id stringlengths 1 6 | url stringlengths 16 1.82k | content stringlengths 37 9.64M |
|---|---|---|
12600 | https://www.gauthmath.com/solution/aIGeCHaGRGf/Given-an-isosceles-triangle-whose-one-angle-is-120-and-radius-of-its-incircle-is | Question
Solution
a) Did not utilize the sentences from question $$3^{a}$$3a. b) Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'. c) Did not utilize the sentences from question $$3^{a}$$3a. d) Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'. e) Did not utilize the sentences from question $$3^{a}$$3a. f) Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'.
C. The core claim of the question is to rewrite the given sentences using the negative form of the verb "to be" in the simple past tense.
Sentence: "Utilize as frases da questão $$3^{a}$$3a"
Negative form in simple past: "Did not utilize the sentences from question $$3^{a}$$3a."
Sentence: "reescrevendo-as para a FORMA NEGATIVA do verbo 'to be' no 'simple past'."
Negative form in simple past: "Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'."
Therefore, the correct answers are:
a) Did not utilize the sentences from question $$3^{a}$$3a.
b) Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'.
c) Did not utilize the sentences from question $$3^{a}$$3a.
d) Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'.
e) Did not utilize the sentences from question $$3^{a}$$3a.
f) Did not rewrite them to the negative form of the verb 'to be' in the 'simple past'.
contact@gauthmath.com |
12601 | https://www.mathbitsnotebook.com/JuniorMath/FractionsDecimals/FDadd.html | Adding Fractions - MathBitsNotebook(Jr)
Adding FractionsMathBitsNotebook.com Topical Outline | Jr Math Outline | MathBits' Teacher Resources Terms of Use Contact Person:Donna Roberts Rules for Adding fractions: 1. The bottom numbers (denominators) must be the same. 2. Add the top numbers (numerators) Put this answer over the denominator. 3. Simplify the fraction, if possible. Remember: "common denominators" ("common denominators" means the "same" denominators) Same Denominators: "Common Denominators"When the fractions have the same denominators, addition and subtraction are easy. When adding (or subtracting) fractions with the same denominators, just add (or subtract) the numerators. Since the denominator are the same, the answer can be found by adding the numerators (tops). This answer is in simplest form. Since there are no numbers that divide exactly into both 5 and 7, we cannot simplify further. (5 and 7 are prime numbers) These fractions have the same denominators, so add the numerators. This answer can be simplified further. Both 6 and 8 can be divided by 2. This is the answer in simplest form. Two Methods for Dealing with Mixed Numbers: Method 1: Separate parts horizontally Remember what you say when you read a mixed number aloud. For example, 4½ is read as "4 and one-half". The "and" means addition. Method 2: Line up vertically Whether you use method 1 or method 2 to solve this problem, the final answer can be reduced Both 6 and 9 can be divided by 3. (simplified) to Different Denominators:When fractions have different denominators, you must find a common denominator. When adding (or subtracting) fractions with different denominators, you must find a common denominator before adding (or subtracting). When you are looking for the least common denominator, you are looking for the smallest number that both denominators will divide into exactly. Remember, you can always multiply the two denominators together to get a common denominator. This number may not be the "least" common denominator, but it will do the job. The least common denominator for 4 and 3 is 12. If an answer is an improper fraction, you should rewrite the answer as a mixed number, unless told otherwise.The least common denominator for 6 and 4 is 12. In this problem, we realized that both 6 and 4 divided exactly into 12, making 12 the least common denominator. NOT using least common denominaor: Let's look at that second problem again. This time we do not recognize 12 as the least common denominator. We will multiply the two denominators together to get a common denominator. The common denominator for 6 and 4 is 24. We still arrived at the correct answer. We just had to simplify one additional time.Dealing with Mixed Numbers: The least common denominator for 8 and 5 is 40. A dding fractions depends upon whether a common denominator is present. The Fundamental Law of Fractions(applied to adding fractions) discusses this need of a common denominator. Fundamental Law of Fractions (applied to adding fractions): (Note: bd will not necessarily be the "least" common denominator) In plain English, this law is telling us that we can find a common denominator by finding the product (the multiplication) of the two existing denominators. The numerators must then be adjusted accordingly. (bd is the product of the two denominators) Adding Fractions on the Number Line: We are going to be looking at the addition of fractions using number lines. When and are plotted on number lines, the subdivisions (sections) on the number lines are different. One of the subdivision is "thirds" and the other is "fourths". Because of this difference in the size of these sections, we have no way of combining (or adding) them together. It would be like adding apples and dogs and wondering what you actually had when you were done. An "appledoodle", maybe???? Smile! There must be a way to "fix" the subdivisions (sections) so that they could be the same size. The trick is to make more equal subdivisions until we have the same equal number on each line. .(Finding the needed number of subdivisions, will create our common denominator.) On the "thirds" line, if we cut each "third" into four equal sections we will get 12 in total. On the "fourths" line, if we cut each "fourth" into three equal sections we will get 12 in total. (This number 12 will be called our common denominator.) Add: Now that both lines contain subdivisions of the same size (both have 12 sections), we can express our fractions using this new subdivision size of 1/12. Then we can add the fractions together (since they are representatives of the same size sections). Now we are adding apples to apples. Let's combine our number lines onto one line to demonstrate this sum. Notice that the, when multiplied top and bottom by 4, creates our new fraction . (4 was the number of subdivisions we created for each "third" section) Notice that the , when multiplied top and bottom by 3, creates our new fraction . (3 was the number of subdivisions we created for each "fourth" section) For help with fractions on your calculator, click here. NOTE:There-posting of materials(in part or whole) from this site to the Internet is copyright violation and is not considered "fair use" for educators. Please read the "Terms of Use". Topical Outline | Jr Math Outline | MathBitsNotebook.com | MathBits' Teacher ResourcesTerms of UseContact Person:Donna Roberts Copyright © 2012-2025 MathBitsNotebook.com. All Rights Reserved. |
12602 | https://physics.stackexchange.com/questions/451770/multiple-triple-points | thermodynamics - Multiple Triple Points - Physics Stack Exchange
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Multiple Triple Points
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I was reading Chandler's Introduction to Modern Statistical Mechanics and noticed a strange feature in one of the figures. The phase diagram in the image has two triple points; however, according to the Gibbs phase rule, a one-component system with three coexisting phases should have zero degrees of freedom.
The last paragraph in the image talks about single-component systems, so I assume that is what is shown in the figure. I have two questions about this:
Is a discrete variable (only two points in (p,T)(p,T) space) considered a true degree of freedom?
If yes to the first question, how are two triple points possible (as in Chandler's Figure 2.4)?
thermodynamics
statistical-mechanics
phase-transition
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edited Jan 2, 2019 at 20:51
Qmechanic♦
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The number of thermodynamic degrees of freedom is the number of independent intensive state variables which can be varied over their domain in the presence of a fixed number of phases.
It is not a triple point or a couple of triple points which represent degrees of freedom. Instead, in a one component system, a triple point corresponds to a situation where one is left with zero degrees of freedom: no intensive variable can be changed (and that's the reason one speaks about triple point). Therefore, there is nothing preventing a complex system to have more than one triple point, as in the example of Chandler's fig. 2.4. In thermodynamics is much more frequent to call this situation as zero-degrees of freedom, more than "discrete degrees of freedom", even though a set of isolated point can always be described as a manyfold of dimension zero.
The reason for the possibility of isolated triple points, while quadruple or higher multiplicity points are impossible, in the case of a one component systems, are all rooted in the proof of the Gibbs' phase rule, i.e. in the counting of the number of variables and the number of equations required to describe phase equilibria.
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answered Jan 2, 2019 at 21:05
GiorgioP-DoomsdayClockIsAt-89GiorgioP-DoomsdayClockIsAt-89
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12603 | https://stackoverflow.com/questions/75409539/algorithm-to-flip-n-elements-of-an-array-to-get-sum-to-zero | Algorithm to flip N elements of an array to get sum to zero - Stack Overflow
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Algorithm to flip N elements of an array to get sum to zero
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Asked 2 years, 7 months ago
Modified2 years, 7 months ago
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Imagine you have an accounting report that consists of multiple hundreds of rows with the following structure:
| AccountID | Value. |
--- |
| 1. | -11775,61 |
| 2. | -11538,40 |
| 3. | 0,05 |
| 4. | 26969,32 |
| ... | ... |
| 500. | 1354,25 |
The total sum in a correct report must be exactly zero, not even a cent off.
The error always comes from some rows having incorrect signs, so we need to find them and flip the sign.
Task
Given a non-zero sum array, the task is to find which rows need to have their signs flipped so that the sum becomes 0. Correct answer is an array of values to be flipped.
Specifics:
Sum must be exactly zero. This is not a minimization algorithm. Missing by 0.01 and by 10000000.00 — these are equally wrong, neither is better.
-> this means, for example, if mismatch is 102.04 — then correct answer must contain at least one number with non-zero second decimal place to zero out the .04 part.
Number of possible sign flips is restricted. Let's say — correct answer is no more than 10 flips.
More frequently than not, flips come in islands/groups of rows located near each other rather than being randomly scattered. Example of flipped row indexes: [1,2,3,4,400,402]
Algorithm must stop after 1 second is elapsed. Not found => we show "Sorry we can't guess" to the client
My attempts
I decided to ask for guidance here as I'm feeling a bit lost in the direction of my research.
I've tried a sort + bruteforce solution:
Sort rows by difference between Total mismatch and the number
Brute-force using bit mask like 0001 0010 0011 meaning "Try flip row 1", "Try flip row 2", "Try flip row 1 and 2", etc. — because given they usually come as groups/islands, I don't want neither breadth-first nor depth-first.
This works on many cases, but fails when there are 2 islands or 1 island and 1-2 outlier rows far from the original island.
Research
In general, I found SSP and many different algorithms, but, obviously, O(2^(n/2)) on 500 rows is an impossible task. I found out that given this is a JS snippet to be run in client's browser — to be under 1 second of computation time, the algorithm can try ~2^22 combinations before it must stop.
I haven't tried going into dynamic programming methods to achieve pseudo-polynomial time here yet.
Expectations
I have a feeling, that, given these very specific restrictions on the task above — there should be something simpler.
The correct answers always are 1-2 islands + 1-2 outliers, never anything complex. 99% cases — 5-6 flips solve the sum. So "10 flips" is the worst case, in reality average is ~5.
Like there must be some specific approach I'm missing that could drastically lower the average case complexity by utilizing these restrictions.
Maybe I'm wrong.
arrays
algorithm
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asked Feb 10, 2023 at 10:03
John AJohn A
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2 While I don't know what algorithm might be best, I can tell you that the first "specific" means you can't use floating-point arithmetic. At least something to think about for the implementation.Some programmer dude –Some programmer dude 2023-02-10 10:14:22 +00:00 Commented Feb 10, 2023 at 10:14
1 Oh yes, you are absolutely right. In my current and any future algorithm — I convert everything to cents first and operate on ints.John A –John A 2023-02-10 11:29:46 +00:00 Commented Feb 10, 2023 at 11:29
How big can an island be?btilly –btilly 2023-02-10 15:41:28 +00:00 Commented Feb 10, 2023 at 15:41
Regular island is ~3-4 rows. Up to, let's say, 10 max. Sometimes there are islands "with holes", e.g. [1,2,3,4,500,502,503] — two islands [1,4], [500,503] but 501 is missing.John A –John A 2023-02-10 21:14:36 +00:00 Commented Feb 10, 2023 at 21:14
We can simplify such tasks saying these are two islands [1,4] and [502,503] with as a single outlier. But in general by "islands" I mean not that they are necessarily strictly contiguous, but rather that mistakes are concentrated around specific points of attraction.John A –John A 2023-02-10 21:18:21 +00:00 Commented Feb 10, 2023 at 21:18
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I think the approach with the bitmap is fine, in the sense that in a limited area you would look at all on-off combinations for flipping.
To support two islands like that, store the effect of each island configuration in a Map keyed by that amount-difference, and associated with that key store the leftmost index of that island together with the flip-bitmap.
Of course, if any of those islands solves the problem, return that solution.
If not, then iterate these islands, and determine which amount you need to combine it with to get the zero sum. Look in the map for that missing amount (which must be a key in that map). If found, you combine the two locations and return that result.
Some care can to be taken to not combine two overlapping islands, as you would always find an equivalent combination where the islands are disjunct.
To demonstrate this idea, I made a function that creates a random array with 500 amounts in the range of (-100 000,...,100 000), and which applies a maximum number of flips in at most 2 islands. I defined "island" as a set of flips where the two extreme indices are not more than 12 apart. So for example, an island can be 12 consecutive flips, or can be 2 flips at indices 3 and 15, or any variation in between.
The demo below will keep producing new data and calculate the solution for it. If the solution cannot be found within 1 second (which never happend on my PC), or the solution could not be found with at most 2 islands, the output is "not found". Otherwise it is an array with indices where the flips should be applied.
Each time the solution is verified by actually summing up the amounts taking into account the flips and asserting that this sum is zero.
Here is the snippet which keeps running random tests:
```javascript
const randint = (end) => Math.floor(Math.random() end);
function randomInput(size, maxFlips, maxIslands) {
let amounts = [];
let sum = 0;
while (size-- > 1) {
let amount = (Math.floor(Math.random() 2e7) - 1e7) / 100;
amounts.push(amount);
sum += amount;
}
// Make sum zero
while (Math.abs(sum) >= 100000) {
let i = randint(amounts.length);
if (amounts[i] sum > 0) {
sum -= amounts[i] 2;
amounts[i] = -amounts[i];
}
}
amounts.push(-Math.round(sum 100) / 100);
// Prepare islands
let islands = Array(maxIslands).fill(0);
// Determine size of each island (could remain 0)
while (maxFlips-- > 0) {
islands[randint(maxIslands)]++;
}
// Apply the flips (could occasionally overlap, making the problem simpler)
for (let count of islands) {
let i = randint(amounts.length - count + 1);
while (count-- > 0) {
// 80% probability that amount is flipped; so we may get holes in islands
if (randint(100) < 80) {
amounts[i + count] = -amounts[i + count]; // Flip!
}
}
}
return amounts;
}
function solve(amounts) {
let deadline = performance.now() + 900;
const map = new Map;
function memo(sum, loc) {
if (map.has(sum)) map.get(sum).push(loc);
else map.set(sum, [loc]);
}
function recur(i, j, bits, width, sum) {
if (width == 0) return memo(sum, [i, bits]);
recur(i, j + 1, bits 2, width - 1, sum);
recur(i, j + 1, bits 2 + 1, width - 1, sum + amounts[j]2);
}
function expand([i, bits]) {
return Array.from(bits.toString(2), (bit, j) => +bit ? i + j : -1)
.filter(i => i >= 0);
}
// Convert to cents
amounts = amounts.map(val => Math.round(val 100));
let sum = amounts.reduce((a, b) => a + b);
if (sum == 0) return []; // No flips
const WIDTH = 12;
// Collect islands with at least one flip (at i)
for (let i = 0; i < amounts.length; i++) {
recur(i, i + 1, 1, WIDTH - 1, amounts[i]2);
}
// Solution with one island?
if (map.has(sum)) return expand(map.get(sum));
// Look for solutions with two islands...
for (let [sum1, islands] of map) {
if (map.has(sum - sum1)) {
for (let [i, bits1] of map.get(sum - sum1)) {
for (let [j, bits2] of islands) {
if (i >= j + WIDTH) return expand([j, bits2]).concat(expand([i, bits1]));
else if (j >= i + WIDTH) return expand([i, bits1]).concat(expand([j, bits2]));
}
}
}
if (performance.now() >= deadline) break;
}
return "not found";
}
function verify(amounts, flips) {
let sum = Math.round(amounts.reduce((acc, val, i) => acc + Math.round(100val) (flips.includes(i) ? -1 : 1), 0));
if (sum != 0) {
console.log("amounts", JSON.stringify(amounts));
console.log("flips", JSON.stringify(flips));
throw "Wrong solution detected! sum=" + sum;
}
}
// I/O handling
const input = document.querySelector("textarea");
const [output, time] = document.querySelectorAll("span");
const pause = document.querySelector("input");
let pausing = false;
setInterval(function repeat() {
if (pausing) return;
const amounts = randomInput(500, 12, 2);
const start = performance.now();
const flips = solve(amounts);
time.textContent = Math.ceil(performance.now() - start);
verify(amounts, flips);
input.value = amounts;
output.textContent = flips;
}, 1000);
pause.onchange = () => pausing = pause.checked;
```
css
textarea { width: 100%; height: 6em }
xml
500 amounts: <textarea readonly></textarea><br>
Flips found: <span></span><br>
Time elapsed: <span></span> ms<br>
<input type="checkbox">Pausing
Run code snippet Edit code snippet
Hide Results Copy Expand
Note that a new test is executed every second. This is not the time it takes to produce the result. The time needed for the processing is displayed in the output.
Secondly, there are often many solutions. This algorithm doesn't look for the solution that needs the least flips, although it will give precedence to one-island solutions. But as there are often many solutions, the solution will often include the very first index, as that index is included in the first islands, which are combined first with all other islands.
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edited Feb 11, 2023 at 0:59
answered Feb 11, 2023 at 0:49
trincottrincot
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John A
John AOver a year ago
This ended up as the best solution for my case and it has a very decent guess rate on real examples we have.
2023-02-11T20:36:44.483Z+00:00
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Comparison to subset sum
As you rightfully point out, the general case of your problem is equivalent to subset sum, which is NP-complete. The conversion is simple, we know how far away from 0 we are, so we have to find a subset of rows with value equal to exactly half that difference. Flipping those will make all data sum to zero.
However, given the general (constraints on the) form of the solution, we can come up with a solution that may be able to usually find a solution in reasonable time.
Constrained solution
First off, you should use integers representing cents (by multiplying by 100) so we don't have to deal with floating points.
Now, we can compute and store every value we can make with at most 2 of the values. This is only 500^2 combinations, which is easily fast enough. We store these in a datastructure that allows for O(1) lookup, such as a hashmap, and we also store the indices we used.
Then we iterate over all contiguous "islands" of at most, say, size 10. There are about 10500 of these, which is still not a problem. You could even completely relax this constraint and simply look at all ~500500/2 islands. We subtract the sum of these islands from the target number to find the amount we still need to flip, and check the hashmap if that value is possible to make with at most 2 rows. If it is, we finally need to check if the latter rows are included within the island, and reject the solution in that case.
This strategy allows for efficient search over all possible combinations of an island and one or two stray flips.
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answered Feb 10, 2023 at 11:56
ADdVADdV
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Since there is no requirement to make a specific number of flips, we can store only the minimal number of flips necessary to create a sum up to the ith element,
for each element:
for each sum stored in the previous round:
if the number of its flips is less than 10:
make a new record of sum with flipped value
if this sum can now be associated with
less flips, update that
make a new record of sum with given value
if this sum can now be associated with
less flips, update that
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edited Feb 10, 2023 at 18:13
answered Feb 10, 2023 at 16:13
גלעד ברקןגלעד ברקן
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btilly
btillyOver a year ago
Don't forget that we want to find a solution rather than just know that it exists.
2023-02-10T16:52:37.663Z+00:00
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גלעד ברקן
גלעד ברקןOver a year ago
@btilly sure, just add the element to a single list that tracks which was this particular path that led to the minimal flips for the sum.
2023-02-10T18:11:31.353Z+00:00
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I tried the dynamic programming thing. Performance is..not as desired. The problem is that the combinatorial explosion in possibilities goes too fast.
But if you only have 2-3 groups of flips, the following may be good enough?
```
function findFlips (accounts) {
for (let i = 0; i < accounts.length; i++) {
accounts[i]['cents'] = Math.round(accounts[i].value 100);
}
let sum = accounts.reduce((a, b) => a+b.cents, 0);
// -2 is a hack to make the start of the next group be 0.
const start = {"sum": sum, "flips": 0, "i": -2, "decisions": 0, "prev": null};
let best = {}
best[start.sum] = [start];
let todo = [start];
let d = new Date;
let startTime = d.getTime();
let count = 0;
while (0 < todo.length) {
let current = todo.shift();
if (current.sum == 0) {
let answer = [];
while (current.prev != null) {
answer.unshift(current.i);
current = current.prev;
}
return answer;
}
// Did we take too long?
count++;
if (0 == count%10000) {
d = new Date;
if (999 < d.getTime() - startTime) {
return null;
}
}
for (let i = current.i+2; i < accounts.length; i++) {
let next = {
"sum": current.sum - 2accounts[i].cents,
"flips": current.flips + 1,
"i": i,
"decisions": current.decisions+1,
"prev": current};
let j = i;
while (j < accounts.length && next.flips < 11 && next.decisions < 5) {
let existing = (best[next.sum] || [])
let k = 0;
while (k < existing.length) {
if (existing[k].i <= next.i && existing[k].flips <= next.flips) {
break;
}
k++;
}
if (k == existing.length) {
existing.push(next);
best[next.sum] = existing;
todo.push(next);
}
j++;
if (j < accounts.length && j < i + 4) {
let next2 = next;
next = {
"sum": next2.sum - 2accounts[j].cents,
"flips": next2.flips + 1,
"i": j,
"decisions": next2.decisions,
"prev": next2};
}
}
}
}
return null;
}
let accounts = [
{'accountId': 1, 'value': 24.12},
{'accountId': 2, 'value': 128.16},
{'accountId': 3, 'value': -545.43},
{'accountId': 4, 'value': 191.92},
{'accountId': 5, 'value': 238.01},
{'accountId': 6, 'value': -500.36},
{'accountId': 7, 'value': -81.22},
{'accountId': 8, 'value': 132.50},
{'accountId': 9, 'value': 246.80},
{'accountId': 10, 'value': 93.80},
{'accountId': 11, 'value': 389.00},
{'accountId': 12, 'value': -331.00},
{'accountId': 13, 'value': -332.00},
{'accountId': 14, 'value': 321.00},
{'accountId': 15, 'value': 321.06},
{'accountId': 16, 'value': -336.04},
{'accountId': 17, 'value': -491.02},
{'accountId': 18, 'value': -119.06},
{'accountId': 19, 'value': 245.69},
{'accountId': 20, 'value': 404.07},
];
console.log(findFlips(accounts));
// Do some flips.
[2, 3, 11, 18].forEach((i) => accounts[i].value = - accounts[i].value);
console.log(findFlips(accounts));
accounts.value += 0.03;
console.log(findFlips(accounts));
```
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answered Feb 10, 2023 at 19:42
btillybtilly
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12604 | https://www.robots.ox.ac.uk/~az/lectures/b1/lect3.pdf | Lecture 3 1B Optimization Michaelmas 2018 A. Zisserman Linear Programming • Extreme solutions • Simplex method • Interior point method • Integer programming and relaxation The Optimization Tree A linear function to be maximized maximize f(x) = c1x1 + c2x2 + … + cnxn Problem constraints subject to a11x1 + a12x2 + … + a1nxn < b1 a21x1 + a22x2 + … + a2nxn < b2 … am1x1 + am2x2 + … + amnxn < bm Non-negative variables e.g. x1, x2 > 0 Linear Programming The name is historical, it should really be called Linear Optimization.
The problem consists of three parts: Linear Programming The problem is usually expressed in matrix form and then it becomes: (Linear programming problems are convex, so a local optimum is the global optimum.) A farmer has an area of A square kilometers to be planted with a combination of wheat and barley. A limited amount F of fertilizer and P of insecticide can be used, each of which is required in different amounts per unit area for wheat (F1, P1) and barley (F2, P2). Let S1 be the selling price of wheat, and S2 the price of barley, and denote the area planted with wheat and barley as x1 and x2 respectively.
The optimal number of square kilometers to plant with wheat vs. barley can be expressed as a linear programming problem.
Example 1 x1 x2 A Example 1 cont.
Maximize S1x1 + S2x2 (the revenue – this is the “objective function”) subject to x1 +x2 < A (limit on total area) F1x1 + F2x2 < F (limit on fertilizer) P1x2 + P2x2 < P (limit on insecticide) x1 >= 0, x2 > 0 (cannot plant a negative area) which in matrix form becomes maximize subject to Example 2: Max flow Given: a weighted directed graph, source s, destination t Interpret edge weights as capacities Goal: Find maximum flow from s to t • Flow does not exceed capacity in any edge • Flow at every vertex satisfies equilibrium [ flow in equals flow out ] e.g. oil flowing through pipes, internet routing Example 3 cont.
Slide: Robert Sedgewick and Kevin Wayne Example 2 cont.
Example 3: shortest path Given: a weighted directed graph, with a single source s Distance from s to v: length of the shortest part from s to v Goal: Find distance (and shortest path) to every vertex e.g. plotting routes on Google maps Application Minimize number of stops (lengths = 1) Minimize amount of time (positive lengths) LP – why is it important?
We have seen examples of: • allocating limited resources • network flow • shortest path Others include: • matching • assignment … It is a widely applicable problem-solving model because: • non-negativity is the usual constraint on any variable that represents an amount of something • one is often interested in bounds imposed by limited resources.
Dominates world of industry Linear Programming – 2D example Inequality constraints: Cost function: 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Example – feasible region from inequalities 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Inequality constraints: Feasible Region Inequality constraints: LP example Cost function: feasible region extreme point 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 LP example Inequality constraints: Cost function: feasible region contour lines extreme point 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 LP example – change of cost function Inequality constraints: Cost function: feasible region contour lines extreme point c Linear Programming – optima at vertices • The key point is that for any (linear) objective function the optima only occur at the corners (vertices) of the feasible polygonal region (never on the interior region).
• Similarly, in 3D the optima only occur at the vertices of a polyhedron (and in nD at the vertices of a polytope).
• However, the optimum is not necessarily unique: it is possible to have a set of optimal solutions covering an edge or face of a polyhedron.
-5 0 5 10 15 -5 0 5 10 15 Cf Constrained Quadratic Optimization feasible region extreme point Sketch solutions for optimization methods 1.
Simplex method • Tableau exploration of vertices based on linear algebra 2.
Interior point method • Continuous optimization with constraints cast as barriers We will look at 2 methods of solution: Simplex algorithm – solution idea How to find the maximum?? • Try every vertex? But there are too many in large problems.
• Instead, simply go from one vertex to the next increasing the cost function each time, and in an intelligent manner to avoid having to visit (and test) every vertex.
