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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http://www.cpp.edu/~jrfisher/www/prolog_tutorial/2_11pl.txt | 1,436,130,340,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-27/segments/1435375097710.59/warc/CC-MAIN-20150627031817-00194-ip-10-179-60-89.ec2.internal.warc.gz | 385,450,353 | 897 | /* prolog tutorial 2.11 Chess queens challenge puzzle */ perm([X|Y],Z) :- perm(Y,W), takeout(X,Z,W). perm([],[]). takeout(X,[X|R],R). takeout(X,[F|R],[F|S]) :- takeout(X,R,S). solve(P) :- perm([1,2,3,4,5,6,7,8],P), combine([1,2,3,4,5,6,7,8],P,S,D), all_diff(S), all_diff(D). combine([X1|X],[Y1|Y],[S1|S],[D1|D]) :- S1 is X1 +Y1, D1 is X1 - Y1, combine(X,Y,S,D). combine([],[],[],[]). all_diff([X|Y]) :- \+member(X,Y), all_diff(Y). all_diff([X]). | 194 | 445 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-27 | latest | en | 0.555708 |
https://www.physicsforums.com/threads/work-done-on-two-blocks.819400/ | 1,531,941,882,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590314.29/warc/CC-MAIN-20180718174111-20180718194111-00035.warc.gz | 961,531,434 | 17,780 | # Homework Help: Work Done on Two Blocks
1. Jun 17, 2015
### Staff: Mentor
1. The problem statement, all variables and given/known data
A mass m1 = 5.1 kg rests on a frictionless table and connected by a massless string over a massless pulley to another mass m2 = 5 kg which hangs freely from the string. When released, the hanging mass falls a distance d = 0.83 m.
2. Relevant equations
3. The attempt at a solution
I don't need help with a specific answer, instead I need help understanding something.
The work done by gravity on the system is 40.71 J. (W = M2gh, W = 5*9.81*0.83 = 40.71)
The work done by the tension of the string on M1 is 20.57 J.
Why is the work done by tension on M1 half of the work done by gravity? I only used the mass of M2 to find the work done on the system by gravity, so shouldn't the work done by the tension be more since M1 has more mass?
(On the other hand, gravity is the one pulling M2 in the first place, which is supplying the tension on M1. So confused...)
2. Jun 17, 2015
### Nathanael
Be more than what? More than half? (It is more than half, not by much, but m1 is not larger by much, either.)
3. Jun 17, 2015
### SammyS
Staff Emeritus
The tension also does work on m2 . (This is a negative quantity.)
4. Jun 17, 2015
### Staff: Mentor
More than the work done by gravity.
But of course that's impossible, as gravity is doing the work to begin with...
Am I correct in assuming that if M2 wasn't bound to M1, the work done by gravity would accelerate M2 to a higher velocity than when M1 and M2 are connected, but the work done would still equal 40.71 J?
I'm sorry if this is confusing. It's a multi-step problem that I've gotten almost every single step wrong the first time through.
5. Jun 17, 2015
### SammyS
Staff Emeritus
The tension is an internal force as regards the entire system. Therefore, it does no net Work to the system as a whole.
6. Jun 17, 2015
### Staff: Mentor
Okay, so gravity is pulling down on M2. M2 wants to accelerate under the force of gravity, but the tension in the rope only allows it to accelerate if M1 also accelerates. The force from M2is transferred through the string to M1 which accelerates under the force from tension. The acceleration of both blocks is the same and is equal to (M2g)/(M1+M2), or 4.86 m/s2.
The work done on each block is equal to their final kinetic energy and added together equals 40.71 J, which is the total work done on the system.
7. Jun 17, 2015
### Nathanael
The tension does negative work on m2 (it slows it down) and positive work on m1 (speeds it up). You don't need to consider it to solve any problems because (as Sammy just said) it is zero for the entire system (it does the same amount of positive work on m1 as it does negative work on m2).
If (out of curiosity) you want to find the work done by tension (call it WT) this is how I would find it:
(KE)1=WT
(KE)2=Wg-WT
(Wg is the work done by gravity and (KE)n is the kinetic energy of block 1 and 2 respectively)
First note about this system of equations: If you look at the total energy, (KE)1+(KE)2, it is equal to the work done by gravity (tension does no work on the system).
Second note: if m1=m2 then (KE)1=(KE)2 (because they have the same speed) which leads to WT=0.5Wg
(If the masses are the same the work done by tension is half the work done by gravity.)
Since they have the same speed, the ratio of kinetic energies will be the ratio of their masses. Thus you have $\frac{m_2}{m_1}=\frac{W_g-W_T}{W_T}$
In this case, this equation leads to $W_T=\frac{W_g}{\frac{5}{5.1}+1}=\frac{W_g}{1.98}=$ slightly more than half the work done by gravity.
Again, none of this is important information. The fact that the work done by tension essentially makes it ignorable. I just wrote this to explain your observation that the work done by tension is (slightly more than) half.
8. Jun 17, 2015
### Nathanael
Yes I'd say this is a satisfactory understanding. Calling tension "the force from M2" is a little iffy to me but you have the right idea.
(You said in your OP M2 was "supplying the tension" which is also a little iffy to me.)
9. Jun 17, 2015
### Staff: Mentor
I'm sorry, I don't know how else to describe it. How would you?
10. Jun 17, 2015
### Nathanael
Yeah you have the right idea it doesn't really matter.
Instead of saying "The force from M2 is transferred through the string to M1" I would say something like "The tension is the same throughout the rope, so it is the same on M_1 and M_2"
Sorry I probably shouldn't have complained because I don't really have a good way to say it either
11. Jun 17, 2015
### SammyS
Staff Emeritus
That looks very satisfactory to me.
12. Jun 17, 2015
Thanks guys! | 1,299 | 4,726 | {"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} | 3.6875 | 4 | CC-MAIN-2018-30 | latest | en | 0.946627 |
http://jean-pierre.moreau.pagesperso-orange.fr/Cplus/trk4_cpp.txt | 1,555,757,892,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-18/segments/1555578529606.64/warc/CC-MAIN-20190420100901-20190420121839-00070.warc.gz | 94,928,914 | 2,178 | /************************************************************************* * Solve a first order DE system (N=2) of the form: * * y' = F(x,y,z), z'=G(x,y,z) with conditions: y(x0)=a and z(x0)=b from * * x=x0 to x=x1 using a Runge-Kutta integration method. * * ---------------------------------------------------------------------- * * SAMPLE RUN: * * Integrate first order DE systemy: * * y' = y*z + cos(x) - 0.5*sin(2*x) * * z' = y*y + z*z - (1 + sin(x)) * * from x=0 to x=2 with initial conditions: y(0)=0 and z(0)=0 * * (25 integration steps). * * * * X Y estimated Z estimated * * ------------------------------------ * * 0.0000 0.0000000 0.0000000 * * 0.0800 0.0765518 -0.0828580 * * 0.1600 0.1452904 -0.1700792 * * 0.2400 0.2050556 -0.2597237 * * 0.3200 0.2550329 -0.3500533 * * 0.4000 0.2948004 -0.4396008 * * 0.4800 0.3243379 -0.5271915 * * 0.5600 0.3440043 -0.6119278 * * 0.6400 0.3544874 -0.6931461 * * 0.7200 0.3567373 -0.7703644 * * 0.8000 0.3518908 -0.8432315 * * 0.8800 0.3411943 -0.9114874 * * 0.9600 0.3259299 -0.9749377 * * 1.0400 0.3073506 -1.0334409 * * 1.1200 0.2866242 -1.0869048 * * 1.2000 0.2647908 -1.1352886 * * 1.2800 0.2427308 -1.1786041 * * 1.3600 0.2211465 -1.2169165 * * 1.4400 0.2005537 -1.2503410 * * 1.5200 0.1812830 -1.2790362 * * 1.6000 0.1634888 -1.3031952 * * 1.6800 0.1471644 -1.3230351 * * 1.7600 0.1321612 -1.3387869 * * 1.8400 0.1182104 -1.3506867 * * 1.9200 0.1049465 -1.3589674 * * 2.0000 0.0919314 -1.3638530 * * * * ---------------------------------------------------------------------- * * REFERENCE: "Méthode de calcul numérique- Tome 2 - Programmes en Basic * * et en Pascal By Claude Nowakowski, Edition du P.S.I., 1984"* * [BIBLI 04]. * * * * C++ Release By J-P Moreau, Paris. * * (www.jpmoreau.fr) * *************************************************************************/ #include #include #define SIZE 100 double X[SIZE],Y[SIZE],Z[SIZE]; double xl,x0,h; int k,kl,l; // y'=yz + cos(x) - 0.5sin(2x) double F(double x,double y,double z) { return (y*z + cos(x) - 0.5*sin(2*x)); } // z'=yy + zz -(1+sin(x)) double G(double x,double y,double z) { return (y*y + z*z -(1.0+sin(x))); } // Integrate sytem from x to x+h using Runge-Kutta void RK4(double x,double y,double z,double h,double *x1,double *y1, double *z1) { double c1,c2,c3,c4,d1,d2,d3,d4,h2; c1=F(x,y,z); d1=G(x,y,z); h2=h/2.0; c2=F(x+h2,y+h2*c1,z+h2*d1); d2=G(x+h2,y+h2*c1,z+h2*d1); c3=F(x+h2,y+h2*c2,z+h2*d2); d3=G(x+h2,y+h2*c2,z+h2*d2); c4=F(x+h,y+h*c3,z+h*d3); d4=G(x+h,y+h*c3,z+h*d3); *x1=x+h; *y1=y+h*(c1+2.0*c2+2.0*c3+c4)/6.0; *z1=z+h*(d1+2.0*d2+2.0*d3+d4)/6.0; } void main() { printf(" Adjust window size...\n"); getchar(); x0=0.0; // starting x xl=2.0; // ending x kl=25; // number of steps in x h=(xl-x0)/kl; // integration step X[0]=x0; Y[0]=0.0; Z[0]=0.0; // starting values // integration loop for (k=0; k | 1,200 | 2,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-18 | latest | en | 0.271554 |
https://www.teacherspayteachers.com/Product/Playful-Penguins-Math-and-Literacy-Centers-178133 | 1,638,114,929,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00372.warc.gz | 1,143,885,358 | 29,913 | # Playful Penguins Math and Literacy Centers
Anna Brantley
18.7k Followers
1st
Subjects
Resource Type
Formats Included
• PDF
Pages
99 pages
Anna Brantley
18.7k Followers
### Description
This penguin themed math and literacy center packet includes 15 hands-on printable centers, plus a bubble map, tree map, and "All About Penguins" writing paper. It would make a wonderful addition to any winter theme in your classroom!! All recording sheets come with a coordinating recording sheet. Check out the downloadable preview and the list of all the activities below for a sneak peek of everything!
Math Centers
1. Penguin Fact Families- Students show the related addition and subtraction facts for each fact family card in the boxes below.
2. Penguin Place Value- Students match up the place value cards and show their work on the recording sheet.
3. Penguin Problem Solving- Students solve each penguin problem and show their work on the recording sheet.
4. Penguin Number Neighbors- Students use a hundred's chart to help the penguins find what is 1 more, 1 less, 10 more, and 10 less than the numbers on their tummies. They show their work on the recording sheet.
5. Penguin Time- Students match the clocks to the correct time (hour & half hour) and show their work on the recording sheet.
6. Three Addend Penguin Addition Sort- Students solve and sort the addition problems by their sums. They show their work on the recording sheet.
7. Penguin Fair Shares- Students solve each fair share problem using the penguins and fish. They show their work on the recording sheet.
Literacy Centers
1. Read the Room- Students read the /en/ and /in/ words on the walls and sort them on the recording sheet.
2. Playful Penguins Making Words- Students make words with the letters in playful penguins and show their work on the recording sheet.
3. Penguin Friends Making Words- Students make words with the letters in penguin friends and show their work on the recording sheet.
4. Penguin Punctuation- Sort the sentences into two groups, those needing periods and those needing question marks. Then they show their work on the recording sheet.
5. Penguin Phonics- Students drop a counter on each board. They use the beginning {onset} and ending {rime} sound the counters landed on to make a real or nonsense word. They show their work on the recording sheet.
6. Penguin Facts and Opinions- Students sort the penguin facts and opinions and show their work on the recording sheet.
7. Penguin ABC Order -Students put the words in abc order (1st & 2nd letter) and show their work on the recording sheet.
8. Playful Penguin Sentences- Students unscramble each set of words to make a complete sentence.
9. Penguin Facts Bubble Map- Students use the bubble map to write facts about penguins.
10. Penguin Tree Map- Students use the tree map to organize important information about penguins.
11. All About Penguins Writing Paper- Students use the information in their bubble and brace maps to write about penguins. *Extra blank sheet of lined paper provided*
Anna Brantley
www.fun-n-first.blogspot.com
Total Pages
99 pages
N/A
Teaching Duration
2 Weeks
Report this Resource to TpT
Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TpT’s content guidelines. | 725 | 3,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.888884 |
https://codereview.stackexchange.com/questions/288723/roman-numerals-look-and-say-sequence | 1,718,494,917,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861618.0/warc/CC-MAIN-20240615221637-20240616011637-00680.warc.gz | 155,979,880 | 43,746 | # Roman numerals look and say sequence
I was practicing question 1 of the 2020 british informatics olympiad past paper:
The Roman look-and-say description of a string (of Is, Vs, Xs, Ls, Cs, Ds and Ms) is made by taking each block of adjacent identical letters and replacing it with the number of occurrences of that letter, given in Roman numerals (*), followed by the letter itself. A block of adjacent identical letters is never broken into smaller pieces before describing it.
Write a program that reads in a Roman numeral representing a number between 1 and 3999 inclusive, followed by n (1 ≤ n ≤ 50). You should apply the Roman look-and-say description n times and then output the number of Is in the final description, followed by the number of Vs.
I wrote this python code to attempt to solve this:
ROMAN_NUMERALS = {
1000: "M",
900: "CM",
500: "D",
400: "DC",
100: "C",
90: "XC",
50: "L",
40: "XL",
10: "X",
9: "IX",
5: "V",
4: "IV",
1: "I"
}
def convert_to_roman_numeral(n):
s = ""
while n > 0:
for key, value in ROMAN_NUMERALS.items():
if n - key >= 0:
s += value
n -= key
break
return s
def get_look_and_say_string(initial_string):
output = ""
while initial_string != "":
for i in range(len(initial_string)):
if initial_string[i] != initial_string[0]:
output += convert_to_roman_numeral(i) + initial_string[i - 1]
initial_string = initial_string[i:]
break
elif len(initial_string) == i + 1:
output += convert_to_roman_numeral(i + 1) + initial_string[i]
return output
s, apply_times = input().split()
for i in range(int(apply_times)):
s = get_look_and_say_string(s)
print(s.count("I"), s.count("V"))
For example, given the input MMXX 1, it outputs 4 0.
(A list of test cases can be found in the mark scheme here)
I would be grateful to hear about any improvements that could be made to this code.
• The code requires that the order of dict entries in ROMAN_NUMERALS is preserved. This is true for CPython since 3.6 and is part of the standard since Python 3.7. This should be specified explicitly or, as you don't use ROMAN_NUMERALS as dict but only as sequence, you could instead use a list or tuple of tuples (key, value) to avoid any possible problems. Commented Jan 2 at 14:55
Michael Butscher is correct to suggest that you should not be using a dictionary; just use a tuple of tuples. Otherwise, for a light refactor:
Replace your incremental string concatenation with ''.join(generator) style, which performs better under some scenarios (though you will want to measure this, and I haven't).
Replace for i in range(len(initial_string)): with a call to enumerate so that you get both the current symbol and its index.
Your elif len(initial_string) == i + 1: is a smell, because it's written to execute at the end of a loop - so just... move it out of the loop, under an else.
Separate an outer input(); print() function from a testable, non-console function; and write unit tests.
"""
"""
from typing import Iterator
ROMAN_NUMERALS = (
(1000, 'M'),
(900, 'CM'),
(500, 'D'),
(400, 'DC'),
(100, 'C'),
(90, 'XC'),
(50, 'L'),
(40, 'XL'),
(10, 'X'),
(9, 'IX'),
(5, 'V'),
(4, 'IV'),
(1, 'I'),
)
def convert_to_roman_numeral(n: int) -> Iterator[str]:
while n > 0:
for increment, symbol in ROMAN_NUMERALS:
new_n = n - increment
if new_n >= 0:
yield symbol
n = new_n
break
def get_look_and_say_string(initial_string: str) -> Iterator[str]:
while initial_string != '':
for i, str_i in enumerate(initial_string):
if str_i != initial_string[0]:
yield from convert_to_roman_numeral(i)
yield initial_string[i - 1]
initial_string = initial_string[i:]
break
else:
yield from convert_to_roman_numeral(len(initial_string))
yield initial_string[-1]
return
def problem(input_str: str) -> tuple[int, int]:
s, apply_times = input_str.split()
for _ in range(int(apply_times)):
s = ''.join(get_look_and_say_string(s))
return s.count('I'), s.count('V')
def test_op() -> None:
assert problem('MMXX 1') == (4, 0)
def test_official() -> None:
assert problem('MMXX 3') == (6, 2)
assert problem('C 1') == (1, 0)
assert problem('V 1') == (1, 1)
assert problem('III 2') == (2, 1)
assert problem('I 7') == (6, 2)
assert problem('MMXX 8') == (30, 12)
assert problem('II 20') == (101, 37)
assert problem('IV 20') == (121, 46)
assert problem('MDCLX 30') == (1795, 695)
assert problem('M 50') == (2858, 1103)
assert problem('V 50') == (2858, 1104)
assert problem('MMDCCCLXXXVIII 50') == (19013, 7333)
def main() -> None:
i, v = problem(input())
print(i, v)
if __name__ == '__main__':
test_op()
test_official()
# main()
## Logarithmic lookup
Your inner convert_to_roman_numeral should probably be changed so that it does a bisection search to get digits:
import bisect
from typing import Iterator
ROMAN_VALUES = (
1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000,
)
ROMAN_SYMBOLS = (
'I', 'IV', 'V', 'IX', 'X', 'XL', 'L', 'XC', 'C', 'DC', 'D', 'CM', 'M',
)
def convert_to_roman_numeral(n: int) -> Iterator[str]:
hi = len(ROMAN_VALUES)
while n > 0:
i_low = bisect.bisect_right(a=ROMAN_VALUES, x=n, hi=hi) - 1
value = ROMAN_VALUES[i_low]
repeat = n // value
yield ROMAN_SYMBOLS[i_low] * repeat
n -= value * repeat
hi = i_low
## Direct logarithmic bound
You can fairly straightforwardly prove that for some n input to convert_to_roman_numeral, the index i into the ROMAN arrays is bounded by
$$\lceil 4 \log_{10} \frac n 4 \rceil \le i \le \lfloor 4 \log_{10} n \rfloor$$
To demonstrate, here are values from 10 through 100 and their index bounds:
From this, you can write an alternative convert_to_roman_numeral that starts on a bound and loops linearly through a very short (1-2) sequence of values to find the target. Whereas in theory this is O(1) compared to bisection's O(log(n)), it still performs slightly better than this; to compare:
import bisect
import math
import timeit
from typing import Iterator
import numpy as np
import pandas as pd
import seaborn
from matplotlib import pyplot as plt
ROMAN_NUMERALS_DICT = {
1000: "M",
900: "CM",
500: "D",
400: "DC",
100: "C",
90: "XC",
50: "L",
40: "XL",
10: "X",
9: "IX",
5: "V",
4: "IV",
1: "I"
}
def convert_roman_op(n):
s = ""
while n > 0:
for key, value in ROMAN_NUMERALS_DICT.items():
if n - key >= 0:
s += value
n -= key
break
return s
ROMAN_NUMERALS_TUPLE = (
(1000, 'M'),
(900, 'CM'),
(500, 'D'),
(400, 'DC'),
(100, 'C'),
(90, 'XC'),
(50, 'L'),
(40, 'XL'),
(10, 'X'),
(9, 'IX'),
(5, 'V'),
(4, 'IV'),
(1, 'I'),
)
def convert_roman_inner(n: int) -> Iterator[str]:
while n > 0:
for increment, symbol in ROMAN_NUMERALS_TUPLE:
new_n = n - increment
if new_n >= 0:
yield symbol
n = new_n
break
def convert_roman_gen(n: int) -> str:
return ''.join(convert_roman_inner(n))
ROMAN_VALUES = (
1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000,
)
ROMAN_SYMBOLS = (
'I', 'IV', 'V', 'IX', 'X', 'XL', 'L', 'XC', 'C', 'DC', 'D', 'CM', 'M',
)
def convert_roman_inner_bisect(n: int) -> Iterator[str]:
hi = len(ROMAN_VALUES)
while n > 0:
i_low = bisect.bisect_right(a=ROMAN_VALUES, x=n, hi=hi) - 1
value = ROMAN_VALUES[i_low]
repeat = n // value
yield ROMAN_SYMBOLS[i_low] * repeat
n -= value * repeat
hi = i_low
def convert_roman_bisect(n: int) -> str:
return ''.join(convert_roman_inner_bisect(n))
def convert_roman_inner_log(n: int) -> Iterator[str]:
max_index = len(ROMAN_VALUES) - 1
while n > 0:
top = min(max_index, int(4*math.log10(n)))
value = ROMAN_VALUES[top]
while value > n:
top -= 1
value = ROMAN_VALUES[top]
repeat = n // value
yield ROMAN_SYMBOLS[top] * repeat
n -= value * repeat
def convert_roman_log(n: int) -> str:
return ''.join(convert_roman_inner_log(n))
methods = (
convert_roman_op,
convert_roman_gen,
convert_roman_bisect,
convert_roman_log,
)
def benchmark() -> None:
results = []
for size in (
10**np.linspace(0, 5, 101)
).astype(int):
reps = max(1, 100 // size)
ref = None
for method in methods:
def run() -> None:
nonlocal ref
output = method(size)
if ref is None:
ref = output
else:
assert ref == output
for _ in range(reps):
t = timeit.timeit(stmt=run, number=1)
results.append((size, method.__name__, t))
df = pd.DataFrame(
data=results,
columns=['size', 'method', 'dur'],
)
fig, ax = plt.subplots()
ax: plt.Axes
seaborn.lineplot(ax=ax, data=df, x='size', y='dur', hue='method')
ax.set_xscale('log')
ax.set_yscale('log')
plt.show()
if __name__ == '__main__':
benchmark()
## Iterating in look-and-say
AJNeufeld has an excellent suggestion to use itertools (or more_itertools which I don't demonstrate). It both cleans up and speeds up your main loop.
"""
"""
import bisect
import itertools
import time
from typing import Iterator
ROMAN_VALUES = (
1, 4, 5, 9, 10, 40, 50, 90, 100, 400, 500, 900, 1000,
)
ROMAN_SYMBOLS = (
'I', 'IV', 'V', 'IX', 'X', 'XL', 'L', 'XC', 'C', 'DC', 'D', 'CM', 'M',
)
def convert_to_roman_numeral(n: int) -> Iterator[str]:
hi = len(ROMAN_VALUES)
while n > 0:
i_low = bisect.bisect_right(a=ROMAN_VALUES, x=n, hi=hi) - 1
value = ROMAN_VALUES[i_low]
repeat = n // value
yield ROMAN_SYMBOLS[i_low] * repeat
n -= value * repeat
hi = i_low
def get_look_and_say_op(initial_string):
output = ""
while initial_string != "":
for i in range(len(initial_string)):
if initial_string[i] != initial_string[0]:
output += ''.join(convert_to_roman_numeral(i)) + initial_string[i - 1]
initial_string = initial_string[i:]
break
elif len(initial_string) == i + 1:
output += ''.join(convert_to_roman_numeral(i + 1)) + initial_string[i]
return output
def problem_op(input_str):
s, apply_times = input_str.split()
for _ in range(int(apply_times)):
s = get_look_and_say_op(s)
return s.count('I'), s.count('V')
def get_look_and_say_gen(initial_string: str) -> Iterator[str]:
while initial_string != '':
for i, str_i in enumerate(initial_string):
if str_i != initial_string[0]:
yield from convert_to_roman_numeral(i)
yield initial_string[i - 1]
initial_string = initial_string[i:]
break
else:
yield from convert_to_roman_numeral(len(initial_string))
yield initial_string[-1]
return
def problem_gen(input_str: str) -> tuple[int, int]:
s, apply_times = input_str.split()
for _ in range(int(apply_times)):
s = ''.join(get_look_and_say_gen(s))
return s.count('I'), s.count('V')
def get_look_and_say_ajneufeld(initial_string: str) -> Iterator[str]:
for letter, repeats in (
(k, sum(1 for _ in g))
for k, g in itertools.groupby(initial_string)
):
yield from convert_to_roman_numeral(repeats)
yield letter
def problem_ajneufeld(input_str: str) -> tuple[int, int]:
s, apply_times = input_str.split()
for _ in range(int(apply_times)):
s = ''.join(get_look_and_say_ajneufeld(s))
return s.count('I'), s.count('V')
def benchmark() -> None:
largest_test_case = 'MMDCCCLXXXVIII 50'
for method in (
problem_op,
problem_gen,
problem_ajneufeld,
):
start = time.perf_counter()
result = method(largest_test_case)
end = time.perf_counter()
assert result == (19013, 7333)
print(method.__name__, end - start)
if __name__ == '__main__':
benchmark()
problem_op 0.12485683699924266
problem_gen 0.09267444600118324
problem_ajneufeld 0.0854689560001134
• I cringe when I see your get_look_and_say_string(initial_string) implementation. Use: for letter, repeats in more_itertools.run_length.encode(initial_string): yield from convert_to_roman_numeral(repeats) yield letter. Otherwise, superb answer. Commented Jan 3 at 21:33
• @AJNeufeld I appreciate that. I can only assume, however, that the Olympiad doesn't support third-party libraries. Commented Jan 3 at 22:50
• True, but the OP was only doing it “for practice”, so I thought it would be allowed. Since it is part of the standard library, itertools should be allowed: for letter, repeats in ((k, sum(1 for _ in g)) for k, g in itertools.groupby(iterable)): Commented Jan 4 at 3:43
• @AJNeufeld I like it! I've included it in a comparison. Commented Jan 4 at 16:21
• I feel like get_look_and_say_string would be greatly simplified by re.search(r'((.)\2*)',initial_string). Commented Jan 4 at 19:55 | 3,580 | 11,869 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.805389 |
https://pdfkul.com/march-meeting_5a30147e1723dd5c0d51d0f8.html | 1,670,252,478,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711017.45/warc/CC-MAIN-20221205132617-20221205162617-00210.warc.gz | 481,035,867 | 11,175 | M. Markham, D.J. Twitchen
C. Abellán, W. Amaya, V. Pruneri, M.W. Mitchell
B. Hensen, H. Bernien, A.E. Dréau, A. Reiserer, N. Kalb, M.S. Blok, J. Ruitenberg, R.F.L. Vermeulen, R.N. Schouten, D. Elkouss, S. Wehner, T.H. Taminiau, R. Hanson
Experimental loophole-free violation of a Bell inequality using entangled electron spins separated by 1.3 km arXiv:1508.05949
• Introduction • Setup at TU Delft • Data evaluation • Discussion
• Introduction • Setup at TU Delft • Data evaluation • Discussion
Bell’s theorem
1. 2. 3.
Communication is limited to speed of light Settings a,b can be freely chosen Local hidden variable model (LHVM) governs the experiment
3’. Quantum theory predicts
Bell’s game Game score 1 if −1 𝑎𝑎𝑎𝑎 𝑥𝑥𝑥𝑥 = 1 C= 0 else 1. 2. 3.
Communication is limited to speed of light Settings a,b can be freely chosen uniformly at random Local hidden variable model (LHVM) governs the experiment
P(𝐶𝐶 = 1) ≤ 3/4
3’. Quantum theory predicts
P(𝐶𝐶 = 1) ≤(√2+2)/4≈0.85
Unless…
Locality loophole
Time t
x ∈ {−1,1}
A
a ∈ {0,1}
y ∈ {−1,1} Distance L
B
b ∈ {0,1}
- If L is so short that Alice can signal to Bob before y is produced then y=f(a,b) - Strategy with winning probability one: x=1, 𝑦𝑦 = −1
𝑎𝑎𝑏𝑏
Detection loophole
A
a ∈ {0,1}
y ∈ {−1,1} Distance L
B
b ∈ {0,1}
- Outcomes are not always conclusive data discarded - Strategy with winning probability one: if a=0 output x=1, if b=0 output y=1 else no output
efficiency η
Time t
x ∈ {−1,1}
Bell experiments Locality loophole addressed (photons)
Aspect (1982), Weihs (1998), Tittel (1999)
Detection loophole closed (ions, atoms, superconducting qubits, photons)
Rowe (2001), Matsukevich (2008), Ansmann (2009), Giustina (2013), Christensen (2013)
Detection loophole closed AND locality loophole addressed = “loophole-free” Bell test
• Introduction • Setup at TU Delft • Data evaluation • Discussion
Time
Setup for closing the loopholes
Space
Bell, J. Phys. (Paris) Colloq. C2, 41 (1981).
GO / NOT GO
𝑥𝑥 ∈ {−1,1}
𝑥𝑥 ∈ {−1,1}
Matter qubits: “easy” high fidelity readout
𝑎𝑎 ∈ {0,1}
L=? limited by 3 us readout L>1 km
𝑏𝑏 ∈ {0,1}
Phys. Rev. Lett. 91, 110405 (2003).
Loophole-free Bell test campus ✗
Bob lab Alice lab ✗
Beamsplitter lab
Experimental scheme Alice
time
space
Beam splitter station
Bob
Experimental scheme Initialization Initialization of the qubits into a well defined state Nature 477, 574 (2011)
time
space
Alice
Beam splitter station
Bob
Experimental scheme Spin-photon entanglement Create spin-photon entanglement on both sides. Send photons to beam splitter.
Alice
Beam splitter station
Bob
Experimental scheme Alice Entanglement generation
* click!
Photon detection projects spins into entangled state Barrett and Kok, PRA 71, 060310 (2005) Nature 497, 86 (2013)
time
space
Beam splitter station
Bob
Experimental scheme Random basis choice
1! Fresh Quantum Random Number Generator Optics Express 22, 1645 (2014) arXiv:1506.02712 (2015)
time
space
Alice
Beam splitter station
Bob
Experimental scheme Readout Distance such that signaling is not possible:
Alice
Beam splitter station
Bob
end
Detection loophole closed Nature 477, 574 (2011)
start
• Introduction • Setup at TU Delft • Data evaluation • Discussion
Statistical analysis LHVM imply limitations in probability or expectation
Finite number of events
Hypothesis test - Null hypothesis: the experiment was governed by a LHVM - p-value: Probability of data at least as extreme as that observed given a LHVM maximized over all LHVMs. - p-value is not the probability that our experiment is ruled by a LHVM
Rigorous bound • No unfounded assumptions: i.i.d. • No approximations: Gaussian
Gill, Found. of Prob. and Phy., 179-206 (2003) Zhang, Glancy, Knill, PRA 84 062118 (2011) Bierhorst, Found. Phys. 44, 736-761 (2014)
Challenges for modelling the experiment t=1(YES) or 0(NO)
T
go?
1 if t −1 C= 0 else
𝑎𝑎𝑎𝑎 𝑥𝑥𝑥𝑥
=1
Local hidden variable model (LHVM) governs the experiment: 1. Setting choices are independent conditioned on the history 2. Inputs have a bias from uniform 𝝉𝝉 (𝟏𝟏𝟏𝟏−𝟓𝟓 ) 3. Settings and outcomes at both sites are conditionally independent on the history 4. Arbitrary statistics at heralding station
P(𝐶𝐶 = 1) ≤ 3/4+3(𝜏𝜏 + 𝜏𝜏 2 )
Upper bound on the p-value Experimental Data - n = number of GO rounds - k = number of wins - 𝜏𝜏 = bias of the RNGs
p-value ≔ max 𝑃𝑃 𝑎𝑎𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑘𝑘 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑛𝑛 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿) 𝐿𝐿𝐻𝐻𝐻𝐻𝑀𝑀
Remarks - Upper bound is the tail of an i.i.d. distribution -
Elkouss, Wehner in preparation (2015)
Evaluation of the experiment Gathered data - 1 event per hour - 245 events over more than 220 hours - 196 “wins” over the 245 events
How do we interpret it?
Hypothesis test - Null hypothesis: the experiment was governed by a LHVM - p-value: probability of data at least as extreme as that observed given a LHVM maximized over all LHVMs.
p-value ≔ max 𝑃𝑃 𝑎𝑎𝑎𝑎 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 196 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 245 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿) ≤ 0.039 𝐿𝐿𝐻𝐻𝐻𝐻𝑀𝑀
• Introduction • Setup at TU Delft • Data evaluation • Discussion
Conclusions
First loophole-free Bell inequality violation arXiv:1508.05949 (2015)
Next: • Independent confirmation • Proof-of-principle of device-independent security … • Quantum networks
Thank you!
どうもありがとうございます！
## march meeting
Experimental loophole-free violation of a Bell inequality ... Communication is limited to speed of light. 2. Settings a,b can ... addressed = âloophole-freeâ Bell test.
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Mar 22, 2018 - Special Meeting of Future Nillumbik Committee minutes. 22 March 2018. Nillumbik Shire Council. Contents. 1. Welcome and apologies. 1. 2. | 2,565 | 7,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-49 | latest | en | 0.616468 |
http://aslstem.cs.washington.edu/topics/view/8135 | 1,529,768,062,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865098.25/warc/CC-MAIN-20180623152108-20180623172108-00303.warc.gz | 25,590,807 | 5,812 | ## Viewing topic: Heaviside Step Function
### Highest Rated Sign
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#### Heaviside Step Function
• Definition: The Heaviside step function is a mathematical function denoted H(x), or sometimes theta(x) or u(x) (Abramowitz and Stegun 1972, p. 1020), and also known as the "unit step function." The term "Heaviside step function" and its symbol can represent either a piecewise constant function or a generalized function. When defined as a piecewise constant function, the Heaviside step function is given by H(x)={0 when x<0; 1/2 when x=0; 1 when x>0 . When defined as a generalized function, it can be defined as a function theta(x) such that integral[theta(x)phi^'(x)dx]=-phi(0) for phi^'(x) the derivative of a sufficiently smooth function phi(x) that decays sufficiently quickly (Kanwal 1998).
Source: http://mathworld.wolfram.com
• There are no comments for this topic. | 237 | 907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.827142 |
https://math.stackexchange.com/questions/1906585/motivation-for-studying-nilpotent-and-solvable-lie-algebras | 1,726,375,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00088.warc.gz | 349,047,429 | 40,476 | # Motivation for Studying Nilpotent and Solvable Lie Algebras
I have started to learn about Lie Algebras from Humphreys book of Lie Algebras. Currently I am studying about nilpotent and solvable Lie Algebras. I think I understand the definition and I also understand the proofs of Engel's theorem and Lie's Theorem, but I don't understand the motivation for studying solvable and nilpotent Lie Algebras?
Is there any result that says that any 'good' Lie algebra decomposes into direct sum of solvable and nilpotent Lie Algebras? (or some similar result)
• Lie theory is an inmmense theory, and any modern exposition is the result of years of collective experience in gauging the importance of its many parts. This usually means that you have to withhold your need for motivation till later in the book. It sort of sucks, yes, but you will need to learn to do this eventually. Commented Aug 28, 2016 at 20:52
• @Mariano Suárez-Álvarez this sort of attitude mostly comes from the lack of familiarity with the origins of the theory. Solvability was called "integrability" by Lie himself and had to do with literally being able to solve a system of first-order PDE's in quadratures if it admits a symmetry described by a solvable Lie algebra, by analogy with Galois theory. This is extremely easy to see and would be very useful for all students of Lie theory to see early on. It's also no accident that Lie formalized these concepts just 2 years into his studies, long before abstract Lie groups even existed. Commented Jul 11 at 15:10
There is an important result, the Levi decomposition theorem that says that under good conditions a Lie algebra is a semidirect product of a semisimple algebra and a solvable one.
Nilpotent and solvable algebras appear quite naturally when one studies the structure theory of Lie algebras, leading for example to the theorem of Levi I mentioned. But quite independently of that, nilpotent and solvable algebras are interesting because lots of algebras are nilpotent or solvable, algebras that we encounter in real life. This may not be very obvious when one is starting to learn abut this, of course, and it can be appreciated only with experience and lots of examples.
• Thank you sir! Is there any 'good' book of Lie Algebras which gives some motivation before defining the concepts? Commented Aug 29, 2016 at 5:27
• Your motivation should be that you want to learn about Lie algebras and you should trust respected authors that they made a sensible choices of topics. If you browse a couple of books and they all touch on solvable algebras and nilpotent ones, then that should count as evidence that it is a useful subject. When you start learning advanced stuff, you need to exercise patience. Commented Aug 29, 2016 at 6:12
• In other words: presumably you do have already some motivation to learn about Lie algebras, for you are doing it. Given that, learn the subject, entrusting such great expositors as Humphreys to lead you into the subject. Commented Aug 29, 2016 at 6:14
• Thank you.I hope I will get much motivations with experience and by solving examples/exercises.Thank you for your great advice. Commented Aug 29, 2016 at 6:18
• Infact your last comment is also a motivation for keep on reading the fantastic book of Humphreys.Thanks again : ) Commented Aug 29, 2016 at 6:28
I know it's been over 7 years, but I'll post this excerpt from the book "Emergence of the theory of Lie groups" by Thomas Hawkins because I find the accepted answer quite dismissive of the original motivation behind these concepts, which is in fact quite natural and simple. This "teleological" attitude of treating fundamental concepts as a mere means to an end (such as proving some highbrow theorems) is unfortunately plaguing exact and natural sciences, whereas I actually find the ability to clearly answer questions like yours very important for mathematics education. As explained in the excerpt, "solvability" referred literally to the complete integrability of a system of linear PDE's. Lie himself exclusively thought of continuous groups not as abstract objects, but as collections of invertible transformations acting on the independent variables in a system of PDE's. It turns out that if the symmetry is solvable, then you can construct a complete set of solutions by solving a sequence of ODE's.
• Can you give any examples of such a system of PDEs? I've always been curious about Lie's original motivations for studying Lie theory but I don't know a single example of the types of systems to which it was supposed to apply. Commented Jul 11 at 15:41
• @QiaochuYuan not anything I can type in a comment, but page 156 of this book describes a system of two 4-th order ODE's with a solvable 3-dimensional symmetry. Commented Jul 11 at 17:35
• @QiaochuYuan also every 2-dimensional Lie algebra is solvable, so, for example, a system of 3-rd order ODE's with a 2-dimensional symmetry group is always solvable. Commented Jul 11 at 17:38
• @QiaochuYuan the exercises at the end of the chapter I linked also contain a lot more elementary examples that are very instructive. Commented Jul 11 at 17:42
• Nice, thanks for the reference! Commented Jul 11 at 17:47 | 1,225 | 5,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-38 | latest | en | 0.96246 |
https://codedump.io/share/xc9PcOaUw4tM/1/r---boolean-operators-ampamp-and- | 1,521,839,810,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648594.80/warc/CC-MAIN-20180323200519-20180323220519-00396.warc.gz | 551,975,782 | 8,784 | SFun28 - 1 year ago 65
R Question
# R - boolean operators && and ||
According to the R language definition the difference between & and && (correspondingly | and ||) is that the former is vectorized while the later is not.
According to this site:
http://www.stat.psu.edu/~dhunter/R/html/base/html/Logic.html
I read the difference akin to the difference between an "And" and "AndAlso" (correspondingly "Or" and "OrElse")...meaning that not all evaluations if they don't have to be (i.e. A or B or C is always true if A is true, so stop evaluating if A is true)
Could someone shed light here? Also, is there an AndAlso and OrElse in R?
Answer Source
The shorter ones are vectorized, meaning they can return a vector, like this:
``````((-2:2) >= 0) & ((-2:2) <= 0)
# [1] FALSE FALSE TRUE FALSE FALSE
``````
The longer form evaluates left to right examining only the first element of each vector, so the above gives
``````((-2:2) >= 0) && ((-2:2) <= 0)
# [1] FALSE
``````
As the help page says, this makes the longer form "appropriate for programming control-flow and [is] typically preferred in if clauses."
So you want to use the long forms only when you are certain the vectors are length one.
You should be absolutely certain your vectors are only length 1, such as in cases where they are functions that return only length 1 booleans. You want to use the short forms if the vectors are length possibly >1. So if you're not absolutely sure, you should either check first, or use the short form and then use `all` and `any` to reduce it to length one for use in control flow statements, like `if`.
The functions `all` and `any` are often used on the result of a vectorized comparison to see if all or any of the comparisons are true, respectively. The results from these functions are sure to be length 1 so they are appropriate for use in if clauses, while the results from the vectorized comparison are not. (Though those results would be appropriate for use in `ifelse`.
One final difference: the `&&` and `||` only evaluate as many terms as they need to (which seems to be what is meant by short-circuiting). For example, here's a comparison using an undefined value `a`; if it didn't short-circuit, as `&` and `|` don't, it would give an error.
``````a
# Error: object 'a' not found
TRUE || a
# [1] TRUE
FALSE && a
# [1] FALSE
TRUE | a
# Error: object 'a' not found
FALSE & a
# Error: object 'a' not found
``````
Finally, see section 8.2.17 in The R Inferno, titled "and and andand".
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download | 678 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-13 | latest | en | 0.923573 |
https://scicomp.stackexchange.com/questions/26152/good-numerical-method-for-solving-the-kadomtsev-petviashvili-equations-is-there | 1,716,183,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058222.5/warc/CC-MAIN-20240520045803-20240520075803-00081.warc.gz | 460,316,069 | 38,831 | # Good numerical method for solving the Kadomtsev Petviashvili equations. Is there an analytical solution?
I need to solve the Kadomtsev Petviashvili (KP) equations
$$\partial_x(\partial_t u+u \partial_x u+\epsilon^2\partial_{xxx}u)+\lambda\partial_{yy}u=0$$
where $$\lambda=\pm 1 \;.$$ My questions are:
1) Which is the best numerical method to solve them? Or maybe a good method that is simple?
2) What does wikipidia mean with: "the KP equation is completely integrable"? I don't think it means KP has a general analytical solution, otherwise why are people still coming out with numerical methods for solving it? See for example this paper.
• If you are solving KP and haven't yet learned what an integrable system is, it's worth asking what your motivation is. I'm not being condescending and I'll provide you with references to help, but I can do that better if you explain your end goal. Which numerical method to use also depends partly on what kind of solutions you care about. Feb 10, 2017 at 17:59
• Maybe edit that comment? I don't think there is a way to write "If you are doing X and haven't learned Y, then..." that doesn't sound condescending. I agree with the general point, though. Answers are probably going to be different if you care about KP as integrable system, KP as numerical methods example, or KP as a model for a real-world problem.
– AJK
Feb 10, 2017 at 23:27
• I need a good numerical method with the aim of using KP as a model for real world problems. If you could also explain what a completely integrable system is that would be nice. Thanks a lot. Feb 11, 2017 at 3:41
• David if you have some hints on some good numerical approach that would be great. thanks Apr 8, 2017 at 12:55 | 440 | 1,721 | {"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} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.960886 |
https://math.answers.com/questions/How_do_you_writ_11_to_the_3rd_power_in_standard_form | 1,709,599,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00502.warc.gz | 381,260,844 | 45,769 | 0
# How do you writ 11 to the 3rd power in standard form?
Updated: 9/26/2023
Wiki User
7y ago
11^3 = 11 x 11 x 11 = 1331
Wiki User
7y ago
Anonymous
Lvl 1
3y ago
11
121
1 331 | 79 | 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} | 2.796875 | 3 | CC-MAIN-2024-10 | latest | en | 0.777813 |
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Question
Updated 12/22/2015 11:47:58 AM
Confirmed by Janet17 [12/22/2015 11:47:58 AM]
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User: Which type of meter is used in these lines from a poem? I could not love thee, Dear, so much Loved I not honor more
Weegy: The type of meter that is used on the line " I could not love thee, Dear, so much Loved I not honor more" is iambic meter.
emdjay23|Points 65820|
User: Which is the opposite of trochaic meter?
Weegy: The opposite of trochaic meter is iambic meter.
User: Which is the opposite of dactylic meter?
Weegy: The opposite of the dactylic meter is the anapaest meter.
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Updated 12/22/2015 11:47:58 AM
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Viewed: 86 - Published at: 9 months ago
Introduction
Sorting is one of the fundamental techniques used in solving problems, especially in those related to writing and implementing efficient algorithms.
Usually, sorting is paired with searching - meaning we first sort elements in the given collection, then search for something within it, as it is generally easier to search for something in a sorted, rather than an unsorted collection, as we can make educated guesses and impose assumptions on the data.
There are many algorithms that can efficiently sort elements, but in this guide we'll be taking a look at how to implement Heap Sort in Java.
In order to understand how Heap Sort works, we first need to understand the structure it's based on - the heap. In this article we'll be talking in terms of a binary heap specifically, but with minor adjustments the same principals can be generalized to other heap structures as well.
We'll be doing another implementation without heaps - but rather, `PriorityQueue`s, which boil the algorithm down to a single line.
`````` <h3 id="heapasadatastructure">Heap as a Data Structure</h3>
``````
A heap is a specialized tree-based data structure which is a complete binary tree that satisfies the heap property, that is, for each node all of it's children are in a relation to it. In a max heap, for a given parent P and a child C, the value of P is greater and or equal to the value of the child C. Analogously, in a min heap, the value of P is lesser than or equal to the value of it's child C. The node at the "top" of the heap (i.e. the node that has no parents) is called the root. Here's an example of a min heap (left) and a max heap (right):
As we mentioned earlier, we see the heap as a tree-based data structure. However, we'll represent it with a simple array and just define how each node (child) relates to it's parent. Assuming our array begins from an index `0`, we can represent the max heap from the illustration above with the following array:
``````53, 25, 41, 12, 6, 31, 18
``````
We can also explain this representation as reading the graph level by level, from left to right. Essentially, we have defined some kind of a relation between a parent node and a child node. For the `k-th` element of the array, we can find it's children on the positions `2*k+1` and `2*k+2`, assuming the indexing starts from `0`. Similarly, we can find the parent of the `k-th` element on the position `(k-1)/2`. Earlier we mentioned that heap is a complete binary tree. A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled and all nodes are left-aligned.
Note: A complete binary tree can be the same as a full binary tree, but at it's core is a different concept, where a full binary tree represents a tree in which every node other than the leaves have exactly two children.
`````` To explain the concept of a complete binary tree a bit further let's look at an example of the max heap from the illustration earlier. If we remove the nodes 12 and 6 we get the following binary tree:
``````
This tree will be represented in an array as:
``````53, 25, 41, -, -, 31, 18
``````
We can see that this isn't a complete binary tree, since the nodes on level `2` (if the root node is at level `0`), aren't left-aligned. While on the other hand, the following binary tree would represent a complete binary tree:
The array for this tree would be:
``````53, 25, 41, 12, 6
``````
From the short example above, we can see that intuitively a complete binary tree is represented with an array that has no "gaps" in it, that is, the positions we represented in the first array above as `-`. Continuing on our explanation of the heap - the process of inserting and deleting elements from it is a crucial step in Heap Sort.
Note: We'll be focusing on a max heap, but keep in mind that everything that applies to the max heap also applies to the min heap as well.
`````` <h4 id="insertinganelementintothemaxheap">Inserting an Element Into the Max Heap</h4>
``````
Using the same max heap we previously had, let's say we want to add element `60`. On first look, it's evident that `60` would be the largest element in our heap, so it should become the root element. But that raises another question: how do we simultaneously keep the form of a complete binary tree and add `60` at the same time? Let's begin by placing the element at the last position in our heap array, and get something like this:
``````// 0 1 2 3 4 5 6 7
53, 25, 41, 12, 6, 31, 18, 60
``````
The numbers in the row above represent the index positions of the array
`````` As discussed earlier, children of the k-th node are located at positions 2*k+1 and 2*k+2, while the parent of every node is at (k-1)/2. Following the same pattern, 60 would be a child of 12.
``````
Now, this disturbs the form of our max heap, as comparing and checking if `60` is lesser than or equal than `12` yields a negative answer. What we'll do is swap these two, as we're sure that there are no lesser numbers than `60` down the binary tree, as `60` was a leaf. After the swap, we get the following:
``````// 0 1 2 3 4 5 6 7
53, 25, 41, 60, 6, 31, 18, 12
``````
We repeat the same step as earlier until `60` is at it's right spot. The parent element of `60` would now be `25`. We swap these two, after which the parent element of `60` is `53`, after which we swap them as well, ending up with a max heap:
``````// 0 1 2 3 4 5 6 7
60, 53, 41, 25, 6, 31, 18, 12
``````
Deleting an Element From the Max Heap
Now, let us discuss removing an element. We'll be using the same max heap as earlier (without the addition of `60`). When talking about removing an element from the heap, the standard deletion operation implies that we should be only removing the root element. In case of the max heap, this is the largest element, and in the case of min heap the smallest. Removing an element from the heap is as simple as removing it from the array. However, this creates a new problem as the removal creates a "gap" in our binary tree, making it not complete. Luckily for us, the solution is just as simple - we replace the deleted root element with the element that's farthest-right on the lowest level in the heap. Doing this guarantees us that we'll have a complete binary tree once again, but yet again creates a new potential problem: whilst our binary tree is now complete, it's may not be a heap. So how do we go about solving this? Let's discuss removing an element on the same max heap as earlier (before adding `60`). After we remove our root, and we move our farthest-right element in it's spot, we have the following:
``````// 0 1 2 3 4 5 6
18, 25, 41, 12, 6, 31
``````
Note: The element at the position 6 is left empty on purpose - this will be important later.
`````` Represented like this, our array isn't a max heap. What we should do next is compare 18 to it's children, specifically to the bigger of the two, and in this case that is 41. If the bigger of the two children is larger than the parent, we swap the two.
``````
After doing this, we get the following array:
``````// 0 1 2 3 4 5 6
41, 25, 18, 12, 6, 31
``````
As `18` is now at the position `2`, it's only child is `31`, and since the child is yet again bigger than the parent, we swap them:
``````// 0 1 2 3 4 5 6 41, 25, 31, 12, 6, 18
``````
And just like that we have a max heap again!
Time Complexity of Insertion and Deletion
Let's take a look at the time complexity of inserting and deleting elements from a heap before implementing the algorithm. Since we're working with a binary tree-like structure it's natural that the time complexity of both the insertion and deletion is `O(logn)`, where `n` represents the size of our array. This is because for a binary tree of height `h`, given the binary nature of the heap - when traversing down the tree, you'll only even get to choose between two options, cutting down the possible paths by two on each step. In the worst-case, when traversing down to the bottom of the tree - the height of the tree, `h`, will be `logn`. With this we wrap up the explanation about heap as a data structure and move on to the main topic of the article - Heap Sort.
Heap Sort in Java
By taking advantage of the heap and it's properties, we've expressed it as an array. We can just as easily max heapify any array. Max heapify-ing is a process of arranging the elements in a correct order so they follow the max heap property. Similarly, you can min heapify an array. For each element, we need to check if any of its children are smaller than itself. If they are, swap one of them with the parent, and recursively repeat this step with the parent (because the new large element might still be bigger than its other child). Leaves have no children, so they're already max heaps on their own. Let's look at the following array:
``````// 0 1 2 3 4 5 6
25, 12, 6, 41, 18, 31, 53
``````
Let's quickly run the heapify algorithm through it and make a heap out of this array, manually, and then implement the code in Java to do that for us. We start from the right and go all the way to the left:
``````25 12 *6* 41 18 **31** **53**
``````
Since both `31 > 6` and `53 > 6`, we take the bigger of the two (in this case `53`) and swap it with their parent, and we get the following: 25 12 53 41 18 31 6.
``````25 *12* 6 **41** **18** 31 6
``````
Once again, `18 > 12` and `41 > 12`, and since `41 > 18`, we swap `42` and `12`.
``````*25*, **41**, **53** 12, 18, 31, 6
``````
In this last step of the way, we see that `41 > 25` and `53 > 25`, and since `53 > 41`, we swap `53` and `25`. After that, we recursively heapify for `25`.
``````53, 41, *25*, 12, 18, **31**, **6**
``````
`31 > 25`, so we swap them.
``````53, 41, 31, 12, 18, 25, 6
``````
We got a max heap! This process may seem daunting, though - when implemented in code, it's actually fairly simple. The process of heapyfing is crucial to Heap Sort, which follows three steps: 1. Build a max heap array using the input array.
2. Since the max heap stores the largest element of the array at the top (that is, the beginning of the array), we need to swap it with the last element within the array, followed by reducing the size of the array (heap) by `1`. After that, we heapify the root.
3. We repeat step 2 as long as the size of our heap is bigger than 1. With a good intuition of how the algorithm works, we can get to implementing it. Generally, since we'll be calling a `heapify()` method multiple times - we implement it separately from the `heapsort()` method, and call it within it. This makes the implementation cleaner and easier to read. Let's start out with the `heapify()` method:
``````public static void heapify(int[] array, int length, int i) {
int left = 2 * i + 1;
int right = 2 * i + 2;
int largest = i;
if (left < length && array[left] > array[largest]) {
largest = left;
}
if (right < length && array[right] > array[largest]) {
largest = right;
}
if (largest != i) {
int tmp = array[i];
array[i] = array[largest];
array[largest] = tmp;
heapify(array, length, largest);
}
}
``````
The `heapify()` method is what does most of the heavy lifting, and it just consists of three `if` statements. The flow of the Heap Sort algorithm itself is fairly simple as well, and relies mainly on `heapify()`:
``````public static void heapSort(int[] array) {
if (array.length == 0) {
return;
}
int length = array.length;
// Moving from the first element that isn't a leaf towards the root
for (int i = length / 2 - 1; i >= 0; i--) {
heapify(array, length, i);
}
for (int i = length - 1; i >= 0; i--) {
int tmp = array[0];
array[0] = array[i];
array[i] = tmp;
heapify(array, i, 0);
}
}
``````
That's about it! We can now supply an array to the `heapSort()` method, which sorts it in-place:
``````public static void main(String[] args){
int[] array = {25, 12, 6, 41, 18, 31, 53};
heapSort(array);
System.out.println(Arrays.toString(array));
}
``````
This results in:
``````[6, 12, 18, 25, 31, 41, 53]
``````
Implementing Heap Sort with a Priority Queue
A Priority Queue is a data structure that is actually a specific type of a queue, in which elements are added with a priority one by one, hence the name. The removal of elements begins with the one with the highest priority. The definition itself is really similar to that of a heap, so it's only natural that you can also implement Heap Sort using this very convenient data structure. Java has a built-in `PriorityQueue` residing in the `util` package:
``````import java.util.PriorityQueue;
``````
The `PriorityQueue` has quite a few of it's own and inherited methods from the `Queue` interface, but for our purposes we'll only need to use a few:
• `boolean add(E e)` - inserts the element `e` into the priority queue.
• `E poll()` - retrieves and removes the head of the priority queue, or returns `null` if it's empty.
• `int size()` - returns the number of elements in the priority queue.
With these, we can really implement Heap Sort through a single `while()` loop.
First of all, we'll create and add the elements to the priority queue, after which we simply run a `while` loop as long as our priority queue `pq` has at least `1` element within it. In every single iteration, we use the `poll()` method to retrieve and remove the head of the queue, after which we print it out and produce the same output as earlier:
``````Queue<integer> pq = new PriorityQueue<>();
int[] array = new int[]{25, 12, 6, 41, 18, 31, 53};
while(pq.size() > 0){
System.out.print(pq.poll() + " ");
}
</integer>
``````
This results in:
``````6 12 18 25 31 41 53
``````
Time complexity of Heapsort
Let's discuss the time complexity of both approaches we've covered. We've discussed earlier that adding and removing elements from a heap requires `O(logn)` time, and since our for loop runs `n` times where `n` is the number of the elements in the array, the total time complexity of Heapsort implemented like this is `O(nlogn)`. On the other hand, both adding and removing the elements from a priority queue takes up `O(logn)` as well, and doing this `n` times also produces `O(nlogn)` time complexity. What about space complexity? Well, since in both approaches we're only using the starting array to sort the array, that means the additional space required for Heap Sort is `O(1)`, making Heap Sort an in-place algorithm.
Conclusion
In conclusion, this article has covered both the theory and the implementation behind the Heap Sort algorithm. We've started out with an explanation of how it works, with an intuitive manual iteration, followed by two implementations. While not as fast as compared to something like Quick Sort or Merge Sort, Heap Sort is often used when the data is partially sorted or when there's a need for a stable algorithm. The in-place aspect of Heap Sort also allows us for better memory usage, when memory is of concern.
Reference: stackabuse.com | 4,011 | 15,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.948793 |
http://www.jiskha.com/display.cgi?id=1254963326 | 1,498,381,481,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320476.39/warc/CC-MAIN-20170625083108-20170625103108-00009.warc.gz | 571,200,211 | 4,071 | chemistry
posted by .
a buffer is prepared by combining 10.0g of NaH2PO4 wiht 150mL of 0.20M NaOH and diluting to 2.0L. What is the pH of the buffer, Ka H2PO4-= 6.2x10-8
• chemistry -
The reaction between the added solid and NaOH is:
NaH2PO4 + NaOH --> Na2HPO4 + H2O
10.0gNaH2PO4 /120.0g/mol = 0.0833 moles NaH2PO4 initially.
In the rection, the limiting reactant is NaOH.
(0.150 L)(0.20 mol/L) = 0.030 mol NaOH
0.030 moles of NaOH ---> 0.030 moles Na2HPO4 produced, and 0.030 moles NaH2PO4 used up.
After the reaction is complete, 0.0833-0.030 = 0.0533 moles of NaH2PO4 is remaining.
In the final mixture, the acid is H2PO4-(aq). The base is HPO4^2- Use the equation:
pH = pKa + log(Base/Acid)
• chemistry -
Write the equation.
NaH2PO4 + NaOH ==> Na2HPO4 + H2O
Convert 10 g NaH2PO4 to moles.
Convert 150 mL of 0.2 M NaOH to moles.
React and calculate the concn of H2PO4 and HPO4 in the 2.0 L container.
Then pH = pKa + log[(base)/(acid)]
Post your work if you get stuck.
• chemistry - | 377 | 992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2017-26 | latest | en | 0.790313 |
https://blogs.sas.com/content/sgf/2023/02/22/data-step-or-proc-it-depends/ | 1,721,286,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00568.warc.gz | 118,368,915 | 12,049 | As you know, almost every SAS programming problem has many very different solutions. I’m going to solve a very simple problem using two different approaches.
The problem: Compute the sum of integers from 1 to 1,000,000.
I bet most of you thought of a solution almost immediately. Let me guess that you thought of one of the solutions shown below:
## Solution 1
```options fullstimer; ❶ data _null_; ❷ do i = 1 to 1000000; Sum + i; ❸ end; file print; ❹ put Sum=; run;```
The option fullstimer will give you more complete timing information.
❷ To be more efficient use data _null_.
❸ The SUM statement does several things: 1) the variable Sum is retained (not set back to missing) for each iteration of the DATA Step. 2) Sum it initialized at zero. 3) If you had an expression instead of the constant (1) missing values would be ignored.
❹ Use FILE PRINT to send the output to the output window instead of the default location, the LOG.
## Solution 2
```data Integers; ❶ do i = 1 to 10000000; output; ❷ end; run; title "Sum of Integers"; proc means data=Integers n sum; ❸ var i; run;```
Create a data set called Integers.
❷ Output an observation for each iteration of the DATA Step. Note that the OUTPUT statement is inside the DO Loop.
❸ Use PROC MEANS to compute the sum.
Although both programs work, there is a difference in CPU time. Does that mean you should always seek a DATA Step solution? Not really. It depends on several factors, such as how often the program is to be run and which method you feel most comfortable with.
Here is a partial listing of the SAS Log showing timing information:
```NOTE: 1 lines were written to file PRINT.
NOTE: DATA statement used (Total process time):
real time 1.00 seconds
user cpu time 0.21 seconds
system cpu time 0.32 seconds
memory 7875.03k
OS Memory 16876.00k
Timestamp 02/16/2023 08:23:18 AM
Step Count 1 Switch Count 0
NOTE: The data set WORK.INTEGERS has 10000000 observations and 1
variables.
NOTE: DATA statement used (Total process time):
real time 0.21 seconds
user cpu time 0.20 seconds
system cpu time 0.00 seconds
memory 410.03k
OS Memory 17392.00k
Timestamp 02/16/2023 08:23:18 AM
Step Count 2 Switch Count 0
NOTE: There were 10000000 observations read from the data set
WORK.INTEGERS.
NOTE: PROCEDURE MEANS used (Total process time):
real time 0.25 seconds
user cpu time 1.07 seconds
system cpu time 0.03 seconds
memory 8471.84k
OS Memory 25116.00k
Timestamp 02/16/2023 08:23:19 AM
Step Count 3 Switch Count 0
```
Do you care about CPU time? Unless this is a production program, I think you should program in a way that is most comfortable (unless you are a compulsive programmer and want the “best” program). By the way, if you remove the FILE PRINT statement from solution 1, the System CPU time is 0.0. I guess there is some overhead to sending the results to your output device.
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Dr. Ron Cody was a Professor of Biostatistics at the Rutgers Robert Wood Johnson Medical School in New Jersey for 26 years. During his tenure at the medical school, he taught biostatistics to medical students as well as students in the Rutgers School of Public Health. While on the faculty, he authored or co-authored over a hundred papers in scientific journals. His first book, Applied Statistics and the SAS Programming Language, was first published by Prentice Hall in 1985 and is now in its fifth edition. Since then, he has published over a dozen books on SAS programming and statistical analysis using SAS. His latest book, A Gentle Introduction to Statistics Using SAS Studio was published this year. Ron has presented numerous papers at SAS Global forums, regional conferences, as well as local user groups. He is presently a contract instructor for SAS Institute and continues to write books on SAS and statistical topics.
1. Wow, I had to Google that one. Never came across that formula before. However, I hope you see where I was going in this blog. Best, Ron
2. Hello Ron, I would say suggest:
data _null_;
sum = 500000*1000001;
put sum=;
run;
But maybe this one is out of competition 🙂
Eric
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,095 | 4,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-30 | latest | en | 0.884847 |
https://nl.mathworks.com/matlabcentral/answers/1633300-function-solve-return-unable-to-perform-assignment-because-the-left-and-right-sides-have-a-differen | 1,653,368,912,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662564830.55/warc/CC-MAIN-20220524045003-20220524075003-00753.warc.gz | 368,084,244 | 25,134 | # function solve return 'Unable to perform assignment because the left and right sides have a different number of elements.'
1 view (last 30 days)
Federico Nebuloni on 21 Jan 2022
Commented: KSSV on 21 Jan 2022
Hi everyone,
trying to run this code where every parameter is known, only 'gamma' is my variable and 'hs_data' is a known function of 't_data'
hs_data = hs_Lderiv./1000; %meter
t_data = t; %seconds
dh = diff(hs_data);
dt = diff(t_data);
syms gamma
dhdt = zeros(size(t_data)-1);
gammaL = zeros(size(hs_data)-1);
gammaH = zeros(size(hs_data)-1);
for n = 1:length(t_data)-1
dhdt(n)= dh(n)./dt(n);
gammaL(n) = solve(dhdt(n) == -(1.2269.*gamma.*a_channel.^7.*hs_data(n).^4)./(pi().*mu.*L.*(a_source.^2+hs_data(n).^2).^5), gamma);
gammaH(n) = solve(dhdt(n) == -((2.*gamma)./(26.08.*pi().*mu.*a_channel.*L.*(a_source.^2+hs_data(n).^2))).*((((2.*a_channel.^2.*hs_data(n))./(a_source.^2+hs_data(n).^2))-((dRho.*g.*a_channel.^2.*hs_data(n))./(2.*gamma))).^4), gamma);
end
the first function 'solve' (the one returnin 'gammaL') works perfectly but the second one always returns the error 'Unable to perform assignment because the left and right sides have a different number of elements.'
is there a very trivial error i cannot find?
thank you very much in advance
KSSV on 21 Jan 2022
hs_data = hs_Lderiv./1000; %meter
t_data = t; %seconds
dh = diff(hs_data);
dt = diff(t_data);
syms gamma
dhdt = zeros(size(t_data)-1);
gammaL = cell(size(hs_data)-1,1);
gammaH = cell(size(hs_data)-1,1);
for n = 1:length(t_data)-1
dhdt(n)= dh(n)./dt(n);
gammaL{n} = solve(dhdt(n) == -(1.2269.*gamma.*a_channel.^7.*hs_data(n).^4)./(pi().*mu.*L.*(a_source.^2+hs_data(n).^2).^5), gamma);
gammaH{n} = solve(dhdt(n) == -((2.*gamma)./(26.08.*pi().*mu.*a_channel.*L.*(a_source.^2+hs_data(n).^2))).*((((2.*a_channel.^2.*hs_data(n))./(a_source.^2+hs_data(n).^2))-((dRho.*g.*a_channel.^2.*hs_data(n))./(2.*gamma))).^4), gamma);
end
##### 2 CommentsShowHide 1 older comment
KSSV on 21 Jan 2022
Thanks is accepting/ voting the asnwer. :)
R2021a
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Start Hunting! | 713 | 2,147 | {"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} | 2.59375 | 3 | CC-MAIN-2022-21 | latest | en | 0.367977 |
https://www.aqua-calc.com/one-to-one/pressure/dyne-per-square-nautical-mile/newton-per-square-micron/1 | 1,632,172,349,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00200.warc.gz | 691,880,626 | 7,811 | # 1 dyne per square nautical mile in newtons per square micron
## dynes/nautical mile² to newton/micron² unit converter of pressure
1 dyne per square nautical mile [dyn/nmi²] = 2.915533496 × 10−24 newton per square micron [N/µ²]
### dynes per square nautical mile to newtons per square micron pressure conversion cards
• 1
through
25
dynes per square nautical mile
• 1 dyn/nmi² to N/µ²
= 2.915533496 × 10-24 N/µ²
• 2 dyn/nmi² to N/µ²
= 5.831066992 × 10-24 N/µ²
• 3 dyn/nmi² to N/µ²
= 8.746600488 × 10-24 N/µ²
• 4 dyn/nmi² to N/µ²
= 1.166213398 × 10-23 N/µ²
• 5 dyn/nmi² to N/µ²
= 1.457766748 × 10-23 N/µ²
• 6 dyn/nmi² to N/µ²
= 1.749320098 × 10-23 N/µ²
• 7 dyn/nmi² to N/µ²
= 2.040873447 × 10-23 N/µ²
• 8 dyn/nmi² to N/µ²
= 2.332426797 × 10-23 N/µ²
• 9 dyn/nmi² to N/µ²
= 2.623980146 × 10-23 N/µ²
• 10 dyn/nmi² to N/µ²
= 2.915533496 × 10-23 N/µ²
• 11 dyn/nmi² to N/µ²
= 3.207086846 × 10-23 N/µ²
• 12 dyn/nmi² to N/µ²
= 3.498640195 × 10-23 N/µ²
• 13 dyn/nmi² to N/µ²
= 3.790193545 × 10-23 N/µ²
• 14 dyn/nmi² to N/µ²
= 4.081746894 × 10-23 N/µ²
• 15 dyn/nmi² to N/µ²
= 4.373300244 × 10-23 N/µ²
• 16 dyn/nmi² to N/µ²
= 4.664853594 × 10-23 N/µ²
• 17 dyn/nmi² to N/µ²
= 4.956406943 × 10-23 N/µ²
• 18 dyn/nmi² to N/µ²
= 5.247960293 × 10-23 N/µ²
• 19 dyn/nmi² to N/µ²
= 5.539513642 × 10-23 N/µ²
• 20 dyn/nmi² to N/µ²
= 5.831066992 × 10-23 N/µ²
• 21 dyn/nmi² to N/µ²
= 6.122620342 × 10-23 N/µ²
• 22 dyn/nmi² to N/µ²
= 6.414173691 × 10-23 N/µ²
• 23 dyn/nmi² to N/µ²
= 6.705727041 × 10-23 N/µ²
• 24 dyn/nmi² to N/µ²
= 6.99728039 × 10-23 N/µ²
• 25 dyn/nmi² to N/µ²
= 7.28883374 × 10-23 N/µ²
• 26
through
50
dynes per square nautical mile
• 26 dyn/nmi² to N/µ²
= 7.58038709 × 10-23 N/µ²
• 27 dyn/nmi² to N/µ²
= 7.871940439 × 10-23 N/µ²
• 28 dyn/nmi² to N/µ²
= 8.163493789 × 10-23 N/µ²
• 29 dyn/nmi² to N/µ²
= 8.455047138 × 10-23 N/µ²
• 30 dyn/nmi² to N/µ²
= 8.746600488 × 10-23 N/µ²
• 31 dyn/nmi² to N/µ²
= 9.038153838 × 10-23 N/µ²
• 32 dyn/nmi² to N/µ²
= 9.329707187 × 10-23 N/µ²
• 33 dyn/nmi² to N/µ²
= 9.621260537 × 10-23 N/µ²
• 34 dyn/nmi² to N/µ²
= 9.912813886 × 10-23 N/µ²
• 35 dyn/nmi² to N/µ²
= 1.020436724 × 10-22 N/µ²
• 36 dyn/nmi² to N/µ²
= 1.049592059 × 10-22 N/µ²
• 37 dyn/nmi² to N/µ²
= 1.078747394 × 10-22 N/µ²
• 38 dyn/nmi² to N/µ²
= 1.107902728 × 10-22 N/µ²
• 39 dyn/nmi² to N/µ²
= 1.137058063 × 10-22 N/µ²
• 40 dyn/nmi² to N/µ²
= 1.166213398 × 10-22 N/µ²
• 41 dyn/nmi² to N/µ²
= 1.195368733 × 10-22 N/µ²
• 42 dyn/nmi² to N/µ²
= 1.224524068 × 10-22 N/µ²
• 43 dyn/nmi² to N/µ²
= 1.253679403 × 10-22 N/µ²
• 44 dyn/nmi² to N/µ²
= 1.282834738 × 10-22 N/µ²
• 45 dyn/nmi² to N/µ²
= 1.311990073 × 10-22 N/µ²
• 46 dyn/nmi² to N/µ²
= 1.341145408 × 10-22 N/µ²
• 47 dyn/nmi² to N/µ²
= 1.370300743 × 10-22 N/µ²
• 48 dyn/nmi² to N/µ²
= 1.399456078 × 10-22 N/µ²
• 49 dyn/nmi² to N/µ²
= 1.428611413 × 10-22 N/µ²
• 50 dyn/nmi² to N/µ²
= 1.457766748 × 10-22 N/µ²
• 51
through
75
dynes per square nautical mile
• 51 dyn/nmi² to N/µ²
= 1.486922083 × 10-22 N/µ²
• 52 dyn/nmi² to N/µ²
= 1.516077418 × 10-22 N/µ²
• 53 dyn/nmi² to N/µ²
= 1.545232753 × 10-22 N/µ²
• 54 dyn/nmi² to N/µ²
= 1.574388088 × 10-22 N/µ²
• 55 dyn/nmi² to N/µ²
= 1.603543423 × 10-22 N/µ²
• 56 dyn/nmi² to N/µ²
= 1.632698758 × 10-22 N/µ²
• 57 dyn/nmi² to N/µ²
= 1.661854093 × 10-22 N/µ²
• 58 dyn/nmi² to N/µ²
= 1.691009428 × 10-22 N/µ²
• 59 dyn/nmi² to N/µ²
= 1.720164763 × 10-22 N/µ²
• 60 dyn/nmi² to N/µ²
= 1.749320098 × 10-22 N/µ²
• 61 dyn/nmi² to N/µ²
= 1.778475433 × 10-22 N/µ²
• 62 dyn/nmi² to N/µ²
= 1.807630768 × 10-22 N/µ²
• 63 dyn/nmi² to N/µ²
= 1.836786102 × 10-22 N/µ²
• 64 dyn/nmi² to N/µ²
= 1.865941437 × 10-22 N/µ²
• 65 dyn/nmi² to N/µ²
= 1.895096772 × 10-22 N/µ²
• 66 dyn/nmi² to N/µ²
= 1.924252107 × 10-22 N/µ²
• 67 dyn/nmi² to N/µ²
= 1.953407442 × 10-22 N/µ²
• 68 dyn/nmi² to N/µ²
= 1.982562777 × 10-22 N/µ²
• 69 dyn/nmi² to N/µ²
= 2.011718112 × 10-22 N/µ²
• 70 dyn/nmi² to N/µ²
= 2.040873447 × 10-22 N/µ²
• 71 dyn/nmi² to N/µ²
= 2.070028782 × 10-22 N/µ²
• 72 dyn/nmi² to N/µ²
= 2.099184117 × 10-22 N/µ²
• 73 dyn/nmi² to N/µ²
= 2.128339452 × 10-22 N/µ²
• 74 dyn/nmi² to N/µ²
= 2.157494787 × 10-22 N/µ²
• 75 dyn/nmi² to N/µ²
= 2.186650122 × 10-22 N/µ²
• 76
through
100
dynes per square nautical mile
• 76 dyn/nmi² to N/µ²
= 2.215805457 × 10-22 N/µ²
• 77 dyn/nmi² to N/µ²
= 2.244960792 × 10-22 N/µ²
• 78 dyn/nmi² to N/µ²
= 2.274116127 × 10-22 N/µ²
• 79 dyn/nmi² to N/µ²
= 2.303271462 × 10-22 N/µ²
• 80 dyn/nmi² to N/µ²
= 2.332426797 × 10-22 N/µ²
• 81 dyn/nmi² to N/µ²
= 2.361582132 × 10-22 N/µ²
• 82 dyn/nmi² to N/µ²
= 2.390737467 × 10-22 N/µ²
• 83 dyn/nmi² to N/µ²
= 2.419892802 × 10-22 N/µ²
• 84 dyn/nmi² to N/µ²
= 2.449048137 × 10-22 N/µ²
• 85 dyn/nmi² to N/µ²
= 2.478203472 × 10-22 N/µ²
• 86 dyn/nmi² to N/µ²
= 2.507358807 × 10-22 N/µ²
• 87 dyn/nmi² to N/µ²
= 2.536514142 × 10-22 N/µ²
• 88 dyn/nmi² to N/µ²
= 2.565669476 × 10-22 N/µ²
• 89 dyn/nmi² to N/µ²
= 2.594824811 × 10-22 N/µ²
• 90 dyn/nmi² to N/µ²
= 2.623980146 × 10-22 N/µ²
• 91 dyn/nmi² to N/µ²
= 2.653135481 × 10-22 N/µ²
• 92 dyn/nmi² to N/µ²
= 2.682290816 × 10-22 N/µ²
• 93 dyn/nmi² to N/µ²
= 2.711446151 × 10-22 N/µ²
• 94 dyn/nmi² to N/µ²
= 2.740601486 × 10-22 N/µ²
• 95 dyn/nmi² to N/µ²
= 2.769756821 × 10-22 N/µ²
• 96 dyn/nmi² to N/µ²
= 2.798912156 × 10-22 N/µ²
• 97 dyn/nmi² to N/µ²
= 2.828067491 × 10-22 N/µ²
• 98 dyn/nmi² to N/µ²
= 2.857222826 × 10-22 N/µ²
• 99 dyn/nmi² to N/µ²
= 2.886378161 × 10-22 N/µ²
• 100 dyn/nmi² to N/µ²
= 2.915533496 × 10-22 N/µ²
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Price conversions and cost calculator for materials and substances | 3,647 | 6,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2021-39 | latest | en | 0.270528 |
https://www.neilbickford.com/blog/category/astronomy/ | 1,660,249,012,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00472.warc.gz | 779,796,621 | 19,640 | Picture yourself standing in the middle of a large field at night. There are sounds all around you, though they may be quiet and distant, such as a river, or an orchestra of crickets.
Suppose you’d like to create a map showing the amount of sound coming from each direction, with as much precision as you can (imagine if you could determine the location of each individual cricket, for instance).
If you had a directional microphone, you could point it in a particular direction and record the average volume output from the microphone in that direction. Then you could rotate it a bit, measure the recorded volume again, and repeat this process until you completed a full circle. (You could also rotate the microphone to cover a full sphere, but in this example there may not be as much interesting audio above or below you unless a plane’s passing overhead, for instance).
The quality of this map depends on the directionality of the microphone, though. If your microphone was omnidirectional (i.e. records the same audio regardless of which direction it’s pointed), you wouldn’t get any useful information – your map would show the same levels in every direction! A cardioid microphone improves on this, but your map would still be incredibly blurry. Further changing the shape of the microphone, we could use a shotgun microphone – but this would still reduce us to a resolution of ten degrees or so (and would also conflate sounds coming from four different directions).
You could also try using a parabolic reflector to bounce all of the audio coming from a particular direction into a single focal point – where you’ll place the microphone. If a sound wave came from a different direction, its reflection would miss the microphone, and be attenuated. However, although you could adjust some factors, such as the size and shape of the parabolic reflector, parabolic reflectors tend to have about the same angular resolution as shotgun microphones, and thus would still produce a blurry map in this case. Here’s the polar plot of Wildtronics’ Professional Mono Parabolic Microphone, for instance:
Side Note: Using Deconvolution
Since the shape of the polar plot (usually) doesn’t zero out any spatial frequencies, we could hypothetically attempt to deconvolve the polar plot of the microphone with the blurry recorded map we have. We discuss this below – unfortunately, this technique has its limits, and while it can sharpen the image, it’s not quite enough to get us a several-order-of-magnitude resolution improvement.
But what if you used two microphones? Standing in the middle of a field at night, you can sort of tell where the river flows and where the group of crickets chirps using your two ears, even though each individual ear acts more like a cardioid microphone than a shotgun microphone or a parabolic reflector. Your brain uses several tricks to infer directions from stereo audio: for instance, sound coming from your right reaches your right ear before it reaches your left ear (and the sound that reaches your left ear has to travel around your head, which affects the sound), and things in front of you sound different than things behind you, thanks to the shape of your head and ears.
Interferometry works on similar principles: Instead of constructing a single mega-telescope to observe the stars, which in this analogy are like the crickets chirping far away in the field, we construct multiple telescopes and combine their results to produce a more accurate image than the first telescope could. (Indeed, some optical telescopes use parabolic reflectors, and then have to contend with the limits of diffraction.) In this way, we can produce beautiful images of the night sky, and in some sense, take photos without lenses.
Hi! This is an introduction to astronomical interferometry and radio telescopy! This article might assume that you have some familiarity with Fourier transforms, but that’s about it.
Continue reading “Interferometry and the Very Large Array” | 801 | 4,002 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | longest | en | 0.957588 |
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### Statistical Analysis Software
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Please visit the Feedbackpage to comment or give suggestions on ArcGIS Desktop Help.Our consultants can help you with conducting data analysis for your research using SPSS, STATA or SAS.Assignment 4: Statistical Analyses Assignment Description In this assignment, you will practice methods you learned in class for descriptive and.Students require help in explaining their work through work of statistical data.Statistical Analysis Help can offer a vast array of services.
Statwing is designed solely for analyzing tables of data, so users finish days of analysis in minutes. | 456 | 2,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-09 | latest | en | 0.876258 |
https://zrna.org/docs/module/sum-integrate-two-reset | 1,669,656,104,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00803.warc.gz | 1,159,941,439 | 2,862 | ### SumIntegrateTwoReset
SumIntegrateTwo with voltage-controlled reset.
