partner.
Differentiation:
The problems are sorted by rule, and then from easiest to most difficult from top to bottom. You can choose to give struggling students only half a set, or one rule at a time. The problem and answer cards are in the same order to make it easier to modify.
Included:
*Instructions and Standards Correlation Sheet
*Large Label for File Folder/Centers Organization
*32 Exponent Problem Cards
Please leave feedback with any suggestions for improvements! :)
Total Pages
7
Included
Teaching Duration
45 Minutes
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\$4.00
User Rating: 4.0/4.0
(297 Followers)
\$4.00 | 363 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-17 | longest | en | 0.890311 |
https://www.physicsforums.com/threads/cold-water.759733/ | 1,601,357,664,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00771.warc.gz | 895,219,481 | 17,189 | # Cold Water
Hello ,
when i take out a cold bottle from the refrigerator , after a few seconds (lets say at room temperature) the bottle is surrounded by water.
the question is - why is it? where did the water come from ?
is it should happen with any cold body ?
(feel free to speak high phycis )
thank u !
Related Other Physics Topics News on Phys.org
Nugatory
Mentor
There's some amount of water vapor floating around in the room air. The air next to the cold bottle is cooled by the bottle, and when it cools some of that water vapor condenses onto the bottle.
Any cold object will do the trick, but a bottle of water is particularly good at keeping its surface cool (because as the surface warms, water circulation brings more cold water from deep inside the bottle to the edge) so you may get a more pronounced effect with a bottle of water than, for example, an apple.
1 person
HallsofIvy
Homework Helper
Note that the amount of water that can be held in the air increases with the temperature of the air. Since the air is, as Nugatory says, cooled by the bottle the amount of water that it can hold is reduced. If the humidity (amount of water already in the air) it high the water will condense out of the air.
1 person
sophiecentaur
Gold Member
Note that the amount of water that can be held in the air increases with the temperature of the air. Since the air is, as Nugatory says, cooled by the bottle the amount of water that it can hold is reduced. If the humidity (amount of water already in the air) it high the water will condense out of the air.
That expression is a bit misleading but it's commonly used. In fact, it's all a matter of partial pressures. Air isn't a sponge (although that's what it seems). The maximum proportion of water vapour in the air is totally due to the vapour pressure at the ambient temperature. You would get exactly the same amount of water vapour in a container if the container were evacuated and then the pressure allowed to increase as water evaporates off a water surface. The equilibrium situation would always involve 'saturated' air (there- I've done it myself!)
1 person
so I have another question - what makes the water act like a gas ? I know that the evaporation temp. at room pressure is ~ 100 C , so if the temperture is < 100 and the pressure is ~ 1 atm why the water are gas?
thank you
sophiecentaur
Gold Member
so I have another question - what makes the water act like a gas ? I know that the evaporation temp. at room pressure is ~ 100 C , so if the temperture is < 100 and the pressure is ~ 1 atm why the water are gas?
thank you
There is still a pressure of vapour from the water surface at temperatures less than 100C. The reason that boiling occurs is that the vapour pressure at 100C is equal to atmospheric. Up on a mountain top, water will boil rapidly in the low 90s (Hopeless for cooking) but, in a pressure cooker, the water will boil at a higher temperature.
Below boiling point, there is a constant exchange of molecules of and onto the surface. Don't confuse the equilibrium situation (in closed box after a long time) with what happens when there is a constant supply of fresh air over the surface.
so lets say I take a glass of water and put it in a closed box ( vacccum)- there will be no vapours?
sophiecentaur | 765 | 3,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.954801 |
https://analyticphysics.com/Runge%20Vector/Visualizing%20the%20Generalized%20Runge%20Vector.htm | 1,718,289,127,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00011.warc.gz | 82,859,697 | 5,417 | In a previous presentation the two choices for the generalized Runge vector were given as
$R ={ RL [ pcosα r -rcosφ p ] RL [ −psinα r +rsinφ p ]$
where the φ measures the angle between the radius vector and a fixed direction, while α measures the angle between the linear momentum vector and the same fixed direction. This short presentation will exhibit explicitly the constancy of these two vectors for power potentials with arbitrary real exponents.
If the angle between the radius and linear momentum vectors is designated by γ, then as pointed out at the end of the previous presentation there is a simple geometric relationship α = φ + γ among the three angles. The angle γ can be evaluated quite easily as
$γ =cos−1 [(r ·p) rp]$
using dynamic quantities available from a numerical integration of the equations of motion. The results of the numerical integration will be shown as interactive graphics.
Since the overall magnitude of the two vectors is not important in the context of this discussion, the vectors
$pcos(φ +γ) r -rcosφ p −psin(φ +γ) r +rsinφ p$
will be evaluated numerically and displayed for bound states. The vectors are constant as well for states that are not bound, but display of these states is less convenient because the evolution is not resticted in radial extent.
