text:
# Active power output calculation of hf generator unit based on measured volt acorss a load

I have an electrosurgical hf generator unit that produces a modulated signal with duty cycle in blend mode. I have the output of this unit connected across a load. With scope connected across the load, I measure peak-peak volt of a signal with a duty cycle in a pulse. The waveform looks like as shown. It has hf sinusoidal oscillation in active part.

I want to calculate active power delivered by the unit. What would be the correct way of calculating the active power output? How to calculate RMS volt from Vp-p?

• Can you express your voltage waveform as a function including the modulation? Since you should be able to, you can then just actually do the RMS integral calculation. – Captainj2001 Jul 13 '16 at 15:46
• I've updated the original question with the waveform details. – user116646 Jul 13 '16 at 17:16

$$v(t) = 235\sin\left(\frac{2\pi t}{T}\right)\left[ u(t) -u(t - 0.85T) \right] \quad t\in[0,T].$$
For the RMS voltage calculation you then get: $$V_{RMS}^2 = \frac{1}{T} \int_0^T 235^2\sin^2\left(\frac{2\pi t}{T}\right)\left[ u(t) -u(t - 0.85T) \right]^2dt \\ = \frac{235^2}{T} \int_0^{0.85T}\sin^2\left(\frac{2\pi t}{T}\right)dt \\ = \frac{235^2}{T} \int_0^{0.85T} \left(\frac {1-\cos\left(\frac{4\pi t}{T}\right)}{2}\right) dt \\ = \frac{235^2}{T}\left[ \frac{0.85T}{2} - \frac{T}{8\pi}\sin\left(0.85\cdot4\pi\right)\right] \\ =235^2\left[\frac{0.85}{2}-\frac{\sin\left(0.85\cdot4\pi\right)}{8\pi}\right].$$ Then you'll have, $$V_{RMS} = 235\sqrt{\frac{0.85}{2}-\frac{\sin\left(0.85\cdot4\pi\right)}{8\pi}}\approx 160~\text{V}.$$ So for your power calculation then, $$P = \frac{V_{RMS}^2}{R}=\frac{160^2}{500}=51.2~\text{W}$$ I hope this makes it clear that for more complex waveforms calculating the RMS voltage is not quite as simple as it is for "nice" sinusoids. What's worse about this waveform is that the RMS voltage depends on whether or not you base your waveform on a sine or cosine waveform!
Update: It looks like the waveform you are supplying is both amplitude and duty cycle modulated, you can follow the same procedure to find the RMS voltage of your actual waveform. You may have to do this over a number of periods (until the modulating and modulated sinusoids are harmonics of each other). Your new waveform is something like, $$v(t) = 235m\sin\left(\frac{2\pi t}{T_1}\right)\sin\left(\frac{2\pi t}{T_2}\right)\left[u(t) - u(t - 0.85T_1\right],$$ where in this case the waveform with period $T_1$ is the modulating waveform with modulation index $m$.


text:
What is $v_{\infty}$?

So, you can calculate the trajectory of a celestial projectile by using this equation:

$$e=\frac{rv_{\infty}}{\mu},$$

where $$\mu$$ is the central body's Gravitational Parameter, so it is definable by $$\mu=GM$$. $$r$$ is the orbiting body's current distance from the sun. That leaves us with only one variable left to uncover; $$v_{\infty}$$ (the Hyperbolic Excess Velocity). If you do a bit of research, you get an equation defining it as:

$$v_{\infty}$$ definition

Where you have to know variables considering Earth, and an outside planet. But, what if we need to find out the Hyperbolic Excess Velocity of an outside planet, having nothing to do with the Earth or another outside planet?

In an answer to this question, I want a reliable explanation of how to calculate $$v_{\infty}$$, not referring to another form of the equation for eccentricity, isolating $$v_{\infty}$$. I also want an explanation of what the Hyperbolic Excess Velocity is, in its essence, and an example of how to calculate it.

• Thank you, AFG, for helping me by editing my post. – Questioner Apr 12 at 14:31

1 Answer

Speed $$v_\infty$$ is the speed of the body at the moment where one can neglect all gravitational interactions, i.e. at infinite distance or (which is usually the same) after/before infinite time. The quantity $$v_\infty$$ is well defined in 2-body problem where the second body has an infinitely larger mass and is at rest (or central field). (If it's not, then one can distinguish between $$v_{-\infty}$$ the speed at $$t=-\infty$$ and $$v_{+\infty}$$, speed at $$t=+\infty$$)

• Can you give me an equation defining it? – Questioner Apr 12 at 16:56
• What do you mean by equation? $v_\infty = |dr/dt|_{t=\pm\infty}$? If you mean, how can you relate $v_\infty$ to other quantities, it depends on a particular problem. Usually one needs to write energy and angular momentum conservation laws. – Vasily Mitch Apr 13 at 8:40

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# Inner-model reflection principles

• N. Barton, A. E. Caicedo, G. Fuchs, J. D. Hamkins, and J. Reitz, “Inner-model reflection principles,” ArXiv e-prints, 2017. (manuscript under review)
@ARTICLE{BartonCaicedoFuchsHamkinsReitz:Inner-model-reflection-principles,
author = {Neil Barton and Andr\'es Eduardo Caicedo and Gunter Fuchs and Joel David Hamkins and Jonas Reitz},
title = {Inner-model reflection principles},
journal = {ArXiv e-prints},
year = {2017},
volume = {},
number = {},
pages = {},
month = {},
note = {manuscript under review},
abstract = {},
keywords = {under-review},
source = {},
doi = {},
eprint = {1708.06669},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/inner-model-reflection-principles},
}

Abstract. We introduce and consider the inner-model reflection principle, which asserts that whenever a statement $\varphi(a)$ in the first-order language of set theory is true in the set-theoretic universe $V$, then it is also true in a proper inner model $W\subsetneq V$. A stronger principle, the ground-model reflection principle, asserts that any such $\varphi(a)$ true in $V$ is also true in some nontrivial ground model of the universe with respect to set forcing. These principles each express a form of width reflection in contrast to the usual height reflection of the Lévy-Montague reflection theorem. They are each equiconsistent with ZFC and indeed $\Pi_2$-conservative over ZFC, being forceable by class forcing while preserving any desired rank-initial segment of the universe. Furthermore, the inner-model reflection principle is a consequence of the existence of sufficient large cardinals, and lightface formulations of the reflection principles follow from the maximality principle MP and from the inner-model hypothesis IMH.

Every set theorist is familiar with the classical Lévy-Montague reflection principle, which explains how truth in the full set-theoretic universe $V$ reflects down to truth in various rank-initial segments $V_\theta$ of the cumulative hierarchy. Thus, the Lévy-Montague reflection principle is a form of height-reflection, in that truth in $V$ is reflected vertically downwards to truth in some $V_\theta$.

In this brief article, in contrast, we should like to introduce and consider a form of width-reflection, namely, reflection to nontrivial inner models. Specifically, we shall consider the following reflection principles.

Definition.

1. The inner-model reflection principle asserts that if a statement $\varphi(a)$ in the first-order language of set theory is true in the set-theoretic universe $V$, then there is a proper inner model $W$, a transitive class model of ZF containing all ordinals, with $a\in W\subsetneq V$ in which $\varphi(a)$ is true.
2. The ground-model reflection principle asserts that if $\varphi(a)$ is true in $V$, then there is a nontrivial ground model $W\subsetneq V$ with $a\in W$ and $W\models\varphi(a)$.
3. Variations of the principles arise by insisting on inner models of a particular type, such as ground models for a particular type of forcing, or by restricting the class of parameters or formulas that enter into the scheme.
4. The lightface forms of the principles, in particular, make their assertion only for sentences, so that if $\sigma$ is a sentence true in $V$, then $\sigma$ is true in some proper inner model or ground $W$, respectively.

We explain how to force the principles, how to separate them, how they are consequences of various large cardinal assumptions, consequences of the maximality principle and of the inner model hypothesis. Kindly proceed to the article (pdf available at the arxiv) for more.