##### Parameters
scale_factor_input1
scale_factor_input2
##### Options
input1_polarity $$\in \{$$ NONINVERTING $$,$$ INVERTING $$,$$ INPUT_DISABLED $$\}$$
input2_polarity $$\in \{$$ NONINVERTING $$,$$ INVERTING $$,$$ INPUT_DISABLED $$\}$$
output_phase $$\in \{$$ PHASE1 $$,$$ PHASE2 $$\}$$
compare_to $$\in \{$$ GROUND $$,$$ SECOND_INPUT $$,$$ VARIABLE_REFERENCE $$\}$$
control_signal_polarity $$\in \{$$ ACTIVE_HIGH $$,$$ ACTIVE_LOW $$\}$$
opamp_mode $$\in \{$$ DEFAULT $$,$$ CHOPPER_STABILIZED $$\}$$
##### Inputs
input1 Half-Cycle
input2 Half-Cycle
control Half-Cycle
##### Outputs
output Continuous
reset_out Continuous
##### Transfer Function
$$\frac{\Delta V_{out}}{\Delta t} = \sum_{i=1}^{2} \pm K_i V_{i} \\ \text{reset if V_{ctrl} > V_{ref}}$$
##### Analog Resource Usage
$$\begin{array}{|c|c|} \hline \text{Opamps} & \text{1 of 8} \\ \hline \text{Comparators} & \text{1 of 4} \\ \hline \text{Capacitors} & \text{3 of 32} \\ \hline \end{array}$$ | 354 | 1,047 | {"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": 2, "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} | 2.53125 | 3 | CC-MAIN-2022-49 | longest | en | 0.277837 |
https://byjus.com/question-answer/abcd-is-a-rectangle-with-diagonals-ad-and-bc-intersecting-at-point-o-if-the-1/ | 1,638,640,341,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362999.66/warc/CC-MAIN-20211204154554-20211204184554-00356.warc.gz | 221,829,431 | 18,095 | Question
# ABCD is a rectangle with diagonals AD and BC intersecting at point O. If the length of AO is 6 units, find the length of CB. [3 MARKS]
Solution
## Concept: 1 Mark Application: 2 Marks Diagonals of a rectangle are of equal length. [Given ABCD is a rectangle] Length of AO is 6 units. [Given] Length of AD is 12 units. [Since O is the midpoint] Length of CB = Length of AD = 12 units.
Suggest corrections | 116 | 419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-49 | latest | en | 0.902067 |
https://www.hireresearcher.co.uk/samples/problem-4-medical-associates-is-a-large-for-profit-group-practice-its-dividends-are-expected-to/ | 1,600,924,773,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00473.warc.gz | 892,193,813 | 6,781 | PROBLEM 4 Medical Associates is a large for-profit group practice. Its dividends are expected to
17 / 01 / 2019 Research Papers
This paper circulates around the core theme of PROBLEM 4 Medical Associates is a large for-profit group practice. Its dividends are expected to together with its essential aspects. It has been reviewed and purchased by the majority of students thus, this paper is rated 4.8 out of 5 points by the students. In addition to this, the price of this paper commences from £ 99. To get this paper written from the scratch, order this assignment now. 100% confidential, 100% plagiarism-free.
PROBLEM 4 Medical Associates is a large for-profit group practice. Its dividends are expected to… 1 answer below » PROBLEM 4 Medical Associates is a large for-profit group practice. Its dividends are expected to grow at a constantrate of 7 percent per year into the foreseeable future. The firm’s last dividend (DO) was \$2, and its current stock price is \$23. The firm’s beta coefficient is 1.6; the rate of return on 20-year T-bonds is 9 percent; and the expected rate of return on the market, as reported by a large financial services firm, is 13 percent. The firm’s target capital structure calls for 50 percent debt financing, the interest rate required on the business’s new debt is 10 percent, and its tax View complete question » PROBLEM 4 Medical Associates is a large for-profit group practice. Its dividends are expected to grow at a constantrate of 7 percent per year into the foreseeable future. The firm’s last dividend (DO) was \$2, and its current stock price is \$23. The firm’s beta coefficient is 1.6; the rate of return on 20-year T-bonds is 9 percent; and the expected rate of return on the market, as reported by a large financial services firm, is 13 percent. The firm’s target capital structure calls for 50 percent debt financing, the interest rate required on the business’s new debt is 10 percent, and its tax rate is 40 percent. a. What is Medical Associate’s cost of equity estimate according to the DCF method? b. What is the cost of equity estimate according to the CAPM? c. On the basis of your answers to Parts a and b, what would be your final estimate for the firm’s cost of equity? d. What is your estimate for the firm’s corporate cost of capital? Must show work in Excel View less » Sep 18 2015 12:50 PM
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FLAT 25% OFF ON EVERY ORDER. Use "FLAT25" as your promo code during checkout | 803 | 3,251 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-40 | latest | en | 0.951778 |
http://mathhelpforum.com/algebra/281683-rectangular-garden-2-a.html | 1,542,134,117,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00353.warc.gz | 208,785,937 | 9,817 | # Thread: Rectangular Garden 2
1. ## Rectangular Garden 2
A gardener has 46 feet of fencing to be used to enclose a rectangular garden that has a border 2 feet wide surrounding it.
A. If the length of the garden is to be twice the width, what will be the dimensions of the garden?
B. What is the area of the garden?
Length of garden = 2x
Width of garden = x
Dimensions = 2x + 4 for the length and x + 4 for the width.
Part A Set Up:
P = 2L + 2W
2(2x + 4) + 2(x + 4) = 46
Part B Set Up:
After finding the dimensions of the garden, I then use the formula A = L•W to find its area.
Correct?
2. ## Re: Rectangular Garden 2
Yes. What do you get as answers?
3. ## Re: Rectangular Garden 2
Once more, please post your answers right away...less "running around"...
4. ## Re: Rectangular Garden 2
Part A
2(2x + 4) + 2(x + 4) = 46
4x + 8 + 2x + 8 = 46
6x + 16 = 46
6x = 46 - 16
6x = 30
x = 30/6
x = 5
Length = 2x or 10
Width = 5
Part B:
A = L•W
A = 10(5)
A = 50 ft^2 | 351 | 987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | longest | en | 0.84943 |
https://www.mathway.com/examples/calculus/applications-of-integration/finding-the-root-mean-square?id=764 | 1,717,071,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00154.warc.gz | 740,092,848 | 24,408 | # Calculus Examples
,
Step 1
The Root Mean Square (RMS) of a function over a specified interval is the square root of the arithmetic mean (average) of the squares of the original values.
Step 2
Substitute the actual values into the formula for the root mean square of a function.
Step 3
Evaluate the integral.
Step 3.1
Let . Then , so . Rewrite using and .
Step 3.1.1
Let . Find .
Step 3.1.1.1
Differentiate .
Step 3.1.1.2
By the Sum Rule, the derivative of with respect to is .
Step 3.1.1.3
Evaluate .
Step 3.1.1.3.1
Since is constant with respect to , the derivative of with respect to is .
Step 3.1.1.3.2
Differentiate using the Power Rule which states that is where .
Step 3.1.1.3.3
Multiply by .
Step 3.1.1.4
Differentiate using the Constant Rule.
Step 3.1.1.4.1
Since is constant with respect to , the derivative of with respect to is .
Step 3.1.1.4.2
Step 3.1.2
Substitute the lower limit in for in .
Step 3.1.3
Simplify.
Step 3.1.3.1
Multiply by .
Step 3.1.3.2
Step 3.1.4
Substitute the upper limit in for in .
Step 3.1.5
Simplify.
Step 3.1.5.1
Multiply by .
Step 3.1.5.2
Step 3.1.6
The values found for and will be used to evaluate the definite integral.
Step 3.1.7
Rewrite the problem using , , and the new limits of integration.
Step 3.2
Combine and .
Step 3.3
Since is constant with respect to , move out of the integral.
Step 3.4
By the Power Rule, the integral of with respect to is .
Step 3.5
Substitute and simplify.
Step 3.5.1
Evaluate at and at .
Step 3.5.2
Simplify.
Step 3.5.2.1
Raise to the power of .
Step 3.5.2.2
Combine and .
Step 3.5.2.3
Raising to any positive power yields .
Step 3.5.2.4
Multiply by .
Step 3.5.2.5
Multiply by .
Step 3.5.2.6
Step 3.5.2.7
Multiply by .
Step 3.5.2.8
Multiply by .
Step 3.5.2.9
Cancel the common factor of and .
Step 3.5.2.9.1
Factor out of .
Step 3.5.2.9.2
Cancel the common factors.
Step 3.5.2.9.2.1
Factor out of .
Step 3.5.2.9.2.2
Cancel the common factor.
Step 3.5.2.9.2.3
Rewrite the expression.
Step 4
Simplify the root mean square formula.
Step 4.1
Multiply by .
Step 4.2
Step 4.3
Reduce the expression by cancelling the common factors.
Step 4.3.1
Factor out of .
Step 4.3.2
Factor out of .
Step 4.3.3
Cancel the common factor.
Step 4.3.4
Rewrite the expression.
Step 4.4
Rewrite as .
Step 4.5
Simplify the numerator.
Step 4.5.1
Rewrite as .
Step 4.5.2
Pull terms out from under the radical, assuming positive real numbers.
Step 4.6
Multiply by .
Step 4.7
Combine and simplify the denominator.
Step 4.7.1
Multiply by .
Step 4.7.2
Raise to the power of .
Step 4.7.3
Raise to the power of .
Step 4.7.4
Use the power rule to combine exponents.
Step 4.7.5
Step 4.7.6
Rewrite as .
Step 4.7.6.1
Use to rewrite as .
Step 4.7.6.2
Apply the power rule and multiply exponents, .
Step 4.7.6.3
Combine and .
Step 4.7.6.4
Cancel the common factor of .
Step 4.7.6.4.1
Cancel the common factor.
Step 4.7.6.4.2
Rewrite the expression.
Step 4.7.6.5
Evaluate the exponent.
Step 5
The result can be shown in multiple forms.
Exact Form:
Decimal Form:
Step 6 | 1,092 | 3,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.711778 |
https://www.geeksforgeeks.org/tag/factorial/page/7/ | 1,571,061,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00046.warc.gz | 876,060,322 | 19,495 | ## Count Divisors of Factorial
Given a number n, count total number of divisors of n!. Examples: Input : n = 4 Output: 24 4! is 24. Divisors of 24… Read More »
## Count digits in a factorial | Set 2
Given an integer n (can be very large), find the number of digits that appear in its factorial, where factorial is defined as, factorial(n) =… Read More »
## Count digits in a factorial | Set 1
Given an integer n, find the number of digits that appear in its factorial, where factorial is defined as, factorial(n) = 1*2*3*4……..*n and factorial(0) =… Read More »
## No of Factors of n!
Given a positive integer n, find the no of factors in n! where n <= 105. Examples : Input : n = 3 Output :… Read More »
## Compute n! under modulo p
Given a large number n and a prime p, how to efficiently compute n! % p? Examples : Input: n = 5, p = 13… Read More »
## Count factorial numbers in a given range
A number F is a factorial number if there exists some integer I >= 0 such that F = I! (that is, F is factorial… Read More »
## Legendre’s formula (Given p and n, find the largest x such that p^x divides n!)
Given an integer n and a prime number p, find the largest x such that px (p raised to power x) divides n! (factorial) Examples:… Read More »
## Factorial of a large number
Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is… Read More »
## Efficient Program to Compute Sum of Series 1/1! + 1/2! + 1/3! + 1/4! + .. + 1/n!
Given a positive integer n, write a function to compute sum of the series 1/1! + 1/2! + .. + 1/n! A Simple Solution solution… Read More »
## C program to calculate the value of nPr
nPr represents n permutation r and value of nPr is (n!) / (n-r)!. filter_none edit close play_arrow link brightness_4 code #include<stdio.h> int fact(int n)… Read More »
## Java Program for factorial of a number
Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is… Read More »
## C Program for factorial of a number
Factorial of a non-negative integer is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is… Read More »
## Python Program for factorial of a number
Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is… Read More »
## Program for factorial of a number
Factorial of a non-negative integer, is multiplication of all integers smaller than or equal to n. For example factorial of 6 is 6*5*4*3*2*1 which is… Read More »
## Count trailing zeroes in factorial of a number
Given an integer n, write a function that returns count of trailing zeroes in n!. Examples : Input: n = 5 Output: 1 Factorial of… Read More » | 793 | 2,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-43 | latest | en | 0.742885 |
https://exercism.org/tracks/python/exercises/matrix | 1,726,662,125,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651895.12/warc/CC-MAIN-20240918100941-20240918130941-00854.warc.gz | 219,653,173 | 9,776 | # Matrix
Easy
## Instructions
Given a string representing a matrix of numbers, return the rows and columns of that matrix.
So given a string with embedded newlines like:
``````9 8 7
5 3 2
6 6 7
``````
representing this matrix:
`````` 1 2 3
|---------
1 | 9 8 7
2 | 5 3 2
3 | 6 6 7
``````
your code should be able to spit out:
• A list of the rows, reading each row left-to-right while moving top-to-bottom across the rows,
• A list of the columns, reading each column top-to-bottom while moving from left-to-right.
The rows for our example matrix:
• 9, 8, 7
• 5, 3, 2
• 6, 6, 7
And its columns:
• 9, 5, 6
• 8, 3, 6
• 7, 2, 7
In this exercise you're going to create a class. Don't worry, it's not as complicated as you think!
Edit via GitHub
### Ready to start Matrix?
Sign up to Exercism to learn and master Python with 17 concepts, 140 exercises, and real human mentoring, all for free. | 304 | 915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.832836 |
https://or.stackexchange.com/questions/599/algorithmic-gap-for-hochbaums-greedy-algorithm-for-metric-uncapacitated-fac | 1,718,203,112,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861173.16/warc/CC-MAIN-20240612140424-20240612170424-00629.warc.gz | 410,805,821 | 30,528 | # Algorithmic gap for Hochbaum's (greedy) algorithm for (metric) uncapacitated facility location
In Jain et al. (2003)1, at the bottom of page 801, they construct an instance of (metric) uncapacitated facility location for which they claim the greedy (Hochbaum's) algorithm has gap $$\Omega\left(\frac{\log n}{\log \log n}\right)$$. The algorithm is as follows:
Construct an instance of set cover, where the ground set of elements is the set of clients $$D$$ and the set system contains all sets of the form $$(i,A)$$ where $$i \in F$$ is a facility and $$A \subseteq D$$ is a set of clients. Set $$(i,A)$$ covers the elements in $$A$$ and $$\operatorname{cost}(i,A) = f_i + \sum{d_{ij}}$$. The greedy algorithm is to choose the set $$(i,A)$$ that minimizes $$\frac{\operatorname{cost}(i,A)}{|A\,\cap\, \text{uncovered}|}$$. This can be done in polytime by iterating through the facilities $$i \in F$$ and through $$s = 1,\ldots,|\text{uncovered}|$$ and considering $$A$$ to be the $$s$$ closest uncovered clients to facility $$i$$.
Part of the proof of Jain et al.1 is to claim that (on the instance they construct) the above greedy algorithm will open all $$k$$ facilities, while the optimal thing to do is to open only one of the facilities. I can't seem to see why the greedy algorithm will open all facilities. I can see how this would be the case if the cost of each element in set $$S_i$$ is $$\sum\limits_{j=1}^ip^{j-1}$$, but in this case opening one facility will have cost $$p^k + \sum_{i=1}^{k-1}p^{k-i+1}\sum_{j=1}^ip^{j-1} = \Theta(kp^k),$$ while they claim that the optimal cost (that of opening one facility) is $$p^k + \sum_{i=1}^{k-1}\sum_{j=1}^ip^{j-1} = \Theta(p^k).$$ A central part of their proof seems to be that since the greedy algorithm described above will open all $$k$$ facilities, it will have cost $$\Omega(kp^k)$$ and therefore, the gap between the algorithms performance and the optimum is $$\Omega(k)$$. Since there are $$n = \Theta(p^k)$$ clients, $$k = \log_p n = \ln n/\ln p$$, and therefore for $$p = \ln n$$, we have that the gap is $$\Omega\left(\frac{\log n}{\log \log n}\right)$$.
Another question I have is that for $$p = O(1)$$, wouldn't this result in a gap of $$\Omega(\log n)$$ which is a stronger result? Why doesn't this work? The last sentence of their paragraph is "We do not know whether the approximation factor of Hochbaum’s algorithm on metric instances is strictly less than $$\log n$$." So I assume I'm missing something.
Reference
[1] Jain, K., Mahdian, M., Markakis, E. et al. (2003). Greedy Facility Location Algorithms Analyzed Using Dual Fitting with Factor-Revealing LP. Journal of the ACM. 50(6):795–824.
## 1 Answer
I'm looking at the algorithm as it's described in Hochbaum (1982), which works like this: Suppose we have enumerated all $$2^n-1$$ subsets of the customers. Subset $$P_m$$ has cost $$C_m = \min_{j\in J} \left(\sum_{i\in P_m} c_{ij} + f_j\right),$$ i.e., the fixed plus transportation cost if we choose the best facility for the set $$P_m$$ of customers. At each iteration, we choose the $$P_k$$ that maximizes the ratio $$\frac{|P_m|}{C_m},$$ open the best facility for that $$P_k$$, remove the customers in $$P_k$$ from all of the other $$P_m$$ sets, and repeat.
I'm not exactly sure how that translates to the algorithm described in Jain et al. (2003); they say the two algorithms are equivalent, so let's take their word for it.
Their example has $$k$$ facilities with a fixed cost of $$p^k$$ for some $$p$$, all located at the same spot. (They don't give any restrictions on $$p$$ but the example only makes sense if $$p$$ is a positive integer.) There are $$k-1$$ groups of customers, called $$S_1,\ldots,S_{k-1}$$, oriented as rings around the facilities. $$S_i$$ contains $$p^{k-i+1}$$ customers, each at a distance of $$\sum_{j=1}^i p^{j-1}$$ from the facilities.
Consider the first iteration of Hochbaum's algorithm. Suppose $$P_1=S_1$$, $$P_2=S_2$$, and $$P_{1,2}=S_1\cup S_2$$. The relevant ratios are: \begin{align} \frac{|P_1|}{C_1} & = \frac{p^k}{p^k+p^kp^0} = \frac12 \\ \frac{|P_2|}{C_2} & = \frac{p^{k-1}}{p^k + p^{k-1}(p^0+p^1)} = \frac{1}{2p+1} < \frac12 \\ \frac{|P_{1,2}|}{C_{1,2}} & = \frac{p^k+p^{k-1}}{p^k + p^kp^0 + p^{k-1}(p^0+p^1)} = \frac{p+1}{3p+1} < \frac12. \end{align} And presumably the pattern will repeat, in the sense that if we either add more "rings" to the subset, or we replace $$S_1$$ with a different "ring", the ratio will go down. Therefore, $$P_1 = S_1$$ attains the maximum ratio, so we open a facility (doesn't matter which one) and assign $$S_1$$ to it. Then remove $$S_1$$ from all of the other subsets, and repeat. The same logic will hold at the next iteration, so we will open a facility and assign $$S_2$$ to it, and so on. This continues until we have opened all $$k$$ facilities.
• Ah, I see, I misunderstood them as saying the sum of the distances of the clients in group $S_i$ from the facilities was $\sum\limits_{j=1}^i p^{j-1}$, but actually each client in group $S_i$ has distance $\sum\limits_{j=1}^i p^{j-1}$ from the facilities. This makes sense, thanks for the clear and concise response. Jun 19, 2019 at 2:36
• Sure, glad to help. Jun 19, 2019 at 2:38
• Also to clarify the relationship b/w the above greedy method to the algorithm described in Jain et. al (2003)—referring to Algorithm 1 from the paper—they are actually not equivalent. They differ in that Alg. 1 no longer accounts for the fixed cost $f_i$ in measuring the effectiveness of covering subset $P_m$ with facility $j$, i.e. $|P_m|/\sum_{i \in P_m} c_{ij}$, if that facility has already been chosen by the algorithm. Their algorithm is able to achieve a constant-factor approximation for metric uncapacitated facility location, while the above greedy procedure is not. Jun 19, 2019 at 2:44
• I was referring to where it says that Hochbaum’s greedy algorithm for facility location is equivalent to the set cover algorithm applied on the set of stars. Maybe I misunderstood that point though. Jun 19, 2019 at 2:48
• Actually after looking again at the Jain et. al (2003) paper, they do say that the optimum solution, opening just one facility, has value $p^k + \sum\limits_{i=1}^{k+1}\sum\limits_{j=1}^i p^{j-1}$. So I believe the sum of the distances of the clients in group $S_i$ from the facilities is $\sum\limits_{j=1}^i p^{j-1}$ (i.e. $d(s) = \frac{1}{p^{k-i+1}}\sum\limits_{j=1}^i p^{j-1}$ for all $s \in S_i$ where $d(a)$ is the distance of client $a$ from the facilities). Jun 19, 2019 at 3:43 | 1,993 | 6,523 | {"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": 59, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-26 | latest | en | 0.896489 |
https://codebase64.org/doku.php?id=base:perspective&rev=1429237998&do=diff | 1,627,891,150,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00651.warc.gz | 189,780,686 | 4,208 | base:perspective
# Differences
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— base:perspective [2015-04-17 04:33] (current) Line 1: Line 1: + ==== Perspective ==== + + by Bitbreaker/Oxyron/Nuance + + Best is to calculate the perspective during multiply with the rotation matrix. As soon as you get the value for Z, lookup a corresponding factor in a table and multply the results for X and Y with that factor: + + + ... matrix multipplication for Z ... + tay + lda z_fact,y + sta z1 + eor #\$ff + sta z2 + + ... matrix multiplication for X ... + + tay + ;multiply with z_fact + lda (z1),y + sec + sbc (z2),y + sta final_x + + ... matrix multiplication for Z ... + + tay + ;multiply with z_fact + lda (z1),y + sec + sbc (z2),y + sta final_y + + + The factor-table you could generate like the following: + + ... + d = 280.0; z0 = 5.0; + for (i = 0; i < 0x100; i++) { + z = i; + //make things signed + if(z > 127) z = z - 256; + q = round(d/(z0-z/64.0)); + //take care that values are sane + if(q > 127) q = 127; + if(q < -127) q = -127; + + if(q < 0) q = 256 + q; + + result[i] = q; + } + ... + | 365 | 1,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-31 | latest | en | 0.725925 |
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# Tempurature Dependence of Salt Solubility
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
I don’t understand how to make this graph...what is it supposed to look like? I’ve tried and just can’t make my computer work it...I also don’t understand what my experimental mass and official masses are supposed to be... please help!
Data: List all measurements and include proper labels.
Test tube Crystallizing Temperature (degrees C)
Solute (g/10ml) KCl NH4Cl
3.5 23.94 degrees C -----
4.0 39.69 degrees C 23.12 degrees C
4.5 57.06 degrees C 34.87 degrees C
5.0 72.22 degrees C 47.55 degrees C
6.0 --- 68.53 degrees C
Calculations/Interpretations: Show all math performed (give the formula, show your setup, and give the result. Use proper significant figures and include proper labels. Also, answer ALL questions asked in the lab.
Record your data.
Test tube
Solute (g/10ml) Solute (g/ 100ml)
3.5
4.0
4.5
5.0
6.0
Official points to plot.
Temperature (C) KCl (g/100ml) NH4Cl (g/100ml)
30 37.1 41.6
50 42.9 50.4
70 48.5 59.9
90 53.8 70.4
Plot these 4 sets of data points on a graph:
1. Experimental KCl solubility
2. Experimental NH4Cl solubility
3. Official KCl solubility
4. Official NH4Cl solubility
Plot temperature (deg C) on the Y-axis (vertical) and solubility (g/100ml) across the X-axis (horizontal).
The temperature range should be from 0 to 100 deg C. The solubility range should be from 0g/100ml to 100 g/100ml.
From Graphs:
KCl NH4Cl
Experimental mass at 60 C:
Official mass at 60 C:
% difference:
(a) Why is it important to not allow the test solutions to boil?
It would be a poor variable to cause a change the volume of the solution, boiling would alter this far more than evaporation from an 80 degree solution.
(b) Why is it better to determine the initial crystallization temperature during cool down rather than the temperature at which all of the salt dissolves?
It is harder to see the point at which all the salt has dissolved and it dissolved through agitation as well as increase in temperature, i.e. as you would warm it up it doesn't all dissolve for the saturation point of that temperature immediately. It is far easier to note the crystallization point when cooling.
© BrainMass Inc. brainmass.com October 10, 2019, 12:11 am ad1c9bdddf
https://brainmass.com/chemistry/stoichiometry/tempurature-dependence-salt-solubility-282085
#### Solution Summary
This solution explains how to solve severla problems dealing with temperature dependence of salt solubility.
\$2.19 | 714 | 2,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-51 | latest | en | 0.874526 |
https://in.mathworks.com/matlabcentral/cody/problems/26-determine-if-input-is-odd/solutions/2095936 | 1,585,412,643,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491998.11/warc/CC-MAIN-20200328134227-20200328164227-00201.warc.gz | 535,524,063 | 15,680 | Cody
# Problem 26. Determine if input is odd
Solution 2095936
Submitted on 17 Jan 2020 by Eric Kelso
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
tf = logical 1
2 Pass
n = 2; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
tf = logical 0
3 Pass
n = 28; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
tf = logical 0
4 Pass
n = 453; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
tf = logical 1
5 Pass
n = 17; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
tf = logical 1
6 Pass
n = 16; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
tf = logical 0 | 259 | 839 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-16 | latest | en | 0.495406 |
https://forum.blockland.us/index.php?topic=326914.0 | 1,611,147,502,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703520883.15/warc/CC-MAIN-20210120120242-20210120150242-00406.warc.gz | 342,292,146 | 6,794 | ### Author Topic: a riddle (Read 594 times)
#### The Resonte!
how many sides does a circle have?
#### Mega-Bear
many sides, they are very well-rounded folk
#### foulprairiedog
the hypothetical perfect math circle has one side
a "circle" in some vertex-based rendering would have anywhere from 4 to infinitely many sides
yes i am counting a square rotated 45 degrees as a circle
the hypothetical perfect math circle has one side
if i look at the front, and someone else looks at the back, are we looking at the same side?
#### Drydess
this riddle sucks!
#### foulprairiedog
if i look at the front, and someone else looks at the back, are we looking at the same side?
hypothetical perfect circle is also a hypothetical perfect plane that is infinitely thin and only has one side, so yes, it is the same one side from all angles
#### TheArmyGuy
Depends on if you mean side as an edge or side as a plane.
Edge: 1
Plane: 2
#### Aide33
a riddle that relies on ambiguous terms that can be defined any way you want is not a riddle
diddle my riddle
#### TheArmyGuy
A better question would be: does a circle have no vertices or an infinite amount of vertices?
#### The Resonte!
a circle has two sides- the inside and the outside!
uh, that's all the riddles i have
#### TlTO
get off the stage
#### Khaz
i believe the actual technical answer is infinity
#### TableSalt
A better question would be: does a circle have no vertices or an infinite amount of vertices?
trick question. It has a single vertex.
#### Drydess
trick question. It has a single vertex. | 385 | 1,576 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.929838 |
https://help.smartsheet.com/function/int | 1,618,936,093,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039476006.77/warc/CC-MAIN-20210420152755-20210420182755-00609.warc.gz | 402,506,986 | 16,756 | INT Function
Returns the integer portion of a number.
Sample Usage
INT(425.99)
Syntax
INT(
• value
)
• value
The value, typically within a cell, to check.
Examples
Here are some examples for reference.
Example Data Formula Description Result
1.4 =INT(1.4) Returns integer portion of 1.4 1
-1.9 =INT(-1.9) Returns integer portion of -1.9 -1
0.5 =INT(0.5) Returns integer portion of 0.5 0
Referenced cell Integer1 that contains the value 425.99 =INT(Integer1) Returns integer portion of 425.99 425 | 150 | 498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | longest | en | 0.635581 |
https://www.weegy.com/?ConversationId=ZTLCIA1J | 1,604,194,205,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00268.warc.gz | 931,739,608 | 11,304 | 2. Which of the following is a sentence fragment? A. Knowing he would be late, she took a nap. B. Realizing you are getting older is sobering. C. Hiding in the tree fort he had built as a kid. D. Checking his mail each day was always fun.
s
The following is a sentence fragment: Hiding in the tree he had built as a kid.
Score 1
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Updated 2/26/2019 10:42:33 AM
Confirmed by yumdrea [2/26/2019 10:42:33 AM]
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For a popular Broadway musical, the theater box office sold 356 tickets at \$80 apiece, 275 tickets at \$60 apiece, and 369 tickets at \$45 apiece. Tickets sold = 356 *80+ 275 *60 + 369 *45 = 28480 + 16500 + 16605 = \$61,585.
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14 + 17 = 31; The proper inverse operation to check the answer: 31 - 17 =14.
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## Application of Thermal Energy Transfer MCQ Quiz : SAT Physics MCQs
MCQ 1: From the following, the very poor conductor of heat is
A) water
B) iron
C) copper
D) air
MCQ 2: The vacuum flask is designed to keep liquids hot by minimizing heat loss in
A) 2 possible ways
B) 4 possible ways
C) 3 possible ways
D) 5 possible ways
MCQ 3: Where direct heating is involved, the utensils are usually made up of
A) iron
B) glass
C) plastic
D) stainless steel/aluminum
MCQ 4: The dense feathers and a layer of fat under the skin allows the penguins to withstand temperatures below
A) −10°C
B) 40°C
C) −40°C
D) 100°C
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www.desk-cycle.com | 1,386,826,607,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164540183/warc/CC-MAIN-20131204134220-00000-ip-10-33-133-15.ec2.internal.warc.gz | 301,267,945 | 19,742 | ## DeskCycle Calorie Calculator
### Displays are not Accurate
Many exercise bikes display calories. Unfortunately the majority of them are not accurate. The reason for the inaccuracy is that they don't measure the actual power you put into the bike. Without knowing power, the calorie estimates can be off by as much as 10 times.
### Why don't they Measure Power?
Because it's expensive to measure power. This is why you only see this feature in the more expensive bikes.
### Our Solution
We calibrate our bikes at the factory. The resistance is known for each position of the adjustment dial at all pedal speeds.
We take it a step further than the expensive exercise bikes. We include the physical characteristics of the person using the bike in the calorie calculation.
### how to use the calorie calculator
• Select your Gender, Height, Weight and Age.
• Then select the Resistance Dial setting that the bike is currently set to.
• Then enter the Time and Distance values from your display.
• Then press the Calculate Calories button.
Note! The time on the display goes back to zero after it reaches 100 minutes. If you use the bike for more than 100 minutes, you will need to add 100 minutes to your time. If you use the bike for more than 200 minutes you will need to add 200 minutes to the time. etc...
### How to Reset the Display
The display accumulates Time and Distance. You can reset the Time and Distance values to zero before each workout. To reset the display simply hold the button down for about 5 seconds.
Gender Male Female
Height
Weight
Age
Resistance Dial Setting
Time
Distance
Enter your settings and press the Calculate Calories button.
### Definitions
Calories Burned Sitting: This is the number of calories that you would have burned for the selected time if you had not used the bike.
Additional Calories Burned Using the Bike: These are the calories burned from pedaling the bike.
Total Calories Burned: The total number of calories that you burned during the workout. This includes the number of calories that you burned using the bike and the number of calories that you would have burned just sitting at your desk for the time that you rode the bike. This is equal to the Calories Burned Sitting + the Additional Calories Burned Using the Bike.
Total Calories Per Hour: The total number of calories that you are burning per hour. This includes the number of calories that you would burn during an hour of using the bike and the number of calories that you would have burned just sitting at your desk for that hour. | 527 | 2,566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.914682 |
http://www.indiastudychannel.com/experts/42495-find-the-total-number-of-pages-in-the-book.aspx | 1,539,693,937,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510749.55/warc/CC-MAIN-20181016113811-20181016135311-00507.warc.gz | 493,148,993 | 9,565 | You must Sign In to post a response.
• # Find the total number of pages in the book.
Searching for answers to a a mathematical problem? Check out this page for response to your queries.
Sabina read 5/8 of a book consisting of 160 pages in one day. Her friend Alia has read 40 pages more than Sabia. Find the number of pages read by Alia. Express it as a fraction of the total number of pages in the book.
Experts: do provide the answer for the above specified word problem.
• Given that Sabia read 5/8 of 160 pages of the book.
Therefore 5/8 x 160 = 100 pages read by Sabina.
Alia read 40 pages more than Sabina.
That means 100 + 40 = 140 pages.
In fraction
(The number of pages read by Alia) / Total pages in the book) = 140/160
=7/8
That means number of pages read by Alia express as 7/8 in fraction.
• 5/8 means here the book is divided into 8 parts and Sabina has read 5 parts out of 8.
total pages=160
160 pages divided into 8 parts = 160/8=20 pages for 1 part
since 20 pages makes 1 part Alia has read 40 pages i.e 2 parts more than Sabina
so Alia has read 7 parts out of 8 =7/8.
• As per question- Sabina read 5/8 of 160 pages of the book.
Therefore,5/8 of 160= 5/8 x 160
= 100 pages read by Sabina.
Alia read 40 pages more than Sabina.
So,100 + 40 = 140 pages read by Alia.
In fraction, the number of pages read by Alia,
=(The number of pages read by Alia) / Total pages in the book)
= 140/160
=7/8
Ans- Alia has read 7 parts out of 8=7/8(87.5%)
• Sign In to post your comments | 454 | 1,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-43 | latest | en | 0.900945 |
http://www.traditionaloven.com/tutorials/angle/convert-gradian-angle-unit-to-full-circle.html | 1,506,294,837,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690228.57/warc/CC-MAIN-20170924224054-20170925004054-00304.warc.gz | 610,947,911 | 11,764 | # angle conversion
## Amount: 1 gradian (grad) of angle Equals: 0.0025 full circles (rev) in angle
Converting gradian to full circles value in the angle units scale.
TOGGLE : from full circles into gradians in the other way around.
## angle from gradian to full circle Conversion Results:
### Enter a New gradian Amount of angle to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other angle measuring units - complete list.
Conversion calculator for webmasters.
## Angles
This calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.
Convert angle measuring units between gradian (grad) and full circles (rev) but in the other reverse direction from full circles into gradians.
conversion result for angle: From Symbol Equals Result To Symbol 1 gradian grad = 0.0025 full circles rev
# Converter type: angle units
This online angle from grad into rev converter is a handy tool not just for certified or experienced professionals.
Second: full circle (rev) is unit of angle.
## 0.0025 rev is converted to 1 of what?
The full circles unit number 0.0025 rev converts to 1 grad, one gradian. It is the EQUAL angle value of 1 gradian but in the full circles angle unit alternative.
How to convert 2 gradians (grad) into full circles (rev)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.0025 * 2 (or divide it by / 0.5)
QUESTION:
## Other applications for this angle calculator ...
With the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:
1. in practicing gradians and full circles ( grad vs. rev ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with angle's values and properties.
International unit symbols for these two angle measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for gradian is: | 560 | 2,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2017-39 | longest | en | 0.842393 |
https://www.onlinemathpro.com/tutorials/graphing-quadratic-functions-1---locating-the-vertex | 1,716,592,538,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00081.warc.gz | 801,972,922 | 18,382 | # Graphing Quadratic Functions 1 - Locating the Vertex
Recall:
1) The graph of a quadratic function is called a Parabola.
2) The Vertex of a parabola is the point at which the parabola stops falling and begins rising (or vise versa). In other words, it is the point at which the parabola officially “turns the corner”. {It can also be thought of as the only point on the parabola at which the slope is equal to zero.}
3) There are three common forms of the equation of a quadratic function:
· Vertex form: y = a(xh)2 + k
· Standard form: y = ax2 + bx + c
· Intercept form: y = a(xp)(xq)
When graphing a parabola, the first step is to locate and plot the vertex.
HOW TO LOCATE THE VERTEX – EQUATION IN VERTEX FORM:
1) The x-coordinate is the OPPOSITE of the constant term of the binomial inside of the parentheses.
2) The y-coordinate is the SAME as the constant term being added (generally to the right) to the squared expression. If no such term can be seen, the y-coordinate of the vertex is 0.
EXAMPLES WITH EXPLANATIONS: (Click on image to enlarge)
HOW TO LOCATE THE VERTEX – EQUATION IN STANDARD FORM:
1) Use the formula: -b / 2a to determine the x-coordinate of the vertex
2) Substitute this value into the function and evaluate in order to determine the y-coordinate of the vertex.
EXAMPLES WITH EXPLANATIONS: (Click on images to enlarge)
HOW TO LOCATE THE VERTEX - EQUATION IN INTERCEPT FORM:
1) The x-coordinate is halfway between p and q.
2) Substitute this value into the function and evaluate in order to determine the y-coordinate of the vertex.
EXAMPLES WITH EXPLANATIONS: (Click on images to enlarge) | 413 | 1,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.849043 |
https://pwntestprep.com/wp/tag/solve-for-expression/ | 1,563,651,838,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526670.1/warc/CC-MAIN-20190720194009-20190720220009-00256.warc.gz | 539,856,594 | 27,905 | ## Posts tagged with: solve for expression
How’s everyone else doing on this quiz? …
This content is for Math Guide owners only.
If you’re on the East Coast, I hope that you and yours are able to weather the storm OK. I’m not sure whether I’ll have power or not on Monday, but I’ll try to get the solution posted then if possible.
This week’s prize will again be access to the Math Guide Beta, which is looking radder and radder by the day. I’ve been doing a TON of work on it. Anyhoo, first correct (and not anonymous) answer in the comments gets access.
If (x2 + 5x)2 – 36 = (x + m)(x + n)(x + p)(x + q), what is the median of the set {m, n, p, q}?
Good luck, kiddos. And stay safe.
UPDATE: Nice work, Ammad. I hope you enjoy the book, which I’ve now shared with you in Google Docs.
Solution below the cut.
The trick to getting this one right (or at least, to getting it right without the aid of a powerful calculator) is recognizing that the left side of the original equation is actually the difference of two squares:
(x2 + 5x)2 – 36
(x2 + 5x + 6)(x2 + 5x – 6)
Of course, each of those can be factored as well:
(x + 3)(x + 2)(x + 6)(x – 1)
So although we don’t know which one is which, we do know that {m, n, p, q} = {-1, 2, 3, 6}, which has a median of 2.5.
As usual, this was a bit tougher than you’d see on the SAT, but if you figured out that, despite its complexity, this was really just a difference of two squares question, then you deserve a cookie. No…two cookies.
Very sorry for the disappearing and reappearing blog the past 24 hours or so. Blogger has had some difficulties, but I’ve used it for the better part of a decade without incident, so this one won’t send me running for the hills. I’m assured that the super long post I wrote and posted yesterday about obsessive vocabulary studying will be back shortly. Or I will seriously freak out.
Prize for answering this weekend’s challenge question: I’ll use an image of your choice in a future post. Image can’t be copyrighted, profane, or have anything to do with the Philadelphia Phillies. I reserve the right not to post anything I think sucks.
If p and q are positive integers, what is p + q?
UPDATE: Congrats to JD for getting it the long way, and then the short way. Solution below the cut.
It’s funny: I write these questions in a vacuum and sometimes I don’t realize exactly how hard they’ll be until I discuss them with a few people. This one, as these challenge questions go, was especially difficult. I knew it was devious to use 243 in the exponent and the denominator, since the exponent literally could have been anything, but I just couldn’t help myself. So I’m sorry if this one drove you nuts.
There are two insights required to solve this:
1. 15243 = 32435243. This is easier to see when you’re dealing with variables (ex: (xy)2 = x2y2), but you can always factor numbers that are raised to exponents, if that’ll help you towards a solution. In this case we’re trying to get to 3p5q; that’s a clue that you’re going to want to factor the 15 out.
2. 243 = 35. Yeah. That’s gonna be important.
Given those things, let’s attack the problem (remember, every step you take should bring you closer, somehow, to 3p5q.
Now you can cancel the 35 out of the denominator by subtracting 5 from the exponent in 3243 in the numerator. (For a review of this and other exponent rules, click here.)
And there you have it. If p and q are positive integers, then they have to be 238 and 243, respectively. So p + q = 481.
If I asked told you it was my birthday and I wanted a cake, what would you do? You’ve got two choices: buy a bunch of ingredients and start baking, or go to a different aisle in the grocery store and just buy the cake.
Baking the cake yourself is not only more time consuming than just buying one; it also gives you more opportunities to screw up (like, say, mistake salt for sugar and bake the grossest cake of all time). Since you know I’m a shameless crybaby who will never let you forget it if you ruin my birthday, you should just buy the cake in the cake aisle, and then use your time to do something more fun than baking.
And so it is with the SAT. What do I mean? WHAT DO I MEAN??? Read on, young squire.
Here’s a pretty common question type on the SAT:
1. If 3x – y = 17 and 2x – 2y = 6, what is the value of x + y?
(A) 8
(B) 9
(C) 11
(D) 12
(E) 14
The SAT is asking you for a cake here. Baking it yourself will still result in a cake, but it will also give you opportunity to screw up, and take longer than just buying one. They don’t give a rat-turd if you buy the ingredients (x and y), so don’t waste time on them! All that matters is the finished cake, (the value of the expression x + y), and we should be able to solve for that directly without ever finding x or y individually.
To do so, first stack up the equations we’re given and the expression we want:
3x – y = 17
2x – 2y = 6
x + y = ?
Do you see it yet? How about now:
3x – y = 17
–[2x – 2y = 6]
x + y = ?
That’s right. All we need to do is subtract one equation from the other (I’ve already distributed the negative here):
3x – y = 17
-2x + 2y = -6
x + y = 11
Note that we didn’t have to do any algebra here (aside from distributing that negative). This is not the exception on the SAT, this is the rule. When you’re given two equations and asked to solve for an expression, you almost NEVER have to do algebra. So instead of jumping into an algebraic quagmire as soon as you see questions like this, ask yourself “How do I quickly go from what they gave me to what they want?” Let’s look at one more example together:
1. If (x – y)2 = 25 and xy = 10, then what is x2 + y2?
(Let’s call this one a grid-in, so no multiple choice.)
Here, we aren’t going to get anywhere by simply adding or subtracting our two equations, but do you see anything else going on? Do you see that you’ve been provided with all the pieces of a particular puzzle?Let’s start by expanding what we were given (FOIL works here of course, but you should really make an effort to memorize the basic binomial squares and difference of two squares to save you time on the test; they all show up frequently):
(x – y)2 = 25
x2 – 2xy y2 = 25
And would you look at that—we’re pretty much already done. Just substitute the value you were given for xy, and do a little subtraction:
x2 – 2(10) y2 = 25
x2 – 20 y2 = 25
x2 y2 = 45
Note again that we didn’t need to solve for x or y individually to find this solution; all we needed to do was move some puzzle pieces around. Note also that actually solving for the individual variables would have been a huge pain.
Again, your mantra: “How do I quickly go from what they gave me to what they want?
##### Now YOU try:
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Also try this question I posted a while back (contains explanation).
1. For real numbers a, b, and c, ab = 1.5, bc = 6, and ac = 25. So abc =(A) 9
(B) 12
(C) 15
(D) 100
(E) 225
Answer and explanation below the fold…
If I told you it was my birthday and I wanted a cake, and you liked me enough to provide one (pretty please?), you’d have a couple options. You could go to the grocery store and buy all the ingredients, and then go home and start baking. Or, you could go to the same store and go to a different aisle, buy a pre-made cake, and spend the time you would have been baking watching awesome videos on the internet. Listen, I’m glad you like me and all, but you don’t really need to take all that time making me a cake. Internet videos. Get some.
What I mean to say is: when the SAT asks you to find an expression (by which I mean anything other than just a plain old variable like x) you should look for a way to solve directly for that expression. Your mantra for a question like this: “How do I go from what they gave me to what they want?
Often, you’ll be adding or subtracting two equations. That doesn’t do us any good with this problem, though. For this one, we’re going to have to get a little more creative. Note that we are given three equations containing parts of what we want: abc. How do we go from what they gave us to what they want?
Try multiplying everything you have together:
(ab)(bc)(ac) = (1.5)(6)(25)
abbcac = 225
Now simplify a little bit:
aabbcc = 225
a2b2c2 = 225
(abc)2 = 225
And take the square root of both sides:
abc = 15. | 2,284 | 8,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-30 | latest | en | 0.943195 |
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2017-01-30, 15:36 #1 R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 101111110112 Posts February 2017 The new Ibm puzzle is out: https://www.research.ibm.com/haifa/p...ruary2017.html
2017-03-02, 23:13 #2 R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 5FB16 Posts The official solution is here: https://www.research.ibm.com/haifa/p...ruary2017.html Just a small note: since the total search space (2^42) is not that large, semi-bruteforce routine with a sample c code (without dirty optimizations, avx2 etc.) is possible: use the mod 7 trick, precompute 6+s*(5+s*s*s*(2+s*(3+s))) + b*(5+s*(5+s*(6+s*s*(3+s*6)))) for b=0..1,s=0..6, then the problem is solvable in roughly 3 days on a a fast i7 using 1 core, or in say a week on a slow core i3. Ofcourse the distribution of the problem to all cores/threads is trivial if you want a faster (bruteforce) solution. My sent solution (not for the star): The answer is: 3271652365264 A dynamic programming solution in PARI-GP: Code: fun(L)={A=matrix(7,2*L+1,i,j,0);A[2,L+1]=1; for(i=1,L,B=matrix(7,2*L+1,i,j,0); for(b=0,1,for(k=1,7,for(l=1,2*L+1,if(A[k,l]!=0, s=k%7;r=l-(L+1); if(s==b,r=r+2*b-1); s=(6+s*(5+s*s*s*(2+s*(3+s)))+b*(5+s*(5+s*(6+s*s*(3+s*6)))))%7; k2=s%7;if(k2==0,k2=7); l2=r+(L+1); B[k2,l2]+=A[k,l]))));A=B); return(2^L-sum(i=1,7,A[i,L+1]))} Here fun(42) gives the answer for 42 bits. The complexity of the above algorithm is O(L^2) for L bits, much faster then the almost trivial O(2^L*L) algorithm. In general if we need the answer for L bits, then for every sequence and in each step r will be in [-L,L] interval and it is enough to know s mod 7. For a given v if we know r and (s mod 7) when we process the bit's of v, then we can get the value of r at the end of the function. So count only the number of possible sequences of length i bits for that we have r and (s mod 7), for the next bit we have 2 possibilities b=0 or 1, for all these sequences we have the same new r2 and (s2 mod 7) values if we fix the bit. After i=L we get the number of sequences with L bits for each (possible) r and (s mod 7). In the code: we store the number of sequences in the matrix A, in A[k,l] we see the number of sequences that has s=k mod 7 and r=l-(L+1), in this way the size of matrix is 7X(2*L+1). At the end, after we processed all L bits, the number of sequences that return a zero value is just sum(i=1,7,A[i,L+1]), so the number of sequences for a non-zero return value is just 2^L-sum(i=1,7,A[i,L+1]).
Similar Threads Thread Thread Starter Forum Replies Last Post Xyzzy Puzzles 27 2018-03-08 06:31 R. Gerbicz Puzzles 42 2017-06-06 02:23 Xyzzy Puzzles 1 2016-03-07 02:48 Xyzzy Puzzles 1 2015-03-02 19:01 robreid Software 2 2004-03-02 12:40
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When 12 marbles are added to a rectangular aquarium, the water
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When 12 marbles are added to a rectangular aquarium, the water
by BTGModeratorVI » Wed Oct 07, 2020 7:20 am
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When 12 marbles are added to a rectangular aquarium, the water in the aquarium rises 1 1/2 inches. In total, how many marbles must be added to the aquarium to raise the water 2 3/4 inches?
A. 16
B. 18
C. 20
D. 22
E. 24
Source: Magoosh
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Re: When 12 marbles are added to a rectangular aquarium, the water
by [email protected] » Sat Oct 10, 2020 7:15 am
BTGModeratorVI wrote:
Wed Oct 07, 2020 7:20 am
When 12 marbles are added to a rectangular aquarium, the water in the aquarium rises 1 1/2 inches. In total, how many marbles must be added to the aquarium to raise the water 2 3/4 inches?
A. 16
B. 18
C. 20
D. 22
E. 24
Source: Magoosh
We can solve this question using equivalent ratios
We'll use the ratio: #of marbles/rise (in inches)
Let x = the number of marbles required to raise the water 2 3/4 inches (aka 2.75 inches).
We can write: 12/1.5 = x/2.75
Cross multiply to get: (1.5)(x) = (12)(2.75)
Simplify: 1.5x = 33
Solve: x = 33/1.5 = 22
Cheers,
Brent
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Re: When 12 marbles are added to a rectangular aquarium, the water
by [email protected] » Sat Oct 17, 2020 7:59 am
BTGModeratorVI wrote:
Wed Oct 07, 2020 7:20 am
When 12 marbles are added to a rectangular aquarium, the water in the aquarium rises 1 1/2 inches. In total, how many marbles must be added to the aquarium to raise the water 2 3/4 inches?
A. 16
B. 18
C. 20
D. 22
E. 24
Solution:
We can create the proportion:
12/(3/2) = x/(11/4)
24/3 = 4x/11
24(11) = 12x
2(11) = x
22 = x
Alternate Solution:
We observe that the addition of each marble raises the water (1 1/2)/12 = (3/2)/12 = 1/8 inch. Thus, to raise the water 2 3/4 = 11/4 inches, we need (11/4)/(1/8) = (11/4) x (8/1) = 11 x 2 = 22 marbles.
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Re: When 12 marbles are added to a rectangular aquarium, the water
by gentvenus » Sat Oct 17, 2020 9:37 am
I've lost my marbles on this one
• Page 1 of 1 | 1,022 | 2,891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2023-23 | latest | en | 0.87207 |
http://www.physics.udel.edu/~watson/phys345/class/17-pos-example.html | 1,539,919,658,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00048.warc.gz | 539,056,501 | 1,806 | ## PHYS345 Electricity and Electronics
Example of Karnaugh Mapping
Boxing Zeroes for Product of Sums:
Consider again the following truth table for three inputs:
A B C O
0 0 0 0
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 1
1 1 1 0
The truth table expressed in a form suitable for Karnaugh mapping:
BC 00 01 11 10
A\
0 0 0 1 1
1 1 0 0 1
Boxing the zeroes:
BC 00 01 11 10
A\
0 0 0 1 1
1 1 0 0 1
The red cells form a 1x2 supercell represented by A'.B'.
The green cells form a 1x2 supercell represented by A.C.
The resulting Boolean expression for this truth table is the complement of the sum of these cells, (A'.B' + A.C)'.
This still looks like a sum of products, just NOTted!
Several applications of De Morgan's theorems will yield the desired product of sums:
(A'.B' + A.C)' = (A'.B')' . (A.C)' by one application of De Morgan's theorem. = (A+B) . (A.C)' by application of the other De Morgan's theorem. = (A+B) . (A'+C') again by application of De Morgan's theorem. Thus results the product of sums!
This realization again requires five gates as shown below:
NOTE: The minimum realization requires only three gates.
This results from the next to last line in the application of De Morgan's theorems above. | 439 | 1,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.663883 |
https://db0nus869y26v.cloudfront.net/en/Stochastic_calculus | 1,723,467,620,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00394.warc.gz | 147,305,836 | 15,274 | Stochastic calculus is a branch of mathematics that operates on stochastic processes. It allows a consistent theory of integration to be defined for integrals of stochastic processes with respect to stochastic processes. This field was created and started by the Japanese mathematician Kiyosi Itô during World War II.
The best-known stochastic process to which stochastic calculus is applied is the Wiener process (named in honor of Norbert Wiener), which is used for modeling Brownian motion as described by Louis Bachelier in 1900 and by Albert Einstein in 1905 and other physical diffusion processes in space of particles subject to random forces. Since the 1970s, the Wiener process has been widely applied in financial mathematics and economics to model the evolution in time of stock prices and bond interest rates.
The main flavours of stochastic calculus are the Itô calculus and its variational relative the Malliavin calculus. For technical reasons the Itô integral is the most useful for general classes of processes, but the related Stratonovich integral is frequently useful in problem formulation (particularly in engineering disciplines). The Stratonovich integral can readily be expressed in terms of the Itô integral, and vice versa. The main benefit of the Stratonovich integral is that it obeys the usual chain rule and therefore does not require Itô's lemma. This enables problems to be expressed in a coordinate system invariant form, which is invaluable when developing stochastic calculus on manifolds other than Rn. The dominated convergence theorem does not hold for the Stratonovich integral; consequently it is very difficult to prove results without re-expressing the integrals in Itô form.
## Itô integral
The Itô integral is central to the study of stochastic calculus. The integral ${\displaystyle \int H\,dX}$ is defined for a semimartingale X and locally bounded predictable process H. [citation needed]
## Stratonovich integral
The Stratonovich integral or Fisk–Stratonovich integral of a semimartingale ${\displaystyle X}$ against another semimartingale Y can be defined in terms of the Itô integral as
${\displaystyle \int _{0}^{t}X_{s-}\circ dY_{s}:=\int _{0}^{t}X_{s-}dY_{s}+{\frac {1}{2))\left[X,Y\right]_{t}^{c},}$
where [XY]tc denotes the quadratic covariation of the continuous parts of X and Y. The alternative notation
${\displaystyle \int _{0}^{t}X_{s}\,\partial Y_{s))$
is also used to denote the Stratonovich integral.
## Applications
An important application of stochastic calculus is in mathematical finance, in which asset prices are often assumed to follow stochastic differential equations. For example, the Black–Scholes model prices options as if they follow a geometric Brownian motion, illustrating the opportunities and risks from applying stochastic calculus.
## Stochastic integrals
Besides the classical Itô and Fisk–Stratonovich integrals, many different notion of stochastic integrals exist such as the Hitsuda–Skorokhod integral, the Marcus integral, the Ogawa integral and more. | 670 | 3,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.894815 |
http://guidescroll.com/2011/09/illyriad-power-of-sovereignty-guide/ | 1,558,746,355,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257845.26/warc/CC-MAIN-20190525004721-20190525030721-00349.warc.gz | 85,453,212 | 8,153 | # Illyriad Power of Sovereignty Guide
Illyriad Power of Sovereignty Guide by Beengalas
Introduction
There was a post a long time ago, when sovereignty was introduced, on the effect it might have. But it seems as most people doesn’t know or understand how powerful sovereignty can be if used correctly. Sadly, I didn’t bother doing the math back then and all my cities beside one is currently not using it. Anyone, I’ll make a post with the numbers to help other players for smarter placement on the future cities.
First of all, there is one thing I cannot stress enough: Only, and absolutely only, build new cities on 7 tile food. No exception, whatsoever. So, if you read this line, and place a new city on a 5 food tile, you fail. The end. :)
This game is pretty much about one thing: food vs population ratio. More food equals more population, which equals more gold. And with gold, you can get bigger and cuter armies. Lets break it down in numbers:
Farm production
A farm tile at level 20 produces 2014 food per hour.
With 5 farm tiles at level 20 you get 2014 * 5 = 10070 food per hour.
Also, lets add a lvl 20 flourmill, usage of prestige and the spell. The bonuses from these are 40%, 20% and 8%, which gives us:
10070 * 1,4 * 1,2 * 1,08 = 18271.
So, pretty much 18.3k is the max amount of food a city can get. But wait, these are depending on your taxation also. With 0 % taxation you gain another 25% to production ( 18271*1,25 = 22838 ).
So, a normal city can, even with prestige, just gain 22.3k food. It cannot even maximize its population (26k). And since its effected by taxation, your maximum income(troop) will be based on your food production.
City at 5k pop
Lets break it down even further and use a 5k pop city as an example and see how much troops it can get and for the sake of argument, all food is maximized:
At 0 % taxation it will produce 22.8k food. But 5k are spent on the population and therefore it would produce 17,8k food per hour. That may sound like a lot, but it isn’t since it get heavily effected by the taxation change once it produces troops.
So, say we have 90% taxation:
18271 * 0,35 – 5000 = 1,4k food over.
Sounds decent? Remember this is a 5k pop city only and it will at 90% taxation only produce 18k gold per hour. That equals 4.5k Tier 2 cavalry. That isn’t much.
City at 10k pop
Say, we have a 10k population city and with 70% taxation:
FOOD: 18271 * 0,55 – 10 000 = 49 food/hour.
GOLD: 10 000 * 4 * 0,7 = 28000 gold/hour
Tier 2 Cavalry = 7k.
City at 15k pop
Taxation at 40%
FOOD: 18271 * 0,85 – 15 000 = 530
GOLD: 15 000 * 4 * 0,4 = 24 000 gold/hour
Tier 2 cavalry = 6k
Normal city conclusion
I know the exact number for maximizing the gold production, but there is no need to go into detail on that. But 7k tier 2 cavalry is close enough at 10k population. Going much higher in pop will only result in less gold per hour and therefore less space for armies. This was mostly to demonstrate the importance between gold and food. Also it might be profitable to gain some more pop just in order to get that 8th city, since it can cover up the loss from other cities. However, the 9th city, with only normal cities, will NOT cover up the cost it requires.
7 food tile cities
A 7 tile city will produce 2014 * 7 = 14098, and with spell, prestige and flourmill: 25579. Compare that to the normal city 18271 food production. This will result in us having higher pop and therefore more army.
Lets start directly with the 10k city.
7 food tile city at 10k pop
At 80 % taxation:
FOOD: 25579 * 0,45 – 10000 = 1510 food/hour
GOLD: 10 000 * 4 *0,8 = 32000
Tier 2 cavalry = 8 000
And there you go folks, at 10k pop, having 7 food tiles gives you 1k more tier 2 cavalry!
7 food tile city at 15k pop
At 60% taxation:
FOOD: 25579 * 0,65 – 15000 = 1626 food/hour
GOLD: 15 000 * 4 * 0,6 = 36000
Tier 2 Cavalry = 9 0000
That right, 3k more Tier 2 cavalry than the 5 food tile city!
Conclusion on 7 tile food city
As shown, having the two extra food tiles can make a huge difference. It allows you to have bigger cities which makes it more profitable to have 8 or 9 cities, and it also allows you to gain bigger army. In conclusion, no matter what, get those 7 tile cities. Other basic resources are worth next to nothing in this game. But now onwards to the important and cool stuff, sovereignty!
Sovereignty
With sovereignty we can have max populated cities, without loosing food, and also have a great army. In this case, I will use one of my cities to illustrate the power of sovereignty. I will call it “City X”, as I for some reason don’t want to reveal it here. Can’t explain why thou. :)
City X
My city X got two food sov. tiles. The first one has 19 in food and the second one has 15. And this result in a 34% bonus to food production, at the cost of 1580 * 2 = 3160 gold/hour. And of course this city has 7 farm tiles as basic. Therefore the food production looks like this:
Basic food production:
(2014 * 7) = 14098
Bonuses:
Flourmill: 40%
Spell: 8%
Prestige: 20%
Sov: 34%
Total food production is:
14098 * 1,4 * 1,08 * 1,2 * 1,34 = 34276
That is, those two sov. tiles gives almost 10k more food production to the city! And the max size of a city is 26980, which can be supported easily. But how much Tier 2 cavalry can I gain?
At 40% taxation:
FOOD: 34276 * 0,85 – 26980 = 2154 food/hour
GOLD: 26980 * 4 * 0,4 = 43168 – 3160 (sov. cost) = 40008 gold/hour
Tier 2 Cavalry = 10k.
So, at maximal population, this city can still support 10k tier 2 cavalry. And that is not max, at lower population it can support up to 12k tier 2 cavalry. 12k tier 2 cavalry in one city, that is not feed at all by other cities! Next issue is to actually get 12k tier 2 cavalry, but thats another matter. And my next location for city would be even better, which will have food production reaching over 40k/hour. Higher cost of sov. tiles of course but that city could support up to 14k tier 2 cavalry.
Big Conclusion
I hope this has given the people of Illyriad a better insight on how to actually pick their location for cities. Because it always boils down to food vs population ratio. This little post have shown the power of sov. tiles, the importance of 7 tiles of food in a city and that there is a breaking point were bigger cities gives less room for an army, althought I didn’t show exactly where it is, even thou I know it :) . Hopefully you have liked this post and that it will help you placing your future cities.
But do remember that long distance between your cities make them more insecure and takes longer time to send stuff between them. Oh, and this pretty much demands usage of prestige, or the time required to actually reach these numbers can almost be calculated in years. | 1,901 | 6,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-22 | latest | en | 0.921461 |
https://docs.microsoft.com/es-es/dotnet/api/system.intptr.op_subtraction?view=netframework-4.8 | 1,643,377,100,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00174.warc.gz | 255,221,415 | 10,136 | # IntPtr.Subtraction(IntPtr, Int32) Operador
## Definición
Resta un desplazamiento del valor de un puntero.Subtracts an offset from the value of a pointer.
``````public:
static IntPtr operator -(IntPtr pointer, int offset);``````
``public static IntPtr operator - (IntPtr pointer, int offset);``
``static member ( - ) : nativeint * int -> nativeint``
``Public Shared Operator - (pointer As IntPtr, offset As Integer) As IntPtr``
#### Parámetros
pointer
IntPtr
Puntero del que se va a restar el desplazamiento.The pointer to subtract the offset from.
offset
Int32
Desplazamiento que se va a restar.The offset to subtract.
#### Devoluciones
IntPtr
Nuevo puntero que es el resultado de restar `offset` de `pointer`.A new pointer that reflects the subtraction of `offset` from `pointer`.
## Comentarios
El Subtraction método define la operación de resta de IntPtr objetos.The Subtraction method defines the subtraction operation for IntPtr objects. Permite el código como el siguiente.It enables code such as the following.
``````int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16, 18, 20};
unsafe {
fixed(int* parr = &arr[arr.GetUpperBound(0)])
{
IntPtr ptr = new IntPtr(parr);
for (int ctr = 0; ctr <= arr.GetUpperBound(0); ctr++)
{
IntPtr newPtr = ptr - ctr * sizeof(Int32);
}
}
}
// The example displays the following output:
// 20 18 16 14 12 10 8 6 4 2
``````
``````Dim arr() As Integer = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20 }
Dim ptr As IntPtr = Marshal.UnsafeAddrOfPinnedArrayElement(arr, arr.GetUpperBound(0))
For ctr As Integer= 0 To arr.GetUpperBound(0)
Dim newPtr As IntPtr = ptr - ctr * Len(arr(0)) | 476 | 1,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.294824 |
https://math.stackexchange.com/questions/3097764/determine-conditions-on-pn-such-that-the-probability-that-a-random-graph-has-at | 1,556,266,570,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578762045.99/warc/CC-MAIN-20190426073513-20190426095513-00255.warc.gz | 481,865,899 | 34,079 | # Determine conditions on pn such that the probability that a random graph has at least one triangle goes to zero as n increases
I'm currently struggling with an exercise about random graphs where is requested to determine the conditions on $$p_n$$ such that the probability that $$G(n, p_n)$$ has at least one triangle goes to zero as $$n → +∞$$, also assuming that the probability that three vertices forms a triangle is $$p_n^3$$. I know I have to apply the first order method but I can't relate the theory with general properties.
• Is the step where you're getting stuck "how do I compute the expected number of triangles in $G(n,p$)"? – Misha Lavrov Feb 3 at 0:36
Let $$T_n$$ be the number of triangles in a $$G(n,p_n)$$ random graph. Then you should be able to get a formula for the expected number of triangles, $$ET_n$$. In order to ensure no triangles, or usually none, you should make (a) $$ET _n$$ large, or (b) $$ET_n$$ small. Can you take it from there?
I don't know much about graph theory but you could work out the probability of that no triangle exists. Starting with one random point A and a random point B, the probability that no triangle exists of the form ABx is given by: $$(1-p_n^3)^{n-2}$$. Now pick a different point C and the probability that no triangle exists of the form ACx conditional on what we already know is given by $$(1-p_n^3)^{n-3}$$. Using this strategy you can calculate the probability of A not being in a triangle which will be of the order $$(1-p_n^3)^{n^2}$$. Can you continue?
• Ah you're right of course. The probability of such a triangle is dependent on the existence of the edge AB. By conditioning on this event, we are still able to calculate this. It should be: $p_n\cdot\left(1-(1-p_n^2)^{n-2}\right) +(1-p_n)\cdot0$. Thanks for finding the mistake. I think it is possible to work this out to the end. But since apparently in the second step the independence assumption is also not valid (sorry for that), it will definitely become more messy than I thought at first glance. – Stan Tendijck Feb 3 at 20:51 | 524 | 2,063 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-18 | latest | en | 0.937327 |
www.careydobson.com | 1,652,914,795,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00114.warc.gz | 794,701,750 | 15,735 | ### How Can I determine the Potential Money Value of My Claim?
Let’s make this clear from the beginning. There is no such thing as a personal injury settlement formula. There is no insurance settlement calculator. There is no magic equation of “x times specials” that determines the proper settlement payout for your case.
You can keep Googling until you are blue in the face. But you will never find it. Any lawyer that suggests otherwise does not understand civil tort cases.
But there is real information that can help you better understand the settlement value of your case.
Valuing cases is hard. Our system of civil justice requires juries to monetize injuries in accident cases. But injuries are just not readily converted to a specific monetary value. So, ultimately, the best that we can hope for is narrowing down a settlement range that your case will fall into within a 90% confidence interval.
Many people are desperate for a settlement formula and the urge is to assume that the appropriate payout for settlement may be some multiple of the special damages. But if it is, it is largely a coincidence. A settlement that works is because it is fair based on what a jury would award you because of what you went through from the accident – your pain, your suffering, your bills – not because it is the result of some pre-ordained mathematical formula. The real end game to value is not a calculation but an estimate of what a jury might award in the case. There is one thing we can tell you without equivocation: juries don’t use math algorithms to compute value.
So, in some cases, a settlement that is three times medical bills and lost wages (called together “specials” is an awful “they stole from you” resolution. In other claims, it is a fantastic result. This is how it should be. Let’s beat the dead horse one more time before we move on: The idea that you can use some multiplier of the medical bills to determine how much money a judge or jury would award simply makes no sense.
Below, we explain just how cases get valued. If you want a lawyer for your claim, call us. We have an undeniable record of success. If you want to discuss your claim with us, call (404) 635-1112 (Atlanta) or you can get free on-line consultation. Keep in mind we handle only serious injury and wrongful death cases. Most of our cases are in Georgia. We accept about a handful of cases outside of Georgia each year.
### Why Settlement Multiplier Talk Actually Decreases the Value of Your Case
Attorneys trying to get the best compensation possible for their clients need to avoid the temptation to engage in the slippery slope of talking to insurance adjusters in “___ times specials” terms. Why? Because there is no formula to calculate settlement value. A traffic collision case is worth what a judge or jury would give the Plaintiff and there are about 10 variables far more important to this calculus than any settlement calculator or formula.
Okay, but why it is actually harmful to talk to insurance companies like this? Insurance adjusters deal with a lot of lawyers. Some who know exactly what they are doing and others who do not have a clue how to properly position an accident case to maximize the value of that case. (You can guess for yourself the relative proportions of those who know what they are talking about and those without a clue. The real truth is depressing.)
When attorneys use formula talk, they are telegraphing to the claims adjuster that you are looking at your client as simply a number that is a part of a formula, not a person. Attorneys that talk in “times specials” math like there is some theorem are not the same advocates that read and evaluate the medical records… or take cases to a jury. Insurance companies take advantage of that. Quick. We are experienced trial lawyers.
Sure, in very small whiplash injury type cases, a multiplier type formula will be correct in many of these claims. Still, even in those small soft tissue whiplashes type cases, tossing around formulas and calculators will send a loud message to the adjuster that will forever mark the lawyer as an inexperienced advocate who is not able to fight the fight their clients need. This makes for a smaller settlement offer in this case and the next one.
Settlement Formulas in Disguise
One thing insurance companies – thinking particularly of Nationwide nd State Farm Insurance – have been doing of late is pitching offers in terms of how much more they are paying than the medical bills and lost wages. Someone smart came up with this because it is like calling something \$9.99 instead of \$10.00. It just sounds more palatable to say, “We are offering \$50,000 more than the special damages” than saying “this offer is 1.8 times the medical bills plus lost wages.” But, really, it is the same thing as a monetary compensation formula. Regardless of how it is packaged, it is nonsense.
### Is There a Way to Figure Out the Value of My Case?
You are not going to find the exact value of your case on-line. You need a professional evaluation of the facts of your case to get an ideal of the reasonable value of your claim. There are a lot of variables at play and there may be reasons why your case is worth much more – or much less – than these statistics suggest. The take home message is clear: median and average settlement value data is great. But it will not calculate your settlement for you and does not take the place of a professional analysis of your case. Also keep in mind that many times, the insurance company will believe you are entitled to a full payout of their entire policy. But unless they know you can file a lawsuit – which almost invariably means retaining counsel – they have little incentive to make a fair offer for your injuries. So if you want to get the true amount of compensation the case is worth, you may have to fight for it.
### Finding a Lawyer for Your Serious Injury Case
Our attorneys look at cases on the merits of each case. We handle only serious injury claims. If you or a loved one has suffered a serious injury or wrongful death as the result of the negligence of someone else, we would be glad to talk to you about your case to see if we believe we can add real value to your claim. Call our lawyers at (404) 635-1112 (Atlanta) or get a free Internet consultation. We are advocates for justice and it is our mission to serve you no matter who or where you are! Contact us 24/7.
### What is the value of a serious personal injury case?
There are many factors that can affect the value of a case, and more importantly, how much money you will actually get to compensate you for what you have gone through. The settlement value of a case is a prediction of what a jury will decide your claim is worth. Different juries will arrive at different verdicts under the same set of circumstances.
Jurors have their own personal biases that can either help or hurt you. This unpredictably leads to a wide range of results. So, for settlement purposes, we guess at what we think the jury would award. The settlement value of a claim increases over time. More than 95% of the time, the value of a case will increase from the time before suit is filed to the time of trial. We have had cases where the pretrial offer was more than 50 times the pre-suit offer.
This is particularly true in medical mistake cases. These are virtually impossible to settle without filing a lawsuit. In other injury cases, we have settled cases for over a million dollars when the pre-suit offer was zero.
Who decides whether to settle the case? Our attorneys present the facts to the client and let them make the call. There are two choices: (1) settle the case, or (2) move forward with the litigation process.
Studies Looking at Value of Personal Injury Cases
There is data available to give victims some idea of the size of personal injury awards in Georgia. The nationwide median jury award in a personal injury case was approximately \$40,000 and plaintiffs win 48% of jury trials. The average personal injury verdict, as opposed to the median verdict, is \$985,675.
Verdicts in Georgia Personal Injury Lawsuits
We have earned millions of dollars for our clients by fighting every single case like it was the Super Bowl. Can we help you? Call 404-635-1112 (Atlanta) or get a free online consultation.
Jury verdicts may vary from one county to the next using the same facts and injuries. However, clearly the larger jury awards in Georgia are in our biggest urban city and counties (City of Atlanta, Fulton and DeKalb County).
Note that our very best cases generally settle out-of-court. Insurance companies like settling the best cases and letting the weaker cases got to trial.
Having quality counsel does often make a difference. Insurance companies need to understand there are repercussions for offering a deal that is not fair. You need an advocate they know is
coming after them and looking for a big verdict. Paradoxically, attorneys that frequently try cases and get big verdicts are the same ones getting the best confidential settlements that no one ever hears about.
So if you have a good case, get experienced counsel with a track record who is willing to fight for you. This is your best change of pushing past the verdict and settlement averages listen above. That data include cases handled by lawyers who do not have a clue about how to handle a tort case and, maybe worse, people without law degrees acting as their own attorneys.
The Factors That Matter in Personal Injury Cases
As we have said above, verdict statistics and our firms success in obtaining favorable settlements or trial judgments is not predictive of the results in any individual case. It sounds cliché but…. your case is unique. Figuring out what a settlement payout or verdict would be is not something you can determine using some advanced calculus formula.
You need to discuss with your lawyer the specific facts of your case and the possible defenses to your claim. Then, you can come to a conclusion about what is acceptable to resolve your case.
• the type of injuries you suffered
• the economic damages (lost wages and medical bills)
• the amount of available insurance
• how clear the causal connection is between your injuries and the accident
• how strong your liability case is
• the quality of all of the witnesses
• the expected jury perceptions of you and the defendant
• how your pain and suffering from the injuries have impacted your life
What does not matter? A lot of things that you really care about should count for settlement value but simply do not:
• You could have died (what matters is what did happen)
• You have gotten behind in paying your bills
• How you couldn’t buy a new car with the money you got for your property damage
• All of the things that could have happened during surgery but didn’t
• Medical problems you could have that are not likely to occur
These things are not admissible at trial and the adjuster will not consider them in determining how much money they should offer. Is this fair? No. What do you do? You fight harder to get compensation for those things that you can get compensated for at trial: your medical bills (even if they have been paid), your lost wages, and, most importantly in every case we handle, your pain and suffering.
We already touched on the other huge variable that the insurance company considers right from the beginning: the location of the trial. In Georgia, Fulton and DeKalb’s County are favorable venues for plaintiffs. Conversely, in more rural areas of Georgia the awards are more conservative. This is true not just in Georgia but throughout the United States — jury awards are typically higher in urban areas than in rural areas.
### How the Insurance Companies Determine Value in Accident Cases?
In motor vehicle crash claims, most of the insurance companies our lawyers deal with use computer software to determine the value of claims. They don’t care about you, your agony, your pain and suffering, or the tragedy of you never being yourself again. They care about hanging on to their money. There is no sense getting mad about it. You just have to find someone willing to buckle up their chin straps and fight for you.
Most insurers use a computer system called Colossus or a similar program. The insurance adjuster inputs the data from your case and it spits out a range of settlement outcomes.
Colossus then specifically looks to the causes of your pain as described in your medical records. One of the most important questions is whether the injuries are permanent. The computer gives higher values for objective injuries like broken bones and herniated discs than for soft tissue injuries. Colossus also gives greater value when the patient went to the hospital for initial treatment immediately after the collision.
While Colossus and similar programs do have some value, the problem is they cannot grasp the complexity of a human’s pain and suffering. There is no computer that can ascertain pain and suffering or how an injury really impacted a person’s life. How much is it worth to not be able to pick up your newborn baby without extreme pain? There is no way a computer can answer this question. Colossus does not really try to bake that into its settlement payout. It just assumes pain is not really a big inconvenience.
This is why your counsel must fully explain to the insurance company why your injuries are different or be prepared to file a lawsuit to obtain fair value for your case. When your lawyer files a lawsuit, the insurance companies will sometimes take a second look at the real trial value of the case, particularly when they know that the attorneys handling the case are willing to go to trial.
While Colossus cannot appreciate pain, suffering and the true impact of the injury on the victim, judges and juries tend to listen to and consider many of the factors that Colossus ignores because it does not understand them. It is only a formula. Juries make distinctions on how much your case is worth based upon whether or not they think the plaintiff is an honest, good person who has suffered as a result of their injuries.
Clearly, the true value of the same injury can vary. If your injury is a scar on your face, the value will depend on the Plaintiff. Sex, age, pride in appearance, are going to matter in determining how much money you are going to get for your case.
The same goes for ankle and leg injuries where the victim can no longer run. If you are a couch potato, the value is less than the client who can tell a jury they ran a marathon last year. These details matter to every juror. But the computer tunes them out completely.
Juries are actually the polar opposite of computers. Juries respond to human pain and agony. A algebraic equation responds to raw data.
Let’s take the mom who used to be a shopaholic with her three kids but now cannot get off a bench at the mall — or even get to the mall — because of the harm that has been caused to her. A computer cannot understand or calculate this kind of impact on a person. Thankfully, the real decider is not a computer but human beings. They understand all too well what real human suffering is (and what it is not).
Picking a Lawyer: Why It Matters So Much Even for Settlement
This bears repeating. Claims adjusters and even their computers consider whether your attorney has a history of winning at trial. They also look to see if plaintiff’s counsel settles all of their personal injury cases (or has no history at all). This is an important reason why the quality of your attorney matters even if you intend to settle your case without filing a lawsuit.
Our law firm handles serious personal injury cases. If you have been hurt, or a loved one has been tragically killed by someone else’s negligence, call to speak to one of our attorneys at (404) 635-1112 (Atlanta) or get for a free no obligation consultation.
### How will my damages be affected if I was partially to blame for the accident?
By statute, Georgia follows the theory of comparative negligence which means that the plaintiff’s fault proportionally reduces his/her recovery. Plaintiff’s own fault bars any recovery if his/her relative fault exceeds 50%.
### How much time do I have to bring my claim without it being forever barred?
The Georgia statute of limitations for most personal injury auto accident related claims is two years. (One notable exception: claims against the state of Georgia or local municipalities can be 1 year or shorter.) This rule applies to medical malpractice claims and product liability claims in Georgia. While there is discovery rule in Georgia, there is a five year statute of repose in medical malpractice cases. The statute of repose is harsh: it can bar a claim before the claim even accrues because it is before the injury occurs.
### What damages can I recover for the wrongful death of a loved one?
The damages recoverable in a Georgia wrongful death action include the victim’s expected earnings, or the value of the victim’s services, from the date of the victim’s death due to negligence to the statistically projected date of natural death (using life tables), plus an intangible element representing the full value of the life to the victim.
### Is there a cap on the amount of personal injury damages I can recover for my injuries?
Like many states during the “medical malpractice crisis” in the middle of this decade, the Georgia Legislature in 2005 imposed a cap on noneconomic damages in medical malpractice lawsuits of \$350,000. A trial judge in Fulton County, Georgia has struck down Georgia’s cap in medical malpractice cases, if in fact in the legislature’s cap of \$350,000 for noneconomic damages was unconstitutional in its affording of special protections to doctors. At the time of this writing, the Georgia Supreme Court had heard arguments in this case and is expected to render a ruling. In Georgia auto accident and product liability cases, Georgia does not have a cap on noneconomic damages.
### Can I punish a person for their reckless conduct that lead to my injuries?
Georgia law provides for punitive damages for wanton conduct: “wanton conduct is that which is so reckless or so charged with indifference to the consequences as to be the equivalent in spirit to actual intent.” This includes conscious indifference to consequences, which is an intentional disregard of the rights of another, or knowingly or willfully disregarding such rights. Georgia law caps punitive damages at \$250,000.00 unless the party acted with specific intent to harm. Punitive damages may only be awarded in auto accident cases or other tort claims when proven by clear and convincing evidence (as opposed to the preponderance of evidence).
### Can I recover damages for my emotional distress related to my accident?
Under Georgia’s law, a plaintiff seeking damages for emotional distress must demonstrate that: (1) he/she suffered a physical impact; (2) the impact caused him/her physical injury; and (3) the injury caused his/her mental suffering or emotional distress.
### I was involved in a hit and run accident, can I recover for my damages?
Under Georgia’s uninsured motorist statute, there must be physical contact between the vehicles unless the evidence is corroborated by an eyewitness to the pedestrian or car accident other than the claimant. Under this interesting rule, the Plaintiff cannot allege a phantom vehicle caused an accident where there was no impact with the phantom vehicle, unless there is an independent eye witness. The purpose of the rule is to eliminate “he said/”phantom said” uninsured motorist cases where there is no evidence beyond the word of the victim.
### How does the fact that my insurance company has paid all of my medical bills affect my claim?
Georgia has a collateral source rule that allows for the introduction of all damages regardless if payment has been made by collateral sources. (A Georgia statute that allowed for the introduction of collateral sources was found by the Georgia Supreme Court to be unconstitutional.
### What are the elements of a Medical Malpractice Lawsuit?
There are three elements of a medical malpractice lawsuit in Georgia: the doctor’s duty to his patient; the doctor’s breach of that duty through the failure to exercise the requisite degree of skill and care; and an injury proximately caused by the doctor’s failure.