The Hamiltonian for the evolution of dynamic quantities is
$H=p22m +ark =pr2 2m +L2 2mr2 +ark =pr2 2m +Veffective =E$
with coupling constant a and arbitrary real exponent k on the power potential. The system can only have bound states when the effective potential, consisting of the power potential plus the angular momentum term, has a concavity or potential well. The state is bound when the value of the energy energy touches the potential well on both sides.
The conditions for bound states are exhibited in the following interactive graphic. The mass parameter has been taken as unity, which is equivalent to absorbing it into the energy and coupling constant. The light blue line is the effective potential and the green line is the value of the energy:
Manipulation of the parameters in the graphic indicates that there are bound states for all positive exponent values as long as the coupling constant and energy are also positive. It also shows that there are bounds states for negative exponents between 0 and −2 as long as the coupling constant and energy are also negative. There are bounds states of negative energy possible for exponents more negative than −2, but then the effective potential well includes the origin and orbits for these states would need to pass through the center of interaction for the problem: these states will not be considered in what follows.
In summary, positive energy states have positive exponent and coupling constant, and negative energy states have negative exponent and coupling constant with a lower bound on the exponent. The two cases will be presented separately for convenience.
Numerical integration of the equations of motion is simplest in the Cartesian coordinates of the two-dimensional plane of orbit. With the usual designations
$r2 =x2 +y2 p2 =px2 +py2$
for the two pairs of conjugate variables in the plane, the equations of motion are
$x· =pxm p·x =−kax (x2 +y2 )k2 -1 y· =pym p·y =−kay (x2 +y2 )k2 -1$
The remaining dynamic quantity is the angle between the radius and Runge vectors, which can be evaluated from the statement of conservation of angular momentum:
$φ· =Lmr2 =Lm(x2 +y2)$
It is now straightforward to integrate this set of five coupled differential equations using a simple Euler method. The mass parameter will again be taken as unity, and the evolution will begin with the conditions r = [ 1 , 0 ] and p = [ 0 , L ].
From a practical point of view one does not need to perform any integration for the angular variable, since the given intial conditions are equivalent to
$φ=cos−1 [xx2 +y2] =sin−1 [yx2 +y2]$
but this approach has the sense of presupposing the final vector will be constant without allowing the parts to come together a bit independently. One could also explicitly integrate the temporal derivative $α·$ of the angle between the linear momentum and Runge vectors, but given the simple relationship among dynamic angles this would merely add processing overhead with no gain.
The interactive graphics have restrictions placed on input parameters in order to reduce the possibility of significant error in the numerical integration. What is critical is to allow enough variation around known soluble cases to see that variation of these parameters still leads to resultant constant vectors. For negative energy states the variation occurs around the Kepler solution, which is the historical source for the Runge vector. For positive energy states the variation occurs around the solution for the simple harmonic oscillator, a case that is fully soluble because of its well-known mathematical relationship to the Kepler problem.
The two choices of the invariant Runge vector are displayed in red and blue, with their constituent portions in lighter shades of each color. The radius vector is shown in black, with the momentum vector in gray extending from its tip as is appropriate for an orbiting body. Two periods of the orbit appear in light gray to emphasize that the constant vectors exist even for nonclosed orbits.
The interactive graphic for bound states of negative energy is
and the interactive graphic for bound states of positive energy is
In both graphics one can see a portion of each choice for Runge vector lying along the radius vector, and a second portion extending from the first that is parallel to the momentum vector. The vector sum of the two portions is in all cases a constant vector for the motion.
The results of this presentation are in a definite sense trivial, since it was pointed out in another presentation that the Runge vector is essentially a rearrangement of the components of the angular momentum tensor. The two vectors displayed are really
$[0, −L] and [−L, 0]$
which are of course constant. The fun in this little presentation is evaluating the two complicated parts and adding them together for a trivial result. It should also be stressed that doing this shows that the Runge vector is a constant for all spherically symmetric potentials, in this case arbitrary powers of the radial separation, and is not restricted to the Kepler potential or the simple harmonic oscillator.
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# aphanumeric to integer?
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## aphanumeric to integer?
(OP)
Hello!
I'd like to have an IF Statement:
IF DATE LE &TODATE AND GE &FROMDATE
Unfortunately all the variables and the field DATE are alphanumeric and I get the error:
(FOC280) COMPARISON BETWEEN COMPUTATIONAL AND ALPHA VALUES IS NOT ALLOWED
What can I do here?
Thanx
Eva
### RE: aphanumeric to integer?
Try this if it's in a DEFINE or COMPUTE;
IF (DATE LE '&TODATE') AND (DATE GE '&FROMDATE')
the () are not required, but clarify the logic. If this is a screening statement in a TABLE, try:
WHERE (DATE LE '&TODATE') AND (DATE GE '&FROMDATE');
or
IF DATE FROM '&FROMDATE' TO '&TODATE'
HTH
### RE: aphanumeric to integer?
(OP)
Hi HTH,
unfortunately it doesn't work...