• This is the idea of the simplex algorithm Simplex method • Optimum must be at the intersection of constraints • Intersections are easy to find, change inequalities to equalities • Intersections are called basic solutions 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 a b c d e f • some intersections are outside the feasible region (e.g. f) and so need not be considered • the others (which are vertices of the feasible region) are called basic feasible solutions Worst complexity … 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 a b c d e f The Simplex Algorithm • Start from a basic feasible solution (i.e. a vertex of feasible region) • Consider all the vertices connected to the current one by an edge • Choose the vertex which increases the cost function the most (and is still feasible of course) • Repeat until no further increases are possible In practice this is very efficient and avoids visiting all vertices Matlab LP Function linprog Linprog() for medium scale problems uses the Simplex algorithm Example Find x that minimizes f(x) = –5x1 – 4x2 –6x3, subject to x1 – x2 + x3 ≤ 20 3x1 + 2x2 + 4x3 ≤ 42 3x1 + 2x2 ≤ 30 0 ≤ x1, 0 ≤ x2, 0 ≤ x3.
f = [-5; -4; -6]; >> A = [1 -1 1 3 2 4 3 2 0]; >> b = [20; 42; 30]; >> lb = zeros(3,1); >> x = linprog(f,A,b,[],[],lb); >> Optimization terminated.
x x = 0.0000 15.0000 3.0000 Interior Point Method • Solve LP using continuous optimization methods • Represent inequalities by barrier functions • Follow path through interior of feasible region to vertex c Barrier function method Approximation via logarithmic barrier functon Algorithm for Interior Point Method Diagram copied from Boyd and Vandenberghe Example: interior point algorithm Integer programming There are often situations where the solution is required to be an integer or have boolean values (0 or 1 only). For example: • assignment problems • scheduling problems • matching problems Linear programming can also be used for these cases Example: assignment problem agents tasks 1 2 3 4 5 1 2 3 4 5 Assignment constraints: • each agent assigned to one task only • each task assigned to one agent only Objective: • assign n agents to n tasks to minimize total cost Example: assignment problem agents tasks 1 2 3 4 5 1 2 3 4 5 Assignment constraints: • each agent assigned to one task only • each task assigned to one agent only Objective: • assign n agents to n tasks to minimize total cost cost matrix a g e n t s tasks tasks a g e n t s Example: assignment problem 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 each agent i assigned to one task only • only one entry in each row each task j assigned to one agent only • only one entry in each column Example solution for xij tasks j a g e n t s i Example: assignment problem Linear Programme formulation 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 tasks j a g e n t s i -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 Agents to tasks assignment problem agents tasks -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 Optimal agents to tasks assignment cost: 11.735499 agents tasks Cost of assignment = distance between agent and task e.g. application: tracking, correspondence Example application: tracking pedestrians Multiple Object Tracking using Flow Linear Programming, Berclaz, Fleuret & Fua, 2009 Example application: tracking pedestrians Multiple Object Tracking using Flow Linear Programming, Berclaz, Fleuret & Fua, 2009 What is next?
• Convexity (when does a function have only a global minimum?) • Robust cost functions • Stochastic algorithms |
12605 | https://www.youtube.com/watch?v=m7KZcZdjLPE | Using Desmos to find intersections and for linear regression
Josiah Hartley
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Posted: 20 Sep 2021
Transcript:
Intro let me show you quickly how to use desmos to do some of the same things that you'll do with the calculator as far as graphing with linear models and so on so if you go to desmos.com d-e-s-m-o-s dot com and you open the graphing calculator you'll see a window like this one so one example of what we did with the calculator was to find the intersection of two lines by graphing them both so here if i want to graph a line i can click here on the left where there's places to enter Finding intersections equations so in the example we did we graphed y equals 3 plus 0.5 x so there's one line and then we wanted to see where this intersected the horizontal line y equals 26. now notice that we don't see that in the graph because we're too zoomed in so if i move over the graph and use the mouse wheel i can scroll to zoom out and then click and drag to move it around you can also use the plus and minus buttons up here on the right to zoom in and out now if i want to find where they intersect if i click on one of them notice that it highlights that point 46 26 and that just like on the calculator tells me that when the amount reaches 26 that's at the time value 46 so just like with the calculator you can see where two lines intersect like that so that's one thing we did the other thing we did on the calculator was linear regression and you can do that here as well it just takes a little more effort to type things in Adding a table but if we erase things there if i want to enter a list of data like we did in the calculator i'm going to hit this plus button up here and i'm going to add a table so on the under the x values i'm going to enter the time values and if you look back at that example with the calculator we start time at 0. so i'm going to enter 0 for the first one and then if we're doing the example with the gasoline consumption we go from 1995 to 2004. so 0 would be 1995 and so on from there so then under the first y value i'm going to enter 116 and then under the first or next x value and enter 118 and then 119 and so on and then on the x values i have to finish entering like so Linear regression so then if i want to view this data i can zoom out and i could readjust the window to view this a little better but if i want to do the regression line notice how it's labeled this second column y1 so i need to type white y and then use an underscore so i'm going to hit shift and the minus button to do the underscore which moves it down and y1 and then i'm going to hit the right arrow key equals and then i want the general form of a linear equation so you can use whatever variables you want here let's say i use a times x1 so again i use x1 plus b now notice this way using an equals sign it doesn't like that but if i instead use a tilde which is just to the left of the one on the number row at the top of the keyboard it's the second function there so if i hit shift and then that button to the left of the one i can get that tilde which says y is related to and that's just the notation that desmos decided to use to set up regression so the notation is the hard part here once you enter the data you just have to use this appropriate notation but then it gives you a bunch of information including the values of a and b so it gives you the value of a which is multiplied by x and then the value of b which is the the y intercept or the constant so you can do linear regression using your calculator if you don't have a calculator you can use excel or you could also use desmos you have a variety of options and any one of them is fine for any of the problems you need to do |
12606 | https://online.stat.psu.edu/stat501/book/export/html/1007 | Topic 3: Poisson & Nonlinear Regression
Topic 3: Poisson & Nonlinear Regression
Topic 3: Poisson & Nonlinear Regression
Overview
Multiple linear regression can be generalized to handle a response variable that is categorical or a count variable. This lesson covers the basics of such models, specifically logistic and Poisson regression, including model fitting and inference.
Multiple linear regression, logistic regression, and Poisson regression are examples of generalized linear models, which this lesson introduces briefly.
The lesson concludes with some examples of nonlinear regression, specifically exponential regression and population growth models.
Objectives
Upon completion of this lesson, you should be able to:
Apply logistic regression techniques to datasets with a binary response variable.
Apply Poisson regression techniques to datasets with a count response variable.
Understand the basics of fitting and inference for nonlinear regression methods when the regression function acting on the predictors is not linear in the parameters.
Topic 3 Code Files
Below is a zip file that contains all the data sets used in this lesson:
STAT501_Topic 3.zip
leukemia_remission.txt
poisson_simulated.txt
toxicity.txt
us_census.txt
T.3.1 - Poisson Regression
T.3.1 - Poisson Regression
The Poisson distribution for a random variable Y has the following probability mass function for a given value Y = y:
P(Y=y|λ)=e−λ λ y y!,
for y=0,1,2,…. Notice that the Poisson distribution is characterized by the single parameter λ, which is the mean rate of occurrence for the event being measured. For the Poisson distribution, it is assumed that large counts (concerning the value of λ) are rare.
In Poisson regression, the dependent variable (Y) is an observed count that follows the Poisson distribution. The rate λ is determined by a set of p−1 predictors X=(X 1,…,X p−1). The expression relating to these quantities is
λ=exp{X β}.
Thus, the fundamental Poisson regression model for observation i is given by
P(Y i=y i|X i,β)=e−exp{X i β}exp{X i β}y i y i!.
That is, for a given set of predictors, the categorical outcome follows a Poisson distribution with rate exp{X β}. For a sample of size n, the likelihood for a Poisson regression is given by:
L(β;y,X)=∏i=1 n e−exp{X i β}exp{X i β}y i y i!.
This yields the log-likelihood:
ℓ(β)=∑i=1 n y i X i β−∑i=1 n exp{X i β}−∑i=1 n log(y i!).
Maximizing the likelihood (or log-likelihood) has no closed-form solution, so a technique like iteratively reweighted least squares is used to find an estimate of the regression coefficients, β^. Once this value of β^ has been obtained, we may proceed to define various goodness-of-fit measures and calculated residuals. The residuals we present serve the same purpose as in linear regression. When plotted versus the response, they will help identify suspect data points.
To illustrate consider this example (Poisson Simulated data), which consists of a simulated data set of size n = 30 such that the response (Y) follows a Poisson distribution with the rate λ=exp{0.50+0.07 X}. A plot of the response versus the predictor is given below.
The following gives the analysis of the Poisson regression data in Minitab:
Select Stat > Regression > Poisson Regression > Fit Poisson Model.
Select "y" for the Response.
Select "x" as a Continuous predictor.
Click Results and change "Display of results" to "Expanded tables."
This results in the following output:
Coefficients
| Term | Coef | SE Coef | 95% CI | Z-Value | P-Value | VIF |
--- --- ---
| Constant | 0.308 | 0.289 | (-0.259, 0.875) | 1.06 | 0.287 | |
| x | 0.0764 | 0.0173 | (0.0424, 0.1103) | 4.41 | 0.000 | 1.00 |
Regression Equation
y = exp(Y')
Y' = 0.308 + 0.0764 x
As you can see, the Wald test p-value for x of 0.000 indicates that the predictor is highly significant.
Deviance Test
Changes in the deviance can be used to test the null hypothesis that any subset of the β's is equal to 0. The deviance, D(β^), is −2 times the difference between the log-likelihood evaluated at the maximum likelihood estimate and the log-likelihood for a "saturated model" (a theoretical model with a separate parameter for each observation and thus a perfect fit). Suppose we test that r<p of the β's are equal to 0. Then the deviance test statistic is given by:
G 2=D(β^(0))−D(β^),
where D(β^) is the deviance of the fitted (full) model and D(β^(0)) is the deviance of the model specified by the null hypothesis evaluated at the maximum likelihood estimate of that reduced model. This test statistic has a χ 2 distribution with p−r degrees of freedom. This test procedure is analogous to the general linear F test procedure for multiple linear regression.
To illustrate, the relevant Minitab output from the simulated example is:
Deviance Table
| Source | DF | Seq Dev | Contribution | Adj Dev | Adj Mean | Chi-Square | P-Value |
--- --- --- --- |
| Regression | 1 | 20.47 | 42.37% | 20.47 | 20.4677 | 20.47 | 0.000 |
| x | 1 | 20.47 | 42.37% | 20.47 | 20.4677 | 20.47 | 0.000 |
| Error | 28 | 27.84 | 57.63% | 27.84 | 0.9944 | | |
| Total | 29 | 48.31 | 100.00% | | | | |
Since there is only a single predictor for this example, this table simply provides information on the deviance test for x (p-value of 0.000), which matches the earlier Wald test result (p-value of 0.000). (Note that the Wald test and deviance test will not in general give identical results.) The Deviance Table includes the following:
The null model in this case has no predictors, so the fitted values are simply the sample mean, 4.233. The deviance for the null model is D(β^(0))=48.31, which is shown in the "Total" row in the Deviance Table.
The deviance for the fitted model is D(β^)=27.84, which is shown in the "Error" row in the Deviance Table.
The deviance test statistic is therefore G 2=48.31−27.84=20.47.
The p-value comes from a χ 2 distribution with 2−1=1 degrees of freedom.
Goodness-of-Fit
The overall performance of the fitted model can be measured by two different chi-square tests. There is the Pearson statistic
X 2=∑i=1 n(y i−exp{X i β^})2 exp{X i β^}
and the deviance statistic
D=2∑i=1 n(y i log(y i exp{X i β^})−(y i−exp{X i β^})).
Both of these statistics are approximately chi-square distributed with n - p degrees of freedom. When a test is rejected, there is a statistically significant lack of fit. Otherwise, there is no evidence of a lack of fit.
To illustrate, the relevant Minitab output from the simulated example is:
Goodness-of-Fit
| Test | DF | Estimate | Mean | Chi-Square | P-Value |
--- --- --- |
| Deviance | 28 | 27.84209 | 0.099436 | 27.84 | 0.473 |
| Pearson | 28 | 26.09324 | 0.93190 | 26.09 | 0.568 |
The high p-values indicate no evidence of a lack of fit.
Overdispersion means that the actual covariance matrix for the observed data exceeds that for the specified model for Y|X. For a Poisson distribution, the mean and the variance are equal. In practice, the data rarely reflects this fact and we have overdispersion in the Poisson regression model if (as is often the case) the variance is greater than the mean. In addition to testing goodness-of-fit, the Pearson statistic can also be used as a test of overdispersion. Note that overdispersion can also be measured in the logistic regression models that were discussed earlier.
Pseudo R 2
The value of R 2 used in linear regression also does not extend to Poisson regression. One commonly used measure is the pseudo R 2, defined as
R 2=ℓ(β 0^)−ℓ(β^)ℓ(β 0^)=1−D(β^)D(β 0^),
where ℓ(β 0^) is the log-likelihood of the model when only the intercept is included. The pseudo R 2 goes from 0 to 1 with 1 being a perfect fit.
To illustrate, the relevant Minitab output from the simulated example is:
Model Summary
| Deviance R-Sq | Deviance R-Sq(adj) | AIC |
---
| 42.37% | 40.30% | 124.50 |
Recall from above that D(β^)=27.84 and D(β^(0))=48.31, so:
R 2=1−27.84 48.31=0.4237.
Raw Residual
The raw residual is the difference between the actual response and the estimated value from the model. Remember that the variance is equal to the mean for a Poisson random variable. Therefore, we expect that the variances of the residuals are unequal. This can lead to difficulties in the interpretation of the raw residuals, yet it is still used. The formula for the raw residual is
r i=y i−exp{X i β}.
Pearson Residual
The Pearson residual corrects for the unequal variance in the raw residuals by dividing by the standard deviation. The formula for the Pearson residuals is
p i=r i ϕ^exp{X i β},
where
ϕ^=1 n−p∑i=1 n(y i−exp{X i β^})2 exp{X i β^}.
ϕ^ is a dispersion parameter to help control overdispersion.
Deviance Residuals
Deviance residuals are also popular because the sum of squares of these residuals is the deviance statistic. The formula for the deviance residual is
d i=sgn(y i−exp{X i β^})2{y i log(y i exp{X i β^})−(y i−exp{X i β^})}.
The plots below show the Pearson residuals and deviance residuals versus the fitted values for the simulated example.
These plots appear to be good for a Poisson fit. Further diagnostic plots can also be produced and model selection techniques can be employed when faced with multiple predictors.
Hat Values
The hat matrix serves the same purpose as in the case of linear regression - to measure the influence of each observation on the overall fit of the model. The hat values, h i,i, are the diagonal entries of the Hat matrix
H=W 1/2 X(X W X)−1 X W 1/2,
where W is an n×n diagonal matrix with the values of exp{X i β^} on the diagonal. As before, a hat value (leverage) is large if h i,i>2 p/n.
Studentized Residuals
Finally, we can also report Studentized versions of some of the earlier residuals. The Studentized Pearson residuals are given by
s p i=p i 1−h i,i
and the Studentized deviance residuals are given by
s d i=d i 1−h i,i.
To illustrate, the relevant Minitab output from the simulated example is:
Fits and Diagnostics for Unusual Observations
| Obs | y | Fit | SE Fit | 95% CI | Resid | Std Resid | Del Resid | HI | Cook's D |
--- --- --- --- --- |
| 8 | 10.000 | 4.983 | 0.452 | (4.171, 5.952) | 1.974 | 2.02 | 2.03 | 0.040969 | 0.11 |
| 21 | 6.000 | 8.503 | 1.408 | (6.147, 11.763) | -0.907 | -1.04 | -1.02 | 0.233132 | 0.15 |
Fits and Diagnostics for Unusual Observation
| Obs | DFITS | | |
--- |
| 8 | 0.474408 | R | |
| 21 | -0.540485 | | X |
R large residual
X Unusual X
The default residuals in this output (set under Minitab's Regression Options) are deviance residuals, so observation 8 has a deviance residual of 1.974 and studentized deviance residual of 2.02, while observation 21 has leverage (h) of 0.233132.
T.3.2 - Polytomous Regression
T.3.2 - Polytomous Regression
Note! that the material in this section is more technical than the preceding Lessons. It is offered as an introduction to more advanced topics and, given the technical nature of the material, it could be considered optional in the context of this course.
In binary logistic regression, we only had two possible outcomes. For polytomous logistic regression, we will consider the possibility of having k> 2 possible outcomes. (Note: The word polychotomous is sometimes used, but note that this is not actually a word!)
Nominal Logistic Regression
The multiple nominal logistic regression model (sometimes called the multinomial logistic regression model) is given by the following:
π j={exp(X β j)1+∑j=2 k exp(X β j)j=2,...,k 1 1+∑j=2 k exp(X β j)j=1
where again π j denotes a probability and not the irrational number. Notice that k - 1 of the groups has its own set of β values. Furthermore, since ∑j=1 k π j=1, we set the β values for group 1 to be 0 (this is what we call the reference group). Notice that when k = 2, we are back to binary logistic regression.
π j is the probability that an observation is in one of the k categories. The likelihood for the nominal logistic regression model is given by:
L(β;y,X)=∏i=1 n∏j=1 k π i,j y i,j(1−π i,j)1−y i,j,
where the subscript (i,j) means the i th observation belongs to the j th group. This yields the log-likelihood:
ℓ(β)=∑i=1 n∑j=1 k y i,j π i,j.
Maximizing the likelihood (or log-likelihood) has no closed-form solution, so a technique like iteratively reweighted least squares is used to find an estimate of the regression coefficients, β^.
An odds ratio (θ) of 1 serves as the baseline for comparison. If θ=1, then there is no association between the response and predictor. If θ>1, then the odds of success are higher for the indicated level of the factor (or for higher levels of a continuous predictor). If θ<1, then the odds of success are less for the indicated level of the factor (or for higher levels of a continuous predictor). Values farther from 1 represent stronger degrees of association. For nominal logistic regression, the odds of success (at two different levels of the predictors, say X(1) and X(2)) are:
θ=(π j/π 1)|X=X(1)(π j/π 1)|X=X(2).
Many of the procedures discussed in binary logistic regression can be extended to nominal logistic regression with the appropriate modifications.
Ordinal Logistic Regression
For ordinal logistic regression, we again consider k possible outcomes as in nominal logistic regression, except that the order matters. The multiple ordinal logistic regression model is the following:
∑j=1 k∗π j=exp(β 0,k∗+X β)1+exp(β 0,k∗+X β)
such that k∗≤k, π 1≤π 2,≤…,≤π k, and again π j denotes a probability. Notice that this model is a cumulative sum of probabilities which involves just changing the intercept of the linear regression portion (so β is now (p - 1)-dimensional and X is n×(p−1) such that first column of this matrix is not a column of 1's). Also, it still holds that ∑j=1 k π j=1.
π j is still the probability that an observation is in one of the k categories, but we are constrained by the model written in the equation above. The likelihood for the ordinal logistic regression model is given by:
L(β;y,X)=∏i=1 n∏j=1 k π i,j y i,j(1−π i,j)1−y i,j,
where the subscript (i, j) means the i th observation belongs to the j th group. This yields the log-likelihood:
ℓ(β)=∑i=1 n∑j=1 k y i,j π i,j.
Notice that this is identical to the nominal logistic regression likelihood. Thus, maximization again has no closed-form solution, so we defer to a procedure like iteratively reweighted least squares.
For ordinal logistic regression, a proportional odds model is used to determine the odds ratio. Again, an odds ratio (θ) of 1 serves as the baseline for comparison between the two predictor levels, say X(1) and X(2). Only one parameter and one odds ratio is calculated for each predictor. Suppose we are interested in calculating the odds of X(1) to X(2). If θ=1, then there is no association between the response and these two predictors. If θ>1, then the odds of success are higher for the predictor X(1). If θ<1, then the odds of success are less for the predictor X(1). Values farther from 1 represent stronger degrees of association. For ordinal logistic regression, the odds ratio utilizes cumulative probabilities and their complements and is given by:
θ=∑j=1 k∗π j|X=X(1)/(1−∑j=1 k∗π j)|X=X(1)∑j=1 k∗π j|X=X(2)/(1∑j=1 k∗π j)|X=X(2).
T.3.3 - Generalized Linear Models
T.3.3 - Generalized Linear Models
All of the regression models we have considered (including multiple linear, logistic, and Poisson) belong to a family of models called generalized linear models. (In fact, a more "generalized" framework for regression models is calledgeneral regression models, which includes any parametric regression model.) Generalized linear models provide a generalization of ordinary least squares regression that relates the random term (the response Y) to the systematic term (the linear predictor X β) via a link function (denoted by g(⋅)). Specifically, we have the relation
E(Y)=μ=g−1(X β),
so g(μ)=X β. Some common link functions are:
Identity Link
The identity link:
g(μ)=μ=X β,
which is used in traditional linear regression.
Logit Link
The logit link:
g(μ)=log(μ 1−μ)=X β⇒μ=e X β 1+e X β,
which is used in logistic regression.
Log Link
The log link:
g(μ)=log(μ)=X β⇒μ=e X β,
which is used in Poisson regression.
Probit Link
The probit link:
g(μ)=Φ−1(μ)=X β⇒μ=Φ(X β),
where Φ(⋅) is the cumulative distribution function of the standard normal distribution. This link function is also sometimes called the normit link. This also can be used in logistic regression.
Complementary Log-Log Link
The complementary log-log link:
g(μ)=log(−log(1−μ))=X β⇒μ=1−exp{−e X β},
which can also be used in logistic regression. This link function is also sometimes called the gompit link.
Power Link
The power link:
g(μ)=μ λ=X β⇒μ=(X β)1/λ,
where λ≠0. This is used in other regressions which we do not explore (such as gamma regression and inverse Gaussian regression).
Also, the variance is typically a function of the mean and is often written as
Var(Y)=V(μ)=V(g−1(X β)).
The random variable Y is assumed to belong to an exponential family distribution where the density can be expressed in the form
q(y;θ,ϕ)=exp{y θ−b(θ)a(ϕ)+c(y,ϕ)},
where a(⋅), b(⋅), and c(⋅) are specified functions, θ is a parameter related to the mean of the distribution, and ϕ is called the dispersion parameter. Many probability distributions belong to the exponential family. For example, the normal distribution is used for traditional linear regression, the binomial distribution is used for logistic regression, and the Poisson distribution is used for Poisson regression. Other exponential family distributions lead to gamma regression, inverse Gaussian (normal) regression, and negative binomial regression, just to name a few.
The unknown parameters, β, are typically estimated with maximum likelihood techniques (in particular, using iteratively reweighted least squares), Bayesian methods, or quasi-likelihood methods. The quasi-likelihood is a function that possesses similar properties to the log-likelihood function and is most often used with count or binary data. Specifically, for a realization y of the random variable Y, it is defined as
Q(μ;y)=∫y μ y−t σ 2 V(t)d t,
where σ 2 is a scale parameter. There are also tests using likelihood ratio statistics for model development to determine if any predictors may be dropped from the model.
T.3.4 - Nonlinear Regression
T.3.4 - Nonlinear Regression
All of the models we have discussed thus far have been linear in the parameters (i.e., linear in the beta's). For example, polynomial regression was used to model curvature in our data by using higher-ordered values of the predictors. However, the final regression model was just a linear combination of higher-ordered predictors.
Now we are interested in studying the nonlinear regressionmodel:
Y=f(X,β)+ϵ,
where X is a vector of p predictors, β is a vector of k parameters, f(⋅) is some known regression function, and ϵ is an error term whose distribution may or may not be normal. Notice that we no longer necessarily have the dimension of the parameter vector simply one greater than the number of predictors. Some examples of nonlinear regression models are:
y i=e β 0+β 1 x i 1+e β 0+β 1 x i+ϵ i y i=β 0+β 1 x i 1+β 2 e β 3 x i+ϵ i y i=β 0+(0.4−β 0)e−β 1(x i−5)+ϵ i.
However, there are some nonlinear models which are actually called intrinsically linear because they can be made linear in the parameters by a simple transformation. For example:
Y=β 0 X β 1+X
can be rewritten as
1 Y=1 β 0+β 1 β 0 1 X=θ 0+θ 1 1 X,
which is linear in the transformed parameters θ 0 and θ 1. In such cases, transforming a model to its linear form often provides better inference procedures and confidence intervals, but one must be cognizant of the effects that the transformation has on the distribution of the errors.
Nonlinear Least Squares
Returning to cases in which it is not possible to transform the model to a linear form, consider the setting
Y i=f(X i,β)+ϵ i,
where the ϵ i are iid normal with mean 0 and constant variance σ 2. For this setting, we can rely on some of the least squares theories we have developed over the course. For other nonnormal error terms, different techniques need to be employed.
First, let
Q=∑i=1 n(y i−f(X i,β))2.
In order to find
β^=argmin β Q,
we first find each of the partial derivatives of Q with respect to β j. Then, we set each of the partial derivatives equal to 0 and the parameters β k are each replaced by β^k. The functions to be solved are nonlinear in the parameter estimates β^k and are often difficult to solve, even in the simplest cases. Hence, iterative numerical methods are often employed. Even more difficulty arises in that multiple solutions may be possible!
Algorithms for nonlinear least squares estimation include:
Newton's method is a classical method based on a gradient approach but which can be computationally challenging and heavily dependent on good starting values.Gauss-Newton algorithm is a modification of Newton's method that gives a good approximation of the solution that Newton's method should have arrived at but which is not guaranteed to converge.Levenberg-Marquardt method can take care of computational difficulties arising from the other methods but can require a tedious search for the optimal value of a tuning parameter.
T.3.5 - Exponential Regression Example
T.3.5 - Exponential Regression Example
One simple nonlinear model is the exponential regression model
y i=β 0+β 1 exp(β 2 x i,1+…+β p+1 x i,1)+ϵ i,
where the ϵ i are iid normal with mean 0 and constant variance σ 2. Notice that if β 0=0, then the above is intrinsically linear by taking the natural logarithm of both sides.
Exponential regression is probably one of the simplest nonlinear regression models. An example where an exponential regression is often utilized is when relating the concentration of a substance (the response) to elapsed time (the predictor).
To illustrate, consider the example of long-term recovery after discharge from the hospital from page 514 of Applied Linear Regression Models (4th ed) by Kutner, Nachtsheim, and Neter. The response variable, Y, is the prognostic index for long-term recovery, and the predictor variable, X, is the number of days of hospitalization. The proposed model is the two-parameter exponential model:
Y i=θ 0 exp(θ 1 X i)+ϵ i,
where the ϵ i are independent normal with constant variance.
We'll use Minitab's nonlinear regression routine to apply the Gauss-Newton algorithm to estimate θ 0 and θ 1. Before we do this, however, we have to find initial values for θ 0 and θ 1. One way to do this is to note that we can linearize the response function by taking the natural logarithm:
log(θ 0 exp(θ 1 X i))=log(θ 0)+θ 1 X i.
Thus we can fit a simple linear regression model with the response, log(Y), and predictor, X, and the intercept (4.0372) give us an estimate of log(θ 0) while the slope (−0.03797) gives us an estimate of θ 1. (We then calculate exp(4.0372)=56.7 to estimate θ 0.)
Minitab: Nonlinear Regression Model
Now we can fit the nonlinear regression model:
Select Stat > Regression > Nonlinear Regression, select prog for the response, and click "Use Catalog" under "Expectation Function."
Select the "Exponential" function with 1 predictor and 2 parameters in the Catalog dialog box and click OK to go to the "Choose Predictors" dialog.