### Can my case be resolved with expert testimony?
In cases involving causation issues that can be resolved solely by testimony from a medical expert, the testimony must be based, at a minimum or reasonable probability.
### How much can I rely on the information contained in this website?
Please remember that only individuals who have you have signed a retainer agreement with our law firm are our clients or have an attorney-client relationship with our law firm or any individual personal injury lawyers. The data is informational and may have changed or may be inaccurate given the nuances of your individual case. These general rules cited above often have exceptions that could and do fill an entire book. The only way to verify whether the general information on Georgia accident law applies to your case is to contact a personal injury lawyer in Georgia (we are Georgia lawyers) and lay out all of the facts of your case. | 4,314 | 21,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-21 | latest | en | 0.946624 |
https://control.com/forums/threads/inertia-ratio.982/ | 1,718,645,698,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00054.warc.gz | 166,845,448 | 38,920 | # Inertia ratio
G
#### Guy H. Looney
I would like to get some feedback on the topic of inertia ratio. I have recently been trying to gather information as to the importance of this specification. For those of you unsure of what I
mean, let me elaborate:
The inertia ratio is equal to the reflected load inertia divided by the motor's rotational inertia. The specifics of the calculation are not
important for this conversation. There are many recommended guidelines to address this topic. The most common is the 10:1 or 5:1 rule-of-thumb. Some manufacturers state that a stepper
system should have less than a 5:1 ratio, while a servo can have a 10:1 ratio.
The question is why does inertia ratio matter? If the motor/drive combination can handle the torque, speed, & regen shouldn't that be enough? I have an opinion on this topic but am curious as to what others think.
Let me throw out another idea. Let's assume that it is critial to maintain a certain inertia ratio. Every motor manufacturer spec's out the rotational inertia of its motors. If you'll look carefully, you'll see that the inertia spec increases if an absolute encoder &/or brake is added to the motor. Why is this? Are we to assume that by increasing the mass of the motor shaft we can effectively increase the motor's rotational inertia? If this is the case, is there
a limit to how much we can increase it or a set of guide-lines stipulating how the mass must be attached? If your answer states that the brake &/or absolute encoder are contained w/ in the motor housing & that's the reason the increase is relavant, you need to look at the construction of the motor. The brake &/or absolute encoder are not subjected to the magnetic field...therefore
they could be added outside the housing & the effect should be the same (w/ respect to increased rotor inertia). That being said, why
can't we simply put a mass on the shaft of a motor (outside the casing) to increase its inertia.
Some have said that it's a question of compliance & rigidity. Assume the load could be pressed on the shaft & that the bearing structure of the motor is sufficient to handle radial & axial loadings. If the drive/motor can handle the torque, speed, & regen is the inertia ratio important? If so, at what point does it matter?
This subject has long been a topic of conversation within our industry. I would really appreciate any feedback that is deemed relevant. I hope that I receive qualtity feedback, and if you do not feel comfortable with this topic, sit back and enjoy the responses.
Sincerely,
Guy H. Looney
Sales Engineer
Regan Controls, Inc.
475 Metroplex Dr.
Suite 212
Nashville, TN 37211
phone: (615) 333-1940, ext. 322
fax: (615) 333-1941
[email protected]
www.regancontrols.com
D
#### David Zishuk
Guy;
Let me first say that with any engineering appplication there is give and take - Regeneration is no exception.
>Let me throw out another idea. Let's assume that it is critial to maintain a certain inertia ratio. Every motor >manufacturer spec's out the
rotational inertia of its motors. If you'll look carefully, you'll see that the inertia >spec increases if an absolute encoder &/or brake is added to the motor. Why is this?
>Are we to assume that by increasing the mass of the motor shaft we can
effectively increase the motor's rotational >inertia? If this is the case,
is there a limit to how much we can increase it or a set of guide-lines
stipulating how >the mass must be attached? If your answer states that the
brake &/or absolute encoder are contained w/ in the motor >housing & that's
the reason the increase is relavant, you need to look at the construction
of the motor. The brake >&/or absolute encoder are not subjected to the
magnetic field...therefore they could be added outside the housing & >the
effect should be the same (w/ respect to increased rotor inertia). That
being said, why can't we simply put a >mass on the shaft of a motor
(outside the casing) to increase its inertia.
NO!
The ratio should be relative to the motor's rotor, not encoder or brake attached to the motor, that is part of the load. Because most manufactures (Yaskawa Included) do not give the rotor only. You can safely assume the
"standard motor", even though it includes an encoder in our case, is rotor inertia. If the Servo Motor's rotor is so light that that the encoder inertia is significant, then it should be shown separately. Even Yaskawa's tiny 10W sigma mini, still has a rotor inertia very much larger than the encoder it comes with.
>Some have said that it's a question of compliance & rigidity. Assume the
load could be pressed on the shaft & that >the bearing structure of the
can handle the >torque, speed, & regen is the inertia ratio important? If
so, at what point does it matter?
This is more complicated, and I can't give a full [perfect] answer.
1) Inertia and Bandwidth
Inertia is a low pass filter, the closer to 1:1 the better your bandwidth (Stopping and Starting time) will be. If you have a large system, and you want 1000Hz response from your load, 1:1 or less, you can forget about it. Smaller systems have quicker response. If you size a system, and you have the RMS. and Peak torque for the application, it will work.
2) Stored Energy
Any Rotating inertia has kinetic energy "stored" in the rotating load. This energy, is quantified by Ke = 1/2*J*w^2. Where J is the total system
inertia (in Kg-M^2), w is the rotational speed in rad/sec.
If you slow it down, then energy is removed from the system and has to go somewhere. In the case of the Servo Motor system, it is "regenerated" onto the DC BUSS- and starts to raise the BUSS voltage. The first device to "absorb" the energy are the BUSS capacitors. If the capacitors can absorb this extra energy- you are home free. The Got-ya is the CAP(s) may not absorb all of it, and the next step is to add more capacitors or "burn" it up with resistance. Some Servo vendors allow more capacitance to be added, Yaskawa prefers the resistor route.
3) Got the Power??
If the Servo Amp has built in resistor(s) to "burn" the regen energy that the caps cannot. You must then check if the resistors wattage is high enough. Both RMS and Peak. Wattage is En / time (sec). The decel time in seconds is used in this calculation. If the built in resistor is not
enough or is no resistor, then a provision for an external resistor should be given.
4) Resistors
One other little got-ya is the Ohmic value of the resistor: it should be low enough to quickly burn up the energy, but not so low to allow a large
destructive current through the regen switching circuit.
The Servo AMP manufacture should give guidelines for resistor Wattage, Ohmic value, and derating
5) Dynamic Brake Resistor
Another reason for the inertia ratio spec is because of DB resistors in Servo AMPS. These are the resistors that the motor leads are shorted to
when a power loss occurs. If you inertia is big, I mean really big, and you may just "POP" this normally small resistors. A good rule of thumb
(sorry this thumb thing is back) is your RMS. Regen Wattage exceeds the servo AMPs RMS capacity, there is a good chance the DB resistor will not handle the "E-STOP"
When a load is fighting gravity, and moves down (no up) Potential energy is changed to Kinetic and, in the case of Servo motor system, ends up on the DC BUSS as higher voltage. this must also be considered in sizing regen. Simple solution, use counterbalance.
7) Read this all more than once, tabulate your questions and call me with them.
David
#### PhilCorso
In response to Guy Looney's query, hopefully I can reply in general terms which do not cover a specific application, nor motor size:
As you already know inertia or moment of inertia or ratio, are not required for normal operating speed, i.e., the torque supplied by the motor equals the torque required by the driven machine, even if connected via a gear box, jack shaft, pulley and belts etc. In other words, if correctly sized, a motor's rated power output should correspond to the power absorbed by the driven machine.
Transient Operation
Important parameters are required to determine operational performance during speed changes, i.e., run-up, braking, reversing. They are the
mass (more importantly moment of inertia) of the motor, any additional significant masses (gear, drive shaft, coupling), the load torque during
run-up, the available torque produced by the motor during run-up, and the torque imparted to the entire drive-train during braking (where
required). Thus, the Equivalent Moment of Inertia of the drive-train is equal to GD^2 (Metric units) or equal to WR^2 (English units). It is
needed to determine the time to effect a desired speed change, and the amount of power to effect that change. The selected units are important as will be shown later in this discussion.
Speed Change
In general, every speed change is associated with a change in kinetic energy. This means that for a change in speed per unit time a certain amount of power is required.
Inertia Ratio
The moment of Inertia of each of the drive train elements having significant mass, must be referred or converted or, in Guy's terms,
"reflected", to the motor shaft speed. The rotating system's equivalent moment of inertia is calculated from the sum of the individual moments
of inertia of all the elements. This includes the motor, and each additional element converted to the motor speed using the "inertia ratio" as follows, in metric units:
GD^2(equivalent, referred to motor shaft speed, Nm) equals
GD^2(mtr)+Sum of {GD^2(i)x[Ni/Nm]^2},
as (i) increases from i=1 to i=n, where the term [Ni/Nm}]^2 represents the individual "inertia ratio's", i.e., the ratio of the (i)th element's
speed to the motor's speed, squared.
Most Common Error
The most common error I have encountered in my experience is the conversion from Metric to English units and vice versa. One vendor supplied a 600 Hp motor, when a 60 Hp was required. The error was two fold:
A) They overlooked the fact that 'D' in the metric equation, GD^2 (kg-mt^2), is the diameter of gyration, while 'R' in the English equation, WR^2 (lb-ft^2), is the radius of rotation.
B) They neglected to square the speed ratio.
The reasons listed above should illustrate the importance of moment of inertia and the inertia ratio. It is a major concern for large motors
because the electrical losses in the motor's rotor during run-up can be significant. It often is the most salient parameter when determining the
number of consecutive starts permitted in one hour. Of course, if rotating mass is negligible, as in small servo motors, then moment of inertia can usually be ignored.
I can supply additional detail, if anyone is interested.
Regards,
Phil Corso, PE
Trip-A-Larm Corp
G
#### George Kaufman
Even if the load inertia is ideally coupled to the motor inertia such that the total inertia is the sum of the motor and load inertia, you'll notice that the total inertia inversely lowers the velocity loop gain. From a service standpoint it is good to have the load inertia equal to or less than the motor inertia so that the inertia of the motor primarily determines the
stability. The servo will be stable when the motor is attached to the machine and when the motor is removed from the machine (such as when doing troubleshooting). If the load inertia is much larger than the motor inertia then the servo system will most likely be unstable when the motor is operated off the machine. Matching inertias can avoid misdiagnosis of a problem when a service technician faces an unstable servo when the motor is operated off the machine.
However, in most systems, the load inertia is not ideally coupled to the motor inertia. The belt drives, gearboxes, ballscrews, couplings, and other transmission elements introduce compliance, damping, friction, backlash, possible variable inertia, and sometimes even non-linearities. Once again, the more dominant you can make the motor inertia the easier it will be to tune the system for the desired performance. It's better to have the dog wagging the tail then to have the tail wagging the dog.
Finally, the required bandwidth of the velocity loop is important when considering the inertia mismatch. If high bandwidth is needed, then it is
important to have a good stiff connection between the load and motor and to keep the load inertia close in value to the motor inertia (no more than a 5:1 mismatch). If the bandwidth can be low, I have seen successful applications where the load inertia is several hundred or even several
thousand times larger than the motor inertia. The drive will typically need features such as programmable gains that can be very large (to counter the large total inertia value), low pass filtering on the current commands, and even notch filters on the current commands.
The Handbook of AC Servo Systems provides more details on the above (available as a free download from www.MotionOnline.com).
Regards,
George Kaufman
J
#### Johan Bengtsson
Wow, a good question, I have been asking myself that one too I can not se any reason either that it would be of any interest to the controller if the inertia is located inside or outside the motor.
I do however have some ideas where this might come from:
1. If the load are not really tight put to the motor you will have a mecanical system where the potors position and the loads position doesn't necesarily be exactly the same and the connection between these will be close to a spring. This will give you mechanical oscillations between the motor and the load.
2. The controller optimisation calculations is based on a model where the motor inertia is known and the load inertia is given as a factor, these calculations might simply bail out totaly when the ratio it too high but a model and optimisation formula based on another assumption
might work a lot better for big ratios.
It is really going to be interesting to study this in the servo simulation we will do in the near future. You did give me some additional thoughts about things to test....
I do definitely look forward to the answers from the rest of the list...
/Johan Bengtsson
----------------------------------------
Box 252, S-281 23 H{ssleholm SWEDEN
Tel: +46 451 49 460, Fax: +46 451 89 833
E-mail: [email protected]
Internet: http://www.pol.se/
----------------------------------------
S
#### Simon Martin
Hi Guy,
>From basic mechanics, inertia ratio of 1 implies the greatest power
transfer.
Empirically, inertia mismatch makes system tuning a nightmare. One of the greatest problems with hydraulic systems is that the inertia of the power
unit (i.e. hydraulic ram) is so low with respect the load inertia, that the tuning is usually the hardest part of the job. I have never enjoyed working on a servo system with an inertia mismatch as high as 10:1, they work but it is difficult to get the correct dynamic response.
Think of it this way, any system like this can be represented as two discs each with a given moment of inertia, representing the motor and load
respectively:
----- -----
| | | |
| |---------| |
| a | | | b |
| |---------| |
| | | |
----- -----
Let Ja be the inertia of a and Jb be the inertia of b.
Case 1 - Ja = Jb
================
Release the coupling between a and b and move a by hand. See the dynamic characteristics of the system. Reconnect the coupling, and apart from the
fact that the system now accelerates at half the rate for a given force, the dynamic characterstics of the system are unchanged. Why? Each load (a and b) reacts in the same way to a given force (conservation of rotational
momentum, energy, etc).
Case 2 - Ja >> Jb
=================
The dynamic characteristics of the system are governed by a, as the highest proportion of the energy required to accelerate, decelerate the system is absorbed by a. If we consider a to be the motor and b the load, then this system is highly inefficient, as the majority of the energy required to move the system is absorbed by the power-unit itself and so is unavailable for the system itself.
Case 3 - Ja << Jb
==================
The dynamic characteristics of the system are governed by b. If we consider a to be the power-unit and b the load, then the dynamic characteristics of the system are governed by the load, the power-unit was optimised to handle its own inertia, but is now having to control a system that is way outside its' spec.
My reasoning in case 3 lacks theoretical background, I have an intuitive feel for it but I don't have a degree in mechanical engineering and I have't really studied in this area. Can anyone fill in the theory here (in words, not maths)?
When you add an encoder, brake, etc you are adding additional load to the system. Why this is considered part of the motor and not the load I do not know. Anyone else?
Debian GNU User
Simon Martin
Project Manager
Isys
mailto: [email protected]
B
#### Bruce Durdle
Guy H. Looney wrote:
> Are we to assume that
>by increasing the mass of the motor shaft we can effectively
>increase the motor's rotational inertia? If this is the case, is there
>a limit to how much we can increase it or a set of guide-lines
>stipulating how the mass must be attached? ....That being said, why
>can't we simply put a mass on the shaft of a motor (outside the
>casing) to increase its inertia.
Adding mass to a rotating shaft will increase its Moment of Inertia, regardless of how or where it is attached. But you also need to look at things like the balance of the rotating parts - just sticking a lump of metal on will not be acceptable.
>Some have said that it's a question of compliance & rigidity.
>Assume the load could be pressed on the shaft & that the bearing
> If the drive/motor can handle the torque, speed, & regen is the
>inertia ratio important? If so, at what point does it matter?
The total inertia connected to a motor is not generally going to affect the steady-state
torque or current of the motor. However, it will significantly affect the dynamic performance.
If inertia is too high, the motor will be slow in accelerating and decelerating, and speed control
may suffer. In addition, with inertia connected via a drive train, the natural frequency of the
load and train as a torsional spring needs to be considered - hit this frequency with eg your stepping rate and you are in BIG trouble.. Just as the resonant frequency of a spring-mass combination is dependent on mass and spring constant, the resonant frequency of a torsional assembly depends on the MI at the end and the torsional rigidity of the shaft.
Bruce.
K
#### Ken Brown
I waited a few days to see the responses roll in on this one. You have touched on a topic that very few people understand and other than "rules
of thumb", very few helpful thoughts have been put forth by anyone in the motion industry. The best thoughts on this I have read come from a
fellow named Curt Wilson with Delta Tau Data Systems. I will let him spell them out in all their mathematical glory in some upcoming Motion
mag article.
or velocity loops when inertia ratios are such that the load inertia is greater than the motor inertia. Issues dealing with inertia relative to
efficiency and power optimization are easily understood and trivial by comparison.
Every newbie to motion control looks at their tuning charts with perfect step response and 30 - 40 Hz system response and says "wow! . . .this is
going to be cake!" . . . the next thing they do is hook the motor up to a high inertia load and all hell breaks loose and they wonder if they
will ever be able to meet the settling time requirements to do the application.
This is not a nice linear problem and other than the aforementioned author, I have not read one shred of logical evidence for why the inertia ratio should matter one hoot. Lots of words, but no real "stability criteria" based on some mathematical model related to inertia.
Then again, there are the "rules of thumb" that are observable. These "rules" have been changing with time, the better the controllers and
drivetrain components get, the higher the acceptable ratio of load to motor inertia.
The best theory that I have seen published deals with compliance of the drive train. Specifically, the compliance between the part of the motor that you are actually applying controlled torque to "the rotor" AND the part of the system that is reading position "the position feedback device" AND the greatest consumer of this torque . . . in this situation, typically "the load".
One common solution to dealing with high load/motor inertia ratios is to put a gearbox between the motor and load. Immediately you have inserted a spring with backlash between the motor and the load. Unless you plan on always running the motor in one direction with an appreciable
friction load, you will pay dearly for this additional mechanical slop. If your feedback device is on the motor, you could arguably say that things are better . . . more stable, however you have not a clue where the load is when accelerating or coming to a stop, the backlash and
torsional windup in the gearbox makes this a great guessing game. Every time you reverse torque, the load will continue merrily on its way to release the torsional wind-up in the gearbox and then remove the backlash and then re-apply the torsional wind-up. And if you have any
overshoot, you have this cycle repeated with each torque reversal. If the feedback is on the load . . . hysteresis and backlash will keep your
gains low and final position hunting will ensue. This is what "deadband" windows and gains are for.
If instead, you are able to directly couple a motor to the load with suitable torque and speed characteristics with only a very stiff
coupling . . . you have reduced the compliance in the system substantially and will fare much better with settling times and tuning issues. Chances are that the motor with these torque speed
characteristics will also be much higher inertia . . . . hmmmm, where does the credit usually go? This is one reason why AC Vector technology
has so much to offer in motion control applications. Ever compared the shaft diameter of an AC motor rated at 60 lb-inches (typical 5HP motor) and a 60 lb-inch servo motor? You will be surprised, especially when you consider TENV AC motors.
Also . . . putting the "extra" inertia directly on the motor shaft in effect increases the motor inertia . . . nothing to do with the part of
the motor that is exposed to the magnetic field, it just needs to be a very tight coupling of inertia between rotor and inertia hub. This is
why you can order motors with inertia hubs installed in front of and behind the magnet stacks on the rotor. The issue is compliance and
torsional stiffness.
One example of how knowledge of this theory has been applied with a relatively high degree of success with a large inertia miss-match is a
contra-torque system designed specifically for high inertia CNC machines. Using a pair of servo motors driving against a common rack gear, the motors and subsequent drivetrains are preloaded against each other with a balanced torque offset. The torque command from the servo PID loop is split to two DACs and offsets applied at this point, all positive commands go to the Forward DAC /CW drive and motor and all
negative commands go to the Reverse DAC /CCW drive and motor. About 30 - 40% of the system torque is used up in the "contra-torque" counter
balance. At zero speed this chews up a bit of current, but the drives are common bussed, so when running at speed, the regen current from the
holdback drive is fed right back into the motoring drive. What is the result??? Torsional wind-up and drivetrain backlash are effectively
eliminated. A 70 ton bridge is positioned accurately to within .0005 inches quickly and accurately. 1/4 G accels can be had with less than .005" of following error. All with an inertia miss-match of greater than 30:1. In this case, the feedback device was mounted directly to
the load in the form of a .5um laser interferometer.
My money is on the theories that deal with drivetrain compliance between the torque producer inertia and torque consumer inertia within the
positioning system. This along with the placement of the feedback device(s) 'position and velocity' - I think - is where the bulk of the
progress will be made in providing quantifiable answers to the question of inertia matching.
Modeling of all these components is ugly at best. Perhaps the final answer will be uniform torsional stiffness, hysteresis and natural
frequency numbers for all commercially available drivetrain components that can be entered directly into the servo loop's autotune algorithm .
. . . I don't see it happening any time soon.
H. Kenneth Brown
Applied Motion Systems, Inc.
http://www.kinemation.com
A
#### andy eccles
The real issue here and also the reason that the feedback device (encoder, resolver,etc) is considered part of the motor instead of considered as part of the load is mechanical resonance. Even in the simplest mechanical configuration there is always going to be a mechanical resonance between the motor and the load. The finite stiffness of the coupling between the motor/feedback device and the load is the spring for this resonance. The stiffer that you make the coupling, the higher the frequency of the resonance. The higher that the inertia ratio is the lower the frequency of the resonance.
When the frequency of the mechanical resonance gets down to a frequency where it can cause the velocity (typically) loop closed loop gain to
increase such that the loop goes unstable you run into trouble.
Andy Eccles
#### PhilCorso
In response to Simon Martin's reply no the subject:
I believe that your perception of the role played by inertia in a dynamic analysis is in error.
Consider the case of a motor accelerating a load (including auxiliaries) from an initial speed, N(i), to a final speed, N(f). Then the time, t(a), it takes to accomplish this is expressed by the acceleration-time curve:
[1] t(a) = K x [integral dN/T(e)] as speed, N, varies from N(i) to N(f).
T(e) is the net torque available to accelerate the total load, i.e., the torque developed by the motor minus the torque required by the load
elements:
[3] T(e) = T(m)-T(l)
K is the inertia constant. It is equal to:
[4] K = 2(Pi)J/60g, where J = moment of inertia and g = acceleration constant.
Note that K is a constant. The only variable is the net torque available to accelerate the load, T(e). If speed dependent then the motor and load
speed-torque curves are used. If T(e) is a constant then the time to accelerate is:
[5] ta = [J/308]x[N(f)-N(i)]/[T(e)]
In conclusion acceleration time is the product of a constant (inertia dependant) and a variable (net torque). The individual moments of inertia of the rotating elements (motor, load, encoder, brake, coupling, etc, reflected to motor shaft speed) are additive.
The inertia ratio is a guideline that says the motor's design torque is "probably" capable of driving an assembly of rotating elements whose
equivalent moment of inertia is "X" times that of the motor's.
Regards,
Phil Corso,PE
Trip-A-Larm Corp
#### PhilCorso
Responding to Ken Brown's reply on the subject, I pose the following:
I do not profess to be expert on servo's. Also, the electrical rotational problems I've experienced were generally void of feedback
concerns and at least two orders of magnitude in size greater than those likely to encounter factors like compliance, rigidity, stability, etc.
But, isn't "compliance" the reciprocal to "stiffness?" Then, if stiffness is defined as a system's ability to resist deviations resulting from loading "forces", how is compliance related to inertia?
Regards,
Phil Corso
Trip-A-Larm Corp
D
#### David Kane
For what its worth, I thought I'd try to explain things in different way.
Inertia is the way that power is reflected or absorbed into the load, under ideal conditions.
At a 1:1 match everything is perfect.
A very high inertia load looks like a mechanical short circuit. The "servo system" is attempting to control this mechanical short circuit, but may not do well.
A low inertia load looks like an open circuit. The servo system has no problem with the load, but the truth is this. The power transfer factor is poor.
Servo motors are typically designed to keep the inertia as low as possible, while producing maximum torque for a given motor design. The combination of inertia and torque (and a few other things thrown in) provide bandwidth. The real target, is high bandwidth. It is most common to use inertia matching as the method to get high bandwidth into the servo designed system.
If a brake added to a servo motor IS included as motor inertia, the motor & drive system would have a lower bandwidth than the same motor without the brake. SO put the brake with the load (not the motor) if you want the system to have the bandwidth of a non-braked motor.
In both electronic and mechanical, maximum power transfer occurs when the power producing inertia or impedence matches the load inertia or impedence.
This is true of audio systems, antennas etc. as well as mechanical
Incidently, in hydraulic systems inertia matching is not typically used. Instead this is calculated as the "Natural Frequency of the Cylinder / Mass
System". The end requirement is the same.
David Kane,
Kane Engineering Group Inc.
AIME
CMCS Certified Motion Control Specialist
[email protected]
B
#### Bill Sturm
> The question is why does inertia ratio matter? If the motor/drive
> combination can handle the torque, speed, & regen shouldn't that
> be enough? I have an opinion on this topic but am curious as to
> what others think.
There is an article on this web page called: Load to inertia mismatch: unveiling the truth.
I have always wondered about inertia ratio also. I had thought that as long as you have enough torque, then you can disregard inertia. I assumed that a motor with zero inertia would be ideal. I also could not understand why anyone would add mass to a motor to increase the inertia. It seems like a waste of energy. I had always looked more carefully at total inertia
The reason that the motor to load inertia ratio is important is that a a typical motion system is not made up of one solidly coupled mass, but 2 or
more loosely coupled masses. You have motor inertia and load inertia, and they may not always be in phase with each other. Or due to backlash, there is a switch from motor inertia only to combined motor and load inertia. In other words, the load inertia can vary. With a close inertia match, the varying load inertia is not as large of a change as it would be if the load inertia dominated the system.
With a tight, stiff system, such as a direct coupled ballscrew, you can get away with higher load inertia ratios. If you are using a belt drive or a loose gear train, you would need to have a closer match. Another conclusion that can be drawn from this is that a pulley mounted rigidly to the motor shaft could generally be counted as motor inertia, and would improve the
motor to load inertia ratio, instead of making it worse. How many times have you spec'd a small motor pulley to keep the inertia ratio within range? This may not be necessary, if you have enough torque to accelerate the total inertia at the desired rate.
To summarize: For performance, look at total inertia and total torque. For stability and ease of tuning, you must also look the motor to load inertia ratio and drivetrain compliance.
Bill Sturm
Livonia, Michigan
K
#### Ken Brown
Phil,
With regard to my comments and relative to the original post, I addressed the situation where feedback devices are used specifically in servo applications for the purpose of Motion Control. By definition, if no feedback device is used, referring to the system as a servo is a
misnomer.
My personal observation with regard to stiffness and compliance is that it is indeed a significant contributor to instability in servo systems of all sizes. This includes servo systems with ratings exceeding 300 hp used in chassis dynamometer applications. With the high level of
instrumentation and drive control involved in these systems, it is possible to quantify the contribution of torsional deflections in each
part of the drivetrain and correlate this compliance "or lack of stiffness" to factors that contribute to position or velocity loop
instability.
Regarding the concept of having a system with two orders of magnitude greater stiffness than those likely to encounter factors like compliance. Nothing is infinitely stiff and when dealing with
structures and practical design / component specifications some compliance will always be present when dealing with a high performance
positioning system. I think that a concept of stiffness or compliance is typically a relative concept, what one persons idea of stiff might be
another's idea of a rubber band, especially when dealing with high bandwidth positioning requirements.
Best Regards,
Ken Brown
Applied Motion Systems, Inc.
http://www.kinemation.com
S
#### Simon Martin
Hi Phil,
Your analysis is correct for a given mechanical system. The idea though is to compare DIFERRENT mechanical systems, where the load is changed, so that we can see the effects of different ratios. Effectively we are analysing a class of problems where the rotational acceleration (alpha) is equal to the torque (tau) times the moment of inertia (J). J is the composite moment of
inertia of the system, comprised by Ja (motor) and Jb (load). In all cases we get:
tau = alpha * J
The idea is to compare when the cases where Ja = Jb; Ja << Jb; Ja >> Jb, i.e. 3 different mechanical systems
Debian GNU User
Simon Martin
Project Manager
Isys
mailto: [email protected]
K
#### Kaufman, George
"Compliance" is the same as "stiffness" in this discussion (not the reciprocal). The motor inertia is connected to the load inertia with a
"spring". The spring is defined with a stiffness coefficient (torque/radian). Compliant and stiff are just relative terms (such as a less stiff spring is compliant). A true measure of the spring is the stiffness coefficient.
Best regards,
George Kaufman
Automation Intelligence
www.MotionOnline.com
#### PhilCorso
Ken,
I'm not questioning your expertise. I am aware (see "aside below) of the impact that the factors you list have on performance. As I understand
it, "stiffness" is that mechanical characteristic of a translational motion system to resist an inherent weakness in the strength of an element. For example, the torsional twisting of a shaft in a rotating system, or the bending moment of a push-rod in a linear system, or cable stretching in a cable/pulley system.
I should have worded my question differently. Is the "stiffness" characteristic accounted for as an adjustment to the system's moment of
inertia? Or as an adjustment to the net torque (force) available acting on the system?
An Aside.
An analogous characteristic, perturbation, is responsible for our ability to see about 53% (I think) of the moon's surface.
Regards,
Phil Corso
#### PhilCorso
Re: George Kaufman's 4-May, 5:05 pm response,
I disagree. Technically Speaking (borrowing IEEE Spectrum jargon) "compliance" in the context of this thread means "to yield," while "stiffness" connotates "to resist." Hence, the reciprocal relationship.
Re: Ken Brown's 3-May, 3:36 pm response,
My statement "... size" referred to system, i.e., hundreds or thousands of Hp... not to stiffness.
Look fellas, I'm a great believer, and practitioner, in dimensional analysis. All I want is for someone to explain the following:
If, using English units, the moment of inertia has the dimension, ft-lb sec^2, and the dimension for torque is ft-lb, then what is the dimension
for "stiffness?" Or "compliance?" Or for that matter any deleterious "factor." If dimensionless, then, are they used to adjust one or the other of the dimensioned variables?
Regards,
Phil Corso, PE
Trip-A-Larm Corp
A
#### Andy Eccles
The units of torsional stiffness are units of torque/angular displacement such as:
ft-lb/degree
or
Torsional stiffness (the main culprit that makes "inertia ratio" such a big deal) has nothing to do with inertia.
Andy Eccles
Pacific Scientific
G
#### George Kaufman
Stiffness has units of torque/angle or force/distance. You can use any units as long as you are consistent.
For example, the stiffness of a servo is measured by applying a load torque or force and seeing how much position movement has occurred. You can do
this directly to the motor shaft or the load that is driven by the motor.
Static stiffness is measured with a constant torque or force applied. A typical servo with PI velocity control and P position control has infinite static stiffness.
Dynamic stiffness is measured with a step change in load torque or force and the amplitude of dynamic position error is recorded. A typical servo as above can have a dynamic stiffness equal to 15,000 lb-in/rad (but this figure is very dependent on the tuning of the servo).
Please download the Handbook of AC Servo Systems from www.MotionOnline.com for a more complete review of the above principles.
Best regards,
George Kaufman
Automation Intelligence | 8,266 | 37,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-26 | latest | en | 0.951859 |
http://lbartman.com/worksheet/8th-grade-math-exponents-worksheets.php | 1,579,278,581,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589861.0/warc/CC-MAIN-20200117152059-20200117180059-00135.warc.gz | 98,115,544 | 15,244 | ## lbartman.com - the pro math teacher
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# 8th Grade Math Exponents Worksheets
Public on 07 Oct, 2016 by Cyun Lee
### eighth grade exponents to a power worksheet 05 one page worksheets
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multiplication chart worksheets | 1,590 | 4,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-05 | longest | en | 0.159164 |
https://www.neetprep.com/questions/1843-Physics/687-Thermodynamics?courseId=1277&testId=1339018-Padma-Shri-H-C-Verma-Objective-Exercises-Based-MCQs | 1,716,115,377,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00809.warc.gz | 810,956,452 | 57,865 | The first law of thermodynamics is a statement of:
1 conservation of heat 2 conservation of work 3 conservation of momentum 4 conservation of energy
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If heat is supplied to an ideal gas in an isothermal process,
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Figure shows two processes $$A$$ and $$B$$ on a system. Let $$\Delta Q_1$$ and $$\Delta Q_2$$ be the heat given to the system in processes $$A$$ and $$B$$ respectively. Then:
1. $$\Delta Q_1>\Delta Q_2$$
2. $$\Delta Q_1=\Delta Q_2$$
3. $$\Delta Q_1<\Delta Q_2$$
4. $$\Delta Q_1\leq \Delta Q_2$$
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Refer to the given figure. Let ΔU1 and ΔU2 be the changes in internal energy of the system in processes A and B. Then:
1. ΔU1 > ΔU2
2. ΔU1 = ΔU2
3. ΔU1 < ΔU2
4. ΔU ≠ ΔU2
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Consider the process on a system shown in the figure. During the process, the work done by the system:
1 continuously increases 2 continuously decreases 3 first increases then decreases 4 first decreases then increases
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Consider the following two statements.
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1 Both A and B are correct. 2 A is correct but B is wrong. 3 B is correct but A is wrong. 4 Both A and B are wrong.
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An ideal gas goes from the state $$i$$ to the state $$f$$ as shown in figure given below. The work done by the gas during the process,
1 is positive 2 is negative 3 is zero 4 cannot be obtained from this information
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Consider two processes on a system as shown in figure (26-Q4).The volumes in the initial states are the same in the two processes and the volumes in the final states are also the same. Let ΔW1 and ΔW2 be the work done by the system in the processes A and B respectively.
1. ΔW1 > ΔW2
2. ΔW1 = ΔW
3. ΔW1 < ΔW2
4. Nothing can be said about the relation between ΔW1 and ΔW
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A gas is contained in a metallic cylinder fitted with a piston. The piston is suddenly moved in to compress the gas and is maintained at this position. As time passes the pressure of the gas in the cylinder
1. increases
2. decreases
3. remains constant
4. increases or decreases depending on the nature of the gas
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The pressure $$p$$ and volume $$V$$ of an ideal gas both increase in a process.
a. such a process is not possible. b. the work done by the system is positive. c. the temperature of the system must increase. d. heat supplied to the gas is equal to the change in internal energy.
Choose the correct option:
1. (a), (b)
2. (b), (c)
3. (c), (d)
4. (a), (d)
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NEET 2023 - Target Batch - Aryan Raj Singh | 1,417 | 5,129 | {"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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.82533 |
http://www.ask.com/question/weight-of-gravel-per-cubic-yard | 1,411,395,662,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657137056.60/warc/CC-MAIN-20140914011217-00248-ip-10-234-18-248.ec2.internal.warc.gz | 356,529,300 | 16,327 | # Weight of Gravel Per Cubic Yard?
The weight of gravel per cubic yard is estimated by the specific weight of size based gravel. Generally gravel ranges from 85 to 125 lb. per cubic foot. Then one can calculate the weight of the gravel.
Q&A Related to "Weight of Gravel Per Cubic Yard?"
1. Approximate the specific weight of your gravel based on the particle size. Gravel ranges from 85 to 125 lb. per cubic foot. Gravel with high percentages of small particles will http://www.ehow.com/how_8730973_calculate-per-cubi...
It depends on the size of your gravel (the amount of air trapped between the rocks) If you have small gravel (decomposed granite) it'll weigh more - if you have larger rocks, (river http://wiki.answers.com/Q/What_does_a_cubic_yard_o...
A cubic yard of gravel weighs approximately 2750 lbs. That is over one ton! Did you know that it would take nine loads in a three cubic foot wheelbarrow to carry one cubic yard? http://www.ask.com/web-answers/Business/Constructi...
Crushed Aggregate will weigh on the average 1.4 tons http://www.chacha.com/question/what-is-the-average... | 264 | 1,100 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2014-41 | latest | en | 0.899266 |
https://www.unitconverters.net/power/kilowatt-to-horsepower-550-ft-lbf-s.htm | 1,563,264,673,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524517.31/warc/CC-MAIN-20190716075153-20190716101153-00464.warc.gz | 868,411,636 | 4,065 | Home / Power Conversion / Convert Kilowatt to Horsepower (550 Ft*lbf/s)
Convert Kilowatt to Horsepower (550 Ft*lbf/s)
Please provide values below to convert kilowatt [kW] to horsepower (550 ft*lbf/s), or vice versa.
From: kilowatt To: horsepower (550 ft*lbf/s)
Kilowatt to Horsepower (550 Ft*lbf/s) Conversion Table
Kilowatt [kW]Horsepower (550 Ft*lbf/s)
0.01 kW0.0134102209 horsepower (550 ft*lbf/s)
0.1 kW0.134102209 horsepower (550 ft*lbf/s)
1 kW1.3410220896 horsepower (550 ft*lbf/s)
2 kW2.6820441792 horsepower (550 ft*lbf/s)
3 kW4.0230662688 horsepower (550 ft*lbf/s)
5 kW6.705110448 horsepower (550 ft*lbf/s)
10 kW13.410220896 horsepower (550 ft*lbf/s)
20 kW26.8204417919 horsepower (550 ft*lbf/s)
50 kW67.0511044798 horsepower (550 ft*lbf/s)
100 kW134.1022089595 horsepower (550 ft*lbf/s)
1000 kW1341.022089595 horsepower (550 ft*lbf/s)
How to Convert Kilowatt to Horsepower (550 Ft*lbf/s)
1 kW = 1.3410220896 horsepower (550 ft*lbf/s)
1 horsepower (550 ft*lbf/s) = 0.7456998716 kW
Example: convert 15 kW to horsepower (550 ft*lbf/s):
15 kW = 15 × 1.3410220896 horsepower (550 ft*lbf/s) = 20.1153313439 horsepower (550 ft*lbf/s) | 393 | 1,145 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-30 | longest | en | 0.542248 |
https://www.jiskha.com/display.cgi?id=1267398759 | 1,498,654,750,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323680.18/warc/CC-MAIN-20170628120308-20170628140308-00130.warc.gz | 903,867,170 | 3,573 | # algebra
posted by .
Evaluate and tabulate the values of y for the given values of x. Tabulate your answer in the form of ordered pairs (x, y).
y = 2x, for x = -2, -1, 0, 1, 2
• algebra -
Substitute the values for x and solve for y.
y = 2(-2) = -4
And so on.... | 90 | 269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-26 | longest | en | 0.684058 |
https://fr.maplesoft.com/support/help/maplesim/view.aspx?path=length&L=F | 1,695,291,828,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00779.warc.gz | 310,330,985 | 21,404 | length - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
length
length of an object
Calling Sequence length(expr) length[shortname](expr)
Parameters
expr - expression
Description
• The length(expr) function returns the length of expr.
• If expr is an integer, the number of decimal digits is returned (where the length of zero is defined to be zero and the length of a negative integer is defined to be the same as the length of the absolute value of the integer).
• If expr is a string, the number of characters in expr is returned.
• If expr is a rtable object, the number of words of storage occupied by the rtable is returned (including the data, the attributes, and the data structure itself).
• For other objects, the length of each operand of expr is computed recursively and added to the number of words used to represent expr. In this way, the measure of the size of expr is returned.
• If the shortname literal is specified, then the length of the expression is computed with all symbols (not strings) in the expression being considered to be 1 character in length.
• To find the number of elements of a list, set, or Array, use the numelems command.
• The length command is thread-safe as of Maple 15.
Examples
> $\mathrm{length}\left(0\right)$
${0}$ (1)
> $\mathrm{length}\left(11\right)$
${2}$ (2)
> $\mathrm{length}\left(14.5\right)$
${7}$ (3)
> $\mathrm{length}\left(\mathrm{abc}\right)$
${3}$ (4)
> $\mathrm{length}\left(x+2y\right)$
${9}$ (5)
> $\mathrm{length}\left(\mathrm{x_initial}+2\mathrm{y_initial}\right)$
${25}$ (6)
> ${\mathrm{length}}_{'\mathrm{shortname}'}\left(\mathrm{x_initial}+2\mathrm{y_initial}\right)$
${9}$ (7)
Compatibility
• The length command was updated in Maple 2022.
• The shortname option was introduced in Maple 2022.
• For more information on Maple 2022 changes, see Updates in Maple 2022. | 508 | 1,934 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.689312 |
https://www.aqua-calc.com/what-is/density/microgram-per-us-teaspoon | 1,696,417,479,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511364.23/warc/CC-MAIN-20231004084230-20231004114230-00702.warc.gz | 680,333,461 | 17,884 | # What is a microgram per US teaspoon (unit)
## The microgram per US teaspoon is a unit of measurement of density
The microgram per US teaspoon density measurement unit is used to measure volume in US teaspoons in order to estimate weight or mass in micrograms
• What is densityInstant conversionsConversion tables
• 1 µg/tsp = 0.0002028841364 µg/mm³
• 1 µg/tsp = 0.2028841364 µg/cm³
• 1 µg/tsp = 202.884136 µg/dm³
• 1 µg/tsp = 202 884.136 µg/m³
• 1 µg/tsp = 0.2028841364 µg/ml
• 1 µg/tsp = 202.884136 µg/l
• 1 µg/tsp = 1.01442068 µg/metric tsp
• 1 µg/tsp = 3.04326205 µg/metric tbsp
• 1 µg/tsp = 50.7210341 µg/metric c
• 1 µg/tsp = 3.32467533 µg/in³
• 1 µg/tsp = 5 745.03897 µg/ft³
• 1 µg/tsp = 155 116.052 µg/yd³
• 1 µg/tsp = 3.00000001 µg/tbsp
• 1 µg/tsp = 6.00000001 µg/fl.oz
• 1 µg/tsp = 48.0000001 µg/US c
• 1 µg/tsp = 96.0000001 µg/pt
• 1 µg/tsp = 192 µg/US qt
• 1 µg/tsp = 768 µg/US gal
• 1 µg/tsp = 2.028841364×10-7 mg/mm³
• 1 µg/tsp = 0.0002028841364 mg/cm³
• 1 µg/tsp = 0.2028841364 mg/dm³
• 1 µg/tsp = 202.884136 mg/m³
• 1 µg/tsp = 0.0002028841364 mg/ml
• 1 µg/tsp = 0.2028841364 mg/l
• 1 µg/tsp = 0.001014420682 mg/metric tsp
• 1 µg/tsp = 0.003043262045 mg/metric tbsp
• 1 µg/tsp = 0.05072103409 mg/metric c
• 1 µg/tsp = 0.003324675327 mg/in³
• 1 µg/tsp = 5.74503897 mg/ft³
• 1 µg/tsp = 155.116052 mg/yd³
• 1 µg/tsp = 0.001 mg/tsp
• 1 µg/tsp = 0.003000000006 mg/tbsp
• 1 µg/tsp = 0.006000000012 mg/fl.oz
• 1 µg/tsp = 0.04800000014 mg/US c
• 1 µg/tsp = 0.09600000007 mg/pt
• 1 µg/tsp = 0.1920000001 mg/US qt
• 1 µg/tsp = 0.7679999998 mg/US gal
• 1 µg/tsp = 2.02884136365×10-10 g/mm³
• 1 µg/tsp = 2.028841364×10-7 g/cm³
• 1 µg/tsp = 0.0002028841364 g/dm³
• 1 µg/tsp = 0.2028841364 g/m³
• 1 µg/tsp = 2.028841364×10-7 g/ml
• 1 µg/tsp = 0.0002028841364 g/l
• 1 µg/tsp = 1.014420682×10-6 g/metric tsp
• 1 µg/tsp = 3.043262045×10-6 g/metric tbsp
• 1 µg/tsp = 5.072103409×10-5 g/metric c
• 1 µg/tsp = 3.324675327×10-6 g/in³
• 1 µg/tsp = 0.005745038967 g/ft³
• 1 µg/tsp = 0.1551160521 g/yd³
• 1 µg/tsp = 1.0×10-6 g/tsp
• 1 µg/tsp = 3.000000006×10-6 g/tbsp
• 1 µg/tsp = 6.000000012×10-6 g/fl.oz
• 1 µg/tsp = 4.800000014×10-5 g/US c
• 1 µg/tsp = 9.600000007×10-5 g/pt
• 1 µg/tsp = 0.0001920000001 g/US qt
• 1 µg/tsp = 0.0007679999998 g/US gal
• 1 µg/tsp = 2.02884136365×10-13 kg/mm³
• 1 µg/tsp = 2.02884136365×10-10 kg/cm³
• 1 µg/tsp = 2.028841364×10-7 kg/dm³
• 1 µg/tsp = 0.0002028841364 kg/m³
• 1 µg/tsp = 2.02884136365×10-10 kg/ml
• 1 µg/tsp = 2.028841364×10-7 kg/l
• 1 µg/tsp = 1.014420682×10-9 kg/metric tsp
• 1 µg/tsp = 3.043262045×10-9 kg/metric tbsp
• 1 µg/tsp = 5.072103409×10-8 kg/metric c
• 1 µg/tsp = 3.324675327×10-9 kg/in³
• 1 µg/tsp = 5.745038963×10-6 kg/ft³
• 1 µg/tsp = 0.000155116052 kg/yd³
• 1 µg/tsp = 1.0×10-9 kg/tsp
• 1 µg/tsp = 3.000000006×10-9 kg/tbsp
• 1 µg/tsp = 6.00000003×10-9 kg/fl.oz
• 1 µg/tsp = 4.800000014×10-8 kg/US c
• 1 µg/tsp = 9.600000007×10-8 kg/pt
• 1 µg/tsp = 1.920000004×10-7 kg/US qt
• 1 µg/tsp = 7.680000016×10-7 kg/US gal
• 1 µg/tsp = 2.02884136365×10-16 t/mm³
• 1 µg/tsp = 2.02884136365×10-13 t/cm³
• 1 µg/tsp = 2.02884136365×10-10 t/dm³
• 1 µg/tsp = 2.028841364×10-7 t/m³
• 1 µg/tsp = 2.02884136365×10-13 t/ml
• 1 µg/tsp = 2.02884136365×10-10 t/l
• 1 µg/tsp = 1.01442068183×10-12 t/metric tsp
• 1 µg/tsp = 3.04326204548×10-12 t/metric tbsp
• 1 µg/tsp = 5.07210340914×10-11 t/metric c
• 1 µg/tsp = 3.3246753272×10-12 t/in³
• 1 µg/tsp = 5.745038963×10-9 t/ft³
• 1 µg/tsp = 1.55116052×10-7 t/yd³
• 1 µg/tsp = 1.0×10-12 t/tsp
• 1 µg/tsp = 3.00000000609×10-12 t/tbsp
• 1 µg/tsp = 6.00000003043×10-12 t/fl.oz
• 1 µg/tsp = 4.8000000138×10-11 t/US c
• 1 µg/tsp = 9.6000000073×10-11 t/pt
• 1 µg/tsp = 1.9200000041×10-10 t/US qt
• 1 µg/tsp = 7.68000001599×10-10 t/US gal
• 1 µg/tsp = 7.15652730641×10-12 oz/mm³
• 1 µg/tsp = 7.156527306×10-9 oz/cm³
• 1 µg/tsp = 7.156527306×10-6 oz/dm³
• 1 µg/tsp = 0.007156527296 oz/m³
• 1 µg/tsp = 7.156527306×10-9 oz/ml
• 1 µg/tsp = 7.156527306×10-6 oz/l
• 1 µg/tsp = 3.578263654×10-8 oz/metric tsp
• 1 µg/tsp = 1.073479096×10-7 oz/metric tbsp
• 1 µg/tsp = 1.789131827×10-6 oz/metric c
• 1 µg/tsp = 1.17274471×10-7 oz/in³
• 1 µg/tsp = 0.0002026502859 oz/ft³
• 1 µg/tsp = 0.005471557721 oz/yd³
• 1 µg/tsp = 3.527396182×10-8 oz/tsp
• 1 µg/tsp = 1.058218858×10-7 oz/tbsp
• 1 µg/tsp = 2.116437722×10-7 oz/fl.oz
• 1 µg/tsp = 1.693150174×10-6 oz/US c
• 1 µg/tsp = 3.386300361×10-6 oz/pt
• 1 µg/tsp = 6.7726007×10-6 oz/US qt
• 1 µg/tsp = 2.709040277×10-5 oz/US gal
• 1 µg/tsp = 4.47282956676×10-13 lb/mm³
• 1 µg/tsp = 4.47282956676×10-10 lb/cm³
• 1 µg/tsp = 4.472829567×10-7 lb/dm³
• 1 µg/tsp = 0.0004472829563 lb/m³
• 1 µg/tsp = 4.47282956676×10-10 lb/ml
• 1 µg/tsp = 4.472829567×10-7 lb/l
• 1 µg/tsp = 2.236414783×10-9 lb/metric tsp
• 1 µg/tsp = 6.70924435×10-9 lb/metric tbsp
• 1 µg/tsp = 1.118207392×10-7 lb/metric c
• 1 µg/tsp = 7.329654437×10-9 lb/in³
• 1 µg/tsp = 1.266564287×10-5 lb/ft³
• 1 µg/tsp = 0.0003419723575 lb/yd³
• 1 µg/tsp = 2.204622614×10-9 lb/tsp
• 1 µg/tsp = 6.613867862×10-9 lb/tbsp
• 1 µg/tsp = 1.322773576×10-8 lb/fl.oz
• 1 µg/tsp = 1.058218859×10-7 lb/US c
• 1 µg/tsp = 2.116437726×10-7 lb/pt
• 1 µg/tsp = 4.232875437×10-7 lb/US qt
• 1 µg/tsp = 1.693150173×10-6 lb/US gal
• 1 µg/tsp = 3.130980697×10-9 gr/mm³
• 1 µg/tsp = 3.130980697×10-6 gr/cm³
• 1 µg/tsp = 0.003130980697 gr/dm³
• 1 µg/tsp = 3.13098069 gr/m³
• 1 µg/tsp = 3.130980697×10-6 gr/ml
• 1 µg/tsp = 0.003130980697 gr/l
• 1 µg/tsp = 1.565490349×10-5 gr/metric tsp
• 1 µg/tsp = 4.696471044×10-5 gr/metric tbsp
• 1 µg/tsp = 0.0007827451741 gr/metric c
• 1 µg/tsp = 5.130758105×10-5 gr/in³
• 1 µg/tsp = 0.08865950014 gr/ft³
• 1 µg/tsp = 2.3938065 gr/yd³
• 1 µg/tsp = 1.543235835×10-5 gr/US tsp
• 1 µg/tsp = 4.629707515×10-5 gr/US tbsp
• 1 µg/tsp = 9.25941503×10-5 gr/fl.oz
• 1 µg/tsp = 0.0007407532031 gr/US c
• 1 µg/tsp = 0.001481506403 gr/pt
• 1 µg/tsp = 0.002963012805 gr/US qt
• 1 µg/tsp = 0.01185205121 gr/US gal
• 1 µg/tsp = 1.39019792585×10-14 sl/mm³
• 1 µg/tsp = 1.39019792585×10-11 sl/cm³
• 1 µg/tsp = 1.390197926×10-8 sl/dm³
• 1 µg/tsp = 1.390197932×10-5 sl/m³
• 1 µg/tsp = 1.39019792585×10-11 sl/ml
• 1 µg/tsp = 1.390197926×10-8 sl/l
• 1 µg/tsp = 6.95098962915×10-11 sl/metric tsp
• 1 µg/tsp = 2.08529688946×10-10 sl/metric tbsp
• 1 µg/tsp = 3.475494815×10-9 sl/metric c
• 1 µg/tsp = 2.27812623816×10-10 sl/in³
• 1 µg/tsp = 3.936602144×10-7 sl/ft³
• 1 µg/tsp = 1.062882579×10-5 sl/yd³
• 1 µg/tsp = 6.85217658332×10-11 sl/tsp
• 1 µg/tsp = 2.05565296891×10-10 sl/tbsp
• 1 µg/tsp = 4.11130597028×10-10 sl/fl.oz
• 1 µg/tsp = 3.289044754×10-9 sl/US c
• 1 µg/tsp = 6.578089529×10-9 sl/pt
• 1 µg/tsp = 1.315617909×10-8 sl/US qt
• 1 µg/tsp = 5.262471629×10-8 sl/US gal
• 1 µg/tsp = 2.23641478338×10-16 short tn/mm³
• 1 µg/tsp = 2.23641478338×10-13 short tn/cm³
• 1 µg/tsp = 2.2364147835×10-10 short tn/dm³
• 1 µg/tsp = 2.236414782×10-7 short tn/m³
• 1 µg/tsp = 2.23641478338×10-13 short tn/ml
• 1 µg/tsp = 2.2364147835×10-10 short tn/l
• 1 µg/tsp = 1.1182073915×10-12 short tn/metric tsp
• 1 µg/tsp = 3.354622175×10-12 short tn/metric tbsp
• 1 µg/tsp = 5.59103696×10-11 short tn/metric c
• 1 µg/tsp = 3.6648272185×10-12 short tn/in³
• 1 µg/tsp = 6.332821435×10-9 short tn/ft³
• 1 µg/tsp = 1.709861788×10-7 short tn/yd³
• 1 µg/tsp = 1.102311307×10-12 short tn/US tsp
• 1 µg/tsp = 3.306933931×10-12 short tn/US tbsp
• 1 µg/tsp = 6.61386788×10-12 short tn/fl.oz
• 1 µg/tsp = 5.291094295×10-11 short tn/US c
• 1 µg/tsp = 1.058218863×10-10 short tn/pt
• 1 µg/tsp = 2.1164377185×10-10 short tn/US qt
• 1 µg/tsp = 8.465750865×10-10 short tn/US gal
• 1 µg/tsp = 1.99679891373×10-16 long tn/mm³
• 1 µg/tsp = 1.99679891373×10-13 long tn/cm³
• 1 µg/tsp = 1.99679891384×10-10 long tn/dm³
• 1 µg/tsp = 1.996798912×10-7 long tn/m³
• 1 µg/tsp = 1.99679891373×10-13 long tn/ml
• 1 µg/tsp = 1.99679891384×10-10 long tn/l
• 1 µg/tsp = 9.98399456696×10-13 long tn/metric tsp
• 1 µg/tsp = 2.99519837054×10-12 long tn/metric tbsp
• 1 µg/tsp = 4.99199728571×10-11 long tn/metric c
• 1 µg/tsp = 3.27216715937×10-12 long tn/in³
• 1 µg/tsp = 5.654304853×10-9 long tn/ft³
• 1 µg/tsp = 1.52666231×10-7 long tn/yd³
• 1 µg/tsp = 9.84206524107×10-13 long tn/US tsp
• 1 µg/tsp = 2.95261958125×10-12 long tn/US tbsp
• 1 µg/tsp = 5.90523917857×10-12 long tn/fl.oz
• 1 µg/tsp = 4.72419133482×10-11 long tn/US c
• 1 µg/tsp = 9.44838270536×10-11 long tn/pt
• 1 µg/tsp = 1.88967653437×10-10 long tn/US qt
• 1 µg/tsp = 7.55870612946×10-10 long tn/US gal
• 1 µg/tsp = 3.19487826197×10-14 st/mm³
• 1 µg/tsp = 3.19487826197×10-11 st/cm³
• 1 µg/tsp = 3.194878262×10-8 st/dm³
• 1 µg/tsp = 3.194878259×10-5 st/m³
• 1 µg/tsp = 3.19487826197×10-11 st/ml
• 1 µg/tsp = 3.194878262×10-8 st/l
• 1 µg/tsp = 1.59743913071×10-10 st/metric tsp
• 1 µg/tsp = 4.79231739286×10-10 st/metric tbsp
• 1 µg/tsp = 7.987195657×10-9 st/metric c
• 1 µg/tsp = 5.235467455×10-10 st/in³
• 1 µg/tsp = 9.046887764×10-7 st/ft³
• 1 µg/tsp = 2.442659696×10-5 st/yd³
• 1 µg/tsp = 1.57473043857×10-10 st/US tsp
• 1 µg/tsp = 4.72419133×10-10 st/US tbsp
• 1 µg/tsp = 9.44838268571×10-10 st/fl.oz
• 1 µg/tsp = 7.558706136×10-9 st/US c
• 1 µg/tsp = 1.511741233×10-8 st/pt
• 1 µg/tsp = 3.023482455×10-8 st/US qt
• 1 µg/tsp = 1.209392981×10-7 st/US gal
• 1 µg/tsp = 6.52287645208×10-12 oz t/mm³
• 1 µg/tsp = 6.522876452×10-9 oz t/cm³
• 1 µg/tsp = 6.522876452×10-6 oz t/dm³
• 1 µg/tsp = 0.006522876438 oz t/m³
• 1 µg/tsp = 6.522876452×10-9 oz t/ml
• 1 µg/tsp = 6.522876452×10-6 oz t/l
• 1 µg/tsp = 3.261438227×10-8 oz t/metric tsp
• 1 µg/tsp = 9.784314675×10-8 oz t/metric tbsp
• 1 µg/tsp = 1.630719113×10-6 oz t/metric c
• 1 µg/tsp = 1.068907939×10-7 oz t/in³
• 1 µg/tsp = 0.000184707292 oz t/ft³
• 1 µg/tsp = 0.004987096875 oz t/yd³
• 1 µg/tsp = 3.215074656×10-8 oz t/US tsp
• 1 µg/tsp = 9.64522399×10-8 oz t/US tbsp
• 1 µg/tsp = 1.929044798×10-7 oz t/fl.oz
• 1 µg/tsp = 1.54323584×10-6 oz t/US c
• 1 µg/tsp = 3.086471673×10-6 oz t/pt
• 1 µg/tsp = 6.172943344×10-6 oz t/US qt
• 1 µg/tsp = 2.469177335×10-5 oz t/US gal
• 1 µg/tsp = 5.43573037674×10-13 troy/mm³
• 1 µg/tsp = 5.43573037674×10-10 troy/cm³
• 1 µg/tsp = 5.435730377×10-7 troy/dm³
• 1 µg/tsp = 0.0005435730365 troy/m³
• 1 µg/tsp = 5.43573037674×10-10 troy/ml
• 1 µg/tsp = 5.435730377×10-7 troy/l
• 1 µg/tsp = 2.717865189×10-9 troy/metric tsp
• 1 µg/tsp = 8.153595563×10-9 troy/metric tbsp
• 1 µg/tsp = 1.358932594×10-7 troy/metric c
• 1 µg/tsp = 8.907566155×10-9 troy/in³
• 1 µg/tsp = 1.539227433×10-5 troy/ft³
• 1 µg/tsp = 0.0004155914063 troy/yd³
• 1 µg/tsp = 2.67922888×10-9 troy/US tsp
• 1 µg/tsp = 8.037686658×10-9 troy/US tbsp
• 1 µg/tsp = 1.607537332×10-8 troy/fl.oz
• 1 µg/tsp = 1.286029866×10-7 troy/US c
• 1 µg/tsp = 2.572059727×10-7 troy/pt
• 1 µg/tsp = 5.144119453×10-7 troy/US qt
• 1 µg/tsp = 2.05764778×10-6 troy/US gal
• 1 µg/tsp = 1.30457529042×10-10 dwt/mm³
• 1 µg/tsp = 1.30457529×10-7 dwt/cm³
• 1 µg/tsp = 0.000130457529 dwt/dm³
• 1 µg/tsp = 0.1304575288 dwt/m³
• 1 µg/tsp = 1.30457529×10-7 dwt/ml
• 1 µg/tsp = 0.000130457529 dwt/l
• 1 µg/tsp = 6.522876454×10-7 dwt/metric tsp
• 1 µg/tsp = 1.956862935×10-6 dwt/metric tbsp
• 1 µg/tsp = 3.261438225×10-5 dwt/metric c
• 1 µg/tsp = 2.137815877×10-6 dwt/in³
• 1 µg/tsp = 0.003694145839 dwt/ft³
• 1 µg/tsp = 0.0997419375 dwt/yd³
• 1 µg/tsp = 6.430149313×10-7 dwt/US tsp
• 1 µg/tsp = 1.929044798×10-6 dwt/US tbsp
• 1 µg/tsp = 3.858089596×10-6 dwt/fl.oz
• 1 µg/tsp = 3.08647168×10-5 dwt/US c
• 1 µg/tsp = 6.172943346×10-5 dwt/pt
• 1 µg/tsp = 0.0001234588669 dwt/US qt
• 1 µg/tsp = 0.0004938354671 dwt/US gal
#### Foods, Nutrients and Calories
TWINKLE SILVER SUGAR CRYSTALS, UPC: 011225347252 weigh(s) 203 grams per metric cup or 6.8 ounces per US cup, and contain(s) 375 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
837 foods that contain Tocopherol, beta. List of these foods starting with the highest contents of Tocopherol, beta and the lowest contents of Tocopherol, beta
#### Gravels, Substances and Oils
CaribSea, Freshwater, Eco-Complete Planted, Red weighs 865 kg/m³ (54.00019 lb/ft³) with specific gravity of 0.865 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Coal, anthracite broken, white-ash, pea-size weighs 856.988 kg/m³ (53.50001 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-410A, liquid (R410A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
Rem (Roentgen Equivalent Man) is the derived unit of any of the quantities expressed as the radiation absorbed dose (rad) equivalent [ millirem fission fragments ]
A force that acts upon an object can cause the acceleration of the object.
lbf/mm² to lbf/a conversion table, lbf/mm² to lbf/a unit converter or convert between all units of pressure measurement.
#### Calculators
Annulus and sector of an annulus calculator | 6,791 | 12,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-40 | latest | en | 0.243201 |
https://socratic.org/questions/joint-probability-mean-and-variance-pl-refer-image-attached | 1,526,809,967,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863277.18/warc/CC-MAIN-20180520092830-20180520112830-00200.warc.gz | 660,072,702 | 86,755 | # Joint probability- mean and variance? Pl refer image attached
Then teach the underlying concepts
Don't copy without citing sources
preview
?
#### Explanation
Explain in detail...
#### Explanation:
I want someone to double check my answer
4
### This answer has been featured!
Featured answers represent the very best answers the Socratic community can create.
Dec 26, 2017
The answer is $\text{(d) } \frac{3}{4} {x}_{2} \mathmr{and} \frac{3}{80} {x}_{2}^{2}$.
#### Explanation:
Part 1: Conditional Mean
By definition, $\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{A} {x}_{1} \cdot \textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} {\mathrm{dx}}_{1}$,
where
$\textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{f \left({x}_{1} , {x}_{2}\right)}{\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)}}$,
where
$\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)} = {\int}_{A} f \left({x}_{1} , {x}_{2}\right) {\mathrm{dx}}_{1}$.
In this case, since ${x}_{1}$ is bounded by $0 < {x}_{1} < {x}_{2}$, our integration interval is $A = \left(0 , {x}_{2}\right) .$ Thus:
$\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)} = {\int}_{0}^{{x}_{2}} 21 \text{ "x_1^2" "x_2^3" } {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 21 \text{ "x_2^3int_0^(x_2)x_1^2" } {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 21 \text{ } {x}_{2}^{3} {\left[\frac{1}{3} {x}_{1}^{3}\right]}_{0}^{{x}_{2}}$
$\textcolor{w h i t e}{{f}_{{X}_{2}} \left({x}_{2}\right)} = 7 \text{ } {x}_{2}^{6}$
Thus,
$\textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{f \left({x}_{1} , {x}_{2}\right)}{\textcolor{red}{{f}_{{X}_{2}} \left({x}_{2}\right)}}$
$\textcolor{w h i t e}{f \left({x}_{1} | {x}_{2}\right)} = \left(21 \text{ "x_1^2" "x_2^3)/(7" } {x}_{2}^{6}\right)$
$\textcolor{w h i t e}{f \left({x}_{1} | {x}_{2}\right)} = \frac{3 \text{ } {x}_{1}^{2}}{{x}_{2}^{3}}$.
Finally,
$\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{A} {x}_{1} \cdot \textcolor{b l u e}{f \left({x}_{1} | {x}_{2}\right)} {\mathrm{dx}}_{1}$
$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = {\int}_{A} {x}_{1} \cdot \frac{3 {x}_{1}^{2}}{x} _ {2}^{3} {\mathrm{dx}}_{1}$
Since we are once again integrating with respect to ${x}_{1}$, the integration interval is the same as before:
$\text{E} \left({X}_{1} | {X}_{2}\right) = {\int}_{0}^{{x}_{2}} {x}_{1} \cdot \frac{3 {x}_{1}^{2}}{x} _ {2}^{3} {\mathrm{dx}}_{1}$
color(white)("E"(X_1|X_2))=3/x_2^3 int_0^(x_2)x_1^3" "dx_1
$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{x} _ {2}^{3} {\left[{x}_{1}^{4} / 4\right]}_{0}^{{x}_{2}}$
$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{4 {x}_{2}^{3}} \left[{x}_{2}^{4}\right]$
$\textcolor{w h i t e}{\text{E} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{4} {x}_{2}$
Part 2: Conditional Variance
The conditional variance is
"Var"(X_1|X_2)="E"(X_1^2|X_2)-["E"(X_1|X_2)]^2
I'll leave the calculation of $\text{E} \left({X}_{1}^{2} | {X}_{2}\right)$ as an exercise. (Hint: just replace ${x}_{1}$ with ${x}_{1}^{2}$ in the formula for $\text{E} \left({X}_{1} | {X}_{2}\right)$.)
The result is:
$\text{Var} \left({X}_{1} | {X}_{2}\right) = \frac{3 {x}_{2}^{2}}{5} - {\left[\frac{3 {x}_{2}}{4}\right]}^{2}$
$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = \frac{3 {x}_{2}^{2}}{5} - \frac{9 {x}_{2}^{2}}{16}$
$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = 3 {x}_{2}^{2} \left[\frac{1}{5} - \frac{3}{16}\right]$
$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = 3 {x}_{2}^{2} \left[\frac{16 - 15}{80}\right]$
$\textcolor{w h i t e}{\text{Var} \left({X}_{1} | {X}_{2}\right)} = \frac{3}{80} {x}_{2}^{2}$.
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• 26 minutes ago | 1,754 | 4,015 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 32, "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} | 4.21875 | 4 | CC-MAIN-2018-22 | longest | en | 0.532758 |
https://programmatic.solutions/qjvd49/consequences-of-lower-bounds-for-epsilon-nets-on-approximation | 1,669,948,351,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00028.warc.gz | 493,587,089 | 16,076 | Theoretical Computer Science
cg.comp-geom set-cover hypergraphs approximation-hardness epsilon-nets
Updated Sat, 21 May 2022 18:22:19 GMT
# Consequences of lower bounds for $\epsilon$-nets on approximation
Many here are probably aware of Alon's recent super-linear lower bounds for $\epsilon$-nets in a natural geometric setting [PDF]. I would like to know what, if anything, such a lower bound implies about the approximability of the associated Set Cover/Hitting Set problems.
To be slightly more specific, consider a family of range spaces, for example, the family:
$\big\{(X,\mathcal{R})$ : $X$ is a finite planar point set, $\mathcal{R}$ contains all intersections of $X$ with lines$\big\}$
If, for some function $f$ that is linear or super-linear, the family contains a range space that does not admit $\epsilon$-nets of size $f(1/\epsilon)$, what, if anything, does this imply about the Minimum Hitting Set problem restricted to this family of range spaces?
## Solution
If a range space has $\epsilon$-net of size $f(1/\epsilon)$, then the integrality gap of the fractional hitting set (or set cover) is $f(1/\epsilon)/(1/\epsilon)$. See the work by Philip Long (here [The Even etal. work is later than this work, and rediscover some of his stuff]). See also the slides 13-16 here.
In short, having non-linear $\epsilon$-nets, indicates that approximating the relevant hitting-set/set cover problem within better than a constant factor is going to be very challenging.
• +0 – Which section of the first paper is relevant to this particular problem? Or equivalently, in the second link, you say "In geometric settings, there is an $\epsilon$-net of size $O(K/\epsilon)$ iff the integrality gap is $K$." I am having trouble understanding that. — Oct 23, 2015 at 19:22 | 442 | 1,781 | {"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} | 3.078125 | 3 | CC-MAIN-2022-49 | latest | en | 0.90175 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-7-exponents-and-exponential-functions-7-3-multipying-powers-with-the-same-base-practice-and-problem-solving-exercises-page-429/20 | 1,537,657,865,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158766.53/warc/CC-MAIN-20180922221246-20180923001646-00438.warc.gz | 742,258,369 | 14,254 | ## Algebra 1
Published by Prentice Hall
# Chapter 7 - Exponents and Exponential Functions - 7-3 Multipying Powers with the Same Base - Practice and Problem-Solving Exercises - Page 429: 20
#### Answer
$b^3$
#### Work Step by Step
We start with the given expression: $b^{-2}\times b^4\times b$ To multiply powers with the same base, we add the exponents: $b^{-2+4+1}=b^3$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 144 | 537 | {"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} | 3.625 | 4 | CC-MAIN-2018-39 | longest | en | 0.780647 |
https://codegolf.stackexchange.com/questions/51402/each-step-of-the-levenshtein-distance | 1,660,757,389,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573029.81/warc/CC-MAIN-20220817153027-20220817183027-00535.warc.gz | 182,400,689 | 69,585 | Each step of the Levenshtein distance
In this challenge you will write a program that takes two newline-separated strings, s1 (the first line) and s2 (the second line), as input (STDIN or closest). You can assume that the length of s1 will always be smaller than 30 and bigger than the length of s2. The program should then output each step in the levenshtein distance from s1 to s2.
To clarify what each step in the levenshtein distance means, the program will print n strings, where n is the levenshtein distance between s1 and s2, and the levenshtein distance between two adjacent strings will always be one. The order doesn't matter. The output should be newline-separated and not include s1, only the in-betweens and s2. The program should also run in under one minute on a modern computer.
Examples:
Input:
Programming
Codegolf
Output:
rogramming
Cogramming
Coramming
Coamming
Codmming
Codeming
Codeging
Codegong
Codegolg
Codegolf
Input:
Questions
Output:
uestions
Aestions
Anstions
Ansions
Answons
Answens
Input:
Offline
Online
Output:
Ofline
Online
Input:
Saturday
Sunday
Output:
Sturday
Surday
Sunday
Here is a link to a python script that prints out the distance and the steps.
This is code-golf so keep you code short; shortest code wins!
• For my edit, I rather presumed that the input would be of the form s1(newline)s2, however, having looked over the question again, I am wondering if instead you intended for the program to select s1 and s2 based on the length of 2 inputed strings, coming in either order, would you mind clarifying this point? That is, do we assume the input is s1 followed by s2, or do we select s1 and s2 based on the length of the two inputs? Jun 7, 2015 at 17:25
• Does an answer have to run in a reasonable amount of time?
– KSab
Jun 7, 2015 at 18:30
• Camper - Ampere, distance 2, the python script runs forever ... Jun 9, 2015 at 13:47
• How strict is "take input from STDIN or closest"? Can I write a function that takes the input via function argument? The currently accepted answer does so.
– nimi
Mar 5, 2016 at 23:57
Haskell, 201 194 bytes
l=length
g[]n u=map(\_->"")n
g(b:c)[]u=(u++c):g c[]u
g(b:c)n@(o:p)u|b==o=g c p(u++[o])|1<2=((u++o:c):g c p(u++[o]))!((u++c):g c n u)
a!b|l a<l b=a|1<2=b
p[a,n]=g a n""
f=interact\$unlines.p.lines
Longer than expected. Maybe I can golf it down a little bit ...
Usage example:
*Main> f -- call via f
Questions -- User input
Answers -- no newline after second line!
uestions -- Output starts here
Aestions
Anstions
Ansions
Answons
Answens
It's a brute force that decides between changing and deleting if the initial characters differ.
• How long does it take to run? Jun 9, 2015 at 5:12
• How can I test (maybe ideone)? Jun 9, 2015 at 13:49
• @Loovjo: shorter strings like your examples are calculated instantly, worst case is about 1:30min. I've interpreted the "should" in "should run in under one minute" not as a strict limit (should vs. must). If this is a must, I can add a "performance pack" for about 20 bytes.
– nimi
Jun 9, 2015 at 15:09
• @edc65: yes, ideone, but it expects the function to be executed to be called "main". Try: ideone.com/CUgU8W
– nimi
Jun 9, 2015 at 15:21
APL (Dyalog 18.0), 66 bytes
{⍺<Ö≢⍵:⍺∇⎕←0~⍨0@(⊃(⍸~p)~⍳⊃⌽⍸p←⍵=⍺↑⍨≢⍵)⊢⍵⋄j←⊃⍸⍺≠⍵⋄⍺≢⍵:⍺∇⎕←⍺[j]@j⊢⍵}
Try it online!
Works correctly now.
A recursive function submission, done with a lot of input from Adám and Bubbler.
Input is taken as <s2> f <s1>.
Explanation
⍺<⍥≢⍵: if the strings have different lengths:
0@(...)⊢⍵ put a zero at the following index:
⊃ first element of:
(⍸~p) characters which do not match
~ without
⍳⊃⌽⍸p←⍵=⍺↑⍨≢⍵ all indices before last character match
0~⍨ remove the zero(thereby removing the element)
⍺∇⎕← display the result and recurse
⋄⍺≢⍵: Otherwise, till the strings match:
i←⊃⍸⍺≠⍵ find first non-matching element
⍺[i]@i⊢⍵ and replace the element at that index
⍺∇⎕← display the result, and recurse | 1,337 | 4,023 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.640625 | 3 | CC-MAIN-2022-33 | longest | en | 0.899148 |
http://www.mathisfunforum.com/viewtopic.php?pid=16421 | 1,386,789,959,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164044626/warc/CC-MAIN-20131204133404-00093-ip-10-33-133-15.ec2.internal.warc.gz | 416,395,634 | 4,664 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2005-10-23 15:44:25
ryos
Power Member
Offline
### Find cubic function from extrema
Find a cubic function of the form f(x) = ax^3 + bx^2 + cx which has a local maximum at f(1) = 0 and a local minimum at f(3) = -2.
I'm stumped out of my mind. I'm pretty sure that both b and c must be negative, but beyond that, no matter how I munge and funge that function, I can't get it to behave. Is there some voodoo calculus trick that I'm missing here?
(I've tried solving the coefficients as a system of equations, but that didn't work. Every time I got it down to one variable with substitutions, the variable would cancel out, leaving me with something stupid like -2=-2. But, maybe I did it wrong.)
Thanks guys.
El que pega primero pega dos veces.
## #2 2005-10-23 16:50:03
MathsIsFun
Offline
### Re: Find cubic function from extrema
f(x) = ax^3 + bx^2 + cx
f'(x) = 3ax^2 + 2bx + c
Maxima and Minima are where slope=0, ie f'(x)=0
Where slope=0:
x=1: 3a + 2b + c = 0
x=3: 27a + 6b + c = 0
Now, we also know that f(1) = 0 and f(3) = -2:
a + b + c = 0
27a + 9b + 3c = -2
So, now it is substitution time
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #3 2005-10-23 17:14:41
MathsIsFun
Offline
### Re: Find cubic function from extrema
3 unknowns and 4 equations ...
27a + 6b + c = 0
27a + 9b + 3c = -2
reduces to 3b+2c=-2, and then b=-2/3 - (2/3)c
a + b + c = 0
==> a = -b-c
==> a = -(-2/3 - (2/3)c)-c
==> a = 2/3 + (2/3)c - c
==> a = 2/3 - (1/3)c
3a + 2b + c = 0
==> 3(2/3 - (1/3)c) + 2(-2/3 - (2/3)c) + c = 0
==> (2 - c) + (-4/3 - (4/3)c) + c = 0
==> 2 - c -4/3 - (4/3)c + c = 0
==> 2 -4/3 - (4/3)c = 0
==> 2/3 - (4/3)c = 0
==> 2 - 4c = 0
==> 2 = 4c
==> c = 1/2
Now, I haven't made a mistake (and my mistake rate is about 10%, so it is likely )
b = -2/3 - (2/3)c = -2/3 - (1/3) = -1
a = 2/3 - (1/3)c = 2/3 - (1/6) = 1/2
Hmm, must have made a miscalc ... this seems odd, not working out ...
Maybe we are missing the "d" term: f(x) = ax^3 + bx^2 + cx + d
I will work on it later today, ryos.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #4 2005-10-23 18:40:27
MathsIsFun
Offline
### Re: Find cubic function from extrema
OK, I did it all on paper, and you do need the "d" term.
The results are:
a=1/2
b=-3
c=4.5
d=-2
I have to go now, but can do a better posting later.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #5 2005-10-23 19:39:12
MathsIsFun
Offline
### Re: Find cubic function from extrema
OK, this is what I had on paper:
27a + 9b + 3c + d = -2
a + b + c + d = 0
Subtract and get: 26a + 8b + 2c = -2
Now you have 3 equations and 3 unknowns:
(1) 3a + 2b + c = 0
(2) 27a + 6b + c = 0
(3) 26a + 8b + 2c = -2
Using (1): 3a + 2b + c = 0 ==> c = -3a -2b
Combine with (2): 27a + 6b + (-3a -2b) = 0
==> 24a + 4b = 0
==> a = -(1/6)b
Substitute into (3) gets:
26a + 8b + 2c = -2
==> (-26/6)b + 8b + 2c = -2
==> (22/6)b + 2c = -2
==> c = -1 - (22/12) b
Putting all that in (1) gets:
3a + 2b + c = 0
==> 3(-1/6) b + 2b + (-1 - (22/12) b) = 0
==> (-3/6 + 2 - 22/12)b = 1
==> b = -3
Knowing b, we can calculate
a = -(1/6) b = 1/2
c = -1 - (22/12) b = 4.5
and then calculate d:
a + b + c + d = 0
0.5 + (-3) + 4.5 + -2 = 0
DONE
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #6 2005-10-24 13:07:59
ryos
Power Member
Offline
### Re: Find cubic function from extrema
Thanks, you're a lifesaver. I had forgotten that you need at least as many equations as unknowns, and didn't think to use the derivatives.
El que pega primero pega dos veces. | 1,588 | 3,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | longest | en | 0.88935 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-2-1st-edition/chapter-12-sequences-and-series-12-3-analyze-geometric-sequences-and-series-12-3-exercises-skill-practice-page-814/28 | 1,696,436,261,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00406.warc.gz | 882,815,035 | 15,771 | ## Algebra 2 (1st Edition)
Published by McDougal Littell
# Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Skill Practice - Page 814: 28
See below
#### Work Step by Step
The sequence is geometric with first term $a_1=5$ and common ratio $r=3$. So, a rule for the nth term is:$$a_n=a_1r^{n-1}=5*3^{n-1}$$ The first 6 terms are: $a_1=5\\a_2=15\\a_3=45\\a_4=135\\a_5=405\\a_6=1215$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 197 | 593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | latest | en | 0.767186 |
https://elisabethjordancalligraphy.com/qa/quick-answer-what-does-magic-number-mean.html | 1,620,548,735,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988961.17/warc/CC-MAIN-20210509062621-20210509092621-00236.warc.gz | 248,546,696 | 8,330 | # Quick Answer: What Does Magic Number Mean?
## Why is 12 a magic number?
The number twelve carries religious, mythological and magical symbolism, generally representing perfection, entirety, or cosmic order in traditions since antiquity..
## Why is 8 a magic number?
“Eight” is a Magic Number in Physics. atomic Nuclei of Magic Numbers (ironically there are seven of them) have a greater binding power and thus are stronger and more stable.
## Why is 7 the perfect number?
Seven is the number of completeness and perfection (both physical and spiritual). It derives much of its meaning from being tied directly to God’s creation of all things. … The word ‘created’ is used 7 times describing God’s creative work (Genesis 1:1, 21, 27 three times; 2:3; 2:4).
## Is 2 a magic number?
The 2 is a magic number and the code needs to be refactored. By introducing a new variable, we can communicate the meaning the value 2 . We could also re-use the variable elsewhere in the code.