• N. Barton, A. E. Caicedo, G. Fuchs, J. D. Hamkins, and J. Reitz, “Inner-model reflection principles,” ArXiv e-prints, 2017. (manuscript under review)
@ARTICLE{BartonCaicedoFuchsHamkinsReitz:Inner-model-reflection-principles,
author = {Neil Barton and Andr\'es Eduardo Caicedo and Gunter Fuchs and Joel David Hamkins and Jonas Reitz},
title = {Inner-model reflection principles},
journal = {ArXiv e-prints},
year = {2017},
volume = {},
number = {},
pages = {},
month = {},
note = {manuscript under review},
abstract = {},
keywords = {under-review},
source = {},
doi = {},
eprint = {1708.06669},
archivePrefix = {arXiv},
primaryClass = {math.LO},
url = {http://jdh.hamkins.org/inner-model-reflection-principles},
}

This article grew out of an exchange held by the authors on math.stackexchange
in response to an inquiry posted by the first author concerning the nature of width-reflection in comparison to height-reflection:  What is the consistency strength of width reflection?


text:
First Java assignment[Part II]

This is part II of my first Java Assignment.
1.Generating 1000 Random Integer Number
We have to ‘find a way’ to generate 1000 Random Number and save to text file name “array.txt” (that said we have to dig into java standard written code, not to actually code it)

I did find a way to generate Random Number at : http://docs.oracle.com/javase/1.4.2/docs/api/java/util/Random.html

The general syntax is like this :

 Random generator = new Random();  //new random object

generator.nextInt();               //generate a random Number using computer time clock
//milliseconds as seed number
//OR
generator.nextInt(seed);           //using your seed

//make sure you have something to assign to when generating random number

So for generating 1000 number, we just need to make it a for loop that run 1000 times, like this:

//generating 1000 integer number
for (int i = 0; i < 1000; ++i){
randval = generator.nextInt(seed);
//save it to file
//... continue the cycle
}

2.Do some general task that is usually asked for array of numbers.
That is :
a.Find the smallest Number.
b.Find the largest Number.
c.Find the Average.
d.Find the mode.
e.Find the median.
f.Sort the array

Let’s shoot one at a time.
a.b.f.Sort + Largest Number + Smallest number.
It is better to sort the array first so then you automatically know the smallest number(first number : arr[0]) and last number to be the largest number(arr[arr.length-1]).

I use the simple and easy kind of sort : BUBBLE SORT. This sorting scheme is easy but performance suffer from it. Basically, starting from the beginning element, you compare it to the second element, swap the two if it is smaller, then compare first element to the third, the fourth,……nth. Then we do it again starting from second element until you start off with the nth-1 element. In general the number of comparison is $(n-1)^2$, in which n is the number of element in the array, which is bad in performance.

You would think that computer is so fast that we can still do it provided it is easy. In reality, it is not like that. Say for example, you would not want your computer to wasting your time in business world. You have to be fast. Say for example, if 1 comparison takes 1 millisecond then the whole BUBBLE SORT would takes $((n-1)^2)*1$ milliseconds to sort. For example if n = 10,000 , it would take $(10,000 - 1)^2 = 99980001$ milliseconds or 27.7722225 hour or roughly a day and 3 hour to sort just 10,000 number. Can you take this seriously now? Thus, for this assignment, bubble sort is used just for showing, not for doing seriously. Therefore, in a function that generate 1000 random number, I would pass in howMany to generate, just for checking, because if it works for 20 numbers, it would work for 1000 or 10,000. By the way, using this scheme, sorting 1000 numbers would just needs 30 minutes – not bad.

/**
* Sort the array using BUBBLE SORT.
* @param arr - array of number.
*/
public static void sort(int[]arr){

//implement simple and easy 'BUBBLE SORT'
//reimplement using C code MATH201
int tmp = 0;
for (int i = 0; i < arr.length; ++i){
for (int j = i + 1; j < arr.length;++j){
//comparing 2 consecutive number
if (arr[i] >= arr[j]){     //change this to '<=' for descending order
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
}

Oh and I would let the user choose their seed value also, because they would not want to wait when I prints off 1000 numbers on the screen and check for everyone of them. Anyhow, by entering this seed value, it helps me do the mode later on.

System.out.print("Please enter your seed number : ");
Scanner sc = new Scanner(System.in);
userSeed = sc.nextInt(); //pass in howManyRandomNumber to generate.
//pass in seed value --> helpful at later time.
generateRandom(howManyRandomNumber, userSeed);

After that, in the main program I just need to write out simple lines :

smallestNumber = Arr[0];
biggestNumber = Arr[Arr.length - 1];

System.out.println("The smallest number is : " + smallestNumber);
System.out.println("The biggest number is : " + biggestNumber);

c.e.AVG + Median
Finding average is easy : $Average = \frac{array[o] + array[1] + ... + array[length-1]}{array.length}$.
Thus all you have to do is loop through the array to find the total number value and divide by array.length.

/**
* Find the average from the records of number
* @param arr - the number array
* @return    - avg the average number
*/
public static int findAvg(int[]arr){
int tmpAvg = 0;

for (int i = 0; i < arr.length; ++i){
tmpAvg += arr[i];
}

tmpAvg /= arr.length;

return tmpAvg;
}

Median is the middle number that divides the array number into 2 halves: the left half and the right half. If the array length is odds, the median is just array[array.length/2]. If the array length is even, the median would be (array[array.length/2] + array[(array.length/2)-1])/2 – the average of the last number of first half and first number of second half.

So, here is code :

/**
* Find the median of the numbers
* @param arr -array of int numbers
* @return  -median numbers
*/

public static int findMedian(int[]arr){

//even case
if ((arr.length % 2) == 0)
return (arr[arr.length/2] + arr[(arr.length/2)-1])/2;

//odd case
else return arr[arr.length/2];

}

d.Find the mode
That is the numbers that appear the most in the sequence, having the greatest frequency. This is when seedNumber helps me.
To count the mode, we would have to count every number and get the number that appear the most. Thus, when generating random numbers, user would enter a seed number so that no matter what how big the array size is, the biggest number cannot get pass seed-1(because it will generate the number in the range 0 – seed – 1). Therefore, I create an array name frequency, whose size is seed to count the mode.
Having done that, all you have to is when you get the number from the input file, increment the value of frequency at that number index or frequency[array[i]]++; so you have a records of all number(which is frequency index)’s mode.

/**
* Get all the number from text to array
* @param arr  -the number array
* @param freg  -the frequency array to count the mode
*/
public static void TextToArray(int[]arr, int []freg){

try{
BufferedReader br = new BufferedReader(
new FileReader("E:\\PortableApps\\eclipse\\workspace\\Hw1\\src\\array.txt"));

String line = br.readLine();
int i = 0;

//reading through the list
while(line != null){
//have to parse the Integer using object
arr[i] = Integer.parseInt(line);

//frequency.. ehhhh..
freg[arr[i]]++;
i++;
line = br.readLine();
}
br.close();
}catch(Exception e){}
}

After that, just loop through frequency array to find the mode, which you have to loop 2 times, the first to find the largest mode, the second time to find numbers that have the same mode:

/**
* Find the mode of the numbers
* @param freg - the frequency array that stores the records of 'howMany'
*       numbers and theirs occurence.
*/
public static void findMode(int[]freg){
int appear_the_most = freg[0];
int noMode = 0;

//finding the number that appear the mode
for (int i = 0; i < freg.length-1; ++i){
if (freg[i] > appear_the_most)
appear_the_most = freg[i];
}

//rescanning for the same-mode number
System.out.print("The mode numbers are[null means none] :");

for (int i = 0; i < freg.length; ++i){
if ((freg[i] == appear_the_most) && (freg[i] > 1))
{
System.out.print(i + " ");
}
else noMode++;
}

if (noMode == freg.length)
System.out.print("null");
System.out.print("\n");
}

Well… Every task done … This is the full code :

/**
* Name   : Duc Ho.
* Class  : MATH410.
* Prof.  : Lonnie Bowe.
* Obj.   :   +Generate 1000 Random Number and save to text file name "array.txt"
*            +Sort the array
*            +Find the Smallest Number
*            +Find the Largest Number
*            +Find the Average
*            +Find the mode
*            +Find the median
*             ?NOTE : User has to enter the seed value to generate in the range (0,seed_number)
* IDE    : Eclipse
* Compiler : Java SE 1.7.0.
* Date  : 02/05/2012.
*/
import java.util.*;
import java.io.*;

public class DucHoA1P2 {

/**
* Generate 'howMany' random number with seed
* @param howMany - howMany random number to generate
* @param seed  - seed value to generate from 0 - seed - 1
*/
public static void generateRandom(int howMany, int seed){

try{
//create a file to save number
//whoever use this file should rewrite the directory to the file
BufferedWriter bw = new BufferedWriter(new FileWriter("E:\\PortableApps\\eclipse\\workspace\\Hw1\\src\\array.txt"));
Random generator = new Random();  //new random object
int randVal;       //take the random value

if (seed > 1000) seed = 1000;
//generate the random Numbers

for (int i = 0; i < howMany; ++i){

randVal = generator.nextInt(seed);  //random val from 0...seed

while (randVal < 0)
randVal = generator.nextInt(seed);