Select days to be the "Actual predictor" and click OK to go back to the Catalog dialog box, where you should see "Theta1 exp( Theta2 days )" in the "Expectation Function" box.
Click "Parameters" and type "56.7" next to "Theta1" and "-0.038" next to "Theta2" and click OK to go back to the Nonlinear Regression dialog box.
Click "Options" to confirm that Minitab will use the Gauss-Newton algorithm (the other choice is Levenberg-Marquardt) and click OK to go back to the Nonlinear Regression dialog box.
Click "Graphs" to confirm that Minitab will produce a plot of the fitted curve with data and click OK to go back to the Nonlinear Regression dialog box.
Click OK to obtain the following output:
Nonlinear Regression: Prog = Theta1 exp(Theta2 days)
Method| Algorithm | Gauss-Newton |
--- |
| Max iterations | 200 |
| Tolerance | 0.00001 |
Starting Values for Parameters| Parameter | Value |
--- |
| Theta1 | 56.7 |
| Theta2 | -0.038 |
Equation
prog = 58.6066 exp(-0.0395865 days)
Parameter Estimates| Parameter | Estimate | SE Estimate |
---
| Theta1 | 58.6066 | 1.47216 |
| Theta2 | 0.0396 | 0.00171 |
prog = Theta1 exp(Theta2 days)
Summary| Iterations | 5 |
| Final SSE | 49.4593 |
| DFE | 13 |
| MSE | 3.80456 |
| S | 1.95053 |
T.3.6 - Population Growth Example
T.3.6 - Population Growth Example
Census Data
A simple model for population growth towards an asymptote is the logistic model
y i=β 1 1+exp(β 2+β 3 x i)+ϵ i,
where y i is the population size at time x i, β 1 is the asymptote towards which the population grows, β 2 reflects the size of the population at time x = 0 (relative to its asymptotic size), and β 3 controls the growth rate of the population.
We fit this model to Census population data (US Census data) for the United States (in millions) ranging from 1790 through 1990 (see below).
| year | population |
--- |
| 1790 | 3.929 |
| 1800 | 5.308 |
| 1810 | 7.240 |
| 1820 | 9.638 |
| 1830 | 12.866 |
| 1840 | 17.069 |
| 1850 | 23.192 |
| 1860 | 31.443 |
| 1870 | 39.818 |
| 1880 | 50.156 |
| 1890 | 62.948 |
| 1900 | 75.995 |
| 1910 | 91.972 |
| 1920 | 105.711 |
| 1930 | 122.775 |
| 1940 | 131.669 |
| 1950 | 150.697 |
| 1960 | 179.323 |
| 1970 | 203.302 |
| 1980 | 226.542 |
| 1990 | 248.710 |
The data are graphed (see below) and the line represents the fit of the logistic population growth model.
To fit the logistic model to the U. S. Census data, we need starting values for the parameters. It is often important in nonlinear least squares estimation to choose reasonable starting values, which generally requires some insight into the structure of the model. We know that β 1 represents asymptotic population. The data in the plot above show that in 1990 the U. S. population stood at about 250 million and did not appear to be close to an asymptote; so as not to extrapolate too far beyond the data, let us set the starting value of β 1 to 350. It is convenient to scale time so that x 1=0 in 1790, and so that the unit of time is 10 years.
Then substituting β 1=350 and x=0 into the model, using the value y 1=3.929 from the data, and assuming that the error is 0, we have
3.929=350 1+exp(β 2+β 3(0)).
Solving for β 2 gives us a plausible start value for this parameter:
exp(β 2)=350 3.929−1 β 2=log(350 3.929−1)≈4.5.
Finally, returning to the data, at time x = 1 (i.e., at the second Census performed in 1800), the population was y 2=5.308. Using this value, along with the previously determined start values for β 1 and β 2, and again setting the error to 0, we have
5.308=350 1+exp(4.5+β 3(1)).
Solving for β 3 we get
exp(4.5+β 3)=350 5.308−1 β 3=log(350 5.308−1)−4.5≈−0.3.
So now we have starting values for the nonlinear least squares algorithm that we use. Below is the output from fitting the model in Minitab using Gauss-Netwon:
Select Stat > Regression > Nonlinear Regression
Select "population" for the Response
Type the following into the "Edit directly" box under Expectation Function: beta1/(1+exp(beta2+beta3(year-1790)/10))
Click Parameters and type in the values specified above (350, 4.5, and –0.3)
As you can see, the starting values resulted in convergence with values not too far from our guess.
Equation
population = 389.166/(1+exp(3.99035 - 0.22662 (year - 1790) / 10))
Parameter Estimates
| Paraemeter | Estimate | SE Estimate |
---
| beta1 | 389.166 | 30.8120 |
| beta2 | 3.990 | 0.0703 |
| beta3 | -0.227 | 0.0109 |
population = beta1/(1 + exp(beta2 + beta3 (year - 1790) / 10))
Lack of Fit
There are no replicates.
Minitab cannot do the lack of fit test based on pure errror.
Summary
| Iterations | 8 |
| Final SSE | 356.400 |
| DFE | 18 |
| MSE | 19.8000 |
| S | 4.44972 |
Here is a plot of the residuals versus the year.
As you can see, the logistic functional form that we chose did catch the gross characteristics of this data, but some of the nuances appear to not be as well characterized. Since there are indications of some cyclical behavior, a model incorporating correlated errors or, perhaps, trigonometric functions could be investigated.
Software Help: Poisson & Nonlinear Regression
Software Help: Poisson & Nonlinear Regression
The next two pages cover the Minitab and R commands for the procedures in this lesson.
Below is a file that contains the data set used in this help section:
us_census.txt
poisson_simulated.txt
Minitab Help: Poisson & Nonlinear Regression
Minitab Help: Poisson & Nonlinear Regression
Minitab®
Hospital recovery (exponential regression)
Select Calc > Calculator to create a log(prog) variable.
Obtain starting values for nonlinear model parameters from fitting a simple linear regression model of log(prog) vs days.
Select Stat > Regression > Nonlinear Regression to fit a nonlinear regression model to data using these starting values. Put prog in the "Response" box and type "Theta1 exp(Theta2 days)" into the box labeled "Edit directly." Before clicking OK, click "Parameters" and enter 56.7 as the starting value for Theta1 and -0.038 as the starting value for Theta2.
A scatterplot of prog vs days with a fitted line based on the nonlinear regression model is produced by default.
U.S. Census population (population growth nonlinear regression)
Obtain starting values for nonlinear model parameters from observing features of a scatterplot of population vs year.
Select Stat > Regression > Nonlinear Regression to fit a nonlinear regression model to data using these starting values. Put population in the "Response" box and type "beta1 / (1 + exp(beta2 + beta3 (year - 1790) / 10))" into the box labeled "Edit directly." Before clicking OK, click "Parameters" and enter 56.7 as the starting value for Theta1 and -0.038 as the starting value for Theta2.
A scatterplot of population vs year with a fitted line based on the nonlinear regression model is produced by default.
Before clicking "OK" in the Regression Dialog, click "Graphs" and put the year in the "Residuals versus the variables" box.
Poisson example (Poisson regression)
Select Graph > Scatterplot to create a scatterplot of the data.
Select Stat > Regression > Poisson Regression > Fit Poisson Model, put y in the "Response" box, and put x in the "Continuous predictors" box. Before clicking "OK," click "Results" and select "Expanded tables" for "Display of results."
Before clicking "OK" in the Regression Dialog, click "Graphs" and select "Residuals versus Fits" to create residual plots using deviance residuals. To change to Pearson residuals, click "Options" in the Regression Dialog and select "Pearson" for "Residuals for diagnostics."
R Help: Poisson & Nonlinear Regression
R Help: Poisson & Nonlinear Regression
Hospital recovery (exponential regression)
Load the recovery data.
Create log(prog) variable.
Obtain starting values for nonlinear model parameters from fitting a simple linear regression model of log(prog) vs days.
Fit nonlinear regression model to data using these starting values.
Create a scatterplot of prog vs days and add a fitted line based on the nonlinear regression model.
```r
recovery <- read.table("~/path-to-data/recovery.txt", header=T)
attach(recovery)
logprog <- log(prog)
summary(lm(logprog ~ days))
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.037159 0.084103 48.00 5.08e-16
days -0.037974 0.002284 -16.62 3.86e-10
exp(4.037159) # 56.66513
model.1 <- nls(prog ~ theta1 exp(theta2 days),
start=list(theta1=56.7, theta2=-0.038))
summary(model.1)
Estimate Std. Error t value Pr(>|t|)
theta1 58.606532 1.472159 39.81 5.70e-15
theta2 -0.039586 0.001711 -23.13 6.01e-12
Residual standard error: 1.951 on 13 degrees of freedom
plot(x=days, y=prog,
panel.last = lines(sort(days), fitted(model.1)[order(days)]))
detach(recovery)
```
U.S. Census population (population growth nonlinear regression)
Load the census data.
Obtain starting values for nonlinear model parameters from observing features of a scatterplot of population vs year.
Fit nonlinear regression model to data using these starting values.
Create a scatterplot of population vs year and add a fitted line based on the nonlinear regression model.
Create a residual plot.
```r
census <- read.table("~/path-to-data/us_census.txt", header=T)
attach(census)
plot(x=year, y=population)
log(350/3.929-1) # 4.478259
log(350/5.308-1) - log(350/3.929-1) # -0.3048229
model.1 <- nls(population ~ beta1 / (1 + exp(beta2 + beta3 (year - 1790) / 10)),
start=list(beta1=350, beta2=4.5, beta3=-0.3))
summary(model.1)
Estimate Std. Error t value Pr(>|t|)
beta1 389.16551 30.81196 12.63 2.2e-10
beta2 3.99035 0.07032 56.74 < 2e-16
beta3 -0.22662 0.01086 -20.87 4.6e-14
Residual standard error: 4.45 on 18 degrees of freedom
plot(x=year, y=population,
panel.last = lines(sort(year), fitted(model.1)[order(year)]))
plot(x=year, y=residuals(model.1),
panel.last = abline(h=0, lty=2))
detach(census)
```
Poisson example (Poisson regression)
Load the poisson data.
Create a scatterplot of the data.
Fit a Poisson regression model of y vs x.
Calculate 95% confidence intervals for the regression parameters based on asymptotic normality and based on profiling the least-squares estimation surface.
Create a scatterplot of y vs x and add a fitted line based on the Poisson regression model.
Conduct a likelihood ratio (or deviance) test for x.
Calculate the sum of squared deviance residuals and the sum of squared Pearson residuals and calculate p-values based on chi-squared goodness-of-fit tests.
Calculate pseudo R 2 for Poisson regression.
Create residual plots using Pearson and deviance residuals.
Calculate hat values (leverages) and studentized residuals.
```r
poisson <- read.table("~/path-to-data/poisson_simulated.txt", header=T)
attach(poisson)
plot(x=x, y=y)
model.1 <- glm(y ~ x, family="poisson")
summary(model.1)
Estimate Std. Error z value Pr(>|z|)
(Intercept) 0.30787 0.28943 1.064 0.287
x 0.07636 0.01730 4.413 1.02e-05
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1
(Dispersion parameter for poisson family taken to be 1)
Null deviance: 48.310 on 29 degrees of freedom
Residual deviance: 27.842 on 28 degrees of freedom
AIC: 124.5
confint.default(model.1) # based on asymptotic normality
confint(model.1) # based on profiling the least-squares estimation surface
plot(x=x, y=y,
panel.last = lines(sort(x), fitted(model.1)[order(x)]))
anova(model.1, test="Chisq")
Df Deviance Resid. Df Resid. Dev Pr(>Chi)
NULL 29 48.310
x 1 20.468 28 27.842 6.065e-06
sum(residuals(model.1, type="deviance")^2) # 27.84209
model.1$deviance # 27.84209
pchisq(model.1$deviance, 28, lower.tail=F) # p-value = 0.4728389
sum(residuals(model.1, type="pearson")^2) # 26.09324
pchisq(sum(residuals(model.1, type="pearson")^2), 28, lower.tail=F) # p-value = 0.5679192
1-model.1$deviance/model.1$null.deviance # Pseudo R-squared = 0.423676
plot(fitted(model.1), residuals(model.1, type="pearson"))
plot(fitted(model.1), residuals(model.1, type="deviance"))
summary(influence.measures(model.1))
dfb.1_ dfb.x dffit cov.r cook.d hat
10 -0.22 0.30 0.37 1.25_ 0.08 0.18
21 0.37 -0.48 -0.57 1.30_ 0.15 0.23_
residuals(model.1) # 1.974329
rstudent(model.1) # 2.028255
detach(poisson)
```
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NCERT Solutions for Class 11 Biology Chapter 6 - Anatomy of Flowering Plants
NCERT Solutions
Class 11
Biology
Anatomy of Flowering Plants
Grab access to free NCERT Solutions at Aasoka which are designed by subject matter experts. Keeping the latest CBSE syllabus in mind, these experts have drafted the solutions and ensured to include crucial topics. The NCERT Solutions for Class 11 serves as excellent study material to kick-start your exam preparation. Overall performance and time management can be improved by practicing these questions. Start your preparation now.
The chapter “Anatomy of Flowering Plants” of Biology Class 11 includes a detailed explanation of the anatomy of Monocotyledonous and Dicotyledonous plants, the tissues, functional organization of higher plants, the tissue system, and so much more.
Question 1:
State the location and function of different types of meristems.
Answer:
Meristem. A meristematic tissue or a meristem consists of a group of cells which have the power of active cell division. These cells are living, thin walled and retain the power of cell division. The word meristem has its origin from the Greek word ‘meristos’ which means divisible.
Types of meristem. According to location in the plant body, meristems are of three types :
(a) Apical meristem (b) Lateral meristem (c) Intercalary meristem.
Show Answer
Question 2:
Cork cambium forms tissues that form the cork. Do you agree with this statement ? Explain.
Answer:
Cork. The peripheral water proof tissue of mature woody stem is called cork. It is composed of dead cells with thick walls impregnated with a waxy material called suberin.
Cork is formed by cork cambium (phellogen), a lateral meristem which replaces the epidermis in stems and roots of older woody plants. Thus it is true that cork cambium forms cork.
Cork formation
Show Answer
Question 3:
Explain the process of secondary growth in a stem of woody angiosperm with the help of schematic diagram. What is its significance ?
Answer:
Secondary growth. Secondary growth can be defined as an increase in diameter of the stem due to the activity of the lateral meristems i.e. vascular cambium and cork cambium (Figs. 6.3 and 6.4). The process of secondary growth in a dicot stem can be studied under headings given below :
Activity of Vascular Cambium. The active vascular cambium forms a ring. This ring is formed by fascicular cambium and interfascicular cambium. Vascular cambium exhibits two types of cell divisions namely additive and multiplicative. Additive divisions result in the formation of secondary phloem towards outside and secondary xylem towards inner side. The cambium has two types of cells. (i) Fusiform initials which are elongated and produce fibres, sieve cells, sieve tubes, tracheids and other vascular elements. (ii) Ray initials which produce parenchyma cells of the rays in wood and phloem. The secondary phloem is made up of vertical rows of sieve tubes, companion cells, fibres and phloem parenchyma. The horizontal rows of growth in the phloem form, the phloem rays. The phloem rays may be one cell thick or many layered thick. In the secondary phloem, the phloem sclerenchyma is formed and is known as bast fibres and this gives rise to textile fibres of jute, flax and hump etc. The phloem rays translocate the food to the cambium, xylem and other inner living cells.
Diagrams showing stages of secondary growth in Dicot stem The secondary xylem is the main wood of the plant. The secondary xylem consists of vertical rows of tracheids, vessel, xylem parenchyma and xylem fibres. Inbetween radial rows of xylem rays are produced. These rays resemble the phloem ray in structure but transport water and minerals from xylem to the cambium and phloem. Vessels in the secondary xylem are of pitted type and give mechanical support.
Annual rings. The activity of the cambium varies due to the seasonal changes. During favourable conditions i.e. spring the cambium produces the secondary xylem (wood) and is called the spring wood. In the autumn the activity of cambium is less and produces wood of smaller diameter and is called autumn wood. These layers form the rings. The growth rings of spring wood and autumn wood constitute the annual ring.
Acitivity of Cork Cambium. Cork cambium or phellogen develops in the cortex. Cork cambium cuts off cells towards outside which form cork tissue to phellem and towards inner side tissue called phelloderm. Phellem, phellogen and phelloderm, these three together form periderm. Lenticel (pore) appears in the periderm of woody stem. It is packed with loose aggregate cells called comple-mentary cells derived from the periderm and acts as an organ of gaseous exchange.
A portion of dicot stem showing formation of cork or periderm
Show Answer
Question 4:
Cut a transverse section of young stem of a plant from your school garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or a dicot stem ? Give reasons.
Answer:
The following features are helpful in ascertaining its type :
Based upon the observable features under the microscope, one can easily ascertain whether it is a monocot stem or a dicot stem.
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Question 5:
The transverse section of a plant material shows the following anatomical features :
(a) the vascular bundles are conjoint, scattered and surrounded by a sclerenchyma- tous bundle sheaths. (b) phloem parenchyma is absent. What will you identify it as ?
Answer:
The plant material in question is monocot stem.
Show Answer
Question 6:
Why are xylem and phloem called complex tissues ?
Answer:
Both xylem and phloem are made up of more than one type of cells and are, therefore, called complex tissues.
Xylem. It is a permanent complex conducting tissue. It is concerned with upward conduction of water and minerals. It is a group of cells which are similar in origin and function but of more than one type in structure.
Elements of xylem.
Xylem is composed of following four different kinds of elements :
Tracheids
Vessel members
Xylem parenchyma (wood parenchyma)
Xylem fibres (wood fibres).
Phloem. It is a permanent complex tissue. It is meant for translocation of food materials. Phloem occurs throughout the plant body along with xylem.
Phloem is made up four types of cells :
(1) Sieve tubes (2) Companion cells
(3) Phloem parenchyma (4) Phloem fibres.
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Question 7:
What is stomatal apparatus ? Explain the structure of stomata with labelled diagrams.
Answer:
The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus. Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two bean-shaped cells known as guard cells. In grasses, the guard cells are dumb-bell shaped. The outer walls of guard cells (away from the stomatal pore) are thin and the inner walls (towards the stomatal pore) are highly thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary cells (Fig. 6.21).
Stomatal apparatus (a) stomata with bean-shaped guard cells (b) stomata with dumb-bell shaped guard cell.
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Question 8:
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Answer:
The three basic tissue systems in the flowering plants along with their respective tissue names are :
Epidermal tissue system com-prising the epidermal cells, stomata and the epidermal appendages-the trichomes and hairs.
Ground tissue system comprising simple tissues such as parenchyma, collenchyma and sclerenchyma.
Vascular tissue system consisting of complex tissues i.e. the phloem and the xylem.
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Question 9:
How is the study of plant anatomy useful to us ?
Answer:
Value of study of plant anatomy
It enriches our knowledge of plant structure.
Anatomical studies can solve many taxonomic problems.
It helps in determining adulteration in spices, coffee, tea, saffron, Asafoetida (hing).
The pharmacognosy and pharma-cology are dependant upon the anatomical studies to know about the drug yielding plants and their actions.
The study of anatomy of wood help in differentiating the type of wood used in preparing various articles.
Forensic experts require the knowledge of plant anatomy to identify small piece of plants or their products for solving criminal cases.
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Question 10:
What is periderm ? How does periderm formation take place in the dicot stems ?
Answer:
As the secondary growth continues for several years, the stem constantly grows in thickness due to increase in secondary vascular tissues. This increase may result in cracking and breaking of outer tissues. To avoid such a breaking of external tissues, the plants develop a new cambium ring, called cork cambium (phellogen) in the outer region.
The cork cambium (phellogen) is a secondary lateral meristem which develops from permanent tissues in the region of epidermis, hypodermis, cortex or even in outer layers of phloem. It produces dead cells or cork (phellem) towards outer side and living cells of secondary cortex (phelloderm) towards inner side. These layers of cork, cork cambium and secondary cortex constitute the protective covering called periderm. The cork cells are highly lignified and check the loss of water from the general surface of stem. Sometimes the layers of cork are interrupted by lenticels.
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Question 11:
How does the structure and location of bulliform cells help in performing their specialised function ?
Answer:
Bulliform cells are adaxial epidermal cells along the veins of isobilateral leaf in grasses. These are large, empty, colourless cells. On absorption of water, they become turgid and expose the leaf surface. Due to loss of water, they turn flaccid and make the leaves curl inwards in order to minimise water loss.
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12610 | https://www.cisco.com/en/US/docs/security/ise/1.0/user_guide/ise10_ui_reference.pdf | A-1 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 A P P E N D I X A User Interface Reference This chapter is a reference for Identity Services Engine (ISE) user interface elements, and covers the following main elements: • Monitor, page A-1 • Policy, page A-57 • Administration, page A-61 Monitor This section covers the following subtabs: • Authentications, page A-1 • Alarms, page A-3 • Reports, page A-16 • Troubleshoot, page A-43 Authentications Go to Monitor > Authentications to display the Authentications page. Authentications data categories are described in the following table.
T able A-1 Authentications Option Description Time Shows the time that the log was received by the monitoring and troubleshooting collection agent. This column is required and cannot be deselected.
Status Shows if the authentication was successful or a failure. This column is required and cannot be deselected. Green is used to represent passed authentications. Red is used to represent failed authentications.
Details Brings up a report when you click the magnifying glass icon, allowing you to drill down and view more-detailed information on the selected authentication scenario. This column is required and cannot be deselected.
Username Shows the username that is associated with the authentication. A-2 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Optionally, you can choose to show the following categories: Calling Station ID Shows the unique identifier for an endpoint, usually a MAC or IP address.
IP Address Shows the IP address of the endpoint device.
NAD IP address of the network access device.
T able A-1 Authentications (continued) Option Description T able A-2 Optional Authentications Categories Option Description Server Indicates the policy service ISE node from which the log was generated.
NAS Port ID Network access server (NAS) port at which the endpoint is connected.
Failure Reason Shows a detailed reason for failure, if the authentication failed.
SGA Security Group Shows a security profile for the authentication.
Authorization Profiles Shows an authorization profile that was used for authentication.
Auth Method Shows the authentication method that is used by the RADIUS protocol, such as Microsoft Challenge Handshake Authentication Protocol version 2 (MSCHAPv2), IEE 802.1x or dot1x, and the like.
Authentication Protocol Shows the authentication protocol used, such as Protected Extensible Authentication Protocol (PEAP), Extensible Authentication Protocol (EAP), and the like.
SGA Security Group Shows the trust group that is identified by the authentication log.
Identity Group Shows the identity group that is assigned to the user or endpoint, for which the log was generated.
Posture Status Shows the status of posture validation and details on the authentication. A-3 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Alarms This section covers the following: • Alarms Inbox, page A-3 • Rules, page A-5 • Schedules, page A-14 Alarms Inbox This section covers the following: • Inbox, page A-3 • Edit > Alarm, page A-4 • Edit > Status, page A-4 Inbox The Monitor > Alarms > Inbox options are as follows: T able A-3 Inbox Option Description Severity Display only. Indicates the severity of the associated alarm. Options are: • Critical • Warning • Info Name Indicates the name of the alarm. Click to display the Alarms: Properties page and edit the alarm.
Time Display only. Indicates the time of the associated alarm generation in the format Ddd Mmm dd hh:mm:ss timezone yyyy, where: • Ddd = Sun, Mon, Tue, Wed, Thu, Fri, Sat.
• Mmm = Jan, Feb, Mar, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec.
• dd = Day of the month, from 01 to 31.
• hh = Hour of the day, from 00 to 23.
• mm = Minute of the hour, from 00 to 59.
• ss = Second of the minute, from 00 to 59.
• timezone = The time zone.
• yyyy = A four-digit year.
Cause Display only. Indicates the cause of the alarm.
Assigned To Display only. Indicates who is assigned to investigate the alarm. A-4 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Edit > Alarm Click Edit on the Inbox to bring up this page that provides information on the event that triggered the alarm. You cannot edit any of the fields in the Alarm tab. The options are shown in the following table.
Edit > Status Click Edit on the Inbox to bring up this page that allows you to edit the status of the alarm and add a description to track the event. The options are shown in the following table.
Status Display only. Indicates the status of the alarm. Options are: • New—The alarm is new.
• Acknowledged—The alarm is known.
• Closed—The alarm is closed.
Edit Check the check box next to the alarm that you want to edit, and click Edit to edit the status of the alarm and view the corresponding report.
Close Check the check box next to the alarm that you want to close, and click Close to close the alarm. You can enter closing notes before you close an alarm.
Note Closing an alarm only removes the alarm. It does not delete the alarm.
Delete Check the check box next to the alarm that you want to delete, and click Delete to delete the alarm.
T able A-3 Inbox (continued) Option Description T able A-4 Edit Alarm Option Description Occurred At Date and time when the alarm was triggered.
Cause The event that triggered the alarm.
Detail Additional details about the event that triggered the alarm. ISE usually lists the counts of items that exceeded the specified threshold.
Report Links Wherever applicable, one or more hyperlinks are provided to the relevant reports that allow you to further investigate the event.
Threshold Information on the threshold configuration.
T able A-5 Edit Status Option Description Status Status of the alarm. When an alarm is generated, its status is New. After you view the alarm, change the status of the alarm to Acknowledged or Closed to indicate the current status of the alarm.
Assigned To (Optional) Specify the name of the user to whom this alarm is assigned.
Notes (Optional) Enter any additional information about the alarm that you want to record. A-5 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Rules Click Monitor > Alarms > Rules to specify parameters for the following alarm rules: • Passed Authentications, page A-6 • Failed Authentications, page A-8 • Authentication Inactivity, page A-9 • ISE Configuration Changes, page A-9 • ISE System Diagnostics, page A-10 • ISE Process Status, page A-10 • ISE System Health, page A-11 • ISE AAA Health, page A-11 • Authenticated But No Accounting Start, page A-12 • Unknown NAD, page A-12 • External DB Unavailable, page A-13 • RBACL Drops, page A-13 • NAD-Reported AAA Downtime, page A-14 A-6 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Passed Authentications Modify the fields described in the following table, to create a threshold with the passed authentication criteria.
T able A-6 Passed Authentications Option Description Passed Authentications Greater than in the past for a , where: • values can be the absolute number of occurrences or percent. Valid values are: – count must be in the range 0 to 99 for greater than.
– count must be in the range 1 to 100 for lesser than.
• value can be occurrences or %.
• values can be 1 to 1440 minutes, or 1 to 24 hours.
• value can be Minutes or Hours.
• values can be any of the following: – ISE Instance – User – Identity Group – Device IP – Identity Store – Allowed Protocol – NAD Port – AuthZ Profile – AuthN Method – EAP AuthN – EAP Tunnel Note In a distributed deployment, if there are two instances, the count is calculated as an absolute number or as a percentage for each of the instances. An alarm is triggered only when the individual count of any instance exceeds the threshold.
Filter ISE Instance Choose a valid ISE instance for the threshold.