## Which are the magic numbers?
The seven most widely recognized magic numbers as of 2019 are 2, 8, 20, 28, 50, 82, and 126 (sequence A018226 in the OEIS). For protons, this corresponds to the elements helium, oxygen, calcium, nickel, tin, lead and the hypothetical unbihexium, although 126 is so far only known to be a magic number for neutrons.
## Is 7 a magic number?
The “magical number 7” and working memory capacity The number of chunks a human can recall immediately after presentation depends on the category of chunks used (e.g., span is around seven for digits, around six for letters, and around five for words), and even on features of the chunks within a category.
## Is 0 a magic number?
0, 1, -1, 2 These are not generally considered magic numbers, and it’s ok to have them in your code. However, it’s often possible to add more context to explain why you’re using these numbers.
## Why is 6174 a magic number?
6174 is known as Kaprekar’s constant after the Indian mathematician D. R. Arrange the digits in descending and then in ascending order to get two four-digit numbers, adding leading zeros if necessary. … Subtract the smaller number from the bigger number.
## Why should you avoid magic numbers in programming?
A magic number is a number in the code that seems arbitrary and has no context or meaning. This is considered an anti-pattern because it makes code difficult to understand and maintain. One of the most important aspects of code quality is how it conveys intention. Magic numbers hide intention so they should be avoided.
## Why is Magic Number bad?
A magic number is a numeric literal (for example, 8080 , 2048 ) that is used in the middle of a block of code without explanation. It is considered bad practice to use magic numbers because: A number in isolation can be difficult for other programmers to understand.
## Why is 7 the luckiest number?
In the creation story, God made the world in six days and rested on the seventh day. Scholars have found that the number seven often represents perfection or completeness in the Bible. … Seven is also a prime number, which means it can only be divided by itself and one.
## Is 9 the most powerful number?
Which is the strongest number? Number 9 is the strongest number among all numbers. Number 9 stands for initiation, wisdom, and spiritual enlightenment. When number 9 is multiplied with other numbers, the result will always reduce to 9. | 773 | 3,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.913288 |
https://converterin.com/area/manzana-nicaragua-to-square-gigameter-gm2.html | 1,601,331,767,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401614309.85/warc/CC-MAIN-20200928202758-20200928232758-00001.warc.gz | 330,974,754 | 9,666 | # MANZANA [NICARAGUA] TO SQUARE GIGAMETER CONVERTER
FROM
TO
The result of your conversion between manzana [Nicaragua] and square gigameter appears here
## MANZANA [NICARAGUA] TO SQUARE GIGAMETER (manzana TO gm2) FORMULA
To convert between Manzana [Nicaragua] and Square Gigameter you have to do the following:
First divide 70.44*100 / 1.0E+18 = 0.
Then multiply the amount of Manzana [Nicaragua] you want to convert to Square Gigameter, use the chart below to guide you.
## MANZANA [NICARAGUA] TO SQUARE GIGAMETER (manzana TO gm2) CHART
• 1 manzana [Nicaragua] in square gigameter = 0. manzana
• 10 manzana [Nicaragua] in square gigameter = 0. manzana
• 50 manzana [Nicaragua] in square gigameter = 0. manzana
• 100 manzana [Nicaragua] in square gigameter = 0. manzana
• 250 manzana [Nicaragua] in square gigameter = 0. manzana
• 500 manzana [Nicaragua] in square gigameter = 0. manzana
• 1,000 manzana [Nicaragua] in square gigameter = 0. manzana
• 10,000 manzana [Nicaragua] in square gigameter = 0. manzana
Symbol: manzana
No description
Symbol: gm2
No description | 386 | 1,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-40 | latest | en | 0.597328 |
http://algebratesthelper.com/what-every-math-student-needs-to-know/ | 1,495,681,098,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607963.70/warc/CC-MAIN-20170525025250-20170525045250-00099.warc.gz | 16,529,306 | 12,518 | ## What Every Student Needs To Know Before Taking A Math Test
By understanding how your teacher grades tests you can significantly raise your math grades even if you have not completely mastered the concepts.
As a middle and high school math teacher I graded thousands of math tests and all along the way I had to determine how many points I would reward a student for their efforts. The process was fairly subjective and I often used my “gut” or best judgment to make a decision on how many points I would award a student for the work on their test. Of course each teacher will grade tests slightly different, but I can assure you most math teachers are looking for certain indicators before they award points. Students can really help improve their grade if they know what their teacher is looking for and it’s much more than the correct answer.
2. Simplified solution
Give me the correct answer as the fraction and I will take points off! Every math teacher I know will deduct points for correct answers that are not simplified. Examples of correct solutions that are not simplified are unreduced fractions, answers that don’t have the proper of units of measure (like an area or volume problem) and variable or number expressions that are not finished. So when it comes to racking up easy points on your tests, finishing and simplifying your solutions go a long way.
3. How much work you show
The best thing a student can do for their grade beyond getting the correct answer is show as much work as possible…even if you don’t know what you’re doing! Of course the work you show has to be a serious effort at trying to figure out the problem, but if you provide written evidence that you really tried to answer a question you will most likely earn some “charity” points—it all adds up in the end and can make a major difference in your overall grade.
4. How many questions you attempt
Never, never, never leave a question blank—this is the kiss of death for your math test score! I would like to stress a few points here. When taking a test, first invest the majority of time on the problems you feel that you can answer correctly. However, you always need to leave a little time to attempt questions that you doubt you can figure out. I’d like to refer you back to my 3rd point “how much work you show” as what you need to be focusing on with these last ditch problems. Always try your best to leave no question blank. If you show your teacher that you at least tried each problem, often you will discover they will “kick” you a few extra points for your effort.
5. How much understanding you demonstrated
If you show that you understand some parts of a problem your teacher will reward you even if you’re not able to complete the problem. For example, let’s say you’re working on an equation and wrote each step correctly until you were stumped. Guess what…you have earned some points! Now let’s take the same problem with the attitude, “I can’t do this, so it’s a waste of my time” this approach will always yield in no points. The main factor in collecting points on your test score is demonstrating what you know even if you can’t get the problem right! Partial credit is what I’m talking about and as teacher I can tell you that partial credit can often be the difference between a B- and an A-. Be smart and get credit for what you know!
6. Whether your work is clear, neat and understandable
Even if your work is correct if your teacher can’t understand it, you’re at risk of not getting points! Neatness is a struggle for many students, but overcoming sloppiness can result in big points. A way to get on a teacher’s good side (which will always improve your score) is to write your work so neatly that it’s easy to read and understand. Now, frustrate your teacher by complicated “chicken scratch” and they won’t be as motivated to stick around and find points from your answer. Remember that increasing your math test scores will be the sum of a lot of little things you do and writing neatly is definitely one of them! | 871 | 4,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-22 | longest | en | 0.96377 |
https://schoolworkhelper.net/laplaces-law-surface-tension/ | 1,695,524,559,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00087.warc.gz | 559,773,297 | 138,152 | ## Surface Tension (N/m² = J/m³)
• Fluid property associated with the presence of a surface toward air
• Interfacial tension- is used to describe analogous phenomena for fluid having interfaces with solids or other liquids (can be +/-)
• Water molecules are forced toward the surface of a fluid due to placement on other molecules and attractive forces. (the attractive force at equilibrium in water but when pushed to the surface; attractive equilibrium is lost and pulls molecules away from the surface, but they can’t go anywhere because surface CAN’T shrink.
• This creates some internal pressure and forces liquid surfaces to contract to the minimal area.
• A force is required to hold the molecules at the surface area (Ơ) [high energy particles out the exterior with no “neighbor” molecules to hold it at equilibrium.]
• A fluid will shrink to the minimal surface area to maintain low energy
Ơ= ∆E/ ∆A (in J/m^2 = N/m=(kg/s^2) ***ALWAYS POSITIVE
W= ∆E = Ơ x ∆A
Ơ= F/lx
• WORK MUST BE DONE TO INCREASES SURFACE OF A FLUID (either force stretch a film or energy to increase surface area)
• Surface tension and pressure are the same (can be regarded as a FORCE, or ENERGY)
• Can be used on thin soap films due to its high cohesiveness
• Cannot use same set up for water; must place flat solid interface on it and determine the force needed to left solid off of the fluid.
## Bubbles and Droplets
• Raindrops are an example of an open system
• Take the most geometrical shape to have the least energy to form surfaces
• Least energy/ surface area for a fixed volume…a SPHERE (alveoli take this shape)
• Energy is given off and raindrop ascertains sphere space;
## Laplace Law
∆p= (4 x ơ)/ r (r= radius)
• Transmural pressure: ∆p= pinside-poutside (In lungs; the difference between alveolar & pleural pressure)
• The pressure inside the bubble is greater to stop it from imploding
• One Surface: Droplets, homogeneous cylinders
• Two Surfaces: Bubbles, hollow tubes
• Hollow/ Homogeneous tubes; finite curvature in only ONE direction across their surfaces
• Bubbles/ Droplets; finite curvature in only TWO directions across their surfaces
The pressure difference between the inside and outside of a fluid with a curved surface is INVERSELY proportional to the radius of curvature of the curved surface.
Smaller bubble, droplet, cylinder has a larger pressure difference ∆p
Formulas
∆p= pinside – poutside= (4 x ơ)/r [BUBBLES]
∆p= pinside – poutside= (2 x ơ)/r [HOLLOW CYLINDERS/ DROPLET]
∆p= pinside – poutside= (2 x ơ)/r [SOLID CYLINDERS]
***Surface tension for a capillary of small radius must be smaller than the surface tension of an arteriole with a larger radius:
rarteriole > rcapillary -> ơarteriole > ơcapillary
• Allows walls of capillaries to be thinner; this, in turn, improves the efficiency to be thinner; improves diffusion of O2 and transport of small ions [Small alveoli are more effective at gas exchange]
• Pulmonary Surfactants: wet the alveolar surface to counterbalance radius effect
• Neonatal Respiratory Distress Syndrome: premature baby can’t make surfactants and lung is stiff; alveoli collapse.
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Notify of | 782 | 3,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | longest | en | 0.847511 |
https://sailboat.guide/nevins/sound-interclub-mower | 1,653,102,116,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534773.36/warc/CC-MAIN-20220521014358-20220521044358-00105.warc.gz | 576,449,880 | 14,883 | # Sound Interclub (Mower)
1926 — 1927
Designer
Charles D. Mower
Builder
Nevins Inc.
Associations
Rosenfeld photo collection
Sound Interclub Class (1926)
# Built
27
Hull
Monohull
Keel
Fin
Rudder
?
Construction
Wood
### Dimensions
Length Overall
28 8 / 8.8 m
Waterline Length
18 11 / 5.8 m
Beam
7 6 / 2.3 m
Draft
4 5 / 1.4 m
Displacement
6,000 lb / 2,722 kg
Ballast
2,500 lb / 1,134 kg (lead)
• 1 / 1
### Rig and Sails
Type
Sloop
Reported Sail Area
425′² / 39.5 m²
Total Sail Area
?
Sail Area
?
P
?
E
?
Air Draft
?
Sail Area
?
I
?
J
?
Forestay Length
?
Make
?
Model
?
HP
?
Fuel Type
?
Fuel Capacity
?
### Accomodations
Water Capacity
?
Holding Tank Capacity
?
?
Cabins
?
Hull Speed
5.6 kn
Classic: 5.84 kn
### Hull Speed
The theoretical maximum speed that a displacement hull can move efficiently through the water is determined by it's waterline length and displacement. It may be unable to reach this speed if the boat is underpowered or heavily loaded, though it may exceed this speed given enough power. Read more.
Formula
Classic hull speed formula:
Hull Speed = 1.34 x √LWL
A more accurate formula devised by Dave Gerr in The Propeller Handbook replaces the Speed/Length ratio constant of 1.34 with a calculation based on the Displacement/Length ratio.
Max Speed/Length ratio = 8.26 ÷ Displacement/Length ratio.311
Hull Speed = Max Speed/Length ratio x √LWL
5.63 knots
Classic formula: 5.84 knots
Sail Area/Displacement
20.6
>20: high performance
### Sail Area / Displacement Ratio
A measure of the power of the sails relative to the weight of the boat. The higher the number, the higher the performance, but the harder the boat will be to handle. This ratio is a "non-dimensional" value that facilitates comparisons between boats of different types and sizes. Read more.
Formula
SA/D = SA ÷ (D ÷ 64)2/3
• SA: Sail area in square feet, derived by adding the mainsail area to 100% of the foretriangle area (the lateral area above the deck between the mast and the forestay).
• D: Displacement in pounds.
20.59
<16: under powered
16-20: good performance
>20: high performance
Ballast/Displacement
41.7
>40: stiffer, more powerful
### Ballast / Displacement Ratio
A measure of the stability of a boat's hull that suggests how well a monohull will stand up to its sails. The ballast displacement ratio indicates how much of the weight of a boat is placed for maximum stability against capsizing and is an indicator of stiffness and resistance to capsize.
Formula
Ballast / Displacement * 100
41.66
<40: less stiff, less powerful
>40: stiffer, more powerful
Displacement/Length
390.8
>350: ultraheavy
### Displacement / Length Ratio
A measure of the weight of the boat relative to it's length at the waterline. The higher a boat’s D/L ratio, the more easily it will carry a load and the more comfortable its motion will be. The lower a boat's ratio is, the less power it takes to drive the boat to its nominal hull speed or beyond. Read more.
Formula
D/L = (D ÷ 2240) ÷ (0.01 x LWL)³
• D: Displacement of the boat in pounds.
• LWL: Waterline length in feet
390.83
<100: ultralight
100-200: light
200-300: moderate
300-400: heavy
>400: very heavy
Comfort Ratio
28.6
20-30: coastal cruiser
### Comfort Ratio
This ratio assess how quickly and abruptly a boat’s hull reacts to waves in a significant seaway, these being the elements of a boat’s motion most likely to cause seasickness. Read more.
Formula
Comfort ratio = D ÷ (.65 x (.7 LWL + .3 LOA) x Beam1.33)
• D: Displacement of the boat in pounds
• LWL: Waterline length in feet
• LOA: Length overall in feet
• Beam: Width of boat at the widest point in feet
28.64
<20: lightweight racing boat
20-30: coastal cruiser
30-40: moderate bluewater cruising boat
40-50: heavy bluewater boat
>50: extremely heavy bluewater boat
Capsize Screening
1.7
<2.0: better suited for ocean passages
### Capsize Screening Formula
This formula attempts to indicate whether a given boat might be too wide and light to readily right itself after being overturned in extreme conditions. Read more.
Formula
CSV = Beam ÷ ³√(D / 64)
• Beam: Width of boat at the widest point in feet
• D: Displacement of the boat in pounds
1.65
<2: better suited for ocean passages
>2: better suited for coastal cruising
### Notes
Original class name is LONG ISLAND SOUND INTERCLUB.
The result of efforts made by a group headed by Carroll B. Alker of the Seawanhaka Corinthian YC and developed for racing on Long Island Sound.
In 1938, a significant number of the original boats were sold to residents at Lake George NY, where organized racing is said to have continued until the late 1950’s.
An article appeared in ‘Wooden Boat’ magazine, (Jan.,2015) describing the complete restoration of 2 original boats.
### For Sale
Have a sailboat to sell? | 1,327 | 4,800 | {"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} | 2.90625 | 3 | CC-MAIN-2022-21 | latest | en | 0.793407 |
http://amburnclasses.org/main/learn_chemistry/chapter_7/Lesson_7.2.html | 1,623,735,654,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00446.warc.gz | 3,677,689 | 4,660 | ↑ home
# Lesson 7.2: Effective Nuclear Charge
← Lesson 7.1: Lesson 7.3: →
Electrons in higher energy shells are on average further from the nucleus. A crude way to visualize these shells, when we are not concerned about the details of the orbital shape studied in lesson 6, is to imagine the electrons as existing in increasingly distant layers from the nucleus. Given this characterization, it does not come as a surprise that lithium with two layers of electrons, two electrons in the first and 1 electron in the second, would be larger that hydrogen with just 1 electron in the first layer. Here is a picture of their relative sizes in pm.
The same trend continues as we move down the periodic table for group 1 elements. Notice that we have drawn the second shell as if it were a fixed size – the same size in lithium and sodium.
This way of looking at things would incline to say that shells have a fixed size, and that when we move across a period we would predict that the size of the atoms remains constant. However, this is not the case. Here is a picture with beryllium included.
Although beryllium contains more stuff – an extra electron, proton and neutron – it is smaller! How are we to make sense of this?
When trying to make sense of the size of atoms, along with other important properties, we need to consider three factors. (Keep in mind that the size of an atom is just the region of space where the electrons exist. A single electron far from the nucleus of an atom can make an atom very large, just like an outer planet can make a solar system very large.)
1. The charge on the nucleus - how much positive charge is attracting a given electron. The greater nuclear charge, the greater the attraction the electron and the smaller the size contribution of that electron.
2. The number of electrons between the nucleus and the electron. The nucleus attracts all of the electrons in an atom, but the electrons between the nucleus and a given electron will repel it.
3. The distance from the nucleus to the given electron.
We keep track of the first two of these factors with the concept of effective nuclear charge. The effective nuclear charge on an electron is just the amount of charge it experiences if we keep track of the repulsions of inner electrons. We will say that inner electrons shield or screen the electron from the charge of the nucleus, so that the charge that is effective at attracting the electron is the total charge minus the charge of the electrons that shield it. As an equation we can write $$Z_{eff} = Z - S$$
Zeff is the effective nuclear charge, Z the total nuclear charge, and the S the shielding - the repulsion caused by electrons that exist between the electron and the nucleus.
Using this concept, lets return to the comparison of beryllium and lithium. A crude method for finding the effective nuclear charge on the outer electron of lithium is to subtract the charge of the inner shell electrons from the charge on the nucleus. Z = 3, the charge of the nucleus, and S = 2 since the electrons in the first shell exist between it and the nucleus and repel it. So the effective nuclear charge on the outer electron is \begin{align*} Z_{eff} & = Z - S \\\\ Z_{eff} & = 3 - 2 \\\\ Z_{eff} & = 1 \end{align*}
For beryllium we again subtract the charge of the inner shell electrons from the charge on the nucleus. Z = 4 , the charge of the nucleus, and S = 2 since the electrons in the first shell exist between it and the nucleus and repel it. So the effective nuclear charge on the outer electron is \begin{align*} Z_{eff} & = Z - S \\\\ Z_{eff} & = 4 - 2 \\\\ Z_{eff} & = 2 \end{align*}
With a greater effective nuclear charge the electrons in beryllium are drawn closer to the nucleus resulting in a smaller atom.
This trend continues as we move from left to right across the periodic table. The electrons are drawn ever closer to the nucleus due to the increase in effective nuclear charge. The outer electrons nonetheless remain further out than those of hydrogen or helium since they occupy a higher energy level.
The process starts again once we have come to the end of a row on the periodic table. With neon the 2nd energy level is filled and the last electron of sodium is forced to occupy the third energy level. This last electron therefore has 10 electrons between it and the nucleus, giving it an effective nuclear charge of 1. \begin{align*} Z_{eff} & = Z - S \\\\ Z_{eff} & = 11 - 10 \\\\ Z_{eff} & = 1 \end{align*}
Moving from left to right across the periodic table we see the same trend as in the second period. The electrons are drawn ever closer to the nucleus due to the increase in effective nuclear charge.
While this is an easy way to approximate S (the number of core electrons) it does fully make sense. For example, it does not make sense to say that core electrons are 100% effective at shielding valence electrons, and that valence electrons do not shield each other at all. To make sense of these more subtle distinctions we need a more sophisticated method for finding S - a more sophisticated method for thinking about electron-electron repulsion in atoms.
Consider the diagram for beryllium shown below. First, the drawing makes it look like the 1s electrons are restricted to the small volume near the nucleus which is not the case, but even as drawn, it is clear that the two core 1s electrons are not in the same location and thus not able to equally screen a valence electron in the 2s shell. Only if the two electrons were in the same location as the proton would they screen with 100 % efficiency. This condition is only approximated if the valence electrons are sufficiently far from the core electrons existing in a relatively small space around the nucleus. Second, it is clear that the valence electrons can shield each other – will exert repulsive forces on each other that will vary as they move about the 2s orbital.
## Slater's Rules
How can we capture these considerations in an effort to generate a better approximation of S and thus Zeff? A set of rules has been devised called Slater's Rules which are easy to apply, captures these insights, and yields Zeff values close to those found with the most sophisticated methods. Here are the rules.
### Rule 1: Electrons in the same energy level of s and p orbitals are 35% effective at screening electrons.
For example, how much is an electron in a 3p electron of Cl , 1s22s22p63s23p5, repelled or screened by the other electrons in the 3s or 3p orbitals? There are 7 electrons in the 3rd energy level of chlorine. Any one of the 7 electrons in the 3s and 3p orbitals is repelled by 6 electrons with 0.35 screening efficiency. This makes a total screening of 0.35 (6) = 2.1 for an electron of chlorine in the 3rd energy level. Notice that this screening of 2.1 would be ignored by the crude method of using the core electrons to find S.
### Rule 2: Electrons in the next lower energy level, n-1, of a s or p electron, screen with 85% efficiency.
To find their contribution to S simply add up the number of electrons in the next lowest energy level as the one under consideration and multiply by 0.85. For example, a 3p electron in Cl, 1s22s22p63s23p5, is screened by the 8 electrons in the 2nd energy level. The contribution to S is 0.85 x 8 = 6.8 units of charge. Notice that in the crude method we would assume that these screen by 8.
### Rule 3: Electrons in the next lower level, n-2, and below, are considered 100% screening electrons.
To find their contribution to S simply add up their number. For example, a 3p electron in Cl , 1s22s22p63s23p5, is then screened by the 2 electrons in the 1st energy level by 2 units of charge, the same amount as we would calculate with the crude method.
That gives us a final value of \begin{align*} Z_{eff} & = Z - S \\\\ S & = 2.1 + 6.8 + 2 = 10.9 \\\\ Z & = 17\\\\ Z_{eff} &= 17 – 10.9 = 6.1. \end{align*} The crude method gave us Zeff = 17 – 10 = 7.
### Rule 4: If the electron under consideration is an f or d orbital electron, the electrons that occupy that subshell are assumed to screen by 35% and all the rest by 100%.
For example, Fe has 6 electrons in the 3d orbital. How much is one of these screened by other 5 d electrons? 5 x (0.35). Note that the 4s electrons do not screen at all and that the rest of the electrons in the 3rd, 2nd , and 1st shield 100%. For Fe this gives us a final value of \begin{align*} Z_{eff} & = Z - S \\\\ S & = 5(0.35) + 18 = 19.75 \\\\ Z & = 26 \\\\ Z_{eff} &= 26 - 19.75\\\\ Z_{eff} &= 6.25 \end{align*}
### Rule 5: A 1s electron is assumed to screen another 1s electron by 30%.
For example, Fe, like all atoms past hydrogen, has 2 electrons in the 1s orbital. How much does the one electron in that 1s orbital screen the other 1s electron? With 30% efficiency or with S = 1 x (0.30). Note that none of the other electrons in hgher energy levels screen the electron in the 1s orbital. For a 1s electron in Fe this gives us an effective nuclear charge of \begin{align*} Z_{eff} & = Z - S \\\\ S & = 1(0.30) = 0.30 \\\\ Z & = 26 \\\\ Z_{eff} &= 26 - 0.30\\\\ Z_{eff} &= 25.7 \end{align*} | 2,274 | 9,144 | {"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} | 2.65625 | 3 | CC-MAIN-2021-25 | latest | en | 0.939518 |
http://www.google.com/patents/US7840623?ie=ISO-8859-1 | 1,405,207,949,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776435471.2/warc/CC-MAIN-20140707234035-00031-ip-10-180-212-248.ec2.internal.warc.gz | 328,842,546 | 64,223 | Patents
Publication number US7840623 B2 Publication type Grant Application number US 11/523,351 Publication date Nov 23, 2010 Filing date Sep 19, 2006 Priority date Sep 26, 2005 Fee status Paid Also published as Publication number 11523351, 523351, US 7840623 B2, US 7840623B2, US-B2-7840623, US7840623 B2, US7840623B2 Inventors Original Assignee Dai Nippon Printing Co., Ltd. Export Citation Patent Citations (12), Classifications (4), Legal Events (2) External Links:
Interpolator and designing method thereof
US 7840623 B2
Abstract
Interpolation of signed values A and B is efficiently performed by simple circuitry. To calculate an interpolated value C based on a 4-bit values A (bits a3a2a1a0) and B (bits b3b2b1b0) expressing a negative number by twos complement notation and a 4-bit interpolation rate D (bits d3d2d1d0) consisting of only a decimal part, a basic expression of C=(1−D)*A+D*B is transformed into an expression composed of an unsigned part that includes a sum of products with a bit di or a logically inverted value ei of the bit di (i=0, 1, 2, and 3), and indicates an absolute value of the interpolated value C, and a signed part indicating a sign of the interpolated value C. Then, 7 bits of bits c6 through c0 are generated from an arithmetic operation of the unsigned part, and logic judgement of the signed part is performed by considering a carry from the digit of the bit c6 of the arithmetic operation of the unsigned part to generate a bit c7. Significant digits of the obtained 8-bit value (bits c7 through c0) are outputted as an interpolated value.
Images(44)
Claims(2)
1. A linear interpolator which performs operations to calculate a linear interpolated value C expressed by an expression of C=(1−D)*A+D*B based on two signed interpolation target values A and B and an interpolation rate D (0≦D<1), comprising:
first interpolation target value input means for inputting an interpolation target value A including bits an−1, an−2, . . . , a1 and a0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
second interpolation target value input means for inputting an interpolation target value B including bits bn−1, bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
interpolation rate input means for inputting an interpolation rate D including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side as digital data of n bits indicating only a decimal part;
unsigned part arithmetic means for generating digital data of (2n−1) bits in total including bits c2n−2, c2n−3, . . . , c1, and c0 in order from the most significant bit side by performing arithmetic operations based on an arithmetic expression of bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σj=0˜n−2ai2ji=0˜n−1di2(i−n)·Σj=0˜n−2bi2j (ei is a logically inverted bit of di);
signed part arithmetic means which calculates a binary number obtained by summing up a result of operation of an−1+bn−1 (any of 0, 1, and 10 in binary expression) and a carry (either 0 or 1 in binary expression) from a digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means, and when a calculated binary number is 1 bit, calculates this bit, and when a calculated binary number is 2 bits, calculates a lower-order bit as the value of the bit c2n−1; and
calculated value output means for outputting “digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures of the digital data of the 2n bits” as digital data indicating a linear interpolated value C,
wherein
the unsigned part arithmetic means comprises:
a selector which selects and outputs either one of the bit ai and bi (i=0, 1, 2, . . . (n−1)) based on a logical value of a predetermined bit of digital data indicating the interpolation rate D; and
a counter which inputs the bit ai or bi (i=0, 1, 2, . . . (n−1)), an output value of the selector or an output value of another counter, and outputs a result of addition of the inputted values.
2. The linear interpolator according to claim 1, wherein
the signed part arithmetic means comprises:
a first XOR circuit which outputs an exclusive logical sum of the bit an−1 and the bit bn−1; and
a second XOR circuit which outputs an exclusive logical sum of a bit indicating a carry (either 0 or 1 in binary expression) from a digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means, and an output bit of the first XOR circuit.
Description
BACKGROUND OF THE INVENTION
The invention disclosed in the present application relates to a linear interpolator and a designing method thereof. More specifically, the present invention relates to a technique for calculating a linear interpolated value of two signed interpolation target values by a hardware circuit, and to a technique for realizing a simple hardware circuit which can calculate a high-accuracy linear interpolated value.
As a method for obtaining an intermediate value of two values by means of interpolation, linear interpolation is most representative. Recently, in particular, as a method for image processing, this linear interpolation is popularly utilized. For example, the method for image synthesis generally called “α-blend” is a method in which an image A and an image B are synthesized at a ratio of α:(1−α)(0≦α≦1), and processing for obtaining an intermediate value of a pixel value of the image A and a pixel value of the image B by means of linear interpolation is performed. Also when enlarging or reducing an image at an arbitrary magnification, linear interpolation is used.
The basic principle of linear interpolation is that calculation is performed for obtaining a linear interpolated value C expressed as C=(1−D)*A+D*B (* sign representing multiplication) based on two interpolation target values A and B and an interpolation rate D (a value corresponding to said α, and 0≦D≦1). When performing arithmetic operations based on this expression by using a digital computing unit, A, B, C, and D are all handled as digital data, and normally, the interpolation rate D is expressed by a bit sequence having finite digits showing only a decimal part. In this case, the interpolation rate D is handled as a value satisfying “0≦D<1” which does not include 1. U.S. Pat. No. 5,113,362 discloses a configuration of a computing unit which performs such linear interpolation (α-blend operation) by pipeline processing, and U.S. Pat. No. 5,517,437 discloses a configuration of a computing unit which performs linear interpolation by processing based on parallel processing.
The computing units disclosed in the patent documents and other conventional general linear interpolators assume that the interpolation target values are positive values. Normally, since pixel values of individual pixels composing an image are expressed as positive values, as the “α-blend” processing for general images, linear interpolation based on two positive pixel values is sufficient, and calculation can be performed by the computing units disclosed in the patent documents. However, recently, various image processes are applied, and an opportunities to handle an image including negative pixel values are quite many in number. For example, when it is necessary to subtract a pixel value of an image B from a pixel value of an image A in some process, an image C obtained as a result of this subtraction may include pixels with negative pixel values. As a matter of course, the use of the linear interpolation is not limited to the image processing, so that there are many other events that linear interpolation considering both positive and negative values becomes necessary.
To meet this demand, there have been proposed some computing units which can execute linear interpolation for two signed interpolation target values. However, all the signed linear interpolators which are conventionally proposed employ complicated circuitries for realizing a function to handle signed digital data, and cannot perform efficient calculation.
The linear interpolators which are conventionally proposed hardly satisfy both of calculation accuracy and circuit simplicity, and either one must be sacrificed. For example, the above-mentioned linear interpolator disclosed in U.S. Pat. No. 5,517,437 is comparatively simple in circuitry, however, an approximate value is obtained as a result of calculation, and instead of the original interpolation rate α (0≦α≦1), handling using a value D (0≦D<1) that does not include 1 is performed, so that a result of calculation when α=1 cannot be correctly obtained. Therefore, to make it possible to correctly obtain a result of calculation when α=1, it is necessary that a comparator for monitoring whether an input value of the interpolation rate α is 1 is given and an extra circuit for outputting an input value B as it is, and the circuitry inevitably becomes complicated.
On the other hand, a method in which a highly accurate result of calculation is obtained by performing division is also proposed, however, the circuitry is still complicated.
Recently, an opportunity to synthesize many images has increased. Therefore, a linear interpolator which always obtains a high-accuracy result of calculation has been demanded. On the other hand, downsizing and cost reduction of a circuit are also demanded, so that a linear interpolator whose circuitry is made as simple as possible has been demanded.
The present application also discloses a technique relating to a linear interpolation device which performs not only the above-described linear interpolation but also more complicated interpolation. Particularly, a technique relating to an interpolation device which performs interpolation of pixels by cubic spline interpolation when scaling display is performed by enlarging or reducing an image has been disclosed.
When displaying a digital image, scaling of enlargement and reduction often becomes necessary. An image given as digital data is a two-dimensional pixel array including a plurality of pixels having predetermined pixel values arrayed at predetermined pitches, so that in the case of enlargement and reduction, new pixels must be obtained by interpolation. As an interpolation method conventionally used for this image enlargement and reduction scaling, nearest neighbor and bilinear interpolation are known.
The nearest neighbor is a method for selecting a nearest sample pixel, so that in the case of reduction, by using a kind of pixel skipping, an image does not become blurred in scaling display. To the contrary, in the case of enlargement, several neighbor identical images are continuously spread, and this results in a flat mosaic image display.
On the other hand, according to bilinear interpolation, linear interpolation is performed by using pixel values of sample pixels on both ends of an interpolation interval, so that pixel values become different from original pixel values in the case of reduction and may result in collapsed image display. To the contrary, in the case of enlargement, pixel values subjected to linear interpolation are filled, so that a blurred image may be displayed. Thus, nearest neighbor and bilinear interpolation are poor in interpolation accuracy, so that it is considered that these defects arise.
Therefore, as interpolation with improved accuracy, for example, a basic concept of cubic convolution is shown in Robert G. Keys, “Cubic Convolution Interpolation for Digital Images Processing” IEEE Trans. on ASSP-29, No. 6, December 1981, pp. 1153-1160. In image scaling display of a digital television, as a bicubic image interpolation calculation with more improved accuracy, bicubic convolution is used. For example, in a medical-use image, scale display with much higher accuracy is required. Therefore, to employ higher-accurate interpolation, bicubic spline interpolation is proposed. A basic concept of cubic spline interpolation that is the base of the bicubic spline interpolation is shown in Hsieh S. Hou and Harry C. Andrews, “Cubic Splines for Image Interpolation and Digital Filtering” IEEE Trans. on ASSP-26, No. 6, December 1978, pp. 508-517, and Michael Unser, “Splines A perfect Fit for Signal and Image processing” IEEE Signal Processing Magazine, November 1999, pp. 22-38.
In the above-described cubic convolution, by using pixel values of (horizontal 4)*(vertical 4)=16 sample pixels, vertical 4-pixel interpolation is performed after horizontal 4-pixel interpolation, and cubic interpolation is performed two times in total per one pixel to be obtained. Thus, in convolution, four cubic curves are used and these are smoothly connected at boundaries among these. In actuality, this unknown coefficients equation has not been completely solved, and 1 coefficient is left undetermined. Therefore, −1 and −½ is experimentally used as the undetermined coefficient value, and meaning and accuracy of the interpolation are ambiguous.
On the other hand, in cubic spline interpolation, four cubic curves are used similarly to the convolution in the interpolation interval, and this interpolation is different in meaning and composition of the expressions from the convolution because the cubic spline interpolation attempts to realize faithful approximation by blending these curves. The number of sample pixels to be used for interpolation is not necessarily four, and may be more than four. Namely, the number of weighing factors when blending four cubic curves called B-splines in the interpolation interval is four. Therefore, the cubic spline interpolation has approximation accuracy higher than that of the convolution, however, cubic interpolation equations are calculated through complicated processes. Particularly, a process for calculating respective cubic spline functions from pixel values of S in total of sample pixels (S≧4) to be referred for interpolation and a process for obtaining pixel values of interpolation points by setting positions of the interpolation points and performing bicubic spline interpolation, are extremely complicated, and at the present, in the case of image enlargement and reduction scaling display being performed, a device which performs practicable cubic spline interpolation suitable for commercial use has not been realized.
SUMMARY OF THE INVENTION
In the present application, the following three inventions are disclosed in order.
<<<<<First Aspect of the Invention>>>>>
An object of a first aspect of the invention is to provide a linear interpolator which can efficiently execute linear interpolation of two signed interpolation target values by a simple constitution, and to provide a method for designing such an interpolator.
(1) The first feature of the first aspect of the invention resides in a linear interpolator which performs operations to calculate a linear interpolated value C expressed by an expression of C=(1−D)*A+D*B based on two signed interpolation target values A and B and an interpolation rate D (0≦D<1), comprising:
first interpolation target value input means for inputting an interpolation target value A including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
second interpolation target value input means for inputting an interpolation target value B including bits bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
interpolation rate input means for inputting an interpolation rate D including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side as digital data of n bits indicating only a decimal part;
unsigned part arithmetic means for generating digital data of (2n−1) bits in total including bits c2n−2, c2n−3, . . . , c1, and c0 in order from the most significant bit side by performing arithmetic operations based on an arithmetic expression of bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σj=0˜n−2aj2ji=0˜n−1di2(i−n)·Σj=0˜n−2bj2j (ei is a logically inverted bit of di);
signed part arithmetic means which calculates a binary number obtained by summing up a result of operation of an−1+bn−1 (any of 0, 1, and 10 in binary expression) and a carry (either 0 or 1 in binary expression) from a digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means, and when a calculated binary number is 1 bit, calculates this bit, and when a calculated binary number is 2 bits, calculates a lower-order bit as the value of the bit c2n−1; and
calculated value output means for outputting “digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures of the digital data of the 2n bits” as digital data indicating a linear interpolated value C.
(2) The second feature of the first aspect of the invention resides in the linear interpolator according to the first feature, wherein
the unsigned part arithmetic means comprising:
a selector which selects and outputs either one of the bit ai and bi (i=0, 1, 2, . . . (n−1)) based on a logical value of a predetermined bit of digital data indicating the interpolation rate D; and
a counter which inputs the bit ai or bi (i=0, 1, 2, . . . (n−1)), an output value of the selector or an output value of another counter, and outputs a result of addition of the inputted values.
(3) The third feature of the first aspect of the invention resides in the linear interpolator according to the first or second feature, wherein
the signed part arithmetic means comprising:
a first XOR circuit which outputs an exclusive logical sum of the bit an−1 and the bit bn−1; and
a second XOR circuit which outputs an exclusive logical sum of a bit indicating a carry (either 0 or 1 in binary expression) from a digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means, and an output bit of the first XOR circuit.
(4) The fourth feature of the first aspect of the invention resides in a method for designing a linear interpolator which performs operations to calculate a linear interpolated value C expressed by an expression of C=(1−D)*A+D*B based on two signed interpolation target values A and B and an interpolation rate D (0≦D<1), comprising the steps of:
designing first interpolation target value input means for inputting an interpolation target value A including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
designing second interpolation target value input means for inputting an interpolation target value B including bits bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
designing interpolation rate input means for inputting an interpolation rate D including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side as digital data of n bits indicating only a decimal part;
designing unsigned part arithmetic means for generating digital data of (2n−1) bits in total including bits c2n−2, c2n−3, . . . , c1, and c0 in order from the most significant bit side by performing arithmetic operations based on an arithmetic expression of bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σj=0˜n−2aj2ji=0˜n−1di2(i−n)·Σj=0˜n−2bj2j (ei is a logically inverted bit of di);
designing signed part arithmetic means which calculates a binary number obtained by summing up a result of operation of an−1+bn−1 (any of 0, 1, and 10 in binary expression) and a carry (either 0 or 1 in binary expression) from a digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means, and when a calculated binary number is 1 bit, calculates this bit, and when a calculated binary number is 2 bits, calculates a lower-order bit as the value of the bit c2n−1; and
designing calculated value output means for outputting “digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures of the digital data of the 2n bits” as digital data indicating a linear interpolated value C.
(5) The fifth feature of the first aspect of the invention resides in the method for designing a linear interpolator according to the fourth feature, wherein
at the step of designing unsigned part arithmetic means,
a design using:
a selector which selects and outputs either one of the bit ai or bi (i=0, 1, 2, . . . (n−1)) based on a logical value of a predetermined bit of digital data indicating the interpolation rate D; and
a counter which inputs the bit ai or bi (i=0, 1, 2, . . . (n−1)), an output value of the selector or an output value of another counter, and outputs a result of addition of the inputted values, is made.
(6) The sixth feature of the first aspect of the invention resides in the method for designing a linear interpolator according to the fourth or fifth feature, wherein
at the step of designing signed part arithmetic means,
a design using:
a first XOR circuit which outputs an exclusive logical sum of the bit an−1 and the bit bn−1; and
a second XOR circuit which outputs an exclusive logical sum of a bit indicating a carry (either 0 or 1 in binary expression) from a digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means, and an output bit of the first XOR circuit,
(7) The seventh feature of the first aspect of the invention resides in a method for designing a linear interpolator which performs operations to calculate a linear interpolated value expressed by an expression of C=(1−D)*A+D*B based on two signed interpolation target values A and B and an interpolation rate D (0≦D<1), wherein
a linear interpolator is constituted so as to obtain a linear interpolated value C by performing multiplications and additions defined by an arithmetic expression of C=−(an−1+bn−1)2n−1+bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σj=0˜n−2aj2ji=0˜n−1di2(i−n)·Σj=0˜n−2bj2j (ei is a logically inverted bit of di) by:
defining an interpolation target value A including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2);
defining an interpolation target value B including bits bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation, composed of a signed part consisting of the most significant bit that indicates a sign and an unsigned part consisting of (n−1) bits that indicate an absolute value (n≧2); and
defining an interpolation rate D including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side as digital data of n bits indicating only a decimal part.
As described above, the linear interpolator according to the first aspect of the invention has circuitry according to arithmetic expressions in ideal forms, so that it becomes possible to efficiently execute linear interpolation of two signed interpolation target values by a simple constitution.
<<<<<Second Aspect of the Invention>>>>>
An object of a second aspect of the invention is to provide a linear interpolator which can calculate a linear interpolated value of two interpolation target values with high accuracy and has simple circuitry, and to provide a method for designing such an interpolator.
(1) The first feature of the second aspect of the invention resides in a linear interpolator which performs operations to calculate a linear interpolated value C expressed by an expression of C=(1−α)*A+α*B based on two interpolation target values A and B and an interpolation rate a (0≦α≦1), comprising:
first interpolation target value input means for inputting an interpolation target value A as digital data of n bits in total including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side;
second interpolation target value input means for inputting an interpolation target value B as digital data of n bits in total including bits bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side;
interpolation rate determining value input means for inputting an interpolation rate determining value D having a relationship of “α=D/(2n−1)” with the interpolation rate α, as digital data of n bits in total including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side;
arithmetic means for generating digital data of 2n bits in total composed of, in order from the most significant bit side, n bits in total of c2n−1, c2n−2, . . . , cn+1, and cn composing an integer part, and n bits in total of cn−1, cn−2, . . . , c1, and c0 composing a decimal part by performing arithmetic operations based on an arithmetic expression of C=(Σj=0˜n−1ej2(j−n)+en−12−n)*Σi=0˜n−1ai2i+(Σj=0˜n−1dj2(j−n)+dn−12−n)*Σi=0˜n−1bi2i (ei is a logically inverted bit of di); and
calculated value output means for outputting “digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures of the digital data of 2n bits” as digital data indicating a linear interpolated value C.
(2) The second feature of the second aspect of the invention resides in the linear interpolator according to the first feature, wherein
the arithmetic means comprising:
a selector which selects and outputs either bit ai or bi (i=0, 1, 2, . . . (n−1)) based on a logical value of a predetermined bit of digital data indicating an interpolation rate determining value D; and
a counter which inputs the bit ai or bi (i=0, 1, 2, . . . (n−1)), an output value of the selector or output value of another counter, and outputs a result of addition of the inputted values.
(3) The third feature of the second aspect of the invention resides in the linear interpolator according to the first or second feature, wherein
to calculate a bit c2n−i−2 (i=0, 1, 2, . . . 2(n−1)) indicating a value of a digit of 2(n−i−2) of the linear interpolated value C, the arithmetic means includes (2n−1) computing units which sum up coefficients of the digit and all carries from lower-order digits, output a sum as a value of the bit c2n−i−2, and output carries to a higher-order digit, and outputs a carry from a computing unit which calculates the bit c2n−2 as a value of the bit c2n−1.
(4) The fourth feature of the second aspect of the invention resides in a method for designing a linear interpolator which performs arithmetic operations to calculate a linear interpolated value C expressed by an expression of C=(1−α)*A+α*B based on two interpolation target values A and B and an interpolation rate α (0≦α≦1), comprising the steps of:
designing first interpolation target value input means for inputting an interpolation target value A as digital data of n bits in total including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side;
designing second interpolation target value input means for inputting an interpolation target value B as digital data of n bits in total including bits bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side;
designing interpolation rate determining value input means for inputting an interpolation rate determining value D having a relationship of “α=D/(2n−1)” with the interpolation rate α, as digital data of n bits in total including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side;
designing arithmetic means for generating digital data of 2n bits in total composed of, in order from the most significant bit side, n bits in total of c2n−1, c2n−2, . . . , cn+1, and cn composing an integer part, and n bits in total of cn−1, cn−2, . . . , c1, and c0 composing a decimal part by performing arithmetic operations based on an arithmetic expression of C=(Σj=0˜n−1ej2(j−n)+en−12−n)*Σi=0˜n−1ai2i+(Σj=0˜n−1dj2(j−n)+dn−12−n)*Σi=0˜n−1bi2i (ei is a logically inverted bit of di); and
designing calculated value output means for outputting “digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures of the digital data of 2n bits” as digital data indicating a linear interpolated value C.
(5) The fifth feature of the second aspect of the invention resides in the method for designing a linear interpolator according to the fourth feature, wherein
at the step of designing the arithmetic means,
a design using:
a selector which selects and outputs either bit ai or bi (i=0, 1, 2, . . . (n−1)) based on a logical value of a predetermined bit of digital data indicating an interpolation rate determining value D; and
a counter which inputs the bit ai or bi (i=0, 1, 2, . . . (n−1)), an output value of the selector or output value of another counter, and outputs a result of addition of the inputted values,
(6) The sixth feature of the second aspect of the invention resides in the method for designing a linear interpolator according to the fourth or fifth feature, wherein
at the step of designing the arithmetic means, to calculate a bit c2n−i−2 (i=0, 1, 2, . . . 2(n−1)) indicating a value of a digit of 2(n−i−2) of the linear interpolated value C, (2n−1) computing units which sum up coefficients of the digit and all carries from lower-order digits, output a sum as a value of the bit c2n−i−2, and output carries to a higher-order digit, are designed so that a carry from a computing unit which calculates the bit c2n−2 is outputted as a value of the bit c2n−1.
(7) The seventh feature of the second aspect of the invention resides in a method for designing a linear interpolator which performs arithmetic operations to calculate a linear interpolated value C expressed by an expression of C=(1−α)*A+α*B based on two interpolation target values A and B and an interpolation rate α (0≦α≦1), wherein
an interpolator is constituted so as to obtain a linear interpolated value C by performing multiplications and additions defined as an arithmetic expression of C=(Σj=0˜n−1ej2(j−n)+en−12−n)*Σi=0˜n−1ai2i+(Σj=0˜n−1dj2(j−n)+dn−12−n)*Σi=0˜n−1bi2i (ei is a logically inverted bit of di) by:
defining an interpolation target value A as digital data of n bits in total including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side;
defining an interpolation target value B as digital data of n bits in total including bits bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side; and
defining an interpolation rate determining value D having a relationship of “α=D/(2n−1)” with the interpolation rate α, as digital data of n bits in total including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side.
As described above, the linear interpolator according to the second aspect of the invention has circuitry according to unique arithmetic expressions obtained by replacing division with addition, so that it becomes possible to obtain a linear interpolated value with high accuracy while simple circuitry is used.
<<<<<Third Aspect of the Invention>>>>>
An object of a third aspect of the invention is to provide an interpolation device which can efficiently perform cubic spline interpolation.
(1) The first feature of the third aspect of the invention resides in an interpolation device which calculates a pixel value f(x) of a new pixel Q defined at an arbitrary position in an interpolation interval both ends of which are at two adjacent pixels by performing interpolation using a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning a position x based on a one-dimensional pixel array obtained by arranging a plurality of pixels having predetermined pixel values at predetermined pitches, comprising:
a data input unit which inputs pixel values of sample pixels and the position x of the new pixel Q when the two adjacent pixels are defined as interval terminal pixels, “a predetermined number of pixel or pixels to be used for interpolation” continuously arranged adjacent to each other on the left of the interpolation interval are defined as left side interpolation pixels, “a predetermined number of pixel or pixels to be used for interpolation” continuously arranged adjacent to each other on the right of the interpolation interval are defined as right side interpolation pixels, and S (S≧4) in total of pixels consisting of the interval terminal pixels, the left side interpolation pixel or pixels, and the right side interpolation pixel or pixels are defined as sample pixels;
a weighting coefficient storage unit which stores weighting factors for the sample pixels, indicating weighting according to the distances from the interpolation interval;
a left side influence factor arithmetic unit which calculates a left side influence factor indicating influences of a pixel value or values of the left side interpolation pixel or pixels on coefficients of the cubic polynomial based on the pixel value or values of the left side interpolation pixel or pixels and the weighting factors;
a right side influence factor arithmetic unit which calculates a right side influence factor indicating influences of a pixel value or values of the right side interpolation pixel or pixels on coefficients of the cubic polynomial based on the pixel value or values of the right side interpolation pixel or pixels and the weighting factors; and
an interpolated value arithmetic unit which calculates a pixel value f(x) by performing an arithmetic operation by determining the coefficients A, B, C, and D of the cubic polynomial by using the left side influence factor, the right side influence factor, and the weighting factors and substituting the position x for the cubic polynomial.
(2) The second feature of the third aspect of the invention resides in the interpolation device according to the first feature, wherein
the interpolated value arithmetic unit comprising:
a square/cube computing unit which calculates a square x2 and a cube x3 of the position x;
a cubic coefficient computing unit which calculates a cubic coefficient A of the cubic polynomial based on the pixel values of the interval terminal pixels, the left side influence factor, the right side influence factor, and the weighting factor;
a quadratic coefficient computing unit which calculates a quadratic coefficient B of the cubic polynomial based on the pixel values of the interval terminal pixels, the left side influence factor, the right side influence factor, and the weighting factor;
a primary coefficient computing unit which calculates a primary coefficient C of the cubic polynomial based on the pixel values of the interval terminal pixels, the left side influence factor, the right side influence factor, and the weighting factor;
a cubic coefficient multiplier which calculates Ax3 by multiplying the cubic coefficient A and the cube x3;
a quadratic coefficient multiplier which calculates Bx2 by multiplying the quadratic coefficient B and the square x2;
a primary coefficient multiplier which calculates Cx by multiplying the primary coefficient C and the position x; and
an adder which performs addition of Ax3+Bx2+Cx+D by defining a pixel value of one interval terminal pixel as the coefficient D.
(3) The third feature of the third aspect of the invention resides in the interpolation device according to the first or second feature, wherein
when the 0-th pixel P0 and the first pixel P1 adjacent to each other in a one-dimensional pixel array are defined as interval terminal pixels, and concerning a predetermined integer i (i≧1), i in total of pixels from the (−i)th pixel P−i to the (−1)st pixel P−1 are defined as a left side interpolation pixel or pixels, i in total of pixels from the second pixel P2 to the (i+1)th pixel Pi+1 are defined as a right side interpolation pixel or pixels, S=2(i+1) sample pixels are defined, a position of the interval terminal pixel P0 is set to 0 and a position of the interval terminal pixel P1 is set to 1, and a position x of a new pixel Q is defined in a range of 0≦x<1,
the data input unit inputs pixel values f0 and f1 of the interval terminal pixels P0 and P1, pixel values f−i through f−1 of the left side interpolation pixels P−i through P−1, pixel values f2 through fi+1 of the right side interpolation pixels P2 through Pi+1, and the position x of the new pixel Q,
the weighting factor storage unit stores values of weighting factors w1, w2, w3, . . . , wi expressed by a recurrence formula of wi=4−(1/wi−1) (wherein w0=2),
the left side influence factor arithmetic unit calculates a left side influence factor g−1 based on an arithmetic expression of:
g −1=−1/w i−1( . . . (−1/w 2(−1/w 1(−¼·f −1 +f −i+1)+f −i+2) . . . )+f −i+(i−2))+f −i+(i−1)
the right side influence factor arithmetic unit calculates a right side influence factor g2 based on an arithmetic expression of:
g 2=−1/w i−1( . . . (−1/w 2(−1/w 1(−¼·f i+1 +f i)+f i−1) . . . )+f i+(3−i))+f i+(2−i)
and the interpolated value arithmetic unit calculates the pixel value f(x) based on an arithmetic expression (w=wi−1):
f(x)=1/(3w−1){−6g −1+(3w+1)(f 0 −f 1)+6g 2 }x 3+3/((5w−1)(3w−1)){6(4w−1)g −1−(9w 2+2w−1)f 0+6w 2 f 1−6wg 2 }x 2+3/((5w−1)(3w−1)){−2(7w−2)g −1−(w 2−4w+1)f 0+2w(2w−1)f 1−2(2w−1)g 2 }x+f 0
(4) The fourth feature of the third aspect of the invention resides in the interpolation device according to the first feature, wherein
the interpolated value arithmetic unit determines auxiliary coefficients a, b, c, and d defined so as to satisfy A=(a−b), B=(b−c), C=(c−d), and D=d instead of the coefficients A, B, C, and D by using the pixel values of the interval terminal pixels, the left side influence factor, the right side influence factor, and the weighting factor, and calculates the pixel value f(x) by an arithmetic operation by a cubic polynomial of “f(x)=[{ax+b(1−x)}x+c(1−x)]x+d(1−x)” for which the position x is substituted instead of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D.”
(5) The fifth feature of the third aspect of the invention resides in the interpolation device according to the fourth feature, wherein
the interpolated value arithmetic unit comprising:
a first auxiliary coefficient computing unit which calculates an auxiliary coefficient b based on the pixel values of the interval terminal pixels, the left side influence factor, the right side influence factor, and the weighting factor;
a second auxiliary coefficient computing unit which calculates an auxiliary coefficient c based on the pixel values of the interval terminal pixels, the left side influence factor, the right side influence factor, and the weighting factor;
a first linear interpolator which calculates a value of {ax+b(1−x)} by inputting a pixel value of one interval terminal pixel as an auxiliary coefficient a and performing linear interpolation based on the auxiliary coefficient a, the auxiliary coefficient b, and the position x;
a second linear interpolator which calculates a value of [{ax+b(1−x)}x+c(1−x)] by performing linear interpolation based on the value of {ax+b(1−x)}, the auxiliary coefficient c, and the position x; and
a third linear interpolator which calculates a value of f(x)=[{ax+b(1−x)}x+c(1−x)]x+d(1−x) by inputting a pixel value of the other interval terminal pixel as the auxiliary coefficient d and performing linear interpolation based on the value of [{ax+b(1−x)}x+c(1−x)], the auxiliary coefficient d, and the position x.
(6) The sixth feature of the third aspect of the invention resides in the interpolation device according to the fourth or fifth feature, wherein
when the 0-th pixel P0 and the first pixel P1 adjacent to each other in a one-dimensional pixel array are defined as interval terminal pixels, and concerning a predetermined integer i (i≧1), i in total of pixels from the (−i)th pixel P−i to the (−1)st pixel P−1 are defined as a left side interpolation pixel or pixels, i in total of pixels from the second pixel P2 to the (i+1)th pixel Pi+1 are defined as a right side interpolation pixel or pixels, S=2(i+1) sample pixels are defined, a position of the interval terminal pixel P0 is set to 0 and a position of the interval terminal pixel P1 is set to 1, and a position x of a new pixel Q is defined in a range of 0≦x<1,
the data input unit inputs pixel values f0 and f1 of the interval terminal pixels P0 and P1, pixel values f−i through f−1 of the left side interpolation pixels P−i through P−1, pixel values f2 through fi+1 of the right side interpolation pixels P2 through Pi+1, and the position x of the new pixel Q,
the weighting factor storage unit stores values of weighting factors w1, w2, w3, . . . , wi expressed by a recurrence formula of wi=4−(1/wi−1) (where w0=2),
the left side influence factor arithmetic unit calculates a left side influence factor g−1 based on an arithmetic expression of:
g −1=−1/w i−1( . . . (−1/w 2(−1/w 1(−¼·f −1 +f −i+1)+f −i+2) . . . )+f −i+(i−2))+f −i+(i−1)
the right side influence factor arithmetic unit calculates a right side influence factor g2 based on an arithmetic expression of:
g 2=−1/w i−1( . . . (−1/w 2(−1/w 1(−¼·f i+1 +f i)+f i−1) . . . )+f i+(3−i))+f i+(2−i)
and the interpolated value arithmetic unit determines auxiliary coefficients a, b, c, and d by an arithmetic expression (w=wi−1):
a=f1,
b=(6g −1−(3w+1)f 0+6wf 1−6g 2)/(3w−1),
c=1/((3w−1)(5w−1))(−6(7w−2)g −1+2(6w 2+2w−1)f 0+6w(2w−1)f 1−6(2w−1)g 2),
d=f0
(7) The seventh feature of the third aspect of the invention resides in the interpolation device according to the first to the sixth features, further comprising:
an image input unit which inputs a two-dimensional pixel array obtained by arranging a plurality of pixels having predetermined pixel values, respectively, at intersections between row lines arranged parallel to each other at predetermined pitches and column lines orthogonal to the row lines;
an image storage unit which stores the two-dimensional pixel array;
a new pixel position input unit which inputs a position of a new pixel T defined at a position different from the positions of the pixels on the two-dimensional pixel array;
an interpolation target intersection determining unit which defines a reference line R that passes through the position of the new pixel T and is parallel to the column lines, and determines a predetermined number of intersections near the new pixel T among intersections between the reference line R and the row lines as interpolation target intersections;
a row direction calculation control unit which makes the interpolation device calculate pixel values of the interpolation target intersections by giving one-dimensional pixel arrays on row lines, to which the interpolation target intersections belong, to the interpolation device; and
a column direction calculation control unit which makes the interpolation device calculate the pixel value of the new pixel T by giving a one-dimensional pixel array consisting of the interpolation target intersections arranged on the reference line R to the interpolation device,
to perform interpolation for a two-dimensional image.
(8) The seventh feature of the third aspect of the invention resides in the interpolation device according to the first to the seventh features,
one or a plurality of the left side influence factor arithmetic unit, the right side influence factor arithmetic unit, and the interpolated value arithmetic unit are constituted by a plurality of computing units, and computing units on a middle stage calculate results of operations of bits obtained without carry propagation and carry information of the bits and gives these to a later computing unit, and in a last computing unit, carry propagation is performed by considering the carry information.
As described above, according to the third aspect of the invention, it becomes possible to efficiently perform cubic spline interpolation, and an interpolation device suitable for image enlargement/reduction scaling display can be provided.
BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is a graph showing a basic concept of linear interpolation;
FIG. 2 is a diagram illustrating a method for expressing a negative number by means of twos complement notation;
FIG. 3 is a diagram showing expressions that define interpolation target values A and B and an interpolation rate D by using bits of binary numbers;
FIG. 4 is a diagram showing a basic expression of interpolation to be executed by a linear interpolator according to a first aspect of the invention;
FIG. 5 is a diagram showing a transformation process of the basic expression shown in FIG. 4;
FIG. 6 is another diagram showing the transformation process of the basic expression shown in FIG. 4;
FIG. 7 is a diagram showing an arithmetic expression characterizing the first aspect of the invention obtained by transforming the basic expression shown in FIG. 4;
FIG. 8 is a diagram showing an expression obtained by substituting n=4 into the arithmetic expression shown in FIG. 7;
FIG. 9 is a diagram showing bit configuration of digital data relating to an arithmetic operation based on the expression shown in FIG. 8;
FIG. 10 is a diagram showing expressions in which the arithmetic operator Σ in the expression shown in FIG. 8 is developed;
FIG. 11 is a table showing coefficients of members composing an unsigned part 2 of the expression shown in FIG. 10, sorted by digits;
FIG. 12 is a table showing coefficients of members composing the expression shown in FIG. 10, sorted by digits;
FIG. 13 are diagrams showing basic components of a linear interpolator according to an embodiment of the first aspect of the invention;
FIG. 14 is a circuit diagram showing a first portion of the linear interpolator according to an embodiment of the first aspect of the invention;
FIG. 15 is a circuit diagram showing a second portion of the linear interpolator according to an embodiment of the first aspect of the invention;
FIG. 16 is a circuit diagram showing a third portion of the linear interpolator according to an embodiment of the first aspect of the invention;
FIG. 17 is a block diagram showing an entire constitution of the linear interpolator according to the first aspect of the invention;
FIG. 18 is a graph showing another basic concept of linear interpolation;
FIG. 19 is a diagram showing expressions defining interpolation target values A and B and interpolation rate determining value D as 8-bit digital data;
FIG. 20 is a diagram showing a state in that the interpolation target values A and B, the interpolation rate determining value D, and the linear interpolated value C shown in FIG. 19 are expressed as bit sequences of binary numbers;
FIG. 21 is a diagram showing a relational expression between the interpolation rate determining value D composed of 8-bit digital data and an interpolation rate α;
FIG. 22 is a diagram showing a method for defining an approximate value α′ of the interpolation rate α based on the relational expression of FIG. 21;
FIG. 23 is a diagram showing an expression for defining the approximate value α′ of the interpolation rate α based on the relational expression of FIG. 21;
FIG. 24 is a diagram showing a basic expression of linear interpolation when 8-bit digital data are used for the interpolation target values A and B and the interpolation rate determining value D;
FIG. 25 is a diagram showing logical formulas to be used for transforming the basic expression shown in FIG. 24;
FIG. 26 is a diagram showing a transformation process of the part of (1−α′) of the basic expression shown in FIG. 24;
FIG. 27 is a diagram showing a characteristic expression according to a second aspect of the invention obtained by substituting the expression of FIG. 26 for the part of (1−α′) of the basic expression shown in FIG. 24;
FIG. 28 is a table showing coefficients of members composing the expression shown in FIG. 27, sorted by digits;
FIG. 29 is a diagram showing general formulas defining the interpolation target values A and B and the interpolation rate determining value D as n-bit digital data;
FIG. 30 is a diagram showing a state in that the interpolation target values A and B, the linear interpolated value C, and the interpolation rate determining value D shown in FIG. 29 as bit sequences of binary numbers;
FIG. 31 is a diagram showing a relational expression between the interpolation rate determining value D composed of n-bit digital data and an interpolation rate α;
FIG. 32 is a method for defining an approximate value α′ of the interpolation rate α based on the relational expression of FIG. 31;
FIG. 33 is a diagram showing an expression defining the approximate value α′ of the interpolation rate α based on the relational expression of FIG. 31;
FIG. 34 is a diagram showing a basic expression of linear interpolation when n-bit digital data are used for the interpolation target values A and B and the interpolation rate determining value D;
FIG. 35 is a diagram showing a transformation process of the part of (1−α′) of the basic expression shown in FIG. 34;
FIG. 36 is a diagram showing a characteristic expression according to a second aspect of the invention obtained by substituting the expression shown in FIG. 35 for the part of (1−α′) of the basic expression shown in FIG. 34;
FIG. 37 is an expression obtained by substituting n=4 into the characteristic expression of FIG. 36;
FIG. 38 is a table showing coefficients composing the expression shown in FIG. 37, sorted by digits;
FIG. 39 are diagrams showing basic components of a linear interpolator according to an embodiment of the second aspect of the invention;
FIG. 40 is a circuit diagram showing a first portion of the linear interpolator according to an embodiment of the second aspect of the invention;
FIG. 41 is a circuit diagram showing a second portion of the linear interpolator according to an embodiment of the second aspect of the invention;
FIG. 42 is a circuit diagram showing a third portion of the linear interpolator according to an embodiment of the second aspect of the invention;
FIG. 43 is a block diagram showing an entire constitution of the linear interpolator according to the second aspect of the invention;
FIG. 44 is a diagram showing a concept of simple linear interpolation based on pixel values of two sample pixels on both ends of an interpolation interval;
FIG. 45 is a diagram showing a concept of an interpolation method based on pixel values of four in total of sample pixels;
FIG. 46 is a diagram showing a concept of an interpolation method based on pixel values of six in total of sample pixels;
FIG. 47 is a diagram showing a concept of an interpolation method based on pixel values of 2(i+1) in total of sample pixels;
FIG. 48 is a diagram showing a correlation of numerical expressions to be used as a basic principle of an interpolation device according to a third aspect of the invention;
FIG. 49 is a block diagram showing a constitution of the interpolation device according to a basic embodiment of the third aspect of the invention;
FIG. 50 is a block diagram showing a constitution example of an interpolated value arithmetic unit 840 in the interpolation device shown in FIG. 49;
FIG. 51 is a block diagram showing another constitution example of the interpolated value arithmetic unit 840 in the interpolation device shown in FIG. 49;
FIG. 52 is a block diagram showing a constitution of an interpolation device having a function to perform interpolation about a two-dimensional image;
FIG. 53 is a plan view showing a two-dimensional pixel array for describing operations of the interpolation device of FIG. 52; and
FIG. 54 is another plan view showing a two-dimensional pixel array for describing operations of the interpolation device of FIG. 52.
DISCLOSURE OF PREFERRED EMBODIMENTS Section 1 Embodiment of First Aspect of the Invention
Here, an embodiment illustrating a first aspect of the invention will be described. The first aspect of the invention proposes a technique for efficiently executing linear interpolation of two signed interpolation target values by a simple constitution.
Section 1-1 Linear Interpolation and Negative Number Expression
First, a basic concept of general linear interpolation and general negative number expression in digital data will be briefly described. FIG. 1 is a graph showing a basic concept of linear interpolation. In the illustrated example, an interval 0-1 is set on the X axis and a method for calculating a function value at an arbitrary position x in the interval 0-1 when a value of the function f(x) is defined only at both ends of the interval is shown. In detail, when f(0)=A and f(1)=B, a value of a function f(x)=C about arbitrary x in the range of 0≦x≦1 is calculated by interpolation using values A and B.
In linear interpolation, as illustrated, a straight line is defined between the point A and the point B, and a value of C is calculated as a vertical coordinate value of the point C on this straight line. Herein, when the position of the arbitrary point x is defined as “a position dividing the interval 0-1 at a ratio of [D:(1−D)],” and D is referred to as interpolation rate, the linear interpolated value C can be calculated by an arithmetic operation expressed by the basic expression of C=(1−D)*A+D*B by using the two interpolation target values A and B and the interpolation rate D (* sign representing multiplication). In this application, an interpolation rate in the range of (0≦D<1) that does not include 1 with respect to a value α (0≦α≦1) showing the original interpolation rate will be indicated as a value D. In this embodiment, for the sake of performing an arithmetic operation using digital data, the value D excluding 1 will be used as the interpolation rate.
For example, when concrete values of A=10, B=20, and D=0.4 are given, by performing an arithmetic operation of C=(1−0.4)*10+(0.4*20) based on the basic expression, an interpolated value of C=14 is calculated. As a matter of course, said basic expression is effective even if either one or both of A and B are negative numbers. For example, when concrete values of A=−10, B=20, and D=0.4 are given, by performing an arithmetic operation of C=(1−0.4)*(−10)+(0.4*20) based on the basic expression, an interpolated value of C=+2 is calculated.
Next, general negative number expression in digital data will be described. FIG. 2 is a diagram describing a method for expressing a negative number by means of twos complement notation. The table on the left in FIG. 2 shows correspondence between numerical values in the decadal system and numerical values in the binary system. As illustrated, by using 4-bit binary numbers, 15 numerical values from +7 to −7 including 0 can be expressed. Expression in the binary system of positive numbers can be realized by adding ones to “0000” in sequence, and expression in the binary system of negative numbers can be realized by subtracting ones from “0000” one by one. This negative number expression method is called “twos complement notation,” and this is a general negative number expression method in digital data.
On the right in FIG. 2, correspondence between bits of a positive number and bits of a negative number is shown. Namely, when a positive number “A” is expressed by a 4-bit sequence of “a3a2a1a0,” the negative number “−A” is expressed as a binary number to which 1 has been added after logically inverting this 4 bits “a3a2a1a0.” For example, binary expression of “+7” is “0111,” so that the binary expression of “−7” becomes “1001” which is a result of addition of 1 to “1000” obtained by inverting all bits of “0111.” A positive number and a negative number whose absolute values are the same always satisfy this relationship when they are binary-expressed.
As clearly understood from the table on the left in FIG. 2, when a negative number is binary-expressed, the most significant bit of the binary expression is always “1.” In other words, in the case of a signed binary number, the most significant bit thereof can be recognized as “signed part” showing a sign, and other bits can be recognized as “unsigned part” showing an absolute value. As illustrated, in a case of a binary number having a sign consisting of four bits in total, the leftmost one bit is the “signed part” and the remaining 3 bits are the “unsigned part.” When the “signed part” is “0,” the binary number is 0 or a positive number, and when the “signed part” is “1,” the binary number thereof is a negative number.
Thus, a signed binary number always includes “signed part” and “unsigned part,” so that handling thereof is slightly troublesome. As described above, interpolators which can perform linear interpolation of two signed interpolation target values have been conventionally proposed. However, these conventional interpolators employ complicated circuitry for handling signed binary numbers, so that they cannot efficiently perform an arithmetic operation. The first aspect of the invention proposes a new method for efficiently performing linear interpolation of signed interpolation target values by a simple constitution.
Section 1-2 Transformation of Basic Expression
It is described above that the linear interpolation of two interpolation target value A and B is performed as an arithmetic operation based on the basic expression of C=(1−D)*A+D*B when the interpolation rate is defined as D (0≦D<1). Then, first, let us consider application of this basic expression to digital data expressed by binary numbers.
FIG. 3 is a diagram showing expressions in which signed interpolation target values A and B and an interpolation rate D in the range of 0≦D<1 are defined by using bits of binary numbers consisting of n bits. On the top of FIG. 3, first,
$( Expression 1 - 1 ) : A = - a n - 1 2 n - 1 + a n - 2 2 n - 2 + … + a 1 2 1 + a 0 2 0 = - a n - 1 2 n - 1 + ∑ j = 0 _n - 2 a j 2 j$
is shown (in the claims and specification of the present application, due to a constraint based on the functions of a word processor, the parameter a that should be entered under the arithmetic operator Σ indicating a sum and the parameter β that should be entered above Σ are entered as subscripts like Σα−β). an−1, an−2, . . . , a1, and a0 in this (Expression 1-1) correspond to the respective bits (0 or 1) when the numerical value A is expressed as a binary number composed of n bits, and an−1 is the most significant bit (the leftmost bit), and a0 is the least significant bit (the rightmost bit). As described above, the most significant bit composes a signed part, so that “−an−12n−1” as the first member of the left-hand side of the (Expression 1-1) is a signed part, and the remaining member “Σj=0˜n−2aj2j” is an unsigned part.
On the lower right of the (Expression 1-1), a bit configuration when n=4, that is, when the numerical value A is expressed as a binary number composed of 4 bits, is shown. In this case, the most significant bit “a3” becomes a signed part, and the part of the remaining 3 bits “a2a1a0” becomes an unsigned part. The meaning of this “Expression 1-1” will be easily understood by applying concrete numerical values. For example, when A=+7 is binary-expressed, it becomes “0111” from the table of FIG. 2. Namely, a3=0, a2=1, a1=1, and a0=1, and when these are applied to the (Expression 1-1) to which n=4 is applied, it becomes:
A=−0*23+1*22+1*21+1*20=+7
On the other hand, in a case of a negative number, when A=−7 is binary-expressed, it becomes “1001” from the table of FIG. 2, so that a3=1, a2=0, a1=0, and a0=1, and when these are applied to the (Expression 1-1) to which n=4 is applied, it becomes:
A=−1*23+0*22+0*21+1*20=−7
Marking the member of the signed part with a minus sign is to obtain a correct value in the case of such a negative number.
Similarly, on the middle of FIG. 3, (Expression 1-2):
$B = - b n - 1 2 n - 1 + b n - 2 2 n - 2 + … + b 1 2 1 + b 0 2 0 = - b n - 1 2 n - 1 + ∑ j = 0 _n - 2 b j 2 j$
is shown. bn−1, bn−2, . . . , b1, b0 in this (Expression 1-2) correspond to the respective bits (0 or 1) when the numerical value B is expressed as a binary number composed of n bits, and bn−1 is the most significant bit (the leftmost bit), and b0 is the least significant bit (the rightmost bit). The most significant bit sill composes a signed part, so that “−bn−12n−1” as the first member of the left-hand side of (Expression 1-2) becomes a signed part, and the remaining members “Σj=0˜n−2bj2j” becomes an unsigned part.
On the lower right of the (Expression 1-2), bit configuration when n=4, that is, when the numerical value B is expressed as a binary number composed of 4 bits is shown. In this case, the most significant bit “b3” becomes a signed part, and the part of the remaining 3 bits “b2b1b0” becomes an unsigned part.
On the other hand, on the lowest stage of FIG. 3, (Expression 1-3):
$D = d n - 1 2 - 1 + d n - 2 2 n - 2 + … + d 0 2 - n = ∑ i = 0 _n - 1 d i 2 ( i - n )$
is shown. dn−1, dn−2, . . . , d0 in this “Expression 1-3” correspond to the respective bits (0 or 1) when the numerical value D is expressed as a binary number composed of n bits that indicates only the decimal part, and dn−1 is the most significant bit (the leftmost bit) and do is the least significant bit (the rightmost bit).
On the lower right of the (Expression 1-3), bit configuration when n=4, that is, when the numerical value D is expressed as a binary number composed of 4 bits that indicate only a decimal part is shown. As illustrated, this 4 bits become bits of digits after the decimal point, and for example, the bit d3 is a value of the digit of the first place of the decimal place (the digit of 2−1), and bit d2 is a value of the digit of the second place of the decimal place (the digit of 2−2). The numerical value D is an interpolation rate in the range of 0≦D<1, so that it is expressed as an unsigned binary number. For example, a binary number of 4 bits “1111” normally indicates a decimal number “15,” however, when the numerical value D is expressed as a binary number of 4 bits “1111,” this 4 bits indicates only a decimal part, and in actuality, it corresponds to a binary number of “0.1111” (corresponding to ½+¼+⅛+ 1/16=0.9375). Therefore, when n=4, D takes a value of 0 to 0.9375.
By (Expression 1-1), (Expression 1-2), and (Expression 1-3) of FIG. 3, the interpolation target values A and B and the interpolation rate D can be defined as digital data, so that when these values are substituted for the basic expression “C=(1−D)*A+D*B,” (Expression 1-4) shown in the upper stage of FIG. 4 is obtained. The part enclosed by a dashed line in this (Expression 1-4) corresponds to the members of the basic expression, in which the (Expression 1-1) through (Expression 1-3) are applied. By developing and rearranging this (Expression 1-4), (Expression 1-5) is obtained. Therefore, by constituting an arithmetic circuit which successively performs arithmetic operations based on the (Expression 1-5), an interpolator for calculating an interpolated value C can be manufactured.
However, if an interpolator for calculating an interpolated value C based on (Expression 1-5) is designed, the constitution becomes complicated, and efficient arithmetic operations cannot be expected. A first reason for this is that some device becomes necessary for properly processing the signed part. As described above, the members an−1 and bn−1 are signed parts of the interpolation target values A and B, and do not indicate absolute values of the interpolation target values. Therefore, proper handling as signed parts is necessary. A second reason is that complicated circuitry using many multipliers and adders becomes necessary since a product of arithmetic operators Σ indicating sums is included. As long as a serial operation based on this (Expression 1-5) is performed, the circuitry dramatically becomes complicated as bit numbers of the interpolation target values A and B, that is, the number of n increases.
The inventor of the present invention considered that transformation of this (Expression 1-5) into another form would make it possible to efficiently perform the arithmetic operation by simple circuitry. Therefore, taking concrete circuitry as an interpolator into consideration, the inventor examined how to transform the (Expression 1-5) to constitute an efficient interpolator, and through trial and error, conceived that the following transformation made it possible to constitute a very efficient interpolator.
First, the (Expression 1-6) of FIG. 5 is considered. This (Expression 1-6) in itself is a formula known mathematically from a long time ago For example, when n=4, the (Expression 1-6) becomes as follows:
$2 - 4 + 2 - 3 + 2 - 2 + 2 - 1 + 2 - 4 = 1 / 16 + 1 / 8 + 1 / 4 + 1 / 2 + 1 / 16 = 1$
As a matter of course, this (Expression 1-6) is a formula satisfied concerning an arbitrary n. By shifting 1 on the right-hand side of this (Expression 1-6) to the left-hand side, the (Expression 1-7) is obtained.
Next, let us consider the (Expression 1-8). The first line of this (Expression 1-8) is a perfectly natural expression of “−D=0−D,” and when the left-hand side of the (Expression 1-7) is substituted for the part of “0” and the (Expression 1-3) is substituted for the part of “D,” second and third lines of (Expression 1-8) are obtained. By transforming these, the form of the second line is further derived from the first line of the (Expression 1-9). di is each bit for indicating only the decimal part of the interpolation rate D, so that as an actual value, either 0 or 1 is taken. Therefore, when ei=(1−di), ei is a logically inverted bit of di (when di=1, ei=0, and when di=0, ei=1). Therefore, the (Expression 1-9) is transformed into the (Expression 1-10) by rewriting the logically inverted bit of di as ei.
Herein, focusing attention on the part being “−D,” enclosed by the dashed line in the (Expression 1-5) of FIG. 4, when the right-hand side of the (Expression 1-10) is substituted for this part, the (Expression 1-11) of FIG. 6 is obtained. In this (Expression 1-11), the part enclosed by the dashed line corresponds to “−D,” and corresponds to the right-hand side of the (Expression 1-10). On the other hand, the part enclosed by an alternate long and short dashed line shown on the first line of the (Expression 1-11) can be transformed into the form of the member T1 shown on the lowest left of the diagram (the member 2n−1 can be incorporated in the inside of the Σ sign). The product of two parts enclosed by the alternate long and short double dashed line shown on the third line of the (Expression 1-11) can be transformed into the form of the member T2 shown on the lowest right part of the diagram (also, the member 2n−1 is incorporated in the inside of the Σ sign).
By thus transforming the (Expression 1-11) of FIG. 6, the (Expression 1-12) of FIG. 7 is obtained. This (Expression 1-12) is equivalent to the basic expression of linear interpolation shown as (Expression 1-4) in FIG. 4, and the form of this (Expression 1-12) is very convenient for designing the interpolator. Hereinafter, this (Expression 1-12) is referred to as “arithmetic expression according to the first aspect of the invention.” A concrete interpolator constituted so as to obtain a linear interpolated value C based on the arithmetic expression of this (Expression 1-12) will be illustrated in Section 1-4 later.
Section 1-3 Features of Arithmetic Expression According to the First Aspect of the Invention
The arithmetic expression (Expression 1-12) according to the first aspect of the invention is composed of three parts of a signed part, an unsigned part 1, and an unsigned part 2 as distinguished by circling these by dashed lines in the diagram, and as described later, these perform individually unique functions. First, observing the entire composition of this (Expression 1-12), it is an expression of the interpolated value C to be calculated as a sum of a plurality of members, and it is understood that each member is in a form of the power of two multiplied by some coefficient. Herein, the power of two indicates a specific digit of a binary number, and the coefficient to multiply the power of two indicates a numerical value of this digit. For example, 2n−1 indicates a digit of the most significant bit of the binary number indicating the interpolated value C, corresponding to the signed part of the interpolated value C.
In FIG. 7, the parts enclosed by the dashed lines shown as “signed part” are members in the form of “−(an−1+bn−1)2n−1,” and this indicates that the bit value of the digit shown as “n−1” (digit of the most significant bit of the interpolated value C) is “−(an−1+bn−1).” Herein, the minus sign attached to the head is for making this bit to function as a signed part indicating a sign as in the case of the (Expression 1-1) and (Expression 1-2) of FIG. 3. Namely, when a negative number is expressed by twos complement notation, this negative number is obtained by subtracting a numerical value indicated by the most significant bit from the numerical value indicated by the remaining bits. For example, when “−7” as a decimal number is binary-expressed, it becomes “1001,” however, the numerical value indicated by the most significant bit “1” is 23=8, and the numerical value indicated by the remaining 3 bits “001” is 1, so that −8+1=−7 is obtained.
Meanwhile, an−1 is a signed part (most significant bit) of the interpolation target value A, and bn−1 is a signed part (most significant bit) of the interpolation target value B, so that the part enclosed by a dashed line marked “signed part” in FIG. 7 functions to set the sum of the signed parts of both interpolation target values A and B to a signed part (most significant bit) of the interpolated value C.