//write out the String representation of Integer
bw.write(String.valueOf(randVal));
bw.newLine();
}

//close the file
bw.close();
}catch(Exception e){}

}

/**
* Get all the number from text to array
* @param arr  -the number array
* @param freg  -the frequency array to count the mode
*/
public static void TextToArray(int[]arr, int []freg){

try{
BufferedReader br = new BufferedReader(
new FileReader("E:\\PortableApps\\eclipse\\workspace\\Hw1\\src\\array.txt"));
String line = br.readLine();
int i = 0;

//reading through the list
while(line != null){
//have to parse the Integer using object
arr[i] = Integer.parseInt(line);
//frequency.. ehhhh..
freg[arr[i]]++;
i++;
line = br.readLine();
}

br.close();
}catch(Exception e){}
}

/**
* Sort the array using BUBBLE SORT.
* @param arr - array of number.
*/
public static void sort(int[]arr){

//implement simple and easy 'BUBBLE SORT'
//reimplement using C code MATH201
int tmp = 0;

for (int i = 0; i < arr.length; ++i){
for (int j = i + 1; j < arr.length;++j){
//comparing 2 consecutive number
if (arr[i] >= arr[j]){
tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
}
}

}

/**
* Find the average from the records of number
* @param arr - the number array
* @return    - avg the average number
*/
public static int findAvg(int[]arr){

int tmpAvg = 0;
for (int i = 0; i < arr.length; ++i){
tmpAvg += arr[i];
}

tmpAvg /= arr.length;

return tmpAvg;
}

/**
* Find the median of the numbers
* @param arr -array of int numbers
* @return  -median numbers
*/
public static int findMedian(int[]arr){

//even case
if ((arr.length % 2) == 0)
return (arr[arr.length/2] + arr[(arr.length/2)-1])/2;

//odd case
else return arr[arr.length/2];
}

/**
* Find the mode of the numbers
* @param freg - the frequency array that stores the records of 'howMany'
*       numbers and theirs occurence.
*/
public static void findMode(int[]freg){
int appear_the_most = freg[0];
int noMode = 0;

//finding the number that appear the mode
for (int i = 0; i < freg.length-1; ++i){ if (freg[i] > appear_the_most)
appear_the_most = freg[i];
}

//rescanning for the same-mode number
System.out.print("The mode numbers are[null means none] :");
for (int i = 0; i < freg.length; ++i){
if ((freg[i] == appear_the_most) && (freg[i] > 1)){
System.out.print(i + " ");
}
else noMode++;
}

if (noMode == freg.length)
System.out.print("null");
System.out.print("\n");
}

/**
* Main methods
* Ask user to input seed value
* Allocate the 2 array : numbers and frequency
* Get the value from text and count the frequency
* Sort the Array
* Output the smallest, largest, median, mode numbers.
* @param args - the string array of arguments.
*/
public static void main(String[] args){

//generating random number and save to "array.txt"
//change it to howmany you want
final int howManyRandomNumber = 10;
double biggestNumber = 0;
double smallestNumber = 0;
int avg = 0;          //huuuhhhh AVG
int median = 0;
int userSeed;
System.out.print("Please enter your seed number : ");

Scanner sc = new Scanner(System.in);
userSeed = sc.nextInt();

//allocate the array of 'howManyRandomNumber' size
int[] Arr = new int[howManyRandomNumber];

//frequency array to .... count the frequency
int[] freg = new int[userSeed];

//generate randomNumber function
generateRandom(howManyRandomNumber, userSeed);

TextToArray(Arr,freg);

sort(Arr);

System.out.println("After sort the array becomes:");
for (int i = 0; i < Arr.length; ++i){
System.out.println(Arr[i]);
}

smallestNumber = Arr[0];
biggestNumber = Arr[Arr.length - 1];

System.out.println("The smallest number is : " + smallestNumber);
System.out.println("The biggest number is : " + biggestNumber);

avg = findAvg(Arr);

System.out.println("After scanning AVG, it finds itself: " + avg);

median = findMedian(Arr);

System.out.println("The median is : " + median);

findMode(freg);
}
}

Phew…. Sample run :

Please enter your seed number : 20
After sort the array becomes:
2
2
9
9
10
14
16
17
18
19
The smallest number is : 2.0
The biggest number is : 19.0
After scanning AVG, it finds itself: 11
The median is : 12
The mode numbers are[null means none] :2 9

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text:
<4004>

## An Accompt of a New Catadioptrical Telescope invented by Mr. Newton, Fellow of the R. Society, and Professor of the Mathematiques in the University of Cambridge.

THis Excellent Mathematician having given us, in the Transactions of February last, an account of the cause, which induced him to think upon Reflecting Telescopes, instead of Refracting ones, hath thereupon presented the Curious World with an Essay of what may be performed by such Telescopes; by which it is found, that Telescopical Tubes may be considerably shortned without prejudice to their magnifying effect.

This new instrument is composed of two Metallin Speculum's, the one Concave, (instead of an Object-glass) the other Plain; and also of a small plano-convex Eye-Glass.

By Figure I. of Tab. I. the structure of it may be easily imagined; viz. That the Tube of this Telescope is open at the end which respects the object; that the other end is close, where the said Concave is laid, and that near the open end there is a flat oval speculum, made as small as may be, the less to obstruct the entrance of the rays of Light, and inclined towards the upper part of the Tube, where is a little hole furnish't with the said Eye-glass. So that the rays coming from the object, do first fall on the Concave placed at the bottome of the Tube; and are thence reflected toward the other end of it, where they meet with the flat speculum, obliquity posited, by the reflection of which they are directed to the little plano-convex Glass, and so to the spectators Eye, who looking downwards sees the Object, which the Telescope is turned to.

To understand this more distinctly and fully, the Reader may please to look upon the said Figure, in which

A B is the Concave speculum, of which the radius or semidiameter is 12$\frac{2}{3}$ or 13 inches.

C D another metalline speculum, whose surface is flat, and the circumference oval.

<4005>

G D an Iron wire, holding a ring of brass, in which the speculum C D is fixed.

F, a small Eye-glass flat above, and convex below, of the twelfth part of an inch radius, if not less; forasmuch as the metal collects the Sun's rays at 6$\frac{2}{3}$ inches distance, and the Eye-glass at less than $\frac{1}{6}$ of an inch distance from its vertex: Besides that the Author (as he informs us) knew their dimensions by the tools to which they were ground, and particularly measuring the diameter of the hemi-spherical Concave, in which the Eye-glass was wrought, found it the sixth part of an inch.

G G G, the fore part of the Tube fastn'd to a brass-ring H I, to keep it immoveable.

P Q K L, the hind-part of the Tube, fastn'd to another brass-ring P Q.

O, an Iron hook fastn'd to the Ring P Q, and furnish't with a screw N, thereby to advance or draw back the hind-part of the Tube, and so by that means to put the specula in their due distance.

M Q G I a crooked Iron sustaining the Tube, and fastned by the nail R to the Ball and Socket S, whereby the Tube may be turned every way.

The Center of the flat speculum C D, must be placed in the same point as the Tube's Axe, where falls the perpendicular to this Axe, drawn to the same from the center of the little Eye-glass: which point is here marked at T.

And to give the Reader some satisfaction to understand, in what degree it represents things distinct, and free from colours, and to know the aperture by which it admits light; he may compare the distances of the focus E from the vertex's of the little Eye-glass and the Concave speculum, that is, E F, $\frac{1}{6}$ of an inch, and E T V, 6$\frac{1}{3}$ inches; and the ratio will be found as 1 to 38; whereby it appears, that the Objects will be magnified about 38 times. To which proportion is very consentaneous, an Observation of the Crown on the weather-cock, about 300 feet distant. For the scheme X fig. 2. represents it bigger by 2$\frac{1}{2}$ times in diameter, when seen <4006> through this, than through an ordinary Telescope of about 2 foot long. And so supposing this ordinary one to magnifie 13 or 14 times, as by the description it should, this new one by the Experiment must magnifie near as much as hath been assigned.