User Choose or enter a valid username for the threshold.
Identity Group Choose a valid identity group name for the threshold.
Device Name Choose a valid device name for the threshold.
Device IP Choose or enter a valid device IP address for the threshold.
Device Group Choose a valid device group name for the threshold.
Identity Store Choose a valid identity store name for the threshold. A-7 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Allowed Protocol Choose a valid allowed protocol name for the threshold.
MAC Address Choose or enter a valid MAC address for the threshold. This filter is available only for RADIUS authentications.
NAD Port Choose a port for the network device for the threshold. This filter is available only for RADIUS authentications.
AuthZ Profile Choose an authorization profile for the threshold. This filter is available only for RADIUS authentications.
AuthN Method Choose an authentication method for the threshold. This filter is available only for RADIUS authentications.
EAP AuthN Choose an EAP authentication value for the threshold. This filter is available only for RADIUS authentications.
EAP Tunnel Choose an EAP tunnel value for the threshold. This filter is available only for RADIUS authentications.
Protocol Configure the protocol that you want to use for your threshold. T able A-6 Passed Authentications (continued) Option Description A-8 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Failed Authentications Modify the fields described in the following table, to create a threshold with the passed authentication criteria.
T able A-7 Failed Authentications Option Description Failed Authentications Greater than in the past for a , where: • values can be the absolute number of occurrences or percent. Valid values must be in the range 0 to 99.
• value can be occurrences or %.
• values can be 1 to 1440 minutes, or 1 to 24 hours.
• value can be Minutes or Hours.
• values can be any of the following: – ISE Instance – User – Identity Group – Device IP – Identity Store – Allowed Protocol – NAD Port – AuthZ Profile – AuthN Method – EAP AuthN – EAP Tunnel Note In a distributed deployment, if there are two instances, the count is calculated as an absolute number or as a percentage for each of the instances. An alarm is triggered only when the individual count of any instance exceeds the specified threshold.
Filter Failure Reason Enter a valid failure reason name for the threshold.
ISE Instance Choose a valid ISE instance for the threshold.
User Choose or enter a valid username for the threshold.
Identity Group Choose a valid identity group name for the threshold.
Device Name Choose a valid device name for the threshold.
Device IP Choose or enter a valid device IP address for the threshold.
Device Group Choose a valid device group name for the threshold.
Identity Store Choose a valid identity store name for the threshold.
Allowed Protocol Choose a valid allowed protocol name for the threshold. A-9 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Authentication Inactivity Define threshold criteria based on authentications that are inactive, modifying the following fields necessary. ISE Configuration Changes Define threshold criteria based on system diagnostics in the ISE instance. MAC Address This filter is available only for RADIUS authentications.
NAD Port This filter is available only for RADIUS authentications.
AuthZ Profile This filter is available only for RADIUS authentications.
AuthN Method This filter is available only for RADIUS authentications.
EAP AuthN This filter is available only for RADIUS authentications.
EAP Tunnel This filter is available only for RADIUS authentications.
Protocol Configure the protocol that you want to use for your threshold. T able A-7 Failed Authentications (continued) Option Description T able A-8 Authentication Inactivity Option Description ISE Instance Choose a valid instance for the threshold.
Device Choose a valid device for the threshold.
Protocol Choose the protocol for threshold. Inactive for Select one of the following options: • Hours—Number of hours, from 1 to 744.
• Days—Number of days, from 1 to 31.
T able A-9 ISE Configuration Changes Option Description Administrator Choose a valid administrator username for the threshold.
Object Name Enter the name of the object for the threshold.
Object Type Choose a valid object type for the threshold.
Change Select a administrative change for the threshold: • Any • Create—Includes “duplicate” and “edit” administrative actions.
• Update • Delete Filter ISE Instance Choose a valid ISE instance for the threshold. A-10 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor ISE System Diagnostics Define threshold criteria based on system diagnostics in the ISE instance. ISE Process Status Define rule criteria based on ISE process status.
T able A-10 ISE System Diagnostics Option Description Severity at and above Choose the severity level for the threshold. This setting captures the indicated severity level and those that are higher within the threshold: • Fatal • Error • Warning • Info • Debug Message Text Enter the message text for the threshold. Maximum character limit is 1024.
Filter ISE Instance Choose a valid ISE instance for the threshold.
T able A-11 ISE Process Status Option Description Monitor Processes ISE Database Adds the ISE database to the configuration.
ISE Database Listener Adds the ISE management to the configuration.
ISE Application server Adds the ISE runtime to the configuration.
ISE M&T Session Monitors this process. If this process goes down, an alarm is generated.
ISE M&T Log Collector Monitors this process. If this process goes down, an alarm is generated.
ISE M&T Alert Process Monitors this process. If this process goes down, an alarm is generated.
ISE M&T Log Processor Monitors this process. If this process goes down, an alarm is generated.
Filter ISE Instance Choose a valid ISE instance for the threshold. A-11 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor ISE System Health Define threshold criteria for ISE system health.
ISE AAA Health Define threshold criteria for ISE AAA Health.
T able A-12 ISE System Health Option Description Average over the past Select the amount of time, where minutes values are: 15, 30, 45, 60 CPU Enter the percentage of CPU usage. The valid range is from 1 to 100.
Memory Enter the percentage of memory usage (greater than or equal to the specified value). The valid range is from 1 to 100.
Disk I/O Enter the percentage of disk usage (greater than or equal to the specified value). The valid range is from 1 to 100.
Disk Space Used/opt Enter the percentage of /opt disk space (greater than or equal to the specified value). The valid range is from 1 to 100.
Disk Space Used/local disk Enter the percentage of local disk space (greater than or equal to the specified value). The valid range is from 1 to 100.
Disk Space Used/ Enter the percentage of the / disk space (greater than or equal to the specified value). The valid range is from 1 to 100.
Disk Space Used/tmp Enter the percentage of temporary disk space (greater than or equal to the specified value). The valid range is from 1 to 100.
Filter ISE Instance Choose a valid ISE instance.
T able A-13 ISE AAA Health Option Description Average over the past Select the amount of time, where minutes values are: 15, 30, 45, 60 RADIUS Throughput Enter the number of RADIUS transactions per second (lesser than or equal to the specified value). The valid range is from 1 to 999999.
RADIUS Latency Enter the number in milliseconds for RADIUS latency (greater than or equal to the specified value). The valid range is from 1 to 999999.
Filter ISE Instance Choose a valid ISE instance for the threshold. A-12 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Authenticated But No Accounting Start Define the threshold rule criteria for a specified number of authenticated sessions for a device IP Unknown NAD Define threshold criteria based on authentications that have failed because of an unknown NAD.
T able A-14 Authentication But No Accounting Start Option Description More than authenticated sessions in the past 15 minutes, where accounting start event has not been received for a Device IP —A count of authenticated sessions in the past 15 minutes.
Filter ISE Instance Choose a valid ISE instance.
Device IP Choose or enter a valid device IP address.
T able A-15 Unknown NAD Option Description Unknown NAD count Greater than in the past for a , where: • values can be any five-digit number greater than or equal to zero (0).
• values can be 1 to 1440 minutes, or 1 to 24 hours.
• value can be Minutes or Hours.
• values can be: – ISE Instance – Device IP Filter ISE Instance Choose a valid ISE instance.
Device IP Choose or enter a valid device IP address .
Protocol Select a protocol for the threshold. Valid options are: RADIUS A-13 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor External DB Unavailable Define threshold criteria based on an external database that ISE is unable to connect to. RBACL Drops Define the RBACL Drops threshold.
T able A-16 External DB Unavailable Option Description External DB Unavailable greater than in the past for a , where: • value can be Percent or Count.
• values can be any one of the following: – 0 to 99 for percent – 0 to 99999 for count • values can be 1 to 1440 minutes, or 1 to 24 hours.
• value can be Minutes or Hours.
• values can be: – ISE Instance – Identity Store Filter ISE Instance Choose a valid ISE instance.
Identity Group Choose a valid identity group name.
Identity Store Choose a valid identity store name.
Allowed Protocol Choose a valid allowed protocol name.
Protocol Select a protocol. Valid options are: RADIUS T able A-17 RBACL Drops Option Description RBACL drops Greater than in the past by a , where: • values can be any five-digit number greater than or equal to zero (0).
• values can be 1 to 1440 minutes, or 1 to 24 hours.
• value can be Minutes or Hours.
• values can be: – NAD – SGT – DGT – DST_IP A-14 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor NAD-Reported AAA Downtime Define threshold criteria based on the AAA downtime that a network access device reports.
Schedules Click Monitor > Alarms > Schedules to establish schedules for alarm rules.
Filter Device IP Choose or enter a valid device IP address.
SGT Choose or enter a valid source group tag.
DGT Choose or enter a valid destination group tag.
Destination IP Choose or enter a valid destination IP address.
T able A-17 RBACL Drops (continued) Option Description T able A-18 NAD-Reported AAA Downtime Option Description AAA down Greater than in the past by a , where: • values can be any five-digit number greater than or equal to zero (0).
• values can be 1 to 1440 minutes, or 1to 24 hours.
• value can be Minutes or Hours.
• values can be: – Device IP – Device Group Filter ISE Instance Choose a valid ISE instance.
Device IP Choose or enter a valid device IP address.
Device Group Choose a valid device group name.
T able A-19 Schedules Option Description Filter Enter a text string on which to filter for a schedule.
Go Click to filter on the text string.
Clear Filter Click to clear the filter field.
Name The name of the schedule. Click the name link to view and/or edit schedule details.
Description Description of the schedule. A-15 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Create Click to create a new schedule. Specify the following: • Name • Description • Schedule—Click a square to select/deselect that hour.
• Select All—Click to select all hours.
• Clear All—Click to clear all selected hours.
• Undo All—Click to clear all fields on this page.
• Submit—Click to create the schedule.
• Cancel—Click to cancel to exit without saving the schedule.
Edit Select a schedule and click Edit to make changes to the schedule. Edit options are the same as the Create options.
Delete Select a schedule and click Delete to delete the schedule. Confirm you choice by clicking Yes in the Confirm Deletion dialog, or No to exit without deleting the schedule.
T able A-19 Schedules (continued) Option Description A-16 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Reports This section covers the following user interface elements: • Catalog, page A-16 • Favorites, page A-25 • Data Formatting, page A-25 • Filters, page A-39 Catalog Select Monitor > Reports > Catalog. Preconfigured system reports are grouped in categories, as shown in Report Type by Category, page A-16.
Report Type by Category T able A-20 Report T ype by Category Report Name Description Logging Category AAA Protocol AAA diagnostics Provides AAA diagnostic details based on severity for a selected time period.
Policy diagnostics, identity stores diagnostics, authentication flow diagnostics, RADIUS diagnostics Authentication Trend Provides RADIUS authentication summary information for a selected time period; along with a graphical representation.
Passed authentications, failed attempts RADIUS Accounting Provides user accounting information based on RADIUS for a selected time period.
RADIUS accounting RADIUS Authentication Provides RADIUS authentication details for a selected time period.
Passed authentications, failed attempts Allowed Protocol Allowed Protocol Authentication Summary Provides RADIUS authentication summary information for a particular allowed protocol for a selected time period; along with a graphical representation.
Passed authentications, failed attempts Top N Authentications By Allowed Protocol Provides the top N passed, failed, and total authentication count for RADIUS authentications with respect to the allowed protocol for a selected time period.
Passed authentications, failed attempts A-17 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Policy Service Instance ISE Administrator Logins Provides access-related events for administrators that includes login, logout, events, and reasons for failed login attempts.
Administrative and operational audit ISE Instance Authentication Summary Provides RADIUS authentication summary information for a particular ISE instance for a selected time period; along with a graphical representation.
This report could take several minutes to run depending on the number of records in the database.
Note When you reload this report, if rate of incoming syslog messages is around 150 messages per second or more, the total number of passed and failed authentications that appear above the graph and the passed and failed authentication count that is displayed in the table do not match.
Passed authentications, failed attempts ISE Configuration Audit Provides all the configuration changes done in ISE by the administrator for a selected time period.
Administrative and operational audit ISE Health Summary Provides the CPU, memory utilization, RADIUS and throughput (in tabular and graphical formats) and also process status, process downtime, and disk space utilization for a particular ISE instance in a selected time period.
System statistics ISE Operations Audit Provides all the operational changes done in ISE by the administrator for a selected time period.
Administrative and operational audit T able A-20 Report T ype by Category (continued) Report Name Description Logging Category A-18 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor ISE System Diagnostics Provides system diagnostic details based on severity for a selected time period.
Internal Operations Diagnostics, distributed management, administrator authentication and authorization Top N Authentication by ISE Instance Provides the top N passed, failed, and total authentication count for RADIUS protocol with respect to a particular ISE instance for a selected time period.
Passed authentications, failed attempts User Change Password Audit Provides the username of the internal user, identity store name, name of the ISE instance, and time when the user password was changed. Helps to keep track of all changes made to internal user passwords across all ISE interfaces.
Administrative and operational audit Endpoint Endpoint MAC Authentication Summary Provides the RADIUS authentication summary information for a particular MAC or MAB for a selected time period; along with a graphical representation.
Passed authentications, failed attempts Endpoint Profiler Summary Provides the endpoint profiler summary information for a particular MAC address for a selected time period.
Profiler Top N Authentications By Endpoint Calling Station ID Provides the top N passed, failed, and total authentication count with respect to endpoint calling station IDs.
Passed authentications, failed attempts Top N Authentications By Endpoint MAC Address Provides the top N passed, failed, and total authentication count for RADIUS protocol with respect to MAC or MAB address for a selected time period.
Passed authentications, failed attempts Top N Authentications By Machine Provides the top N passed, failed, and total authentication count for RADIUS protocol with respect to machine information for a selected time period.
Passed authentications, failed attempts Failure Reason T able A-20 Report T ype by Category (continued) Report Name Description Logging Category A-19 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Authentication Failure Code Lookup Provides the description and the appropriate resolution steps for a particular failure reason.
— Failure Reason Authentication Summary Provides the RADIUS authentication summary information for a particular failure reason; along with a graphical representation for a selected time period.
Failed attempts Top N Authentications By Failure Reason Provides the top N failed authentication count for RADIUS protocols with respect to Failure Reason for a selected time period.
Failed attempts Network Device AAA Down Summary Provides the number of AAA unreachable events that a NAD logs within a selected time period.
— Network Device Authentication Summary Provides the RADIUS authentication summary information for a particular network device for a selected time period, along with the graphical representation.
Passed authentications, failed attempts Network Device Log Messages Provides you the log information of a particular network device, for a specified time period.
— Session Status Summary Provides the port sessions and status of a particular network device obtained by SNMP.
— Top N AAA Down By Network Device Provides the number of AAA down events encountered by each of the network devices.
— Top N Authentications by Network Device Provides the top N passed, failed, and total authentication count for RADIUS with respect to network device for a selected time period.
Passed authentications, failed attempts User T able A-20 Report T ype by Category (continued) Report Name Description Logging Category A-20 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Client Provisioning Provides a summary of successful and unsuccessful client provisioning evaluation and download events, displayed according to the associated User ID.
Client provisioning Guest Provisioning Provides session (login and log out) information for selected guests over a specified time period.
Passed authentications, RADIUS accounting Guest Activity Provides guest information for a selected time period.
Passed authentications Guest Sponsor Summary Provides sponsor information along with a graphical representation, for a selected time period.
Passed authentications Top N Authentications By User Provides top N passed, failed, and total authentication count for RADIUS with respect to users for a selected time period.
Passed authentications, failed attempts Unique Users Provides the count for the number of unique users.
User User Authentication Summary Provides RADIUS authentication summary information for a particular user for a selected time period; along with the graphical representation.
Passed authentications, failed attempts Security Group Access RBACL Drop Summary Provides a summary of RBAC drop events. — SGT Assignment Summary Provides a summary of SGT assignments for a selected time period.
Passed authentications Top N RBACL Drops By Destination Provides the top N RBACL drop event count with respect to destination for a selected time period.
— Top N RBACL Drops By User Provides the top N RBACL drop event count with respect to the user for a selected time period.
— Top N SGT Assignments Provides the top N SGT assignment count for a selected time period.
Passed authentications Session Directory T able A-20 Report T ype by Category (continued) Report Name Description Logging Category A-21 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Report Type Page Select a category name from the Reports navigation pane. The Reports Type page appears.
RADIUS Active Sessions Provides information on RADIUS authenticated, authorized, and started sessions.
Dynamically control active RADIUS sessions. Send a reauthenticate or disconnect request to a NAD to: • Reauthenticate the user • Terminate the session • Terminate the session and restart the port • Terminate the session and shut down the port Passed authentications, RADIUS accounting RADIUS Session History Provides a summary of RADIUS session history, such as total authenticated, active, and terminated sessions and total and average session duration and throughput for a selected time period.
Passed authentications, RADIUS accounting RADIUS Terminated Sessions Provides all the RADIUS terminated session information for a selected time period.
Passed authentications, RADIUS accounting Posture Posture Detail Assessment Provides the posture authentication summary information for a particular user for a selected time period.
Posture and Client Provisioning Audit, Posture and Client Provisioning Diagnostics Posture Trend Provides the count of passed or failed, as well as status information for a particular policy for a selected time period; along with the graphical representation.
Posture and Client Provisioning Audit, Posture and Client Provisioning Diagnostics T able A-20 Report T ype by Category (continued) Report Name Description Logging Category T able A-21 Report T ype Page Option Description Report Name A list of available report names for the category you selected.
Type The type of report. A-22 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Report Name Page Not all options listed in the following table are used in all reports.
Modified At The time the report was last modified by an administrator, in the format Ddd Mmm dd hh:mm:ss timezone yyyy, where: • Ddd = Sun, Mon, Tue, Wed, Thu, Fri, Sat.
• Mmm = Jan, Feb, Apr, May, Jun, Jul, Aug, Sep, Oct, Nov, Dec.
• dd = A two-digit numeric representation of the day of the month, from 01 to 31.
• hh = A two-digit numeric representation of the hour of the day, from 00 to 23.
• mm = A two-digit numeric representation of the minute of the hour, from 00 to 59.
• ss = A two-digit numeric representation of the second of the minute, from 00 to 59.
• timezone = The time zone.
• yyyy = A four-digit representation of the year.
Filter Enter a text string to search for on in the field and click Filter.
T able A-21 Report T ype Page Option Description T able A-22 Report Name Page Option Description User Enter a username or click Select to enter a valid username on which to configure your threshold.
MAC Address Enter a MAC address or click Select to enter a valid MAC address on which to run your report.
Identity Group Enter an identity group name or click Select to enter a valid identity group name on which to run your report.
Device Name Enter a device name or click Select to enter a valid device name on which to run your report.
Device IP Enter a device IP address or click Select to enter a valid device IP address on which to run your report.
Device Group Enter a device group name or click Select to enter a valid device group name on which to run your report.
Allowed Protocol Enter an allowed protocol name or click Select to enter a valid allowed protocol name on which to run your report Identity Store Enter an identity store name or click Select to enter a valid identity store name on which to run your report.
ISE Instance Enter an ISE instance name or click Select to enter a valid ISE instance name on which to run your report. A-23 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Failure Reason Enter a failure reason name or click Select to enter a valid failure reason name on which to run your report.
Protocol Use the drop down list box to select which protocol on which you want to run your report. RADIUS is the only option at this time.
Authentication Status Use the drop down list box to select which authentication status on which you want to run your report. Valid options are: • Pass Or Fail • Pass • Fail Radius Audit Session ID Enter the RADIUS audit session identification name on which you want to run a report.
ISE Session ID Enter the ISE session identification name on which you want to run a report.
Severity Use the drop down list box to select the severity level on which you want to run a report. This setting captures the indicated severity level and those that are higher within the threshold. Valid options are: • Fatal • Error • Warning • Info • Debug End Point IP Address Enter the end point IP address on which you want to run a report.
Command Accounting Only Check the check box to enable your report to run for command accounting.
Top Use the drop down list box to select the number of top (most frequent) authentications by allowed protocol on which you want to run your report. Valid options are: • 10 • 50 • 100 • 500 • 1000 • All T able A-22 Report Name Page Option Description A-24 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor By Use the drop down list box to select the type of authentications on which you want to run your report. Valid options are: • Passed Authentications • Failed Authentications • Total Authentications Administrator Name Enter the administrator username, or click Select to select the administrator username, for which you want to run your report.
Object Type Enter a valid object type on which you want to run your report.
Object Name Enter the name, or click Select to select the object name, of the object on which you want to run your report.
Authorization Status Use the drop down list box to select which authentication status on which you want to run your report. Valid options are: • Pass Or Fail • Pass • Fail Time Range Use the drop down list box to select the time range on which you want to run your report. Valid options are: • Last Hour (for the ISE Health Summary report only) • Today • Yesterday • Last 7 Days • Last 30 Days • Custom—You must configure a Start Date and End Date, or a Day.
Note Some options are not valid for some Time Range entries of the various reports.
Start Date Enter a date, or click the date selector icon to select a start date for running your report.
End Date Enter a date, or click the date selector icon to select an end date for running your report.
Day Enter a date, or click the date selector icon to select an end date for running your report.
Clear Click to delete the contents of an associate text box.
Run Click to run the report for which you have made selections.
T able A-22 Report Name Page Option Description A-25 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Favorites Select Monitor > Reports > Favorites to display a list of favorite reports. Favorites allows you to bookmark frequently used reports by saving them as favorite reports.
For a list of all available reports, see Report Type by Category, page A-16.
Favorites Page Data Formatting Data Types and Formats T able A-23 Favorites Page Option Description Favorite Name The name of the favorites report. Click to open a summary of an associated report.
Report Name The report name associated with a Catalog (Report) type.
Report Type The general category name associated with the report.
T able A-24 Data T ypes and Formats Data type Option Description Date and Time Unformatted The data retains the default format set by the template or theme.
General Date June 5, 2006 12:00:00 AM GMT +00:00 Long Date June 5, 2006 Medium Date Jun 5, 2006 Short Date 6/5/06 Long Time 12:00:00 AM GMT +00:00 Medium Time 12:00:00 AM Short Time 12:00 Custom The format depends on a format code you type. For example, typing yyyy/mm results in 2006/10. You learn more about custom formatting later in this chapter. A-26 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Custom Number Format Patterns Number Unformatted The number retains the default format set by the template or theme.
General Number 6066.88 or 6067, depending on the decimal and thousands separator settings Currency $6,067.45 or ¥6067, depending on the locale and optional settings Fixed 6067 or 6,067 or 6067.45, depending on optional settings Percent 45% or 45.8%, depending on optional settings Scientific 2E04 or 2.67E04, where the number after the E represents the exponent of 10, depending on optional settings. For example, 2.67E04 means 2.67 multiplied by 10 raised to the fourth power.
Custom The format depends on a format code you type. For example, typing #,### results in a format with a comma as a thousands separator and no decimal points. You learn more about custom formats later in this chapter.
String Unformatted The string retains the default format set by the template or theme.
Uppercase The string displays in all uppercase, for example GREAT NEWS.
Lowercase The string displays in all lowercase, for example great news.
Custom The format depends on the format code you type. Use custom formatting for postal codes, telephone numbers, and other data that does not match standard formats.
T able A-25 Custom Number Format Patterns Format pattern Data in the data set Result of formatting 0000.00 12.5 124.5 1240.553 0012.50 0124.50 1240.55 #.000 100 100.25 100.2567 100.000 100.250 100.257 $#,### 2000.00 20000.00 $2,000 $20,000 ID # 15 ID 15 T able A-24 Data T ypes and Formats (continued) Data type Option Description A-27 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Symbols for Defining Custom String Formats Results of Custom String Format Patterns Results of Custom Date Formats Symbol Description @ Character placeholder. Each @ character displays a character in the string. If the string has fewer characters than the number of @ symbols that appear in the format pattern, spaces appear. Placeholders are filled from right to left, unless you specify an exclamation point (!) at the beginning of the format pattern. & Same as @, except that if the string has fewer characters, spaces do not appear. !
Specifies that placeholders are to be filled from left to right. > Converts string characters to uppercase.
< Converts string characters to lowercase.
T able A-26 Results of Custom String Format Patterns Format pattern Data in the data source Results of formatting (@@@) @@@-@@@@ 6175551007 5551007 (617) 555-1007 ( ) 555-1007 (&&&) &&&-&&&& 6175551007 5551007 (617) 555-1007 () 555-1007 !(@@@) @@@-@@@@ 6175551007 5551007 (617) 555-1007 (555) 100-7 !(&&&) &&&-&&&& 6175551007 5551007 (617) 555-1007 (555) 100-7 !(@@@) @@@-@@@@ + ext 9 5551007 (555) 100-7 + ext 9 !(&&&) &&&-&&&& + ext 9 5551007 (555) 100-7 + ext 9 >&&&-&&&&&-&& D1234567xy D12-34567-XY <&&&-&&&&&-&& D1234567xy d12-34567-xy T able A-27 Results of Custom Date Formats Format Result of formatting MM-dd-yy 04-15-06 E, M/d/yyyy Fri, 4/15/2006 MMM d Apr 15 MMMM April yyyy 2006 W 3 (the week in the month) A-28 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Supported Calculation Functions w 14 (the week in the year) D 105 (the day in the year) T able A-28 Supported Calculation Functions Function Description Example of use ABS(num) Displays an absolute value for the data in a column.
ABS([TemperatureCelsius]) ADD_DAY (date, daysToAdd) Adds a specified number of days to a date value and displays the result as a date value. ADD_DAY([ClosingDate], 30) ADD_HOUR (date, hoursToAdd) Adds a specified number of hours to a time value and displays the result as a time value.
ADD_HOUR([OpenHour], 8) ADD_MINUTE (date, minutesToAdd) Adds a specified number of minutes to a time value and displays the result as a time value.
ADD_MINUTE([StartTime], 60) ADD_MONTH (date, monthsToAdd) Adds a specified number of months to a date value and displays the result as a date value.
ADD_MONTH([InitialRelease], 2) ADD_QUARTER (date, quartersToAdd) Adds a specified number of quarters to a date value. ADD_QUARTER([ForecastClosing], 2) ADD_SECOND (date, secondsToAdd) Adds a specified number of seconds to a time value.
ADD_SECOND([StartTime], 30) ADD_WEEK (date, weeksToAdd) Adds a specified number of weeks to a date value and displays the result as a date value.
ADD_WEEK([askByDate], 4) ADD_YEAR (date, yearsToAdd) Adds a specified number of years to a date value.
ADD_YEAR([HireDate], 5) AND Combines two conditions and returns records that match both conditions. For example, you can request records from customers who spend more than $50,000 a year and also have a credit rank of A.
This function is used to connect clauses in an expression and does not take arguments.
AVERAGE(expr) Displays an average value for the column. AVERAGE([CostPerUnit]) AVERAGE (expr, groupLevel) Displays the average value at the specified group level.
AVERAGE([TotalCost], 2) T able A-27 Results of Custom Date Formats Format Result of formatting A-29 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor BETWEEN(value, upperBound, lowerBound) For a specified column, displays true if a value is between two specified values and false otherwise. String values and date or time values must be enclose in quotation marks. For dates and times, use the short date and short time formats.