The part enclosed by the dashed line marked “unsigned part 1” in FIG. 7 is also a part composed by an arithmetic operation based on an−1 and bn−1 (signed parts of the interpolation target values A and B), and the bits after an−2 and bits after bn−2 have no relation thereto. The member of the power of two included in this unsigned part 1 is a member of 2n−2 at a maximum (member of 2(i−1) in the case of i=n−1 in an arithmetic operation for calculating the sum total of Σ), so that the signed part (digit of 2n−1) of the interpolated value is not directly defined by the arithmetic operation of this unsigned part 1, however, when carrying from the digit of 2n−2 occurs, it influences the signed part (digit of 2n−1) of the interpolated value C. In this view, this unsigned part 1 functions to perform an arithmetic operation which may influence the signed part of the interpolated value C.
On the other hand, the part marked “unsigned part 2” in FIG. 7 is all composed of only an arithmetic operation concerning bits after an−2 and bits after bn−2, and includes no arithmetic operation concerning an−1 and bn−1 (signed parts of the interpolation target values A and B). In other words, this unsigned part 2 functions to perform an arithmetic operation concerning only the unsigned parts of the interpolation target values A and B.
In the (Expression 1-12) shown in FIG. 7, the functions of the three parts of the signed part, the unsigned part 1, and the unsigned part 2 are clearly distinguished, so that if an interpolator is constituted based on this expression, the signed part can be properly processed. This is the first feature of the (Expression 1-12).
A second feature of the (Expression 1-12) is that di and ei are complementarily incorporated in the form of the expression. For example, the unsigned part 1 includes an expression in the form of “an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1),” and the unsigned part 2 includes an expression in the form of “Σi=0˜n−1ei2(i−n)·Σj=0˜n−2aj2ji=0˜n−1di2(i−n)·Σj=0˜n−2bj2j.” As described above, ei is a logically inverted bit of di, so that when di equals 1, ei is always 0, and when di is 0, ei is always 1. Therefore, when calculating the respective members of the expression, in the case of di=1, the member including ei is always 0, so that only the member including di is only calculated, and in the case of di=0, the member including di is always 0, so that only the member including ei is only calculated. Thereby, an efficient arithmetic operation using a simplified arithmetic circuit becomes possible.
Next, to make clearer the advantage of the arithmetic expression according to the first aspect of the invention, a concrete example when n=4 in the (Expression 1-12) shown in FIG. 7 is considered. The (Expression 1-13) shown in FIG. 8 is obtained by substituting 4 for n in the (Expression 1-12) shown in FIG. 7. This expression corresponds to an arithmetic expression when the interpolation target values A and B and the interpolation rate D are all given as 4-bit digital data. Namely, the interpolation target value A is given as 4-bit data “a3a2a1a0,” the interpolation target value B is given as 4-bit data “b3b2b1b0,” and the interpolation rate D is given as 4-bit data “d3d2d1d0” (see the diagrams on the lower rights of the expressions of FIG. 3). In this (Expression 1-13), focusing attention on the members of the power of two, the maximum member is the member of 23, and the minimum is the member of 2−4, and the interpolated value C calculated as a result of arithmetic operation is given as data of 8 bits in total indicating the values of the digits of 23, 22, 21, 20, 2−1, 2−2, 2−3, and 2−4. Herein, the individual data of 8 bits are expressed as c7, c6, c5, c4, c3, c2, c1, and c0.
FIG. 9 is a diagram showing bit configuration of digital data concerning an arithmetic operation based on the expression shown in FIG. 8. Namely, the arithmetic operation shown in FIG. 8 is for calculating a linear interpolated value to be given as 8-bit data “c7c6c5c4c3c2c1c0” based on the interpolation target value A given as 4-bit data “a3a2a1a0,” the interpolation target value B given as 4-bit data “b3b2b1b0,” and the interpolation rate D given as 4-bit data “d3d2d1d0” (data showing only a decimal part including a decimal point positioned on the left of the bit d3). As described above, the bit c4 is a value of the digit of 20, and the bit c3 is the value of the digit of 2−1, so that a decimal point is positioned between the bit c4 and the bit c3. Therefore, as shown on the lower stage of FIG. 9, among the data of 8 bits of the interpolated value C, the bit “c7” is a signed part, the bits “c6c5c4” are an integer part, and the bits “c3c2c1c0” is a decimal part.
The linear interpolated value C thus obtained does not need to be outputted as a result of arithmetic operation left as 8-bit data, and only necessary significant digits are properly selected and outputted. In practical use, the interpolation target values A and B are given as 4-bit data, so that in many cases, the interpolated value C to be finally outputted being also 4-bit data is sufficient. In this case, the decimal part is rounded off (or rounded up) and 4-bit data composed of only the signed part and the integer part is outputted.
Next, it will be described in detail how efficient arithmetic operation the designing of an interpolator based on the arithmetic expression shown in FIG. 8 can be realized. The (Expression 1-14) shown in FIG. 10 is a result of development of the arithmetic operator Σ in the (Expression 1-13) of FIG. 8. The signed part of the (Expression 1-14) is the same as that of the (Expression 1-13), however, the unsigned part 1 is developed to three lines as illustrated. Herein, comparing the second line and the third line, these have a difference that the second line is an expression concerning a3 and di (i=0, 1, 2, 3), and on the other hand, the third line is an expression concerning b3 and ei (i=0, 1, 2, 3), however, their forms of expressions are completely the same. The unsigned part 2 is developed to the circled numbers 1 through 9 as illustrated. From the comparison between the circled numbers 2 and 6, the circled numbers 3 and 7, the circled numbers 4 and 8, and the circled numbers 5 and 9, respectively, they have a difference that the former numbers are expressions concerning aj (j=0, 1, 2) and ei (i=0, 1, 2, 3), and the latter numbers are expressions concerning bj (j=0, 1, 2) and di (i=0, 1, 2, 3), however, their forms of expressions are completely the same.
FIG. 11 is a table sorting the coefficients of the members of the unsigned part 2 (expressions of the circled numbers 1 through 9) by digits. The columns arranged horizontally in this table correspond to scaling positions of 23, 22, 21, 20, 2−1, 2−2, 2−3, and 2−4, respectively, and the bits c7, c6, c5, c4, c3, c2, c1, and c0 indicate the respective bits of an interpolated value C calculated by the arithmetic operation. As described above, a decimal point is positioned between the bit c4 and the bit c3, and in 8-bit data of the interpolated value C, the bit “c7” is a signed part, the bits “c6c5c4” are an integer part of an unsigned part, and the bits “c3c2c1c0” are a decimal part of the unsigned part.
On the first row of the table of FIG. 11, coefficients of the expression of the circled number 5 of the (Expression 1-14) of FIG. 10 are described. Namely, the coefficient “a0e3” in the expression “a0e32−1+a1e320+a2e321” of the circled number 5 is described on the digit of 2−1, and the coefficient “a1e3” is described on the digit of 20, and the coefficient “a2e3” is described on the digit of 21. Similarly, the coefficients of the expression of the circled number 9 are described on the second row of the table of FIG. 11, the coefficients of the expression of the circled number 4 are described on the third row, the coefficients of the expression of the circled number 8 are described on the fourth row, the coefficients of the expression of the circled number 3 are described on the fifth row, and the coefficients of the expression of the circled number 7 are described on the sixth row, the coefficients of the expression of the circled number 2 are described on the seventh row, the coefficients of the expression of the circled number 6 are described on the eighth row, and the coefficients of the expression of the circled number 1 are described on the ninth row.
The coefficients a0, a1, a2, b0, b1, and b2 described in the table of FIG. 11 are any bits of the unsigned parts of the interpolation target values A and B, and can be regarded as bits indicating absolute values (in other words, the signed parts a3 and b3 of the interpolation target values A and B are not included in the table of FIG. 11). Therefore, concerning these bits, the part of the absolute value of the interpolated value C can be calculated by performing only addition without considering signs.
Conveniently, in the coefficients of the same digit, many pair of complementary coefficients are included. For example, on the digit of 2−4, three coefficients of a0e0 on the seventh row, b0d0 on the eighth row, and a0 on the ninth row are described, and the bit c0 can be calculated by summing up these three coefficients. However, ei is logically inverted bit of di, a0e0 on the seventh row and b0d0 on the eighth row compose a pair of complementary coefficients, and either one is always 0. For example, when e0=1, d0=0, so that the bit c0 is calculated only by performing an arithmetic operation to add a0 on the seventh row to a0 on the ninth row. To the contrary, when e0=0, d0=1, so that the bit c0 is calculated only by performing an arithmetic operation to add b0 on the eighth row to a0 on the ninth row.
Resultantly, to calculate the bit c0, either the coefficient a0 on the seventh row or the coefficient b0 on the eighth row is selected according to the value of d0 (in detail, when d0=0, a0 is selected, and when d0=1, b0 is selected), and the result of selection is added to the coefficient a0 on the ninth row. Such a pair of complementary coefficients can be found for all digits, for example, on the fifth row and the sixth row, and the seventh row and the eighth row of the digit of 2−3, the third row and the fourth row, the fifth row and the sixth row, and the seventh row and the eighth row of the digit of 2−2, and so on. Therefore, the constitution of the arithmetic circuit which performs addition for each digit is significantly simplified.
The table shown in FIG. 12 sorts all coefficients of the respective members composing the (Expression 1-14) shown in FIG. 10 by digits. In this table, “−(a3+b3)” described on the digit of 23 corresponds to the coefficient of the signed part of the (Expression 1-14), and the coefficients described in the parts enclosed by thick lines correspond to the respective coefficients of the unsigned part 1 of the (Expression 1-14). In other words, the table shown in FIG. 12 is a table obtained by further adding the coefficients of the signed part and the unsigned part 1 to the table of FIG. 11. The coefficients added in FIG. 12 do not relate to the absolute value of the interpolated value C, but relate to the sign of the interpolated value C. Hereinafter, the way of determination of the sign of the interpolated value C will be described.
On the digit of 23 of FIG. 12, the signed coefficient “−(a3+b3)” is described, and this is a reprint of the coefficient of the signed part of the (Expression 1-14) without change for convenience, and it does not indicate an original bit to be used in an actual operation. In the original bit operation, there is no concept of the sign “minus.” Therefore, in the table of FIG. 12, the bits c6, c5, c4, c3, c2, c1, and c0 corresponding to the unsigned part of the interpolated value C can be determined by normal arithmetic operation of summing the coefficients described on the respective digits (columns), and carrying to the higher-order digit as appropriate, however, the value of the bit c7 corresponding to the signed part is determined by the following method.
First, the result of operation of (a3+b3) is calculated. a3 and b3 are both 1-bit values (0 or 1), so that the result of operation of (a3+b3) becomes any one of 0, 1, and 10 in binary expression. Next, carrying from the digit of the bit c6 (digit of 22) (either 0 or 1 in binary expression: 1 when carrying occurs, and 0 when carrying does not occur) is calculated by performing an arithmetic operation relating to the unsigned part (addition to obtain the bits c6 to c0 shown in FIG. 12) and added to the result of operation of (a3+b3). Then, a binary number obtained as a result of this addition is calculated, and when the calculated binary number is 1 bit, this bit is set as the value of the bit c7, and when the calculated binary number is 2 bits, the lower-order bit is set as the value of the bit c7.
The meaning of determination of the value of bit c7 by this method will be described. Herein, for the sake of convenience of explanation, description is dividedly given based on the signs of the two interpolation target values A and B.
First, the case where both values A and B are positive is considered. In this case, a3=0 and b3=0, so that the sum of (a3+b3) is also 0. In this case, carrying from the unsigned part also results in 0 (that is, carrying from the unsigned part does not occur). This is because the parts enclosed by the thick lines in FIG. 12 (unsigned parts 1) are all 0 since a3=0 and b3=0, and carrying caused by these results in 0. Concerning the parts on the right of the parts enclosed by the thick lines (unsigned parts 2: parts described in the table of FIG. 11), carrying caused by these parts to the digit of the bit c7 always results in 0.
The reason for this is that this operation is interpolation of the two interpolation target values A and B. The unsigned part indicates an absolute value of the interpolated value C, and this absolute value always becomes an intermediate value of the absolute value of the value A and the absolute value of the value B, so that it never exceeds the absolute value of the value A and the absolute value of the value B. Therefore, as long as addition concerning the parts described in the table of FIG. 11 is performed, carrying to the digit of bit c7 always results in 0. This can be easily understood by considering that, when two positive numbers less than 100 are set as interpolation target values A and B, an interpolated value C thereof is always a positive value less than 100. In this case, it is impossible that the interpolated value C becomes a three-digit number. As long as the interpolation target values A and B are both two-digit numbers, the interpolated value C thereof also becomes a two-digit number, and a carrying from the second digit to the third digit is always 0.
Resultantly, when both values A and B are positive, by adding the carry (=0) from the unsigned part to (a3+b3)=0, a bit c7=0 is obtained. This indicates the obtained interpolated value C is a positive number, and means that correct sign handling was performed.
Next, the case where the values A and B are both negative is considered. In this case, a3=1 and b3=1, so that the sum of (a3+b3) becomes 10. In this case, a carry from the unsigned part also always becomes 1 (that is, carrying from the digit of the bit c6 to the digit of the bit c7 always occurs). This is easily understood by considering the arithmetic operation of the parts (unsigned parts 1) enclosed by the thick lines in FIG. 12. First, the sum of the unsigned part (part enclosed by the thick line) described on the digit of the bit c3 is “a3d0+b3e0+b3,” and a3=1 and b3=1, and either d0 or e0 is 1, so that the sum total of the digit of the bit c3 becomes at least “10,” and carrying from the digit of bit c3 to the digit of bit c4 always occurs.
Next, the sum of the unsigned part 1 (part enclosed by the thick line) described on the digit of bit c4 is “a3d1+b3e1,” and a3=1 and b3=1, and either one of d1 and e1 is 1 and the other is 0, so that “a3d1+b3e1=1.” Herein, a carry from the digit of the bit c3 is added thereto, so that the sum total of the digit of the bit c4 becomes at least “10,” and carrying from the digit of the bit c4 to the bit c5 always occurs.
Similarly, carrying from the digit of the bit c5 to the digit of the bit c6 also always occurs, and carrying from the digit of the bit c6 to the digit of the bit c7 also always occurs. Resultantly, when both values A and B are negative, by adding the carry (=1) from the unsigned part to (a3+b3)=10, the result “11” of addition is obtained. The bit c7 is defined as a lower-order bit of this result of addition, so that bit c7=1 is obtained. This indicates that the obtained interpolated value C is a negative, and means that correct sign handling was performed.
Last, the case where either one of the values A and B is positive and the other is negative is considered. In this case, the sum of (a3+b3) becomes 1, so that the value of the bit c7 becomes 0 or 1 depending on the carry from the unsigned part. Herein, detailed description is omitted, however, when the obtained interpolated value C becomes zero or a positive number, carrying from the unsigned part always occurs, and when the obtained interpolated value C becomes a negative number, carrying from the unsigned part does not occur. Therefore, when the interpolated value C is zero or a positive number, by adding the carry of 1 to 1 that is a result of operation of (a3+b3), a result “10” of addition is obtained, and the bit c7 defined as a lower-order bit of the result of addition becomes 0. This indicates that the obtained interpolated value C is zero or a positive number, and means that correct sign handling was performed. To the contrary, when the interpolated value C is negative, by adding the carry 0 to the result 1 of operation of (a3+b3), a result “1” of addition is obtained, and the bit c7 becomes 1. This indicates that the obtained interpolated value is a negative number, and means that correct sign handling was performed.
By thus determining the value of the bit c7 by the above-described method in the arithmetic operation based on the table of FIG. 12, correct sign handling becomes possible.
For the sake of convenience of explanation, the concrete case where n=4 is described as an example, however, as a matter of course, n can be set to an arbitrary number as long as it is an integer not less than 2. In short, the first aspect of the invention is characterized in that a linear interpolator which performs operations for calculating a linear interpolated value C expressed as C=(1−D)*A+D*B based on two signed interpolation target values A and B and an interpolation rate D (0≦D<1) is constituted so as to obtain the linear interpolated value C by performing multiplications and additions defined by the arithmetic expression shown in the (Expression 1-12) of FIG. 7, that is, C=−(an−1+bn−1)2n−1+bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σj=0˜n−2aj2ji=0˜n−1di2(i−1)·Σj=0˜n−2bj2j (ei is a logically inverted bit of di).
In this case, the interpolation target value A to be given to this interpolator is composed of a signed part consisting of a most significant bit indicating a sign and an unsigned part consisting of (n−1) bits indicating an absolute value (n≧2), and has bits of an−1, an−2, . . . , a1, and a0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation.
The interpolation target value B to be given to this interpolator is composed of a signed part consisting of a most significant bit indicating a sign and an unsigned part consisting of (n−1) bits indicating an absolute value (n≧2), and has bits of bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation.
Furthermore, the interpolation rate D to be given to this interpolator has bits of dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side as digital data of n bits in total indicating only a decimal part.
In the above-described arithmetic expressions, the part of “bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σj=0˜n−2aj2ji=0˜n−1di2(i−n)·Σj=0˜n−2bj2j” (ei is a logically inverted bit of di) indicates an arithmetic operations relating to the unsigned part, and by performing this arithmetic operation relating to the unsigned part 2, digital data of (2n−1) bits in total including c2n−2, c2n−3, . . . , c1, and c0 in order from the most significant bit side can be generated.
On the other hand, in the above-described arithmetic operations, for the part of “−(an−1+bn−1)2n−1,” a binary number obtained by summing up the result of operation an−1+bn−1 (any of 0, 1, and 10 in binary expression) and carry (either 0 or 1 in binary expression) from the digit of bit c2n−2 obtained as a result of operation relating to the unsigned part is calculated, and when the calculated binary number is 1 bit, this bit is calculated as the value of the bit c2n−1, and when the calculated binary number is 2 bits, the lower-order bit is calculated as the value of the bit c2n−1.
When digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence is obtained, this digital data of 2n bits or digital data corresponding to a necessary number of significant figures in the digital data of 2n bits is outputted as digital data indicating the linear interpolated value C.
By thus designing a linear interpolator based on the arithmetic expression shown as the (Expression 1-12) of FIG. 7, very rational sign handling becomes possible, and when performing addition of the same digit, an efficient operation using a pair of complementary coefficients becomes possible, so that a linear interpolator which can efficiently perform linear interpolation of two signed interpolation target value by a simple constitution can be realized.
Section 1-4 Concrete Constitution Example of Linear Interpolator According to First Aspect of the Invention
Next, a concrete constitution example of a linear interpolator having a function to perform interpolation based on the arithmetic expression (Expression 1-12) according to the first aspect of the invention shown in FIG. 7 will be described. The arithmetic expression of the (Expression 1-12) is mainly composed of a signed part and unsigned parts (unsigned parts 1 and 2), so that the part to perform an arithmetic operation relating to the signed part will be referred to as a signed part arithmetic means, and the part to perform an arithmetic operation relating to the unsigned part will be referred to as unsigned part arithmetic means. The unsigned part arithmetic means can be constituted by a combination of a selector and a counter. As a selector, a device which has a function to select and output either the bit ai or bi (i=0, 1, 2, . . . (n−1)) based on a logical value of a predetermined bit of digital data indicating an interpolation rate D is used, and as a counter, a device having a function to input the bit ai or bi (i=0, 1, 2, . . . (n−1)), an output value of the selector, or an output value of another counter and output a result of addition of the inputted values is used.
FIG. 13 are diagrams showing a concrete example of the selector and counter having the above-described functions, and these selector and counter are used as detailed basic components of the linear interpolator described later.
First, the selector 100 shown in FIG. 13( a) is a component having function to output either one of the two input values X and Y (both are 1-bit values) as a selected value S. It is determined which input value is selected depending on the logical state of a control value D of 1 bit. Namely, when the control value D=0, the input value X is selected (S=X), and when the control value D=1, the input value Y is selected (S=Y).
Next, the 2-2 counter 200 shown in FIG. 13( b) is a component having a function to perform addition of the two input values X and Y (both are 1-bit values), outputs the value of the digit of 20 that is a result of operation as a sum S, and outputs the value of the digit of 21 that is a result of operation as carry C (carried to the digit on the higher-order). Both sum S and carry C are 1-bit data.
The 3-2 counter 300 shown in FIG. 13( c) is a component having a function to perform addition of three input values X, Y, and Z (all are 1-bit values), outputs a value of the digit of 20 that is a result of operation as a sum S, and outputs the value of the digit of 21 of the result of operation as a carry C (carried to the higher-order digit). The sum S and the carry C is still 1-bit data.
FIG. 14 through FIG. 16 are circuit diagrams showing a linear interpolator constituted by combining the selector 100, the 2-2 counter 200, and the 3-2 counter 300 shown in FIG. 13. For the sake of convenience of illustration, they are dividedly shown by three diagrams, however, one linear interpolator is constituted by the entire circuit drawn in these three diagrams. In the diagrams, the components 101 through 116 attached with numbers in the range from 100 to 199 are components equivalent to the selector 100 shown in FIG. 13( a), the components 201 through 204 attached with numbers in the range from 200 to 299 are components equivalent to the 2-2 counter 200 shown in FIG. 13( b), and the components 301 through 312 attached with numbers in the range from 300 to 399 are components equivalent to the 3-2 counter 300 shown in FIG. 13( c). The components 401 and 402 shown in FIG. 16 are XOR circuits (exclusive OR circuits).
For the sake of convenience of explanation, the selected values outputted from the selectors are indicated by reference numerals including S attached to the heads of the selector numbers, and the sums or carries to be outputted from the counters are indicated by reference numerals including S or C attached to the heads of the counter numbers. For example, S101 shown on the right side of FIG. 14 denotes a selected value outputted from the selector 101, S201 denotes a sum outputted from the counter 201, and C201 denotes a carry (to be added to the higher-order digit) outputted from the counter 201. S401 and S402 shown on the left side of FIG. 16 are exclusive logical sums of the respective XOR circuits 401 and 402.
The linear interpolator shown in FIG. 14 through FIG. 16 performs arithmetic operations to calculate an interpolated value C based on the arithmetic expression (expression when n=4) shown as (Expression 1-13) of FIG. 8. The constitution thereof corresponds to the table of FIG. 12. Namely, the bit c0 arithmetic unit, the bit c1 arithmetic unit, and the bit c2 arithmetic unit shown in FIG. 14, the bit c3 arithmetic unit and the bit c4 arithmetic unit shown in FIG. 15, and the bit c5 arithmetic unit and the bit c6 arithmetic unit shown in FIG. 16 are components for performing arithmetic operations to calculate the bits c0 through c6 in the table of FIG. 12, respectively, and the bit c7 arithmetic unit is a component for performing an arithmetic operation to calculate the bit c7 in the table of FIG. 12. In other words, unsigned part arithmetic means is constituted by the bit c0 arithmetic unit through the bit c6 arithmetic unit, and a signed part arithmetic means is constituted by the bit c7 arithmetic unit.
The values a0, a1, a2, and a3 shown in FIG. 14 through FIG. 16 are bits of the interpolation target value A, the values b0, b1, b2, and b3 are bits of the interpolation target value B, and the values d0, d1, d2, and d3 are bits of the interpolation rate D. The output values S0 through S7 drawn under the respective bit arithmetic units correspond to the bits c0 through c7 in the table of FIG. 12.
For example, focusing attention on the constitution of the bit c0 arithmetic unit shown in FIG. 14, it is understood that the constitution is made for performing an arithmetic operation to calculate the bit c0 in the table of FIG. 12. The selector 101 selects either one of the two input values a0 and b0 based on the control value d0, and outputs the selected input value as the selected value S101, and this is processing corresponding to addition of the coefficients “a0e0” and “b0d0” shown on the digit of the bit c0 in the table of FIG. 12. As described above, “a0e0” and “b0d0” are in a complementary relationship, and when d0=1, a0e0=0 and b0d0=b0, and when d0=0, a0e0=a0 and b0d0=0, so that it is possible that by the selector 101, S101=b0 is outputted as the selected value when d0=1, and S101=a0 is outputted as the selected value when d0=0 instead of the processing corresponding to addition of the coefficients “a0e0” and “b0d0.” Last, if the selected value S101 and a0 are summed up by the counter 201, and as a result, the operation to calculate the bit c0 in the table of FIG. 12 is performed.
At this time, when there is carrying to the digit of the bit c1, carry C201=1 is transmitted to the bit c1 arithmetic unit. As a matter of course, when there is no carrying, carry C201=0 is transmitted to the bit c1 arithmetic unit. Operations for other digits are the same as this.
Only the bit c7 arithmetic unit on the left side of FIG. 16 must determine the bit c7 to function as a signed part, so that the unit has a constitution different from that of other arithmetic units. As processing for determining the bit c7, as described in Section 1-3 above, a binary number that is obtained by adding a carry from the bit c6 arithmetic unit to the sum of (a3+b3) (any of 0, 1, and 10 in binary expression) is calculated, and when the calculated binary number is 1 bit, this bit is set as the value of the bit c7, and when the calculated binary number is 2 bits, the lower-order bit is set as the value of the bit c7. The illustrated circuit constituted by the XOR circuits 401 and 402 are circuits for performing this processing.
Namely, the output value S401 of the XOR circuit 401 becomes 0 when a3 and b3 are equal to each other, or becomes 1 when a3 and b3 are different, so that it becomes 0 when a3=0 and b3=0, becomes 1 when a3=1 and b3=0, becomes 1 when a3=0 and b3=1, and becomes 0 when a3=1 and b3=1. The output value S402 of the XOR circuit 402 becomes an exclusive logical sum of the output values S401 and C312 (carry from the bit c6 arithmetic unit), so that it becomes 0 when S401=0 and C312=0, becomes 1 when S401=1 and S312=0, becomes 1 when S401=0 and C312=1, and becomes 0 when S401=1 and C312=1. Resultantly, the output value S402 of the XOR circuit 402 is the lower-order bit of a binary number obtained by adding a carry from the bit c6 arithmetic unit to the sum of (a3+b3).
When n is set to an arbitrary value, the signed part arithmetic means is constituted by a first XOR circuit which outputs an exclusive logical sum of the bit an−1 and the bit bn−1 and a second XOR circuit which outputs an exclusive logical sum of a bit indicating a carry (either 0 or 1 in binary expression) from the digit of the bit c2n−2 obtained as a result of operation of the unsigned part arithmetic means and the output bit of the first XOR circuit.
The signed part arithmetic means is not always necessarily constituted by using the XOR circuits. For example, instead of the XOR circuits 401 and 402 shown in FIG. 16, it is also possible to use the 2-2 counter 200 shown in FIG. 13( b). In this case, the value of the sum S of the 2-2 counter is used as the output values S401 and S402 and the carry C is ignored. In this constitution, the carry C becomes wasteful, so that in practical use, it is preferable to use the XOR circuits as shown in FIG. 16.
A constitution example of the major portion of the linear interpolator according to the first aspect of the invention is described above with reference to the circuit diagrams of FIG. 14 through FIG. 16, however, in actuality, to this major portion, a component for inputting the digital data A, B, and D and a component for outputting the digital data C from the major portion must be added. FIG. 17 is a block diagram showing an entire constitution of a linear interpolator according to the first aspect of the invention, obtained by adding the components for said inputting and outputting.
As illustrated, this linear interpolator comprises a first interpolation target value input means 10 for inputting n-bit digital data A, second interpolation target value input means 20 for inputting n-bit digital data B, an interpolation rate input means 30 for inputting n-bit digital data D, and arithmetic means 40.
The first interpolation target value input means 10 is means for inputting an interpolation target value A including bits an−1, an−2, . . . , a1, and a0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation composed of a signed part consisting of a most significant bit an−1 indicating a sign and an unsigned part consisting of (n−1) bits (an−2, . . . , a1, and a0) indicating an absolute value (n≧2).
The second interpolation target value input means 20 is means for inputting an interpolation target value B including bits bn−1, bn−2) . . . , b1, and b0 in order from the most significant bit side as digital data of n bits in total expressing a negative number by twos complement notation composed of a signed part consisting of a most significant bit bn−1 indicating a sign and an unsigned part consisting of (n−1) bits (bn−2, . . . , b1, and b0) indicating an absolute value (n≧2).
On the other hand, the interpolation rate input means 30 is means for inputting an interpolation rate D including bits dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side as digital data of n bits indicating only a decimal part.
The arithmetic means 40 comprises signed part arithmetic means 41, unsigned part arithmetic means 42, and calculated value output means 43 as illustrated. Herein, the unsigned part arithmetic means 42 functions to generate digital data of (2n−1) bits in total including bits c2n−2, c2n−3, . . . , c1, and c0 in order from the most significant bit side by performing an arithmetic operation based on the arithmetic expression bn−12−1+(an−1Σi=0˜n−1di2(i−1)+bn−1Σi=0˜n−1ei2(i−1))+(2−ni=0˜n−1ei2(i−1))·Σi=0˜n−2aj2ji=0˜n−1di2(i−n)·Σi=0˜n−2bj2j (ei is a logically inverted bit of di). The signed part arithmetic means 41 functions to calculate, as a value of the bit c2n−1, a bit of the digit on the right side of a binary number obtained by adding a result (any of 0, 1, and 10 in binary expression) of operation of an−1+bn−1 and a carry (either 0 or 1 in binary expression) from the digit of the bit c2n−2 obtained as a result of operation by the unsigned part arithmetic means. The calculated value output means 43 functions to output “digital data of 2n bits in total composed by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures in the digital data of 2n bits” as digital data showing a linear interpolated value C.
The digital data of 2n bits in total obtained by arranging the bits c2n−1, c2n−2, c2n−3, . . . , c1, and c0 in sequence is data including a decimal point positioned between the bit cn and the bit cn−1, and the bit c2n−1 indicates a signed part, the bits c2n−2, c2n−3, . . . , cn indicate an integer part, and the bits cn−1, . . . , c1, and c0 indicate a decimal part.
As described above, according to the first aspect of the invention, interpolation of signed values A and B can be efficiently performed by simple circuitry. The summary of the first aspect of the invention is described as follows with reference to FIG. 12. First, to calculate an interpolated value C based on 4-bit values A (bits a3 a2a1a0) and B (bits b3b2b1b0) expressing a negative number by twos complement notation and a 4-bit interpolation rate D (bits d3d2d1d0) composed of only a decimal part, the basic expression of C=(1−D)*A+D*B is transformed into an expression composed of an unsigned part which includes a sum of products with the bit di or a logically inverted value ei of di (i=0, 1, 2, 3), and indicates an absolute value of the interpolated value C and a signed part indicating a sign of the interpolated value C. Then, by an arithmetic operation of the unsigned part, 7 bits c6 through c0 are generated, and the signed part is logically judged by considering a carry from the digit of the bit c6 caused by the arithmetic operation of the unsigned part to generate the bit c7. Necessary significant digits of the obtained 8-bit value C (bits c7 to c0) are outputted as an interpolated value.
Section 2 Embodiment According to Second Aspect of the Invention
Next, the second aspect of the invention will be described based on the illustrated embodiment. The second aspect of the invention realizes a linear interpolator which can calculate a linear interpolated value of two interpolation target values with high accuracy and has simple circuitry.
Section 2-1 Basic Concept of Linear Interpolation
First, a basic concept of general linear interpolation will be briefly described. FIG. 18 is a graph showing the basic concept of linear interpolation. In the illustrated example, an interval 0-1 is set on the x axis, and when a value of the function f(x) is defined only on both ends of the interval, a method for calculating the function value at an arbitrary position α in the interval 0-1 by linear interpolation is shown. In detail, when f(0)=A and f(1)=B, a function value f(α)=C concerning an arbitrary a in the range of 0≦α≦1 is calculated by linear interpolation using the values A and B.
In linear interpolation, as illustrated, a straight line is defined between the point A and the point B, and as a vertical coordinate value of a point C on this straight line, the value of C is calculated. Herein, the position of the arbitrary point α is defined as “position dividing the interval 0-1 at a ratio of [α:(1−α)],” and α is referred to as an interpolation rate, the linear interpolated value C can be calculated by the arithmetic operation expressed as the basic expression C=(1−α)*A+α*B by using the two interpolation target values A and B and the interpolation rate α (* sign representing multiplication).
For example, when concrete values A=10, B=20, and α=0.4 are given, by performing the arithmetic operation of C=(1−0.4)*10+(0.4*20), an interpolated value C=14 is obtained. Such interpolation is generally called “α-blend.”
The principle of such linear interpolation was described in Section 1-1 with reference to FIG. 1. However, in the example shown in FIG. 1, description using a value D as the interpolation rate is given, and on the other hand, in the example shown in FIG. 18, description using a value α as the interpolation rate is given. The difference between the value α and the value D is that the value α is a value indicating the original interpolation rate in the range “0≦α≦1,” and on the other hand, the value D is a value indicating an interpolation rate in the range of “0≦D<1” that does not include 1.
In the first aspect of the invention described in Section 1, the use of the value D as the interpolation rate is due to the operation using digital data. Namely, when an interpolator which performs linear interpolation is constituted by an electronic circuit in actuality, it is necessary to handle values as digital data composed of bit sequences with finite lengths, respectively. Therefore, normally, an arithmetic operation using the value D instead of the value α for the sake of convenience is performed. This point will be described in greater detail.
The original interpolation rate α is a number in the range of 0≦α≦1, and herein, it is assumed that α=0. d7d6d5d4d3d2d1d0 is defined by using the digital value D shown by an 8-bit sequence of d7d6d5d4d3d2d1d0 (d7, d6, . . . , are 1-bit data taking 1 or 0). Similarly, when the interpolation target value A is expressed as an 8-bit sequence “a7a6a5a4a3a2a1a0” and the interpolation target value B is expressed as an 8-bit sequence “b7b6b5b4b3b2b1b0,” the arithmetic operation based on the above-described basic expression C=(1−α)*A+α*B can be performed by repeating addition of the bits in principle, and an interpolator can be constituted by a comparatively simple circuit.
However, the interpolation rate α defined as the form of α=0. d7d6d5d4d3d2d1d0 is not a mathematically accurate value. Namely, when the digital value D takes a minimum value 0, all bits become 0, so that α=0.00000000, and it is mathematically 0. However, when the digital value D takes a maximum value 255, all bits become 1, so that α=0.11111111, and when it is decadally expressed, it becomes approximately 0.9961. Resultantly, the constitution of the interpolator becomes simple by defining the interpolation rate α as the form of α=0. d7d6d5d4d3d2d1d0, however, α is defined only in the range of 0≦α≦approximately 0.9961, and an arithmetic operation concerning α=1 cannot be performed.
To cope with this, an interpolator which is provided with a comparator for monitoring whether the input value of the interpolation rate α is 1, and output the input value B without change when α=1 has also been used, however, it is necessary to provide an extra circuit such as the comparator, and the circuitry inevitably becomes complicated. In addition, interpolation is performed always in the range of 0≦α≦approximately 0.9961, so that an obtained interpolated value includes an error.
Therefore, for more accurate interpolation, a method for determining the interpolation rate α in the form of α=D/255 has been proposed. According to this definition, when the digital value D takes a maximum value 0, α=0/255=0, and when the digital value D takes a maximum value 255, α=255/255=1. Therefore, α in the range of 0≦α≦1 can be defined, and it is possible to cope with the case of α=1, and an accurate interpolated value can be obtained.
However, since the interpolation rate α is defined as the form of α=D/255, the arithmetic operation based on the basic expression C=(1−α)*A+α*B becomes the complicated form of C=(1-(D/255))*A+(D/255)*B including division. An interpolator for calculating such an expression including division inevitably becomes complicated in circuitry. An object of the second aspect of the invention is to provide a linear interpolator which can perform accurate arithmetic operations and can be realized by simple circuitry.
Section 2-2 Unique Point of the Second Aspect of the Invention
Herein, for the sake of convenience, two interpolation target values A and B are given as 8-bit digital data, and a digital value D for determining the interpolation rate α (in the description of the embodiment according to the second aspect of the invention, an example in which this digital data D is referred to as “interpolation rate determining value”) is given as 8-bit digital data, and as a result of arithmetic operation based on the basic expression C=(1−α)*A+α*B by using these digital data, a linear interpolated value C expressed as 16-bit digital data is obtained, is described. Herein, in the 16-bit data of the linear interpolated value C, the higher-order 8 bits are an integer part, and the lower-order 8 bits are a decimal part.
FIG. 19 shows expressions for thus defining the interpolation target values A and B, the interpolation rate determining value D, and the linear interpolated value C.
In FIG. 19, first, (Expression 2-1):
$A = ∑ i = 0 _ 7 a i 2 i = a 0 2 0 + a 1 2 1 + … + a 7 2 7$
is shown. a0, a1, . . . , a7 in this (Expression 2-1) are bits (0 or 1) when the numerical value A is expressed as a binary number consisting of 8 bits, and a0 is the least significant bit (rightmost bit) and a7 is the most significant bit (leftmost bit).
Similarly, in the (Expression 2-2), an expression:
$B = ∑ i = 0 _ 7 b i 2 i = b 0 2 0 + b 1 2 1 + … + b 7 2 7$
is shown, and b0, b1, . . . , b7 in this expression are bits of the value B expressed as a binary number consisting of 8 bits, and b0 is the least significant bit and b7 is the most significant bit.
In the (Expression 2-3), an expression:
$D = ∑ i = 0 _ 7 d i 2 i = d 0 2 0 + d 1 2 1 + … + d 7 2 7$
is shown, and d0, d1, . . . , d7 in this expression are bits of the value D expressed as a binary number consisting of 8 bits, and d0 is the least significant bit and d7 is the most significant bit.
On the other hand, in the (Expression 2-4), an expression:
$C = ∑ i = 0 _ 15 c i 2 ( i - 8 ) = c 0 2 - 8 + c 1 2 - 7 + … + c 7 2 - 1 + c 8 2 0 + c 9 2 1 + … + c 15 2 7$
is shown, and c0, c1, . . . , c15 in this expression are bits when the value C is expressed as a binary number consisting of 16 bits, and c0 is the least significant bit and c15 is the most significant bit.
While the values A, B, and D are defined as 8-bit data, respectively, the reason for definition of the value C as 16-bit digital data two times the 8-bit data is that the value C is a linear interpolated value which is obtained by an operation based on the basic expression C=(1−α)*A+α*B. Namely, the basic expression includes members of multiplication of values expressed with accuracy of 8 bits, so that a product of these values is obtained as a 16-bit value. Therefore, the value C as a final result of operation is defined as a 16-bit value. Here, c0 to c7 corresponding to the lower-order 8 bits of the 16 bits of this value C are a decimal part, and c8 to c15 corresponding to the higher-order 8 bits are an integer part. In practical use, in this 16-bit value C, only predetermined significant digits (normally, 8 digits of the integer part) are outputted as a final linear interpolated value C.
The last (Expression 2-5) shows the relationship between the interpolation rate α and the interpolation rate determining value D. Namely, these satisfy the relationship of “α=1/(28−1)*D=(1/255)*D.” The value D is 8-bit digital data, so that it takes an integer in the range from 0 to 255, and by defining the value α by the (Expression 2-5) by using this value D, the value α in the range from 0 to 1 can be defined.
FIG. 20 is a diagram showing a state in that the interpolation target values A and B, the interpolation rate determining value D, and the linear interpolated value C of FIG. 19 are expressed as bit sequences of binary numbers. When the values A, B, D, and C are defined according to the expressions of FIG. 19, these values are expressed as bit sequences as shown in FIG. 20. As described above, the values A, B, and D are 8-bit data, respectively, and on the other hand, the value C is 16-bit data, and the higher-order 8 bits c15 to c8 are an integer part and the lower-order 8 bits c7 to c0 are a decimal part, and a decimal point is positioned between the bit c8 and c7.
Resultantly, the linear interpolator to be designed is an interpolator having a function to calculate a linear interpolated value C as 16-bit digital data by performing arithmetic operations based on the basic expression C=(1−α)*A+α*B based on the interpolation target values A and B and the interpolation rate determining value D given as 8-bit digital data as shown in FIG. 20.
Herein, the arithmetic expression (Expression 2-5) becomes a problem. When the interpolation rate α is defined as the form of α=(1/255)*D, α can be defined in the whole of the necessary range of 0≦α≦1, so that it becomes possible to cope with the case of α=1, and an accurate interpolated value can be obtained. However, division of (D/255) must be performed, and the circuitry inevitably becomes complicated. The second aspect of the invention is characterized in that this problem is solved based on the following unique point of view.
First, when the value 1/255 is expressed as a binary number, it is expressed in the form of a repeating decimal of “0.000000010000000100000001000 . . . ” (repetition of “1” arranged after seven “0”) as shown in the (Expression 2-6) of FIG. 21. On the other hand, the interpolation rate α is a result of multiplication of this repeating decimal by 8-bit digital data D (a binary number expressed as a bit sequence “d7d6d5d4d3d2d1d0”). However, in the repeating decimal, “1” only appears every 8 digits, and a product of this “1” and the bit sequence “d7d6d5d4d3d2d1d0” is the bit sequence “d7d6d5d4d3d2d1d0” in itself, so that the product of (1/255)*D is expressed by the bit sequence shown in the (Expression 2-7) of FIG. 21. Namely, the interpolation rate α is expressed as a binary fractional value in which a bit sequence “d7d6d5d4d3d2d1d0” is repeatedly arranged after the decimal point.
When D is the minimum value 0, the bit sequence “d7d6d5d4d3d2d1d0” becomes “00000000,” so that α is a binary number in which the bit “0” continues to infinite digits after the decimal point, and this is equal to 0. On the other hand, when D is the maximum value 255, the bit sequence “d7d6d5d4d3d2d1d0” becomes “11111111,” and α is a binary number in which the bit “1” continues to infinite digits after the decimal point, and this is extremely close to “1” mathematically.
However, in a real interpolator, it is impossible to perform operations to infinite digits after the decimal point to define the value of α, so that the value must be rounded off to a predetermined digit. Therefore, as shown on the upper stage of FIG. 22, the ninth decimal place (the digit of 2−9) of the value of the interpolation rate α is rounded off by carrying it to the higher-order digit when the value of this digit is 1 and cutting it when the value is 0 so that an approximate value α′ having significant digits to the eighth decimal place (digit of 2−8) is calculated finally. Herein, let us consider what bit sequence the approximate value α′ is expressed as dividedly between the case where the digit of 2−9 is 0 and cut off and in the case where the digit of 2−9 is 1 and carried to the higher-order digit.
First, the case where the digit of 2−9 is 0 and cut off is considered. As shown on the upper stage of FIG. 22, the bit of the digit of 2−9 is “d7,” so that this digit is cut off when “d7=0.” In this case, the digit of 2−9 is cut off, so that as shown on the middle of FIG. 22, the approximate value α′ is a binary number “0. d7d6d5d4d3d2d1d0” having significant digits to the eighth decimal place (the digit of 2−8).
Next, let us consider the case where the digit of 2−9 is 1 and carried to the higher-order digit. In this case, the digit of d7 is carried to the higher-order digit, so that this is when “d7=1.” Therefore, as shown on the lower stage of FIG. 22, the approximate value α′ is a binary number having significant digits to the eighth decimal place (the digit of 2−8) obtained by adding the binary number “0.00000001” to “0. d7d6d5d4d3d2d1d0.” Herein, the value “0.00000001” to be added is a result of carrying of 1 from the digit of 2−9 to the digit of 2−8.
Considering these two cases, resultantly, the approximate value α′ having significant digits to the eighth decimal place (the digit of 2−8) can be expressed as:
$α ′ = 0. d 7 d 6 d 5 d 4 d 3 d 2 d 1 d 0 + 0.0000000 d 7$
as shown in the (Expression 2-8) of FIG. 23. When the bit d7 is 0, the part “0.0000000d7” in the expression shown above becomes 0, so that the expression shown above becomes α′=0. d7d6d5d4d3d2d1d0, and this is as shown on the middle stage of FIG. 22. On the other hand, when the bit d7 is 1, the part “0.0000000d7” in the expression shown above becomes “0.00000001,” so that the expression shown above becomes α′=0. d7d6d5d4d3d2d1d0+0.00000001, and this is as shown on the lower stage of FIG. 22. Thus, the definitional expression of the approximate value α′ according to the (Expression 2-8) is equivalent to the definition of the approximate value α′ in the two cases shown in FIG. 22.
The fractional value in binary expression “0. d7d6d5d4d3d2d1d0” can be expressed by the expression Σj=0˜7dj2(j−8), and the fractional value in binary expression “0.0000000d7” can be expressed as d72−8, so that the approximate value α′ shown by the (Expression 2-8) can be defined by the following expression:
α′=Σj=0˜7 d j2(j−8) +d 72−8
as shown in the (Expression 2-9).
Resultantly, when the interpolation rate α is defined as the form:
α=(1/255)*D
by the (Expression 2-5) of FIG. 19 by using the 8-bit digital data D (bit sequence “d7d6d5d4d3d2d1d0”), if this is directly calculated, division must be performed, and a dividing circuit must be prepared as a interpolator, however, by defining the approximate value α′ of the interpolation rate α as the following form:
α′=Σj=0˜7 d j2(j−8) +d 72−8
as shown in the (Expression 2-9) and using this, the approximate value α′ can be obtained only by addition, so that only an adding circuit must be prepared as a interpolator. This is the most important point of the second aspect of the invention.
Therefore, linear interpolation in the second aspect of the invention is performed by an arithmetic operation based on the expression “C=(1−α′)*A+α′*B” shown as (Expression 2-10) on the upper stage of FIG. 24 by using the approximate value α′ determined based on the interpolation rate determining value D.
Section 2-3 Transformation of Expression for Constituting More Efficient Interpolator
In “(Expression 2-10) C=(1−α′)*A+α′*B,” by substituting the (Expression 2-9) of FIG. 23 for the part of α′, substituting the (Expression 2-1) of FIG. 19 for the part of A, and substituting the (Expression 2-2) of FIG. 19 for the part of B, as shown on the middle stage and the lower stage of FIG. 24, the expression “C=[1−(Σj=0˜7dj2(j−8)+d72−8)]*Σi=0˜7ai2i+(Σj=0˜7dj2(j−8)+d72−8)*Σj=0˜7bi2i” is obtained. In this expression, no division is included, so that by constituting an interpolator based on this expression, at least a dividing circuit is not necessary.
However, the inventor of the present invention considered that further transformation of the (Expression 2-10) into a more proper form would make it possible to design a more efficient interpolator. Therefore, the inventor examined how to transform the (Expression 2-10) to constitute an efficient interpolator while considering concrete circuitry as an interpolator, and through trial and error, conceived the fact that the following transformation made it possible to constitute an extremely efficient interpolator.
First, (Expression 2-11) of FIG. 25 is considered. This (Expression 2-11) in itself is a formula known from a long time ago, mathematically. For example, when n=8, the (Expression 2-11) becomes (Expression 2-12). If this (Expression 2-12) is developed in actuality, it is expressed as follows: 2−8+2−7+2−6+2−5+2−4+2−3+2−2+2−1+2−8= 1/256+ 1/128+ 1/64+ 1/32+ 1/16+⅛+¼+½+ 1/256=1. As a matter of course, this (Expression 2-11) is a formula for the arbitrary n.
Next, concerning the part [1−(Σj=0˜7dj2(j−8)+d72−8)] shown on the middle stage of the (Expression 2-10) of FIG. 24, let us consider the transformation shown in (Expression 2-13) of FIG. 26. First, when the part “1” described on the first line of the (Expression 2-13) is substituted by the left-hand side of the (Expression 2-12) of FIG. 25, the second line of the (Expression 2-13) is obtained. By combining the first and third members of the expression on the second line and combining the second member and the fourth member, the expression on the third line is obtained.
dj is each bit of the interpolation rate determining value D, so that it takes 0 or 1 as an actual value. Therefore, when assuming ej=(1−dj), ej is a logically inverted bit of dj (ej=0 when dj=1, and ej=1 when dj=0). Therefore, the expression on the third line of the (Expression 2-13) is transformed into the expression on the fourth line by changing dj to the logically inverted bit of ej.
Herein, the part of [1−(Σj=0˜7dj2(j−8)+d72−8)] shown on the middle stage of the (Expression 2-10) of FIG. 24 is substituted by “Σj=0˜7ej2(j−8)+e72−8” shown on the fourth line of FIG. 26, the (Expression 2-14) shown in the upper half of FIG. 27 is obtained. By developing this expression, as shown in the lower half of FIG. 27, the form of the sum of the four members shown by the circled numbers 1 through 4 is obtained. In detail, the linear interpolated value C to be calculated is obtained by calculating the sum of the following four members:
• First member: Σj=0˜7ej2(j−8)i=0˜7ai2i
• Second member: e72−8i=0˜7ai2i
• Third member: Σj=0˜7dj2(j−8)i=0˜7bi2i
• Fourth member: d72−8i=0˜7bi2i
This (Expression 2-14) is equivalent to the basic expression of linear interpolation shown as (Expression 2-10) of FIG. 24, however, this form of (Expression 2-14) is very convenient for designing an interpolator.
Section 2-4 Features of Arithmetic Expression According to the Second Aspect of the Invention
A first feature of the (Expression 2-14) shown in FIG. 27 is that, as originally shown in the (Expression 2-5) of FIG. 19, the relationship between the linear rate determining value D and the linear rate α is defined by using the expression “α=(1-255)*D,” so that α in the range of 0≦α≦1 can be defined, it becomes possible to cope with the case of α=1 (strictly speaking, not α=1 but α extremely close to 1 can be defined as described in Section 2-2), and an accurate interpolated value can be obtained. In addition, the expression using the approximate value α′ is defined based on the unique point described in Section 2-2, so that the arithmetic expression does not include division.
A second feature of the (Expression 2-14) shown in FIG. 27 is that, in the form of the expression, dj and ej are incorporated complementarily. For example, the first member includes the expression form “Σj=0˜7ej2(j−8),” and the third member includes the expression form “Σj=0˜7dj2(j−8).” The second member includes the expression form “e72−8,” and the fourth member includes the expression form “d72−8.”
As described above, ej is a logically inverted bit of dj, so that when dj is 1, ej is always 0, and when dj is 0, ej is always 1. Therefore, when calculating the respective members of the expressions, when dj=1, the member including ej always becomes 0, so that only the members including dj are calculated, and when dj=0, the member including dj always becomes 0, so that only the members including ej are calculated. Thereby, a more simple arithmetic circuit and more efficient operations are realized.
FIG. 28 is a table sorting coefficients of members composing the (Expression 2-14) of FIG. 27 by digits. The columns arranged horizontally in this table correspond to scaling positions of 27, 26, 25, . . . , 2−6, 2−7, and 2−8, respectively, and the bits c15, c14, c13, . . . , c2, c1, and c0 show the bits of the interpolated value C to be calculated by performing operations. As described above, a decimal point is positioned between the bit c8 and the bit c7, and in the 16-bit data of the interpolated value C, the higher-order 8 bits “c15c14 . . . c9c8” are an integer part, and the lower-order 8 bits “c7c6 . . . c1c0” are a decimal part.
The values of “circled numbers” and “j” described on the right side of the rows of the table of FIG. 28 correspond to the values of the members indicated by the circled numbers and j in these members of the expression of FIG. 27. For example, on the right side of the first row of the table of FIG. 28, “circled number 1” and “j=7” are described, and these indicate that the coefficients arranged on the first row of this table correspond to the coefficients when “j=7” in the expression “Σj=0˜7ej2(j−8)j=0˜7ai2i” marked with “circled number 1” in FIG. 27. In actuality, when “j=7” is applied, the part “Σj=0˜7ej2(j−8)” becomes “e72−1,” and coefficients of the digits obtained by a product of “e72−1” and “Σi=0˜7ai2i” are arranged on the first row of the table of FIG. 28. For example, when i=0, a product of “e72−1”*“a020”=“a0e72−1” is obtained. The coefficient “a0e7” entered at the right end (the digit of 2−1) of the first row of the table of FIG. 28 indicates the coefficient of this product.
Resultantly, the (Expression 2-14) of FIG. 27 can be performed as many product-sum operations, and the coefficients of the respective digits are described in any part of the table of FIG. 28. The coefficients a0, a1, a2, . . . b0, b1, and b2 described in the table of FIG. 28 are the bits of the interpolation target values A and B, and the coefficients d0, d1, d2, . . . , are bits of the interpolation rate determining value D. The coefficients e0, e1, e2, . . . , are logically inverted bits of the respective coefficients d0, d1, d2, . . . .
Conveniently, the coefficients of the same digit include many pairs of complementary coefficients. For example, on the digit of 2−8, the coefficients of a0e0, b0d0, a0e7, and b0d7 are described, and the bit c0 can be calculated by summing up these four coefficients. However, ej is the logically inverted bit of dj, so that a0e0 and b0d0 are a pair of complementary coefficients, and either one is always 0. For example, when e0=1, d0=0, so that a0e0=a0 and b0d0=0, and a0e0+b0d0=a0. To the contrary, when e0=0, d0=1, so that a0e0=0 and b0d0=b0, and a0e0+b0d0=b0. Resultantly, the result of the operation “a0e0+b0d0” becomes a0 when d0=0 and becomes b0 when d0=1. Therefore, in place of a computing unit which performs the operation “a0e0+b0d0,” a selector which selects either the input value a0 or b0 can be made to perform the operation “a0e0+b0d0.”
The same applies to the operation “a0e7+b0d7” on the digit of 2−8, and based on the value of d7, the selector which selects either the input value a0 or b0 can be made to perform the operation “a0e7+b0d7.” The digit of 2−8 is illustrated as described above, and such complementary coefficients can be found for all digits. Therefore, the constitution of the arithmetic circuit which performs addition for each digit is remarkably simplified.
Thus, by constituting an interpolator so as to obtain a linear interpolated value by performing multiplications and additions defined by the (Expression 2-14) shown in FIG. 27, operations by which accurate interpolated values are obtained become possible, and the constitution of the arithmetic circuit is remarkably simplified.
Section 2-5 Expansion to General Formula
In the Sections 2-2 through 2-4 described above, for easy understanding, two interpolation target values A and B are given as 8-bit digital data, an interpolation rate determining value D for determining an interpolation rate α is given as 8-bit digital data, and the concrete case where a linear interpolated value C expressed as 16-bit digital data is obtained as a result of execution of operations based on the basic expression of C=(1−α)*A+α*B by using these digital data, is described. However, the second aspect of the invention is not limited to this example.
Herein, the second aspect of the invention is expanded to general formulas, and an expression to obtain an interpolated value C as 2n-bit digital data when the values A, B, and D are given as n-bit digital data is shown. For clear comparison with the above-described expressions, the expressions given in the following description are indicated by adding n to the corresponding expression numbers that have already been given above.
For example, the (Expression 2-1n) shown in FIG. 29 corresponds to expansion of the (Expression 2-1) shown in FIG. 19 to a general formula. Namely, the interpolation target value A given as n-bit digital data is defined by the following expression shown in (Expression 2-1n):
$A = ∑ i = 0 _n - 1 a i 2 i = a 0 2 0 + a 1 2 1 + … + a n - 1 2 ( n - 1 )$
Similarly, the interpolation target value B given as n-bit digital data is defined by the following expression shown as (Expression 2-2n):
$B = ∑ i = 0 _n - 1 b i 2 i = b 0 2 0 + b 1 2 1 + … + b n - 1 2 ( n - 1 )$
and the interpolation rate determining value D given as n-bit digital data is defined by the following expression shown in (Expression 2-3n):
$D = ∑ i = 0 _n - 1 d i 2 i = d 0 2 0 + d 1 2 1 + … + d n - 1 2 ( n - 1 )$
On the other hand, the linear interpolated value C to be calculated as 2n-bit digital data is defined by the following expression shown as (Expression 2-4n):
$C = ∑ i = 0 _ 2 n - 1 c i 2 ( i - n ) = c 0 2 - n + c 1 2 - ( n - 1 ) + … + c n - 1 2 - 1 + c n 2 0 + + c n + 1 2 1 + … + c 2 n - 1 2 ( n - 1 )$
In this interpolated value C of 2n bits, the lower-order n bits c0 through cn−1 indicate a decimal part, and the higher-order n bits cn through c2n−1 indicate an integer part. As a matter of course, in practical use, in the value C of 2n bits, output of only predetermined significant digits (normally, n digits of the integer part) as a final linear interpolated value C is sufficient.
The relationship between the interpolation rate α and the interpolation rate determining value D can be defined by the following expression shown as (Expression 2-5n):
α=(1/(2n−1))*D
When D is given as n-bit digital data, the minimum value of D is 0 and the maximum value is “2n−1,” so that by using the (Expression 2-5n), α can be defined as a value in the range of 0≦α≦1.
Resultantly, the (Expression 2-1) through (Expression 2-5) shown in FIG. 19 are expressions in which the setting of n=8 is substituted for the (Expression 2-1n) through (Expression 2-5n). FIG. 30 is a diagram showing a state in that the interpolation target values A and B, the interpolation rate determining value D, and the linear interpolated value C shown in FIG. 29 are expressed as bit sequences of binary numbers. When defining the values A, B, D, and C according to the expressions of FIG. 29, these values are expressed as bit sequences shown in FIG. 30. As described above, although the values A, B, and D are n-bit data, the value C is 2n-bit data, and the higher-order n bits c2n−1 through cn are an integer part, the lower-order n bits cn−1 through c0 are a decimal part, and a decimal point is positioned between the bit cn and the bit cn−1.
Resultantly, the linear interpolator according to the second aspect of the invention is an interpolator having a function to calculate a linear interpolated value as 2n-bit digital data by performing arithmetic operations based on the basic expression of C=(1−α)*A+α*B based on the interpolation target values A and B and the interpolation rate determining value D given as n-bit digital data shown in FIG. 30.
The division in the arithmetic expression (Expression 2-5n) becomes a problem. As described above, when the interpolation rate α is defined as the form of α=(1/(2n−1))*D, α can be defined in the necessary whole range of 0≦α≦1, so that it becomes possible to cope with the case of α=1, and an accurate interpolated value can be obtained. However, division of (1/(2n−1)) must be performed, so that the circuitry inevitably becomes complicated. To solve this problem, the unique point described in the Section 2-2 is utilized.
First, when the value (1/(2n−1)) is expressed by a binary number, it is expressed as a repeating decimal (repetition of “1” arranged after n−1 “0”) as shown in the (Expression 2-6n) of FIG. 31. On the other hand, the interpolation rate α is a result of multiplication of this repeating decimal by n-bit digital data D (binary number shown as a bit sequence “dn−1dn−2 . . . d0.” However, in the repeating decimal, “1” only appears every n digits, and a product of this “1” and the bit sequence “dn−1dn−2 . . . d0” is the bit sequence “dn−1dn−2 . . . d0” in itself, so that resultantly, the product of (1/(2n−1))*D is expressed as the bit sequence shown in the (Expression 2-7n) of FIG. 31. Namely, the interpolation rate α is expressed as a binary fractional value in which the bit sequence “dn−1dn−2 . . . d0” is repeatedly arranged after the decimal point.
When D takes the minimum value 0, the bit sequence “dn−1dn−2 . . . d0” becomes “00 . . . 0,” so that α is a binary number in which the bit “0” continues to infinite digits after the decimal point, and this is equal to 0. On the other hand, when D is the maximum value (2n−1), the bit sequence “dn−1dn−2 . . . d0” becomes “11 . . . 1,” so that α is a binary number in which the bit “1” continues to infinite digits after the decimal point, and this is a value extremely close to 1, mathematically.
Herein, as shown on the upper stage of FIG. 32, the (n+1)-th decimal place (the digit of 2−(n+1)) of the value of the interpolation rate α is rounded off by carrying it to the higher-order digit when it is 1 or cutting it off when it is 0, and finally, an approximate value α′ having significant digits to the n-th decimal place (the digit of 2−n) is calculated. First, as shown on the middle stage of FIG. 32, when “dn−1=0,” the digit of 2−(n+1) is 0 and cut off, so that the approximate value α′ is a binary number “0.dn−1dn−2 . . . d0” having significant digits to the n-th decimal place (the digit of 2−n). On the other hand, when “dn−1=1,” as shown on the lower stage of FIG. 32, the digit of 2−(n+1) is 1 and carried to the higher-order digit, so that the approximate value α′ is a binary number having significant digits to the n-th decimal place (the digit of 2−n) obtained by adding the binary number “0.00 . . . 001” to “0.dn−1dn−2 . . . d0.” Herein, the value “0.000 . . . 001” to be added is a binary fractional value in which only the n-th decimal place is “1,” and is a carry to the digit of 2−n from the digit of 2−(n+1) that is 1 and carried.
Considering these two cases, resultantly, the approximate value α′ having significant digits to the n-th decimal place (the digit of 2−n) can be expressed by the following expression shown in (Expression 2-8n) of FIG. 33:
$α ′ = 0. d n - 1 d n - 2 d n - 3 … d 0 + 0.000 … d n - 1$
Herein, “dn−1” of the members of the second line is the bit on the n-th decimal place. When the bit “dn−1” is 0, the part “0.000 . . . dn−1” of the expression shown above becomes 0, so that resultantly, the expression shown above becomes α′=0.dn−1dn−2dn−3 . . . d0, and this is as shown on the middle stage of FIG. 32. On the other hand, when the bit “dn−1” is 1, the part “0.000 . . . dn−1” of the expression shown above becomes “0.000 . . . 1 (the terminal 1 is on the n-th decimal place),” so that resultantly, the expression shown above becomes α′=0.dn−1dn−2dn−3 . . . d0+0.000 . . . 1, and this is as shown in the expression on the lower stage of FIG. 32. Thus, the definitional expression of the approximate value α′ based on the (Expression 2-8n) is equivalent to the definition of the approximate value α′ in the two cases shown in FIG. 32.
The binary fractional value “0.dn−1dn−2dn−3 . . . d0” can be expressed by the expression Σj=0˜n−1dj2(j−n), and the binary fractional value “0.000 . . . dn−1” can be expressed as dn−12−n, so that resultantly, the approximate value α′ shown by the (Expression 2-8n) of FIG. 33 can be defined by the following expression as shown in the (Expression 2-9n):
α′=Σj=0˜n−1 d j2(j−n) +d n−12−n
Thus, also when the interpolation rate α is defined in the following form:
α=(1/(2n−1))*D
by the (Expression 2-5n) of FIG. 29 by using the n-bit digital data D (bit sequence “dn−1dn−2dn−3 . . . d0”), if the approximate value α′ of the interpolation rate α is defined in the following form:
α′=Σj=0˜n−1 d j2(j−n) +d n−12−n
and used as shown in the (Expression 2-9n), the division becomes unnecessary and the approximate value α′ can be obtained only by addition.
Resultantly, the interpolated value C should be calculated by the arithmetic operation of “C=(1−α′)*A+α′*B” based on the (Expression 2-10n) of FIG. 34, however, in this expression, by substituting the (Expression 2-9n) of FIG. 33 for the part of α′, substituting the (Expression 2-1n) of FIG. 29 for the part of A, and substituting the (Expression 2-2n) of FIG. 29 for the part of B, as shown on the middle stage and the lower stage of FIG. 34, the expression “C=[1−(Σj=0˜n−1dj2(j−n)+d n−12−n)]*Σi=0˜n−1ai2i+(Σj=0˜n−1dj2(j−n)+dn−12−n)*Σi=0˜n−1bi2i” is obtained. This expression includes no division, so that by constituting an interpolator based on this expression, at least a dividing circuit is not necessary. However, in the second aspect of the invention, to design a more efficient interpolator, this (Expression 2-10n) is further transformed as follows.
First, as shown in (Expression 2-13n) of FIG. 35, the part of [1−(Σj=0˜n−1dj2(j−n)+dn−12−n)] in the (Expression 2-10n) is transformed into the following form:
Σj=0˜n−1 e j2(j−n) +e n−12−n
Herein, ej=(1−dj), and ej is a logically inverted bit of dj (when dj=1, ej=0, and when dj=0, ej=1). The form on the second line of the (Expression 2-13n) is derived by substituting the left-hand side of the (Expression 2-11) of FIG. 25 for the part “1” described on the first line of the (Expression 2-13n), and the process of this is exactly the same as the transformation process of the (Expression 2-13) shown in FIG. 26.
Therefore, when the part of [1−(Σj=0˜n−1dj2(j−n)+dn−12−n)] shown on the middle of the (Expression 2-10n) is substituted by “Σj=0˜n−1ej2(j−n)+en−12−n” shown on the second line of FIG. 35, the (Expression 2-14n) shown in the upper half of FIG. 36 is obtained. When this expression is developed, as shown in the lower half of FIG. 36, it becomes the form of a sum of the four members indicated by the circled numbers 1 through 4. In detail, the linear interpolated value C to be calculated is obtained by calculating the sum of the following four members:
• First member: Σj=0˜n−1ej2(j−n)i=0˜n−1ai2i
• Second member: en−12−ni=0˜n−1ai2i
• Third member: Σj=0˜n−1dj2(j−n)i=0˜n−1bi2i
• Fourth member: dn−12−ni=0˜n−1bi2i
This (Expression 2-14n) is equivalent to the basic expression of linear interpolation shown as (Expression 2-10n) in FIG. 34. A result of setting n=8 in this (Expression 2-14n) is the (Expression 2-14) shown in FIG. 27, and as described in the Section 2-4 above, this form is very convenient for designing an interpolator.
Namely, this (Expression 2-14n) defines the relationship between the linear rate determining value D and the linear rate α by using the expression “α=(1/(2n−1)*D” as shown in the original (Expression 2-5n) of FIG. 29, so that α in the range of 0≦α≦1 can be defined, and it becomes possible to cope with the case of α=1 (strictly speaking, not α=1 but α extremely close to 1 can be defined as described in the Section 2-2), and an accurate interpolated value can be obtained. In addition, based on the unique point described in the Section 2-2, the expression using the approximate value α′ is defined, so that the arithmetic expression includes no division, and dj and ej are complementarily incorporated in the form of the expression, so that a part of the operation can be realized by only a selector or the like, and the arithmetic circuit is simplified and efficient operations become possible.
Resultantly, the second aspect of the invention is characterized in that a linear interpolator which performs operations to calculate a linear interpolated value C expressed as C=(1−α)*A+α*B based on two interpolation target values A and B and an interpolation rate α (0≦α≦1) is constituted so as to obtain the linear interpolated value C by performing multiplications and additions defined by the arithmetic expression shown in the (Expression 2-14n) of FIG. 36, that is, the arithmetic expression of C=(Σj=0˜n−1ej2(j−n)+en−12−n)*Σj=0˜n−1ai2i+(Σj=0˜n−1dj2(j−n)+dn−12−n)*Σi=0˜n−1bi2i)ei is a logically inverted bit of di).
In this case, the interpolation target value A that should be given to this interpolator is defined as digital data of n bits in total including bits of an−1, an−2, . . . , a1, and a0 in order from the most significant bit side, the interpolation target value B is defined as digital data of n bits in total having the bits of bn−1, bn−2, . . . , b1, and b0 in order from the most significant bit side, the interpolation rate determining value D is defined as digital data of n bits in total having the bits of dn−1, dn−2, . . . , d1, and d0 in order from the most significant bit side, and the relationship “α=D/(2n−1)” is defined between the interpolation rate determining value D and the interpolation rate α. As a result of this operation, digital data of 2n bits in total having the bits of c2n−1, c2n−2, . . . , c1 and c0 in order from the most significant bit side is generated, so that this digital data of 2n bits or digital data corresponding to the number of significant figures in the digital data of 2n bits is outputted as digital data indicating the linear interpolated value C.
By thus designing a linear interpolator based on the arithmetic expression shown in the (Expression 2-14n) of FIG. 36, a linear interpolated value of two interpolation target values can be obtained with high accuracy, and a linear interpolator having simple circuitry can be realized.
Section 2-6 Concrete Constitution Example of Linear Interpolator According to the Second Aspect of the Invention
Next, a concrete constitution example of a linear interpolator having a function to perform interpolation based on the arithmetic expression (Expression 2-14n) according to the second aspect of the invention shown in FIG. 36 will be described. However, the concrete constitution example of the interpolator when the number of n is large becomes complicated in circuitry and illustration becomes difficult due to paper-based constraints, so that a constitution example of a comparatively small interpolator when n=4 is set is shown. This interpolator has a function to output an interpolated value C as digital data of 8 bits based on input values A, B, and D given as 4-bit digital data.
The (Expression 2-15) shown in FIG. 37 is an arithmetic expression when n=4 is set in the (Expression 2-14n) of FIG. 36. As described above, this arithmetic expression is in the form of a sum of four members indicated by the circled numbers 1 through 4 as shown in the lower half of FIG. 37. Namely, the linear interpolated value C to be calculated is obtained by calculating the sum of the following four members:
• First member: Σj=0˜3ej2(j−4)i=0˜3ai2i
• Second member: e32−4i=0˜3ai2i
• Third member: Σj=0˜3dj2(j−4)i=0˜3bi2i
• Fourth member: d32−4i=0˜3bi2i
FIG. 38 is a table sorting the coefficients of the members of the (Expression 2-15) of FIG. 37 by digits, and its basic composition is the same as that of the table shown in FIG. 28 (table in the case of n=8). Namely, the columns arranged horizontally in the table of FIG. 38 correspond to scaling positions of 23, 22, 21, 20, 2−1, 2−2, 2−3, and 2−4, and the bits c7, c6, c5, c4, c3, c2, c1, and c0 indicate the bits of the interpolated value C obtained by arithmetic operations. In this case, a decimal point is positioned between the bit c4 and the bit c3, and in the 8-bit data composing the interpolated value C, the higher-order 4 bits “c7, c6, c5, c4” are an integer part and the lower-order 4 bits “c3, c2, c1, c0” are a decimal part.
The values of “circled numbers” and “j” described on the right side of the respective rows of the table of FIG. 38 correspond to the members of the circled numbers in the expression of FIG. 37 and the value of j in these members. For example, on the right side of the first row of the table of FIG. 38, “circled number 1” and “j=3” are described, and these indicate that the coefficients arranged on the first row of this table correspond to coefficients when “j=3” is set in the expression “Σj=0˜3ej2(j−4)i=0˜3ai2i” of the first member marked with the “circled number 1.” In actuality, when “j=3” is set, the part “Σj=0˜3ej2(j−4)” becomes “e32−1,” and the coefficients of the respective digits obtained by a product of “e32−1” and “Σj=0˜3ai2i” are arranged on the first row of the table of FIG. 38. For example, when i=0, a product of “e32−1”*“a020”=“a0e32−1” is obtained. The coefficient “a0e3” described on the right end (the digit of 2−1) of the first row of the table of FIG. 38 indicates the coefficient of this product.
Thus, the (Expression 2-15) of FIG. 37 can be performed as many product-sum operations, and the coefficients of each digit are described on any portion of the table of FIG. 38. In addition, the coefficients of the same digit include many pairs of complementary coefficients (two coefficients either one of which always becomes 0). For example, in the coefficients “a0e0” and “b0d0,” either e0 or d0 is always 0, so that these are complementary coefficients, and the result of the operation of “a0e0+b0d0” becomes either “a0” or “b0.” The point that a selector which selects either one of the input values “a0” and “b0” based on the value of do can be made to perform the operation of “a0e0+b0d0” has already been described above.
Resultantly, the interpolator which performs operations based on the (Expression 2-15) of FIG. 37 can be constituted by a combination of a selector and a counter. As the selector, a device is used which has a function to select and output either the bit ai or bi (i=0, 1, 2, . . . (n−1): n=4 in this example) based on a logical value of a predetermined bit of digital data showing the interpolation rate determining value D, and as a counter, a device is used which has a function to input the bit ai or bi (i=0, 1, 2, . . . (n−1): n=4 in this example), an output value of the selector, or an output value of another counter and outputs the results of addition of inputted values.
FIG. 39 are diagrams showing a concrete example of the selector and the counter having the above-described functions, and these selector and counter are used as basic components of the concrete linear interpolator described later.
First, the selector 500 shown in FIG. 39( a) is a component having a function to output either one of the two input values X and Y (both are 1-bit values) as a selected value S. It is determined which input value is selected according to the logical state of the control value D of 1 bit. Namely, when the control value D=0, the input value X is selected (S=X), and when the control value D=1, the input value Y is selected (S=Y).
Next, the 2-2 counter 600 shown in FIG. 39( b) is a component which functions to perform addition of two input values X and Y (both are 1-bit values), outputs the value of the digit of 20 of the result of addition, and outputs the value of the digit of 21 of the result of addition as a carry C (carry to the higher-order digit). The sum S and the carry C are both 1-bit data.
The 3-2 counter 700 shown in FIG. 39( c) is a component which functions to perform addition of three input values X, Y, and Z (all are 1-bit values), outputs the value of the digit of 20 of the result of addition as a sum S, and outputs the value of the digit of 21 of the result of addition as the carry C (carry to the higher-order digit). The sum S and the carry C are still 1-bit data.
FIG. 40 through FIG. 42 are circuit diagrams showing a linear interpolator constituted by combining the selector 500, the 2-2 counter 600, and the 3-2 counter 700 of FIG. 39. For the sake of convenience of illustration, the interpolator is dividedly shown in three diagrams, however, the whole of the circuits drawn in these three drawings form one linear interpolator. In the diagrams, the components 501 through 520 attached with the reference numerals in the range from 500 to 599 are equivalent to the selector 500 shown in FIG. 39( a), the components 601 through 604 attached with the reference numerals in the range of 600 to 699 are equivalent to the 2-2 counter 600 of FIG. 39( b), and the components 701 through 712 attached with the reference numerals in the range of 700 to 799 are equivalent to the 3-2 counter 700 of FIG. 39( c).
For the sake of convenience of explanation, the selected value to be outputted from each selector is indicated by a reference numeral with S attached to the head, and the sum or carry to be outputted from each counter is indicated by a reference numeral with S or C attached to the head. For example, the reference numeral S501 indicated on the right side of FIG. 40 denotes a selected value to be outputted from the selector 501, the reference numeral S601 is a sum to be outputted from the counter 601, and the reference numeral C601 denotes a carry (to be added to the higher-order digit) to be outputted from the counter 601.
This linear interpolator shown in FIG. 40 through FIG. 42 is for performing arithmetic operations to obtain an interpolated value based on the arithmetic expression (arithmetic expression when n=4) shown in the (Expression 2-15) of FIG. 37, and the constitution thereof corresponds to the table of FIG. 38. Namely, the bit c0 arithmetic unit, the bit c1 arithmetic unit, and the bit c2 arithmetic unit shown in FIG. 40, the bit c3 arithmetic unit and the bit c4 arithmetic unit shown in FIG. 41, and the bit c5 arithmetic unit, the bit c6 arithmetic unit, and the bit c7 arithmetic unit shown in FIG. 42 are components for performing arithmetic operations to obtain the bits c0 to c7 in the table of FIG. 38, respectively. The bit c7 arithmetic unit has only a function to directly output the carry C712 from the bit c6 arithmetic unit.
The values a0, a1, a2, and a3 shown in FIG. 40 through FIG. 42 are bits of the interpolation target value A, the values b0, b1, b2, and b3 are bits of the interpolation target value B, and values d0, d1, d2, and d3 are bits of the interpolation rate determining value D. The outputs values S0 through S7 drawn under the respective bit arithmetic units correspond to the values of the bits c0 through c7 in the table of FIG. 38.
For example, focusing attention on the constitution of the bit c0 arithmetic unit shown in FIG. 40, it is understood that this constitution is for performing arithmetic operations to obtain the bit c0 in the table of FIG. 38. The selector 501 functions to select either one of the two input values a0 and b0 based on the control value d3 and output the selected input value as the selected value S501, and this is processing corresponding to the addition of the coefficients “a0e3” and “b0d3” shown on the digit of the bit c0 in the table of FIG. 38. As described above, “a0e3” and “b0d3” are in a complementary relationship, and when d3=1, a0e3=0 and b0d3=b0, and when d3=0, a0e3=a0 and b0d3=0, so that resultantly, when d3=1, S501=b0 is outputted as a selected value when d3=0, S501=a0 is outputted as a selected value by the selector 501 instead of the processing corresponding to the addition of the coefficients “a0e3” and “b0d3.”
Similarly, the selector 502 performs a function to select either one of the two input values a0 and b0 based on the control value d0 and outputs the selected input value as the selected value S502, and this corresponds to the addition of the coefficients “a0e0” and “b0d0” shown on the digit of the bit c0 in the table of FIG. 38. When S502=b0 is outputted as a selected value when d0=1 and S502=a0 is outputted as the selected value when d0=0 by the selector 502, this substitutes for the processing corresponding to the addition of the coefficients “a0e0” and “b0d0.” Last, by summing up the selected value S501 and the selected value S502 by the counter 601, as a result, the operation for calculating the bit c0 in the table of FIG. 38 is performed.
In this case, when there is a carry to the digit of the bit c1, the carry C601=1 is transmitted to the bit c1 arithmetic unit. As a matter of course, when there is no carry, the carry C601=0 is transmitted to the bit c1 arithmetic unit. Arithmetic operations for other digits are the same as this. For the digit of 23, an actual arithmetic operation is not performed, and a carry from the digit of 22 is outputted without change as the value of the digit of 23. Therefore, as shown on the left end of FIG. 42, a substantial circuit is not prepared in the bit c7 arithmetic unit, and arithmetic units having substantial circuits are the seven bit arithmetic units c6 through c0.
Resultantly, to calculate the bit c6−i (i=0, 1, 2, 3, 4, 5, 6) indicating the value of the digit of 2(2−i) of the linear interpolated value C, the circuit shown in FIG. 40 through FIG. 42 has seven arithmetic units which sums up the coefficients of the digits and the carries from the lower-order digits, outputs the sum as the value of the bit c6−i, and outputs carries to the higher-order digits, and the circuit outputs a carry from the arithmetic unit that performs an operation of the bit c6 as the value of the bit c7.
By expanding this to a general formula using n, to calculate the bit c2n−i−2 (i=0, 1, 2, . . . 2(n−1)) indicating the value of the digit of 2(n−i−2) of the linear interpolated value C, the linear interpolator according to the second aspect of the invention has (2n−1) arithmetic units which sum up the coefficients of the digits and all carries from the lower-order digits, outputs the sum as the value of this bit c2n−i−2, and outputs the carries to the higher-order digits, and functions to output the carry from the arithmetic unit which performs an operation for the bit c2n−2 as the value of the bit c2n−1.
A constitution example of the major portion of the linear interpolator according to the second aspect of the invention is described above with reference to the circuit diagrams of FIG. 40 through FIG. 42, however, in actuality, to this major portion, a component for inputting the digital data A, B, and D and a component for outputting the digital data C from this major portion must be added. FIG. 43 is a block diagram showing an entire constitution of a linear interpolator according to the second aspect of the invention, obtained by adding the components for inputting and outputting.
As illustrated, this linear interpolator comprises first interpolation target value input means 15 for inputting n-bit digital data A, second interpolation target value input means 25 for inputting n-bit digital data B, interpolation rate determining value input means 35 for inputting n-bit digital data D, arithmetic means 45, and calculated value output means 55.
The first interpolation target value input means 15 is means for inputting an interpolation target value A as digital data of n bits in total including bits of an−1, an−2, . . . , a1, a0 in order from the most significant bit side, and the second interpolation target value input means 25 is means for inputting the interpolation target value B as digital data of n bits in total including the bits of bn−1, bn−2, . . . , b1, b0 in order from the most significant bit side, and the interpolation rate determining value input means 35 is means for inputting the interpolation rate determining value D satisfying the relationship “α=D/(2n−1)” with the interpolation rate α, as digital data of n bits in total including the bits of dn−1, dn−2, . . . , d1, d0 in order from the most significant bit side.