Thus far as to the structure of this Telescope. Concerning the Metalline matter, fit for these reflecting Speculums, the Inventor hath also considered the same, as may be seen by two of his Letters, written to the Publisher from Cambridge Jan. 18. and 29. 16$\frac{71}{72}$. to this effect, viz.

1. That for a fit metalline substance, he would give this Caution, that whilest men seek for a white, hard and durable metallin composition, they resolve not upon such an one, as is full of small pores, only discoverable by a Microscope. For though such an one may to appearance take a good polish, yet the edges of those small pores will wear away faster in the polishing than other parts of the metal; and so, however the Metal seem polite, yet it shall not reflect with such an accurate regularity as it ought to do. Thus Tin-glass mixt with ordinary Bell-metall makes it more white and apt to reflect a greater quantity of light; but withall its fumes, raised in the fusion, like so many aerial bubles, fill the metall full of those Microscopical pores. But white Arsenick both blanches the Metall and leaves it solid without any such pores, especially if the fusion hath not been too violent. What the Stellate Regulus of Mars (which I have sometimes used) or other such like substance will do, deserves particular examination.

To this he adds this further intimation, that Putty or other such like powder, with which 'tis polished, by the sharp angles of its particles fretteth the metall, if it be not very fine, and fills it full of such small holes, as he speaketh of. Wherefore care must be taken of that, before judgment be given, whether the metall be throughout the body of it porous or not.

2. He not having tried, as he saith, many proportions of the Arsenick and Metall, does not affirm, which is absolutely best, but thinks, there may conveniently be used <4007> any quantity of Arsenick equalling in weight between a sixt and eight part of the Copper, a greater proportion making the Metal brittle.

The way, which he used, was this. He first melted the Copper alone, then put in the Arsenick, which being melted, he stirred them a little together, bewaring in the mean time, not to draw in breath near the pernicious fumes. After this, he put in Tin, and again so soon as that was melted (which was very suddenly) he stirred them well together, and immediately powred them off.

He saith, he knows not, whether by letting them stand longer on the fire after the Tin was melted, a higher degree of fusion would have made the metall porous; but he thought that way he proceeded to be safest.

He adds, that in that metall, which he sent to London, there was no Arsenick, but a small proportion of Silver; as he remembers, one shilling in three ounces of metall. But he thought withall, that the Silver did as much harm in making the metall soft, and so less fit to be polish't, as good in rendring it white and luminous.

At another time he mixed Arsenick one ounce, Copper six ounces, and Tin two ounces: And this an Acquaintance of his hath, as he intimates, polish't better, than he did the other.

As to the objection, that with this kind of Perspectives, objects are difficultly found, he answers in another letter of his to the Publisher, of Jan. 6. 16$\frac{71}{72}$. that that is the inconvenience of all Tubes that magnifie much; and that after a little use the inconvenience will grow less, seeing that himself could readily enough find any day-Objects, by knowing which way they were posited from the other objects that he accidentally saw in it; but in the night to find Stars, he acknowledges it to be more troublesome; which yet may, in his opinion, be easily remedied by two sights affixed to the Iron rod, by which the Tube is susteined; or by an ordinary perspective glass fastn'd to the same frame with the Tube, and directed towards the same object, as Des-Cartes in his Dioptricks hath described for remedying the same inconvenience of his best Telescopes.

<Tab. I>


text:
OpenStudy (anonymous):

can 9 and 10 be reduced

4 years ago
OpenStudy (anonymous):

what does it mean to reduce $$9$$ ?

4 years ago
OpenStudy (anonymous):

maybe make $\huge 9$ into $9$

4 years ago
OpenStudy (anonymous):

can you post what the question says exactly? it does not make sense as it is written you cannot reduce an integer

4 years ago


text:
# If H is a subgroup of G, then H has no more Sylow subgroups than G

If $H$ is a subgroup of the finite group $G$, then how do I show that $n_p(H) \leq n_p(G)$? Here $n_p(X)$ is the number of Sylow $p$-subgroups in the finite group $X$.

Here is my attempt: Suppose the order of $G$ is $n$, the order of $H$ is $m$, then $m$ divides $n$ by Lagrange's theorem. By Sylow's counting theorem $n_p(H) = 1 \pmod p$ and $n_p(H)$ divides $m$, $n_p(G) =1 \pmod p$, $n_p(G)$ divides $n$.

How do I continue after this?

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It is unclear which facts you are given and what you are trying to prove. Is $p$ a prime? What do you mean by $p(G)$? If you clarify your question you are more likely to get some help solving it. – Mark Bennet Feb 28 '14 at 17:11
yes, p is a prime, and np(g) means the order of sylp(G) – user132177 Feb 28 '14 at 17:17
What does sylp mean? Are you trying to prove that if $H$ is a subgroup of $G$ then the order of $H$ is less than or equal to the order of $G$? – user1729 Feb 28 '14 at 17:20
sylp(G):={sylow p-subgroups of G}, I want to prove if H is a subgroup of G then the order of sylp(H) is less than or equal the order sylp(G) – user132177 Feb 28 '14 at 17:24
Every Sylow $p$-subgroup $Q$ of $H$ is contained in some Sylow $p$-subgroup $P$ of $G$. You need to prove that you cannot have two distinct Sylow subgroups $Q_1$ and $Q_2$ contained in the same Sylow $p$-subgroup $P$ of $G$. To prove that impossible, consider $P \cap H$ would contain both $Q_1$ and $Q_2$. – Derek Holt Feb 28 '14 at 17:29

I like this question, and wanted it to have a bit of a longer answer:

## Surprising

This result should be a little surprising. After all the Sylow $p$-subgroups of $H$ can have smaller order, and even though $G$ may have only a few subgroups of order $p^n$, it might have a ton of order $p^{n-1}$. For example, in $G=A_4$, we have only $n_2(G)=1$ Sylow $2$-subgroup, but it has $3$ subgroups of order $2^1$, and so a subgroup $H$ has $n_2(H) \leq 3$, but that leaves open the possibility of $n_2(H) \in \{2,3\}$ both of which contradict the theorem. There is an analogue of $A_4$ called $G=\operatorname{AGL}(1,p^2)$ with approximately the same behavior: $n_p(G)=1$ but $G$ has $p+1$ subgroups of order $p$. When one actually looks for a subgroup $H$, one runs into a problem: no single $H$ contains all those subgroups of order $p$ (or even two of them in the $A_4$ and AGL cases) unless it contains an entire Sylow $p$-subgroup (making those smaller $p$-subgroups irrelevant).

## Proof

This idea leads to a simple proof (given by Derek Holt in the comments).

We construction a 1-1 function $f$ from $\operatorname{Syl}_p(H)$ to $\operatorname{Syl}_p(G)$ where $\operatorname{Syl}_p(X)$ is the set of Sylow $p$-subgroups of the finite group $X$. Given a Sylow $p$-subgroup $Q$ of $H$, Sylow's containment theorem says that $Q$ (a $p$-subgroup of $G$) is contained in some Sylow $p$-subgroup $f(Q)$ of $G$. If $f(Q_1) = f(Q_2)$, then $Q_1, Q_2 \leq f(Q_1)$ and $Q_1, Q_2 \leq H$, so $Q_1, Q_2 \leq f(Q_1) \cap H$. However, $f(Q_1) \cap H$ is a $p$-subgroup of $H$, and a $p$-subgroup of $H$ can only contain at most a single Sylow $p$-subgroup of $H$, so $Q_1 = Q_2$. Hence $f$ is a 1-1 function, and $n_p(H) \leq n_p(G)$.

## Normal subgroups

If $H$ is a normal subgroup of $G$, then in fact $n_p(H)$ divides $n_p(G)$. Hall (1967) calculated $n_p(G) = n_p(H) \cdot n_p(G/H) \cdot n_p(T)$ where $T=N_{PH}(P \cap H)$ for any Sylow $p$-subgroup $P$ of $G$.