BETWEEN([PostalCode], 11209, 12701) BETWEEN([ReceiptDate], “10/01/06”, “12/31/06”) CEILING (num, significance) Rounds a number up, away from 0, to the nearest specified multiple of significance. For data that has been converted from a double or float to an integer, displays the smallest integer that is greater than or equal to the float or double.
CEILING([PortfolioAverage], 1) COUNT( ) Counts the rows in a table.
COUNT( ) COUNT(groupLevel) Counts the rows at the specified group level.
COUNT(2) COUNTDISTINCT(expr) Counts the rows that contain distinct values in a table.
COUNTDISTINCT([CustomerID]) COUNTDISTINCT([Volume]2) COUNTDISTINCT (expr, groupLevel) Counts the rows that contain distinct values at the specified group level.
COUNTDISTINCT([CustomerID], 3) DAY(date) Displays the number of a day in the month, from 1 to 31, for a date-and-time value.
DAY([forecastShipping]) DIFF_DAY(date1, date2) Displays the difference between two date values, in the number of days.
DIFF_DAY([checkoutDate], [returnDate]) DIFF_HOUR(date1, date2) Displays the difference between two time values, in the number of hours.
DIFF_HOUR([StartTime],[Finish Time]) DIFF_MINUTE (date1, date2) Displays the difference between two time values, in the number of minutes.
DIFF_MINUTE([StartTime], [FinishTime]) DIFF_MONTH (date1, date2) Displays the difference between two date values, in the number of months.
DIFF_MONTH([askByDate], [shipByDate]) DIFF_QUARTER (date1, date2) Displays the difference between two date values, in the number of quarters.
DIFF_QUARTER([PlanClosing], [ActualClosing]) DIFF_SECOND (date1, date2) Displays the difference between two time values, in the number of seconds.
DIFF_SECOND([StartTime], [FinishTime]) T able A-28 Supported Calculation Functions Function Description Example of use A-30 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor DIFF_WEEK(date1, date2) Displays the difference between two weeks as a number. DIFF_WEEK([askByDate], [shipByDate]) DIFF_YEAR(date1, date2) Displays the difference between two years as a number.
DIFF_YEAR([HireDate], [TerminationDate]) false The Boolean false. This function is used in expressions to indicate that an argument is false.
In the following example, false indicates that the second argument, ascending, is false and therefore the values should be returned in descending order.
RANK([Score], false) FIND(strToFind, str) Displays the index of the first occurrence of specified text. The index is zero-based. The search is case sensitive and the search string cannot include wildcards. The value in the strToFind argument must be enclosed in quotation marks. FIND("HQ", [OfficeName]) FIND(strToFind, str, startPosition) Similar to FIND(strToFind, str) but supports providing a start position for the search. The index is zero-based.
FIND("HQ", [OfficeName], 3) FIRST(expr) Places the first value that appears in a specified column into the calculated column. This function supports viewing a row-by-row comparison against a specific value.
FIRST([customerID]) FIRST(expr, groupLevel) Displays the first value that appears in the specified column at the specified group level.
FIRST([customerID], 3) IF(condition, doIfTrue, doIfFalse) Displays the result of an If...Then...Else statement.
IF([purchaseVolume] >5 , 7 , 0) where • [purchaseVolume] is the column name and >5 is the test condition.
• 7 is the value to place in the new column if the condition is true.
• 0 is the value to place in the new column if the condition is false.
T able A-28 Supported Calculation Functions Function Description Example of use A-31 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor IN(value, check) Displays true if a data row contains a value specified by the check argument and false otherwise. String values and date or time values must be enclosed in quotation marks. For dates and times, use the short date and short time formats for your locale.
IN([custID], 101) IN([city], "New Haven") IN([FinishTime], "16:09") IN(value, check1, ..., checkN) Displays true if a data row contains any value specified by the check argument list and false otherwise. String values and date or time values must be enclosed in quotation marks. For dates and times, use the short date and short time formats for your locale.
IN([city], “New Haven”, “Baltimore”, “Cooperstown”) IN([ShipDate], “05/01/06”, “05/10/06”, “05/15/06”) ISBOTTOMN(expr, n) Displays true if the value is within the lowest n values for the expression, and false otherwise.
ISBOTTOMN([OrderTotals], 50) ISBOTTOMN (expr, n, groupLevel) Displays true if the value is within the lowest n values for the expression at the specified group level, and false otherwise.
ISBOTTOMN([OrderTotals], 50, 2) ISBOTTOMNPERCENT (expr, percent) Displays the lowest n percentage.
ISBOTTOMNPERCENT([Sales Total], 5) ISBOTTOMNPERCENT (expr, percent, groupLevel) Displays the lowest n percentage for the expression at the specified group level.
ISBOTTOMNPERCENT([Sales Total], 5, 3) ISNULL(value) Displays true if a row does not display a value. Displays false if a row displays a value.
ISNULL([DepartmentName]) ISTOPN(expr, n) Displays true if the value is within the highest n values for the expression, and false otherwise.
ISTOPN([OrderTotals], 10) ISTOPN(expr, n, groupLevel) Displays true if the value is within the highest n values for the expression at the specified group level, and false otherwise.
ISTOPN([OrderTotals], 10, 3) ISTOPNPERCENT(expr, percent) Displays true if the value is within the highest n percentage, and false otherwise.
ISTOPNPERCENT([SalesTotals], 5) T able A-28 Supported Calculation Functions Function Description Example of use A-32 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor ISTOPNPERCENT(expr, percent, groupLevel) Displays true if the value is within the highest n percentage values for the expression at the specified group level, and false otherwise.
ISTOPNPERCENT([SalesTotals], 5, 3) LAST(expr) Displays the last value in a specified column.
LAST([FinishTime]) LAST(expr, groupLevel) Displays the last value for the expression at the specified group level.
LAST([FinishTime], 3) LEFT(str) Displays the character at the left of the specified string.
LEFT([city]) LEFT(str, n) Displays the specified number of characters in a column’s string, counting from the left.
LEFT([city], 3) LEN(str) Displays the length of a string, including spaces and punctuation marks.
LEN([Description]) LIKE(str) Displays true if the values match, and false otherwise. Use SQL syntax to specify the string pattern.
The following rules apply: • Literal pattern characters must match exactly. LIKE is case-sensitive.
• A percent character (%) matches zero or more characters.
• An underscore character () matches any single character.
• Escape a literal percent, underscore, or backslash character () with a backslash character.
LIKE([customerName], "D%") LIKE([quantityOrdered], "2") LOWER(str) Displays the string in a specified column in lowercase.
LOWER([cityName]) MAX(expr) Displays the highest value in the specified column.
MAX([OrderTotal]) MAX(expr, groupLevel) Displays the highest value for the expression at the specified group level.
MAX([OrderTotal], 2) T able A-28 Supported Calculation Functions Function Description Example of use A-33 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor MEDIAN(expr) Displays the median value in a specified column.
MEDIAN([HomePrices]) MEDIAN (expr, groupLevel) Displays the median value for the expression at the specified group level.
MEDIAN([HomePrices], 2) MIN(expr) Displays the lowest value in the specified column.
MIN([OrderTotal]) MIN(expr, groupLevel) Displays the lowest value for the expression at the specified group level.
MIN([OrderTotal], 1) MOD(num, div) Displays the remainder after a number is divided by a divisor. The result has the same sign as the divisor.
MOD([Salary], 12) MONTH(date) Displays the name of the month for a specified date-and-time value.
MONTH([ForecastShipDate]) MONTH(date, option) Displays the month of a specified date-and-time value, in one of three optional formats: • 1 - Displays the month number of 1 through 12.
• 2 - Displays the complete month name in the user’s locale.
• 3 - Displays the abbreviated month name in the user’s locale.
MONTH([Semester], 2) MOVINGAVERAGE (expr, window) Displays an average value over a specified window, such as an average price or volume over a number of days.
MOVINGAVERAGE([Price], [Days]) NOTNULL(value) For a specified column, displays true if a data value is not empty. Displays false if a data value is empty.
NOTNULL([DepartmentID]) NOW( ) Displays the current time stamp. NOW([PastDueDate]) OR The logical OR operator.
This function is used to connect clauses in an expression and does not take arguments.
T able A-28 Supported Calculation Functions Function Description Example of use A-34 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor PERCENTILE(expr, pct) Displays a percentile value, a value on a scale of 100 that indicates the percent of a distribution that is equal to or below the specified value. Valid pct argument ranges are 0 to 1. 0 returns the minimum value of the series. 1 returns the maximum value of the series.
PERCENTILE([Rank], 1) PERCENTILE (expr, pct, groupLevel) Displays a percentile value for the expression at the specified group level. Valid pct argument ranges are 0 to 1. 0 returns the minimum value of the series. 1 returns the maximum value of the series.
PERCENTILE([Income], 60, 1) PERCENTRANK(expr) Displays the percentage rank of a value.
PERCENTRANK([TestScores]) PERCENTRANK(expr, groupLevel) Displays the percentage rank of a value at the specified group level.
PERCENTRANK([TestScores], 2) PERCENTSUM(expr) Displays a value as a percentage of a total.
PERCENTSUM([OrderTotals]) PERCENTSUM(expr, groupLevel) Displays a value as a percentage of a total at the specified group level.
PERCENTSUM([OrderTotals], 3) QUARTER(date) Displays the quarter number, from 1 through 4, of a specified date-and-time value.
QUARTER([ForecastCloseDate]) QUARTILE(expr, quart) Displays the quartile value, where the quart argument is an integer between 0 and 4.
QUARTILE([OrderTotal], 3) QUARTILE (expr, quart, groupLevel) Displays the quartile value for the expression at the specified group level, where the quart argument is an integer between 0 and 4.
QUARTER([OrderTotal], 2, 3) RANK(expr) Displays the rank of a number, string, or date-and-time value, starting at 1. Duplicate values receive identical rank but the duplication does not affect the ranking of subsequent values.
RANK([AverageStartTime]) T able A-28 Supported Calculation Functions Function Description Example of use A-35 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor RANK(expr, ascending, groupLevel) Displays the rank of a number, string, or date-and-time value in either ascending or descending order, at the specified group level. To display values in ascending order, use true as the second argument. To display values in descending order, use false as the second argument.
RANK([Score], false, 3) RANK([Score], true, 2) RIGHT(str) Displays the character at the right of a string.
RIGHT([name]) RIGHT(str, n) Displays the specified number of characters in a string, counting from the right.
RIGHT([name], 3) ROUND(num) Rounds a number.
ROUND([SalesTarget]) ROUND(num, dec) Rounds a number to the specified number of digits. The default value for dec is 0.
ROUND([StockValue], 2) ROUNDDOWN(num) Rounds a number down.
ROUNDDOWN([StockPrice]) ROUNDDOWN(num, dec) Rounds a number down, away from 0, to the specified number of digits. The default value for dec is 0.
ROUNDDOWN([StockPrice], 2) ROUNDUP(num) Rounds a number up.
ROUNDUP([TotalValue]) ROUNDUP(num, dec) Rounds a number up, away from 0, to the specified number of digits. The default value for dec is 0.
ROUNDUP([TotalValue], 2) RUNNINGSUM(expr) Displays a running total, adding the values in successive data rows.
RUNNINGSUM([StockValue]) SEARCH(pattern, str) Case-insensitive search function that can use wildcard characters.
An asterisk ( ) matches any sequence of characters, including spaces.
A question mark ( ? ) matches any single character.
The following search yields New York, New Haven, and so on from the City column: SEARCH([CustomerData:city], "new") SEARCH (pattern, str, startPosition) Searches for a specified pattern in a string, starting at a specified position in the string. A case-insensitive search function that can use wildcard characters. SEARCH([Location], "new", 1) T able A-28 Supported Calculation Functions Function Description Example of use A-36 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor SQRT(num) Displays the square root of a value.
SQRT([PrincipalValue]) STDEV(expr) Displays the standard deviation.
STDEV([PurchaseFrequency]) SUM(expr) Displays the sum of two specified values.
SUM([Price]+[Tax]) TODAY( ) Displays a time stamp value equal to midnight of the current date.
TODAY([DueDate]) TRIM(str) Displays a string with all leading and trailing blank characters removed. Also removes all consecutive blank characters. Leading and trailing blanks can be spaces, tabs, and so on.
TRIM([customerName]) TRIMLEFT(str) Displays a string with all leading blanks removed. Does not remove consecutive blank characters.
TRIMLEFT([PortfolioName]) TRIMRIGHT(str) Displays a string with all trailing blanks removed. Does not remove consecutive blank characters.
TRIMRIGHT([Comments]) true The Boolean true. This function is used in expressions to indicate that an argument is true.
In the following example, true indicates that the second argument, ascending, is true and therefore the values should be returned in ascending order.
RANK([Score], true) UPPER(str) Displays a string in a specified column in all uppercase.
UPPER([cityName]) UPPER("new haven") VAR(expr) Displays a variance for the specified expression.
VAR([EstimatedCost]) WEEK(date) Displays the number of the week, from 1 through 52, for a date-and-time value.
WEEK([LeadQualifyingDate]) T able A-28 Supported Calculation Functions Function Description Example of use A-37 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Supported Operator Formats WEEKDAY(date, option) Displays the day of the week in one of the following format options: • 1 - Returns the day number, from 1 (Sunday) through 7 (Saturday). 1 is the default option.
• 2 - Returns the day number, from 1 (Monday) through 7 (Sunday).
• 3 - Returns the day number, from 0 (Monday) through 6 (Sunday).
• 4 - Returns the weekday name according to the user’s locale.
• 5 - Returns the abbreviated weekday name according to the user’s locale.
WEEKDAY([DateSold], 4) WEIGHTEDAVERAGE (value, weight) Displays a weighted average of a specified value.
WEIGHTEDAVERAGE([Score], weight) YEAR(date) Displays the four-digit year value for a date-and-time value.
YEAR([ClosingDate]) T able A-29 Supported Operator Formats Operator Description x + y Addition of numeric values x - y Subtraction of numeric values x y Multiplication of numeric values x / y Division of numeric values x% Percentage of a numeric value x & y Concatenation of string values x = y Test for equality of two values x > y Tests whether x is greater than y x < y Tests whether x is less than y x >= y Tests whether x is greater than or equal to y x <= y Tests whether x is less than or equal to y x <> y Tests whether x is not equal to y T able A-28 Supported Calculation Functions Function Description Example of use A-38 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Aggregate Function Formats x AND y Tests for values that meet both condition x and condition y x OR y Tests for values that meet either condition x or condition y NOT x Tests for values that are not x T able A-30 Aggregate Function Formats Aggregate functions Description Average Calculates the average value of a set of data values.
Count Counts the data rows in the column.
Count Value Counts distinct values in the column.
First Returns the first value in the column.
Last Returns the last value in the column.
Max Returns the highest value in the column.
Median Returns the median value in the column.
Min Returns the lowest value in the column.
Mode Returns the most frequently-occurring value in the column.
Quartile Returns one of four equal-sized sets of data, based on the rank you select. For example, you can request the first quartile to get the top quarter of the data set or the fourth quartile to get the fourth quarter of the data set.
Standard Deviation Returns the standard deviation, the square root of the variance.
Sum Adds the values in the column.
Variance Returns a value that indicates the spread around a mean or expected value.
Weighted average Returns the weighted average of a numeric field over a set of data rows. In a weighted average, some numbers carry more importance, or weight, than others.
T able A-29 Supported Operator Formats Operator Description A-39 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Filters Conditions for Filters T able A-31 Conditions for Filters Condition Description Any Of Returns any of the values you specify.
Between Returns values that are between two specified values. When you select Between, a second Value field appears for the second default value. Bottom N Returns the lowest n values in the column.
Bottom Percent Returns the lowest n percent of values in the column.
Equal to Returns values that are equal to a specified value.
Greater Than Returns values that are greater than a specified value.
Greater Than or Equal to Returns values that are greater than or equal to a specified value.
Is False In a column that evaluates to true or false, returns data rows that contain false values. Is Not Null Returns data rows that contain values. Is Null Returns data rows that do not contain values. Is True In a column that evaluates to true or false, returns data rows that contain true values. Less Than Returns values that are less than another value.
Less Than or Equal to Returns values that are less than or equal to another value.
Like Returns strings that match all or part of the specified string. % matches zero or more characters. _ matches one character.
Not Between Returns values that are not between two specified values. When you select Not Between, a second Value field appears for the second default value.
Not Equal to Returns values that are not equal to another value.
Not Like Returns strings that do not match all or part of the specified string. % matches zero or more characters. _ matches one character.
Top N Returns the top n values in the column.
Top Percent Returns the top n percent of values in the column. A-40 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Filter Condition Examples T able A-32 Filter Condition Examples Type of filter condition Description Examples of instructions to data source Comparison Compares the value of one expression to the value of another expression using: • Equal to • Not Equal to • Less Than • Less Than or Equal to • Greater Than • Greater Than or Equal to quantity = 10 custName = 'Acme Inc.' custName > 'P' custState <> 'CA' orderDate > {d '2005-06-30'} Range Tests whether the value of an expression falls or does not fall within a range of values using Between or Not Between. The test includes the endpoints of the range.
price BETWEEN 1000 AND 2000 custName BETWEEN 'E' AND 'K' orderDate BETWEEN {d '2005-01-01'} AND {d '2005-06-30'} Membership Tests whether the value of an expression matches one value in a set of values using Any Of.
officeCode IN (101,103,104) itemType IN ('sofa', 'loveseat', 'endtable', 'clubchair') orderDate IN ({d '2005-10-10'}, {d '2005-10-17'}) Pattern-matching Tests whether the value of a string field matches or does not match a specified pattern using Like or Not Like. % matches zero or more characters. _ matches one character.
custName LIKE 'Smith%' custName LIKE 'Smiths_n' custState NOT LIKE 'CA%' Null value Tests whether a field has or does not have a null, or missing, value using Is Null or Is Not Null.
manager IS NULL shipDate IS NULL shipDate IS NOT NULL A-41 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Report Context Menus Use context menus as shortcuts to performing data formatting and organizing tasks. To bring up a context menu, right click an element in a report. The context menu options that are displayed are unique to the element selected.
T able A-33 Report Context Menus Option Description Aggregation Opens a dialog box that supports creating an aggregate row for this column. Alignment Opens a submenu that contains: • Align Left. Aligns the column data to the left.
• Align Center. Centers the column data.
• Align Right. Aligns the column data to the right.
Calculation Opens a submenu that supports creating a calculated column based on this column.
Chart Opens a submenu that supports inserting a chart.
Column Opens a submenu that contains: • Delete Column. Deletes the selected column.
• Reorder Columns. Opens a dialog box that supports changing the order of columns in the report design.
• Column Width. Opens the Column Properties dialog box, which supports setting the column width.
• Do Not Repeat Values. Suppresses consecutive duplicate data values in a column. If the column is already set to Do Not Repeat Values, this menu item changes to Repeat Values.
Data Fields Opens a dialog box that displays the report columns. Supports adding or removing data fields.
Filter Opens a submenu that contains: • Filter. Opens a dialog box that supports creating filters based on this column.
• Top or Bottom N. Opens a dialog box that supports displaying the highest or lowest n values or the highest or lowest n percent in the column.
Format Data Opens a dialog box that supports formatting the data type. For example, if the column contains numeric data, the Number column format dialog box opens and you can format the data as currency, percentages, and so on. A-42 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Group Opens a submenu that contains: • Add Group. Creates a group based on this column. When you select a grouped column, this menu item changes to Delete Group.
• Add Section. Creates a section based on this column. When you select a section column, this menu item changes to Delete Section.
• Hide Detail. Hides the group's or section's detail rows. If the detail rows are hidden, this menu item changes to Show Detail. This option is available when you select a grouped column or a section column.
• Page Break. Sets a page break before or after a group or section. This option is available when you select a grouped column or a section column.
Sort Opens a submenu that contains: • Sort Ascending. Sorts the column rows in ascending order.
• Sort Descending. Sorts the column rows in descending order.
• Advanced Sort. Opens the Advanced Sort dialog box, which supports performing a sort based on additional columns.
Style Opens a submenu that contains: • Font. Opens the Font dialog box, which supports modifying the font properties of column data. • Conditional Formatting. Opens a dialog box that supports setting conditional formatting rules for data in this column.
T able A-33 Report Context Menus Option Description A-43 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Troubleshoot To bring up Cisco ISE troubleshooting tools, go to Monitor > Troubleshoot > Diagnostic Tools. Use the following tools to solve problems that may appear on your network: • General Tools, page A-43 • Security Group Access Tools, page A-50 General Tools To access the following General Tools for troubleshooting go to Monitor > Troubleshoot > Diagnostic Tools and expand General Tools in the left panel. Choose from the following tools: • Connectivity Tests, page A-43 • RADIUS Authentication Troubleshooter, page A-43 • Execute Network Device Command, page A-46 • Evaluate Configuration Validator, page A-46 • Posture Troubleshooting, page A-48 • TCP Dump, page A-49 Connectivity Tests Perform connectivity tests to troubleshoot failed authentications and other problems.
RADIUS Authentication Troubleshooter Check RADIUS authentication results and troubleshoot problems that may occur.
T able A-34 Connectivity T ests Option Description Hostname or IP Address Enter the hostname or IP address for a connection you want to test. Click Clear to clear the hostname or IP address .
ping Click ping to view the packets sent and received, packet loss (if any) and the time it takes for the test to complete.
traceroute Click traceroute to view the intermediary IP addresses (hops) between the Monitoring persona node and the tested hostname or IP address, and the time it takes for each hop to complete.
nslookup Click nslookup cto view the server and IP address of your tested domain name server hostname or IP address.
T able A-35 RADIUS Authentication T roubleshooter Option Description Search and select a RADIUS authentication for troubleshooting Username Enter the username of the user whose authentication you want to troubleshoot, or click Select to choose the username from a list. Click Clear to clear the username. A-44 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor MAC Address Enter the MAC address of the device that you want to troubleshoot, or click Select to choose the MAC address from a list. Click Clear to clear the MAC address.
Audit Session ID Enter the audit session ID that you want to troubleshoot. Click Clear to clear the audit session ID.
NAS IP Enter the NAS IP address or click Select to choose the NAS IP address from a list. Click Clear to clear the NAS IP address.
NAS Port Enter the NAS port number or click Select to choose a NAS port number from a list. Click Clear to clear the NAS port number.
Authentication Status Choose the status of your RADIUS authentication from the Authentication Status drop-down list box. The available options are: • Pass or Fail • Pass • Fail Failure Reason Enter the failure reason or click Select to choose a failure reason from a list. Click Clear to clear the failure reason.
Time Range Select a time range from the drop-down list. The RADIUS authentication records that are created during this time range are used: • Last hour • Last 12 hours • Today • Yesterday • Last 7 days • Last 30 days • Custom Start Date-Time (Only if you choose Custom Time Range) Enter the start date and time, or click the calendar icon to select the start date and time. The date should be in the mm/dd/yyyy format and time in the hh:mm format.
End Date-Time (Only if you choose Custom Time Range) Enter the end date and time, or click the calendar icon to select the end date and time. The date should be in the mm/dd/yyyy format and time in the hh:mm format.
Fetch Number of Records Choose the number of records that you want to fetch from the drop-down list: 10, 20, 50, 100, 200, or 500.
T able A-35 RADIUS Authentication T roubleshooter Option Description A-45 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor RADIUS Authentication Troubleshooting—Progress Details RADIUS Authentication Troubleshooting—Results Summary T able A-36 RADIUS Authentication T roubleshooting Progress Details Option Description Specify Connection Parameters for Network Device a.b.c.d Username Enter the username for logging in to the network device.
Password Enter the password.
Protocol Choose the protocol from the Protocol drop-down list. Valid options are: • Telnet • SSHv2 Note Telnet is the default option. If you choose SSHv2, you must ensure that SSH connections are enabled on the network device.
Port Enter the port number.
Enable Password Enter the enable password.
Same As Login Password Check this check box if the enable password is the same as the login password.
Use Console Server Select this check box to use the console server.
Console IP Address (If the Use Console Server check box is selected) Enter the console IP address.
Advanced (Use if there is an “Expect timeout error” or the device has non-standard prompt strings) Note The Advanced options appear only for some of the troubleshooting tools.
Username Expect String Enter the string that the network device uses to prompt for username; for example, Username:, Login:, and so on.
Password Expect String Enter the string that the network device uses to prompt for password; for example, Password:.
Prompt Expect String Enter the prompt that the network device uses. For example, #, >, and @.
Authentication Failure Expect String Enter the string that the network device returns when there is an authentication failure; for example, Incorrect password, Login invalid, and so on.
T able A-37 RADIUS Authentication T roubleshooting Results Summary Option Description Diagnosis and Resolution Diagnosis The diagnosis for the problem is listed here.
Resolution The steps for resolution of the problem are detailed here. A-46 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Execute Network Device Command Execute the show command on a network device.
Evaluate Configuration Validator Evaluate the configuration of a network device and identify any configuration problems.
Troubleshooting Summary A step-by-step summary of troubleshooting information is provided here. You can expand any step to view further details.
Note Any configuration errors are indicated by red text.
T able A-37 RADIUS Authentication T roubleshooting Results Summary Option Description T able A-38 Execute Network Device Command Option Description Enter Information Network Device IP Enter the IP address of the network device on which you want to run the command.
Command Enter the show command.
T able A-39 Evaluate Configuration Validator Option Description Enter Information Network Device IP Enter the IP address of the network device whose configuration you want to evaluate.
Select the configuration items below that you want to compare against the recommended template.
AAA This option is selected by default.
RADIUS This option is selected by default.
Device Discovery This option is selected by default.
Logging This option is selected by default.
Web Authentication Select this check box to compare the web authentication configuration.
Profiler Configuration Select this check box to compare the Profiler configuration.
SGA Check this check box if you want to compare Security Group Access configuration.
802.1X Check this check box if you want to compare the 802.1X configuration, and choose one of the following options: • Open Mode • Low Impact Mode (Open Mode + ACL) • High Security Mode (Closed Mode) A-47 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Progress Details Results Summary T able A-40 Progress Details Option Description Specify Connection Parameters for Network Device a.b.c.d Username Enter the username for logging in to the network device.
Password Enter the password.
Protocol Choose the protocol from the Protocol drop-down list. Valid options are: • Telnet • SSHv2 Note Telnet is the default option. If you choose SSHv2, you must ensure that SSH connections are enabled on the network device.
Port Enter the port number.
Enable Password Enter the enable password.
Same As Login Password Check this check box if the enable password is the same as the login password.
Use Console Server Check this check box to use the console server.
Console IP Address (Only if you check the Use Console Server check box) Enter the console IP address.
Advanced (Use these if you see an “Expect timeout error” or you know that the device has non-standard prompt strings) Note The Advanced options appear only for some of the troubleshooting tools.
Username Expect String Enter the string that the network device uses to prompt for username; for example, Username:, Login:, and so on.