On the other hand, the arithmetic means 45 is means for generating digital data of 2n bits in total composed of, in order from the most significant bit side, n bits in total of c2n−1, c2n−2, . . . , cn+1, cn of the integer part and n bits in total of cn−1, cn−2, . . . , c1, c0 of the decimal part by performing arithmetic operations based on the arithmetic expression of C=(Σj=0˜n−1ej2(j−n)+en−12−n)*Σi=0˜n−1ai2i+(Σj=0˜n−1dj2(j−n)+dn−12−n)*Σi=0˜n−1bi2i (ei is a logically inverted bit of di).
The calculated value output means 55 is means for outputting “digital data of 2n bits in total composed by arranging the bits of c2n−1, c2n−2, c2n−3, . . . , c1, c0 in sequence in order from the most significant bit side” or “digital data corresponding to a necessary number of significant figures of the digital data of 2n bits” as digital data showing the linear interpolated value C.
Thus, according to the second aspect of the invention, simple circuitry for calculating a linear interpolated value C concerning interpolation target values A and B with high accuracy is provided. The summary of the second aspect of the invention is described as follows with reference to FIG. 22. First, the interpolated value C is obtained based on the basic expression of C=(1−α)*A+α*B by using an interpolation rate α. Next, by defining α=D/255 by using a 8-bit value D (bits d7d6 . . . d1d0), α satisfying “0≦α≦1” is set. The binary expression of 1/255 becomes a repeating decimal of 0.00000001 . . . , so that the binary expression of α becomes 0.d7d6 . . . d1d0d7d6 . . . . By defining an approximate value α′ obtained by rounding off the digit of 2−9 of α by cutting the digit when it is 0 or by carrying 1 to the higher-order digit of 2−9 when said digit is 1 as α′=Σj=0˜7dj2(j−8)+d72−8, the approximate value α′ can be obtained by addition instead of division of α=D/255, so that a simple circuit including no division can be realized.
Section 3 Embodiment According to Third Aspect of the Invention
Next, an embodiment illustrating a third aspect of the invention will be described. The third aspect of the invention realizes an interpolator which can efficiently perform cubic spline interpolation.
Section 3-1 Basic Concept of Cubic Spline Interpolation
As described above, the cubic spline interpolation itself is known, however, for the sake of convenience, the concept of this interpolation is briefly described based on the description in the aforementioned document Hsieh S. Hou Harry C. Andrews, “Cubic splines for Image Interpolation and Digital Filtering,” IEEE Trans. on ASSP-26, No. 6, December 1978, pp. 508-517. Herein, interpolation for one-dimensional pixel arrays is described, however, by repeatedly performing this one-dimensional interpolation longitudinally and transversely, the same can be applied to two-dimensional pixel array (described in detail in Section 3-5).
First, the concept of the linear interpolation will be described in a point different from that of the Section 1 and Section 2 with reference to FIG. 44. It is assumed that one-dimensional pixel array is defined on the x axis as illustrated. Namely, a plurality of pixels having predetermined pixel values are arranged at the predetermined pitch on the x axis. In the diagram of this application, provided existing pixels are shown by white circles. Herein, when the pixel number k is defined, FIG. 44 shows a state in that the 0-th (k=0) pixel P0 and the first (k=1) pixel P1 are arranged on the x axis. For the sake of convenience, it is assumed that the pixel P0 is arranged at the coordinate position of x=0 and the pixel P1 is arranged at the coordinate position of x=1, and an interval “0≦x<1” which ends at these pixels is defined as an interpolation interval. When an interpolation point Q (herein, referred to also as a new pixel Q) is defined at an arbitrary position x in this interpolation interval, calculation of a pixel value f(x) to be given to this new pixel Q is considered. In other words, an interpolated value of an arbitrary interpolation point Q between the adjacent two pixels P0 and P1 is calculated.
The simplest method for performing this interpolation is linear interpolation using pixel values of the two pixels P0 and P1. Herein, the pixels to be used for interpolation are referred to as sample pixels. In the example of FIG. 44, based on pixel values of the two sample pixels P0 and P1, a pixel value f(x) of the new pixel Q defined at an arbitrary position x in the interpolation interval is determined. Namely, when the pixel values of the sample pixels P0 and P1 are defined as f0 and f1, the pixel value of the new pixel Q can be determined by performing a simple arithmetic expression of f(x)=f0·(1−x)+fi·x.
In this linear interpolation, the pixel value of the interpolation point Q can be obtained by a simple product-sum operation using the pixel values of the pair of sample pixels P0 and P1 on both sides of interpolation point Q, so that the burden of operation is very small. The linear interpolation circuits shown in the Section 1 and Section 2 are also circuits for performing interpolation based on this linear interpolation. However, the interpolation accuracy of this linear interpolation is very low, so that smoothly continuous pixel values cannot be obtained.
On the other hand, in cubic convolution interpolation and cubic spline interpolation with higher accuracy, interpolation using more sample pixels is performed. For example, the example shown in FIG. 45 shows interpolation based on pixel values of four sample pixels in total, and in currently proposed cubic convolution interpolation and cubic spline interpolation, such interpolation using four sample pixels is performed. Namely, by using pixel values f−1, f0, f1, and f2 of four sample pixels of the (−1)st (k=−1) pixel P−1, the 0-th (k=0) pixel P0, the first (k=1) pixel P1, and the second (k=2) pixel P2, a pixel value f(x) of a new pixel Q defined at an arbitrary position x in an interpolation interval (interval between the pixel P0 and the pixel P1) is calculated.
In cubic spline interpolation, a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning the position x is calculated, and an arithmetic operation obtained by substituting the coordinate value x of the new pixel Q with this polynomial is performed to calculate the pixel value f(x). Herein, the polynomial “f(x)=Ax3+Bx2+Cx+D” can be calculated by:
$f ( x ) = ∑ k = - 1 2 c k β 3 ( x - k ) ( Expression 3 - 1 )$
This (Expression 3-1) expresses f(x) as a sum of four functions defined about k=−1, 0, 1, and 2, respectively. Herein, ck is called interpolation coefficient, β3(x−k) is called cubic B spline function.
This function β3(x−k) is defined as follows. That is, by defining the following functions:
$( x - k ) + 3 = { ( x - k ) 3 x > k 0 x ≤ k ( Expression 3 - 2 )$
$ω k 3 ( i ) = ∏ j = k - 2 i ≠ j k + 2 ( i - j ) ( Expression 3 - 3 )$
the function β3(x−k) is calculated as:
$β 3 ( x - k ) = 4 · ∑ i = k - 2 k + 2 ( x - i ) + 3 ω k 3 ( i ) ( Expression 3 - 4 )$
When i=k−2, k−1, k, k+1, and k+2 are substituted for i of this (Expression 3-4) and transformed to a form of a sum, the following expression:
$β 3 ( x - k ) = 1 6 { ( x - k + 2 ) + 3 - 4 · ( x - k + 1 ) + 3 + 6 · ( x - k ) + 3 - 4 · ( x - k - 1 ) + 3 + ( x - k - 2 ) + 3 } ( Expression 3 - 5 )$
is obtained. Herein, when the (Expression 3-5) is substituted for the (Expression 3-1) by considering the definition of the (Expression 3-2) and the condition of 0≦x<1, the following expression:
$f ( x ) = 1 6 { c - 1 [ ( x + 3 ) 3 - 4 · ( x + 2 ) 3 + 6 · ( x + 1 ) 3 - 4 · x 3 ] + c 0 [ ( x + 2 ) 3 - 4 · ( x + 1 ) 3 + 6 · x 3 ] + c 1 [ ( x + 1 ) 3 - 4 · x 3 ] + c 2 x 3 ] } = 1 6 ( ( - c - 1 + 3 c 0 - 3 c 1 + c 2 ) x 3 + ( 3 c - 1 - 6 c 0 + 3 c 1 ) x 2 + ( - 3 c - 1 + 3 c 1 ) x + ( c - 1 + 4 c 0 + c 1 ) ) ( Expression 3 - 6 )$
is obtained.
When x=0 is set in this (Expression 3-6), f(0)=(c−1+4c0+c1), and when the pixel value of the k-th pixel is defined as fk and generalized to the following expression:
$f k = 1 6 ( c k - 1 + 4 c k + c k + 1 ) ( k = 0 , 1 , 2 , 3 ) ( Expression 3 - 7 )$
Herein, the coefficients c−2 and c3 out of the four interpolation coefficients c−1, c0, c1, and c2 are handled as c−2=c0 and c3=c1, the following relational expression:
$[ f - 1 f 0 f 1 f 2 ] = 1 6 [ 4 2 0 0 1 4 1 0 0 1 4 1 0 0 2 4 ] · [ c - 1 c 0 c 1 c 2 ] ( Expression 3 - 8 )$
is obtained.
This is an expression showing the relationship between the pixel values f−1, f0, f1, and f2 of the four sample pixels and the four interpolation coefficients c−1, c0, c1, and c2. Therefore, to calculate the four interpolation coefficients c−1, c0, c1, and c2 based on the pixel values f−1, f0, f1, and f2 of the four sample pixels, this (Expression 3-8) is solved about c−1, c0, c1, and c2. A solution method for this is comparatively complicated, so that intermediate steps are omitted, and only the result of operation becomes the following expression:
$[ c - 1 c 0 c 1 c 2 ] = 1 15 [ 26 - 14 4 - 1 - 7 28 - 8 2 2 - 8 28 - 7 - 1 4 - 14 26 ] · [ f - 1 f 0 f 1 f 2 ] ( Expression 3 - 9 )$
When this (Expression 3-9) is substituted for the (Expression 3-6), the following expression:
$f ( x ) = 1 5 ( ( - 3 f - 1 + 7 f 0 - 7 f 1 + 3 f 2 ) x 3 + ( 7 f - 1 - 13 f 0 + 8 f 1 - 2 f 2 ) x 2 + ( - 4 f - 1 + f 0 + 4 f 1 - f 2 ) x ) + f 0 ( Expression 3 - 10 )$
is obtained.
This (Expression 3-10) is in the form of a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning the position x, and the coefficients A, B, C, and D are given as the pixel values f−1, f0, f1, and f2 of the four sample pixels. Therefore, when the pixel values f−1, f0, f1, and f2 of the four sample pixels P−1, P0, P1, and P2 shown in FIG. 45 and the position x of the new pixel Q (0≦x<1, that is, the new pixel Q is positioned between the sample pixels P0 and P1) are given, the pixel value f(x) of the new pixel Q can be calculated. This is the basic concept of cubic spline interpolation using four sample pixels, and this is a known matter.
Section 3-2 Approach Unique to the Third Aspect of the Invention
The cubic spline interpolation thus using four sample pixels are realized by operations using the (Expression 3-10). However, if the number of sample pixels exceeds 4, the expression corresponding to the (Expression 3-10) becomes very complicated, and in some cases, it becomes very difficult in actuality to solve such an expression.
For example, as shown in FIG. 46, cubic spline interpolation using six in total of (−2)nd (k=−2) to third (k=3) sample pixels P−2, P−1, P0, P2, and P3 is considered. Also in this case, the interpolation interval is 0≦x<1, and a new pixel Q (position: coordinate value x) whose interpolated value is to be calculated must be defined between the sample pixels P0 and P1. When thus performing interpolation using six in total of sample pixels, an interpolation equation including many pixel values f−2, f−1, f0, f1, f2, and f3 of the six sample pixels must be derived instead of (Expression 3-10). The following expression:
$[ f - 2 f - 1 f 0 f 1 f 2 f 3 ] = 1 6 [ 4 2 0 0 0 0 1 4 1 0 0 0 0 1 4 1 0 0 0 0 1 4 1 0 0 0 0 1 4 1 0 0 0 0 2 4 ] · [ c - 2 c - 1 c 0 c 1 c 2 c 3 ] ( Expression 3 - 11 )$
shows the relationship between the six pixel values f−2, f−1, f0, f1, f2, and f3 and the six interpolation coefficients c−2, c−1, c0, c1, c2, and c3, and corresponds to the (Expression 3-8) shown in the case of interpolation using four sample pixels. To find the six interpolation coefficients c−2, c−1, c0, c1, c2, and c3 based on the pixel values f−2, f−1, f0, f1, f2, and f3, this (Expression 3-11) must be solved for c−2, c−1, c0, c1, c2, and c3. Calculation of this becomes very complicated.
Thus, the cubic spline interpolation in which the number of sample pixels is set to 6 or more needs a very complicated operation, and conventionally, it is difficult to efficiently perform this operation. To solve this problem, the inventor of the present invention conceived the following unique approach. Hereinafter, the concept of this unique approach will be described with reference to numerical expressions.
The inventor of the present invention focused attention on the point that more than half the values of the determinant including 6 rows and 6 columns of the (Expression 3-11) are “0,” and adjusted the six polynomials corresponding to the (Expression 3-11) to delete the interpolation coefficients c−2 and c3. Then, the inventor considered that the remaining interpolation coefficients c−1, c0, c1, and c2 could be determined by solving the (Expression 3-11) into the following form of (Expression 3-11):
$[ g - 1 f 0 f 1 f 2 ] = 1 6 [ w 1 0 0 1 4 1 0 0 1 4 1 0 0 1 w ] · [ c - 1 c 0 c 1 c 2 ] ( Expression 3 - 12 )$
In this (Expression 3-12), new variables g−1, g−2, and w are used, and these variables are defined as follows:
${ g - 1 = - 1 4 f - 2 + f - 1 = 1 6 ( 14 4 c - 1 + c 0 ) g 2 = - 1 4 f 3 + f 2 = 1 6 ( 14 4 c 2 + c 1 ) w = 7 2 ( Expression 3 - 13 )$
As shown in this (Expression 3-13), g−1 is a factor to be determined from the pixel values f−2 and f−1 of the left two pixels P−2 and P−1 in FIG. 46, and g2 is a factor to be determined from the pixel values f2 and f3 of the two right pixels P2 and P3 in FIG. 46. As shown in the determinant of (Expression 3-12), w can be regarded as a weighting factor for finding g−1 and g2.
Herein, by solving the (Expression 3-12) for c−1, c0, c1, and c2, the following expression:
$[ c - 1 c 0 c 1 c 2 ] = 6 ( 5 w - 1 ) ( 3 w - 1 ) [ 15 w - 4 1 - 4 w w - 1 1 - 4 w - w ( 1 - 4 w ) - w 2 w w - w 2 - w ( 1 - 4 w ) 1 - 4 w - 1 w 1 - 4 w 15 w - 4 ] · [ g - 1 f 0 f 1 g 2 ] ( Expression 3 - 14 )$
is derived. The interpolation coefficients are only four c−1, c0, c1, and c2, so that by substituting these for the (Expression 3-6), the following expression:
$f ( x ) = 1 ( 3 w - 1 ) { - 6 g - 1 + ( 3 w + 1 ) ( f 0 - f 1 ) + 6 g 2 } x 3 + 3 ( 5 w - 1 ) ( 3 w - 1 ) { 6 ( 4 w - 1 ) g - 1 - ( 9 w 2 + 2 w - 1 ) f 0 + 6 w 2 f 1 - 6 w g 2 } x 2 + 3 ( 5 w - 1 ) ( 3 w - 1 ) { - 2 ( 7 w - 2 ) g - 1 - ( w 2 - 4 w + 1 ) f 0 + 2 w ( 2 w - 1 ) f 1 - 2 ( 2 w - 1 ) g 2 } x + f 0 ( Expression 3 - 15 )$
is obtained. This (Expression 3-15) is in the form of a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning the position x, and the coefficients A, B, C, and D can be found based on the pixel value f−1 and f0, the two factors g−1 and g2 and the weighting factor w shown in the (Expression 3-13). As shown in the (Expression 3-13), the factor g−1 is found from the two pixel values f−2 and f−1, and the factor g2 is found from the two pixel values f2 and f3, and resultantly, the coefficients A, B, C, and D in the (Expression 3-15) can be found from the six pixel values f−2, f−1, f0, f1, f2, and f3 and the weighting factor w. Therefore, by giving the position x (0≦x<1) of the new pixel Q, the pixel value f(x) thereof can be calculated.
This approach in the case of using six simple pixels is also applicable to the case where the number of sample pixels is further increased, and it can be expanded to more generalized expressions. Namely, in the description given above, the (Expression 3-8) is shown as the case where the number of sample pixels is set to 4 and the (Expression 3-11) is shown as the case where the number of sample pixels is set to 6, however, as a general formula when the number of sample pixels is set to “2(i+1)” (i is an integer of 1 or more), the following expression:
$[ f - i f - i + 1 ⋮ f - 1 f 0 f 1 f 2 ⋮ f i f i + 1 ] = 1 6 [ 4 2 0 0 ⋯ 0 0 0 0 0 1 4 1 0 ⋯ 0 0 0 0 0 0 ⋰ ⋰ ⋰ ⋰ ⋮ ⋮ ⋮ ⋰ ⋮ ⋮ ⋯ 1 4 1 0 0 0 ⋰ ⋮ 0 ⋯ 0 1 4 1 0 0 ⋰ 0 0 ⋯ 0 0 1 4 1 0 ⋰ 0 0 ⋯ 0 0 0 1 4 1 0 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋰ ⋰ ⋰ ⋰ ⋮ 0 0 0 0 ⋯ 0 0 1 4 1 0 0 0 0 ⋯ 0 0 0 2 4 ] [ c - i c - i + 1 ⋮ c - 1 c 0 c 1 c 2 ⋮ c i c i + 1 ] ( Expression 3 - 16 )$
is known (for example, the aforementioned document Hsieh S. Hou Harry C. Andrews, “Cubic Splines for Image Interpolation and Digital Filtering,” IEEE Trans. on ASSP-26, No. 6, December 1978, pp. 508-517.)
This (Expression 3-16) is an application to the cubic spline interpolation using “2(i+1)” in total of (−i)th (k=−i) to (i+1)th (k=i+1) sample pixels P−i, P−i+1, . . . , P−1, P0, P1, P2, . . . , Pi, and Pi+1 as shown in FIG. 47. Also in this case, the interpolation interval is absolutely 0≦x<1, and the new pixel Q whose interpolated value is to be found (position is the coordinate value x) must be defined between the sample pixels P0 and P1. Thus, when interpolation using “2(i+1)” in total of sample pixels is performed, as shown in the (Expression 3-16), an expression including “2(i+1)” of pixel values f−1, f−i+1, . . . , f−1, f0, f1, f2, . . . , fi, and fi+1 and “2(i+1)” of interpolation coefficients c−1, c−i+1, . . . , c−1, c0, c1, c2, . . . , ci, and ci+1 is defined.
A result of the solution of a general formula like this (Expression 3-16) for the interpolation coefficients c−1, c−i+1, . . . , c−1, c0, c1, c2, . . . , ci, and ci+1 is not shown in any document as far as the inventor knows. However, the above-described approach according to the idea of the inventor is also applicable to the general formula shown in this (Expression 3-16). Namely, a determinant of this (Expression 3-16) having “2(i+1)” rows and “2(i+1)” columns can be restored to a determinant having “2(i+1)−1” rows and “2(i+1)−1” columns by the same method as that for restoring the (Expression 3-11) to the (Expression 3-12). In this case, for example, the pixel values f−1 and f−i+1 are substituted by factors having values satisfying the following expression:
$- 1 4 f - i + f - i + 1 = 1 6 ( 7 2 c - i + 1 + c - i + 2 ) ( Expression 3 - 17 )$
This (Expression 3-17) corresponds to an expression concerning the factor g−1 of (Expression 3-13). Therefore, similarly to the substitution of the factor g−1 in the (Expression 3-12) for the pixel values f−2 and f−1 of the (Expression 3-11), the pixel values f−i and f−i+1 of the (Expression 3-16) can be substituted by the factor shown in the (Expression 3-17), and the pixel values fi and fi+1 of the (Expression 3-16) can also be substituted by a similar factor.
By repeating this processing, in the end, the (Expression 3-16) can be restored to the form shown in the (Expression 3-12). In other words, the (Expression 3-12) is not only a restored form of the interpolation equation using 6 sample pixels shown in the (Expression 3-11) but also a restored form of a generalized interpolation equation using “2(i+1)” sample pixels as shown in the (Expression 3-16).
However, in this case, the weighting factor w also needs to be generalized by using i. In detail, an initial value of the weighting factor w when i=0 is defined as w0=2, the weighting factor wi with respect to a value of arbitrary i is defined by the following recurrence formula:
$w i = 4 - 1 w i - 1 ( where w 0 = 2 ) ( Expression 3 - 18 )$
In detail, w0=2, w1=7/2, w2=26/7 and so on. Then, when the factors g−1, g2, and w shown in the (Expression 3-12) are defined as follows:
${ g - 1 = - 1 w i - 1 ( … ( - 1 w 2 ( - 1 w 1 ( - 1 4 f - i + f - i + 1 ) + f - i + 2 ) … ) + f - i + ( i - 2 ) ) + f - i + ( i - 1 ) = 1 6 ( ( - 1 w i - 2 + 4 ) c - i + ( i - 1 ) + c - i + 1 ) g 2 = - 1 w i - 1 ( … ( - 1 w 2 ( - 1 w 1 ( - 1 4 f i + 1 + f i ) + f i - 1 ) … ) + f i + ( 3 - i ) ) + f i + ( 2 - i ) = 1 6 ( ( - 1 w i - 2 + 4 ) c - i + ( i + 2 ) + c - i + ( i + 1 ) ) w = w i - 1 ( Expression 3 - 19 )$
the (Expression 3-12) can be handled as a restored form of a generalized interpolation equation using “2(i+1)” sample pixels.
For example, when g−1, g2, and w shown in the (Expression 3-13) correspond to the factors and weighting factor in the case of i=2 (that is, in the case where interpolation is performed by using six sample pixels), and are obtained by substituting 2 for i in the (Expression 3-19).
As described above, when the (Expression 3-12) is solved for c−1, c0, c1, and c2, the (Expression 3-14) is obtained, and the interpolation coefficients are only four of c−1, c0, c1, and c2. Then, by applying these to the (Expression 3-6), the (Expression 3-15) is obtained. Resultantly, the (Expression 3-15) can be applied without change in the case of generalization using “2(i+1)” sample pixels, and in this case, the values of g−1, g2, and w are calculated by using the (Expression 3-19).
The (Expression 3-15) is in the form of a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning the position x, and the coefficients A, B, C, and D are calculated by the two factors g−1 and g2 and the weighting factor w shown in the (Expression 3-19).
Herein, as shown on the upper stage of the (Expression 3-19), the factor g−1 is a factor to be found based on i in total of pixel values f−i, f−i+1, f−i+2, . . . , f−i+(i−2), f−i+(i−1) and the weighting factors w1, w2, . . . , wi−1, and these i pixel values are pixel values of the (−i)th (k=−i) through (−1)st sample pixels P−i, . . . , P−1, that is, i pixels continuously arranged adjacent to each other on the left side of the interpolation interval. In this application, for the sake of convenience, the “predetermined number of pixels to be used for interpolation” continuously arranged adjacent to each other on the left side of the interpolation interval like the sample pixels P−i, . . . , P−1 are referred to as left side interpolation pixels, and g−1 is referred to as a left side influence factor. Even when the (−i−1)th (k=−i−1) pixel is arranged on the further left side of FIG. 47, it is not a sample pixel and does not correspond to the left side interpolation pixel as long as it is not used for interpolation.
On the other hand, as shown on the middle stage of the (Expression 3-19), the factor g2 is found based on i in total of pixel values fi+1, fi, fi−1, . . . , fi+(3−i), fi+(2−i) and the weighting factor w1, w2, . . . , wi−1. These i pixel values are of the (i+1)th (k=i+1) through second (k=2) sample pixels Pi+1, . . . , P2 as shown in FIG. 47, that is, i pixels continuously arranged adjacent to each other on the right side of the interpolation interval. In the present application, for the sake of convenience, like these sample pixels Pi+1, . . . , P2, the “predetermined number of pixels to be used for interpolation” continuously arranged adjacent to each other on the right side of the interpolation interval are referred to as right side interpolation pixels and g2 is referred to as a right side influence factor. As a matter of course, even when the (i+2)th (k=i+2) pixel is further arranged on the right side of FIG. 47, it is not a sample pixel and does not correspond to the right side interpolation pixel as long as it is not used for interpolation.
As shown in FIG. 47, in the present application, the two pixels P0 and P1 positioned on the left and right ends of the interpolation interval are referred to as interval terminal pixels. Resultantly, in the case of expansion to a general formula using “2(i+1)” sample pixels, the pixel value f(x) of the new pixel Q can be calculated based on the (Expression 3-15) and the (Expression 3-19), if the pixel values f0 and f1 of the interval terminal pixels P0 and P1, the pixel values f−i, . . . , f−1, of i left side interpolation pixels P−1, . . . , P−1, the pixel values fi+1, . . . , f2 of i right side interpolation pixels Pi+1, . . . , P2, and the position x of the new pixel Q are given. This is the principle of the interpolation device according to the third aspect of the invention.
FIG. 48 is a diagram showing correlation of the numerical expressions to be used in this basic principle. The (Expression 3-15) shown on the upper stage of the diagram is a basic expression for calculating a pixel value f(x), and it is in the form of a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning the position x. The variables in this (Expression 3-15) are the position x of the new pixel Q, the pixel values f0 and f1 of the interval terminal pixels, the left side influence factor g−1, the right side influence factor g2, and the weighting factor w as shown on the right side of the diagram.
The (Expression 3-19) shown on the middle stage of the diagram is a computation expression for the left side influence factor g−1, the right side influence factor g2, and the weighting factor w used as variables in the (Expression 3-15) on the upper stage. The variables in this (Expression 3-19) are the pixel values f−i through f−1 of the left side interpolation pixels, pixel values f2 through fi+1 of the right side interpolation pixels, and the weighting factor wi (i=1, 2, 3, . . . ) as shown on the right side of the diagram.
Herein, the weighting factor wi is a factor indicating weighting for each sample pixel according to the distance from the interpolation interval. Logically, the weighting is not simple weighting indicating such as “the j-th weighting factor wj indicates weighting of the j-th sample pixel Pj,” and one weighting factor influences weighting of many sample pixels.
For example, the first expression of the (Expression 3-19) shown on the middle stage of FIG. 48 is an expression for calculating the left side influence factor g−1, and focusing attention on the (−i)th pixel value f−i described as a first member, it is successively multiplied by the reciprocals of the weighting factors w1, w2, . . . , wi−1, however, focusing attention on the (−1)st pixel value f−1 described as the last member (described as f−i+(i−1) in the diagram for the sake of explanation), it is not multiplied by the weighting factor at all, and focusing attention on the (−2)nd pixel value f−2 (described as f−i+(i−2) in the diagram for the sake of explanation) described as a member just before the (−1)st pixel value is multiplied by only the reciprocal of the weighting factor wi−1. This means that the farther the pixel value of the pixel is from the interpolation interval, the more times it is multiplied by the reciprocal of the weighting factor, and resultantly, the weighting factor in the third aspect of the invention is a parameter indicating weighting for each sample pixel according to the distance from the interpolation interval.
The (Expression 3-18) shown on the lower stage of FIG. 48 is a computation expression for this weighting factor wi, and is given in the form of a recurrence formula setting w0=2 as an initial value. By using this recurrence formula, the necessary number of weighting factors can be calculated in advance as w1=7/2, w2=26/7 and so on, and the calculated values can be stored in the table. Therefore, in actual interpolation, it is no longer necessary to perform calculation based on this (Expression 3-18), and necessary weighting factors are read from this table.
Resultantly, to calculate the pixel value of the new pixel Q defined at the arbitrary position x in the interpolation interval, first, the parameter i indicating the sum of the sample pixels to be used for interpolation is determined, and as shown in FIG. 47, the pixel values f−i through f−1 of i in total of left side interpolation pixels P−i through P−1 and pixel values f2 through fi+1 of i in total of right side interpolation pixels P2 through Pi+1 are obtained, necessary weighting factors w1, w2, . . . , wi+1 are readout from the table, and based on the (Expression 3-19) of FIG. 48, processing for obtaining the left side influence factor g−1, the right side influence factor g2, and the weighting factor w is performed. Thereafter, by using the pixel values f0 and f1 of the interval terminal pixels P0 and P1, the position x of the new pixel Q, the factors g−1 and g2 that have already been calculated, and the weighting factor w, the pixel value f(x) is calculated based on the (Expression 3-15) of FIG. 48.
Thus, according to the approach unique to the third aspect of the invention, first, as a previous step, as shown in FIG. 47, by using the pixel values f−i through f−1 of the left side interpolation pixels P−i through P−1, the left side influence factor g−1 indicating influences of these pixel values on the respective coefficients of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D” is calculated, and similarly, by using the pixel values f2 through fi+1 of the right side interpolation pixels P2 through Pi+1, the right side influence factor g2 indicating influences of these pixel values on the coefficients of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D” is calculated. Then, as a post-step, the pixel value f(x) is calculated by using the result of operation of said previous step and the pixel values f0 and f1 of the interval terminal pixels P0 and P1 and the position x shown in FIG. 47.
The cubic spline interpolation based on this process is also applicable to the case where the number of sample pixels to be used for interpolation is arbitrarily set, and very efficient operation becomes possible. Particularly, the weighting factor to be used for operations of both the previous step and the post-step can be prepared in advance as a constant stored in the form of a table, so that processing required when performing interpolation is only reading out from this table.
Section 3-3 Interpolation Device According to Basic Embodiment
Subsequently, a constitution of an interpolation device 800 according to a basic embodiment of the third aspect of the invention will be described with reference to the block diagram of FIG. 49. As illustrated, this device comprises a data input unit 810, a left side influence factor arithmetic unit 820, a right side influence factor arithmetic unit 830, an interpolated value arithmetic unit 840, and a weighting factor storage unit 850.
This interpolation device 800 has a function to calculate a pixel value f(x) of a new pixel Q defined at an arbitrary position x in the interpolation interval whose end points are at two adjacent pixels P0 and P1 based on a one-dimensional pixel array including a plurality of pixels having predetermined pixel values arranged at predetermined pitches by using a cubic polynomial “f(x)=Ax3+Bx2+Cx+D” concerning the position x.
On the upper stage of FIG. 49, a conceptional diagram of the one-dimensional pixel array to be subjected to this interpolation is shown. This one-dimensional pixel array is completely the same as that shown in FIG. 47, and cubic spline interpolation to calculate the pixel value f(x) of the new pixel Q defined at the position x on the interpolation interval satisfying “0≦x<1” on the x axis based on the pixel values of “2(i+1)” sample pixels is performed.
Namely, the 0-th pixel P0 and the first pixel P1 adjacent to each other in the one-dimensional pixel array are set as interval terminal pixels, and with respect to a predetermined integer i (i≦1), i in total of pixels from the (−i)th pixel P−i to the (−1)st pixel P−1 are defined as the left side interpolation pixels, and i in total of pixels from the second pixel P2 to the (i+1)th pixel Pi+1 are defined as the left side interpolation pixels. Then, S=2(i+1) sample pixels are defined, and the new pixel Q is defined at the position x in the range of 0≦x<1 provided that the position of the interval terminal pixel P0 is set to 0 and the position of the interval terminal pixel P1 is set to 1.
The data input unit 810 has a function to input the pixel values f0 and f1 of the interval terminal pixels P0 and P1 at both ends of the interpolation interval, the pixel values f−i through f−1 of the left side interpolation pixels P−i through P−1 as “predetermined number of pixels to be used for interpolation” (i pixels in the illustrated example) continuously arranged adjacent to each other on the left side of the interpolation interval, the pixel values f2 through fi+1 of the right side interpolation pixels P2 through Pi+1 as “predetermined number of pixels to be used for interpolation” (i pixels in the illustrated example) continuously arranged adjacent to each other on the right side of the interpolation interval, and the position x of the new pixel Q. In actuality, the data input unit 810 can be constituted by a buffer circuit which retrieves and temporarily stores these data.
On the other hand, the weighting factor storage unit 850 is a component having a function to store the weighting factor indicating weighting according to the distance from the interpolation interval for each sample pixel. Namely, it stores values of the weighting factors w1, w2, w3, . . . , wi expressed by the recurrence formula of wi=4−(1/wi−1), (where w0=2) shown in the (Expression 3-18). In detail, as described above, the weighting factor storage unit 850 can be constituted by a register circuit which stores values of the weighting factors w1=7/2, w2=26/7 and so on calculated in advance by this recurrence formula in the form of a table.
The left side influence factor arithmetic unit 820 is a component having a function to calculate the left side influence factor g−1 based on the pixel values f−i through f−1 of the left side interpolation pixels inputted by the data input unit 810 and the weighting factors w1 through wi−1 readout from the table in the weighting factor storage unit 850. This left side influence factor g−1 is a factor indicating influences of the respective pixel values f−i through f−1 of the left side interpolation pixels on the coefficients of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D,” and can be calculated, as described above, based on the first expression of the (Expression 3-19) of FIG. 48, that is, the following arithmetic expression:
g −1=−1/w i−1( . . . (−1/w 2(−1/w 1(−¼·f −1 +f −i+1)+f −i+2) . . . )+f −i+(i−2))+f −i+(i−1)
The right side influence factor arithmetic unit 830 is a component having a function to calculate the right side influence factor g2 based on the pixel values f2 through fi+1 of the right side interpolation pixels inputted by the data input unit 810 and the weighting factors w1 through wi−1 readout from the table in the weighting factor storage unit 850. This right side influence factor g2 indicates influences of the respective pixel values f2 through fi+1 of the right side interpolation pixels on the coefficients of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D,” and can be calculated, as described above, based on the second expression of the (Expression 3-19) of FIG. 48, that is, the following arithmetic expression:
g 2=−1/w i−1( . . . (−1/w 2(−1/w 1(−¼·f i+1 +f i)+f i−1) . . . )+f i+(3−i))+f i+(2−i)
Herein, the operations to be performed in the left side influence factor arithmetic unit 820 and the right side influence factor arithmetic unit 830 are repetition of product-sum operations as shown in the (Expression 3-19) of FIG. 48, so that these arithmetic units 820 and 830 can be realized by a combination of general multiplier and adder.
On the other hand, the interpolated value arithmetic unit 840 is a component having a function to calculate the pixel value f(x) by determining the coefficients A, B, C, and D of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D” based on the pixel values f0 and f1 of the interval terminal pixels inputted by the data input unit 810, the left side influence factor g−1 calculated by the left side influence factor arithmetic unit 820, the right side influence factor g2 calculated by the right side influence factor arithmetic unit 830, and the weighting factor wi readout from the table stored in the weighting factor storage unit 850, and performing an operation by substituting the position x inputted by the data input unit 810 for this cubic polynomial. As described above, this operation is performed based on the (Expression 3-15) of FIG. 48. As shown in this (Expression 3-15), the coefficient D becomes equal to the pixel value f0 of one interval terminal pixel, so that it can be determined without any operation.
FIG. 50 is a block diagram showing a constitution example of the interpolated value arithmetic unit 840 in the interpolation device 800 shown in FIG. 49. Herein, the cubic coefficient computing unit 841A performs an arithmetic operation to calculate a cubic coefficient A, and the quadratic coefficient computing unit 841B performs an arithmetic operation to calculate a quadratic coefficient B, and the primary coefficient computing unit 841C performs an arithmetic operation to calculate a primary coefficient C. All of these perform the arithmetic operations to calculate the respective coefficients based on the pixel values f0 and f1 of the interval terminal pixels, the left side influence factor g−1, the right side influence factor g2, and the weighting factor w.
In detail, as shown in the (Expression 3-15) of FIG. 48, the cubic coefficient computing unit 841A performs an arithmetic operation of A=1/(3w−1){−6g−1+(3w+1)(f0−f1)+6g2}, the quadratic coefficient computing unit 841B performs an arithmetic operation of B=3/((5w−1)(3w−1)){6(4w−1)g−1−(9w2+2w−1)f0+6w2f1−6wg2}, and the primary coefficient computing unit 841C performs an arithmetic operation of 3/((5w−1)(3w−1)){−2(7w−2)g−1−(w2−4w+1)f0+2w(2w−1)f1−2(2w−1)g2}. As described above, the coefficient D becomes D=f0, so that it does not need any operation.
On the other hand, the square/cube computing unit 842 is a component for calculating a square value x2 and a cube value x3 of a given position x. The cubic coefficient multiplier 843A calculates Ax3 by multiplying the cubic coefficient A calculated by the cubic coefficient computing unit 841 and the cube value x3 calculated by the square/cube computing unit 842. Then, the quadratic coefficient multiplier 843B calculates Bx2 by multiplying the quadratic coefficient B calculated by the quadratic coefficient computing unit 841B and the square value x2 calculated by the square/cube computing unit 842. Furthermore, the primary coefficient multiplier 843C calculates Cx by multiplying the primary coefficient C calculated by the primary coefficient computing unit 841C and the given position x.
Last, addition of Ax3+Bx2+Cx+D is performed by the adder 844 to find the pixel value f(x).
Section 3-4 More Efficient Embodiment
Resultantly, the interpolated value arithmetic unit 840 shown in FIG. 50 performs processing for calculating the pixel value f(x) by calculating the cubic polynomial “f(x)=Ax3+Bx2+Cx+D” without change. Therefore, the coefficients of the third power, the second power, and the first power of x require multiplication, respectively, and the number of computing units inevitably increases. Therefore, herein, an embodiment of the interpolated value arithmetic unit 840 with which more efficient arithmetic operation can be performed will be described.
As described above, the interpolated value arithmetic unit 840 is a component for calculating the cubic polynomial in the form of “f(x)=Ax3+Bx2+Cx+D” based on the (Expression 3-15) of FIG. 48, however, calculation of values directly indicating the four coefficients A, B, C, and D is not always necessary to calculate f(x).
Now, the following expression:
$f ( x ) = [ { ax + b ( 1 - x ) } x + c ( 1 - x ) ] x + d ( 1 - x ) = [ { ( a - b ) x + ( b - c ) } x + ( c - d ) ] x + d = ( a - b ) x 3 + ( b - c ) x 2 ( c - d ) x + d ( Expression 3 - 20 )$
is considered. The first, second, and third expressions of this (Expression 3-20) are all equivalent expressions. Herein, focusing attention on the third expression, it is understood that this expression expresses the cubic polynomial in the form of “f(x)=Ax3+Bx2+Cx+D” by using coefficients shown as the lower case letters a, b, c, and d defined so as to satisfy A=(a−b), B=(b−c), C=(c−d), and D=d instead of the coefficients shown as the upper case letters A, B, C, and D. The coefficients shown as the lower case letters a, b, c, and d are referred to as auxiliary coefficients.
By defining the relationship of A=(a−b), B=(b−c), C=(c−d), and D=d, resultantly, the (Expression 3-20) is completely equivalent to the cubic polynomial “f(x)=Ax3+Bx2+Cx+D,” so that in the interpolated value arithmetic unit 840 shown in FIG. 49, it is allowed that the pixel value f(x) is calculated by using this (Expression 3-20). As a matter of course, the values of the auxiliary coefficients a, b, c, and d are different from the values of the coefficients A, B, C, and D, so that an arithmetic expression for calculating the auxiliary coefficients a, b, c, and d becomes necessary. Considering the relationship of A=(a−b), B=(b−c), C=(c−d), and D=d, the values of the auxiliary coefficients a, b, c, and d can be calculated by the following expression:
$a = f 1 , b = ( 6 g - 1 - ( 3 w + 1 ) f 0 + 6 wf 1 - 6 g 2 ) ( 3 w - 1 ) c = 1 ( 3 w - 1 ) ( 5 w - 1 ) ( - 6 ( 7 w - 2 ) g - 1 + 2 ( 6 w 2 + 2 w - 1 ) f 0 + 6 w ( 2 w - 1 ) f 1 - 6 ( 2 w - 1 ) g 2 ) , d = f 0 ( Expression 3 - 21 )$
by using the pixel values f0 and f1 of the interval terminal pixels, the left side influence factor g−1, the right side influence factor g2, and the weighting factor w.
Thereby, in the interpolated value arithmetic unit 840, instead of the coefficients A, B, C, and D, the auxiliary coefficients a, b, c, and d defined so as to satisfy A=(a−b), B=(b−c), C=(c−d), and D=d are determined based on the (Expression 3-21), and by performing the operation of the cubic polynomial “f(x)=[{ax+b(1−x)}x+c(1−x)]x+d(1−x)” in which the position x is substituted, shown as the first expression of the (Expression 3-20) instead of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D,” the pixel value f(x) is calculated. Thus, employment of the arithmetic operation using the auxiliary coefficients a, b, c, and d instead of the coefficients A, B, C, and D has the following two advantages.
The first advantage is that the operation of the coefficients can be improved in efficiency. Carefully checking the (Expression 3-21), it is understood that the auxiliary coefficient a=f1 and the auxiliary coefficient d=f0. Namely, the auxiliary coefficients a and d are completely the same as the pixel values f1 and f0 of the interval terminal pixels, and no operation is necessary. Resultantly, in the arithmetic operation process for the auxiliary coefficients, performing of the operations to calculate the auxiliary coefficients b and c and the operations to calculate the factors g−1 and g2 and the weighting factor w necessary for said operation are only required. For example, when the parameter i=4 is set, interpolation using 10 in total of sample pixels (pixel values f−4 through f5) is performed, however, the actual operation is performed as follows:
${ b = 1 265 { 3 ( - f - 4 + 4 f - 3 - 14 f - 2 + 52 f - 1 ) - 317 f 0 + 582 f 1 + 3 ( f 2 - 4 f 4 + 14 f 3 - 52 f 2 ) } , c = 1 13515 { 209 ( f - 4 - 4 f - 3 + 14 f - 2 - 52 f - 1 ) + 13516 f 0 + 10864 f 1 + 56 ( f 5 - 4 f 4 + 14 f 3 - 52 f 2 ) } , g - 1 = 1 2 · 26 ( - f - 4 + 4 f - 3 - 14 f - 2 + 52 f - 1 ) , g 2 = - 1 2 · 26 ( f 5 - 4 f 4 + 14 f 3 - 52 f 2 ) , w = 97 26 ( Expression 3 - 22 )$
In this (Expression 3-22), when setting h−1=−f−4+4f−3−14f−2+52f−1 and h2=f5−4f4+14f3−52f2, by performing an operation in the following form:
${ h - 1 = - f - 4 + 4 f - 3 - 14 f - 2 + 52 f - 1 = - f - 4 + 4 f - 3 + ( 2 - 16 ) f - 2 + ( 32 + 16 + 4 ) f - 1 , h 2 = f 5 - 4 f 4 + 14 f 3 - 52 f 2 = f 5 - 4 f 4 + ( 16 - 2 ) f 3 - ( 32 + 16 + 4 ) f 2 , b = 3 265 h - 1 - 317 265 f 0 + 582 265 f 1 + 3 265 h 2 , c = - 209 13515 h - 1 + 13516 13515 f 0 + 10864 13515 f 1 + 56 13515 h 2 ( Expression 3 - 23 )$
h−1 and h2 are calculated, and furthermore, the auxiliary coefficients b and c can be calculated. This means that the auxiliary coefficients b and c can be calculated only by performing shift operation and product-sum operation.
The second advantage is the point that the operation to calculate f(x) by using the auxiliary coefficients a, b, c, and d can also be improved in efficiency. This is because the cubic polynomial concerning f(x) is in the form “f(x)=[{ax+b(1−x)}x+c(1−x)]x+d(1−x)” as shown as the first expression of the (Expression 3-20). This polynomial is equivalent to the “f(x)=Ax3+Bx2+Cx+D” as described above, however, it looks very complicated. However, the expression in this form has a unique advantage in that the value of f(x) can be calculated by connecting three existing linear interpolators in series. The reason for this is as follows.
Now, the case where two pixels Pa and Pb are present on the x axis and a new pixel P is defined between these pixels is considered. Herein, it is assumed that the pixel Pa is arranged at the position of the coordinate value x=0, the pixel Pb is arranged at the position of the coordinate value x=1, and the new pixel P is arranged at a coordinate value α (0≦α≦1). The pixel values of the pixels Pa, Pb, and P are indicated by the same reference symbols Pa, Pb, and P.
In this case, when the pixel value P of the new pixel P is calculated by simple linear interpolation, it can be calculated by the following expression:
$P = ( 1 - α ) P a + α P b = α _ P a + P a 2 - m + α P b ( Expression 3 - 24 )$
Herein, (1−α) and α are weighting factors for the pixel values Pa and Pb, and the second expression of this (Expression 3-24) indicates an expression as a binary operation. Namely, when m denotes the lowest-order decimal position, and a is a bit sequence expressed as:
α=[0. a 1 a 2 . . . a m]2 (Expression 3-25)
the bar of α indicates a value obtained by inverting each bit as:
α=[0. ā 1 ā 2 . . . ā m]2 (Expression 3-26)
The pixel values Pa, Pb, and P are also expressed as the following bit sequences:
Pa=[x1 x2 . . . xm]2, Pb=[y1 y2 . . . ym]2, P=[p1 p2 . . . pm·pm+1pm+2 . . . p2m]2 (Expression 3-27)
of binary numbers. An interpolator which performs interpolation based on this expression can be realized by a comparatively simple constitution. The pixel values Pa and Pb are expressed exclusively by a, so that the selector logic can be applied, and an advantage in that carrying can be reduced is obtained.
Considering that the arithmetic operation of “P=(1−α)Pa+αPb” can be performed by a linear interpolator having a comparatively simple constitution, it is understood that the aforementioned expression “f(x)=[{αx+b(1−x)}x+c(1−x)]x+d(1−x)” can be calculated by connecting three linear interpolators. Namely, focusing attention on the part of “ax+b(1−x),” when a=Pb, b=Pa, and α=x are applied, this operation becomes the same as the operation of “P=(1−α)Pa+αPb” of the linear interpolator. Therefore, the part “ax+b(1−x)” can be calculated by the first linear interpolator. Next, focusing attention on the part “{ax+b(1−x)}x+c(1−x),” when {ax+b(1−x)}=Pb, c=Pa, and α=x are applied, this operation becomes the same as the operation “P=(1−α)Pa+αPb” of the linear interpolator. Therefore, this part can be calculated by the second linear interpolator. Last, focusing on the entire expression of [{ax+b(1−x)}x+c(1−x)]x+d(1−x), when [{ax+b(1−x)}x+c(1−x)]=Pb, d=Pa, and α=x are applied, this operation also becomes the same as the operation “P=(1−α)Pa+αPb” of the linear interpolator. Therefore, this part can be calculated by the third linear interpolator.
Thus, when the auxiliary coefficients a, b, c, and d are determined instead of the coefficients A, B, C, and D, and an arithmetic operation using the cubic polynomial “f(x)=[{ax+b(1−x)}x+c(1−x)]x+d(1−x)” instead of the cubic polynomial “f(x)=Ax3+Bx2+Cx+D” is performed as the operation in the interpolated value arithmetic operation 840, arithmetic processing very efficient in practical use becomes possible. FIG. 51 is a block diagram showing a constitution example of the interpolated value arithmetic unit 840 for performing this arithmetic processing.
As illustrated, the interpolated value arithmetic unit 840 shown in FIG. 51 comprises a first auxiliary coefficient computing unit 845, a second auxiliary coefficient computing unit 846, a first linear interpolator 847, a second linear interpolator 848, and a third linear interpolator 849.
The first auxiliary coefficient computing unit 845 has a function to calculate the auxiliary coefficient b by the expression of b=(6g−1−(3w+1)f0+6wf1−6g2)/(3w−1) as illustrated (expression shown in the aforementioned expression 3-21) by using the pixel values f0 and f1 of the interval terminal pixels, the left side influence factor g−1, the right side influence factor g2, and the weighting coefficient w. The second auxiliary coefficient computing unit 846 has a function to calculate the auxiliary coefficient c by the expression of c=1/((3w−1)(5w−1))(−6(7w−2)g−1+2(6w2+2w−1)f0+6w(2w−1)f1−6(2w−1)g2) as illustrated (expression shown in the aforementioned expression 3-21) by using the pixel values f0 and f1 of the interval terminal pixels, the left side influence factor g−1, the right side influence factor g2, and the weighting factor w.
On the other hand, the linear interpolators 847, 848, and 849 are linear interpolators all having a function to calculate “P=(1−α)Pa+αPb.” Herein, to the first linear interpolator 847, the auxiliary coefficient b that is the result of the arithmetic operation of the first auxiliary coefficient computing unit 845 is given, and to the second linear interpolator 848, the auxiliary coefficient c that is the result of the arithmetic operation of the second auxiliary coefficient computing unit 846 is given. The first linear interpolator 847 inputs the pixel value f1 of one interval terminal pixel P1 as the auxiliary coefficient a, and calculates the value of {ax+b(1−x)} by performing linear interpolation based on the auxiliary coefficients a and b and the position x. The second linear interpolator 848 calculates the value of [{ax+b(1−x)}x+c(1−x)] by performing linear interpolation based on the value of {ax+b(1−x)}, the auxiliary coefficient c, and the position x. Then, the last third linear interpolator inputs the pixel value f0 of the other interval terminal pixel P0 as the auxiliary coefficient d, and calculates the value of f(x)=[{ax+b(1−x)}x+c(1−x)]x+d(1−x) by performing linear interpolation based on the value of [{ax+b(1−x)}x+c(1−x)], the auxiliary coefficient d, and the position x.
Thus, the interpolated value arithmetic unit shown in FIG. 51 can perform a function equivalent to that of the interpolated value arithmetic unit shown in FIG. 50, and its circuitry becomes very simple, so that more efficient operation becomes possible.
Section 3-5 Interpolation of Two-dimensional Image
An embodiment of the third aspect of the invention in which interpolation is performed with respect to a one-dimensional pixel array is described above, however, for scaling display by enlarging or reducing an image, interpolation must be performed for a two-dimensional image, that is, a two-dimensional pixel array. Therefore, herein, a constitution example of an interpolation device having a function to perform interpolation for a two-dimensional pixel array is described with reference to the block diagram of FIG. 52.
The interpolation device shown in FIG. 52 is an interpolation device for two-dimensional pixel arrays, however, it includes the whole of the interpolation device for one-dimensional pixel arrays of the embodiments described above. Namely, the interpolation device 800 shown in FIG. 52 is the interpolation device for one-dimensional pixel arrays described above (for example, a device having the constitution shown in FIG. 49). The interpolation device for two-dimensional pixel arrays shown in FIG. 52 is constituted by further adding an image input unit 910, an image storage unit 920, a new pixel position input unit 930, an interpolation target intersection determining unit 940, a row direction calculation control unit 950, and a column direction calculation control unit 960 to the interpolation device 800.
First, the image input unit 910 is a component having a function to input a general two-dimensional image, and the image storage unit 920 is a component having a function to store the inputted two-dimensional image. In actuality, the image storage unit 920 is constituted by an image buffer memory, etc., and the image input unit 910 is constituted by a device for writing data on this buffer memory. Herein, the following description is given upon regarding the two-dimensional image to be stored in the image storage unit 920 as a two-dimensional pixel array including a plurality of pixels having predetermined pixel values arranged at intersections between row lines arranged parallel at predetermined pitches and column lines orthogonal to the row lines.
FIG. 53 is a plan view of an example of such a two-dimensional pixel array. The nine horizontal lines shown in the figure are the row lines and the nine vertical lines are column lines, and white circles drawn at the intersections are pixels having predetermined pixel values, respectively. Herein, for the sake of convenience of explanation, a pixel array of nine rows and nine columns is shown, however, as a matter of course, in actuality, a larger pixel array is taken as a two-dimensional image.
The new pixel position input unit 930 has a function to input the positions of new pixels T (interpolation points) that should be defined on the two-dimensional image stored in the image storage unit 920. As a matter of course, the new pixels T are defined at positions different from the positions of the pixels on the two-dimensional pixel array stored in the image storage unit 920. Herein, for the sake of convenience of explanation, it is assumed that the new pixel T is defined at the position as shown in FIG. 53. The positions of the new pixels T are designated from another device which displays the image by enlarging or reducing it. Normally, a number of points are designated as the positions of new pixels T, however, herein, as shown in FIG. 53, it is assumed that one new pixel T is designated between the m-th row and the (m+1)th row between the n-th column and the (n+1)th column, and processing for determining the pixel value of this new pixel T by means of interpolation will be described.
When the position of the new pixel T is inputted, first, the interpolation target intersection determining unit 940 performs processing for defining a reference line R that passes through the position of the new pixel T and is parallel to the column lines on the two-dimensional image stored in the image storage unit 920. In FIG. 54, this reference line R is shown by a dashed line. Then, among intersections of the reference line R and the row lines, a predetermined number of intersections near the new pixel T are determined as interpolation target intersections. Herein, the number of intersections to be determined as the interpolation target intersections corresponds to the number of pixels to be used for interpolation, so that the number is determined in advance. In FIG. 54, an example in which six intersections U1 through U6 indicated by black circles are determined as the interpolation target intersections is shown.
As a matter of course, the six interpolation target intersections U1 through U6 are only intersections, and no pixel values are defined at this time. Therefore, to provide these interpolation target intersections U1 through U6 with a function as pixels, processing for giving pixel values to the respective intersections is performed. The row direction calculation control unit 950 is a component for performing this processing. Namely, concerning the interpolation target intersections U1 through U6, this row direction calculation control unit 950 makes the interpolation device 800 to calculate the pixel values of the interpolation target intersections by giving one-dimensional pixel arrays on the row lines, to which the intersections belong, to the interpolation device 800.
For example, in the case of the example shown in FIG. 54, first, concerning the interpolation target intersection U1 belonging to the (m−2)th row, a one-dimensional pixel array on this row line is given to the interpolation device 800 so that the interpolation device interpolates a pixel value of the interpolation target intersection U1. As described above, the interpolation device 800 is for one-dimensional pixel arrays, and can calculate the pixel value of the interpolation target intersection U1 by the method described above. Herein, for the sake of convenience, it is assumed that the interpolation for a one-dimensional pixel array is performed by using the six pixels indicated by cross marks on the (m−2)th row as sample pixels.
The row direction calculation control unit 950 successively calculates the pixel values of the interpolation target intersections U1 through U6 by using the interpolation device 800. As a result, to the interpolation target intersections U1 through U6, predetermined pixel values are given as interpolated values, and a function as a pixel is given.
The column direction calculation control unit 960 performs processing to make the interpolation device 800 to calculate the pixel values of the new pixels T by giving one-dimensional pixel arrays composed of the respective interpolation target intersections arranged on the reference line R to the interpolation device 800. In the case of the illustrated example, the interpolation target intersections U1 through U6 are respectively given as sample pixels to the interpolation device 800, and as a result, the pixel value of the new pixel T is determined.
Resultantly, in the case of the example shown in FIG. 54, the pixel value of the new pixel T is determined by performing an operation while considering the pixel values of 36 in total of sample pixels belonging to the (n−2)th column through the (n+3)th column and the (m−2)th row through the (m+3)th row, and as a result, interpolation for the two-dimensional image is executed.
Section 3-6 Embodiment Involving Carry Save
The interpolation device according to the third aspect of the invention comprises the left side influence factor arithmetic unit 820, the right side influence factor arithmetic unit 830, and the interpolated value arithmetic unit 840 as shown in FIG. 49, and each of these arithmetic units is constituted by a plurality of computing units. In this case, by constituting the computing unit on the middle stage so as to perform an operation without carry propagation, the operation can be made more efficient.
Generally, processing for propagating carries of the respective digits in an operation of binary data, the operation becomes complicated and efficiency lowers. Therefore, in the respective computing units, operations are performed without carry propagation, and the results are outputted as the results of operations to the subsequent computing units. At this time, the carry information of the respective bits are also outputted to the subsequent computing units. This processing is performed in the computing unit on each stage, and in the computing unit on the last stage, by performing carry propagation while considering the carry information, more efficient operation becomes possible.
According to the third aspect of the invention, an interpolation device consisting of compact hardware which can efficiently perform complicated operations of bicubic spline interpolation using S*S sample pixels can be realized. In addition, the number S (S≧4) of sample pixels to be used can be selected, and interpolation is directly performed at a high speed with high accuracy, so that an excellent display can be selected and displayed in real time in enlargement and reduction scaling of moving images. Therefore, this can be industrially used for scaling display of digital televisions, digital zooming display of digital cameras, fine zooming display of medical images, and zooming detection of a target in monitor cameras.
Thus, according to the third aspect of the invention, cubic spline interpolation can be efficiently performed. The summary of the third aspect of the invention can be described as follows with reference to FIG. 49. According to this invention, based on a one-dimensional pixel array composed of sample pixels P−i through Pi+1, a pixel value f(x) of a point Q on an arbitrary position x (0≦x<1) in an interpolation interval is calculated by interpolation using a cubic polynomial “f(x)=Ax3+Bx2+Cx+D.” First, for the sample pixels, weighting factors indicating weighting according to the distances from the point Q are prepared in the storage unit 850. Next, the pixel values f−i through fi+1 of the sample pixels are inputted by the data input unit 810. A left side influence factor g−1 indicating influences of the pixel values f−i through f−1 of left side interpolation pixels on the coefficients of the polynomial is calculated by the arithmetic unit 820, and a right side influence factor g2 indicating influences of the pixel values f2 through fi+1 of right side interpolation pixels on the coefficients of the polynomial is calculated by the arithmetic unit 830. Last, in the arithmetic unit 840, the pixel value f(x) is calculated by determining the coefficients A, B, C, and D by using g−1, f0, f1, g2, and the weighting factor.
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Non-Patent Citations
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Classifications
U.S. Classification708/290
International ClassificationG06F7/38
Cooperative ClassificationG06F7/544
European ClassificationG06F7/544
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2.1: Accelerators
Resolving power
Both nuclear and particle physics experiments are typically performed at accelerators, where particles are accelerated to extremely high energies, in most cases relativistic (i.e., $$v\approx c$$). To understand why this happens we need to look at the role the accelerators play. Accelerators are nothing but extremely big microscopes. At ultrarelativistic energies it doesn’t really matter what the mass of the particle is, its energy only depends on the momentum:
$E = h \nu =\sqrt{m^2c^4+{\vec{p}}^2c^2}\approx pc$
from which we conclude that
$\lambda =\frac{c}{\nu} = \frac{h}{p}.$
The typical resolving power of a microscope is about the size of one wave-length, $$\lambda$$. For an an ultrarelativistic particle this implies an energy of
$E=pc = h \frac{c}{\lambda}$
Table $$\PageIndex{0}$$: Size and energy-scale for various objects
particle scale energy
atom $$10^{-10}\rm m$$ 2 keV
nucleus $$10^{-14}\rm m$$ 20 MeV
nucleon $$10^{-15}\rm m$$ 200 MeV
quark? $$<10^{-18}\rm m$$ $$>$$200 GeV
You may not immediately appreciate the enormous scale of these energies. An energy of 1 TeV ($$=10^{12} \text{eV}$$) is $$3 \times 10^{-7}\text{ J}$$, which is the same as the kinetic energy of a 1g particle moving at 1.7 cm/s. And that for particles that are of submicroscopic size! We shall thus have to push these particles very hard indeed to gain such energies. In order to push these particles we need a handle to grasp hold of. The best one we know of is to use charged particles, since these can be accelerated with a combination of electric and magnetic fields – it is easy to get the necessary power as well.
Types
We can distinguish accelerators in two ways. One is whether the particles are accelerated along straight lines or along (approximate) circles. The other distinction is whether we used a DC (or slowly varying AC) voltage, or whether we use radio-frequency AC voltage, as is the case in most modern accelerators.
DC fields
Acceleration in a DC field is rather straightforward: If we have two plates with a potential $$V$$ between them, and release a particle near the plate at lower potential it will be accelerated to an energy $$\frac{1}{2} m v^2 = e V$$. This was the original technique that got Cockroft and Wolton their Nobel prize.
van der Graaff generator
A better system is the tandem van der Graaff generator, even though this technique is slowly becoming obsolete in nuclear physics (technological applications are still very common). The idea is to use a (non-conducting) rubber belt to transfer charge to a collector in the middle of the machine, which can be used to build up sizable (20 MV) potentials. By sending in negatively charged ions, which are stripped of (a large number of) their electrons in the middle of the machine we can use this potential twice. This is the mechanism used in part of the Daresbury machine.
Other linear accelerators
Linear accelerators (called Linacs) are mainly used for electrons. The idea is to use a microwave or radio frequency field to accelerate the electrons through a number of connected cavities (DC fields of the desired strength are just impossible to maintain). A disadvantage of this system is that electrons can only be accelerated in tiny bunches, in small parts of the time. This so-called “duty-cycle”, which is small (less than a percent) makes these machines not so beloved. It is also hard to use a linac in colliding beam mode (see below).
There are two basic setups for a linac. The original one is to use elements of different length with a fast oscillating (RF) field between the different elements, designed so that it takes exactly one period of the field to traverse each element. Matched acceleration only takes place for particles traversing the gaps when the field is almost maximal, actually sightly before maximal is OK as well. This leads to bunches coming out.
More modern electron accelerators are build using microwave cavities, where standing microwaves are generated. Such a standing wave can be thought of as one wave moving with the electron, and another moving the other wave. If we start of with relativistic electrons, $$v\approx c$$, this wave accelerates the electrons. This method requires less power than the one above.
Figure $$\PageIndex{3}$$: Acceleration by a standing wave
Cyclotron
The original design for a circular accelerator dates back to the 1930’s, and is called a cyclotron. Like all circular accelerators it is based on the fact that a charged particle (charge $$qe$$) in a magnetic field $$B$$ with velocity $$v$$ moves in a circle of radius $$r$$, more precisely
$q v B = \frac{\gamma m v^2}{r},$
here $$\gamma m$$ is the relativistic mass, $$\gamma=(1-\beta^2)^{-1/2}$$, $$\beta=v/c$$. A cyclotron consists of two metal “D”-rings, in which the particles are shielded from electric fields, and an electric field is applied between the two rings, changing sign for each half-revolution. This field then accelerates the particles.
The field has to change with a frequency equal to the angular velocity,
$f = \frac{\omega}{2\pi} = \frac{v}{2\pi r} = \frac{qB}{2\pi \gamma m}.$
For non-relativistic particles, where $$\gamma \approx 1$$, we can thus run a cyclotron at constant frequency, 15.25 MHz/T for protons. Since we extract the particles at the largest radius possible, we can determine the velocity and thus the energy,
$E=\gamma mc^2 = [(qBRc)^2+m^2c^4]^{1/2}$
Synchroton
The shear size of a cyclotron that accelerates particles to 100 GeV or more would be outrageous. For that reason a different type of accelerator is used for higher energy, the so-called synchroton where the particles are accelerated in a circle of constant diameter.
In a circular accelerator (also called synchroton), see Figure $$\PageIndex{5}$$, we have a set of magnetic elements that bend the beam of charged into an almost circular shape, and empty regions in between those elements where a high frequency electro-magnetic field accelerates the particles to ever higher energies. The particles make many passes through the accelerator, at every increasing momentum. This makes critical timing requirements on the accelerating fields, they cannot remain constant.
Using the equations given above, we find that
$f = \frac{q B}{2\pi \gamma m} = \frac{q B c^2}{2\pi E} =\frac{q B c^2}{2\pi (m^2 c^4+q^2B^2R^2c^2)^{1/2}}$
For very high energy this goes over to
$f = \frac{c}{2\pi R},\qquad E = q B R c,$
so we need to keep the frequency constant whilst increasing the magnetic field. In between the bending elements we insert (here and there) microwave cavities that accelerate the particles, which leads to bunching, i.e., particles travel with the top of the field.
So what determines the size of the ring and its maximal energy? There are two key factors:
As you know, a free particle does not move in a circle. It needs to be accelerated to do that. The magnetic elements take care of that, but an accelerated charge radiates – That is why there are synchroton lines at Daresbury! The amount of energy lost through radiation in one pass through the ring is given by (all quantities in SI units) $\Delta E = \frac{4\pi}{3 \epsilon_0} \frac{q^2\beta^3\gamma^4}{R}$ with $$\beta=v/c$$, $$\gamma=1/\sqrt{1-\beta^2}$$, and $$R$$ is the radius of the accelerator in meters. In most cases $$v \approx c$$, and we can replace $$\beta$$ by 1. We can also use one of the charges to re-express the energy-loss in eV: $\Delta E \approx \frac{4\pi}{3 \epsilon_0} \frac{q \gamma^4}{R} \Delta E \approx \frac{4\pi}{3 \epsilon_0} \frac{q}{R} \left( \frac{E}{m c^2} \right)^4.$ Thus the amount of energy lost is proportional to the fourth power of the relativistic energy, $$E=\gamma m c^2$$. For an electron at $$1$$ TeV energy $$\gamma$$ is $\gamma_e = \frac{E}{m_ec^2} = \frac{10^{12}}{511 \times 10^3} =1.9\times 10^6$ and for a proton at the same energy $\gamma_p = \frac{E}{m_pc^2} = \frac{10^{12}}{939 \times 10^6} = 1.1\times 10^3$ This means that a proton looses a lot less energy than an electron (the fourth power in the expression shows the difference to be $$10^{12}$$!). Let us take the radius of the ring to be 5 km (large, but not extremely so). We find the results listed in Table $$\PageIndex{1}$$.
Table $$\PageIndex{1}$$: Energy loss for a proton or electron in a synchroton of radius 5km
proton $$E$$ $$\Delta E$$
1 GeV $$1.5 \times 10^{-11}$$ eV
10 GeV
100 GeV $$1.5 \times 10^{-3}$$ eV
1000 GeV $$1.5 \times 10^{1}$$ eV
electron $$E$$ $$\Delta E$$
1 GeV $$2.2 \times 10^{2}$$ eV
10 GeV $$2.2$$ MeV
100 GeV $$22$$ GeV
1000 GeV $$2.2 \times 10^{15}$$ GeV
The other key factor is the maximal magnetic field. From the standard expression for the centrifugal force we find that the radius $$R$$ for a relativistic particle is related to it’s momentum (when expressed in GeV/c) by
$p = 0.3 B R$
For a standard magnet the maximal field that can be reached is about 1 T, for a superconducting one 5T. A particle moving at $$p=1{\rm TeV/c}=1000{\rm GeV/c}$$ requires a radius of
Table $$\PageIndex{2}$$: Radius $$R$$ of an synchroton for given magnetic fields and momenta.
$$B$$ $$p$$ $$R$$
1 T 1 GeV/c 3.3 m
10 GeV/c 33 m
100 GeV/c 330 m
1000 GeV/c 3.3 km
5 T 1 GeV/c 0.66 m
10 GeV/c 6.6 m
100 GeV/c 66 m
1000 GeV/c 660 m | 2,537 | 9,331 | {"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} | 3.515625 | 4 | CC-MAIN-2022-05 | latest | en | 0.883947 |
http://oeis.org/A121154 | 1,540,341,767,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00089.warc.gz | 264,015,105 | 3,639 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A121154 a(n) = the n-th divisor of r(n)^n, where r(n) = (10^n-1)/9 the repunits and the positive divisors of r(n)^n are written in order from smallest to largest. 0
1, 11, 9, 121, 11111, 13, 21613201, 1111, 333, 4961, 219660193449998401 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Sequence continues: n=12: unknown, n=13: 17530969, n=14: 562529, n=15: unknown, n=16: unknown, n=17: 79066065031254967758669825213076144489, LINKS EXAMPLE 1, 11, 101, 121, 1111, 1331, 10201, 12221, 14641, 112211, 134431,... is the beginning of the sequence of divisors of 1111^4 and 121 is the 4th term of this sequence, so a(4) = 121. CROSSREFS Cf. A121067. Sequence in context: A219746 A220664 A038323 * A256078 A078200 A105034 Adjacent sequences: A121151 A121152 A121153 * A121155 A121156 A121157 KEYWORD more,nonn AUTHOR Jason Earls, Aug 13 2006 STATUS approved
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Last modified October 23 20:19 EDT 2018. Contains 316530 sequences. (Running on oeis4.) | 450 | 1,364 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.694802 |
https://www.physicsforums.com/threads/how-do-you-convert-watts-per-meter-squared-to-decibels.236897/ | 1,579,823,086,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250614086.44/warc/CC-MAIN-20200123221108-20200124010108-00200.warc.gz | 1,029,542,709 | 17,725 | # How do you convert watts per meter squared to decibels?
## Main Question or Discussion Point
Hi.
How do you convert watts per metre squared to decibels?
I know you use logs but can't seems to work it out.
Thanks for any help.
Related Other Physics Topics News on Phys.org
Its ok I figured it out.
[decibels] = 10log + 120
where S = sound intensity in watts per metre squared
I may be comeplete wrong but aren't decibels 20Log(gain)? Also, the argument of the log must be unitless.
$$=\frac{W}{m^2}=\frac{J}{m^2\cdot s}=\frac{N}{m\cdot s} =\frac{kg\cdot m}{s^2}\frac{1}{m\cdot s}=\frac{kg}{s^3}\neq 1$$
I would just look at your equation again.
The power density must be divided by a certain constant power density j0 to obtain dimensionless argument for log. I think they chose the minimum power density that can be heard by humans for j0 (this is supposed to be 10^-12 W/m^2). So 10*log(S)+120 is more consistently written as
10*log(S/(1W/m^2))+120=10*log(S/(10^-12 W/m^2))
20*log() is used for sound pressure amplitudes. The power density is proportional to the square of the pressure amplitude. Since log(x^2)=2*log(x), the multiplication constant must be twice smaller in the first case (so that we the same increase of decibels for the same power/pressure change).
Last edited:
Sound intensity level is defined as:
$$L_J = 10 \log_{10}\left( \frac{J}{J_0} \right) \text{ dB}$$
where $J$ is the sound intensity and $J_0$ is the reference intensity level.
In the case that $J_0 = 10^{-12} \, \frac{ \text{W}}{\text{m}^2}$, the standard sound intensity reference level, the unit dB is sometimes changed to dB (SIL) where SIL stands for Sound Intensity Level.
Sound pressure level on the other hand as a slightly different formula, since sound pressure is related to sound intensity squared:
$$L_p = 10 \log_{10} \left( \frac{p^2}{p_0^2} \right) = 20 \log_{10} \left( \frac{p}{p_0} \right) \text{ dB}$$
where $p$ is the sound pressure and $p_0$ is the reference sound pressure.
Again, in the case that $p_0 = 2\times 10^{-5} \text{ Pa}$ (root-mean-square pressure), the standard reference sound pressure, the unit dB is sometimes changed to dB (SPL) where SPL stands for Sound Pressure Level.
These reference soundpressure/intensity are I believe related to the human threshold of hearing at 1000 Hz. I believe it is about the soundlevel of a mosquito flying at 3m distance.
Note that the unit dB is not actually a unit, it is dimensionless.
Also note that it's perfectly possible to have a negative sound level (-5 dB for example is not uncommon). It simply means that the soundlevel is below the average threshold of human hearing.
rbj
I may be comeplete wrong but aren't decibels 20Log(gain)?
when gain is a voltage ratio or a current ratio (or in the case of acoustics, the ratio of pressure deviation levels) it's 20log10(gain). but if it's the ratio of power or energy (which are proportional to the square of voltage or current or acoustic pressure), then it's 10log10(gain).
Also, the argument of the log must be unitless.
that was just playing fast and loose with the scalers.
actually Ed wasn't play so fast and loose:
[decibels] = 10log + 120
where S = sound intensity in watts per metre squared
the actual meaning of that is
[decibels] = (10dB)log(S/S0) + 120dB
where S0 = 1 W/m2
Last edited:
So, when measuring sound intensity in decibels using the standard for $$J_0$$, by what power does sound fall off? My intuition says it should be $$r^{-2}$$ because it spreads radially outwards but then there is air pressure and stuff like that in there.
rbj
J is a symbol for intensity. if energy is conserved, spherically radiating energy has to have intensity (power per unit area at a right angle to the direction of propagation) that is inverse-square. distance doubles, surface area of sphere centered at the point source increases by a factor of 4, equal power is distributed over 4 times the area, intensity is reduced by a factor of 1/4, and, since this is about power not voltage, that's -6.02 dB.
Sound intensity is proportional to $r^{-2}$ while sound pressure is propertional to $r^{-1}$. | 1,118 | 4,134 | {"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} | 4.03125 | 4 | CC-MAIN-2020-05 | longest | en | 0.841434 |
http://math.stackexchange.com/questions/283355/poisson-distribution-with-poisson-parameter | 1,469,385,051,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00012-ip-10-185-27-174.ec2.internal.warc.gz | 159,090,307 | 16,771 | # Poisson distribution with Poisson parameter
I have a problem with following exercise (it comes from Geoffrey G. Grimmett, David R. Stirzaker, Probability and Random Processes, Oxford University Press 2001, page 161, ex. 3a):
Let $X$ have the Poisson distribution with parameter $Y$ where $Y$ has the Poisson distribution with parameter $\mu$. Show that $$G_{X+Y}(s) = e^{\mu(s e^{s-1} - 1)}$$
So $G_Y(s) = e^{\mu(s-1)}$ Now I want to compute $G_X$ (is this approach correct?) $$G_X(s) = \sum_{x=0}^{\infty} s^x P(X=x) = \sum_{x=0}\sum_{y=0} s^x \frac{e^{-y} y^x}{x!} \frac{e^{-\mu} \mu^y}{y!}$$ $$G_X(s) = e^{-\mu} \sum_{x=0} \frac{s^x}{x!} \sum_{y=0} \frac{y^x e^{-y}\mu^y}{y!}$$
And I don't know how to cope with this summation.
Secondly, are these variables independent? I.e. can I use then following formula? $$G_{X+Y}(s) = G_X(s) G_Y(s)$$
-
1. $X$ and $Y$ are not independent, since $X$ has been defined using $Y$.
2. The easiest way to compute $G_{X+Y}(s)={\Bbb E}s^{X+Y}$ is to first condition on $Y$. So, first, $${\Bbb E}[s^{X+Y}\mid Y]=s^Y{\Bbb E}[s^X\mid Y]. \qquad (*)$$ Since $X$ is Poisson with parameter $Y$, conditioned on $Y$, the pgf of $X$ is $e^{Y(s-1)}$. Plugging this into (*) gives $${\Bbb E}[s^{X+Y}\mid Y]=s^Y e^{Y(s-1)}. \qquad (**)$$ The only remaining step is to take the expectation of (**). Since the right-hand side of (**) is of the form $\alpha^Y$, this can be done by plugging $\alpha$ into the pgf for $Y$. This gives the desired result.
- | 549 | 1,485 | {"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} | 3.859375 | 4 | CC-MAIN-2016-30 | latest | en | 0.728898 |
https://cstheory.stackexchange.com/questions/52872/bin-covering-problem-with-variable-bin-sizes | 1,723,265,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00001.warc.gz | 147,640,472 | 41,462 | # Bin Covering problem with variable bin sizes
I have a decision problem that I cannot seem to map to a standard studied problem, although it seems similar to a few. I am wondering if anyone has come across this problem before, or if someone can identify a standard name for this problem?
The decision problem is to check if a set of items (given size) can be assigned to a set of bins with each bin shrinking the size of the item by a scalar number, such that each bin contains at-least some scaled total size.
Formalizing
Assume we have n items in a list. Each of these has a positive size.
List of items $$L = (a_i, a_2 ,... , a_n)$$
Size of an items $$a_i$$ is given by $$s(a_i) ∈ ℝ⁺, i ∈ [1, n]$$
We have have m bins, each we wish to fill so that it contains a total size of C or more.
Furthermore, each bin has a size compressor associated with it. The size compressor scales down the size of any item in that bin.
Size compressor $$W_i ∈ (0, 1], i ∈ [1, m]$$
Hence if some set of items $$S_i \subseteq L$$ are assigned to bin i, the total size of items in the bin can be given as
$$W_i * \sum_{a∈S_i} s(a)$$.
The decision problem then is, can these n items be assigned to these m bins such that the total size of items in each bin is greater than C?
Similar Problems
It seems to me that this problem is similar to a few standard problems, but I can't seem to map it exactly to a problem I know. Most problems this seems to be similar to are NP Hard, so I think my problem is also NP Hard. I am thus hoping to map it to a known and studied problem so I can leverage a known approximate algorithms.
If $$W_i$$ = 1, i.e. no size compression, the problem could be constructed as a standard bin covering problem. The solution to this standard problem gives us the maximum number of bins that could be filled under this constraint, and comparing it to m would give us a solution to the decision problem. I tried looking for bin covering problems with variable bin sizes, to accommodate $$W_i ∈ (0, 1]$$, but this search came up empty.
Similarly, with the same assumption of $$W_i$$ = 1, one could construct this as a multiway number partitioning problem. Here, we would get the minimum total size that could be assigned to a bin under roughly equal size assignment, and compare it to C to give us a solution to our decision problem. I tried looking for multiway number partitioning problem with unequal partition sizes, to accommodate $$W_i ∈ (0, 1]$$, but this search came up empty.
Finally, it could be constructed as max-min fair item allocation problem. Here each bin assigns a subjective "value" (for us weight) to each item, and the goal is to maximize the minimum value each bin is allocated. While we can map our problem to this one, I feel like our problem should be somewhat easier to solve since the weight assigned to any bin is not arbitrary but some fixed scalar multiple of an inherent value. Here again, I couldn't find similar works.
We can try solving this problem using 0-1 integer linear programming, but I don't know if it is easy to show that the solutions we get are provably approximate.
• What is the context in which this problem arose? Feels like a homework problem since it is quite easy to prove it is NP-Complete with $W_i = 1$ for all $i$ and $m=2$. Commented May 24, 2023 at 0:09
• Thanks for your reply. Unfortunately, it is not a home work problem. It arose in the context of a quantum networking research where a single source is trying to assign resources to pairs of nodes in an undirected weighted graph. Resources can only be assigned in blocks. Each pair needs C resources after loss. Yes, I agree that that with Wi = 1 the problem reduces to a known one (I make mention of this in the original post near where i talk about reduction to the bin covering problem). Commented May 24, 2023 at 0:27
• I am confused. If all $W_i$ are $1$ then the problem is at least as hard as bin covering. By scaling this also shows that for any fixed $\alpha$ where $W_i = \alpha$ for all $i$ the problem is hard. I am not sure what you are asking for. Are the $W_i$ some fixed constants that are not part of the input? Commented May 24, 2023 at 1:52
• Yes, that is correct, $W_i$'s are fixed values I don't have control over. Ultimately, I am trying to find efficient approximate algorithms to solve this problem. E.g. I have 6 items with sizes [1, 3, 5,10, 3, 8] and four bins with $W_i$’s [0.5, 0.9, 0.2, 0.4]. I want to fill each bin so it is filled to at least a scaled size of 5 each. I am looking for an efficient approximate algorithm that could answer this question Commented May 24, 2023 at 14:18
• Since the problem is NP-hard even for $m=2$ and $W_i=1$ for all $i$ you need to relax the constraint. If you want to ensure that each bin is filled to at least $(1-\epsilon)C$ then it may be possible to do it for any fixed $\epsilon > 0$ in polynomial time. The related problem here is to schedule jobs to minimize makespan when processors have different speeds. epubs.siam.org/doi/10.1137/0217033 Commented May 25, 2023 at 19:47 | 1,288 | 5,079 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.954241 |
https://seanet2017.com/matematicas-simplest-180 | 1,674,950,122,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00495.warc.gz | 521,760,600 | 3,882 | # Find the zeros of a function solver
In this blog post, we will take a look at how to Find the zeros of a function solver. We will also look at some example problems and how to approach them.