Eva
### RE: aphanumeric to integer?
You need to post more of your code. Run the FOCEXEC with ECHO=ON and post the code which is echoed back to your screen.
EX focexec ECHO=ON
### RE: aphanumeric to integer?
(OP)
Hi kiddpete,
I oversaw something you wrote:
IF DATE FROM '&FROMDATE' TO '&TODATE'
This worked! But I wonder why this works and the other suggestion doesn't.... Maybe you know why.
And: thanx for your help!
Eva
### RE: aphanumeric to integer?
What are the contents of &FROMDATE and &TODATE, and what is the format of DATE? That may answer the question.
### RE: aphanumeric to integer?
(OP)
They're all alphanumeric. I guess that's why you have to put the variable into ' and ' . But the rest.. I don't know.
### RE: aphanumeric to integer?
No, tell me the actual content of &FROMDATE and &TODATE and the length of DATE. Is &FROMDATE 20031231, 12/31/2003, or whatever it is.
### RE: aphanumeric to integer?
(OP)
Hi kiddpete,
all of the variables and the field DATE have the format 20030228. The field DATE is an A8.
Eva
### RE: aphanumeric to integer?
I'm drawing a blank. I don't see why all three forms don't work, but if the 'FROM TO' formulation works, go with it.
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Already a member? Close this window and log in. | 882 | 3,509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.821178 |
https://discuss.pytorch.org/t/best-way-to-split-a-3d-grid-of-points-into-blocks/70784 | 1,660,827,529,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573197.34/warc/CC-MAIN-20220818124424-20220818154424-00662.warc.gz | 193,155,042 | 5,378 | Best way to split a 3D grid of points into blocks?
Hi all. If I have a batch of uniform 3D grids of coordinate locations, with shape (for example), [1, 32, 32, 32, 3], what is the best way for me to split this up into multiple even chunks, so I could end up with something such as [1, 4096, 2, 2, 2, 3]? In other words, I’m splitting up that one big 32 x 32 x 32 cube where each point is an x, y, z coordinate location into 4096 smaller 2 x 2 x 2 cubes? Does a simple view operation make sense here, or would it throw off the coordinate values? I was looking into operations like torch.chunk, but they require a specific dimension to split along, which I’m not sure applies here.
My use case for this is that I have a smaller [1, 16, 16, 16, 3] cube, so I’m trying to match up points from this smaller shape into the corresponding cubes in the upsampled [1, 32, 32, 32, 3] shape (since a single coordinate point in the 16^3 shape corresponds to 8 points in the 32^3 shape).
For additional context, this is how I generate my 3D grid right now:
``````pxs = torch.linspace(-1, 1, 32)
pys = torch.linspace(-1, 1, 32)
pzs = torch.linspace(-1, 1, 32)
pxs = pxs.view(-1, 1, 1).expand(*shape).contiguous().view(size)
pys = pys.view(1, -1, 1).expand(*shape).contiguous().view(size)
pzs = pzs.view(1, 1, -1).expand(*shape).contiguous().view(size)
points = torch.stack([pxs, pys, pzs], dim=1)
grid_3d = torch.reshape(points, (32, 32, 32, 3))
``````
`tensor.unfold` could be used here.
Have a look at this post for an example.
1 Like
Oh awesome, I think this might be exactly what I need. Thanks!
1 Like
I hope you don’t mind me following up on this question with a related one! Tensor unfolding + your example is definitely what I’m looking for, but I was wondering if you had any ideas on ways to visualize/verify the blocks so I can double check that the blocks include the correct sections that I want (especially to double check if I’m using fold in the proper order of dimensions), as well as how they join together to make the whole block.
You could initialize the input tensor with:
``````x = torch.arange(batch_size*C*H*W)..view(batch_size, C, H, W).float()
``````
so that you can visually inspect each patch and make sure it contains the right values.
I’m not sure I understand the last sentence. Would you like to fold them back to the original input afterwards?
Thanks for your reply and help so far, I really appreciate it! To clarify my last sentence, is there a way to check in what order you combine each of the smaller blocks to generate the larger block? For example, when I split up my 32x32x32 grid into 8 smaller 16x16x16 grids, the returned tensor from the fold operations will just have essentially a list of the blocks, but I don’t know if those blocks were taken by going horizontally across the larger grid, or by going vertically. I could match these up by manually inspecting the values in each of the smaller blocks, and trying to check where they appear in the larger block to see, but that’s a bit tedious. Hopefully I explained that better?
The context here is that I’m trying to match up a coordinates from a separate smaller grid with each block. | 847 | 3,185 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.884344 |
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