Password Expect String Enter the string that the network device uses to prompt for password; for example, Password:.
Prompt Expect String Enter the prompt that the network device uses. For example, #, >, and @.
Authentication Failure Expect String Enter the string that the network device returns when there is an authentication failure; for example, Incorrect password, Login invalid, and so on.
T able A-41 Results Summary Option Description Diagnosis and Resolution Diagnosis The diagnosis for the problem is listed here.
Resolution The steps for resolution of the problem are detailed here. A-48 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Posture Troubleshooting Find and resolve posture problems on the network.
Troubleshooting Summary A step-by-step summary of troubleshooting information is provided here. You can expand any step to view further details.
Note Any configuration errors are indicated by red text.
T able A-41 Results Summary Option Description T able A-42 Posture T roubleshooting Option Description Search and Select a Posture event for troubleshooting Username Enter the username to filter on.
MAC Address Enter the MAC address to filter on, using format: xx-xx-xx-xx-xx-xx Posture Status Select the authentication status to filter on: • Any • Compliant • Noncompliant • Unknown Failure Reason Enter the failure reason or click Select to choose a failure reason from a list. Click Clear to clear the failure reason.
Time Range Select a time range from the drop-down list . The RADIUS authentication records that are created during this time range are used: • Last hour • Last 12 hours • Today • Yesterday • Last 7 days • Last 30 days • Custom Start Date-Time: (Only if you choose Custom Time Range) Enter the start date and time, or click the calendar icon to select the start date and time. The date should be in the mm/dd/yyyy format and time in the hh:mm format. A-49 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor TCP Dump Use the tcpdump utility to monitor the contents of packets on a network interface and troubleshoot problems on the network as they appear.
End Date-Time: (Only if you choose Custom Time Range) Enter the end date and time, or click the calendar icon to select the start date and time. The date should be in the mm/dd/yyyy format and time in the hh:mm format.
Fetch Number of Records Select the number of records to display: 10, 20, 50, 100, 200, 500 Search Result Time Time of the event Status Posture status Username User name associated with the event MAC Address MAC address of the system Failure Reason Failure reason for the event T able A-42 Posture T roubleshooting Option Description T able A-43 TCP Dump Option Description Status: • Stopped—the tcpdump utility is not running • Start—Click to start the tcpdump utility monitoring the network.
• Stop—Click to stop the tcpdump utility Host Name Choose the name of the host to monitor from the drop-down list.
Network Interface Choose the network interface to monitor from the drop-down list.
Promiscuous Mode • On—Click to turn on promiscuous mode (default).
• Off—Click to turn off promiscuous mode.
Promiscuous mode is the default packet sniffing mode. It is recommended that you leave it set to On. In this mode the network interface is passing all traffic to the system’s CPU.
Filter Enter a boolean expression on which to filter. Standard tcpdump filter expressions are supported.
Format Select a format for the tcpdump file from the drop-down list: • Human Readable • Raw Packet Data A-50 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Security Group Access Tools To access the following General Tools for troubleshooting go to Monitor > Troubleshoot > Diagnostic Tools and expand Security Group Access Tools in the left panel. Choose from the following tools: • Egress SGACL Policy, page A-50 • SXP-IP Mappings, page A-51 • IP User SGT, page A-54 • Device SGT, page A-55 Egress SGACL Policy Compare Security Group Access-enabled devices using theEgress policy diagnostic too. Progress Details Dump File Displays data on the last dump file, such as the following: Last created on Wed Apr 27 20:42:38 UTC 2011 by admin File size: 3,744 bytes Format: Raw Packet Data Host Name: Positron Network Interface: GigabitEthernet 0 Promiscuous Mode: On • Download—Click to download the most recent dump file.
• Delete—Click to delete the most recent dump file.
T able A-43 TCP Dump (continued) Option Description T able A-44 Progress Details for Egress SGACL Policy Option Description Specify Connection Parameters for Network Device a.b.c.d Username Enter the username for logging in to the network device.
Password Enter the password.
Protocol Choose the protocol from the Protocol drop-down list. Valid options are: • Telnet • SSHv2 Note Telnet is the default option. If you choose SSHv2, you must ensure that SSH connections are enabled on the network device.
Port Enter the port number.
Enable Password Enter the enable password. A-51 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Results Summary SXP-IP Mappings Compare SXP-IP mappings between a device and its peers.
Peer SXP Devices Same As Login Password Check this check box if the enable password is the same as the login password.
Use Console Server Check this check box to use the console server.
Console IP Address (Only if you check the Use Console Server check box) Enter the console IP address.
Advanced (Use these if you see an “Expect timeout error” or you know that the device has non-standard prompt strings) Note The Advanced options appear only for some of the troubleshooting tools.
Username Expect String Enter the string that the network device uses to prompt for username; for example, Username:, Login:, and so on.
Password Expect String Enter the string that the network device uses to prompt for password; for example, Password:.
Prompt Expect String Enter the prompt that the network device uses. For example, #, >, and @.
Authentication Failure Expect String Enter the string that the network device returns when there is an authentication failure; for example, Incorrect password, Login invalid, and so on.
T able A-44 Progress Details for Egress SGACL Policy Option Description T able A-45 Results Summary for Egress SGACL Policy Option Description Diagnosis and Resolution Diagnosis The diagnosis for the problem is listed here.
Resolution The steps for resolution of the problem are detailed here.
Troubleshooting Summary A step-by-step summary of troubleshooting information is provided here. You can expand any step to view further details.
Note Any configuration errors are indicated by red text.
T able A-46 Peer SXP Devices for SXP-IP Mappings Option Description Peer SXP Devices Peer IP Address IP address of the peer SXP device.
VRF The VRF instance of the peer device. A-52 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Progress Details Peer SXP Mode The SXP mode of the peer device; for example, whether it is a speaker or a listener.
Self SXP Mode The SXP mode of the network device; for example, whether it is a speaker or a listener.
Connection State The status of the connection.
Common Connection Parameters User Common Connection Parameters Check this check box to enable common connection parameters for all the peer SXP devices.
Note If the common connection parameters are not specified or if they do not work for some reason, the Expert Troubleshooter again prompts you for connection parameters for that particular peer device.
Username Enter the username of the peer SXP device.
Password Enter the password to gain access to the peer device.
Protocol • Choose the protocol from the Protocol drop-down list box. Valid options are: – Telnet – SSHv2 Note Telnet is the default option. If you choose SSHv2, you must ensure that SSH connections are enabled on the network device.
Port • Enter the port number. The default port number for Telnet is 23 and SSH is 22.
Enable Password Enter the enable password if it is different from your login password.
Same as login password Check this check box if your enable password is the same as your login password.
T able A-46 Peer SXP Devices for SXP-IP Mappings Option Description T able A-47 Progress Details for SXP-IP Mappings Option Description Specify Connection Parameters for Network Device a.b.c.d Username Enter the username for logging in to the network device.
Password Enter the password. A-53 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Results Summary Protocol Choose the protocol from the Protocol drop-down list. Valid options are: • Telnet • SSHv2 Note Telnet is the default option. If you choose SSHv2, you must ensure that SSH connections are enabled on the network device.
Port Enter the port number.
Enable Password Enter the enable password.
Same As Login Password Check this check box if the enable password is the same as the login password.
Use Console Server Check this check box to use the console server.
Console IP Address (Only if you check the Use Console Server check box) Enter the console IP address.
Advanced (Use these if you see an “Expect timeout error” or you know that the device has non-standard prompt strings) Note The Advanced options appear only for some of the troubleshooting tools.
Username Expect String Enter the string that the network device uses to prompt for username; for example, Username:, Login:, and so on.
Password Expect String Enter the string that the network device uses to prompt for password; for example, Password:.
Prompt Expect String Enter the prompt that the network device uses. For example, #, >, and @.
Authentication Failure Expect String Enter the string that the network device returns when there is an authentication failure; for example, Incorrect password, Login invalid, and so on.
T able A-47 Progress Details for SXP-IP Mappings (continued) Option Description T able A-48 Results Summary for SXP-IP Mappings Option Description Diagnosis and Resolution Diagnosis The diagnosis for the problem is listed here.
Resolution The steps for resolution of the problem are detailed here.
Troubleshooting Summary A step-by-step summary of troubleshooting information is provided here. You can expand any step to view further details.
Note Any configuration errors are indicated by red text. A-54 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor IP User SGT Use the IP User SGT diagnostic tool to compare IP-SGT values on a device with an ISE assigned SGT. Progress Details T able A-49 IP User SGT Option Description Enter Information Network Device IP Enter the IP address of the network device.
Filter Results Username Enter the username of the user whose records you want to troubleshoot.
User IP Address Enter the IP address of the user whose records you want to troubleshoot.
SGT Enter the user SGT value.
T able A-50 Progress Details for IP User SGT Option Description Specify Connection Parameters for Network Device a.b.c.d Username Enter the username for logging in to the network device.
Password Enter the password.
Protocol Choose the protocol from the Protocol drop-down list. Valid options are: • Telnet • SSHv2 Note Telnet is the default option. If you choose SSHv2, SSH connections must be enabled on the network device.
Port Enter the port number.
Enable Password Enter the enable password.
Same As Login Password Check this check box if the enable password is the same as the login password.
Use Console Server Check this check box to use the console server.
Console IP Address (Only if you check the Use Console Server check box) Enter the console IP address.
Advanced (Use these if you see an “Expect timeout error” or you know that the device has non-standard prompt strings) Note Advanced options appear only for some of the troubleshooting tools.
Username Expect String Enter the string that the network device uses to prompt for username; for example, Username:, Login:, and so on.
Password Expect String Enter the string that the network device uses to prompt for password; for example, Password:. A-55 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Results Summary Device SGT Use the Device SGT diagnostic tool to compare the device SGT with the most recently assigned value.
Prompt Expect String Enter the prompt that the network device uses. For example, #, >, and @.
Authentication Failure Expect String Enter the string that the network device returns when there is an authentication failure; for example, Incorrect password, Login invalid, and so on.
T able A-50 Progress Details for IP User SGT (continued) Option Description T able A-51 Results Summary for IP User SGT Option Description Diagnosis and Resolution Diagnosis The diagnosis for the problem is listed here.
Resolution The steps for resolution of the problem are detailed here.
Troubleshooting Summary A step-by-step summary of troubleshooting information is provided here. You can expand any step to view further details.
Note Any configuration errors are indicated by red text.
T able A-52 Device SGT Option Description Enter Information Network Device IPs (comma-separated list) Enter the network device IP addresses (whose device SGT you want to compare with an ISE-assigned device SGT) separated by commas.
Common Connection Parameters A-56 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Monitor Use Common Connection Parameters Select this check box to use the following common connection parameters for comparison: • Username—Enter the username of the network device.
• Password—Enter the password.
• Protocol—Choose the protocol from the Protocol drop-down list box. Valid options are: – Telnet – SSHv2 Note Telnet is the default option. If you choose SSHv2, SSH connections must be enabled on the network device.
• Port—Enter the port number. The default port number for Telnet is 23 and SSH is 22.
Enable Password Enter the enable password if it is different from your login password.
Same as login password Select this check box if your enable password is the same as your login password.
T able A-52 Device SGT (continued) Option Description A-57 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Policy Policy This section covers the following user interface elements: • Authentication, page A-57 Authentication Allowed Protocols Service T able A-53 Allowed Protocols Service Option Description Allowed Protocols Process Host Lookup Check this check box to configure Cisco ISE to process the Host Lookup field (for example, when the RADIUS Service-Type equals 10) and use the System UserName attribute from the RADIUS Calling-Station-ID attribute. Uncheck this check box if you want Cisco ISE to ignore the Host Lookup request and use the original value of the system UserName attribute for authentication. When unchecked, message processing is done according to the protocol (for example, PAP). Authentication Protocols Allow PAP/ASCII This option enables PAP/ASCII. PAP uses cleartext passwords (that is, unencrypted passwords) and is the least secure authentication protocol.
When you check the Allow PAP/ASCII check box, you can check the Detect PAP as Host Lookup check box to configure Cisco ISE to detect this type of request as a Host Lookup (instead of PAP) request.
Allow CHAP This option enables CHAP authentication. CHAP uses a challenge-response mechanism with password encryption. CHAP does not work with Microsoft Active Directory.
Allow MS-CHAPv1 This option enables MS-CHAPv1.
Allow MS-CHAPv2 This option enables MS-CHAPv2.
Allow EAP-MD5 This option enables EAP-based MD5 hashed authentication.
When you check the Allow EAP-MD5 check box, you can check the Detect EAP-MD5 as Host Lookup check box to configure Cisco ISE to detect this type of request as a Host Lookup (instead of EAP-MD5) request. A-58 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Policy Allow EAP-TLS This option enables the EAP-TLS Authentication protocol and configures EAP-TLS settings. You can specify how Cisco ISE will verify the user identity as presented in the EAP identity response from the end-user client. User identity is verified against information in the certificate that the end-user client presents. This comparison occurs after an EAP-TLS tunnel is established between Cisco ISE and the end-user client.
Note EAP-TLS is a certificate-based authentication protocol. EAP-TLS authentication can occur only after you have completed the required steps to configure certificates. Refer to Chapter 12, “Managing Certificates” for more information on certificates.
Allow LEAP This option enables Lightweight Extensible Authentication Protocol (LEAP) authentication.
Allow PEAP This option enables the PEAP authentication protocol and PEAP settings. The default inner method is MS-CHAPv2.
When you check the Allow PEAP check box, you can configure the following PEAP inner methods: • Allow EAP-MS-CHAPv2—Check this check box to use EAP-MS-CHAPv2 as the inner method.
– Allow Password Change—Check this check box for Cisco ISE to support password changes.
– Retry Attempts—Specifies how many times Cisco ISE requests user credentials before returning login failure. Valid values are 1 to 3.
• Allow EAP-GTC—Check this check box to use EAP-GTC as the inner method.
– Allow Password Change—Check this check box for Cisco ISE to support password changes.
– Retry Attempts—Specifies how many times Cisco ISE requests user credentials before returning login failure. Valid values are 1 to 3.
• Allow EAP-TLS—Check this check box to use EAP-TLS as the inner method.
T able A-53 Allowed Protocols Service (continued) Option Description A-59 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Policy Allow EAP-FAST This option enables the EAP-FAST authentication protocol and EAP-FAST settings. The EAP-FAST protocol can support multiple internal protocols on the same server. The default inner method is MS-CHAPv2.
When you check the Allow EAP-FAST check box, you can configure EAP-FAST as the inner method: • Allow EAP-MS-CHAPv2 – Allow Password Change—Check this check box for Cisco ISE to support password changes in phase zero and phase two of EAP-FAST.
– Retry Attempts—Specifies how many times Cisco ISE requests user credentials before returning login failure. Valid values are 1-3.
• Allow EAP-GTC • Allow Password Change—Check this check box for Cisco ISE to support password changes in phase zero and phase two of EAP-FAST.
• Retry Attempts—Specifies how many times Cisco ISE requests user credentials before returning login failure. Valid values are 1-3.
• Use PACs—Choose this option to configure Cisco ISE to provision authorization PACs1 for EAP-FAST clients. Additional PAC options appear.
• Don't use PACs—Choose this option to configure Cisco ISE to use EAP-FAST without issuing or accepting any tunnel or machine PACs. All requests for PACs are ignored and Cisco ISE responds with a Success-TLV without a PAC.
When you choose this option, you can configure Cisco ISE to perform machine authentication.
1.
PACs = Protected Access Credentials.
T able A-53 Allowed Protocols Service (continued) Option Description A-60 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Policy PAC Options T able A-54 PAC Options Option Description Use PAC • Tunnel PAC Time to Live—The TTL1 value restricts the lifetime of the PAC. Specify the lifetime value and units. The default is 90 days. The range is between 1 and 1825 days.
• Proactive PAC Update When: of PAC TTL is Left—The Update value ensures that the client has a valid PAC. Cisco ISE initiates an update after the first successful authentication but before the expiration time that is set by the TTL. The update value is a percentage of the remaining time in the TTL. The default is 90%.
• Allow Anonymous In-band PAC Provisioning—Check this check box for Cisco ISE to establish a secure anonymous TLS handshake with the client and provision it with a PAC by using phase zero of EAP-FAST with EAP-MSCHAPv2.
Note To enable anonymous PAC provisioning, you must choose both of the inner methods, EAP-MSCHAPv2 and EAP-GTC.
• Allow Authenticated In-band PAC Provisioning—Cisco ISE uses SSL server-side authentication to provision the client with a PAC during phase zero of EAP-FAST. This option is more secure than anonymous provisioning but requires that a server certificate and a trusted root CA be installed on Cisco ISE.
When you check this option, you can configure Cisco ISE to return an Access-Accept message to the client after successful authenticated PAC provisioning.
– Server Returns Access Accept After Authenticated Provisioning—Check this check box if you want Cisco ISE to return an access-accept package after authenticated PAC provisioning.
• Allow Machine Authentication—Check this check box for Cisco ISE to provision an end-user client with a machine PAC and perform machine authentication (for end-user clients who do not have the machine credentials). The machine PAC can be provisioned to the client by request (in-band) or by the administrator (out-of-band). When Cisco ISE receives a valid machine PAC from the end-user client, the machine identity details are extracted from the PAC and verified in the Cisco ISE external identity source. After these details are correctly verified, no further authentication is performed.
Note Cisco ISE only supports Active Directory as an external identity source for machine authentication.
When you check this option, you can enter a value for the amount of time that a machine PAC is acceptable for use. When Cisco ISE receives an expired machine PAC, it automatically reprovisions the end-user client with a new machine PAC (without waiting for a new machine PAC request from the end-user client). A-61 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Administration Administration This section covers the following: • System > Settings > Monitoring, page A-61 • System > Operations > Data Management > Monitoring Node, page A-64 System > Settings > Monitoring To access system monitoring tools go to Administration > System > Settings, then expand Monitoring in the left panel. This section covers the user interface elements for the following monitoring tools: • Alarm Syslog Targets, page A-62 • Email Settings, page A-62 • Failure Reasons Editor, page A-62 • System Alarm Settings, page A-63 • Enable Stateless Session Resume—Check this check box for Cisco ISE to provision authorization PACs for EAP-FAST clients and always perform phase two of EAP-FAST (default = enabled).
Uncheck this check box in the following cases: – If you do not want Cisco ISE to provision authorization PACs for EAP-FAST clients – To always perform phase two of EAP-FAST When you check this option, you can enter the authorization period of the user authorization PAC. After this period, the PAC expires. When Cisco ISE receives an expired authorization PAC, it performs phase two EAP-FAST authentication.
• Preferred EAP Protocol—Check this check box to choose your preferred EAP protocols from any of the following options: EAP-FAST, PEAP, LEAP, EAP-TLS, and EAP-MD5. By default, LEAP is the preferred protocol to use if you do not enable this field.
1.
TTL = Time To Live T able A-54 PAC Options (continued) Option Description A-62 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Administration Alarm Syslog Targets Define the destination where alarm syslog messages are sent.
Email Settings Define the email address for the mail server and the name that is shown for messages received from the mail server, such as admin@somedomain.com.
Failure Reasons Editor View and edit failure reasons.
Viewing Failure Reasons T able A-55 Alarm Syslog T argets Option Description Identification Name Name of the alarm syslog target. The name can be 255 characters in length.
Description (Optional) A brief description of the alarm that you want to create. The description can be up to 255 characters in length.
Configuration IP Address IP address of the machine that receives the syslog message. This machine must have the syslog server running on it. It is recommended that you use a Windows or a Linux machine to receive syslog messages.
Use Advanced Syslog Options Port Port in which the remote syslog server listens. By default, it is set to 514. Valid options are from 1 to 65535.
Facility Code Syslog facility code to be used for logging. Valid options are Local0 through Local7.
T able A-56 Email Settings Option Description Mail Server Enter a valid email host server.
Mail From Enter the name that users see when they receive a message from the mail server, such as admin@somedomain.com.
T able A-57 Viewing Failure Reasons Option Description Failure Reasons The name of possible failure reasons. Click a failure reason name to open the Failure Reasons Editor page. A-63 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Administration Editing Failure Reasons Results Summary System Alarm Settings Enable, disable, and configure system alarm notification settings.
T able A-58 Editing Failure Reasons Option Description Failure Reason Display only. The error code and associated failure reason name.
Description Enter a free text description of the failure reason to assist administrators; use the text tools as needed.
Resolution Steps Enter a free text description of possible resolution steps for the failure reason to assist administrators; use the text tools as needed.
T able A-59 Results Summary for Failure Reasons Option Description Diagnosis and Resolution Diagnosis The diagnosis for the problem is listed here.
Resolution The steps for resolution of the problem are detailed here.
Troubleshooting Summary A step-by-step summary of troubleshooting information is provided here. You can expand any step to view further details.
Note Any configuration errors are indicated by red text.
T able A-60 System Alarm Settings Option Description System Alarm Settings Notify System Alarms Check this check box to enable system alarm notification.
System Alarms Suppress Duplicates Designate the number of hours that you want to suppress duplicate system alarms from being sent to the Email Notification User List. Valid options are 1, 2, 4, 6, 8, 12, and 24.
Email Notification A-64 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Administration System > Operations > Data Management > Monitoring Node To monitoring data management tools go to Administration > System > Operations, then expand Data Management > Monitoring Node in the left panel. This section covers the user interface elements for the following tools: • Full Backup On Demand, page A-64 • Scheduled Backup, page A-65 • Data Purging, page A-65 • Data Restore, page A-66 Full Backup On Demand Perform a full backup of the monitoring database on demand.
Email Notification User List Enter a comma-separated list of e-mail addresses or ISE administrator names or both. Do one of the following: • Enter the e-mail addresses.
• Click Select and enter valid administrator names. The administrator is notified by e-mail only if e-mail identification is specified in that administrator’s account. When a system alarm occurs, an e-mail is sent to all the recipients in the Email Notification User List.
Click Clear to clear this field.
Email in HTML Format Select this check box to send e-mail notifications in HTML format, or uncheck to send s plain text.
Syslog Notification Send Syslog Message Select this check box to send a syslog message for each system alarm generates Note To send syslog messages successfully, you must configure Alarm Syslog Targets, which are syslog message destinations. See, Configuring Alarm Syslog Targets, page 22-53 for more information.
T able A-60 System Alarm Settings (continued) Option Description T able A-61 Full Backup On Demand Option Description Data Repository Select a repository from the drop-down list, in which to back up the monitoring database. If no repository is selected, a backup will not occur.
Backup Now Click to perform a full backup of the monitoring database.
Full Backup On Demand Status Shows the Name, Start Time, End Time, and Status of an on demand backup. A-65 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Administration Scheduled Backup Schedule an incremental or full monitoring database backup.
Data Purging Purge data prior to an incremental or full backup.
T able A-62 Scheduled Backup Option Description Incremental Backup On Click the On radio button to enable incremental backup.
Off Click the Off radio button to disable incremental backup.
Configure Incremental Monitor Database Backup Data Repository Select a data repository for the backup files.
Schedule Select the time of the day to perform the incremental backup.
Frequency Choose the frequency of incremental backups: • Daily • Weekly—Typically occurs at the end of every week.
• Monthly—Typically occurs at the end of every month.
Configure Full Monitor Database Backup Data Repository Select a data repositoryused to store the backup files.
Schedule Select the time of the day to perfrom the database backup.
Frequency Choose the frequency of the backups: • Daily—Occurs at the specified time each day.
• Weekly—Occurs on the last day of every week.
• Monthly—Occurs on the last day of every month.
T able A-63 Data Purging Option Description Data Purging Percentage of Disk Space Enter a numerical percentage value for allowed disk space usage. This threshold triggers a purge when disk space usage meets or exceeds this value. The default is 80 percent. The maximum value allowed is 100 percent.
Data Repository Select the data repository to backup data prior to purge.
Maximum Stored Data Period Enter a value in (30-day) months to be utilized when the disk space usage threshold for purging (Percentage of Disk Space) is met. Note For this option, each month consists of 30 days. The default of three months equals 90 days. A-66 Cisco Identity Services Engine User Guide, Release 1.0 OL-22972-01 Appendix A User Interface Reference Administration Data Restore Restore a full or incremental backup.
Submit Click to proceed with the data purge.
Cancel Click to exit without purging data.
T able A-63 Data Purging (continued) Option Description T able A-64 Data Restore Column Description Available Backups to Restore Select the radio button next to the name of the backup you want to restore. The backup filename includes the time stamp. For example, ISEViewBackup-20090618_003400.
Date Shows the date of the backup Repository Shows the name of the repository where the backup is stored.
Type Shows the type of backup, full or incremental Restore Click to restore the selected backup of the monitoring database. |
12611 | https://artofproblemsolving.com/community/c1461_i_guess_this_is_a_blog?srsltid=AfmBOoq7CZv1_6xZ0y_aVT43rfs2V_REwPcJ_mmOhkvZuY6N0YEmCCF- | I guess this is a blog
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I guess this is a blog ======================
Updates (resources link, new website/blog)
by math154, Jul 31, 2014, 5:17 PM
MIT website:
Wordpress blog: (will hopefully have stuff once I figure out how to do
In particular, new shared resources link (handouts, notes, etc., both my own and other googled ones; in particular, I now have most of the stuff I wanted to share from HS):
EDIT 7/26/19: This is broken. Some files may be available upon request (email me), but for many reasons I cannot post a public link. My understanding is that other people have been creating their own repositories, anyways.
(The link will inevitably break once I start moving stuff around, so let me know if that's the case. But I'll try to at least keep it updated on my website.)
but was too lazy to. Note that this is a suboptimal way to share resources. An easy optimization would be to have a globally shared folder for every subject (more realistically, to avoid an organizational nightmare, people could label their files with multiple tags). This is actually sort of the idea behind ( Hopefully Alex Zhu sets up something along these lines for his startup.
(I should also add that just having resources without guidance is suboptimal too... but adding guidance/some direction doesn't seem too hard after collecting resources... also the linked blog post discusses a lot more about contests. Generally, I think better communication/collaboration in the contest community (and in general) could do a lot...)