## Try to find the zeros of a function solver
When you try to Find the zeros of a function solver, there are often multiple ways to approach it. solving equations is a process that involves isolating the variable on one side of the equation. This can be done using inverse operations, which are operations that undo each other. For example, addition and subtraction are inverse operations, as are multiplication and division. When solving an equation, you will use these inverse operations to move everything except for the variable to one side of the equal sign. Once the variable is isolated, you can then solve for its value by performing the inverse operation on both sides of the equation. For example, if you are solving for x in the equation 3x + 5 = 28, you would first subtract 5 from both sides of the equation to isolate x: 3x + 5 - 5 = 28 - 5. This results in 3x = 23. Then, you would divide both sides of the equation by 3 to solve for x: 3x/3 = 23/3. This gives you x = 23/3, or x = 7 1/3. Solving equations is a matter of isolating the variable using inverse operations and then using those same operations to solve for its value. By following these steps, you can solve any multi-step equation.
Algebra is the branch of mathematics that deals with the rules of operations and relations, and the study of quantities which may be either constant or variable. Factoring is a technique used to simplify algebraic expressions. When an expression is factored, it is rewritten as a product of simpler factors. This can be helpful in solving equations and graphing functions. In general, factoring is the process of multiplying two or more numbers to get a product. For example, 6 can be factored as 2 times 3, since 2 times 3 equals 6. In algebra, factoring is often used to simplify equations or to find solutions. For example, the equation x^2+5x+6 can be simplified by factoring it as (x+3)(x+2). This can be helpful in solving the equation, since now it can be seen that the solution is x=-3 or x=-2. Factoring can also be used to find zeroes of polynomials, which are important in graphing functions. In general, polynomials can be factored into linear factors, which correspond to zeroes of the function. For example, the function f(x)=x^2-4x+4 has zeroes at x=2 and x=4. These zeroes can be found by factoring the polynomial as (x-2)(x-4). As a result,factoring is a powerful tool that can be used to simplify expressions and solve equations.
To solve for the domain and range of a function, you will need to consider the inputs and outputs of the function. The domain is the set of all possible input values, while the range is the set of all possible output values. In order to find the domain and range of a function, you will need to consider what inputs and outputs are possible given the constraints of the function. For example, if a function takes in real numbers but only outputs positive values, then the domain would be all real numbers but the range would be all positive real numbers. Solving for the domain and range can be helpful in understanding the behavior of a function and identifying any restrictions on its inputs or outputs.
Math is a difficult subject for many people. It can be frustrating to get stuck on a problem and not know how to proceed. Luckily, there are a number of online Math solver websites that can help. These sites allow users to input a Math problem and receive step-by-step instructions on how to solve it. In addition, many of these sites also provide video tutorials and other resources that can help users understand the underlying concepts. As a result, Math solver websites can be a valuable resource for students who are struggling with Math. | 871 | 3,907 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2023-06 | latest | en | 0.950978 |
https://stdlib.io/develop/docs/api/@stdlib/math/base/dists/beta/median/ | 1,571,387,524,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00518.warc.gz | 717,022,991 | 55,623 | # Median
Beta distribution median.
The median for a beta random variable is the value for which the regularized incomplete beta function I(α,β) is equal to 0.5, i.e.
where α > 0 is the first shape parameter and β > 0 is the second shape parameter.
## Usage
var median = require( '@stdlib/math/base/dists/beta/median' );
#### median( alpha, beta )
Returns the median of a beta distribution with parameters alpha (first shape parameter) and beta (second shape parameter).
var v = median( 1.0, 1.0 );
// returns 0.5
v = median( 4.0, 12.0 );
// returns ~0.239
v = median( 8.0, 2.0 );
// returns ~0.820
If provided NaN as any argument, the function returns NaN.
var v = median( NaN, 2.0 );
// returns NaN
v = median( 2.0, NaN );
// returns NaN
If provided alpha <= 0, the function returns NaN.
var v = median( 0.0, 1.0 );
// returns NaN
v = median( -1.0, 1.0 );
// returns NaN
If provided beta <= 0, the function returns NaN.
var v = median( 1.0, 0.0 );
// returns NaN
v = median( 1.0, -1.0 );
// returns NaN
## Examples
var randu = require( '@stdlib/random/base/randu' );
var EPS = require( '@stdlib/constants/math/float64-eps' );
var median = require( '@stdlib/math/base/dists/beta/median' );
var alpha;
var beta;
var v;
var i;
for ( i = 0; i < 10; i++ ) {
alpha = ( randu()*10.0 ) + EPS;
beta = ( randu()*10.0 ) + EPS;
v = median( alpha, beta );
console.log( 'α: %d, β: %d, Median(X;α,β): %d', alpha.toFixed( 4 ), beta.toFixed( 4 ), v.toFixed( 4 ) );
} | 474 | 1,478 | {"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} | 3.03125 | 3 | CC-MAIN-2019-43 | latest | en | 0.299024 |
https://www.physicsforums.com/threads/calculating-new-velocity-from-velocity-and-momentum-confusing.707757/ | 1,508,434,940,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00801.warc.gz | 974,149,189 | 17,015 | # Calculating new velocity from velocity and momentum? Confusing
1. Aug 29, 2013
### lovesam
Hiya! Sam here,...and I'm new. I need help with this online homework problem. To be honeset, I have no idea where to even begin. I've tried the momentum equation where mass times velocity times a proportionality factor gamma, and I think it's the wrong start. Please help??
Question: A skater with a mass of 85kg is moving with a velocity of <3,1,0> m/s at t=3.2 s. If their momentum changes at a rate of <0, 170, 0> kg m/s^2 until t=3.7s, what is their new velocity?
2. Aug 29, 2013
### janhaa
Maybe you can use this equation...
$\dot{p}=m\frac{\Delta v}{t}=ma$
3. Aug 29, 2013
### CAF123
p=γmv is appropriate for relativistic velocities, this is clearly not the case here so you can use the classical form p=mv. It wouldn't be wrong to use the former eqn but γ would just be 1.
Can you write what 'If their momentum changes at a rate of <0, 170, 0> kg m/s^2' means mathematically?
4. Aug 29, 2013
### lovesam
Thank you! I'll try what you have given me.
Unfortunately, I do no quite understand what you mean in your question...
5. Aug 30, 2013
### CAF123
Do you understand the notation <..,...,...>? It is a vector in 3D space, so here the entries represent x, y and z directions. The rate of change of momentum in x and z directions are zero, while that in y is non zero:$$\frac{dp_x}{dt}= \frac{dp_z}{dt}= 0\,\,\,\text{while}\,\,\,\frac{dp_y}{dt} = 170$$ You can use the last equation to get eqn for momentum in y between the two times given. | 467 | 1,556 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-43 | longest | en | 0.89632 |
http://mathhelpforum.com/trigonometry/163161-r-formula-visualization-2.html | 1,527,063,565,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865456.57/warc/CC-MAIN-20180523063435-20180523083435-00214.warc.gz | 183,693,611 | 10,828 | # Thread: r formula visualization
1. Originally Posted by arccos
Well I already looked at the answer so i would pick 1-sqrt(5)(-1).So the point here is to notice that only by equating it to -1 can we obtain all positive values in the maximum value?
and if so, for the expression to be maximum, there must only be positive values?
I don't understand what you just said. The expression is maximum when the sine is minimum.
2. Originally Posted by arccos
Of course, you can plot the graph and take its maximum value from it but it really isn't necessary.
1 - sqrt(5)sin (x+26.57)
What we can do now is to decide for sin (x+26.57) to be maximum or minimum and that's by equating it to 1 and -1 because the maximum or minimum depends on sin (x+26.57). There is nothing else we can alter in the expression other than that.
3. Originally Posted by mathaddict
Of course, you can plot the graph and take its maximum value from it but it really isn't necessary.
1 - sqrt(5)sin (x+26.57)
What we can do now is to decide for sin (x+26.57) to be maximum or minimum and that's by equating it to 1 and -1 because the maximum or minimum depends on sin (x+26.57). There is nothing else we can alter in the expression other than that.
Sorry for the late reply.I think i kind of get what you were trying to bring across to me. 1-(-1) gives you a bigger value than 1-1 ,am i wrong?
Also,i've heard about cases with ^2(power) and to find the minimum value we have to equate it to 0 as 0^2 = 0 whilst 1 or (-1)^2 will give you a bigger value than if you were to equate it to 0.
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https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/03%3A_Two-Dimensional_Kinematics | 1,721,911,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763858305.84/warc/CC-MAIN-20240725114544-20240725144544-00188.warc.gz | 391,860,582 | 31,277 | # 3: Two-Dimensional Kinematics
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The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a wound, and a puppy chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow curved paths rather than straight lines. Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool table or a skater on an ice rink) is two-dimensional, and thus described by two-dimensional kinematics.
• 3.0: Prelude to Two-Dimensional Kinematics
Motion not confined to a plane, such as a car following a winding mountain road, is described by three-dimensional kinematics. Both two- and three-dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight-line motion in the previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected insights about nature.
• 3.1: Kinematics in Two Dimensions - An Introduction
An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle.
• 3.2: Vector Addition and Subtraction- Graphical Methods
A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.
• 3.3: Vector Addition and Subtraction- Analytical Methods
Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made.
• 3.4: Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.
• 3.5: Addition of Velocities
Velocities in two dimensions are added using the same analytical vector techniques. Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame. Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).
• 3.E: Two-Dimensional Kinematics (Exercises)
This page titled 3: Two-Dimensional Kinematics is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform. | 2,487 | 7,891 | {"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} | 2.96875 | 3 | CC-MAIN-2024-30 | latest | en | 0.198641 |
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;
;
49-43-94
;
93-34-39
,
,
Second 2 tandole wala formula
granty ke sath jis ko yaqeen na
mera pichla record check kare
,
,
Asif Khokher
,
Mobile 03013112583
Mobile 03013112583
,
,
,
,
,
,
,
,
,
,
SECOND BEST AAKRA ROUTEEN
,
,
,
1625 SE AB TAK UNBREAK
,
,
,
162 X 6789 X END = 1209(599)6
SECOND AAKRA PASS 95
,
,
,
,
714 X 6789 X END = 2349(676)3
SECOND AAKRA PASS 76
,
,
,
,
122 X 6789 X END 6860(113)1
SECOND AAKRA PASS 11
,
,
,
,
494 X 6789 X END 1124(774)6
SECOND AAKRA PASS 74
,
,
,
,
872 X 6789 X END 3504(649)4
SECOND AAKRA PASS 64
,
,
,
NOW NOW CALL ONLY 2 TANDOLE
,
,
,
631 X 6789 X END 1835(144)7
SECOND AAKRA WAIT
,
,
B 40000 SECOND AAKRA
,
,
14--41--44
14--41--44
14--41--44
14--41--44
14--41--44
,,
,,
ONLY 2 TANDOLE SECOND
,
ASIF KHOKHER
,
03013112583
03013112583
03013112583
,
,
,
,
,
,
,
,
,
,
FIRST-SECOND AAKRA FORMULA
,
,
7141 SE UNBREAK
,
,
737 X 714 = 526((218))
FIRST AAKRA PASS 12
,
,
,
,
737 X 122 = 89((419))
FIRST AAKRA PASS 49
,
,
,
,
737 X 494 = 634((078))
FIRST AAKRA PASS 87
,
,
,
,
737 X 872 = 642((664))
SECOND AAKRA PASS 64
,
,
,
NOW NOW WAIT WAIT
,
,
,
737 X 631 = 465((047))
FIRST-SECOND AAKRA WAIT
,
,
,
40000 BOND FIRST-SECOND AAKRA
,
,
,
04--07--40
,
47--70--74
,
,
,
,
,
04--07--40
,
47--70--74
,
,
,
FAQAT 2 TANDOLE SECOND GRANTY
FORMULE KE SATH ABHI RABTA KARE
,
,
,
ASIF KHOKHER
,
03013112583
03013112583
03013112583
Posted By:-
asif khokher
28-02-2017
14:53:06
Asif Khokher
-
-
Bond 40000
-
-
City Karachi
-
-
Is dafa faqat second 2 tandole
formule ke sath jise yaqeen na
ae wo pichla record check kare
-
-
-
V-VIP FIRST TANDOLA FORMULA GRANTY
,
,
,
122017 SE UNBREAK
,
,
128844 x XXXX = END
40145((494))4
494==944==449
FIRST TANDOLA PASS 494
,
,
,
,
498844 x XXXX = END
66332((782))5
782==872==827
FIRST TANDOLA PASS 872
,
,
,
,
878844 x XXXX = END
16048((631))5
631==361==613
FIRST TANDOLA PASS 631
,
,
,
,
NEW NAE FORMULE KE LIA CALL
,
,
,
,
638844 x XXXX = END
12668((CALL ME))5
call==call==call
FIRST TANDOLA PASS call
,
,
,
,
First faqat 3 tandole granty
formule ke sath abhi rabta kare
,
,
only asif khokher
,
03013112583
03013112583
-
-
-
-
Special First-Second Aakra
Formula
,
,
,
9369 Se unbreak
,
,
,
9369 x 444 = 41(5983)6
Second aakra Pass 35-83
,
,
,
,
3483 x 444 = 15(4645)2
First aakra Pass 65
,
,
,
,
6560 x 444 = 29(1264)0
Second aakra Pass 21-26-62
,
,
,
,
3852 x 444 = 17(1028)8
Second aakra Pass 20
,
,
,
,
5656 x 444 = 25(1126)4
Second aakra Pass 62
,
,
,
,
3818 x 444 = 16(9519)2
Second aakra Pass 59
,
,
,
,
5272 x 444 = 23(4076)8
Second aakra Pass 07
,
,
,
,
7207 x 444 = 31(9990)8
Second aakra Pass 00.00
,
,
,
,
1625 x 444 = 7(2150)0
Second aakra Pass 15-15
,
,
,
,
7141 x 444 = 3(1706)04
Second aakra Pass 76
,
,
,
,
1220 x 444 = 5(4168)0
Second aakra Pass Fail
,
,
,
,
4947 x 444 = 2(1964)68
Second aakra Pass 96
,
,
,
,
8720 x 444 = 3(8716)80
Second aakra Pass 87
,
,
,
,
6312 x 444 = 2(8025)28
First-Second aakra Pass WAIT
,
,
,
,
40000 Bond Karachi MAKE
AAKRA
,
,
,
80-82-85-08
,
,
02-05-28-20
,
,
25-58-50-52
,
,
Second 2 tandole aur 2 aakre
rabta kare formule ke sath
,
emandar log rabta akre
.
.
.
Asif Khokher
,
03013112583
03013112583
-
-
-
-
V-VIP First Taluk Routeen
,
,
0272 SE UNBREAK
,
,
027 X 2288 X = END
((235))75414
FIRST TALUK PASS 72
,
,
,
720 X 2288 X = END
((271))37949
FIRST TALUK PASS 16
,
,
,
162 X 2288 X = END
((137))38587
FIRST TALUK PASS 71
,
,
,
714 X 2288 X = END
((266))87535
FIRST TALUK PASS 12
,
,
,
122 X 2288 X = END
((779))16906
FIRST TALUK PASS 49
,
,
,
,
494 X 2288 X = END
((127))75147
FIRST TALUK PASS 87
,
,
,
,
872 X 2288 X = END
((398))05676
FIRST TALUK PASS 63
,
,
now now wait wait
,
,
631 X 2288 X = END
((208))43505
FIRST TALUK PASS WAIT
,
,
40000 BOND V-VIP TALUK
,
,
2,,,,,0,,,,,8
2,,,,,0,,,,,8
2,,,,,0,,,,,8
2,,,,,0,,,,,8
2,,,,,0,,,,,8
2,,,,,0,,,,,8
2,,,,,0,,,,,8
-
-
100% Granty First 2 aakra Routeen
aj hi rabta kare only asif
.
.
Asif Khokher
,
03013112583
03013112583
03013112583
,
,
,
,
,
,
,
,
,
,
FIRST -- SECOND AAKRA FORMULA
;
;
;
UNBREAK FROM 1625
;
162569
551655 X = END LOCK = END
((1678))8413
FIRST AAKRA PASS 71
;
;
;
;
714160
557155 X = END LOCK = END
((1729))5299
FIRST AAKRA PASS 12
;
;
;
;
122017
551255 X = END LOCK = END
((1657))1651
SECOND AAKRA PASS 65
;
;
;
;
494769
554955 X = END LOCK = END
((1709))1229
SECOND AAKRA PASS 09
;
;
;
;
872005
558755 X = END LOCK = END
((1744))4730
SECOND AAKRA PASS 74
;
;
NOW NOW WAIT WAIT
;
;
631069
556355 X = END LOCK = END
((1722))0905
FIRST-SECOND AAKRA WAIT
;
;
;
;
FIRST-SECOND AAKRA MAKE 40000 BOND
;
;
12-17-21-22
;
27-71-72
;
;
12-17-21-22
;
27-71-72
;
;
;
FIRST Single aakra tayar hai
abhi rabta kare Asif Khokher
,
Mobile 03013112583
,
,
,
,
,
,
,
,
,
,
V-VIP ONLY FIRST AAKRA FORMULA
,
,
,
1220 SE UNBREAK
;
;
1220 X 7823 = 954
FIRST AAKRA PASS 49
,
,
,
,
4947 X 7823 = 387
FIRST AAKRA PASS
,
,
,
,
8720 X 7823 = 682
FIRST AAKRA PASS 63
,
,
NOW NOW WAIT WAIT
,
,
6312X 7823 = 493
FIRST AAKRA PASS WAIT
,
,
,
BOND 40000 MAKE FIRST AAKRA
,
,
,
49-43-94
;
93-34-39
;
;
49-43-94
;
93-34-39
,
,
Second 2 tandole wala formula
granty ke sath jis ko yaqeen na
mera pichla record check kare
,
,
Asif Khokher
,
Mobile 03013112583
Mobile 03013112583
,
,
,
,
,
,
,
,
,
,
SECOND BEST AAKRA ROUTEEN
,
,
,
1625 SE AB TAK UNBREAK
,
,
,
162 X 6789 X END = 1209(599)6
SECOND AAKRA PASS 95
,
,
,
,
714 X 6789 X END = 2349(676)3
SECOND AAKRA PASS 76
,
,
,
,
122 X 6789 X END 6860(113)1
SECOND AAKRA PASS 11
,
,
,
,
494 X 6789 X END 1124(774)6
SECOND AAKRA PASS 74
,
,
,
,
872 X 6789 X END 3504(649)4
SECOND AAKRA PASS 64
,
,
,
NOW NOW CALL ONLY 2 TANDOLE
,
,
,
631 X 6789 X END 1835(144)7
SECOND AAKRA WAIT
,
,
B 40000 SECOND AAKRA
,
,
14--41--44
14--41--44
14--41--44
14--41--44
14--41--44
,,
,,
ONLY 2 TANDOLE SECOND
,
ASIF KHOKHER
,
03013112583
03013112583
03013112583
,
,
,
,
,
,
,
,
,
,
FIRST-SECOND AAKRA FORMULA
,
,
7141 SE UNBREAK
,
,
737 X 714 = 526((218))
FIRST AAKRA PASS 12
,
,
,
,
737 X 122 = 89((419))
FIRST AAKRA PASS 49
,
,
,
,
737 X 494 = 634((078))
FIRST AAKRA PASS 87
,
,
,
,
737 X 872 = 642((664))
SECOND AAKRA PASS 64
,
,
,
NOW NOW WAIT WAIT
,
,
,
737 X 631 = 465((047))
FIRST-SECOND AAKRA WAIT
,
,
,
40000 BOND FIRST-SECOND AAKRA
,
,
,
04--07--40
,
47--70--74
,
,
,
,
,
04--07--40
,
47--70--74
,
,
,
FAQAT 2 TANDOLE SECOND GRANTY
FORMULE KE SATH ABHI RABTA KARE
,
,
,
ASIF KHOKHER
,
03013112583
03013112583
03013112583 | 5,817 | 13,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-13 | longest | en | 0.23591 |
https://download.racket-lang.org/releases/6.1/doc/reference/parametric-contracts.html | 1,627,499,293,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00539.warc.gz | 235,537,241 | 5,424 | #### 8.3Parametric Contracts
(require racket/contract/parametric) package: base
The most convenient way to use parametric contract is to use contract-out’s #:exists keyword. The racket/contract/parametric provides a few more, general-purpose parametric contracts.
syntax(parametric->/c (x ...) c)
Creates a contract for parametric polymorphic functions. Each function is protected by c, where each x is bound in c and refers to a polymorphic type that is instantiated each time the function is applied.
At each application of a function, the parametric->/c contract constructs a new opaque wrapper for each x; values flowing into the polymorphic function (i.e. values protected by some x in negative position with respect to parametric->/c) are wrapped in the corresponding opaque wrapper. Values flowing out of the polymorphic function (i.e. values protected by some x in positive position with respect to parametric->/c) are checked for the appropriate wrapper. If they have it, they are unwrapped; if they do not, a contract violation is signaled.
Examples:
> (define/contract (check x y) (parametric->/c [X] (boolean? X . -> . X)) (if (or (not x) (equal? y 'surprise)) 'invalid y))
> (check #t 'ok)
'ok
> (check #f 'ignored)
check: broke its contract
promised: X
produced: 'invalid
in: the range of
...
(parametric->/c (X) ...)
contract from: (function check)
blaming: (function check)
at: eval:2.0
> (check #t 'surprise)
'surprise
procedure(new-∀/c [name]) → contract? name : (or/c symbol? #f) = #f
Constructs a new universal contract.
Universal contracts accept all values when in negative positions (e.g., function inputs) and wrap them in an opaque struct, hiding the precise value. In positive positions (e.g. function returns), a universal contract accepts only values that were previously accepted in negative positions (by checking for the wrappers).
The name is used to identify the contract in error messages and defaults to a name based on the lexical context of new-∀/c.
For example, this contract:
(let ([a (new-∀/c 'a)]) (-> a a))
describes the identity function (or a non-terminating function). That is, the first use of the a appears in a negative position and thus inputs to that function are wrapped with an opaque struct. Then, when the function returns, it is checked to determine whether the result is wrapped, since the second a appears in a positive position.
The new-∀/c construct constructor is dual to new-∃/c.
procedure(new-∃/c [name]) → contract? name : (or/c symbol? #f) = #f
Constructs a new existential contract.
Existential contracts accept all values when in positive positions (e.g., function returns) and wrap them in an opaque struct, hiding the precise value. In negative positions (e.g. function inputs), they accepts only values that were previously accepted in positive positions (by checking for the wrappers).
The name is used to identify the contract in error messages and defaults to a name based on the lexical context of new-∀/c.
For example, this contract:
(let ([a (new-∃/c 'a)]) (-> (-> a a) any/c))
describes a function that accepts the identity function (or a non-terminating function) and returns an arbitrary value. That is, the first use of the a appears in a positive position and thus inputs to that function are wrapped with an opaque struct. Then, when the function returns, it is checked to see if the result is wrapped, since the second a appears in a negative position.
The new-∃/c construct constructor is dual to new-∀/c. | 886 | 3,524 | {"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} | 2.71875 | 3 | CC-MAIN-2021-31 | latest | en | 0.693375 |
https://studysoup.com/tsg/25564/college-physics-1-edition-chapter-20-problem-10pe | 1,632,736,438,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00616.warc.gz | 565,101,296 | 13,158 | ×
Get Full Access to College Physics - 1 Edition - Chapter 20 - Problem 10pe
Get Full Access to College Physics - 1 Edition - Chapter 20 - Problem 10pe
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# A clock battery wears out after moving 10,000 C of charge
ISBN: 9781938168000 42
## Solution for problem 10PE Chapter 20
College Physics | 1st Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
College Physics | 1st Edition
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5
Problem 10PE
Problem 10PE
A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second flowed?
Step-by-Step Solution:
Step 1 of 3
Solution 10PE
1. Time t = 231.481 Days
2. 3.12 x 1017 Electrons per second
Step 2 of 3
Step 3 of 3
##### ISBN: 9781938168000
This textbook survival guide was created for the textbook: College Physics , edition: 1. The answer to “A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second flowed?” is broken down into a number of easy to follow steps, and 34 words. College Physics was written by and is associated to the ISBN: 9781938168000. Since the solution to 10PE from 20 chapter was answered, more than 1163 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 10PE from chapter: 20 was answered by , our top Physics solution expert on 07/07/17, 04:39PM. This full solution covers the following key subjects: clock, flowed, Battery, DID, electrons. This expansive textbook survival guide covers 34 chapters, and 3125 solutions.
#### Related chapters
Unlock Textbook Solution | 488 | 1,791 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-39 | latest | en | 0.899898 |
https://www.doityourself.com/forum/greenhouses-sheds-sun-rooms/231610-how-much-would-8-10-wood-shed-weigh.html | 1,679,769,332,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945368.6/warc/CC-MAIN-20230325161021-20230325191021-00145.warc.gz | 821,293,427 | 53,617 | > >
>
# How much would a 8 * 10 wood shed weigh?
#1
09-13-05, 06:57 AM
Member
Join Date: Jun 2004
Location: usa
Posts: 93
How much would a 8 * 10 wood shed weigh?
I'm looking at moving a wood shed, I removed the floor ( dirt now). I nailed 2*4s across the inside of the shed, about 1 foot up into the studs from the inside. (Also braced the door opening) Thinking with 6 guys, using the 2*4s as handles, we could lift and move it (standing on the inside). I also screwed the 2*4s into the studs (3"), and hope that they would not pull out when we lift it.
Want to move it 20 feet. Already made a new base in a dry location. The old floor was rotten ( was on garden ties - and each time it rained, water went into the shed)
Thinking also perhaps using dollys with tires to sit the corners on, then roll it. But how much would this thing weigh? I figure no more then 1000 lb, so two dollys ar 900 lb ratings should work. ( and we lift/pull on the other corners)
Any comments? will this work, or will I be a pain in everone's back!
#2
09-18-05, 04:23 PM
Member
Join Date: Mar 2000
Location: Arlington, WA
Posts: 8,670
Received 1 Upvote on 1 Post
So far, you have told us the shed is 8 X 10, and it's wood. Based on that, there is NO WAY to make a guess as to how much it might weigh.
How tall it it? What kind of siding does it have? What kind of roof? 2X2 walls? 2X12 walls?
Actually, none of this matters. The way to move the shed 20' is to lift one end, place a 3" or larger pipe under that end, lift it farther and move that pipe closer to the center and add a second identical pipe under the end you lifted first, then start moving the shed. You'll need a third, and probably a 4th pipe to keep moving the shed onto. Once the shed is moved off of the first pipe, move the pipe to the oter end and roll the shed onto it. Once you have the shed where you want it, remove all of the pipes. That way, you aren't lifting more than 1/2 of the weight of the shed at any one time, and the pipes are doing most of the work.
#3
10-03-05, 12:38 PM
Member
Join Date: Jun 2004
Location: usa
Posts: 93
Thanks for the reply, I got it moved on 4 dollies
I also disovered that bottle nose jacks are not very stable when trying to lift something with jacks only.! ( floor jacks worked better)
I got it up on 4 dollies with tires on them, connected a chain to the dollies and connected to my truck. Had to put it into 4*4 to get it to move ( all it did was spin the tires!) But it moved, It is now living it its new location. (In one piece also!)
So, with a truck, 4 guys, couple of Dollies and a case of beer, anything is possible! | 746 | 2,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-14 | latest | en | 0.969514 |
https://rdrr.io/rforge/nor1mix/man/norMix.html | 1,521,561,514,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647498.68/warc/CC-MAIN-20180320150533-20180320170533-00329.warc.gz | 684,338,704 | 58,732 | # norMix: Mixtures of Univariate Normal Distributions In nor1mix: Normal (1-d) Mixture Models (S3 Classes and Methods)
## Description
Objects of class `norMix` represent finite mixtures of (univariate) normal (aka Gaussian) distributions. Methods for construction, printing, plotting, and basic computations are provided.
## Usage
``` 1 2 3 4 5 6 7 8 9 10 11 12``` ```norMix(mu, sig2 = rep(1,m), sigma = rep(1,m), w = NULL, name = NULL, long.name = FALSE) is.norMix(obj) m.norMix(obj) var.norMix(x, ...) ## S3 method for class 'norMix' mean(x, ...) ## S3 method for class 'norMix' print(x, ...) ## S3 method for class 'norMix' x[i,j, drop=TRUE] ```
## Arguments
`mu` numeric vector of length K, say, specifying the means μ of the K normal components. `sig2` deprecated! numeric vector of length K, specifying the variances σ^2 of the K normal components. Do specify `sigma` instead! `sigma` numeric vector of length K, specifying the standard deviations σ of the K normal components. `w` numeric vector of length K, specifying the mixture proportions p[j] of the normal components, j = 1,…,K. Defaults to equal proportions `name` optional name tag of the result (used for printing). `long.name` logical indicating if the `name` attribute should use punctuation and hence be slightly larger than by default. `obj,x ` an object of class `norMix`. `i,j,drop` for indexing, see the generic `[` extractor function. `...` further arguments passed to methods.
## Details
The (one dimensional) normal mixtures, R objects of class `"norMix"`, are constructed by `norMix` and tested for by `is.norMix`. `m.norMix()` returns the number of mixture components; the `mean()` method for `class "norMix"` returns the (theoretical / true) mean E[X] and `var.norMix()` the true variance E[(X- E[X])^2] where X ~ <norm.mixt>.
The subsetting aka “extract” method (`x[i,j]`; for generic `[`)—when called as `x[i,]`—will typically return a `"norMix"` object unless matrix indexing selects only one row in which case `x[i, , drop=FALSE]` will return the normal mixture (of one component only).
For further methods (density, random number generation, fitting, ...), see below.
## Value
`norMix` returns objects of class `"norMix"` which are currently implemented as 3-column matrix with column names `mu`, `sigma`, and `w`, and further attributes. The user should rarely need to access the underlying structure directly.
## Note
For estimation of the parameters of such a normal mixture, we provide a smart parametrization and an efficient implementation of the direct MLE or also the EM algorithm, see `norMixMLE()` which includes `norMixEM()`.
## Author(s)
Martin Maechler
`dnorMix` for the density, `pnorMix` for the cumulative distribution and the quantile function (`qnorMix`), and `rnorMix` for random numbers and `plot.norMix`, the plot method.
`MarronWand` has the Marron-Wand densities as normal mixtures.
`norMixMLE()` and `norMixEM()` provide fitting of univariate normal mixtures to data.
## Examples
``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22``` ```ex <- norMix(mu = c(1,2,5))# defaults: sigma = 1, equal proportions ('w') ex plot(ex, p.comp = TRUE)# looks like a mixture of only 2; 'p.comp' plots components ## The 2nd Marron-Wand example, see also ?MW.nm2 ex2 <- norMix(name = "#2 Skewed", mu = c(0, .5, 13/12), sigma = c(1, 2/3, 5/9), w = c(.2, .2, .6)) m.norMix (ex2) mean (ex2) var.norMix(ex2) (e23 <- ex2[2:3,]) # (with re-normalized weights) stopifnot(is.norMix(e23), all.equal(var.norMix(ex2), 719/1080, tol=1e-14), all.equal(var.norMix(ex ), 35/9, tol=1e-14), all.equal(var.norMix(ex[2:3,]), 13/4, tol=1e-14), all.equal(var.norMix(e23), 53^2/(12^3*4),tol=1e-14) ) plot(ex2, log = "y")# maybe "revealing" ```
### Example output
```'Normal Mixture' object ``NM3.125_111''
mu sigma w
[1,] 1 1 0.3333333
[2,] 2 1 0.3333333
[3,] 5 1 0.3333333
Warning message:
In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated.
[1] 3
[1] 0.75
[1] 0.6657407
Warning message:
In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated.
'Normal Mixture' object ``#2 Skewed[2:3,]''
mu sigma w
[1,] 0.500000 0.6666667 0.25
[2,] 1.083333 0.5555556 0.75
Warning messages:
1: In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated.
2: In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated.
3: In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated.
4: In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated.
Warning messages:
1: In mj - mu :
Recycling array of length 1 in vector-array arithmetic is deprecated. | 1,433 | 4,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-13 | longest | en | 0.649561 |
https://hoven-in.appspot.com/Home/Aptitude/Boats-and-Streams/maths-aptitude-quiz-questions-and-mock-test-on-boats-and-streams-002.html | 1,716,793,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00560.warc.gz | 251,812,834 | 6,566 | # Boats and Streams Quiz Set 002
### Question 1
The speed of a boat in still water is 84 km/h. It takes the same time to travel an upstream distance of 9 km as it takes to travel a downstream distance of 19 km. The rate of flow of the stream is?
A
30 km/h.
B
31 km/h.
C
29 km/h.
D
33 km/h.
Soln.
Ans: a
Let v km/h be the rate of flow of the stream. Upstream and downstream times are equal. So \$9/{84 - v}\$ = \$19/{84 + v}\$. Solving for v, or verifying by putting the given options one by one, we get v = 30 km/h.
### Question 2
Two similar boats, A and B, start to move towards each other. If they meet when A has travelled \$(1/8)\$th of the distance, what is the ratio of the upstream speed to the downstream speed of a boat in that river?
A
7.
B
8.
C
6.
D
9.
Soln.
Ans: a
The boats are same, so they have the same speed in still water. Let the upstream speed be u and v be the downstream speed, and let the initial distance between them be L. When they meet they have travelled for the same time. So \$(L/8)/u = ({7L}/8)/v\$. The ratio \$v/u\$ = 7.
### Question 3
The speed of a boat in still water is 10 km/h and rate of flow of the stream is 4 km/h. If it travels upstream for 5 hours, what is the distance travelled by the boat during the journey?
A
30 km.
B
31 km.
C
29 km.
D
33 km.
Soln.
Ans: a
The effective speed of the boat in the stream is 10 - 4 = 6 km/h. The distance is speed × time = 6 × 5 = 30 km.
### Question 4
Two similar boats, A and B, start to move towards each other. If they meet when A has travelled \$(1/5)\$th of the distance, what is the ratio of the upstream speed to the downstream speed of a boat in that river?
A
4.
B
5.
C
3.
D
6.
Soln.
Ans: a
The boats are same, so they have the same speed in still water. Let the upstream speed be u and v be the downstream speed, and let the initial distance between them be L. When they meet they have travelled for the same time. So \$(L/5)/u = ({4L}/5)/v\$. The ratio \$v/u\$ = 4.
### Question 5
The speed of a steamer in still water, the upstream speed of the steamer and the downstream speed of the steamer form an A.P.(Arithmetic Progression) whose common difference is \$1{1/2}\$. What is the speed of the river?
A
\$1{1/2}\$ units.
B
\$2{1/2}\$ units.
C
\$3{1/2}\$ units.
D
\$2{1/4}\$ units.
Soln.
Ans: a
Let the speed of the steamer in still water be u and the speed of the river be v. The three speeds are u - v, u and u + v. They are in an A.P. with a common difference equal to v. Since v is also the speed of the river, the required answer is: \$1{1/2}\$. | 782 | 2,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.924488 |
http://www.best-practice.com/risk-management-best-practices/financial-risk-management/calculating-gear-ratio-or-debt-capital-ratio/calculating-gear-ratio-or-debt-capital-ratio-29122011/ | 1,723,094,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00145.warc.gz | 38,755,980 | 28,337 | ## Calculating Gear Ratio or Debt/ Capital Ratio
The gearing ratio or debt/Capital ratio, calculates the percentage of the total funds that represent the debt of the company. It is very important for best practice in financial management. It is also referred to as the debt-to-capital ratio, or simply – debt ratio. Calculating this ratio is a fundamental best practice that serves as a tool for analyzing how the company gets funded.
## Importance of the Gear Ratio
The figure obtained as the gear ratio is helpful in comparing the liabilities of the company in relation to the total capital. It provides companies the opportunity to review their dependence on debt financing that is usually external. In simpler terms, it defines the risks associated for stakeholders to be able to implement best practices for risk management.
Knowing the gear ratio also allows business owners to know their borrowing capacity. This is a recommended best practice that enables risk management to a great extent. This information also assists in best practices related to scheduling financial payments to settle the debts and leases. The debt/capital ratio is a best practice tool for analysts and bond rating companies to evaluate the creditworthiness of the company.
Assuming that the lowest value of gear ratio is the best achievement is a mistake and goes against risk management. This is what misleads most companies into taking larger loans to expand their business. Usually utilities that have higher capital requirements will have a higher gear ratio. Therefore, those companies that develop or manufacture new products or technology tend to have a higher gear ratio. This not a recommended best practice. It is important to mention that companies with higher debt must have positive earnings and a steady flow of cash to ensure risk management.
## Calculating the Gear Ratio
There are many approaches for calculating the gear ratio with best practices. However, the most common approach is to divide the sum of liabilities (or the long-term debts) by the sum total of assets. The sum total of assets is obtained as a sum of the liabilities and funds of the stockholders.
This is illustrated in the following mathematical equation:
Gear Ratio or debt/capital ratio = Sum of Liabilities ÷ Sum Total of Assets
## An Example of Calculation
Let us assume that the balance sheet of a company reports the total liabilities as \$10,000,000 and the total fund of stockholders as \$13,000,000. To ensure sufficient risk management it needs to know their gear ratio. The gear ratio or debt/capital ratio will be calculated as follows:
10,000,000 ÷ (10,000,000 + 13,000,000) = 10,000,000 ÷ (23,000,000) = 0.434. This can be expressed as a percentage by multiplying it by 100. Therefore the gear ratio will be reported as 43.4%.
Other approaches to calculate the gear ratio separate the different portion of the total of liabilities. However, for best practices this approach is not ideal because it becomes too complicated. Most experts consider those steps of the calculation as pointless. Some business people actually prefer to divide the total liabilities by the total funds.
Whichever approach one takes, the outcome is usually the same. However, ensuring that there are no errors is recommended and therefore sticking to simpler formulas is the ideal best practice.
Further reading: Corporate Governance | Audit | Performance Improvement | 676 | 3,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.95282 |
https://numberworld.info/202220100010 | 1,660,020,549,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570901.18/warc/CC-MAIN-20220809033952-20220809063952-00358.warc.gz | 396,524,397 | 3,902 | # Number 202220100010
### Properties of number 202220100010
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
2f1541d9aa
Base 32:
5sak3mda
sin(202220100010)
0.72436478514595
cos(202220100010)
0.68941689712428
tan(202220100010)
1.0506919516586
ln(202220100010)
26.03262254517
lg(202220100010)
11.305824320837
sqrt(202220100010)
449688.892469
Square(202220100010)
4.0892968848054E+22
### Number Look Up
Look Up
202220100010 which is pronounced (two hundred two billion two hundred twenty million one hundred thousand ten) is a very great figure. The cross sum of 202220100010 is 10. If you factorisate the figure 202220100010 you will get these result 2 * 5 * 20222010001. The number 202220100010 has 8 divisors ( 1, 2, 5, 10, 20222010001, 40444020002, 101110050005, 202220100010 ) whith a sum of 363996180036. 202220100010 is not a prime number. The number 202220100010 is not a fibonacci number. The figure 202220100010 is not a Bell Number. The figure 202220100010 is not a Catalan Number. The convertion of 202220100010 to base 2 (Binary) is 10111100010101010000011101100110101010. The convertion of 202220100010 to base 3 (Ternary) is 201022222001210110011201. The convertion of 202220100010 to base 4 (Quaternary) is 2330111100131212222. The convertion of 202220100010 to base 5 (Quintal) is 11303121321200020. The convertion of 202220100010 to base 8 (Octal) is 2742520354652. The convertion of 202220100010 to base 16 (Hexadecimal) is 2f1541d9aa. The convertion of 202220100010 to base 32 is 5sak3mda. The sine of 202220100010 is 0.72436478514595. The cosine of the number 202220100010 is 0.68941689712428. The tangent of the figure 202220100010 is 1.0506919516586. The root of 202220100010 is 449688.892469.
If you square 202220100010 you will get the following result 4.0892968848054E+22. The natural logarithm of 202220100010 is 26.03262254517 and the decimal logarithm is 11.305824320837. that 202220100010 is very unique number! | 734 | 2,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.667996 |
https://practice.geeksforgeeks.org/problems/unique-bsts-1587115621/1 | 1,623,759,362,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00026.warc.gz | 436,486,045 | 25,800 | Showing:
Handle Score
@Ibrahim Nash 6420
@mb1973 5704
@Quandray 5245
@akhayrutdinov 5111
@saiujwal13083 5046
@sanjay05 3762
@kirtidee18 3673
@marius_valentin_dragoi 3523
@mantu_singh 3510
@sushant_a 3459
Unique BST's
Medium Accuracy: 44.17% Submissions: 24442 Points: 4
Given an integer. Find how many structurally unique binary search trees are there that stores the values from 1 to that integer (inclusive).
Example 1:
Input:
N = 2
Output: 2
Explanation:for N = 2, there are 2 unique
BSTs
1 2
\ /
2 1
Example 2:
Input:
N = 3
Output: 5
Explanation: for N = 3, there are 5
possible BSTs
1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
You don't need to read input or print anything. Your task is to complete the function numTrees() which takes the integer N as input and returns the total number of Binary Search Trees possible with keys [1.....N] inclusive. Since the answer can be very large, return the answer modulo 1000000007.
Expected Time Complexity: O(N2).
Expected Auxiliary Space: O(N).
Constraints:
1<=N<=1000
### Editorial
We strongly recommend solving this problem on your own before viewing its editorial. Do you still want to view the editorial?
Unique BST's | 421 | 1,343 | {"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} | 2.859375 | 3 | CC-MAIN-2021-25 | longest | en | 0.635104 |
http://www.freemathhelp.com/forum/archive/index.php/t-71116.html | 1,386,933,004,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164931675/warc/CC-MAIN-20131204134851-00099-ip-10-33-133-15.ec2.internal.warc.gz | 346,791,502 | 2,336 | PDA
View Full Version : Applications of Quadratic Functions Help?
Zib
06-19-2011, 03:59 PM
The question asks me to find the equation of the quadratic function in standard form that has zeros 1+the square root of 11 and 1-the square root of 11 and that contains the point (4, -6)
Can anyone tell me how to set this up to begin with?
JeffM
06-19-2011, 05:07 PM
The question asks me to find the equation of the quadratic function in standard form that has zeros 1+the square root of 11 and 1-the square root of 11 and that contains the point (4, -6)
Can anyone tell me how to set this up to begin with?
Suppose u and v are zeroes of f(x).
Then f(x) = a(x - u)(x - v). (The proof of that IMPORTANT relationship is messy).
Now can you solve the problem?
Zib
06-19-2011, 05:44 PM
Thank you, but what I'm having trouble with is how 1+the square root of 11 and 1-the square root of 11 could be fit into that form. Can you help me please?
JeffM
06-19-2011, 05:59 PM
Thank you, but what I'm having trouble with is how 1+the square root of 11 and 1-the square root of 11 could be fit into that form. Can you help me please?
They are just numbers. Substitute them into a(x - u)(x - v) and do the algebra.
a(x - u)(x - v) = ax[sup:122sqf83]2[/sup:122sqf83] - ax(u + v) + auv. Original post neglected to insert the x into ax(u + v).
u = 1 + 11[sup:122sqf83]1/2[/sup:122sqf83].
v = 1 - 11[sup:122sqf83]1/2[/sup:122sqf83].
So u + v = 2.
And uv = 1 - 11 = 10.
So put the quadratic into standard form. The only parameter left is in a. So what value of a results in a quadratic that goes through the stipulated point?
Zib
06-19-2011, 06:36 PM
3! Thank you!
Except, it only works when I expand (x - u)(x - v) = ax2 - ax(u + v) + auv? Did you just miss a key, or does that expansion only work by coincidence?
mmm4444bot
06-19-2011, 11:37 PM
I expand (x - u)(x - v) = ax2 - ax(u + v) + auv? Did you just miss a key
Yes -- Jeff intended to type -ax(u + v)
Looks like you got: f(x) = 3x^2 - 6x - 30
Good work. 8-)
PS: Note the caret symbol in red? That's how we text exponent notation. (It's a shift 6, on most keyboards.)
And, we can text "square root of 11" as sqrt(11).
JeffM
06-20-2011, 12:11 AM
Yes I miskeyed. Sorry for any confusion it may have caused.
Zib
06-20-2011, 10:56 PM
Thank you both :D | 760 | 2,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2013-48 | longest | en | 0.872285 |
https://books.google.ie/books?qtid=757e783a&dq=editions:UOM39015065618988&lr=&id=r3QAAAAAMAAJ&output=html_text&sa=N&start=70 | 1,685,431,957,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00669.warc.gz | 172,557,112 | 6,011 | Books Books
If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
The First Six Books: Together with the Eleventh and Twelfth - Page 200
by Euclid - 1781 - 520 pages
## Euclid's Elements of Geometry: Chiefly from the Text of Dr. Simson, with ...
Robert Potts - Geometry, Plane - 1860 - 380 pages
...Wherefore, if two triangles, &c. QED PROPOSITION XXVI. THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz, either the sides adjacent to the equal angles in each, or the sides...
## The examination papers as set for the preliminary literary examination of ...
Royal college of surgeons of England - 1860 - 336 pages
...other, then the sides AB, BC shall lie in one straight line. 3. If two triangles have two angles of the one equal to two angles of the other, each to each ; and one side equal to one side, viz., the sides adjacent to the equal angles in each triangle ; then shall...
## Examination papers used at the examinations for direct commissions [&c.].
War office - 1861 - 714 pages
...another, the sides also which subtend the equal angles shall be equal to one another. 2. If two triangles have two angles of one equal to two angles of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles, or the sides opposite...
## The school Euclid: comprising the first four books, by A.K. Isbister
Euclides - 1862 - 172 pages
...to the angle CFD ; (m. 27) and BFC is double of the angle KFC, and CFD double of CFL ; 145 there are two angles of one equal to two angles of the other, each to each, and the side FC, which is adjacent to the equal angles in each, is common to both ; therefore the other sides...
## Euclid's Elements of geometry, books i. ii. iii. iv
Euclides - 1862 - 140 pages
...Therefore, if two triangles, &c. QED PROPOSITION 26.— THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side; namely, either the side adjacent to the equal angles in sach, or the aide...
## Responsions
University of Oxford - Education, Higher - 1863 - 328 pages
...plane superficies. Write out Euclid's three postulates. 2. If two triangles have two angles of the one equal to two angles of the other, each to each, and the sides adjacent to the equal angles also equal, then shall the other sides be equal, each to each ;...
## The elements of plane geometry; or, The first six books of Euclid, ed. by W ...
Euclides - 1863 - 122 pages
...(I. Ax. 11) to the right angle BFD. Therefore the two triangles E BD and FBD have two angles of the one equal to two angles of the other, each to each ; and the side BD, which is opposite to one of the equal angles in each, is common to both. Therefore their other...
## Euclid's plane geometry, practically applied; book i, with explanatory notes ...
Euclides - 1863 - 72 pages
...LARDXEB.S Euclid, p. 56. PROP. 26.— THEOR. — (Important.) If two triangles have two angles of the one equal to two angles of the other* each to each, and one side equal to one side, viz., either the sides adjacent to the equal angles in each, or the sides... | 850 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-23 | latest | en | 0.857765 |
http://www.popflock.com/learn?s=Lie_group | 1,610,956,028,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514423.60/warc/CC-MAIN-20210118061434-20210118091434-00271.warc.gz | 167,034,561 | 52,601 | Lie Group
Get Lie Group essential facts below. View Videos or join the Lie Group discussion. Add Lie Group to your PopFlock.com topic list for future reference or share this resource on social media.
Lie Group
In mathematics, a Lie group (pronounced "Lee") is a group that is also a differentiable manifold. A manifold is a space that locally resembles Euclidean space, whereas groups define the abstract, generic concept of multiplication and the taking of inverses (division). Combining these two ideas, one obtains a continuous group where points can be multiplied together, and their inverse can be taken. If, in addition, the multiplication and taking of inverses are defined to be smooth (differentiable), one obtains a Lie group.
Lie groups provide a natural model for the concept of continuous symmetry, a celebrated example of which is the rotational symmetry in three dimensions (given by the special orthogonal group ${\displaystyle {\text{SO}}(3)}$). Lie groups are widely used in many parts of modern mathematics and physics.
Lie groups were first found by studying matrix subgroups ${\displaystyle G}$ contained in ${\displaystyle {\text{GL}}_{n}(\mathbb {R} )}$ or ${\displaystyle {\text{GL}}_{n}(\mathbb {C} )}$, the groups of ${\displaystyle n\times n}$ invertible matrices over ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$. These are now called the classical groups, as the concept has been extended far beyond these origins. Lie groups are named after Norwegian mathematician Sophus Lie (1842-1899), who laid the foundations of the theory of continuous transformation groups. Lie's original motivation for introducing Lie groups was to model the continuous symmetries of differential equations, in much the same way that finite groups are used in Galois theory to model the discrete symmetries of algebraic equations.
## Overview
The set of all complex numbers with absolute value 1 (corresponding to points on the circle of center 0 and radius 1 in the complex plane) is a Lie group under complex multiplication: the circle group.
Lie groups are smooth differentiable manifolds and as such can be studied using differential calculus, in contrast with the case of more general topological groups. One of the key ideas in the theory of Lie groups is to replace the global object, the group, with its local or linearized version, which Lie himself called its "infinitesimal group" and which has since become known as its Lie algebra.
Lie groups play an enormous role in modern geometry, on several different levels. Felix Klein argued in his Erlangen program that one can consider various "geometries" by specifying an appropriate transformation group that leaves certain geometric properties invariant. Thus Euclidean geometry corresponds to the choice of the group E(3) of distance-preserving transformations of the Euclidean space R3, conformal geometry corresponds to enlarging the group to the conformal group, whereas in projective geometry one is interested in the properties invariant under the projective group. This idea later led to the notion of a G-structure, where G is a Lie group of "local" symmetries of a manifold.
Lie groups (and their associated Lie algebras) play a major role in modern physics, with the Lie group typically playing the role of a symmetry of a physical system. Here, the representations of the Lie group (or of its Lie algebra) are especially important. Representation theory is used extensively in particle physics. Groups whose representations are of particular importance include the rotation group SO(3) (or its double cover SU(2)), the special unitary group SU(3) and the Poincaré group.
On a "global" level, whenever a Lie group acts on a geometric object, such as a Riemannian or a symplectic manifold, this action provides a measure of rigidity and yields a rich algebraic structure. The presence of continuous symmetries expressed via a Lie group action on a manifold places strong constraints on its geometry and facilitates analysis on the manifold. Linear actions of Lie groups are especially important, and are studied in representation theory.
In the 1940s–1950s, Ellis Kolchin, Armand Borel, and Claude Chevalley realised that many foundational results concerning Lie groups can be developed completely algebraically, giving rise to the theory of algebraic groups defined over an arbitrary field. This insight opened new possibilities in pure algebra, by providing a uniform construction for most finite simple groups, as well as in algebraic geometry. The theory of automorphic forms, an important branch of modern number theory, deals extensively with analogues of Lie groups over adele rings; p-adic Lie groups play an important role, via their connections with Galois representations in number theory.
## Definitions and examples
A real Lie group is a group that is also a finite-dimensional real smooth manifold, in which the group operations of multiplication and inversion are smooth maps. Smoothness of the group multiplication
${\displaystyle \mu :G\times G\to G\quad \mu (x,y)=xy}$
means that ? is a smooth mapping of the product manifold into G. These two requirements can be combined to the single requirement that the mapping
${\displaystyle (x,y)\mapsto x^{-1}y}$
be a smooth mapping of the product manifold into G.
### First examples
${\displaystyle \operatorname {GL} (2,\mathbf {R} )=\left\{A={\begin{pmatrix}a&b\\c&d\end{pmatrix}}:\,\det A=ad-bc\neq 0\right\}.}$
This is a four-dimensional noncompact real Lie group; it is an open subset of ${\displaystyle \mathbb {R} ^{4}}$. This group is disconnected; it has two connected components corresponding to the positive and negative values of the determinant.
• The rotation matrices form a subgroup of , denoted by . It is a Lie group in its own right: specifically, a one-dimensional compact connected Lie group which is diffeomorphic to the circle. Using the rotation angle ${\displaystyle \varphi }$ as a parameter, this group can be parametrized as follows:
${\displaystyle \operatorname {SO} (2,\mathbf {R} )=\left\{{\begin{pmatrix}\cos \varphi &-\sin \varphi \\\sin \varphi &\cos \varphi \end{pmatrix}}:\,\varphi \in \mathbf {R} /2\pi \mathbf {Z} \right\}.}$
Addition of the angles corresponds to multiplication of the elements of , and taking the opposite angle corresponds to inversion. Thus both multiplication and inversion are differentiable maps.
• The affine group of one dimension is a two-dimensional matrix Lie group, consisting of ${\displaystyle 2\times 2}$ real, upper-triangular matrices, with the first diagonal entry being positive and the second diagonal entry being 1. Thus, the group consists of matrices of the form
${\displaystyle A=\left({\begin{array}{cc}a&b\\0&1\end{array}}\right),\quad a>0,\,b\in \mathbb {R} .}$
### Non-example
We now present an example of a group with an uncountable number of elements that is not a Lie group under a certain topology. The group given by
${\displaystyle H=\left\{\left({\begin{matrix}e^{2\pi i\theta }&0\\0&e^{2\pi ia\theta }\end{matrix}}\right):\,\theta \in \mathbb {R} \right\}\subset \mathbb {T} ^{2}=\left\{\left({\begin{matrix}e^{2\pi i\theta }&0\\0&e^{2\pi i\phi }\end{matrix}}\right):\,\theta ,\phi \in \mathbb {R} \right\},}$
with ${\displaystyle a\in \mathbb {R} \setminus \mathbb {Q} }$ a fixed irrational number, is a subgroup of the torus ${\displaystyle \mathbb {T} ^{2}}$ that is not a Lie group when given the subspace topology.[1] If we take any small neighborhood ${\displaystyle U}$ of a point ${\displaystyle h}$ in ${\displaystyle H}$, for example, the portion of ${\displaystyle H}$ in ${\displaystyle U}$ is disconnected. The group ${\displaystyle H}$ winds repeatedly around the torus without ever reaching a previous point of the spiral and thus forms a dense subgroup of ${\displaystyle \mathbb {T} ^{2}}$.
A portion of the group ${\displaystyle H}$ inside ${\displaystyle \mathbb {T} ^{2}}$. Small neighborhoods of the element ${\displaystyle h\in H}$ are disconnected in the subset topology on ${\displaystyle H}$
The group ${\displaystyle H}$ can, however, be given a different topology, in which the distance between two points ${\displaystyle h_{1},h_{2}\in H}$ is defined as the length of the shortest path in the group ${\displaystyle H}$ joining ${\displaystyle h_{1}}$ to ${\displaystyle h_{2}}$. In this topology, ${\displaystyle H}$ is identified homeomorphically with the real line by identifying each element with the number ${\displaystyle \theta }$ in the definition of ${\displaystyle H}$. With this topology, ${\displaystyle H}$ is just the group of real numbers under addition and is therefore a Lie group.
The group ${\displaystyle H}$ is an example of a "Lie subgroup" of a Lie group that is not closed. See the discussion below of Lie subgroups in the section on basic concepts.
### Matrix Lie groups
Let ${\displaystyle \operatorname {GL} (n,\mathbb {C} )}$ denote the group of ${\displaystyle n\times n}$ invertible matrices with entries in ${\displaystyle \mathbb {C} }$. Any closed subgroup of ${\displaystyle \operatorname {GL} (n,\mathbb {C} )}$ is a Lie group;[2] Lie groups of this sort are called matrix Lie groups. Since most of the interesting examples of Lie groups can be realized as matrix Lie groups, some textbooks restrict attention to this class, including those of Hall[3] and Rossmann.[4] Restricting attention to matrix Lie groups simplifies the definition of the Lie algebra and the exponential map. The following are standard examples of matrix Lie groups.
• The special linear groups over ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} }$, ${\displaystyle \operatorname {SL} (n,\mathbb {R} )}$ and ${\displaystyle \operatorname {SL} (n,\mathbb {C} )}$, consisting of ${\displaystyle n\times n}$ matrices with determinant one and entries in ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$
• The unitary groups and special unitary groups, ${\displaystyle {\text{U}}(n)}$ and ${\displaystyle {\text{SU}}(n)}$, consisting of ${\displaystyle n\times n}$ complex matrices satisfying ${\displaystyle U^{*}=U^{-1}}$ (and also ${\displaystyle \det(U)=1}$ in the case of ${\displaystyle {\text{SU}}(n)}$)
• The orthogonal groups and special orthogonal groups, ${\displaystyle {\text{O}}(n)}$ and ${\displaystyle {\text{SO}}(n)}$, consisting of ${\displaystyle n\times n}$ real matrices satisfying ${\displaystyle R^{\mathrm {T} }=R^{-1}}$ (and also ${\displaystyle \det(R)=1}$ in the case of ${\displaystyle {\text{SO}}(n)}$)
All of the preceding examples fall under the heading of the classical groups.
### Related concepts
A complex Lie group is defined in the same way using complex manifolds rather than real ones (example: ${\displaystyle \operatorname {SL} (2,\mathbb {C} )}$), and similarly, using an alternate metric completion of ${\displaystyle \mathbb {Q} }$, one can define a p-adic Lie group over the p-adic numbers, a topological group in which each point has a p-adic neighborhood.
Hilbert's fifth problem asked whether replacing differentiable manifolds with topological or analytic ones can yield new examples. The answer to this question turned out to be negative: in 1952, Gleason, Montgomery and Zippin showed that if G is a topological manifold with continuous group operations, then there exists exactly one analytic structure on G which turns it into a Lie group (see also Hilbert-Smith conjecture). If the underlying manifold is allowed to be infinite-dimensional (for example, a Hilbert manifold), then one arrives at the notion of an infinite-dimensional Lie group. It is possible to define analogues of many Lie groups over finite fields, and these give most of the examples of finite simple groups.
The language of category theory provides a concise definition for Lie groups: a Lie group is a group object in the category of smooth manifolds. This is important, because it allows generalization of the notion of a Lie group to Lie supergroups.
### Topological definition
A Lie group can be defined as a (Hausdorff) topological group that, near the identity element, looks like a transformation group, with no reference to differentiable manifolds.[5] First, we define an immersely linear Lie group to be a subgroup G of the general linear group ${\displaystyle \operatorname {GL} (n,\mathbb {C} )}$ such that
1. for some neighborhood V of the identity element e in G, the topology on V is the subspace topology of ${\displaystyle \operatorname {GL} (n,\mathbb {C} )}$ and V is closed in ${\displaystyle \operatorname {GL} (n,\mathbb {C} )}$.
2. G has at most countably many connected components.
(For example, a closed subgroup of ${\displaystyle \operatorname {GL} (n,\mathbb {C} )}$; that is, a matrix Lie group satisfies the above conditions.)
Then a Lie group is defined as a topological group that (1) is locally isomorphic near the identities to an immersely linear Lie group and (2) has at most countably many connected components. Showing the topological definition is equivalent to the usual one is technical (and the beginning readers should skip the following) but is done roughly as follows:
1. Given a Lie group G in the usual manifold sense, the Lie group-Lie algebra correspondence (or a version of Lie's third theorem) constructs an immersed Lie subgroup ${\displaystyle G'\subset \operatorname {GL} (n,\mathbb {C} )}$ such that ${\displaystyle G,G'}$ share the same Lie algebra; thus, they are locally isomorphic. Hence, G satisfies the above topological definition.
2. Conversely, let G be a topological group that is a Lie group in the above topological sense and choose an immersely linear Lie group ${\displaystyle G'}$ that is locally isomorphic to G. Then, by a version of the closed subgroup theorem, ${\displaystyle G'}$ is a real-analytic manifold and then, through the local isomorphism, G acquires a structure of a manifold near the identity element. One then shows that the group law on G can be given by formal power series;[6] so the group operations are real-analytic and G itself is a real-analytic manifold.
The topological definition implies the statement that if two Lie groups are isomorphic as topological groups, then they are isomorphic as Lie groups. In fact, it states the general principle that, to a large extent, the topology of a Lie group together with the group law determines the geometry of the group.
## More examples of Lie groups
Lie groups occur in abundance throughout mathematics and physics. Matrix groups or algebraic groups are (roughly) groups of matrices (for example, orthogonal and symplectic groups), and these give most of the more common examples of Lie groups.
### Dimensions one and two
The only connected Lie groups with dimension one are the real line ${\displaystyle \mathbb {R} }$ (with the group operation being addition) and the circle group ${\displaystyle S^{1}}$ of complex numbers with absolute value one (with the group operation being multiplication). The ${\displaystyle S^{1}}$ group is often denoted as ${\displaystyle U(1)}$, the group of ${\displaystyle 1\times 1}$ unitary matrices.
In two dimensions, if we restrict attention to simply connected groups, then they are classified by their Lie algebras. There are (up to isomorphism) only two Lie algebras of dimension two. The associated simply connected Lie groups are ${\displaystyle \mathbb {R} ^{2}}$ (with the group operation being vector addition) and the affine group in dimension one, described in the previous subsection under "first examples."
• The group SU(2) is the group of ${\displaystyle 2\times 2}$ unitary matrices with determinant ${\displaystyle 1}$. Topologically, ${\displaystyle {\text{SU}}(2)}$ is the ${\displaystyle 3}$-sphere ${\displaystyle S^{3}}$; as a group, it may be identified with the group of unit quaternions.
• The Heisenberg group is a connected nilpotent Lie group of dimension ${\displaystyle 3}$, playing a key role in quantum mechanics.
• The Lorentz group is a 6-dimensional Lie group of linear isometries of the Minkowski space.
• The Poincaré group is a 10-dimensional Lie group of affine isometries of the Minkowski space.
• The exceptional Lie groups of types G2, F4, E6, E7, E8 have dimensions 14, 52, 78, 133, and 248. Along with the A-B-C-D series of simple Lie groups, the exceptional groups complete the list of simple Lie groups.
• The symplectic group ${\displaystyle {\text{Sp}}(2n,\mathbb {R} )}$ consists of all ${\displaystyle 2n\times 2n}$ matrices preserving a symplectic form on ${\displaystyle \mathbb {R} ^{2n}}$. It is a connected Lie group of dimension ${\displaystyle 2n^{2}+n}$.
### Constructions
There are several standard ways to form new Lie groups from old ones:
• The product of two Lie groups is a Lie group.
• Any topologically closed subgroup of a Lie group is a Lie group. This is known as the Closed subgroup theorem or Cartan's theorem.
• The quotient of a Lie group by a closed normal subgroup is a Lie group.
• The universal cover of a connected Lie group is a Lie group. For example, the group ${\displaystyle \mathbb {R} }$ is the universal cover of the circle group ${\displaystyle S^{1}}$. In fact any covering of a differentiable manifold is also a differentiable manifold, but by specifying universal cover, one guarantees a group structure (compatible with its other structures).
### Related notions
Some examples of groups that are not Lie groups (except in the trivial sense that any group having at most countably many elements can be viewed as a 0-dimensional Lie group, with the discrete topology), are:
• Infinite-dimensional groups, such as the additive group of an infinite-dimensional real vector space, or the space of smooth functions from a manifold ${\displaystyle X}$ to a Lie group ${\displaystyle G}$, ${\displaystyle C^{\infty }(X,G)}$. These are not Lie groups as they are not finite-dimensional manifolds.
• Some totally disconnected groups, such as the Galois group of an infinite extension of fields, or the additive group of the p-adic numbers. These are not Lie groups because their underlying spaces are not real manifolds. (Some of these groups are "p-adic Lie groups".) In general, only topological groups having similar local properties to Rn for some positive integer n can be Lie groups (of course they must also have a differentiable structure).
## Basic concepts
### The Lie algebra associated with a Lie group
To every Lie group we can associate a Lie algebra whose underlying vector space is the tangent space of the Lie group at the identity element and which completely captures the local structure of the group. Informally we can think of elements of the Lie algebra as elements of the group that are "infinitesimally close" to the identity, and the Lie bracket of the Lie algebra is related to the commutator of two such infinitesimal elements. Before giving the abstract definition we give a few examples:
• The Lie algebra of the vector space Rn is just Rn with the Lie bracket given by
[AB] = 0.
(In general the Lie bracket of a connected Lie group is always 0 if and only if the Lie group is abelian.)
• The Lie algebra of the general linear group GL(n, C) of invertible matrices is the vector space M(n, C) of square matrices with the Lie bracket given by
[AB] = AB − BA.
• If G is a closed subgroup of GL(n, C) then the Lie algebra of G can be thought of informally as the matrices m of M(n, R) such that 1 + ?m is in G, where ? is an infinitesimal positive number with ?2 = 0 (of course, no such real number ? exists). For example, the orthogonal group O(n, R) consists of matrices A with AAT = 1, so the Lie algebra consists of the matrices m with (1 + ?m)(1 + ?m)T = 1, which is equivalent to m + mT = 0 because ?2 = 0.
• The preceding description can be made more rigorous as follows. The Lie algebra of a closed subgroup G of GL(n, C), may be computed as
${\displaystyle \operatorname {Lie} (G)=\{X\in M(n;\mathbb {C} )|\operatorname {exp} (tX)\in G{\text{ for all }}t{\text{ in }}\mathbb {\mathbb {R} } \},}$[7][3] where exp(tX) is defined using the matrix exponential. It can then be shown that the Lie algebra of G is a real vector space that is closed under the bracket operation, ${\displaystyle [X,Y]=XY-YX}$.[8]
The concrete definition given above for matrix groups is easy to work with, but has some minor problems: to use it we first need to represent a Lie group as a group of matrices, but not all Lie groups can be represented in this way, and even it is not obvious that the Lie algebra is independent of the representation we use.[9] To get around these problems we give the general definition of the Lie algebra of a Lie group (in 4 steps):
1. Vector fields on any smooth manifold M can be thought of as derivations X of the ring of smooth functions on the manifold, and therefore form a Lie algebra under the Lie bracket [XY] = XY − YX, because the Lie bracket of any two derivations is a derivation.
2. If G is any group acting smoothly on the manifold M, then it acts on the vector fields, and the vector space of vector fields fixed by the group is closed under the Lie bracket and therefore also forms a Lie algebra.
3. We apply this construction to the case when the manifold M is the underlying space of a Lie group G, with G acting on G = M by left translations Lg(h) = gh. This shows that the space of left invariant vector fields (vector fields satisfying Lg*XhXgh for every h in G, where Lg* denotes the differential of Lg) on a Lie group is a Lie algebra under the Lie bracket of vector fields.
4. Any tangent vector at the identity of a Lie group can be extended to a left invariant vector field by left translating the tangent vector to other points of the manifold. Specifically, the left invariant extension of an element v of the tangent space at the identity is the vector field defined by v^g = Lg*v. This identifies the tangent space TeG at the identity with the space of left invariant vector fields, and therefore makes the tangent space at the identity into a Lie algebra, called the Lie algebra of G, usually denoted by a Fraktur ${\displaystyle {\mathfrak {g}}.}$ Thus the Lie bracket on ${\displaystyle {\mathfrak {g}}}$ is given explicitly by [vw] = [v^, w^]e.
This Lie algebra ${\displaystyle {\mathfrak {g}}}$ is finite-dimensional and it has the same dimension as the manifold G. The Lie algebra of G determines G up to "local isomorphism", where two Lie groups are called locally isomorphic if they look the same near the identity element. Problems about Lie groups are often solved by first solving the corresponding problem for the Lie algebras, and the result for groups then usually follows easily. For example, simple Lie groups are usually classified by first classifying the corresponding Lie algebras.
We could also define a Lie algebra structure on Te using right invariant vector fields instead of left invariant vector fields. This leads to the same Lie algebra, because the inverse map on G can be used to identify left invariant vector fields with right invariant vector fields, and acts as −1 on the tangent space Te.
The Lie algebra structure on Te can also be described as follows: the commutator operation
(x, y) -> xyx−1y−1
on G × G sends (ee) to e, so its derivative yields a bilinear operation on TeG. This bilinear operation is actually the zero map, but the second derivative, under the proper identification of tangent spaces, yields an operation that satisfies the axioms of a Lie bracket, and it is equal to twice the one defined through left-invariant vector fields.
### Homomorphisms and isomorphisms
If G and H are Lie groups, then a Lie group homomorphism f : G -> H is a smooth group homomorphism. In the case of complex Lie groups, such a homomorphism is required to be a holomorphic map. However, these requirements are a bit stringent; every continuous homomorphism between real Lie groups turns out to be (real) analytic.[10]
The composition of two Lie homomorphisms is again a homomorphism, and the class of all Lie groups, together with these morphisms, forms a category. Moreover, every Lie group homomorphism induces a homomorphism between the corresponding Lie algebras. Let ${\displaystyle \phi \colon G\to H}$ be a Lie group homomorphism and let ${\displaystyle \phi _{*}}$ be its derivative at the identity. If we identify the Lie algebras of G and H with their tangent spaces at the identity elements then ${\displaystyle \phi _{*}}$ is a map between the corresponding Lie algebras:
${\displaystyle \phi _{*}\colon {\mathfrak {g}}\to {\mathfrak {h}}}$
One can show that ${\displaystyle \phi _{*}}$ is actually a Lie algebra homomorphism (meaning that it is a linear map which preserves the Lie bracket). In the language of category theory, we then have a covariant functor from the category of Lie groups to the category of Lie algebras which sends a Lie group to its Lie algebra and a Lie group homomorphism to its derivative at the identity.
Two Lie groups are called isomorphic if there exists a bijective homomorphism between them whose inverse is also a Lie group homomorphism. Equivalently, it is a diffeomorphism which is also a group homomorphism.
### Lie group versus Lie algebra isomorphisms
Isomorphic Lie groups necessarily have isomorphic Lie algebras; it is then reasonable to ask how isomorphism classes of Lie groups relate to isomorphism classes of Lie algebras.
The first result in this direction is Lie's third theorem, which states that every finite-dimensional, real Lie algebra is the Lie algebra of some (linear) Lie group. One way to prove Lie's third theorem is to use Ado's theorem, which says every finite-dimensional real Lie algebra is isomorphic to a matrix Lie algebra. Meanwhile, for every finite-dimensional matrix Lie algebra, there is a linear group (matrix Lie group) with this algebra as its Lie algebra.[11]
On the other hand, Lie groups with isomorphic Lie algebras need not be isomorphic. Furthermore, this result remains true even if we assume the groups are connected. To put it differently, the global structure of a Lie group is not determined by its Lie algebra; for example, if Z is any discrete subgroup of the center of G then G and G/Z have the same Lie algebra (see the table of Lie groups for examples). An example of importance in physics are the groups SU(2) and SO(3). These two groups have isomorphic Lie algebras,[12] but the groups themselves are not isomorphic, because SU(2) is simply connected but SO(3) is not.[13]
On the other hand, if we require that the Lie group be simply connected, then the global structure is determined by its Lie algebra: two simply connected Lie groups with isomorphic Lie algebras are isomorphic.[14] (See the next subsection for more information about simply connected Lie groups.) In light of Lie's third theorem, we may therefore say that there is a one-to-one correspondence between isomorphism classes of finite-dimensional real Lie algebras and isomorphism classes of simply connected Lie groups.
### Simply connected Lie groups
A Lie group ${\displaystyle G}$ is said to be simply connected if every loop in ${\displaystyle G}$ can be shrunk continuously to a point in ${\displaystyle G}$. This notion is important because of the following result that has simple connectedness as a hypothesis:
Theorem:[15] Suppose ${\displaystyle G}$ and ${\displaystyle H}$ are Lie groups with Lie algebras ${\displaystyle {\mathfrak {g}}}$ and ${\displaystyle {\mathfrak {h}}}$ and that ${\displaystyle f:{\mathfrak {g}}\rightarrow {\mathfrak {h}}}$ is a Lie algebra homomorphism. If ${\displaystyle G}$ is simply connected, then there is a unique Lie group homomorphism ${\displaystyle \phi :G\rightarrow H}$ such that ${\displaystyle \phi _{*}=f}$, where ${\displaystyle \phi _{*}}$ is the differential of ${\displaystyle \phi }$ at the identity.
Lie's third theorem says that every finite-dimensional real Lie algebra is the Lie algebra of a Lie group. It follows from Lie's third theorem and the preceding result that every finite-dimensional real Lie algebra is the Lie algebra of a unique simply connected Lie group.
An example of a simply connected group is the special unitary group SU(2), which as a manifold is the 3-sphere. The rotation group SO(3), on the other hand, is not simply connected. (See Topology of SO(3).) The failure of SO(3) to be simply connected is intimately connected to the distinction between integer spin and half-integer spin in quantum mechanics. Other examples of simply connected Lie groups include the special unitary group SU(n), the spin group (double cover of rotation group) Spin(n) for ${\displaystyle n\geq 3}$, and the compact symplectic group Sp(n).[16]
Methods for determining whether a Lie group is simply connected or not are discussed in the article on fundamental groups of Lie groups.
### The exponential map
The exponential map from the Lie algebra ${\displaystyle M(n;\mathbb {C} )}$ of the general linear group ${\displaystyle GL(n;\mathbb {C} )}$ to ${\displaystyle GL(n;\mathbb {C} )}$ is defined by the matrix exponential, given by the usual power series:
${\displaystyle \exp(X)=1+X+{\frac {X^{2}}{2!}}+{\frac {X^{3}}{3!}}+\cdots }$
for matrices ${\displaystyle X}$. If ${\displaystyle G}$ is a closed subgroup of ${\displaystyle GL(n;\mathbb {C} )}$, then the exponential map takes the Lie algebra of ${\displaystyle G}$ into ${\displaystyle G}$; thus, we have an exponential map for all matrix groups. Every element of ${\displaystyle G}$ that is sufficiently close to the identity is the exponential of a matrix in the Lie algebra.[17]
The definition above is easy to use, but it is not defined for Lie groups that are not matrix groups, and it is not clear that the exponential map of a Lie group does not depend on its representation as a matrix group. We can solve both problems using a more abstract definition of the exponential map that works for all Lie groups, as follows.
For each vector ${\displaystyle X}$ in the Lie algebra ${\displaystyle {\mathfrak {g}}}$ of ${\displaystyle G}$ (i.e., the tangent space to ${\displaystyle G}$ at the identity), one proves that there is a unique one-parameter subgroup ${\displaystyle c:\mathbb {R} \rightarrow G}$ such that ${\displaystyle c'(0)=X}$. Saying that ${\displaystyle c}$ is a one-parameter subgroup means simply that ${\displaystyle c}$ is a smooth map into ${\displaystyle G}$ and that
${\displaystyle c(s+t)=c(s)c(t)\ }$
for all ${\displaystyle s}$ and ${\displaystyle t}$. The operation on the right hand side is the group multiplication in ${\displaystyle G}$. The formal similarity of this formula with the one valid for the exponential function justifies the definition
${\displaystyle \exp(X)=c(1).\ }$
This is called the exponential map, and it maps the Lie algebra ${\displaystyle {\mathfrak {g}}}$ into the Lie group ${\displaystyle G}$. It provides a diffeomorphism between a neighborhood of 0 in ${\displaystyle {\mathfrak {g}}}$ and a neighborhood of ${\displaystyle e}$ in ${\displaystyle G}$. This exponential map is a generalization of the exponential function for real numbers (because ${\displaystyle \mathbb {R} }$ is the Lie algebra of the Lie group of positive real numbers with multiplication), for complex numbers (because ${\displaystyle \mathbb {C} }$ is the Lie algebra of the Lie group of non-zero complex numbers with multiplication) and for matrices (because ${\displaystyle M(n,\mathbb {R} )}$ with the regular commutator is the Lie algebra of the Lie group ${\displaystyle GL(n,\mathbb {R} )}$ of all invertible matrices).
Because the exponential map is surjective on some neighbourhood ${\displaystyle N}$ of ${\displaystyle e}$, it is common to call elements of the Lie algebra infinitesimal generators of the group ${\displaystyle G}$. The subgroup of ${\displaystyle G}$ generated by ${\displaystyle N}$ is the identity component of ${\displaystyle G}$.
The exponential map and the Lie algebra determine the local group structure of every connected Lie group, because of the Baker–Campbell–Hausdorff formula: there exists a neighborhood ${\displaystyle U}$ of the zero element of ${\displaystyle {\mathfrak {g}}}$, such that for ${\displaystyle X,Y\in U}$ we have
${\displaystyle \exp(X)\,\exp(Y)=\exp \left(X+Y+{\tfrac {1}{2}}[X,Y]+{\tfrac {1}{12}}[\,[X,Y],Y]-{\tfrac {1}{12}}[\,[X,Y],X]-\cdots \right),}$
where the omitted terms are known and involve Lie brackets of four or more elements. In case ${\displaystyle X}$ and ${\displaystyle Y}$ commute, this formula reduces to the familiar exponential law ${\displaystyle \exp(X)\exp(Y)=\exp(X+Y)}$
The exponential map relates Lie group homomorphisms. That is, if ${\displaystyle \phi :G\to H}$ is a Lie group homomorphism and ${\displaystyle \phi _{*}:{\mathfrak {g}}\to {\mathfrak {h}}}$ the induced map on the corresponding Lie algebras, then for all ${\displaystyle x\in {\mathfrak {g}}}$ we have
${\displaystyle \phi (\exp(x))=\exp(\phi _{*}(x)).\,}$
In other words, the following diagram commutes,[Note 1]
(In short, exp is a natural transformation from the functor Lie to the identity functor on the category of Lie groups.)
The exponential map from the Lie algebra to the Lie group is not always onto, even if the group is connected (though it does map onto the Lie group for connected groups that are either compact or nilpotent). For example, the exponential map of SL(2, R) is not surjective. Also, the exponential map is neither surjective nor injective for infinite-dimensional (see below) Lie groups modelled on C? Fréchet space, even from arbitrary small neighborhood of 0 to corresponding neighborhood of 1.
### Lie subgroup
A Lie subgroup ${\displaystyle H}$ of a Lie group ${\displaystyle G}$ is a Lie group that is a subset of ${\displaystyle G}$ and such that the inclusion map from ${\displaystyle H}$ to ${\displaystyle G}$ is an injective immersion and group homomorphism. According to Cartan's theorem, a closed subgroup of ${\displaystyle G}$ admits a unique smooth structure which makes it an embedded Lie subgroup of ${\displaystyle G}$--i.e. a Lie subgroup such that the inclusion map is a smooth embedding.
Examples of non-closed subgroups are plentiful; for example take ${\displaystyle G}$ to be a torus of dimension 2 or greater, and let ${\displaystyle H}$ be a one-parameter subgroup of irrational slope, i.e. one that winds around in G. Then there is a Lie group homomorphism ${\displaystyle \varphi :\mathbb {R} \to G}$ with ${\displaystyle \mathrm {im} (\varphi )=H}$. The closure of ${\displaystyle H}$ will be a sub-torus in ${\displaystyle G}$.
The exponential map gives a one-to-one correspondence between the connected Lie subgroups of a connected Lie group ${\displaystyle G}$ and the subalgebras of the Lie algebra of ${\displaystyle G}$.[18] Typically, the subgroup corresponding to a subalgebra is not a closed subgroup. There is no criterion solely based on the structure of ${\displaystyle G}$ which determines which subalgebras correspond to closed subgroups.
## Representations
One important aspect of the study of Lie groups is their representations, that is, the way they can act (linearly) on vector spaces. In physics, Lie groups often encode the symmetries of a physical system. The way one makes use of this symmetry to help analyze the system is often through representation theory. Consider, for example, the time-independent Schrödinger equation in quantum mechanics, ${\displaystyle {\hat {H}}\psi =E\psi }$. Assume the system in question has the rotation group SO(3) as a symmetry, meaning that the Hamiltonian operator ${\displaystyle {\hat {H}}}$ commutes with the action of SO(3) on the wave function ${\displaystyle \psi }$. (One important example of such a system is the Hydrogen atom.) This assumption does not necessarily mean that the solutions ${\displaystyle \psi }$ are rotationally invariant functions. Rather, it means that the space of solutions to ${\displaystyle {\hat {H}}\psi =E\psi }$ is invariant under rotations (for each fixed value of ${\displaystyle E}$). This space, therefore, constitutes a representation of SO(3). These representations have been classified and the classification leads to a substantial simplification of the problem, essentially converting a three-dimensional partial differential equation to a one-dimensional ordinary differential equation.