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Hall's result is really nice. – mesel Feb 28 '14 at 19:19
Just to be clear, $T=N_{PH}(P \cap H)$ is perfectly fine. The quotient seems pretty silly to me. – Jack Schmidt Feb 28 '14 at 20:38
Ah, you are right. For some weird reason Hall states his theorem with $N_{PH}(P \cap H) / P \cap H$. – Mikko Korhonen Feb 28 '14 at 20:57
Yeah, I'm not sure why. I've been trying to put restrictions on $T$ based on $G$, but I don't think any of them would care about normal subgroups of $O_p(T)$. – Jack Schmidt Feb 28 '14 at 21:26
Like imagine $G=S_n \wr S_m$, then $G$ doesn't have too many normal subgroups (certainly we know which simple groups have any chance of occurring in a composition series of $G$), but how bad can $T$ be? $G$ itself contains a bazillion simple groups, but only $A_n$, $A_m$, and $C_2$ are allowed as composition factors of $H$ or $G/H$. Can $T$ have others? – Jack Schmidt Feb 28 '14 at 21:28

If $Q \in \operatorname{Syl}_p(H)$, then $Q \leq P$ for some $P \in \operatorname{Syl}_p(G)$. Then $Q \leq P \cap H \leq H$ and $P \cap H$ is a $p$-group, so it follows that $Q = P \cap H$.

In other words, we have the inclusion $\operatorname{Syl}_p(H) \subseteq \{P \cap H: P \in \operatorname{Syl}_p(G) \}$ which gives the result. This inclusion is an equality when $H$ is a normal subgroup, but not in general.

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This is really a consequence of the pigeonhole principle: Assume that $H$ has more Sylow $p$-subgroups than $G$. Since every Sylow $p$-subgroup of $H$ is contained in a Sylow $p$-subgroup of $G$, there exists a Sylow $p$-subgroup $P$ of $G$ containing two distinct Sylow $p$-subgroups $Q_1$ and $Q_2$ of $H$. Since $Q_1$ and $Q_2$ are distinct, the group generated by them is not a $p$-group; on the other hand, $<Q_1,Q_2>\le P$ which is a $p$-group, a contradiction.

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Why is it true that every Sylow-$p$ subgroup of $H$ is contained in a Sylow-$p$ subgroup of $G$? It's true that such subgroups must be $p$-groups, but I can't see why every $p$-subgroup would need to be contained in a Sylow $p$-subgroup. – Eric Stucky Feb 28 '14 at 19:42
@EricStucky: This is by the definition of Sylow $p$-subgroups: A Sylow $p$-subgroup of $G$ is a maximal (w.r.t. inclusion, not order) $p$-subgroup of $G$. See en.wikipedia.org/wiki/Sylow_theorems. – jpvee Feb 28 '14 at 19:47
Thank you. That was never how they were defined in my classes. But it's great that the second theorem still works in this context! – Eric Stucky Feb 28 '14 at 19:55


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# Maximal ideal space of $C^*$-algebra of Riemann integrable functions

Let $R([0,1])$ be the unital commutative $C^*$-algebra of complex valued Riemann integrable functions on $[0,1]$ with pointwise operations and the supremum norm.

In the 1980 paper The Gelfand space of the Banach algebra of Riemann integrable functions by Jörg Blatter (MR0602719) the author states, that no satisfactory representation is known for the space of maximal ideals of $R([0,1])$.

For the smaller space of regulated functions on $[0,1]$ the maximal ideal space consists of the functionals $\beta_x,\delta_x, \gamma_x$ given by $\beta_x(f)=f(x)$, $\delta_x(f)=\lim_{y\to x^+}f(y)$ and $\gamma_x(f)=\lim_{y\to x^-}f(y)$ where $x\in [0,1]$, see e.g. The character space of the algebra of regulated functions by S. K. Berberian (MR0487932).

I wonder whether some progress has been made concerning the maximal ideal space of $R([0,1])$ since Blatter's paper. In fact I would be very thankful if someone could provide me with an example of a character on $R([0,1])$ which is not of the form $\beta_x$, $\delta_x$, $\gamma_x$ as above or some intuition why the determination of the maximal ideal space of $R([0,1])$ is so much harder than the corresponding problem for the space of regulated functions.

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lvb, just a remark: your last sentence seems to imply that the characters $\beta_x$, $\delta_x$, $\gamma_x$ extend to $R([0,1])$ which I don't think they do. By the way, what's the topology on $[0,1]\sqcup [0,1]\sqcup [0,1]$ regarded as the promitive ideal space of the regulated functions? – Rasmus Jun 26 '11 at 18:51
In fact every maximal ideal of R is an extension of one of $\beta_x$, $\delta_x$, $\gamma_x$. It's just that they do not extend in a unique way. You can construct the maximal ideals via ultrafilters. – George Lowther Jun 26 '11 at 22:36
Thank you. You are right. I should have seen that. Do you know a reference where this construction is carried out? – lvb Jun 27 '11 at 7:50
lvb, if you want to make sure that George Lowther notices your comment, use @George Lowther. – Rasmus Jun 27 '11 at 20:35
May be his can help you. Consider algebra of sets $\Sigma=\{E\subset[0,1]: \lambda(\operatorname{int}(E))=\lambda(\operatorname{cl}(E))\}$. Then $L_\infty([0,1],\Sigma,\lambda|_\Sigma)$ is isomorphic to the space Riemann integrable functions with $\sup$-norm. (see exercise 363Yi in Measure theory D. H. Fremlin Vol 3) – Norbert Dec 9 '13 at 16:48


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Chapter 9 R Console

When iNZight starts, two windows open: one is the main window in which we work, and a second window is an R Console. R code can be directly typed into this window at the prompt >. The datasets that are imported into iNZight are available at the R console. However note that the R installation that comes with iNZight is different from the one you might have installed yourself with Rstudio, so R packages you may have added will not be available.

This manual does not cover the writing of R code, however some simple calculations may be conveniently done at the R Console.

For example, if we have a confidence interval from iNZight, we can compute the margin of error by simply halving the length of the interval. If the confidence interval has been computed by iNZight as, say,

   Lower    Mean   Upper
529.7   584.6   639.5

(this is the confidence interval for the mean school size in the apisrs dataset), then we can compute the margin of error in R by typing

> (639.5-529.7)/2

at the prompt: i.e. MOE=(Upper-Lower)/2. R responds with the answer:

[1] 54.9

Then the SE = MOE/1.96, and RSE=SE/Estimate are given by

> 54.9/1.96
[1] 28.0102

and

> 28.0102/584.6
[1] 0.04791345

i.e. an RSE of 4.8%.


text:
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# xy>0

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xy>0 [#permalink]  27 Nov 2019, 11:17
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$$xy>0$$

 Quantity A Quantity B $$x^4y^3$$ $$0$$

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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Re: xy>0 [#permalink]  27 Nov 2019, 12:11
As xy is greater than 0, both x and y have the same sign,either -ve or +ve. If both are +ve, A is bigger. If both are -ve, A is -ve and B is bigger. Therefore the answer is D.
Re: xy>0   [#permalink] 27 Nov 2019, 12:11
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# 8: Solving Linear Equations

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Teetering high above the floor, this amazing mobile remains aloft thanks to its carefully balanced mass. Any shift in either direction could cause the mobile to become lopsided, or even crash downward. In this chapter, we will solve equations by keeping quantities on both sides of an equal sign in perfect balance.

Figure 8.1 - A Calder mobile is balanced and has several elements on each side. (credit: paurian, Flickr)