This post has been edited 4 times. Last edited by math154, Jul 27, 2019, 1:05 AM
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ramblings, organizing aops posts into more useful format
by math154, Jul 9, 2014, 4:28 AM
these are mostly notes to myself
Click to reveal hidden text
I feel like it could be nontrivially helpful (for me and for others) to re-organize some of my posts into blog posts.
for instance
darn productivity (although i don't know if it actually conveys the right attitude) is too hard. it's no longer like high school where i had on the order of 2-3 actually substantial goals. actually i'm not sure i have many more right now, but there's just so much more to do/learn/explore, and so much less guidance/obviously-good paths (also feels like there's so much less time darn, even though that's probably not true).
in other news yesterday i wasted time making my website less sucky; (but I'm glad I got the reflections on classes down)
hmm so I feel like there's often a conflict between practical advice and advice that would obviously be better in an ideal world. like attempting to do research right now (even though i do basically have a urop (either playing around with quaternion/octonion algebra factorization/number theory or kissing number problem for dimensions 9 to 12 or so, probably in the construction direction) i can work on, i haven't formalized it and i'm not sure if it's worth it (although i do think quaternions/octonions are special enough to merit this kind of attention, and it does seem interesting...) vs. learning more stuff. and even when learning stuff, like whether it's always worth it to try to prove a lemma/theorem first[url][/url] rather than just reading/checking/(understanding? but that's the whole question) the proof. or pushing out more subjects on expii vs. exploring other options (and generally doing stuff that is more interesting/innovative, probably mostly for myself right now but maybe eventually for others too)---i can't say i've done anything truly difficult for expii, even though, in some sense, this is obviously meaningful work (low-hanging fruit?) that (from what we can tell) still needs to be done.
and this is all just in math/education-related stuff. i wonder if i'll ever pick piano up again. also other real life stuff but i don't feel like writing about that now (i'd rather just do math/other stuff). maybe some other time.
at the end of the day the best advice always seems to be to do what it takes to be happy... because over-analyzing things always leads to apparent contradictions in this crazy world.
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notes on weierstrass approximation theorem
by math154, Jul 4, 2014, 10:22 PM
First instinct is to use Lagrange interpolation. Runge's phenomenon says equally spaced nodes are bad for this. More generally even smarter things like Chebyshev nodes are bad. See comments here for some intuition: high degree means greater oscillations in between nodes, as we've only controlled nodes perfectly and it's thus hard to bound stuff between nodes. (On the other hand, I don't see good intuition a priori why something like Chebyshev nodes shouldn't work, it's just that it's more plausible that it won't work than a "smoother/more-averaged-out" approximation. In fact the Wikipedia says all absolutely continuous guys are good with Chebyshev so blah.)
Let's do a continuous/smoother version of Lagrange interpolation instead. Analogous to Lagrange interpolation polynomials ( with and for ), we want interpolation polynomials such that when for some bijection (and quickly decays as leaves ); for simplicity we just use , but perhaps in other contexts we'd need to do something more complicated? We'll then take ; this is apparently related to blurring; actually, this has some issues with endpoints, but if we first reduce to (or write out stuff more carefully) then we're good to go. Anyway here works, where is a fixed positive integer and is just a scale factor to make , with an easy bound of . The rest is not hard, looking at using a uniform continuity bound to bound when is small. (We need uniform continuity since the same bound will apply independently of .)
Some other links:
The probabilistic proof with Bernstein polynomials is interesting and I don't have great intuition for why this works while the Lagrange interpolation doesn't, except that the term has "smooth/well-behaved contribution" that peaks when . (Compared with Lagrange where the contribution is at , at the , and perhaps unpredictable elsewhere. That probably makes stuff around particularly hard to control, given that our main bound will probably come from uniform continuity of around .)
(and Hermite interpolation in general).
Another approach: is that it suffices to approximate the absolute value function on :
By the polynomial interpolation Wikipedia link above though, we should be able to use Chebyshev nodes to do this. Fix . Let for , so for and for . Then with , so on .
But (by differentiating or using roots of unity) we have if , and otherwise. So letting and noting that , we have
For we have , so for fixed , we have a negative number with increasing magnitude as increases in ![Image 61: $(n/2,n]$]( So we have an alternating series, which bounds stuff by
for . Of course by symmetry we get the same bound for , so for all , and we're done.
(Actually note that these are Chebyshev nodes of second kind, but not a huge difference...)
More reading:
Chebyshev equioscillation theorem, and theorem/reference about how well Chebyshev can do in general:
This post has been edited 5 times. Last edited by math154, Jul 5, 2014, 6:32 AM
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why not (Expii)
by math154, Jul 4, 2014, 7:13 AM
[url= wrote:
Hey AoPSers,
I believe anyone can succeed with the right environment---the type AoPS has given all of us. And that's what Expii, a crowd-sourced (interactive) journey-based education-compressing initiative, is all about.
We just released a transformation/intuition-based geometry map (calculus has been up a couple months, but I'd feel bad spamming y'all about that), along with an awesome diagram drawer by v_Enhance, and we need your help to crowd-source the topics, which include basics like the Pythagorean theorem, less basic stuff like Olympiadgeometryfundamentals, practical tools like complexnumbers and not-so-practical tools like Tarski's procedure to solve any geometry problem with Cartesian coordinates, and modern-based foundations of geometry.
(We'll hopefully have other subjects rolling out reasonably soon, including mechanics (physics), algebra, and probability/statistics.)
PM v_Enhance, Mewto55555, abacadaea, or me if you want a referral code (to make an account, although you can still access most things without logging in), and please spread the word to your friends!
Thanks so much for your help!
math154
This post has been edited 1 time. Last edited by math154, Jul 6, 2014, 11:28 PM
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Black MOP handouts and website
by math154, Jun 4, 2014, 5:22 AM
For convenience (both for the students and others), I'm sharing the handouts for my two black MOP classes.
There isn't a solutions file, but instead I've added relevant discussion links to almost all of the problems which will serve just as well (if not better---many of these threads could benefit from further perspectives!).
Edit: If the link breaks, let me know. This will also be updated on website:
EDIT 11/2/18: The old links seem to be broken. For the next 4-5 years, please use the following links:
(including MOP 2018 handouts, which I think are more useful than my MOP 2014 ones)
(older handouts, including MOP 2014 and a Diophantine equations problem list)
This post has been edited 5 times. Last edited by math154, Nov 2, 2018, 4:22 AM
Reason: new links
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random stuff
by math154, Dec 15, 2013, 7:43 AM
I should stop procrastinating on finals studying. Anyway, here are some nice problems from Putnam seminar this semester... (A lot of these are probably Putnam problems even if I didn't label them as such.)
(MIT Problem Solving Seminar, Congruences and Divisibility) Consider with . Suppose that . Prove or disprove that for all .
Click to reveal hidden text
See my Math StackExchange post for a solution and questions about potential alternative approaches.
(MIT Problem Solving Seminar, Abstract Algebra) Let be a noncommutative ring with identity. Show that if an element has more than one right inverse, then has infinitely many right inverses.
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%
%
Assume for the sake of contradiction that has finitely many right inverses, which form the set , and fix some . If denotes , then has the same (finite) size as , which has at least two elements by assumption.
It's clear that . In particular, . But , so and we deduce .
Now . Yet implies , whence
Finally, we see that , so has size , contradiction.
%we might try to prove that from .
(MIT Problem Solving Seminar, ``Hidden'' Independence and Uniformity) A snake on the chessboard is a nonempty subset of the squares of the board obtained as follows: Start at one of the squares and continue walking one step up or to the right, stopping at any time. The squares visited are the squares of the snake. Find the total number of ways to cover an chessboard with disjoint snakes.
Click to reveal hidden text
Place the square in the lattice grid in the Cartesian plane.
First suppose we have a snake partition. For each point , define as follows: if is end of a snake, then set ; otherwise, let be the next point in the snake (either or ).
Then the only restriction on is that for any two points along the same ``main'' diagonal. Indeed, for any such with for all , we get a unique snake partition by drawing snakes according to the rules . (The snakes will start at the non-invertible points of .)
Now let denote the number of valid , restricted to the diagonal , for . Then the total number of snake partitions of is just .
Clearly , and by symmetry (consider rotating everything by ), for . Now fix , and represent by the set of and tiles corresponding to , for , and for any non-invertible square with . Then the restriction for valid is equivalent to having no two of these tiles intersect. Furthermore, any tile representation of (not necessarily valid) covers the points with as well as both the invertible and non-invertiible points with , so is valid \ifff{} it corresponds to a domino tiling of the union of the - and -main diagonals!
It immediately follows that , the number of ways to tile a grid with and blocks. We conclude that the number of snake partitions of is .
\textbf{Comment.} We can think of as really just drawing arrows between successive points in the snakes.
\textbf{Comment.} We can easily generalize to chessboards. When , there will just be a few main diagonals of the same length in the middle, but we still get a product of Fibonacci numbers.
(MIT Problem Solving Seminar, Generating Functions; Putnam 1948 A6) Show that
Click to reveal hidden text
This is probably trivialized by Wilf-Zeilberger pairs, but I haven't yet taken the time to understand how those work.
Let . Then , so differentiating yields , or .
The rest is standard. We have . From and , we conclude that
as desired.
\textbf{Comment.} Since
and , we have
Since , this problem establishes the identity .
(MIT Problem Solving Seminar, Probability) Choose points at random from the unit interval . Let be the probability that for all . Compute the generating function .
Click to reveal hidden text
Define , so that is a polynomial with constant term . Then and for ,. (We may vacuously define .)
Thus we may uniquely determine from (for ) using and . Differentiating once more (for ), we get .
If , then suggests trying for some (formal) power series . Now from , we get , and from we get , yielding .
All of this is reversible, so must equal , and we conclude .
(MIT Problem Solving Seminar, Analysis; Putnam 1996 B6) Let be the vertices of a convex polygon which contains the origin in its interior. Prove that there exist positive reals such that .
Click to reveal hidden text
Let denote our convex -gon. Define .
Fix , and observe that if for some , then for each . But because lies in the \emph{interior} , there exists (indpendent of ) such that each of the four points lies in as well, so by taking suitable weighted geometric means of the , we obtain and . Thus and .
Yet by the extreme value theorem, attains a minimum on the compact set , say at , where means cannot lie on the boundary of . Thus , a differentiable function on , has a local (actually, a global) minimum at , and we conclude that , so we're done.
(MIT Problem Solving Seminar, Analysis; Putnam 1999 A5) Prove that there is a constant such that for every polynomial of degree ,.
Click to reveal hidden text
\begin{lem}
For fixed $z\in\C$ and , the set of satisfying is contained in an interval of length at most .
\end{lem}
\begin{proof}
If , then , so . Clearly we have no solutions for .
If , must lie in an interval of length .
Otherwise, if , then has magnitude at least and the same sign as , and must therefore lie in either or , which both have length .
\end{proof}
Let and write for some complex numbers . Let . By the lemma, the set of satisfying lies in a union of intervals of total length at most , so , allowing us to take .
(MIT Problem Solving Seminar, Recurrences; Putnam 1997 A6) For a positive integer and any real number , define recursively by ,, and for , Fix and then take to be the largest value for which . Find in terms of and ,.
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Fix , and suppose ; then , so for .
Let , so , and
But if is a root of with multiplicity , then it's a root of with multiplicity , and so . Writing (since ), we obtain . Hence , with equality \ifff , so when is maximal (such that ), we have , so for .
(MIT Problem Solving Seminar, Polynomials; Putnam 1956 B7?) The nonconstant polynomials and with complex coefficients have the same set of numbers for their zeros but possibly different multiplicities. The same is true of the polynomials and . Prove that .
Click to reveal hidden text
%(On the original Exam, the assumption that and are nonconstant was inadvertently omitted.)
%\item Let be the set of roots (without multiplicity) of a polynomial . Are there two distinct polynomials such that and .
Go by contradiction and suppose for some and complex ,
and
for constants and positive integers with and .
The sets and must be disjoint ( cannot share any roots), so differentiating shows that
Thus , and similarly we have .
On the other hand,
forces , a clear contradiction.
%Equivalent wording by Alex Zhu: Let and be the sets of roots of and , respectively, and let . Then has at least roots from and at least from
This post has been edited 11 times. Last edited by math154, Dec 15, 2013, 8:45 AM
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random productivity tools
by math154, Aug 20, 2013, 11:25 PM
... from t0rajir0u.
Boomerang: this offers several related and useful features. First, it lets you temporarily remove an email from your inbox and reintroduce it at a specified later time, which is useful for emails that you want a specific future self to deal with. Second, it lets you reintroduce emails you send to your inbox at a specified later time, which is useful for emails that you want to follow up in in case you don't get a response. Third, it lets you send emails at a specified later time.
Workflowy: this is an arbitrarily nested bulleted list, and it is much more useful than it sounds. You can also use Workflowy as a to-do list, and it also has Android and iOS versions with it syncs with.
Remember the Milk: this is a straightforward to-do list app. There is an obvious way to use it and a slightly less obvious way to use it, which is to create a tag called "main" and add a smartlist that searches for items tagged "main." Then you use your RTM inbox to write down notes of any kind about stuff you might want to do, but you use "main" to write down next actions. For example, "do my history project" is a note about stuff you might want to do, but it's not a next action: it's way too vague. A next action here is "look up sources to use for my history project and write them down."
RTM has Android and iOS versions and syncs with them. RTM also allows you to repeat items after a certain number of days.
Beeminder: this is an app that lets you set various goals, e.g. "spend at least half an hour on homework every day," and takes your money if you don't reach them.
Beeminder fights akrasia by fighting hyperbolic discounting. See for some details. You can also make your Beeminder goals public and share them with your friends to take advantage of social commitment effects for extra motivation.
Besides punishing you, Beeminder also shows you a graph of your progress, which is both helpful and motivating; it's like a quantitative version of Be careful not to make goals that are too ambitious at the start, or else you'll be demotivated if you fail to reach them.
Anki: this is an intelligent flashcard program. It uses spaced repetition ( and more or less solves the problem of how to put information into your long-term memory.
Mutually beneficial referral link (+250 words per month).
This post has been edited 8 times. Last edited by math154, Aug 21, 2013, 2:06 AM
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Random Stuff
by math154, Dec 9, 2012, 10:59 PM
So I haven't updated in a long time... Here's some random stuff.
3-D proofs for Brianchon and Pascal.
Click to reveal hidden text
This proof only works for non-degenerate cases (for Brianchon, all distinct points on opposite sides; for Pascal, no two opposite sides parallel)
circumscribes the circle with tangency points ( on , on , etc.)
lift stuff alternately by constant angles so that is above, is below , etc. so that is a plane, etc.
now we get three planes of the form ,, etc. that are not the same, but have pairwise intersections exactly equal to lines ,,, which means that the three planes must intersect in a point, which must lie on all three lines; now project everything onto the plane of to get a proof of brianchon
for pascal, note that are coplanar, so and intersect at a point on the line equal to the intersection of planes and , which also contains and
if ,,, then the plane (possibly degenerate, but then the result's obvious) intersects plane at a line containing ,,, so we're done
Let be positive rational numbers and let be integers greater than 1. If , show that for all .
Click to reveal hidden text
Let and go by contradiction. For motivation we look at the case. WLOG are ``minimal'' in the sense that for . Then our assumption is equivalent to for all .
Note that is the minimal polynomial of over the rationals, so . But then ; by symmetry and we can easily get a contradiction by considering the coefficients.
In a similar vein,
so for some choice of t h roots of unity and (any) nontrivial t h root of unity , we have . But then
contradiction. (We can also take magnitudes and use the triangle inequality: for equality, the must be in the same positive direction since the are all positive, but then would be in .)
(Russia 1998) Each square of a square board contains either or . Such an arrangement is deemed successful if each number is the product of its neighbors. Find the number of successful arrangements.
Click to reveal hidden text
For there are two; we restrict our attention to . Let . Take the of each entry so we're working in instead; we're choosing some of the entires to be 1. Viewing this as the equation (where boundary cases are 0 for convenience), we can set up a matrix equation where iff entries and (there are total) are neighboring, and has a 1 for each entry we include in the successful arrangement.
We show by induction on that (whence must be the all-zero vector and there's exactly one solution)---note that for the matrix equation doesn't actually correspond to the original problem, but this is not an issue. To do this, observe that (as we're working ). Now for a fixed permutation that contributes 1 to the sum (i.e. doesn't vanish), draw an arrow from to for each ; we get a partition of the original board into disjoint directed cycles of size 1 or for some (arrows connect neighboring elements---these directed cycles correspond to permutation cycles). These cycles are nontrivially ``reversible'' iff , so pairing up the permutations with their reversals (this corresponds to ), we're left (mod 2, of course) with the number of tilings of the grid using ,, and tiles. Reflecting over the horizontal axis we're left with horizontally symmetric tilings; now reflecting over the vertical axis we're left with tilings that are both horizontally and vertically symmetric.
For odd it's easy to check that where ,, is Fibonacci, as all dominoes intersecting one of two axes must lie within the axes (by symmetry). By induction we get .
Comment. We can use this kind of symmetry argument directly on the original problem (e.g. consider the product of a successful arrangement with its vertical reflection, which results in another successful arrangement), which is slightly weaker but still works for the particular problem (we don't immediately get ), but the linear algebra setting is perhaps more motivated. It may be possible to do nontrivial general analysis on boards using the same recursive linear algebra ideas, but when both of are even it seems hard to reduce the problem.
(Russia 2010) Let be a connected graph disconnected by the removal of (all of the edges of) any odd cycle. Prove that is 4-partite.
Click to reveal hidden text
Since we need to find a 4-(vertex)-coloring, the following lemma will help:
Lemma. Let be two positive integers. A graph is -colorable iff there exists an edge partition such that and are - and -colorable, respectively.
Proof. Just use product coloring for the ``if'' direction: we can assign the color to a vertex colored in , respectively.
Conversely, if is -colorable, then split the graph into sets , each containing the vertices of exactly colors; we can then take defined by and , which are - and -colorable, respectively. Alternatively, label the colors for and and define by the first and second coordinates, respectively---the induced subgraphs are clearly - and -colorable, and we can then remove duplicate edges from until we get an edge partition. (The idea is that independent sets behave like single vertices, and can easily be split in the desired form.)
In view of the lemma, we try to find an edge partition such that and are both bipartite.
Let be a maximal bipartite subgraph; then we can't add any more edges to without inducing an odd cycle. Now assume for the sake of contradiction that (where ) has an odd cycle ,. By the maximality of , and (indices modulo ) are connected by a (simple) path of even length in (i.e. with edges in ) for all , so in particular are connected in for all . Then since is connected, every vertex is connected in to at least one vertex of , whence must be connected as well, contradicting the problem hypothesis. Hence must be bipartite and since is bipartite by construction, we're done by the lemma.
Comment. In fact, we could've taken to be a maximal forest (any spanning tree) instead of a maximal bipartite graph.
In other news, December TST is this Thursday and MIT decisions ~~should~~ will come out ~~soon~~ Saturday.
This post has been edited 10 times. Last edited by math154, May 9, 2013, 2:29 PM
algebraic comboalgebraic ntcollegecombinatoricsgeometrylinear algebranumber theoryPolynomialsprojectionsprojective geometry
2 Comments
(Comment)
Storage
by math154, Feb 3, 2012, 5:16 AM
Hmm a bit on the easy side I guess, but still nice.
(Russia 2011, 11.4) Ten cars, which do not necessarily start at the same place, are all going one way on a highway which does not loop around. The highway goes through several towns. Every car goes with some constant speed in those towns and with some other constant speed out of those towns. For different cards these speeds can be different. 2011 flags are put in different places next to the highway. Every car went by every flag, and no car passed another right next to any of the flags. Prove that there are at least two flags at which all cars went by in the same order.
Solution
Let denote the time it takes for car to get to the first flag (choose a starting time arbitrarily). Suppose car takes time per unit distance when in and out of town, respectively; let (not both zero) be the distances from flag to flag along in and out of town routes. Then car takes to get to flag . If for some , then we must have for all (or else we violate the "no passing at flags" rule), whence car is essentially a "clone" of car . Since clones are ranked the same at each flag, we can remove all clones to get (WLOG) pairwise distinct cars (WLOG cars through ). Then for all and .
Now note that for , iff
and iff . Graphing in the Cartesian plane for (note the symmetry between and ), we get at most (where is the number of lines) distinct regions (well-known, e.g. this). But and are in the same region iff the orderings at flags are equal, so we're done by pigeonhole.
Comment: This is not ridiculous or anything, but doing the 1-d case helps a bit. Indeed, this generalizes to any number of dimensions, and the bound should be basically tight and not hard to find.
(Russia 2002, 11.8) Show that the numerator of (in reduced form) is infinitely often not a prime power.
Solution
Outline: Bound denominators using powers of and primes from to . Consider for , and note that , so consider as well. Subtracting dudes, get a contradiction.
Alternatively, use the stronger result of d'Aurizio mentioned here.
Edit: Actually, this paper looks pretty incorrect...
This post has been edited 6 times. Last edited by math154, Apr 21, 2012, 4:07 AM
combinatoricsnumber theoryp-adic valuationpigeonhole principle
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(Comment)
More storage
by math154, Jan 3, 2012, 3:29 AM
(Sierpinski) Prove that for all there exists a such that more than prime numbers can be written in the form for some integer , where is a nonconstant monic polynomial.
Solution
Let . For fixed , let denote the number of positive such that is prime, and suppose for the sake of contradiction that there exists some such that for all . Fix a real ; if is the set of primes, then
[\begin{align}
N\sum_{k\ge1}\frac{1}{k^s} &\ge\sum_{k\ge1}\frac{g(k)}{k^s} \
&=\sum_{p\in\mathbb{P}}\sum_{1\le f(T)for some independent of . Any gives the desired contradiction, since the sum of the reciprocals of the primes diverges.
Edit: Thanks.
(ROM TST 1996) Let and consider a set of pairwise distinct positive integers smaller than or equal to . Prove that one can find nine distinct numbers and three nonzero integers such that ,, and .
Solution
The key is to realize that we probably want to use pigeonhole somehow (the and are rather ugly for an inductive argument). But this is not really related to the geometry of numbers since we have at least as many equations as variables, so we want to "fix" some sort of vector and then find the 's from there (the three vectors , etc. are orthogonal to ).
Now we want to get a feel of the most natural way to fix . Note that we can parameterize all solutions to by
This gives us the idea of fixing (all nonzero for pigeonhole to work out).
Intuitively, to get the best bounds for this simplest case, we want to take (or all ). (Furthermore, if we had, say, , then it would be hard to control the requirement that are all nonzero.) Luckily, the rest works out easily: to make as small as possible (for pigeonhole to work optimally, since we know all the are less than or equal to ), let the elements of be , and for the , choose from the first set of indices, for , choose from the second set of indices, and for , choose from the last set of indices. We get a total of (optimal by AM-GM) (note that and ), yet only triples with and are possible in the first place (to count all possible triples, we can assume the smallest number chosen is and then choose two larger values in the range ), so by pigeonhole, some must occur at least times, as desired.
(USA TST 2003) For a pair of integers with , a subset of is called a skipping set for if for any . Let be the maximum size of a skipping set for . Determine the maximum and minimum values of .
Solution
Let .
For the minimum, note that we can keep doing the following: take , which kills at most two others in ( if they're in ), and repeat with the new set . Each time we remove at most elements, so we end up with at least elements, which is realized for .
For the maximum, we first analyze the one variable case to gain some intuition: of course, for fixed , we can just consider each residue modulo separately (and it's just alternating in , out , etc. for a fixed residue), where the best ratio is around , achieved for for .
With the proof and construction for the one variable case in mind, we suspect that for will give the maximum (we know it can't be far). Indeed, we can get with .
It remains to show that always. Assume for the sake of contradiction that for some .
First we get some crude estimations from alone: letting for and , we get
If is odd, this becomes , and if is even, then .
Hence . If , then , so and from the previous paragraph, . Thus , so because equality holds everywhere, we must have , which contradicts .
Otherwise, if , then , so , contradiction.
(Erdős and Selfridge) Find all positive integers with the following property: for any real numbers , knowing the numbers ,, determines the values uniquely.
Solution
The property (say, goodness) holds iff for every two "polynomials" and (with real but not necessarily integer exponents), then .
We'll show that is good iff it's not a power of 2. To do this, we want to do some sort of bounding with values in (maybe ), algebraic manipulation (especially either using to telescope products some way or differentiation), or just special values. Screwing around, it turns out differentiation is pretty helpful here.
If is not a power of 2, then by the general Leibniz rule on and the fact that
for some polynomials (both facts are easily proven by induction), we can show by induction on (base case is just ) that (sorry for sloppiness, but this is assuming the strong inductive hypothesis for )
which implies for all . Thus (by induction/Stirling numbers) for all integers , so by easy bounding considerations .
Otherwise, let . For the construction, we take to be actual polynomials (i.e. with integer exponents), so taking the induction from the previous paragraph up until , we have that for some polynomial . Plugging this in, we have . We can just take (identically); then the sets of numbers corresponding to and suffice.
Edit: There's a cleaner way using rational approximation (assume all the guys are large integers). If is a root of with multiplicity , then for some . But then means that is a power of 2, as desired.
(Brouwer-Schrijver) Prove that the minimal cardinality of a subset of that intersects all hyperplanes is .
Solution
Suppose by way of contradiction that there's a set intersecting all hyperplanes with . WLOG ; now consider the polynomial (over )
where is taken so that . Since ,. Since the coefficient of in is nonzero, CNS guarantees such that (), whence does not intersect the plane (obviously does not belong to it, and because , the second part of vanishes at and so the first cannot, whence does not intersect the plane either), contradiction.
On the other hand, the construction for is obvious: just take the set of with at least of the 's zero (if a plane does not contain the origin, it's WLOG of the form with , and so works).
Comment: This extends directly to any finite field .
This post has been edited 8 times. Last edited by math154, Dec 9, 2012, 11:06 PM
algebraic comboanalytic ntcnscombinatoricsgenerating functionspigeonhole principlePolynomials
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(Comment)
math154
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Good job on 5 problems on USAMO. I know that is a pretty bad score for someone as awesome as you on a normal USAMO, but no one got all 6 this year so it's GREAT!
VICTOR WANG 42 ON IMO 2013 LET'S GO.
See you at MOP this year (probably ).
by yugrey, May 3, 2013, 10:32 PM
you can just write "Solution" instead of "Click to reveal hidden text"
by math154, Feb 6, 2013, 6:33 PM
Hello. May I ask a stupid question: how to turn "Hidden Text" into hidden "Solution" tag?
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sorry, i've decided this blog is just for the random stuff i do. don't take it personally... you can always post things on your own blog
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Math154: It was West Somethign Middle School from Columbia that got 2nd.
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math154 mew and tinytim tied for first at chapter...
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12612 | https://pubmed.ncbi.nlm.nih.gov/33590829/ | Upadacitinib improves patient-reported outcomes vs placebo or adalimumab in patients with rheumatoid arthritis: results from SELECT-COMPARE - PubMed
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Clinical Trial
Rheumatology (Oxford)
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. 2021 Dec 1;60(12):5583-5594.
doi: 10.1093/rheumatology/keab158.
Upadacitinib improves patient-reported outcomes vs placebo or adalimumab in patients with rheumatoid arthritis: results from SELECT-COMPARE
Vibeke Strand1,Namita Tundia2,Martin Bergman3,Andrew Ostor4,Patrick Durez5,In-Ho Song2,Jeffrey Enejosa2,Casey Schlacher2,Yan Song6,Roy Fleischmann7
Affiliations Expand
Affiliations
1 Division of Immunology/Rheumatology, Stanford University, Palo Alto, CA.
2 AbbVie Inc, North Chicago, IL.
3 Drexel University College of Medicine, Philadelphia, PA, USA.
4 Cabrini Medical Centre, Monash University, Melbourne, Australia.
5 Rheumatology, Cliniques Universitaires Saint-Luc-Université Catholique de Louvain-Institut de Recherche Expérimentale et Clinique (IREC), Brussels, Belgium.