The case of a connected compact Lie group K (including the just-mentioned case of SO(3)) is particularly tractable.[19] In that case, every finite-dimensional representation of K decomposes as a direct sum of irreducible representations. The irreducible representations, in turn, were classified by Hermann Weyl. The classification is in terms of the "highest weight" of the representation. The classification is closely related to the classification of representations of a semisimple Lie algebra.
One can also study (in general infinite-dimensional) unitary representations of an arbitrary Lie group (not necessarily compact). For example, it is possible to give a relatively simple explicit description of the representations of the group SL(2,R) and the representations of the Poincaré group.
## Early history
According to the most authoritative source on the early history of Lie groups (Hawkins, p. 1), Sophus Lie himself considered the winter of 1873-1874 as the birth date of his theory of continuous groups. Hawkins, however, suggests that it was "Lie's prodigious research activity during the four-year period from the fall of 1869 to the fall of 1873" that led to the theory's creation (ibid). Some of Lie's early ideas were developed in close collaboration with Felix Klein. Lie met with Klein every day from October 1869 through 1872: in Berlin from the end of October 1869 to the end of February 1870, and in Paris, Göttingen and Erlangen in the subsequent two years (ibid, p. 2). Lie stated that all of the principal results were obtained by 1884. But during the 1870s all his papers (except the very first note) were published in Norwegian journals, which impeded recognition of the work throughout the rest of Europe (ibid, p. 76). In 1884 a young German mathematician, Friedrich Engel, came to work with Lie on a systematic treatise to expose his theory of continuous groups. From this effort resulted the three-volume Theorie der Transformationsgruppen, published in 1888, 1890, and 1893. The term groupes de Lie first appeared in French in 1893 in the thesis of Lie's student Arthur Tresse.[20]
Lie's ideas did not stand in isolation from the rest of mathematics. In fact, his interest in the geometry of differential equations was first motivated by the work of Carl Gustav Jacobi, on the theory of partial differential equations of first order and on the equations of classical mechanics. Much of Jacobi's work was published posthumously in the 1860s, generating enormous interest in France and Germany (Hawkins, p. 43). Lie's idée fixe was to develop a theory of symmetries of differential equations that would accomplish for them what Évariste Galois had done for algebraic equations: namely, to classify them in terms of group theory. Lie and other mathematicians showed that the most important equations for special functions and orthogonal polynomials tend to arise from group theoretical symmetries. In Lie's early work, the idea was to construct a theory of continuous groups, to complement the theory of discrete groups that had developed in the theory of modular forms, in the hands of Felix Klein and Henri Poincaré. The initial application that Lie had in mind was to the theory of differential equations. On the model of Galois theory and polynomial equations, the driving conception was of a theory capable of unifying, by the study of symmetry, the whole area of ordinary differential equations. However, the hope that Lie Theory would unify the entire field of ordinary differential equations was not fulfilled. Symmetry methods for ODEs continue to be studied, but do not dominate the subject. There is a differential Galois theory, but it was developed by others, such as Picard and Vessiot, and it provides a theory of quadratures, the indefinite integrals required to express solutions.
Additional impetus to consider continuous groups came from ideas of Bernhard Riemann, on the foundations of geometry, and their further development in the hands of Klein. Thus three major themes in 19th century mathematics were combined by Lie in creating his new theory: the idea of symmetry, as exemplified by Galois through the algebraic notion of a group; geometric theory and the explicit solutions of differential equations of mechanics, worked out by Poisson and Jacobi; and the new understanding of geometry that emerged in the works of Plücker, Möbius, Grassmann and others, and culminated in Riemann's revolutionary vision of the subject.
Although today Sophus Lie is rightfully recognized as the creator of the theory of continuous groups, a major stride in the development of their structure theory, which was to have a profound influence on subsequent development of mathematics, was made by Wilhelm Killing, who in 1888 published the first paper in a series entitled Die Zusammensetzung der stetigen endlichen Transformationsgruppen (The composition of continuous finite transformation groups) (Hawkins, p. 100). The work of Killing, later refined and generalized by Élie Cartan, led to classification of semisimple Lie algebras, Cartan's theory of symmetric spaces, and Hermann Weyl's description of representations of compact and semisimple Lie groups using highest weights.
In 1900 David Hilbert challenged Lie theorists with his Fifth Problem presented at the International Congress of Mathematicians in Paris.
Weyl brought the early period of the development of the theory of Lie groups to fruition, for not only did he classify irreducible representations of semisimple Lie groups and connect the theory of groups with quantum mechanics, but he also put Lie's theory itself on firmer footing by clearly enunciating the distinction between Lie's infinitesimal groups (i.e., Lie algebras) and the Lie groups proper, and began investigations of topology of Lie groups.[21] The theory of Lie groups was systematically reworked in modern mathematical language in a monograph by Claude Chevalley.
## The concept of a Lie group, and possibilities of classification
Lie groups may be thought of as smoothly varying families of symmetries. Examples of symmetries include rotation about an axis. What must be understood is the nature of 'small' transformations, for example, rotations through tiny angles, that link nearby transformations. The mathematical object capturing this structure is called a Lie algebra (Lie himself called them "infinitesimal groups"). It can be defined because Lie groups are smooth manifolds, so have tangent spaces at each point.
The Lie algebra of any compact Lie group (very roughly: one for which the symmetries form a bounded set) can be decomposed as a direct sum of an abelian Lie algebra and some number of simple ones. The structure of an abelian Lie algebra is mathematically uninteresting (since the Lie bracket is identically zero); the interest is in the simple summands. Hence the question arises: what are the simple Lie algebras of compact groups? It turns out that they mostly fall into four infinite families, the "classical Lie algebras" An, Bn, Cn and Dn, which have simple descriptions in terms of symmetries of Euclidean space. But there are also just five "exceptional Lie algebras" that do not fall into any of these families. E8 is the largest of these.
Lie groups are classified according to their algebraic properties (simple, semisimple, solvable, nilpotent, abelian), their connectedness (connected or simply connected) and their compactness.
A first key result is the Levi decomposition, which says that every simply connected Lie group is the semidirect product of a solvable normal subgroup and a semisimple subgroup.
• Connected compact Lie groups are all known: they are finite central quotients of a product of copies of the circle group S1 and simple compact Lie groups (which correspond to connected Dynkin diagrams).
• Any simply connected solvable Lie group is isomorphic to a closed subgroup of the group of invertible upper triangular matrices of some rank, and any finite-dimensional irreducible representation of such a group is 1-dimensional. Solvable groups are too messy to classify except in a few small dimensions.
• Any simply connected nilpotent Lie group is isomorphic to a closed subgroup of the group of invertible upper triangular matrices with 1's on the diagonal of some rank, and any finite-dimensional irreducible representation of such a group is 1-dimensional. Like solvable groups, nilpotent groups are too messy to classify except in a few small dimensions.
• Simple Lie groups are sometimes defined to be those that are simple as abstract groups, and sometimes defined to be connected Lie groups with a simple Lie algebra. For example, SL(2, R) is simple according to the second definition but not according to the first. They have all been classified (for either definition).
• Semisimple Lie groups are Lie groups whose Lie algebra is a product of simple Lie algebras.[22] They are central extensions of products of simple Lie groups.
The identity component of any Lie group is an open normal subgroup, and the quotient group is a discrete group. The universal cover of any connected Lie group is a simply connected Lie group, and conversely any connected Lie group is a quotient of a simply connected Lie group by a discrete normal subgroup of the center. Any Lie group G can be decomposed into discrete, simple, and abelian groups in a canonical way as follows. Write
Gcon for the connected component of the identity
Gsol for the largest connected normal solvable subgroup
Gnil for the largest connected normal nilpotent subgroup
so that we have a sequence of normal subgroups
1 ? Gnil ? Gsol ? Gcon ? G.
Then
G/Gcon is discrete
Gcon/Gsol is a central extension of a product of simple connected Lie groups.
Gsol/Gnil is abelian. A connected abelian Lie group is isomorphic to a product of copies of R and the circle group S1.
Gnil/1 is nilpotent, and therefore its ascending central series has all quotients abelian.
This can be used to reduce some problems about Lie groups (such as finding their unitary representations) to the same problems for connected simple groups and nilpotent and solvable subgroups of smaller dimension.
## Infinite-dimensional Lie groups
Lie groups are often defined to be finite-dimensional, but there are many groups that resemble Lie groups, except for being infinite-dimensional. The simplest way to define infinite-dimensional Lie groups is to model them locally on Banach spaces (as opposed to Euclidean space in the finite-dimensional case), and in this case much of the basic theory is similar to that of finite-dimensional Lie groups. However this is inadequate for many applications, because many natural examples of infinite-dimensional Lie groups are not Banach manifolds. Instead one needs to define Lie groups modeled on more general locally convex topological vector spaces. In this case the relation between the Lie algebra and the Lie group becomes rather subtle, and several results about finite-dimensional Lie groups no longer hold.
The literature is not entirely uniform in its terminology as to exactly which properties of infinite-dimensional groups qualify the group for the prefix Lie in Lie group. On the Lie algebra side of affairs, things are simpler since the qualifying criteria for the prefix Lie in Lie algebra are purely algebraic. For example, an infinite-dimensional Lie algebra may or may not have a corresponding Lie group. That is, there may be a group corresponding to the Lie algebra, but it might not be nice enough to be called a Lie group, or the connection between the group and the Lie algebra might not be nice enough (for example, failure of the exponential map to be onto a neighborhood of the identity). It is the "nice enough" that is not universally defined.
Some of the examples that have been studied include:
## Notes
### Explanatory notes
1. ^ "Archived copy" (PDF). Archived from the original (PDF) on 2011-09-28. Retrieved .CS1 maint: archived copy as title (link)
### Citations
1. ^ Rossmann 2001, Chapter 2.
2. ^ Hall 2015 Corollary 3.45
3. ^ a b Hall 2015
4. ^ Rossmann 2001
5. ^ T. Kobayashi-T. Oshima, Definition 5.3.
6. ^ This is the statement that a Lie group is a formal Lie group. For the latter concept, for now, see F. Bruhat, Lectures on Lie Groups and Representations of Locally Compact Groups.
7. ^ Helgason 1978, Ch. II, § 2, Proposition 2.7.
8. ^ Hall 2015 Theorem 3.20
9. ^ But see Hall 2015, Proposition 3.30 and Exercise 8 in Chapter 3
10. ^ Hall 2015 Corollary 3.50. Hall only claims smoothness, but the same argument shows analyticity.
11. ^ Hall 2015 Theorem 5.20
12. ^ Hall 2015 Example 3.27
13. ^ Hall 2015 Section 1.3.4
14. ^ Hall 2015 Corollary 5.7
15. ^ Hall 2015 Theorem 5.6
16. ^ Hall 2015 Section 13.2
17. ^ Hall 2015 Theorem 3.42
18. ^ Hall 2015 Theorem 5.20
19. ^ Hall 2015 Part III
20. ^ Arthur Tresse (1893). "Sur les invariants différentiels des groupes continus de transformations". Acta Mathematica. 18: 1-88. doi:10.1007/bf02418270.
21. ^
22. ^ Helgason, Sigurdur (1978). Differential Geometry, Lie Groups, and Symmetric Spaces. New York: Academic Press. p. 131. ISBN 978-0-12-338460-7.
23. ^ Bäuerle, de Kerf & ten Kroode 1997 | 11,740 | 49,665 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 189, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-04 | latest | en | 0.908702 |
https://www.unitconverters.net/flow-molar/picomol-second-to-mol-second.htm | 1,518,928,454,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811655.65/warc/CC-MAIN-20180218042652-20180218062652-00348.warc.gz | 979,733,496 | 3,280 | Home / Flow - Molar Conversion / Convert Picomol/second to Mol/second
# Convert Picomol/second to Mol/second
Please provide values below to convert picomol/second [pmol/s] to mol/second [mol/s], or vice versa.
From: picomol/second To: mol/second
### Picomol/second to Mol/second Conversion Table
Picomol/second [pmol/s]Mol/second [mol/s]
0.01 pmol/s1.0E-14 mol/s
0.1 pmol/s1.0E-13 mol/s
1 pmol/s1.0E-12 mol/s
2 pmol/s2.0E-12 mol/s
3 pmol/s3.0E-12 mol/s
5 pmol/s5.0E-12 mol/s
10 pmol/s1.0E-11 mol/s
20 pmol/s2.0E-11 mol/s
50 pmol/s5.0E-11 mol/s
100 pmol/s1.0E-10 mol/s
1000 pmol/s1.0E-9 mol/s
### How to Convert Picomol/second to Mol/second
1 pmol/s = 1.0E-12 mol/s
1 mol/s = 1000000000000 pmol/s
Example: convert 15 pmol/s to mol/s:
15 pmol/s = 15 × 1.0E-12 mol/s = 1.5E-11 mol/s | 334 | 788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-09 | longest | en | 0.535399 |
https://infinitylearn.com/surge/study-material/cbse-notes/class-12/geography/geography-class-12-important-questions-chapter-3-population-composition/ | 1,726,851,762,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701419169.94/warc/CC-MAIN-20240920154713-20240920184713-00203.warc.gz | 276,655,462 | 29,574 | Geography Class 12 Important Questions Chapter 3 Population Composition
# Geography Class 12 Important Questions Chapter 3 Population Composition
## 1 Mark Questions
Questions 1. Why is the age structure considered an important indicator of population composition? Given one reason. (HOTS; Delhi 2016)
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Answer: Age structure is an important indicator of population composition because it tells about the proportion of population i.e. the percentage of the young and ageing population. This helps in planning for the future.
Questions 2. Why is the sex ratio in some European countries recorded favourable to females? Given one reason. Delhi 2016
Answer: In some European countries, females recorded a favourable sex ratio because women enjoy better socio-economic status.
Questions 3. Name the country where the sex ratio is most unfavourable to women in the world. All Indio 2016
Answer: A country, where the sex ratio is most unfavourable to women, is UAE as there are 468 females per 1000 males.
Questions 4. What is the average sex ratio in the world? All India 2016
Answer: The average sex ratio in the world is 990 females per 1000 males.
Questions 5. Give the meaning of ‘Age structure’. (All Indio 2014,2013)
Answer: The age structure of the population refers to the number of people of different age groups residing in a country.
Questions 6. How is sex ratio calculated in different countries of the world? Delhi 2013
Answer: In some countries the sex ratio is calculated by using a formula:
$$\frac { Male\quad Population }{ Female\quad Population } \times 1000$$
Or the number of males per thousand females.
In India, the sex ratio is calculated by using a formula:
$$\frac { Female\quad Population }{ male\quad Population } \times 1000$$
Or the number of females per thousand males.
Also Check: CBSE Class 12 Date Sheet 2024 Out
Questions 7. Give the meaning of Age-sex structure of a population. (All India 2013)
Answer: The age-sex structure of a population refers to the number of females and males in different age groups.
Questions 8. Define the term ‘sex ratio’. (Delhi 2012)
Answer: The ratio between the number of women and men in the population is called the sex ratio.
Questions 9. Name the country having the highest sex ratio in the world. (All Indio 2012, Delhi 2010)
Answer: The country has the highest sex ratio in the world is Latvia.
Questions 10. Which country has the lowest sex ratio in the world? (Delhi 2011,2008)
Answer: A country having lowest sex ratio in the world is the United Arab Emirates.
Questions 11. Which ag” group of population indicates the largest working population? (All India 2008)
Answer: The age group of population indicating the working population is 15 to 59 years.
## 3 Marks Questions
Questions 12. Divide the population of the world into two groups on the basis of residence. How do they differ from each other? Explain any two points of difference. (Delhi 2011)
Answer: Depending upon the place of residence, the population of a country is divided into two groups:
1. Rural population Characteristics of Rural Population
(a) People living in villages are known as the rural population. The main occupation of rural people is primary activities i.e. hunting, fishing, mining, agriculture, agricultural labour, etc.
(b) Majority of the world’s population reside in rural areas but the density of population is low.
2. Urban population Characteristics of Urban Population
(a) People living in towns and cities are known as an urban population. Urban people are engaged in secondary or tertiary occupations i.e. manufacturing, public and private services, transport and communication fields, etc.
(b) Urban areas have a very high density of population as more job opportunities lead to the in-migration of people.
Questions 13. What do you understand by the occupational structure? Name the specific economic activities or occupational categories as identified by the United Nations? (Delhi 2011)
Answer: Occupational structure refers to the proportional distribution of people under specific economic activities. These activities are as follows:
• Agriculture, forestry, hunting and fishing
• Mining and quarrying
• Manufacturing industry
• Construction
• Electricity
• Unclassified occupations
• Commerce
• Transportation and Communication
There is another group which is not engaged in any economic activity and is also not the part of the active population i.e. children below the working age, old people, housewives and students. There is another classification of occupational structure which is divided into following four major groups:
• Primary activities such as hunting and agriculture.
• Secondary activities such as manufacturing.
• Tertiary activities such as service sector i.e. transport, communication, etc.
• Quaternary activities such as intellectual tasks i.e. research and development activities.
Questions 14. Why is sex ratio unfavourable to women in the world? Explain any three reasons. (HOTS; All Indio 2011)
Answer: The sex ratio is important information about the status of women in a country in a region where gender discrimination is rampant. The sex ratio is bound to be unfavourable to women.
Reason for unfavourable sex ratio for women are as follows:
1. Sex ratio is always unfavourable to women in those countries where gender discrimination is rampant.
2. In such countries or areas, the practice of female foeticide, female infanticide and domestic violence against women are common practices.
3. The main reason for such practices is that women suffer from lower socio-economic status in general and societies are male-dominating in particular.
## 5 Marks Questions
Questions 15. What is the sex ratio? Explain the world pattern of sex ratio with suitable examples. (All India 2015)
Answer: The ratio between the number of women and men in the population is called the sex ratio.
The world pattern of sex ratio is described as under:
1. On average, the world population reflects a sex ratio of 990 females per 1000 males. Highest sex ratio is recorded in Latvia i.e. 1187 females per 1000 males while the lowest is recorded in UAE i.e. 468 females per 1000 males.
2. The pattern of sex ratio does not exhibit variation in the developed region of the world. It is favourable for females in 139 countries and unfavourable in the remaining 72 countries.
3. Many Asian countries have low sex ratios like China, India, Saudi Arabia, Pakistan and Afghanistan that points to low socio-economic status and lower sex ratio of women.
4. Many European countries have a high sex ratio where females outnumber males. This deficit is attributed to a better status of women and excessively male-dominated out-migration to different parts of the world in the past.
Questions 16. What is the sex ratio? Why is sex ratio unfavourable to women in some countries of the world? Explain any four reasons. (All Indio 2010)
Answer: For sex ratio, The ratio between the number of women and men in the population is called the sex ratio.
The world pattern of sex ratio is described as under:
1. On a rage, the world population reflects a sex ratio of 990 females per 1000 males. Highest sex ratio is recorded in Latvia i.e. 1187 females per 1000 males while the lowest is recorded in UAE i.e. 468 females per 1000 males.
2. The pattern of sex ratio does not exhibit variation in the developed region of the world. It is favourable for females in 139 countries and unfavourable in the remaining 72 countries.
3. Many Asian countries have low sex ratio like China, India, Saudi Arabia, Pakistan and Afghanistan that points to low socio-economic status and lower sex ratio of women.
4. Many European countries have a high sex ratio where females outnumber males. This deficit is attributed to better status of women and excessively male dom male-dominated action to different parts of the world in the past.
Sex ratio is unfavourable to women in some countries due to the following reasons:
1. Gender discrimination In many countries of the world, there is widespread gender discrimination. Males are given more preference than females and enjoy greater rights and privileges.
2. Female foeticide Due to gender discrimination, females are not preferred. This results in evil practices like female foeticide, females infanticide, etc which causes unfavourable sex ratio.
3. Low socio-economic status Females have to face social discriminations due to which they are not economically independent or are paid less as compared to males. This lowers their status in society.
4. Domestic violence Low social status also leads to domestic violence like mental and physical tortures. This also leads to unfavourable sex ratio in many countries.
Questions 17. Describe the rural and urban population composition of the world with examples. (Delhi 2009)
Answer: The division of rural and urban population is based on the occupational structure, socio-economic level and level of development. The figure below presents the rural-urban composition of the population:
• The population composition of rural and urban areas of developed and developing countries are different.
• In developed countries like Canada, Finland, New Zealand and other West European countries females outnumber males in urban areas. Females settle in urban areas to avail the vast job opportunities and as farming is highly mechanised, it remains largely a male occupation in rural areas.
• In developing countries like Pakistan, Nepal and other Asian countries, urban areas remain male dominated due to predoa dominance of male migration. Also, female participation in agriculture is fairly high in these countries. Shortage of jobs, housing and lack of security in cities discourage women to migrate from rural to urban areas.
## Value Based Questions
Questions 18. “Population ageing is the process by which the share of the older population becomes proportionally larger. This is a new phenomenon of the 20th century.
In most of the developed countries of the world, the population in higher age groups has increased due to increased life expectancy. With a reduction in birth rates, the proportion of children in the population has declined.”
(i) What leads to population ageing?
(ii) Which value is required while taking care of the ageing population?
(i) Increased life expectancy leads to population ageing.
(ii) Values which are required to take care of the ageing population are:
• Nursing spirit
• Dedication
• Sacrifice
Questions 19. “Proportion of literate population of a country is an indicator of its socio-economic development as it reveals the standard of living, the social status of females, availability of educational facilities and policies of a government. Level of economic development is
both a cause and consequence of literacy”.
(i) What a literate population indicate?
(ii) Which value you have learnt from the given statement? | 2,292 | 11,084 | {"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} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.86563 |
http://www.consultemps.fr/ebooks/asymptotic-methods-in-quantum-mechanics-application-to-atoms-molecules-and | 1,600,526,399,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00598.warc.gz | 152,407,224 | 11,285 | # Read e-book online Asymptotic Methods in Quantum Mechanics: Application to PDF
By Professor S. H. Patil, Professor K. T. Tang (auth.)
ISBN-10: 3642573177
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Asymptotic tools in Quantum Mechanics is an in depth dialogue of the final homes of the wave features of many particle platforms. specific emphasis is put on their asymptotic behaviour, because the outer sector of the wave functionality is so much delicate to exterior interplay. The research of those neighborhood homes is helping in developing easy and compact wave capabilities for classy platforms. It additionally is helping in constructing a large figuring out of other elements of quantum mechanics. As functions, wave services with right asymptotic varieties are used to systematically generate a wide information base for susceptibilities, polarizabilities, interactomic potentials and nuclear densities of many atomic, molecular and nuclear systems.
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Additional resources for Asymptotic Methods in Quantum Mechanics: Application to Atoms, Molecules and Nuclei
Example text
G. laser fields, and in the analysis of dispersion coefficients. 2 Dispersion Coefficients The dominant interaction between two spherically symmetric atoms at large separation R is given in terms of dispersion coefficients. Let two atoms be on the z axis, atom A at the origin and atom B at a distance R from the origin. Let electron 1 be at a distance r from the nucleus of A and electron 2 be at a distance r' from the nucleus of B. "p R 1 IR - rl _ 1 IR + r'l + 1 IR - r + r'l . 38) When there are more than one electrons, this term has to be summed over all the pairs, one from each atom.
I' + m)! 42) m which follows from taking coef. of x 21 in (x + y)2H21' = L[coef. of xHm in (x + y/H'] m [coef. of x l - m in (x + y)lH'] . 43) This leads to (2£ + 2i')! (£ + i')! 44) so that finally we have 8E(2) = + 2£')! (2£' + 1)! 45) When more than one electron is involved, the multipolar potential terms rlyt0 are to be summed over the electrons of the atom. One usually writes the expression for this energy as a two-body, long-range interaction potential in the form 48 4. 46) where n-2 C2n (AB) = (2n - 2)!
24). J k=O (47r)2(n + I)! ] . 2ni +n ;+4-k(n- + 1- £ - k)! 2 Wave Functions Satisfying Cusp, Coalescence & Asymptotic Conditions 25 The calculated values of the coefficients of Ci (L) for L = 1,2, and the polarizabilities are given in table 1. The polarizabilities are again in good agreement with the accurate values. 383. 78). 19) In particular, for £1 + £2 + £3 = 4, we have a hyperpolarizability B defined as B = -2 L t(£l, £2, £3), £1 + £2 + £3 = 4 = -2t(l, 2,1) - 4t(l, 1, 2) . 12). 2 Wave Functions Satisfying Cusp, Coalescence and Asymptotic Conditions The wave function we have described in the previous section has the advantage of great simplicity. | 812 | 2,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-40 | latest | en | 0.873107 |
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# Breakdown of Official Guide 13th Edition vs. 12th Edition
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Magoosh GMAT Instructor
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Re: Breakdown of Official Guide 13th Edition vs. 12th Edition [#permalink]
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10 Apr 2013, 14:22
samg wrote:
Thanks Mike. I will post a reply to this thread, if I come across any further mismatches. The file I had downloaded was from the 1st post of this thread, and I see you have updated it. Were there any other corrections made, other than the one I had pointed out? Just want to know- if there are more than one, then I will have to update my copy.
There were a few corrections, maybe three or four, all right at the end of the DS section. These were the only changes between the previous version and the current version.
Mike
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Re: Breakdown of Official Guide 13th Edition vs. 12th Edition [#permalink]
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17 Apr 2013, 13:36
Eduardowbeck wrote:
Very useful! Thanks.
You are quite welcome.
Mike
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Re: Breakdown of Official Guide 13th Edition vs. 12th Edition [#permalink]
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14 May 2014, 11:31
There is some overlap between OG12 and GMAT Focus. However there is none between OG13 and the GMAT Focus. Please keep that in mind if you are thinking of buying GMAT Focus and have already done OG12.
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16 Aug 2015, 23:37
Has anyone got break down of OG earlier editions like 10th vs 12th ???
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Re: Breakdown of Official Guide 13th Edition vs. 12th Edition [#permalink]
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05 Sep 2016, 12:52
Thank you, Margarette!
Do you also have 12th to 15th breakdown by any chance? Infact, are they even similar?
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Re: Breakdown of Official Guide 13th Edition vs. 12th Edition [#permalink] 05 Sep 2016, 12:52
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# Breakdown of Official Guide 13th Edition vs. 12th Edition
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Home Our Teaching Resources C programming Structures
Contents
# Structures organise our data
Show me your [functions] and conceal your [data structures] and I shall continue to be mystified,
show me your [data structures] and I won't usually need your [functions]; they'll be obvious.
Brooks,
## Structures do for variables what functions do for code
Representation is the essence of programming.
Brooks
We have already discussed the fact that nearly every program is so large that it is completely impossible to hold in all in our head at the same time.
• Functions organise code into self contained units, enabling us to think at a higher level without having to keep track of the details.
• Structures do exactly the same for variables.
Structures allow us to organise our data, letting us think at a higher level without having to keep track of the details
### Composite variables
We have already encountered C's two built-in composite variables: complex numbers and arrays. The latter allow us to think in terms of high-level mathematical concepts such as vectors and matrices and allow us to pass a single memory address to a function to allow it to access and modify any member of the array. However arrays are limited to composite objects of the same type (all ints, all doubles, etc.) C structures remove these limitations.
C structures allow us to define our own composite data types where several "sub-variables" (called "members") can be combined into one composite variable which can be treated as a single unit, similar to the individual elements of an array.
Since structures have members with different types it would be inconvenient to refer to them by number, instead they are referred to by name using a dot (.) rather than square brackets.
An array has elements of the same type referred as array[number], a structure has members of different types referred to as structure.name.
## Defining a new composite variable (structure) type
### Example: a nuclear problem
Imagine a simple problem involving radioactive nuclides. If each nuclide has a name (stored as an array of characters) and a a half life (stored as a double) then to store the data for several nuclides we will need a two-dimensional array of characters and a one-dimensional array of doubles.
This isn't very convenient, but C doesn't define a composite variable type consisting of a string and a double and even if it did, it's likely that a little later on we would want to add more data such as mass and atomic number.
It would be great if C allowed us to define our own variable types, with names such as "Nuclide" or "Isotope", and luckily it does. The following code defines a composite variable type called a "Nuclide":
The structure name (nuclide) is optional but if it is present it's normal to give the structure and the typedef similar names. In our case the typedef name is the same as the structure name but is capitalised
```#define MAXLEN 32
typedef struct nuclide {
double halflife;
char name[MAXLEN];
} Nuclide;
```
The above code defines a new data type called "Nuclide" consisting of a double called halflife and a character array called name. This definition does not create any Nuclide variables. Rather it puts the variable type "Nuclide" on exactly the same basis as the built-in types int, float etc: we can now create (declare) some Nuclides if we need to.
The structure definition does not create any structures it just says what the structure means so we can later create some.
It's also worth noticing that the Nuclide structure contains an array: structures can contain arrays and we shall see below that we can have arrays of structures.
### Optional aside
The above combines two C features: structures, which actually do the work, and typedefs which give user-friendly names to things which already exist.
#### Structures
The following is the bare structure definition (technically referred to as a "specifier") which defines a new variable type, "struct nuclide":
```#define MAXLEN 32
struct nuclide {
double halflife;
char name[MAXLEN];
} ;
```
We can now create variables of type "struct nuclide" just like ints, floats, etc.
#### typedef
Since "struct nuclide" is not the most user-friendly name for a new variable type, we can make things a little neater by using C's "typedef" mechanism to give our structure a neater name:
```typedef struct nuclide Nuclide;
```
#### Combining the two
For convenience we may combine the two as above:
```typedef struct nuclide { double halflife; char name[MAXLEN];
} Nuclide;
```
Now "Nuclide" and "struct nuclide" are synonyms; they mean the same thing.It is a common convention to give the synonym the same name as the structure type but with an initial capital letter.
1. Define a structure to represent a cuboid.
2. The first part of our first use of structures.
3. A solid cuboid can be considered to have three dimensions, a density and a mass. This can be represented by an double array of size three and two individual doubles.
4. Create a new on-line program in a new window, with a suitable title and opening comment.
5. Above main() create a structure definition for a cuboid with the above quantities. Give the structure and its members sensible names but do not at this point declare any actual structures inside main().
6. Build & run. At this point your program does not do anything it didn't do before, we are just checking your code compiles.
## Declaring structures
If we have the above code just once at the top of our file then Nuclide joins the list of variables we can now declare, just like ints, floats, etc, and we can access the individual structures members as varname.member:
```#include <stdio.h>
#include <string.h>
#define MAXLEN 32
typedef struct nuclide {
double halflife;
char name[MAXLEN];
} Nuclide;
int main() {
Nuclide nuc;
nuc.halflife = 3.2;
strncpy(nuc.name, "Mystuff", MAXLEN);
nuc.name[MAXLEN-1] = '\0';
// Now do something...
return 0;
}
```
```Step through this code
```
Once we have defined what a structure means we can then create some just like any other variable type.
In the above example we have assigned a value to nuc.mass and used nuc.name as an argument to strncpy(). We could have also read in the value of nuc.halflife with scanf():
``` printf("Please enter the half-life.\n");
scanf("%lg", &nuc.halflife);
```
Anything we can do with an ordinary variable or an array we can do when it's part of a structure
We can also have arrays of structures, although these are quite uncommon:
```#include <stdio.h>
int main() {
Nuclide nucs[MAXNUCLIDES];
for (int n = 0; n < MAXNUCLIDES; ++n)
scanf("%s %lg", nucs[n].name, &nucs[n].halflife);
...
return 0;
}
```
```Step through this code
```
1. Declare a structure to represent a cuboid
2. To practice structures.
3. Edit your previous Mini-exercise that defined a structure to represent a cuboid.
4. Declare an actual cuboid inside of main(). Read in the value of the density into the structure using scanf() and print that value out to the screen as a check.
5. Build & run. Check the output is correct.
6. Now read in the three dimensions, calculate the mass and print out all five numbers (the three dimensions, the density and the mass).
7. Build & run. Check the output is correct..
### Another example: an ellipse
The following defines a structure which we might use to represent an ellipse:
```typedef struct ellipse {
float centre[2];
float axes[2];
float orientation;
float area;
} Ellipse;
```
We will use this example in the rest of the notes.
### Initialising structures to all zeros
It's possible to initialise structures in much the same way as arrays, by putting the values inside { }, so in theory we can write:
``` Ellipse ellie = { { 0.7, 45.2 }, /* centre */
{ 2.3, 8.4 }, /* axes */
1.87 /* orientation */ };
```
However this is horrifically unclear and will fail completely if we add another member at the start of the structure. The only real use of this is to initialise a structure to all zeros using the rule that missing initialisers are set to zero:
``` Ellipse zero = { 0 }; // All zeros
```
## When to use structures and what to put in them
In general:
Any type of "thing" in the problem we are thinking about whose properties cannot be represented by an existing variable type or array should normally have its own structure type.
The structure should tell us everything we need to know about that thing.
As a general rule we should ask ourselves: "If I had two ellipses (or nuclides etc.) what would I need two of and what would I still just need one of?" Things we would need two of should be part of the structure.
For example, if we had two ellipses they would each have their own centre and axes but would share the same value of PI.
### Structures are "proper variables"
Unlike arrays where the "value" of the name of an array is the address of its first element, structures are "proper variables" that can be copied (although this is quite rare):
``` mystruct = yourstruct;
```
This has a serious consequence when passing structures to functions as we will see in the next example.
## Passing structures to functions
Given that structures are proper variables we may pass copies of them to functions. This rather simple example shows a function whose job is to calculate and print out the area of an ellipse. (In practice this function is so simple it probably isn't worth making a separate function.) Just for fun it also makes a doomed attempt to move the ellipse.
```#include <stdio.h>
#define PI 3.14159265358979
typedef struct ellipse {
float centre[2];
float axes[2];
float orientation;
} Ellipse;
void print_area(Ellipse el) {
float area;
area = PI * el.axes[0] * el.axes[1];
printf("The area is %f\n", area);
el.centre[0] = 123.456; // Move the ellipse
el.centre[1] = -78.9;
}
int main() {
Ellipse ellie;
printf("Centre? (x,y)\n");
scanf("%f %f", &ellie.centre[0], &ellie.centre[1]);
// NB: in a real program we should check
// the axes are actually > 0
printf("Length of axes ( > 0 )?\n");
scanf("%f %f", &ellie.axes[0], &ellie.axes[1]);
printf("Orientation to the vertical?\n");
scanf("%f", &ellie.orientation);
print_area(ellie);
return 0;
}
```
```Step through this code
```
1. Step through the above "Key example".
2. To see that the structure in the called function is a copy of the original structure
3. Step through the above "Key example" in a new window.
4. Fast-forward to the call to print_area() (second line from the bottom) either by clicking on the button to the right of the call or pressing the "Fast forward to the interesting bit" button near the top. (If you find this unclear just reload the page and manually step forward to this point in the program.)
5. Now step forward and note how the function receives a copy of the original ellipse.
6. Notive how when print_area() is moving the ellipse it is moving the local copy, not the original. Any changes made to the local copy are lost when the function returns, the original is unaffected.
This example illustrates why using structures without pointers is only of limited use: the function print_area()receives a copy of the original structure, not the structure itself, so if we change the copy inside the function the original is unchanged. For this reason structures are almost always used with pointers.
#### Changing the values of the members of an ellipse structure
There are several reasons we may wish to allow a function to modify an ellipse, for example:
1. We may wish to move the ellipse. (Or resize it, rotate it, or ...)
2. Some properties of the ellipse, such as the area, may be so useful we may wish to add a member of the structure to store it.
#### Reminder: pointers
We have already encountered this problem in a previous lecture where we wanted a function to be able to manipulate an array declared inside another function. Let's remind ourselves of what we did:
Pointers
Specific requirement: To enable a function to access an array declared inside another function. Solution: A variable called a pointer whose value is the address of another variable (in this case the address of the first element of the array) Wider application: To enable a function to access an object declared (or otherwise created) inside another function To enable a piece of code to act on one set of data and later on another set. And, as in this case, usually both.
The solution when we want a function to be able to manipulate a structure declared inside another function is the same: pass the address of the structure to the function and have a pointer as a parameter to the function to receive the address. Thus every function has a pointer pointing to the same original ellipse structure and is making changes to that structure.
## Pointers to structures
Structures are most useful when combined with pointers.
These work just as we would expect. Suppose, as suggested above, that we don't just want our function to print out the area of the ellipse but to store it for later. To do this we need to add an "area" member to our ellipse structure:
```typedef struct ellipse {
float centre[2];
float axes[2];
float orientation;
float area;
} Ellipse;
```
The enlarged structure now takes twenty four bytes. A particular Ellipse, say called "ellie", might be stored starting at byte 800 in which case its members would be laid out as follows:
``` ----------------------- ------------------------------------------------
Byte number: | 800 - 803 | 804 - 807 | 808 - 811 | 812 - 815 | 816 - 819 | 820 - 823 |
----------------------- ------------------------------------------------
Used for: | centre[0] | centre[1] | axes[0] | axes[1] |orientation| area |
------------------------------------------------------------------------
| <-------------------------- ellie ----------------------------------> |
```
A statement such as:
``` ellie.orientation = 11.4;
```
would mean that the computer would go to the start of ellie (800), go along 16 bytes to byte 816 and write the value 11.4 into the four bytes 816 - 819.
Alternatively, we could declare a pointer to an Ellipse, assign it the address of ellie (800) and use that pointer, as in the following rather contrived example:
```#define PI 3.14159265358979
int main() {
Ellipse ellie, *ep; // ellie is a structure, ep just a pointer
ep = &ellie;
(*ep).orientation = PI/4;
// More code here...
```
The notation (*ep).orientation is rather awkward so pointers to structures are used so often they have their own notation:
``` ep->orientation = PI/4;
```
The fact that they have their own notation should tell us how important pointers to structures are!
If p is a pointer to a structure then p->member accesses a member of that structure.
Our code is now exactly the same as before but with structure. replaced by pointer->, as in this short example:
``` Ellipse ellie, *ep; // ellie is a structure, ep just a pointer
ep = &ellie;
ep->orientation = PI/4;
```
```Step through this code
```
### Dynamically allocating structures
Although the previous code is entirely legal it's a bit of a bodge. Rather than declaring a structure and then having a pointer whose value is the address of that structure it's far more common to dynamically allocate the stucture:
``` Ellipse *e = NULL;
e = malloc( sizeof *e);
if ( e == NULL ) {
fprintf(stderr, "Out of memory\n");
exit(99);
}
e->orientation = PI/4;
```
```Step through this code
```
Structures are usually dynamically allocated via pointers rather than declared as variables.
1. Pointer to a structure
2. The first stage in using pointers to functions.
3. Take your previous code and declare a pointer to a cuboid structure and dynamically allocate it some memory in the same way as above.
4. Remove the old declaration of the cuboid structure so that you are left with the declaration of the pointer to a structure and the dynamic memory allocation.
5. Now replace all the occurances of structurename. with pointername->
6. Build & run. Check the output is correct.
## Passing pointers-to-structures to functions
The old function to print the area of an ellipse had the prototype:
```// Print the area of an elipse to the screen
void print_area(Ellipse el);
```
The new function, which we have renamed calculate_area() now saves the value of the area in the area structure member rather than just printing it. That is it accepts a pointer and modifies the original ellipse structure. It has the prototype:
```// Calculate the area of an ellipse and save it in the structure
void calculate_area(Ellipse *el);
```
The function itself is extremely short (partly because we have put the "move ellipse" code into a separate function where it belongs):
```// Calculate the area of an ellipse and save it in the structure
void calculate_area(Ellipse *e) {
e->area = PI * e->axes[0] * e->axes[1];
}
```
Passing pointers-to-structures to functions allows the function to access and change the members of the structure.
### Passing pointers between functions
In reality the calculate_area() function above is still a little simple to be worth having as a separate function. However wWe may wish to define a new function to read in the values of the ellipse, which we might imaginitively call "read_ellipse()", and another to resize an ellipse by mutliplying each dmension by a constant scaling factor which we might call resize_ellipse(). In this case it would be quite sensible for both read_ellipse() and resize_ellipse() call our new calculate_area() function to update the area. They would do this by passing the pointer they receive to our new calculate_area(). As an example the resize_ellipse() function might look like this:
```// Resize an ellipse by the same scaling factor in each dimension.
void resize_ellipse(Ellipse *el, float scale) {
el->axes[0] *= scale;
el->axes[1] *= scale;
calculate_area(el);
}
```
This illustrates an extremely common practice: functions that receive pointers often pass those pointers on to other functions. The obvious analogy is with a phone number:
If I were to tell you "my phone number is 0314 159 265, ring me any time you have a question about programming", not only could you ring me yourself but could pass that number onto your class-mates who could also ring me.
Similarly, if a function has a pointer to something it can pass the value of that pointer to another function which can then access the original object in memory.
### The complete code
The following code has three top-level functions and one utility function:
• A function to read in the values of the ellipse structure from the keyboard,
• a function to resize an ellipse,
• a function to move an ellipse and
• a utility function to calculate and save the area of the ellipse.
```#include <stdio.h>
#include <stdlib.h>
#define PI 3.14159265358979
typedef struct ellipse {
float centre[2];
float axes[2];
float orientation;
float area;
} Ellipse;
void calculate_area(Ellipse *el);
void resize_ellipse(Ellipse *el, float scale);
void move_ellipse(Ellipse *el, float dx[2]);
int main() {
float moveby[2], resize;
Ellipse *e = NULL;
e = malloc( sizeof *e);
if ( e == NULL ) {
fprintf(stderr, "Out of memory\n");
exit(99);
}
printf("Welcome to the ellipse program\n");
printf("The area of the ellipse is: %f\n", e->area);
printf("Amount to move the ellipse?\n");
scanf("%g %g", &moveby[0], &moveby[1]);
move_ellipse(e, moveby);
printf("Amount to resize the ellipse?\n");
scanf("%g", &resize);
resize_ellipse(e, resize);
return 0;
}
printf("Centre? (x,y)\n");
scanf("%f %f", &el->centre[0], &el->centre[1]);
printf("Length of axes ( > 0 )?\n");
scanf("%f %f", &el->axes[0], &el->axes[1]);
printf("Orientation to the vertical?\n");
scanf("%f", &el->orientation);
calculate_area(el); // Pass the pointer to another function
}
// Resize an ellipse by the same scaling factor in each dimension.
void resize_ellipse(Ellipse *el, float scale) {
el->axes[0] *= scale;
el->axes[1] *= scale;
calculate_area(el);
}
// Calculate the area of an ellipse and save it in the structure
void calculate_area(Ellipse *e) {
e->area = PI * e->axes[0] * e->axes[1];
}
void move_ellipse(Ellipse *el, float dx[2]) {
el->centre[0] += dx[0];
el->centre[1] += dx[1];
}
```
```Step through this code
```
Looking at the above example from the bottom upwards, the first thing we see is that move_ellipse() and calculate_area() are passed a pointer to the original ellipse and so are able to change the values of its position and area respectively.
Then resize_ellipse() is also passed a pointer to the original ellipse. As well as accessing the ellipse itself, it also passes this pointer on to calculate_area() to update the area.
Similarly, for read_ellipse() we pass just one pointer, to the ellipse, rather than three or five individual pointers to ellipse.orientation, ellipse.centre[0], etc. and it again passes the value of the pointer it was given on to the function calculate_area(). As mentioned above this is extremely common.
Functions that receive a pointer to a structure often pass that pointer onto other functions.
1. Step through the above "Key example".
2. To see a small but typical code in action.
3. Step through the above "Key example".
4. Step through until the call to read_ellipse()
5. When read_ellipse() is active practice minimising main() by clicking the button next to the function name.
• You may then need to manually un-minimise it when read_ellipse() returns.
6. Note how at various points there are up to three functions with pointers pointing to the same allocated ellipse (just as there are a number of people with your phone number stored inside their mobile phones).
1. Pass a pointer to a structure to a function.
2. To practice an extremely common programming task.
3. Above main() create a function with a name suitable for a function that will print out all the details about a cuboid. It should accept a pointer to a cuboid but for now the body of the function should be left empty.
4. Now cut-and-paste the code that prints out the cuboid details from main() to the new function.
5. Build & run. Check the output is correct..
6. Now do the same for the code that reads in the structure values such as dimensions, etc.
7. Build & run. Check the output is correct..
## More than one type of structure
The ellipse example has just one type of structure, but the generalisation is straightforward.
### Example: a projectile
Imagine a function to calculate the position and velocity of a projectile thrown into the air. Its position and velocity are known at time t=0 and we need calculate its position and velocity at time t + dt. Its prototype would look something like:
```void move(float x[NDIMS], float v[NDIMS], float mass,
float drag_coeff, float ywind[NDIMS],
float viscosity, float dt);
```
This is a very simple problem but the function has seven arguments. Worse, they are all floats or arrays of floats so it would be very easy to get two in the wrong order and the compiler would not notice. There are one hundred and forty four different legal ways of ordering these arguments (and nearly seven thousand if we allow for the chance of putting the same one in twice), but only one of these is the right one!
If we take a look at the arguments we see they split into three groups: x, v, mass and drag_coeff are properties of the projectile, ywind and viscosity are properties of the air and time is a physical quantity in its own right. This suggests we want two structures, one for the projectile and one for the air, and to leave time as it is.
The following code declares what are in effect two new types of variables. Again, this code does not actually create any structures, it just tells the compiler what we mean by "Projectile" and 'Air'.
```#define NDIMS 3
typedef struct projectile {
float x[NDIMS];
float v[NDIMS];
float mass;
float drag_coeff;
} Projectile;
typedef struct air {
float ywind[NDIMS];
float viscosity;
} Air;
```
We have gone from seven numbers to three things: projectile, air and time.
The function is now declared as:
```void move(Projectile *proj, Air *thisair, float dt);
```
Not only do we now only have three variables rather than seven, all three are of a different type so if we were to call the function with two of its arguments in the wrong order the compiler would notice and tell us.
## Thinking at a higher level
Almost without noticing it, we've made bit of a mental leap. We started by thinking about how we could reduce the number of arrays needed to represent our nuclides or reduce and organise the (floating point) arguments to a function. But we have quickly reached the stage where we have stopped talking about integers, floating-point numbers and strings and have started talking about nuclides, ellipses and projectiles.
This is the point about structures: we identify the types of "things" we are dealing with and typically we define a type of structure to represent that type.
Structures allow us to think at a higher level, in terms of the "things" we are dealing with rather than individual data values.
### Structures allow extensibility
Another thing we did without noticing it was that we added an "area" member to our ellipse structure. All w had to do was to type in the extra member and recompile. Similarly we mentioned that our nuclide information may need to be extended to include its mass and atomic number.
Our simple projectile example only has one dimension, y. But if our program is successful somebody is bound to ask us to extend it to three dimensions in which case y, vy and ywind will be replaced by arrays. If we used the "separate variables" approach we would have to go through each of our functions changing the number (and type) of arguments.
With the "structure" approach, we just add some more members to the structure definition, change the part of the code responsible for calculating the acceleration and recompile.
It's extremely common for a program to start off quite simply but for more features and properties to be added later, so the question of extensibility is hugely important.
Structures can easily be extended to include new members.
### Structures help make our functions more modular
When we added the area to our functions did not need to know about it. Indeed, resize_ellipse() and read_ellipse() jut passed the pointer to calculate_area() to calculate it: they did not even need to know the area member existed.
In our projectile example we might need to deal with the fact that real projectiles spin in the air and have texture. Even if we assume a spherical shape, that's several more variables. And somebody is sure to want to throw non-spherical objects. In this case our height_velocity function is going to split into two parts:
• An numerical algorithm for moving the projectile.
• A function for calculating the force on the object (and hence its acceleration), taking into account shape, spin, etc.
But if we use structures, the function definition remains unchanged: our acceleration function could become very complicated but our function call remains unchanged. Thus,the rest of our code doesn't need to know about it.
Functions don't even need to know of the existence of structure members that don't concern them.
### Reference
Show me your flowcharts and conceal your tables, and I shall continue to be mystified. Show me your tables, and I won't usually need your flowcharts; they'll be obvious.
Brooks, F. R. Jr, The Mythical Man-Month (1975).
## Summary
The text of each key point is a link to the place in the web page.
## For study after the lecture | 6,284 | 27,747 | {"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} | 3.125 | 3 | CC-MAIN-2018-34 | latest | en | 0.939213 |
https://community.smartsheet.com/discussion/128005/find-the-next-date-in-a-list-including-the-same-date | 1,723,104,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00052.warc.gz | 131,492,067 | 109,480 | # Find the next date in a list INCLUDING the same date
Options
I need some help, please! I've been all over discussions and tried everything I can think of but I can't get this to work.
I have a table that tracks status change dates for our projects. The date and time is recorded automatically each time the status changes. I want to show how long each status takes during the process. I am trying to get it to recognize the next date in the process for each project so I can then do the math for how long it was until that date. *of note, due to various factors the dates are not necessarily in order on the table, and all the projects are mixed together, so the formula has to find the next date for that specific project number.
Sometimes the status changes happens on the same day, and I want the formula to register that date and count it as '0 days' between those two statuses. This is the part I'm struggling with!
With all the Min/Collect formulas I've seen and tried it either:
1. returns the next date that isn't the same day and skips over instances where the status changed the same day which makes all the timelines downstream incorrect
OR
2. Always returns itself since it is the next date if you don't exclude same dates
If I were coding this I'd put it in a recordset/array by project number, have it sort by date/time, then run through the list and do calculations but I have no idea how to translate this to Smartsheet!
Here's a shot of my table - for simplicity I am only showing one project:
• ✭✭✭✭✭✭
Options
New helper column formula:
=VALUE(YEAR(Date@row) + "" + RIGHT("0" + MONTH(Date@row), 2) + "" + RIGHT("0" + DAY(Date@row), 2) + "" +
SUBSTITUTE([Time of Change]@row, ":", "."))
New Min/Collect:
=MIN(COLLECT(Date:Date, [Helper Column]:[Helper Column], @cell > [Helper Column]@row, [CCT #]:[CCT #], @cell = [CCT #]@row))
• ✭✭✭✭✭✭
Options
Even if it is on the same date, I assume the time of change would still be different?
• Options
Yes, the time of change would still be different
• ✭✭✭✭✭✭
Options
In that case I would suggest using a helper column to convert the time into a usable number
=VALUE(SUBSTITUTE([Time of Change]@row, ":", "."))
Then you can use a MIN/COLLECT combo along the lines of
=MIN(COLLECT(Date:Date, Date:Date, @cell >= Date@row, [TIme Helper]:[Time Helper], @cell > [Time Helper]@row))
• Options
Hi Paul,
That almost worked! However, when the next status update is on a different day with an earlier time than the previous date, it doesn't register with that equation:
• ✭✭✭✭✭✭
Options
New helper column formula:
=VALUE(YEAR(Date@row) + "" + RIGHT("0" + MONTH(Date@row), 2) + "" + RIGHT("0" + DAY(Date@row), 2) + "" +
SUBSTITUTE([Time of Change]@row, ":", "."))
New Min/Collect:
=MIN(COLLECT(Date:Date, [Helper Column]:[Helper Column], @cell > [Helper Column]@row, [CCT #]:[CCT #], @cell = [CCT #]@row))
• Options
That works brilliantly! Thank you so very much!!
• ✭✭✭✭✭✭
Options
Happy to help. 👍️
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 850 | 3,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.929117 |
http://onetomato.vn/category/uncategorized/ | 1,582,321,645,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145538.32/warc/CC-MAIN-20200221203000-20200221233000-00486.warc.gz | 109,278,721 | 23,200 | # Mathematics that started in China, particularly in Asia
Clear up it separately and completely the idea of unity mathematics is to choose some single occurrence , identify it as one entity and treat it as such.
Mathematics is all about finding patterns, relationships and explanations, not singular entities. A universal picture emerges out of different observations of data. If you see a http://apina.cl/2020/02/07/descriptive-novel-descende-mathematics/ car, or a triangle, or a dog, you can relate them by noticing their similarities, as they are part of a group of objects called groups. But if you make an observation of a bunch of many cars, you may see a similarity of shape, but you will also see a variety of shapes.
This type of objects could be the exact very same condition, or unique shapes, nevertheless all of them have similarities in shape, sizeand weight, speed, etc.. As an example, the shape of a car may be square, rectangle, rectangular, etc.. It is possible to receive harder collections, the full details the shape of a triangle, by way of instance, ellipse, square or circle, that will be likewise connected with some other shapes in various methods.
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Inside this scenario, if you gauge the length of the more compact triangle, you’ll come to realize that greater triangle includes a more and shorter sides, as per the distances. That is, that the proportion of the lengths of those 2 would be exactly like that the proportion of the lengths of the two circles.
Identity is the simplest thing to establish in mathematics. What is identity if you measure two points in two different shapes? The answer is that they are equal.
Unity mathematicians aim to show that the only thing that exists in a single point is one single element, and no other point in the universe is bigger than that single point. There is no singularity or separate point in the universe, but there is a single point in the universe, just like two circles can be either circular or elliptical. Two circles can be the same size, and they can both be equally distant from the center of the universe.
In other words, an object is a’only unit’ so it is’one’ – it will not come close to being a’million’, or’million’, or maybe’billion’, nonetheless it’s not going to be a’trillion’, possibly. All things are’one’ once the math of unity math measures them. After dimensions are taken by you, you’ll discover that what you quantify is’you’ and absolutely practically nothing else.
Now, a few mathematicians attempt to provide something that doesn’t seem to be only one, which is called indeterminate definitions. But they don’t offer some thing that generally seems to be’one’, they feature something which may be described as being one. If either side of a triangle are described, one particular negative is’too far’ different from the other, and it cannot be’just one’. So the two sides are described like getting’three’they cannot function as’just one’.
Unity mathematics tries to show that one can never have two sides that are exactly the same and that is why, as we have seen, in the universe there is no limit to the size of the universe. We would be only ‘one’ as far as we would extend the dimensions. stretch our minds beyond our understanding of the universe.
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## The Positive results involving Globalization That many of us Certainly not Talk About
Globalization has become one of the most debated subjects of recent times. It has been hailed as one of the best things to happen to the world by its proponents, while opponents have labeled it as neo-imperialism. Despite its challenges, the positive role it has played in changing the lives of persuasive essay millions of people around the world is commendable.
Globalization has become one of the most debated subjects of recent times. It has been hailed as one of the best things to happen to the world by its proponents, while opponents have labeled it as neo-imperialism. Despite its challenges, the positive role it has played in changing the lives of millions of people around the world is commendable.
Globalization has changed us into a company that searches the world, not just to sell or to source, but to find intellectual capital � the world�s best talents and greatest ideas.
� Jack Welch
Although there are a host of challenges associated with globalization, it has numerous positive effects as well. Anti-globalization crusaders who represent issues such as environmental degradation, economic inequality, loss of jobs in the host country etc., have made their presence felt at various global conferences. The issues that they raise are real and can have catastrophic effects if not tackled earnestly. However, the advantages of globalization are all-pervasive and a lot of developing countries are benefiting from it. As the subject of this article pertains to the positive effects of globalization, we will focus on the positive changes it has brought to the world � both economically and culturally, and how it can very well become one of the defining factors of the present century.
# Benefits of Globalization
### Increased Competition
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One of the most visible effects is the improved quality of products due to global competition. Customer service and the �customer is the king� approach to production have led to improved quality of products and services. As domestic companies have to fight out foreign competition, they are compelled to raise their standards and customer satisfaction levels in order to survive in the market. Besides, when a global brand enters a new country, it comes in riding on some goodwill, which it has to live up to. This creates competition in the market and a �survival of the fittest� situation.
### Employment
With globalization, companies have forayed into the developing countries and hence generated employment for them. But it can turn out to be either good or bad, depending on the point of view you wish to see it from. It has given an opportunity to invest in the emerging markets and tap the talent which is available there. In developing countries, there is often a lack of capital which hinders the growth of domestic companies and hence, employment. In such cases, due to global nature of the businesses, people of developing countries too can obtain gainful employment opportunities. But the developed countries have lost jobs on account of this shift of jobs to the developing world and hence it is a pinch felt by people in the First World.
### Investment and Capital Flows
A lot of companies have directly invested in developing countries like Brazil and India by starting production units, but what we also need to see is the amount of Foreign Direct Investment (FDI) that flows into the developing countries. Companies which perform well attract a lot of foreign investment and thus push up the reserve of foreign exchange.
While discussing the effects of globalization, how can we forget about the impact of foreign trade on an economy. Comparative advantage has always been a factor, even in during old times. While trade originated in the times of early kingdoms, it has been institutionalized due to globalization. Previously, people had to resort to unfair means and destruction of kingdoms and countries to get what they wanted. Today, it is done in a more humane way, with mutual understanding. People who operate in uncivilized ways have to face the WTO and other world organizations that have been established with a view to control and regulate trade activities of the countries.
While it is generally assumed that all the innovations happen in the Western world, the know-how also comes into developing countries due to globalization. Without it, the knowledge of new inventions and medicines would remain cooped up in the countries that came up with them and no one else would benefit. The spread of know-how can also be expanded to include economic and political knowledge, which too has spread far and wide. The most obvious example of the spread of knowledge is that the Western world today is waking up to the benefits of Ayurveda and Yoga � traditional Indian practices, while the Western antibiotics are flooding
## 100 Trouble Remedy Article Subject areas with Sample Essays
Virginia has been a university English instructor for over 20 years. She specializes in helping people write essays faster and easier.
Struggling to find a great topic? Scan the lists below to find questions my students have generated to find something you want to write about. help on writing a book report Topics are divided into:
2. Social Issues
7. Driving and Transportation
Remember you can adapt these general ideas to fit your own school, town, or situation.
## Relationship Topics
1. How can social media bullying be prevented?
2. What is the best way to help someone who is depressed?
3. What’s the best way to deal with manipulative and domineering people?
4. How can you help a friend or roommate who is making poor life choices?
5. How can you get out of a bad relationship?
6. How can you move out of the “friend zone”?
7. What is the best way for a woman to show a man she is interested? Should she ask him out?
8. How has texting affected face-to-face relationships?
9. What can be done to make sure that people today actually know how to have “real” relationships?
10. How can we get away from stereotypes and racism? What steps can we take as individuals and as a society?
11. How can we learn to understand people who are different from us culturally, racially, or socioeconomically?
12. How can we break down barriers between groups?
## 4 Parts of Problem Solution
Describe a problem vividly.
Propose a solution.
Argue that the solution is practical, feasible, cost-effective, and workable.
Explain why it is better than other solutions.
## Social Problems
1. How can we help homeless people in our community?
2. How can we prevent people from dropping out of high school?
3. What is the best way to prevent teen pregnancy?
4. How can kids be persuaded not to experiment with illegal drugs?
5. What is the best way to prevent deaths from drunk driving?
6. How can teenagers be convinced to drive more safely?
7. What can be done to prevent divorce?
8. How can kids with divorced parents be helped to do well in school, have strong relationships, and build successful lives and marriages?
9. What can be done to eliminate or prevent racism?
10. What is the best way to help people who are victims of family violence?
11. How can we change the welfare system to help people escape generational poverty?
12. How can we deal with illegal immigration?
13. What should we do about the increase in gun violence?
14. How can we best rehabilitate prisoners so that they can be productive members of society?
15. How can healthcare be ensured for everyone around the world?
16. How can we improve literacy?
17. What can be done to prevent human trafficking?
18. How can we prevent children from being negatively influenced by violence and pornography in media such as video games, movies, and the Internet?
19. How can we persuade people to make healthy lifestyle choices like not smoking, exercising, and maintaining a healthy BMI?
20. How can we balance the right for free speech with the right to not be insulted or abused?
21. Choose a city you know. How can that city be made more bicycle and pedestrian-friendly?
22. How can we best reduce the problem of terrorism?
23. Should there be controls in the way the media portrays celebrities?
24. How can we handle the problem of online data mining or the fact that data brokers are selling our information to advertisers, employers, health insurers, and credit rating agencies?
25. What should we do to help people get jobs?
## Athletes and Sports
1. How can steroid use be limited in all sports (or one in particular)?
2. Should college athletes be paid?
3. How can colleges better handle the combination of education, athletics, and business?
4. Do kids train too hard in athletics at young ages?
5. What is the best way for a young athlete to learn a sport? (You might want to talk about a particular sport you know well).
6. How can coaches best encourage their athletes to do their best?
7. Think about your local sports team. What can be done to make that team more effective? How can the fans be made to support the team more?
8. Pick your favorite sport. How can someone best prepare themselves to be excellent at that sport?
9. How can sports media be more effective at covering a sport you like to watch? Should the announcers be biased or not? Which games should be covered by sports networks? What types of interviews and special features make the sport more interesting to watch?
10. What can be done to increase attendance at your favorite sporting event?
11. How should a coach or player handle media interviews when they lose a big game?
12. How can players prepare themselves for the (inevitable) loss?
13. What should professional players
# The essay about leaders provides you with the potential to present your unique opinions on a subject.
This is the essence of it. While many people have the ability to write persuasive and motivating essays, some are just not as good as others in this area.
An important point to ponder is whether you are asking questions or taking a position. You can either begin to ask your own questions about the topic, or make an objection to someone else’s position. Neither choice is wrong, and both are effective. It depends on your response to questions from others in the discussion, as well as your ability to set the tone of the discussion.
An effective essay about leaders can only be written if it presents the whole of the topic and presents your views in a well-organized manner. Be sure that you use the same type of structure for all sections of the essay, and that you use the appropriate styles for the topics, style, content, and punctuation.
## Do not assume that questions can be answered in a paragraph of text.
Sometimes it is easier to paraphrase a question or answer. This is not necessarily a bad thing, but you http://sabinabeasley.com/2020/02/06/betsy-devos-decide-on-to-get-secretary-of-your-practice-is-easily-the-most-jeered/ should make sure that your writing is very interesting and understandable, or else you will be limited in the length of your essay.
Always keep your essay on topic. Do not let it go off on tangents, for fear that it will end up being a non-existent subject. Do not become attached to one side or the other of the issue.
You can encourage your readers to read through your essay and re-read it if they find it hard to comprehend. Readers can go over a portion of the essay more than once. Some do not want to do this, and do not like to re-read any part of the essay again, but keep in mind that people want to have access to as much information as possible, in order to form an informed opinion.
You need to write your essay in such a way that when you start the topic of your new topic, you do not have a hole in the topic, so that you can fill it later. Therefore, you have to have a solid point of view, and that you can expand upon later on.
## It is also important to create a theme in your essay, and let the topic evolves as you change the subject.
After all, the better essay on leaders you are, the more interesting the essay will be. Thus, it is critical to write an essay that will stand the test of time.
The way to write an essay is to take your selected topic, turn it into an essay, and then keep on writing on that topic. While doing this, you need to keep your essay in line with your point of view, without deviating from it. You must follow certain rules if you are to have a well-written essay on a particular topic.
Your essay on leaders should not focus on specific aspects of leadership that you would like to research further. Instead, you should take a general approach to the topic. As a rule, do not write about a specific person, but only on the concept of leadership. Allowing yourself the freedom to move between different topics allows you to become more creative and original.
There is no need to spend hours making the tip sheet for your essay. You do not need to stop while writing your essay, because the article is basically to provide you with a unique and insightful perspective on the topic.
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## What is Actually Going on with Learning Theories School Nursing
Various theorists have various views. Theories can likewise be classified based on their objectives, they may be descriptive or prescriptive. Moreover, scientific theories have to be falsifiable.
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Last, the environment-behavior interaction, external aspects can alter the manner in http://www.enproestruturasmetalicas.com.br/what-is-e-corresponding-into-in-surplus-excessive-fat-physics/ which you display the behavior. On the flip side, ineffective cognitive processes result to learning difficulties that may be seen anytime during the lifetime of someone. Therefore, it’s reaching to the former limit of the person growth potential.
In this early stage of life, it is necessary for children to get consistent care so they can learn how to trust the men and women in the world around them. There are lots of approaches to define family. Throughout psychology history, lots of different child growth theories have emerged to explain the changes which take place during the early portion of life.
Additional development in social learning theory posits that learning will probably occur if there’s a close identification between the observer and the model and should the observer also has a whole lot of http://richardholbrook.com/studying-numerous-nursing-theories/ self-efficacy. The theory is that communication going on in the aArenaa will be, in the majority of circumstances, the absolute best and most effective communication. Functionalist theory interprets each portion of society with regard to the way that it contributes to the stability of the entire society.
## The Tried and True Method for Learning Theories School Nursing in Step by Step Detail
Students also make the most of specific clinical experiences in the surrounding more rural places. Nurses need to be critical thinkers and has to have the capacity to rationalize their decision making to make sure safe nursing practice. School nurses have a number of responsibilities and are a critical part of the educational system.
In this early stage of life, it is necessary for children to get consistent care so they can learn how to trust the men and women in the world around them. Throughout psychology history, plenty of different child development theories have emerged to spell out the changes which happen during the early part of life. There are a lot of approaches to define family.
Definitions are utilized to convey the overall significance of the concepts of the theory. The Social Learning Theory is only one of many that have marked a long-lasting influence on society and the area of criminology. Concepts might be abstract or concrete.
They need to be continuously conscious of the well-being of their patients. http://54.247.55.85/2020/01/24/checking-out-a-range-of-nursing-theories/ Clinical experience is additionally a substantial portion of your training. The following are a couple of the most prominent, together with the nurses who developed them.
## If You Read Nothing Else Today, Read This Report on Learning Theories School Nursing
Information technology for a mode to transform healthcare delivery is going to be stressed. The exact same class can’t be utilized to fulfill requirements for over one major. Many traditional learning methods now are extremely much intertwined.
One other important outcome reflection is that the essential scenarios like this dilemma in the instance of Miss June ought to be taught in nursing curriculum so they can prepare themselves for the situations similar to this. Some folks are born leaders. You are more prone to visit law school in case the individuals who you care about most think that you should.
Learning is considered to occur because salivation was conditioned to another stimulus which didn’t elicit it initially. Learning medical terminology can truly be difficult. Key Concepts are usually quite short articles and each one is dedicated to a certain topic.
There are several different practice models that help determine the way social workers decide to aid people fulfill their objectives. It is a method by which people are able to understand their experiences and, as a consequence, modify their behavior. Individuals will probably must totally finish the health part of such kind of an extensive training course by means of a native healthcare establishment.
Let’s take a close look at a number of the crucial theories of child development. The purpose was supposed to explain the fundamental implications of the profession and to improve the condition of the profession. Therefore, it is reaching to the last limit of the person growth potential.
The theoretical framework might be rooted in a particular theory, in which case, your work is anticipated to check the validity of that current theory regarding certain events, issues, or phenomena. Collective societal change is an indispensable tenet. A rather interesting topic that’s researched regarding the social learning theory is aggression.
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# When I write a response to literature essay questions, my goal is to do what is necessary in order to convince the student that I am an expert.
In other words, I’m trying to convince the student that my thoughts are valid and that I have some special knowledge that they need to use in their own paper.
Here are three things that I do to be a good literature essay questioner. It’s my hope that these suggestions will help you become an excellent literature essay questioner.
* First of all, it’s critical to understand that there are two types of literature essay questions. First of all, there are the type of questions that have specific definitions https://www.infipix.com/top-10-hardest-themes-within-mathematics/ that one can supply. These are often hard and one has to ask very specific questions that will provide specific answers.
## On the other hand, there are responses that you can work with and actually come up with your own.
These are generally far more open and one can answer them as they wish. This way, the student can come up with a response that will satisfy his or her style and also their style of writing.
* Secondly, a great response to literature essay questions can help the student to understand their writing more thoroughly. By being able to take a look at the response you get, the student can better understand where he or she is heading and thus make better decisions in the future.
* Thirdly, one must remember that there are plenty of schools where you can get feedback about what you’ve written. These feedbacks can be very valuable and allow you to start from scratch.
* A good response to literature essay questions helps the student to build on their writing skills. By providing them with a style guide that will enable them to avoid common mistakes, you’ll be helping them to be better writers in the future.
* Great responses to literature essay questions are extremely important for every school. For example, a good response will make the student feel comfortable in writing his or her own papers.
* A great response to literature essay questions will teach the student what works and what doesn’t work. It allows them to build up a foundation and then, when they move on to solving problems and writing essays of their own, they’ll be armed with the right tools.
* A good response to literature essay questions will enable the student to step outside of their comfort zone and makes them use more of the available resources. In addition, it allows them to put their skills to the test.
* A good response to literature essay questions will teach the student how to make better decisions. It will help them to be more informed and also why they should do so.
# A kinesthetic mastering fashion composition is different from the conventional essay in that it places emphasis on physical exercise and movement.
It is based on the premise that student should actively engage in moving through his thoughts, feelings, and ideas in order to comprehend them. If students are able to engage their bodies during the reading process, they can better comprehend their essays and further the conversation.
Kinesthetic exercises are performed while participating in this essay format, such as; picking websites that write essays for you up a pencil or pen, circling a word to make it the focal point of attention, or making a face similar to the expression you see on a page. While students think of the words they will be writing, they should move through their thoughts, not just the actual writing of the essay.
## Using kinesthetic exercises are a great way to speed up the essay writing process for students.
By improving the spatial relations between themselves and their classmates, students can be more productive with their writing.
It is important to understand that kinesthetic exercises are not simply “getting in your own way”. While students should try to keep their physical actions to a minimum, they are an essential part of the essay writing process.
Even while using kinesthetic movements, students should practice being in control of themselves and their body movements. Students should move through their thoughts and feelings in a controlled manner and continue to do so until it is effective. When they slow down, they are taking control of their thinking process, which allows them to achieve mastery over their essay.
Kinesthetic exercises are more important for students who are beginning to write assignments and are not used to the kinesthetic pace. This form of learning style essay is less formal and easier to learn and improve.
## Kinesthetic exercises donot have to be complex.
Simple exercises like grabbing and making faces can be incorporated into an essay writing process.
While basic exercises like circles and drops have already been covered in the essay writing process, there are many more exercises to be included in the process. The simple exercises are going to give students the advantage to succeed in this method of essay writing.
Students should look at two basic examples: sentences like “I sit behind the refrigerator”We walk along the sidewalk. ” Using these, students can practice moving their body through the essay by making a face similar to these sentences, and continuing until they are successful.
Kinesthetic exercises should be done in an environment where students can concentrate on what they are doing. Kinesthetic exercises should be done in the presence of other students so that they can participate in the process.
Kinesthetic exercises are a great way to help improve a student’s ability to think and move. If students use exercises to focus on spatial relations, they can be more productive with their essay writing. | 7,978 | 40,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-10 | latest | en | 0.957525 |
https://cstheory.stackexchange.com/questions/1680/separation-of-limited-nondeterminism-classes/1698 | 1,643,106,784,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304810.95/warc/CC-MAIN-20220125100035-20220125130035-00702.warc.gz | 235,491,813 | 36,296 | Separation of limited nondeterminism classes?
It is interesting to find the best lower bound on the number of nondeterministic bits needed to solve satisfiability problem. Let $\beta_k P$ be the class of problems solvable by a nondeterministic Turing machine in polynomial time and using $O(\log^k n)$ nondeterministic bits.
$P=\beta_1P$. Notice that $O(\log n)$-clique is in $\beta_2P$. It is an open problem whether $\beta_1 P = \beta_2 P$. It is known that $\beta_1P \neq \beta_2P$ implies $P \neq NP$
What is the most serious attempt at separating $\beta_1P$ from $\beta_2P$?
Does $\beta_kP=NP(\log^k n)$ form an infinite hierarchy?
This reference contains a survey of limited nondeterminism:
• I'll refer you to the meta discussion on how to ask a question, and in particular the part about explaining why YOU care. Sep 26 '10 at 8:03
• @Suresh, for me, the $\beta_1P$ vs. $\beta_2P$ problem looks more natural problem than the $P$ vs $NP$ problem. Sep 26 '10 at 13:29
• Yes, but can you summarize what's known first and any conjectures/questions you might have ? Sep 26 '10 at 14:56
• Does $\beta_k P=NP(\log^k n)$ define an infinite hierarchy? Sep 26 '10 at 15:35
• Then the question should be that: it's better to ask more focused questions. Also, please fix the typos in the title Sep 26 '10 at 15:49
For attempts to separate the bounded nondeterminism hierarchy, I think monotone dualization of prime formulas is the most salient topic.
Consider the decision problem
MONOTONE DUAL
Input: two monotone CNF formulas, from which no literals can be removed
Question: is the one formula the dual of the other?
MONOTONE DUAL can be decided with $O((\log\ n)^2/\log\ \log\ n)$ nondeterministic steps. There is also a quasi-polynomial $n^{o(\log\ n)}$ time upper bound for this problem which has stood since 1996.
So MONOTONE DUAL is in co-$\beta_2$P and also doesn't seem to require the full power of this class. On the other hand, MONOTONE DUAL may be a good candidate for a problem that is outside P = co-$\beta_1$P.
This is surveyed in:
• Thomas Eiter, Kazuhisa Makino, and Georg Gottlob, Computational aspects of monotone dualization: A brief survey, DAM 156 2035– 2049, 2008. doi: 10.1016/j.dam.2007.04.017 (preprint)
I am not sure there is more work along these lines. As with many other areas with the potential to separate P from NP, after some promising early progress new ideas now appear to be necessary.
This is not a complete answer, yet I think it is helpful.
The answer is in fact taken from the following paper:
Beigel, R. and Goldsmith, J. 1998. Downward Separation Fails Catastrophically for Limited Nondeterminism Classes. SIAM J. Comput. 27, 5 (Oct. 1998), 1420-1429. DOI= http://dx.doi.org/10.1137/S0097539794277421
The abstract states almost everything:
The $$\beta$$ hierarchy consists of classes $$\beta_k={\rm NP}[\log^k n]\subseteq {\rm NP}$$. Unlike collapses in the polynomial hierarchy and the Boolean hierarchy, collapses in the $$\beta$$ hierarchy do not seem to translate up, nor does closure under complement seem to cause the hierarchy to collapse. For any consistent set of collapses and separations of levels of the hierarchy that respects $${\rm P} = \beta_1\subseteq \beta_2\subseteq \cdots \subseteq {\rm NP}$$, we can construct an oracle relative to which those collapses and separations hold; at the same time we can make distinct levels of the hierarchy closed under computation or not, as we wish. To give two relatively tame examples: for any $$k \geq 1$$, we construct an oracle relative to which
$${\rm P} = \beta_{k} \neq \beta_{k+1} \neq \beta_{k+2} \neq \cdots$$
and another oracle relative to which
$${\rm P} = \beta_{k} \neq \beta_{k+1} = {\rm PSPACE}.$$
We also construct an oracle relative to which $$\beta_{2k} = \beta_{2k+1} \neq \beta_{2k+2}$$ for all k.
This shows that the $$\beta$$ hierarchy has contradictory relativizations, nominating its separation as a "hard" problem.
The literature does not seem to care much about the $$\beta$$ hierarchy, since a regular search does not show up with many relevant results. In particular, there's a very limited (and seemingly irrelevant) number of papers citing the above results.
Unrelated but Worth Noting
You may also see:
The complexity of optimization problems. Journal of Computer and System Sciences, 36:490-509, 1988.
where it's proved that If $$FP^{NP} = FP^{NP[log]}$$ (that is, allowed only a logarithmic number of queries), then $$P = NP$$. | 1,236 | 4,485 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-05 | latest | en | 0.897752 |
https://programsandcourses.anu.edu.au/2022/course/MATH6204 | 1,685,926,641,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00442.warc.gz | 515,985,151 | 11,780 | • Offered by Mathematical Sciences Institute
• ANU College ANU Joint Colleges of Science
Specialist
• Course subject Mathematics
• Areas of interest Mathematics
• Course convener
• Dr Vigleik Angeltveit
• Mode of delivery In Person
• Co-taught Course
• Offered in Second Semester 2022
Algebraic Topology (MATH6204)
In Sem 2 2022, this course is on campus with remote adjustments only for participants with unavoidable travel restrictions/visa delays.
Algebraic topology studies properties of topological spaces and maps between them by associating algebraic invariants (fundamental groups, homology groups, cohomology groups) to each space.
This course gives a solid introduction to fundamental ideas and results that are employed nowadays in most areas of mathematics, theoretical physics and computer science.
This course aims to understand some fundamental ideas in algebraic topology; to apply discrete, algebraic methods to solve topological problems; to develop some intuition for how algebraic topology relates to concrete topological problems.
Topics to be covered include:
• Fundamental group and covering spaces
• Brouwer fixed point theorem
• Fundamental theorem of algebra
• Homology theory and cohomology theory
• Jordan-Brouwer separation theorem
• Lefschetz fixed theorem
• Additional topics (Orientation, Poincare duality, if time permits)
Note: Graduate students attend joint classes with undergraduates but will be assessed separately.
## Learning Outcomes
Upon successful completion, students will have the knowledge and skills to:
On satisfying the requirements of this course, students will have the knowledge and skills to:
1. Explain the fundamental concepts of algebraic topology and their role in modern mathematics and applied contexts.
2. Demonstrate accurate and efficient use of algebraic topology techniques.
3. Demonstrate capacity for mathematical reasoning through analyzing, proving and explaining concepts from algebraic topology.
4. Apply problem-solving using algebraic topology techniques applied to diverse situations in physics, engineering and other mathematical contexts.
## Indicative Assessment
Assessment will be based on:
• Assignment 1 (10%: LO 1-4)
• Assignment 2 (10%; LO 1-4)
• Assignment 3 (10%; LO 1-4)
• Presentation (10%; LO 1-4)
• Final exam (60%; LO 1-4)
The ANU uses Turnitin to enhance student citation and referencing techniques, and to assess assignment submissions as a component of the University's approach to managing Academic Integrity. While the use of Turnitin is not mandatory, the ANU highly recommends Turnitin is used by both teaching staff and students. For additional information regarding Turnitin please visit the ANU Online website.
Three lectures per week and regular workshops.
## Requisite and Incompatibility
Incompatible with MATH4204
You will need to contact the Mathematical Sciences Institute to request a permission code to enrol in this course.
## Fees
Tuition fees are for the academic year indicated at the top of the page.
Commonwealth Support (CSP) Students
If you have been offered a Commonwealth supported place, your fees are set by the Australian Government for each course. At ANU 1 EFTSL is 48 units (normally 8 x 6-unit courses). More information about your student contribution amount for each course at Fees
Student Contribution Band:
1
Unit value:
6 units
If you are a domestic graduate coursework student with a Domestic Tuition Fee (DTF) place or international student you will be required to pay course tuition fees (see below). Course tuition fees are indexed annually. Further information for domestic and international students about tuition and other fees can be found at Fees.
Where there is a unit range displayed for this course, not all unit options below may be available.
Units EFTSL
6.00 0.12500
## Course fees
Domestic fee paying students
Year Fee
2022 \$4200
International fee paying students
Year Fee
2022 \$6000
Note: Please note that fee information is for current year only.
## Offerings, Dates and Class Summary Links
ANU utilises MyTimetable to enable students to view the timetable for their enrolled courses, browse, then self-allocate to small teaching activities / tutorials so they can better plan their time. Find out more on the Timetable webpage.
The list of offerings for future years is indicative only.
Class summaries, if available, can be accessed by clicking on the View link for the relevant class number.
### Second Semester
Class number Class start date Last day to enrol Census date Class end date Mode Of Delivery Class Summary
6064 25 Jul 2022 01 Aug 2022 31 Aug 2022 28 Oct 2022 In Person View | 1,003 | 4,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.848527 |
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