text:
The acute angles of the triangles are $\alpha,$ $90^{\circ} -\alpha,$ $\beta,$ $90^{\circ} - \beta.$ The vertices of the rhombus split the sides of the rectangle into segments of lengths $\cos\alpha,$ $\sin\alpha,$ $\cos\beta,$ $\sin\beta,$ … Click hereto get an answer to your question  Show that the points ( - 3,2),( - 5, - 5),(2, - 3) and (4,4) are the vertices of a rhombus. Rhombus Formulas. Area of Rhombus. Perimeter of Rhombus Using Diagonals Calculator. How to determine the perimeter of a quadrilateral using distance . An example of a quadrilateral that cannot be cyclic is a non-square rhombus. CISCE ICSE Class 9. 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The formula for the area is equal to the product of diagonals of rhombus divided by 2. A rhombus normally has no right angles (at the vertices). Solution: Given, a = 5 cm. Show that (4,-1) , (6,0) , (7,2) and (5,1) are the vertices of a rhombus . Since all the sides of the rhombus are congruent, and the opposite angles are parallel to each other, the area of the rhombus is given as: Area of a rhombus, A = (½) pq square units. Since the side of the rhombus is $\sqrt{41}$, the area of the rhombus is $41\sin\theta$, where $\theta$ is either of the (supplementary) internal angles of the rhombus. Area of Parallelogram Given Coordinates Calculator. So remember, a rhombus is just a parallelogram where all four sides are equal. Important Solutions 3. The congruence of opposite sides and opposite angles is a direct consequence of the Euclidean parallel postulate and neither … Rhombus is a flat shape which has 4 equal straight sides, where opposite sides are parallel and opposite angles are equal. What is a formula of rhombus? Where “p” and “q” are the two diagonals of the rhombus. If we were told that the rhombus is not a square, B would be the correct answer. So the area is somewhere between $0$ (if the rhombus is totally "flat") and $41$ (if the rhombus is a square). A rhombus is a parallelogram with all of its sides being equal. (ii) Length of diagonal must be equal in any square.So we need to prove this the length of diagonals are not same.If the given vertices are satisfying these conditions then we can say the given vertices forms a rhombus. Three of the four vertices for rhombus ABCD are A(–3, –3), B(1, –6), and C(5, –3). Opposite angles are congruent in a rhombus. It doesn’t matter which variables you give them. Properties. Formula of Perimeter of Rhombus. The perimeter of a rhombus is given as: $\large P=4\times a$ Here, a = Length of the side of the rhombus. Quadrilateral is a 2 dimensional closed figure with four sides. Perimeter is a path which surrounds a two dimensional shape. The rhombic dodecahedron is a zonohedron.Its polyhedral dual is the cuboctahedron.The long diagonal of each face is exactly √ 2 times the length of the short diagonal, so that the acute angles on each face measure arccos(1 / 3), or approximately 70.53°.. Textbook Solutions 19009. Every rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides. In Euclidean geometry, a parallelogram is a simple (non-self-intersecting) quadrilateral with two pairs of parallel sides. Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. Area of Rhombus Formula. The diagonals of a rhombus meet at right angles. show that the points (5,-1,1),(7,-4,7),(1,-6,10)and(-1,-3,4) are the vertices of a rhombus by vectors? Sides of Rhombus Formula. Area Ar of a parallelogram may be calculated using different formulas. A square is a rhombus with all of the angles being equal as well as all of the sides. And just to make things clear, some rhombuses are squares, but not all of them. Learn its definition, properties, Formulas, Shapes explained with real life illustrated examples. Being the dual of an Archimedean polyhedron, the rhombic dodecahedron is face-transitive, … 2. How to test whether the given points are vertices of rhombus ? The word cyclic is from the Ancient Greek κύκλος (kuklos) which means "circle" or "wheel". h = height of rhombus. Graph the rhombus with vertices (-2, 9), (3, 9), (7, 6), and (2, 6). Question Papers 10. Again, the length of the side also found if area and a diagonal of a rhombus are given. Since, by definition, all four sides of a rhombus are the same length, the formula is = ... Label the vertices of your rhombus, if they are not already labeled. The rhombus inscribed into a rectangle has side length of $1.$ The rhombus cuts off of the rectangle two pairs of equal right triangles. The length of the side of rhombus can be determined if the diagonals are known. Advertisement Remove all ads. The perimeter of a circle is called its circumference. The formulas for rhombus are defined for two major attributes, such as: Area; Perimeter; Get more: Mathematics formulas. Directions: Given the vertices, determine the quadrilaterois most specific classification: Parallelogram. In a rhombus, the diagonals bisect and are perpendicular, meaning that inside the rhombus there are four right-angle triangles, with side lengths half that of the diagonals, use the pythagoran theorem to find the hypotenuses a² + b² = c² Now we have the base, given any of the angles inside the rhombus, we can find all of them because they are corresponding and all add … Question: Find the perimeter of a rhombus having a side of 5 cm. From the formula of area of a rhombus, we can easily find the diagonal of the rhombus. Ex 7.2 , 10 Find the area of a rhombus if its vertices are (3, 0), (4, 5), ( 1, 4) and ( 2, 1) taken in order. The vertices of the parallelogram: $$\displaystyle A(-1, 2), B(4, 4), C(2, -1), \text{ and }D(-3, -3). 2 / 4. Set up the formula for perimeter of a rhombus. The diagonals of a rhombus bisect each other at 90 degrees. Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) Are the Vertices of a Rhombus. Side of a Rhombus when Diagonals are given calculator uses Side A=sqrt((Diagonal 1)^2+(Diagonal 2)^2)/2 to calculate the Side A, Side of a Rhombus when Diagonals are given can be defined as the line segment that joins two vertices in a rhombus provided the value for both the diagonals are given. The "base times height" method First pick one side to be the base. [Hint : Area of a rhombus = 1/2 (product of its diagonals)] Let the vertices be A(3, 0) , B(4, 5) C( 1, 4) , D( 2, 1) We know that Area of Rhombus = 1/2 (Product of diagonals) = 1/2 AC BD We need to find AC & BD using distance formula Finding AC AC = (( 2 … Choose a formula based on the values you know to begin with. Trapezium, Parallelogram, Squares, Rectangle, Rhombus, Kite and more with Cuemath Transcript. a = measure of any interior angle. Let's go how to derive a formula for finding the length of the side depending on the diagonals. The perimeter of a rhombus = 4 × a Sketch the three vertices on graph paper. Easy to use online calculators to calculate the area Ap, sides, diagonals, height and angles of a parallelogram. Solved Example. It can be represented as: Using distance formula, you can find the length of the sides of parallelogram. Find the area of this rhombus. … d2 = length of diagonal 2. b = length of any side. How to use the distance formula and Pythagorean theorem to determine if three ordered pairs are the vertices of a right triangle Rhombus has two diagonals the longer d 1 , and the smaller d 2 Diagonals of a rhombus formulas: It contains formulas, calculator techniques, ... Side - A side is a line segment which joins any two adjacent vertices. Sum of interior angles will be 360 degree Difference between square and rhombus: Area of rhombus = (1/2) x (d₁ x d₂) Here d₁ and d₂ are diagonals. Many of the area calculations can be applied to them also. If a rhombus has right angles (at the vertices), it is called a square. Given the length of the rhombus' diagonals, use the formula in solving for the area. Diagonal of rhombus is an any line segment that is bounded by two distinct angles of rhombus. Distance formula: √[(x 2 - x 1 )² + (y 2 - y 1 )²] AB = √[(1 + 2)² + (0 - 1)²]$$ We are using the distance formula to find out the measure of sides of the parallelogram. Rectangle, Rhombus, or Square, Justify your answer using the distance formula 1. How to find the Area and Perimeter of a Rhombus – How do you find the perimeter of a rhombus with vertices? The area of the rhombus is the region covered by it in a two-dimensional plane. APPEARS IN. 1. In fact, if all four sides are equal, it has to be a parallelogram. Because you could have a rhombus like this that comes in where the angles aren't 90 degrees. Distance Formula - Exercise 28 [Page 335] Q 15 Q 14 Q 16. All the steps, described above can be performed with … As a consequence of this definition, the point where two lines meet to form an angle and the corners of polygons and polyhedra are vertices. A rhombus is actually just a special type of parallelogram. Show That The Points Are The Vertices Of An Isosceles Triangle The formulas and properties given below are valid in the convex case. Both squares and rhombuses have perpendicular diagonal bisectors that split each diagonal into 2 equal pieces, and also splitting the quadrilateral into 4 equal right triangles. Properties of rhombus: It has four equal sides ; Both diagonals are intersecting each other at right angle. Question Bank Solutions 15191. Find the fourth vertex. Vertex Formula In geometry, a vertex is a point where two or more curves, lines, or edges meet. . It follows that any rhombus has the following properties: Opposite angles of a rhombus have equal measure. It is used for measuring the path or its length. Wheel '' bounded by two distinct angles of a parallelogram, sides diagonals. P ” and “ Q ” are the two diagonals of a parallelogram parallelogram! Can find the area calculations can be determined if the diagonals, the! 2. b = length of the sides sides ; Both diagonals are known given points vertices. 14 Q 16 ’ t matter which variables you give them or square, Justify your answer using distance., calculator techniques,... side - a side is a flat shape which has 4 equal sides! Not vertices of rhombus formula square of any side rhombus like this that comes in the... You know to begin with rhombus are defined for two major attributes, as... The word cyclic is from the Ancient Greek κύκλος ( kuklos ) which means  circle or! A path which surrounds a two dimensional shape of rhombus: it four. You know to begin with 4 equal straight sides, where opposite are! Applied to them also d₂ are diagonals two major attributes, such as: area ; ;! 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Of diagonal 2. b = length of the sides of the sides Q 16 called its circumference point... N'T 90 degrees to be a parallelogram are of equal length and opposite. Well as all of the sides called its circumference if we were told that the is!, height and angles of a rhombus is actually just a special type of parallelogram told that the rhombus the... Use online calculators to calculate the area Ap, sides, where opposite sides are parallel and opposite of! For finding the length of the side also found if area and a of..., calculator techniques,... side - a side of rhombus: it to! Equal straight sides, where opposite sides are equal a a rhombus having a side a! Formula in solving for the area Ap, sides, where opposite sides are parallel and angles! Kuklos ) which means  circle '' or  wheel '' equal sides ; Both diagonals are known equal! All four sides are equal definition, properties, formulas, calculator techniques,... side - side... 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I ) First we have to find the length of the side of?... ( kuklos ) which means  circle '' or  wheel '' Shapes! As well as all of the parallelogram it doesn ’ t matter which variables you give them calculator techniques...... Segment that is bounded by two distinct angles of a rhombus is symmetric across of., it has four equal sides ; Both diagonals are intersecting each other at right angles ( at the )! Two dimensional shape the perimeter of a rhombus – how do you find the of! Calculators to calculate the area and perimeter of a rhombus with all of the angles being equal as as.  we are using the distance formula 1 do you find the perimeter of a circle is called circumference! Fact, if all four sides are equal side depending on the values you know begin! If we were told that the rhombus is just a parallelogram are of length! These diagonals 4 × a a rhombus with all of the sides of parallelogram for the area Ap,,... 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The opposite or facing sides of parallelogram other at right angles ( at vertices. Its length other at 90 degrees a formula based on the values know! Variables you give them the opposite angles of a rhombus is not a square do you find the length the. The side depending on the values you know to begin with a formula for finding the of... 4 × a a rhombus with vertices diagonals of a rhombus is the region covered it.
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# Iron Oxide: Construction, Properties, Strategies And Software

The IOMNPs alone in both low or high doses did not induce CNS pathology either following 4 or 24h after administration. However, administration of IOMNPs in SCI group slightly enhanced the pathological modifications in the CNS after 24h but not 4h after trauma. Cerebrolysin treatment markedly attenuated IOMNP-induced aggravation of SCI-induced wire pathology and induced important neuroprotection.

3 Fe2O3 + H2 → 2 Fe3O4 + H2OIron oxide is insoluble in water however dissolves readily in strong acid, e.g. hydrochloric and sulfuric acids. It also dissolves well in solutions of chelating brokers such as EDTA and oxalic acid. Iron reacts with water within the form of steam to type iron oxide, along with the release of hydrogen.

For instance, iron oxide indicates that iron is in the +2 or ferrous oxidation state, and iron oxide reveals that iron is within the +3 or ferric oxidation state. Ferric oxide ought to be written as iron oxide to avoid ambiguity when writing the name iron oxide. Iron oxides are chemical compounds composed of iron and oxygen. Oxyhydroxides are a related class of compounds, maybe the best identified of which is rust. Several of the important floor reactions, investigated using synchrotron-based methods, are described in numerous chapters of this book.

Iron oxide with suitable surface chemistry is prepared by several techniques, corresponding to moist chemical, dry methods, or microbiological techniques. The iron oxide will respond with the sulphuric corrosive to form water and iron sulphate. Iron oxide reacts with sulfuric acid to type iron sulfate and water.

During rusting of iron, the iron metallic reacts with oxygen within the air within the presence of water to kind hydrated iron $$\left( \right)$$ oxide, $$_2_3_2$$. Iron oxide is an ambiguous name for the collection of ionic compounds that contain iron and oxygen ions. Iron compounds ought to be named with Roman numerals following the iron to avoid this ambiguity.

It is It is a rock-forming mineral which is found in metamorphic, igneous, and sedimentary rocks. Acids readily assault it and subsequently it’s typically referred to as rust. The composition of pure hematite by weight is 70% iron and 30% oxygen. Metals and non-metals may be in the highest oxidation state of oxygen compounds . In addition to alkali steel and alkaline earth metal, transition metals and postal metals form an ion oxide containing O2 anions.

Fe3 could additionally be current as a half of the dissolved iron in pure water at alkaline pH, and Fe 2 may exist at pH 10 and above. The complete solubility of iron at pH ranges from 4 to 9 is proven graphically by seven curves for Eh values from zero.10 to +0.50. O has a charge of -2 and since you’re told Fe has a cost of +4 based mostly of the roman numeral, you want 2 Oxide ions to balance out the charge on Fe.

The oxidizer you choose will determine how fast the iron is oxidized. The stronger your oxidizing agent, the sooner your iron will oxidize. The addition of an oxidizing agent will allow the reaction to proceed a lot red light has a longer wavelength and higher energy than blue light sooner. Times will range primarily based on the ratios of reactants and oxidizer, but you will probably see some change in just a few seconds.

After injection, the particles accumulate within the reticuloendothelial system. Liver tumors have few or no reticuloendothelial cells and so there is a contrast between normal liver and tumor. This article was co-authored by Bess Ruff, MA. Bess Ruff is a Geography PhD student at Florida State University. She acquired her MA in Environmental Science and Management from the University of California, Santa Barbara in 2016.

She has performed survey work for marine spatial planning tasks in the Caribbean and provided research support as a graduate fellow for the Sustainable Fisheries Group. Storing your iron oxide in a dry place is beneficial. Whatever your choice of iron salt, you’ll need to dissolve it in water.

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# The equation of the hyperbola with vertices at $(0,\pm 6)$ and eccentricity $\large\frac{5}{3}$ is _________ and its foci are ___________

$\begin{array}{1 1}\large\frac{y^2}{36}-\frac{x^2}{64}\normalsize =1,(0,\pm 10)\\\large\frac{y^2}{36}+\frac{x^2}{64}\normalsize =-1,(0,\pm 12)\\\large\frac{x^2}{36}-\frac{y^2}{64}\normalsize =-1,(0,\pm 20)\\\large\frac{y^2}{64}-\frac{x^2}{36}\normalsize =1,(0,\pm 15)\end{array}$

• General equation of a hyperbola with its vertices on $y$ axis is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1 • a^2=b^2(e^2-1) The given vertices are (0,\pm 6) and eccentricity e=\large\frac{5}{3} b=6 Foci be =\large\frac{5}{3}$$\times 6=10$
Hence the foci of the hyperbola are $(0,\pm 10)$
$a^2=b^2(e^2-1)$
$\Rightarrow a^2=36(\large\frac{25}{9}$$-1) \Rightarrow 36\times \large\frac{16}{9}$$=64$
Hence the equation of the hyperbola is $\large\frac{y^2}{36}-\frac{x^2}{64}$$=1$


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# Logistic Regression of tick data

I've been given some data (it's financial tick data) and I want to predict based on some observed variables whether the next move will be up, down or unchanged.

So I have been trying to use multinomial logistic regression, this is my first time doing logistic regression so I want to check that I have done this correctly and that my results look reasonable.

Right now I am doing 3 bi-variate logistic regressions.

So I code the data such that I have three new time series labelled up, down and unchanged. These are generated by testing whether the move n steps ahead was up, down or unchanged in the original series and adding a one to the appropriate array and making all other entries zero.

I then do a bi-variate logistic regression of these up, down and unchanged arrays against the regressors individually.

I can then calculate the probability of each using the transformation:

$$\textrm{prob} = 1/(1+\exp[-Bx])$$

where $\beta$ are the betas from the bi-variate logistic regressions. And $x$ is the value of the regressors.

This gives me the probability of up, down and unchanged.

I then simply compare if $\textrm{probability of UP} > \textrm{probability of Down}$ if so the model predicts up and vice versa.

Q.1) Is my methodology correct? Right now I am doing all calculations on the price series (not the returns series)?

Q.2) When I test this accuracy in sample I am getting 70% accuracy in sample (for both up and down moves)? Is that a reasonable test score in sample?

Q.3) The model probability for unchanged is very low typically around 14%. So unchanged is never selected (because the probabilities of up down moves are always much larger). However unchanged is the most commonly observed change with an unconditional probability of 91%. Is there a way I can correct the model so that unchanged is forecast accurately by the model.

Update: Here is the code unfortunately I am getting differences between 2 variable regression results and the 3 variable results!

Possible error between two: 2 variable regression run using mnrfit() and an equivalent 3 variable version on asset returns. The returns have been classified as positive, negative or flat. The logistic regression is then run on the rtns vs the classified returns (this is a simple test to check that the regression functions as intended). When I do this for the 2 variable version i.e. Up rtns v everything else, the regression gives an estimate of the probability that a 0 return is not an up return of 88%. As the return size is increased the probability that it is positive increases eventually converging to 1 (as you would expect). Also as increasingly negative rtns are put into the logistic regression model the probability that the rtn is positive goes to zero. The same is true for the 2 Variable estimate of Down returns v everything else. The figures are similar to those above but with the signs of the returns reversed.

Now when I run the 3 variable version. Things initially look OK: when given a return of zero it estimates the probability that it is zero to be 86% with a prob of a down move =6.4% and an up move =7.6% so very similar to the 2 variable case. Moreover when larger and larger returns are entered the probability that it is a positive return converges to 1 as you would expect but when you put in larger and larger negative returns the probability that the return is negative converges to zero while the probability it is equal to zero increases to 1!!! Which is clearly wrong.

Data1 = LoadMat_SHFE_Range(Contract1, StartDate, EndDate);

rtn=(Data1.Mid(2:end)-Data1.Mid(1:end-1))./(Data1.Mid(1:end-1));

NStep=0;

Up=nan(length(rtn),1);
Down=nan(length(rtn),1);
Flat=nan(length(rtn),1);
RtnClass=nan(length(rtn),1);

for i=1:length(rtn)-NStep

if(rtn(i+NStep)>0)
Up(i)=2;
Down(i)=1;
Flat(i)=1;
elseif(rtn(i+NStep)<0)
Up(i)=1;
Down(i)=2;
Flat(i)=1;
elseif(rtn(i+NStep)==0)
Up(i)=1;
Down(i)=1;
Flat(i)=2;
end

end

[BUp,dev,stats] = mnrfit(rtn,Up);
MatProbUp = mnrval(BUp,0.1);

[BDown,dev,stats] = mnrfit(rtn,Down);
MatProbDown = mnrval(BDown,0.1);

for i=1:length(rtn)

if(rtn(i)>0)
RtnClass(i)=3;
elseif(rtn(i)<0)
RtnClass(i)=2;
elseif(rtn(i)==0)
RtnClass(i)=1;
end

end

[BM,dev,stats] = mnrfit(rtn,RtnClass,'model','ordinal');
[pihat,dlow,hi] = mnrval(BM,0,stats,'model','ordinal');

• It's hard for me to understand what your methodology is exactly, can you share some code? Nov 22 '15 at 22:19

## 2 Answers

I agree with Malick in the sense that an ARMA-GARCH is a better model but I would improve the model by doing a ARFIMA-FIGARCH, the FI stands for Fractional Integrated that are used to deal for long memory processes like HF data. When trying to fit the ARMA the normal approach would be to see the ACF and the PACF which will show a lot of dependency (almost all lags would be significant) and the same with the GARCH model, so a common way to deal with is with FI models.

On the other hand, don't discard the logistic regression. Sure you are going to have some problems with the Volatility clusters, but information is useful. I would suggest that instead a 1,0 model try to perform an order Probit or Logit so you can have a probability for the magnitude of the increase, (it is a great complement to the ARFIMA-FIGARH model). In this model the dependent variable is going to be an integer form 1 to 5 lets say and each number represents an interval of increase or decrease in the price. Lets say 1 if the price is going to be down between 0 and 5, 2 if the price is going up between 0 to 5 and so on.

The code you posted is wrong since you do not model the time series behavior of the up/down process (ie if you have 10 up move and consequently 10 down move it is not the same as the opposite ie 10 down and after 10 up..). I would recommend you to use standards Arma Garch models apply on returns instead of modeling the process of up/down. These are (at least) the main reasons:

1. if you use your binary process you'll have problems to take into account volatility clustering and so on (the conditional variance part.).

2. since you use tick data, you 'll have consistency problems since you may get 1 move in 5 seconds and after 100 moves in 5 seconds. If you use standards models such as Arma Garch models, you can deal with this fact by resampling the midquote every x seconds (or any other aggregation methods).

3. With standards models you 'll be able to achieve your aims by choosing a simple rule: let's say if the one step ahead forecasted conditional mean return is positive with 0.05 confidence level you could assume the next move will be up. Additionally you can add explanatory variables in both the conditional mean and variance processes.

4. it is likely that you are also interested in the magnitude of the move , and it is easier to obtain it via box-Jenkins class of models.

5. You also need to consider the move in the bid-ask spread ! if you use returns it is implicitly considered.
• Thanks Malick, I take your point about the time series and that will be the next step but right now I'm just trying to get a feel for the data. I am using the idea shown here: beyondmicrofoundations.blogspot.co.uk/2011/09/…. Are you saying this approach is completely flawed? Nov 23 '15 at 17:53
• @Bazman Yes this approach is completely wrong because roughly speaking it does not consider the "t" (time) subscript. If you mix the data, (ex: move the first tick to the last position) you'll still obtain the same forecast. If you want to consider the time dimension and keep this kind of model then you should have a look to panel logistic model. Nov 23 '15 at 20:51


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# legend font style, f and | are very close

3 views (last 30 days)
Leia on 8 Jul 2021
Commented: Leia on 8 Jul 2021
In the lower legend f and | are very close, but in the upper one they are far. How can i set it?
dpb on 8 Jul 2021
Obviously the difference between the two is that a blank is inserted between the f and the | in the one and not the other. It's not clear from the post what the Q? really is -- do you not see that or do you want some way to change the spacing besides inserting the blank, I suppose?
You can't change default kerning for a given font...it is what it is in the installed font.

DGM on 8 Jul 2021
Edited: DGM on 8 Jul 2021
If picking a font with a better spacing behavior isn't an option, you can always try to pad it using something other than a regular space.
███ ███ hair space -- thinnest nonzero-width space -- char(8202)
███ ███ thin space -- typically 1/5 em -- char(8201)
███ ███ punctuation space -- width of narrow punctuation -- char(8200)
███ ███ figure space -- tabular width of font -- char(8199)
███ ███ 6-per em space -- 1/6 em -- char(8198)
███ ███ 4-per em space -- 1/4 em -- char(8197)
███ ███ 3-per em space -- 1/3 em -- char(8196)
███ ███ em space -- em -- char(8195)
███ ███ en space -- 1/2 em -- char(8194)
███ ███ normal #32 space -- typically 1/4 em -- char(32) (or just spacebar)
One of the thinner spaces might be narrow enough to suit, but if you have a lot of things to plot, it might be quite the hassle.
EDIT
I guess there's also LaTeX
% just one test label
h = legend({'$$\left|f\right|$$'},'interpreter','latex');
Leia on 8 Jul 2021
thank you so much!

Ankit on 8 Jul 2021
As per as my knowledge this is due to "FontName". Try with other FontName for e.g. Calibri etc you will see the difference.


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# Evaluate the following limits. lim_{(x,y)rightarrow(1,-2)}frac{y^2+2xy}{y+2x}

Question
Limits and continuity
Evaluate the following limits. $$\lim_{(x,y)\rightarrow(1,-2)}\frac{y^2+2xy}{y+2x}$$

2021-03-05
The given expression $$\lim_{(x,y)\rightarrow(1,-2)}\frac{y^2+2xy}{y+2x}$$
Calculation
$$\lim_{(x,y)\rightarrow(1,-2)}\frac{y^2+2xy}{y+2x}=\lim_{(x,y)\rightarrow(1,-2)}\frac{y(y+2x)}{y+2x}$$
$$=\lim_{(x,y)\rightarrow(1,-2)}$$
$$=-2$$

### Relevant Questions

Evaluate the following limits.
$$\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}$$
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{2x^2-xy-y^2}$$
Evaluate the following limits. $$\lim_{(x,y)\rightarrow(2,2)}\frac{y^2-4}{xy-2x}$$
Calculate the following limits using the factorization formula
$$x^{n}-a^{n}=1x-a21x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+...+a^{n-2}x+a^{n-1}2$$, where n is a positive integer a is a real number.
$$\lim_{x\rightarrow 2}\frac{x^{5}-32}{x-2}$$
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(2,0)}\frac{1-\cos y}{xy^2}$$
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(0,\pi/2)}\frac{1-\cos xy}{4x^2y^3}$$
$$\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{yz-xy-xz-x^2}{yz+xy+xz-y^2}$$
$$\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^2+xy-xz-yz}{x-z}$$
Evaluate the following limits. $$\lim_{(x,y)\rightarrow(1,-1)}\frac{10uv-2v^2}{u^2+v^2}$$
$$\lim_{(x,y)\rightarrow(1,2)}\frac{\sqrt{y}-\sqrt{x+1}}{y-x-1}$$