6 Analysis Group, Inc, Boston, MA.
7 University of Texas Southwestern Medical Center, MCRC, Dallas, TX, USA.
PMID: 33590829
PMCID: PMC8645276
DOI: 10.1093/rheumatology/keab158
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Clinical Trial
Upadacitinib improves patient-reported outcomes vs placebo or adalimumab in patients with rheumatoid arthritis: results from SELECT-COMPARE
Vibeke Strand et al. Rheumatology (Oxford).2021.
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. 2021 Dec 1;60(12):5583-5594.
doi: 10.1093/rheumatology/keab158.
Authors
Vibeke Strand1,Namita Tundia2,Martin Bergman3,Andrew Ostor4,Patrick Durez5,In-Ho Song2,Jeffrey Enejosa2,Casey Schlacher2,Yan Song6,Roy Fleischmann7
Affiliations
1 Division of Immunology/Rheumatology, Stanford University, Palo Alto, CA.
2 AbbVie Inc, North Chicago, IL.
3 Drexel University College of Medicine, Philadelphia, PA, USA.
4 Cabrini Medical Centre, Monash University, Melbourne, Australia.
5 Rheumatology, Cliniques Universitaires Saint-Luc-Université Catholique de Louvain-Institut de Recherche Expérimentale et Clinique (IREC), Brussels, Belgium.
6 Analysis Group, Inc, Boston, MA.
7 University of Texas Southwestern Medical Center, MCRC, Dallas, TX, USA.
PMID: 33590829
PMCID: PMC8645276
DOI: 10.1093/rheumatology/keab158
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Abstract
Objective: To evaluate the impact of upadacitinib vs placebo and adalimumab treatment, on patient-reported outcomes (PROs) in SELECT-COMPARE in an active RA population with inadequate responses to MTX (MTX-IR).
Methods: PROs in patients receiving upadacitinib (15 mg QD), placebo, or adalimumab (40 mg EOW) while on background MTX were evaluated over 48 weeks. PROs included Patient Global Assessment of Disease Activity (PtGA) and pain by visual analogue scale (VAS), the HAQ Disability Index (HAQ-DI), the 36-Item Short Form Survey (SF-36), morning (AM) stiffness duration and severity, the Functional Assessment of Chronic Illness Therapy-Fatigue (FACIT-F), and work instability. Least squares mean (LSM) changes and proportions of patients reporting improvements ≥ minimal clinically important differences (MCIDs) and scores ≥ normative values were evaluated.
Results: Upadacitinib and adalimumab resulted in greater LSM changes from baseline vs placebo across all PROs (P < 0.05) at week 12, and pain and AM stiffness severity (P < 0.05) at week 2. More upadacitinib- vs placebo-treated (P < 0.05) and similar percentages of upadacitinib- vs adalimumab-treated patients reported improvements ≥ MCID across all PROs at week 12. Upadacitinib vs adalimumab resulted in greater LSM changes from baseline in PtGA, pain, HAQ-DI, stiffness severity, FACIT-F, and the SF-36 Physical Component Summary (PCS) (all P < 0.05) at week 12. More upadacitinib- vs adalimumab-treated patients reported scores ≥ normative values in HAQ-DI and SF-36 PCS (P < 0.05) at week 12. More upadacitinib- vs adalimumab-treated patients maintained clinically meaningful improvements in PtGA, pain, HAQ-DI, FACIT-F, and AM stiffness through 48 weeks.
Conclusion: In MTX-IR patients with RA, treatment with upadacitinib resulted in statistically significant and clinically meaningful improvements in PROs equivalent to or greater than with adalimumab.
Trial registration: ClinicalTrials.gov, NCT02629159.
Keywords: DMARDs; inflammation; outcome measures; quality of life; rheumatoid arthritis.
© The Author(s) 2021. Published by Oxford University Press on behalf of the British Society for Rheumatology.
PubMed Disclaimer
Figures
Fig . 1
SF-36 domain scores at baseline…
Fig . 1
SF-36 domain scores at baseline and week 12 for all treatment groups relative…
Fig. 1
SF-36 domain scores at baseline and week 12 for all treatment groups relative to age- and gender-matched normative values ADA, adalimumab; A/G norms, age- and gender-matched normative values; BL, baseline; BP, bodily pain; GH, general health; MH, mental health; PBO, placebo; PF, physical functioning; RE, role-emotional; RP, role-physical; SF-36, 36-Item Short-Form Health Survey; SF, social functioning; UPA, upadacitinib; VT, vitality.
Fig . 2
Patients reporting PRO score improvements…
Fig . 2
Patients reporting PRO score improvements ≥ MCID at week 12 a P<…
Fig. 2
Patients reporting PRO score improvements ≥ MCID at week 12 a P< 0.05 for UPA vs PBO; b P< 0.05 for UPA vs ADA. ADA, adalimumab; AM, morning; BP, bodily pain; FACIT-F, Functional Assessment of Chronic Illness Therapy–Fatigue; GH, general health; HAQ-DI, HAQ-Disability Index; MCID, minimal clinically important difference; MCS, Mental Component Summary; MH, mental health; NNT, number needed to treat; PBO, placebo; PCS, Physical Component Summary; PF, physical functioning; PRO, patient-reported outcome; PtGA, Patient’s Global Assessment of Disease Activity; RA-WIS; Work Instability Scale for RA; RE, role-emotional; RP, role-physical; SF, social functioning; SF-36, 36-Item Short-Form Health Survey; UPAD, upadacitinib; VAS, visual analog scale; VT, vitality.
Fig . 3
Patients reporting PRO scores ≥…
Fig . 3
Patients reporting PRO scores ≥ normative values at baseline and week 12 a…
Fig. 3
Patients reporting PRO scores ≥ normative values at baseline and week 12 a P< 0.01 for UPAD vs PBO; b P< 0.05 for UPA vs ADA. ADA, adalimumab; BL, baseline; BP, bodily pain; FACIT-F, Functional Assessment of Chronic Illness Therapy–Fatigue; GH, general health; HAQ-DI, HAQ-Disability Index; MCS, Mental Component Summary; MH, mental health; PBO, placebo; PCS, Physical Component Summary; PF, physical functioning; PRO, patient-reported outcome; RA-WIS, Work Instability Scale for RA; RE, role-emotional; RP, role-physical; SF, social functioning; SF-36, 36-Item Short-Form Health Survey; UPA, upadacitinib; VT, vitality.
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References
Smolen JS, Aletaha D, McInnes IB.. Rheumatoid arthritis. Lancet 2016;388:2023–38. - PubMed
Halls S, Dures E, Kirwan J. et al. Stiffness is more than just duration and severity: a qualitative exploration in people with rheumatoid arthritis. Rheumatology (Oxford) 2015;54:615–22. - PMC - PubMed
Hewlett S, Cockshott Z, Byron M. et al. Patients’ perceptions of fatigue in rheumatoid arthritis: overwhelming, uncontrollable, ignored. Arthritis Rheum 2005;53:697–702. - PubMed
Pollard LC, Choy EH, Gonzalez J. et al. Fatigue in rheumatoid arthritis reflects pain, not disease activity. Rheumatology (Oxford) 2006;45:885–9. - PubMed
Westhoff G, Buttgereit F, Gromnica-Ihle E. et al. Morning stiffness and its influence on early retirement in patients with recent onset rheumatoid arthritis. Rheumatology (Oxford) 2008;47:980–4. - PubMed
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Advanced Logic Puzzle Techniques
Logic puzzles are a lot like Ferris wheels: equal parts relaxing, thrilling, and a lot of fun… until you get stuck on one! We all know the frustration of glaring back and forth between the clues and the grid, checking every word and every box, increasingly becoming convinced that the puzzle creator just MUST have made a mistake.
In the interest of brevity, the puzzle solving instructions you’ll find in a printed logic grid puzzle book typically only cover the most basic of techniques to get you started. The aim of this article is to uncover the remaining secrets standing between you and those taunting puzzles that threaten to hasten the ageing process with just a few confounding sentences!
If you take some time now to master the advanced techniques for solving logic puzzles, then together with some brain power (perhaps a coffee or two) you can systematically solve logic puzzles of all levels of difficulty using the puzzle grid.
At the very least, the next time you find yourself stuck on a logic puzzle, the tips outlined below might just get the wheels moving again.
For most of the examples in this article we'll use a simple puzzle scenario where five students are transferring into a new class. The objective would be to determine which subject each student transfers into, the name of the teacher, and the month they transfer into that class.
Note: If you’ve landed on this article and instead need to start with the basics, check out Solving Logic Puzzles For Beginners.
Let’s get started!
1. Advanced Grid Techniques: It’s One Of Two Options
Notice in the circled area of the below example, for the subjects History and Social Studies the subjects have the same two options available to them for Teacher – Mr Duffy and Miss Savage.
Where you see a pattern like this, there is an opportunity to do some additional elimination. And we all know we’re here because we love crossing things out!
Because these two teacher options are going to be shared between the subjects History and Social Studies, we can eliminate those two teacher names for all other subjects, as below:
That’s it for the first tip, a simple one to look out for!
2. Greater Than And Less Than Clues
Consider the following example clue:
Clue: Jonathan transferred classes one month after the Social Studies student, but sometime before Miss Savage's student.
This is the type of clue we’d initially draw some information from, then return to later when we’ve worked through the other clues. But some of that initial information is often missed, so let’s quickly run through what this clue tells us straight away:
Jonathan can not have transferred classes in the first month available.
Jonathan can not have transferred classes in the last month available.
Jonathan is not the Social Studies student.
Jonathan does not have Miss Savage as a teacher.
The Social Studies student can not have transferred classes in either of the last two months available (because two students need to come after that person – Jonathan and Miss Savage’s student).
The Social Studies student does not have Miss Savage as a teacher.
Miss Savage’s student can not have transferred classes in either of the first two months available.
Because Jonathan has the next sequential month immediately after the Social Studies student, we can check the grid for any months the Social Studies student has crossed out from other clues. Jonathan can not have the next month after any that are crossed out.
Vice versa, if we had already crossed out any other months for Jonathan from other clues, we can cross out the month immediately prior to that month for the Social Studies student.
When you have sequential clues like this, it can help to write a quick note to visualize the relationship, i.e.:
Social Studies >(1 month) Jonathan >? Miss Savage
3. Advanced Grid Techniques: Consecutive Fact Pairs
Say we know from the clues that “Andrew transferred classes one month after Amanda” and the below is the state of our grid from other eliminations.
It has been previously determined that Amanda is the student who transferred in either April or May, and that Andrew is the student who transferred in either May or June. If Amanda transferred classes in April, our clue tells us that Andrew will have transferred classes the following month, May. And if Amanda transferred classes in May, then Andrew transferred classes in June.
This situation gives us an additional piece of information – in either scenario, the month of May has to be used by one of these two students, Amanda or Andrew. Therefore, we can eliminate the month of May for all other students:
4. Advanced Grid Techniques: As Above, So Below… And Across
We have touched on this principle in our guide that covers the basic techniques: Solving Logic Puzzles For Beginners, but it is an essential concept to understand in order to fill in every detail you can in a logic puzzle grid!
Tip: Take note of the symmetry in the way we mark information in the grid and you might find these techniques easier to recognize.
Example 1
Take the below partially filled out grid as an example, where we have highlighted the rows and columns we will be looking at:
Anywhere where there is a tick, the information along the row can be transferred to the column below, and vice versa. Why? Using the student “Bella” as an example in the above grid, we know that Bella changed classes in the month of March. So, we can consider any fact about Bella and any fact about the month of March to be one and the same.
Looking at the above grid, we know that the person who changed classes in the month of March did not transfer into Mathematics nor History, and we also know they have the teacher Mr Garcia. All of this information applies to Bella, since we know Bella is the March student, so we can repeat it down Bella’s column as so:
Note where we have just ticked the intersection of Bella and Mr Garcia, the information in the Subject section below – that the subject is not Mathematics nor History – has also been applied to the row, in the section to the right of Mr Garcia’s name.
If you are new to logic puzzles it may seem redundant to mark this information in multiple places, but we do it because one of the sections of facts may have additional information filled in that the other sections don’t have (perhaps not early on, but later as we work through the clues), and this will get us closer to being able to tick more boxes and solve the puzzle.
Example 2
Below we have a similar example, but in this case some of the information in the columns highlighted needs to be marked in the corresponding rows where there is a box ticked, and vice versa – there is information in the rows that needs to be applied to the columns beneath the ticks.
Once we have filled in all the corresponding information, the grid will look like this:
Example 3
So far we have examined transferring the information below or to the right of a ticked box, but we can also deduct more information by looking above or to the left of a ticked box. Take a look at this example, with some areas highlighted:
First, let’s look at the row highlighted yellow, where we know that the teacher who had a student transfer in March is Mr Garcia. Notice that we also know the March student did not transfer into the subjects Mathematics, History, or Social Studies. Given that the March student and Mr Garcia are one and the same, the same subject eliminations apply to Mr Garcia.
Therefore, in the section where Teacher and Subject intersect, we will be able to cross out Mathematics, History, and Social Studies for Mr Garcia.
Looking at the row highlighted in blue, the same principal applies for the April student who we know transferred into Mr Duffy’s class. We know that student is not Toby, Andrew, or Jonathan. Therefore, in the section where Teacher and Student intersect, we can cross out Toby, Andrew, and Jonathan against Mr Duffy’s name.
Finally, in the purple highlighted row we see that Ms Watson has the subject Biology. We also see that Ms Watson does not have students Amanda nor Andrew. So, in the bottom section of the grid where Subject and Student intersect, we can cross out Amanda and Andrew for Biology.
The result of applying all of this information looks like this:
But that’s not all! Take a look at the tick showing Ms Watson has Biology. Travel from there up the Biology column, and you will see a cross indicating that Biology wasn’t the class a student changed to in June. Because Ms Watson and Biology can be treated as one and the same, we can also deduce from this that Ms Watson did not have the June student. We can cross out June for Ms Watson in the far right section where Month and Teacher intersect.
Staying in the section where Month and Teacher intersect, we can see that Ms Watson does not have the March student nor the April student. This information can be transferred to the section where Month and Subject intersect, because we know that Ms Watson is the Biology teacher.
The result of these changes looks like this:
You’ll also notice that thanks to all those eliminations, we now know that Mr Garcia, who has the March student, has the subject Geography. This can be ticked in two sections. But, it’s time for us to move on to the next technique!
5. Unique List Clues
This is a more complex clue type that initially allows you to eliminate a number of options, but can have additional value later once you’ve filled out more of the puzzle.
Clue: The five students are the one who changed classes in March, the one who has Mrs Knight as their teacher, Andrew, the one who has Ms Watson as their teacher, and the student studying Social Studies.
Part 1: The Initial Eliminations
From the information in the clue, we can extract enough information to fill out the grid as above. Why? The clue lists the 5 unique people in the puzzle by referencing one of their attributes. This means that none of the five people are referenced twice in that clue – therefore the other attributes mentioned are immediately ruled out for each person.
For example, we can determine that:
– The student who changed classes in March can not have Mrs Knight nor Ms Watson as a teacher, is not Andrew, and is not studying Social Studies.
– The student who has Mrs Knight as their teacher can not have changed classes in March, can not be Andrew, and is not studying Social Studies.
– Andrew can not have changed classes in March, does not have Mrs Knight nor Ms Watson as a teacher, and is not studying social studies.
And so on.
Is this clue exhausted now? Not necessarily!
Part 2: Revisiting The Clue
Let’s say you have worked through all the clues in the puzzle and have filled out a lot of the grid, but you are a bit stumped. Your grid happens to look like this:
For reference, here is the clue again:
Clue: The five students are the one who changed classes in March, the one who has Mrs Knight as their teacher, Andrew, the one who has Ms Watson as their teacher, and the student studying Social Studies.
Because this clue represents every person in the puzzle, we know that each person is represented by one and only one of those five facts. Knowing this, we can pick out someone from the puzzle and try to determine where they fit into the clue.
In the grid above, take a look at the student named Toby and consider which person he would be in the clue.
He can’t be Andrew, because we know him by name to be Toby.
He can’t be the one who changed classes in March, because we know that Toby changed classes in July.
There is a cross next to Social Studies for his name, so he can’t be the one studying Social Studies.
The only options left are that his teacher is either Ms Watson or Mrs Knight – however there is a cross next to Mrs Knight for Toby’s column.
By process of elimination, Toby has to be the person with Ms Watson as a teacher.
Keep looking for more ways to use the information in this clue. For example, if we examine the updated grid:
We can see that there are only two Student name options remaining for the teacher Mrs Knight, who was mentioned in our clue. The options are Amanda and Jonathan. However if we look down Amanda’s column, it shows that she is studying Social Studies – which was also mentioned in our clue. Therefore, Amanda can’t be the one who has Mrs Knight as her teacher, as Amanda has already been covered in the list of students in our clue. Jonathan has to be the student with Mrs Knight as his teacher.
Next, if we take a look at the student Bella, we know that she does not have Mrs Knight or Ms Watson as her teacher, and she is not studying Social Studies, and she can’t be the student named Andrew, so looking at our clue there is only one option that remains – Bella must be the student who changed classes in March.
6. Double Pair Clues
For this example, imagine we have the following clue:
Of the student who transferred in June and the student who transferred to Mrs Knight's class, one had the subject Mathematics and the other was Andrew.
This clue type describes two unique people in the puzzle, we just don’t know which facts match up with each other.
Let’s say we already have some information in the grid from other clues:
The initial information we can get from the clue is that Mrs Knight’s student was not the June transfer, and Andrew did not transfer into Mathematics. We have already made those simple eliminations in the above grid, which have been highlighted.
Once we have been through the other clues in a puzzle and filled some information into the grid, we can return to a clue like this and potentially draw more information from it.
The first recommendation upon returning to this kind of clue is to check if any eliminations have been made that solve how the pairs of facts match up. For example, if June was now crossed out for Andrew, we would know that Andrew must be the one who transferred to Mrs Knight’s class, and that the June student has to be the one who transferred into a Mathematics class. The clue would then be solved.
However if we can’t completely solve the clue yet, we may be able to make further eliminations with this clue type by checking the grid to see if any of these pairs have anything in common. Let’s see the clue again:
Of the student who transferred in June and the student who transferred to Mrs Knight's class, one had the subject Mathematics and the other was Andrew.
Firstly, if there are any facts we’ve already determined about both the June transfer and Mrs Knight that they share, we can apply the same information to both the subject Mathematics and Andrew (they are the same two people, we just don’t know which is which). The above grid does not have any information that helps with this scenario though.
Now let’s look at the second half of the clue – are there any facts in the grid shared between both Andrew and Mathematics? They both have the months March and July crossed out! This means we can also cross out these months for Mrs Knight, because she is the teacher of either Andrew or the student who transferred into Mathematics.
It might be easier to digest this reasoning in simple logic statements:
Mrs Knight matches to either Andrew or Mathematics
Neither Andrew nor Mathematics match to March or July
Therefore Mrs Knight does not match to March or July.
Additionally, the grid shows that Andrew did not transfer into Geography or Social Studies. The clue tells us that the June student and Mrs Knight’s student are Andrew and the Mathematics student (without knowing which is which) – so we can extrapolate that neither the June student nor Mrs Knight’s student transferred into Geography or Social Studies.
The above findings allow us to make these grid eliminations:
7. Looking For Elimination Patterns
This is another fun pattern to look out for! Take a look at the below example and see if you can deduce an additional elimination we can make.
As displayed in the grid, previous eliminations show us that we know Bella’s month is either March or April. We also know that Biology’s month is not March nor April.
This tells us that Bella and Biology can not possibly share any of the month options, therefore we can determine that Bella can not be the student who transferred into the subject of Biology, and we can put an X in the grid where Bella and Biology intersect.
Of course, in the real world you’re going to be studying a grid that has many more markings than this simplified example, and it can take some time to spot these patterns.
Here’s a more complex example:
Based on available Month options, we can see that Toby can not be the student who transferred into Geography (The geography transfer was in either March or May, but Toby has both those months crossed out).
As previously determined, Bella can not be the Biology student (we’ve crossed this out already).
The month of April is either Mathematics or Social Studies. Looking at where Teachers and Subjects intersect, we can see that Mr Garcia and Miss Savage both have both those subjects crossed out, therefore neither of those Teachers can match to the month of April.
Again, the month of April is either Mathematics or Social Studies. Looking down Toby’s column, we can see he has both those subjects crossed out. Toby can not be April.
Where Subject and Student intersect, we can see that Geography is either Toby or Bella. Where Month and Student intersect, we can see that May is crossed out for both Toby and Bella. Therefore, May can not match to Geography.
8. Sometimes You Need To Take Notes
Some harder puzzles get so complex that unless your brain is as sharp as a surgeon’s scalpel, you’re going to need to take notes.
Consider the following set of clues:
Andrew transferred classes one month before Ms Watson’s student.
The Social Studies student changed classes one month after Mr Garcia’s student, but one month before Jonathan.
Toby changed classes sometime after the Mathematics student, who changed classes sometime after Bella.
The History student changed classes one month before the Biology student.
Some simple scribbles that lay out the order of each group of facts should help you visualize which consecutive arrangements can and can’t work.
9. Trial And Error Is OK Too !
One thing you can rely on with a logic grid puzzle is that there is only a single solution, and it can be arrived at entirely with logic – no guesswork.
But if you’re stuck, it is okay to test out a possible solution and see if it works – this is a form of logical deduction in itself!
Ready to give a hard logic puzzle a try? Here are a few to choose from:
Market Macarons (Hard Logic Puzzle)
Jigsaw Night (Hard Logic Puzzle)
The Cheesy Chuckle Club (Extreme Logic Puzzle)
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26.2: Energy Flow through Ecosystems
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: Dec 13, 2021
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26.1: Ecology of Ecosystems
26.3: Biogeochemical Cycles
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: 70923
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All living things require energy in one form or another. Energy is required by most complex metabolic pathways (often in the form of adenosine triphosphate, ATP), especially those responsible for building large molecules from smaller compounds, and life itself is an energy-driven process. Living organisms would not be able to assemble macromolecules (proteins, lipids, nucleic acids, and complex carbohydrates) from their monomeric subunits without a constant energy input.
It is important to understand how organisms acquire energy and how that energy is passed from one organism to another through food webs and their constituent food chains. Food webs illustrate how energy flows directionally through ecosystems, including how efficiently organisms acquire it, use it, and how much remains for use by other organisms of the food web.
How Organisms Acquire Energy in a Food Web
Energy is acquired by living things in three ways: photosynthesis, chemosynthesis, and the consumption and digestion of other living or previously living organisms by heterotrophs.
Photosynthetic and chemosynthetic organisms are both grouped into a category known as autotrophs: organisms capable of synthesizing their own food (more specifically, capable of using inorganic carbon as a carbon source). Photosynthetic autotrophs (photoautotrophs) use sunlight as an energy source, whereas chemosynthetic autotrophs (chemoautotrophs) use inorganic molecules as an energy source. Autotrophs are critical for all ecosystems. Without these organisms, energy would not be available to other living organisms and life itself would not be possible.
Photoautotrophs, such as plants, algae, and photosynthetic bacteria, serve as the energy source for a majority of the world’s ecosystems. These ecosystems are often described by grazing food webs. Photoautotrophs harness the solar energy of the sun by converting it to chemical energy in the form of ATP (and NADP). The energy stored in ATP is used to synthesize complex organic molecules, such as glucose.
Chemoautotrophs are primarily bacteria that are found in rare ecosystems where sunlight is not available, such as in those associated with dark caves or hydrothermal vents at the bottom of the ocean (Figure 26.2.1). Many chemoautotrophs in hydrothermal vents use hydrogen sulfide (H2S), which is released from the vents as a source of chemical energy. This allows chemoautotrophs to synthesize complex organic molecules, such as glucose, for their own energy and in turn supplies energy to the rest of the ecosystem.
Productivity within Trophic Levels
Productivity within an ecosystem can be defined as the percentage of energy entering the ecosystem incorporated into biomass in a particular trophic level. Biomass is the total mass, in a unit area at the time of measurement, of living or previously living organisms within a trophic level. Ecosystems have characteristic amounts of biomass at each trophic level. For example, in the English Channel ecosystem, the primary producers account for a biomass of 4 g/m2 (grams per meter squared), while the primary consumers exhibit a biomass of 21 g/m2.
The productivity of the primary producers is especially important in any ecosystem because these organisms bring energy to other living organisms by photoautotrophy or chemoautotrophy. The rate at which photosynthetic primary producers incorporate energy from the sun is called gross primary productivity. An example of gross primary productivity is shown in the compartment diagram of energy flow within the Silver Springs aquatic ecosystem as shown (Figure 26.2.2). In this ecosystem, the total energy accumulated by the primary producers (gross primary productivity) was shown to be 20,810 kcal/m2/yr.
Because all organisms need to use some of this energy for their own functions (like respiration and resulting metabolic heat loss) scientists often refer to the net primary productivity of an ecosystem. Net primary productivity is the energy that remains in the primary producers after accounting for the organisms’ respiration and heat loss. The net productivity is then available to the primary consumers at the next trophic level. In our Silver Spring example, 13,187 of the 20,810 kcal/m2/yr were used for respiration or were lost as heat, leaving 7,632 kcal/m2/yr of energy for use by the primary consumers.
Ecological Efficiency: The Transfer of Energy between Trophic Levels
As illustrated in Figure 26.2.2, large amounts of energy are lost from the ecosystem from one trophic level to the next level as energy flows from the primary producers through the various trophic levels of consumers and decomposers.
The main reason for this loss is the second law of thermodynamics, which states that whenever energy is converted from one form to another, there is a tendency toward disorder (entropy) in the system. In biologic systems, this means a great deal of energy is lost as metabolic heat when the organisms from one trophic level consume the next level. In the Silver Springs ecosystem example, we see that the primary consumers produced 1103 kcal/m2/yr from the 7618 kcal/m2/yr of energy available to them from the primary producers. The measurement of energy transfer efficiency between two successive trophic levels is termed the trophic level transfer efficiency (TLTE) and is defined by the formula:
TLTE=production at present trophic levelproduction at previous trophic level×100
In Silver Springs, the TLTE between the first two trophic levels was approximately 14.8 percent. The low efficiency of energy transfer between trophic levels is usually the major factor that limits the length of food chains observed in a food web. The fact is, after four to six energy transfers, there is not enough energy left to support another trophic level. In the Lake Ontario example, only three energy transfers occurred between the primary producer, (green algae), and the apex consumer (Chinook salmon).
Ecologists have many different methods of measuring energy transfers within ecosystems. Some transfers are easier or more difficult to measure depending on the complexity of the ecosystem and how much access scientists have to observe the ecosystem. In other words, some ecosystems are more difficult to study than others, and sometimes the quantification of energy transfers has to be estimated.
Another main parameter that is important in characterizing energy flow within an ecosystem is the net production efficiency. Net production efficiency (NPE) allows ecologists to quantify how efficiently organisms of a particular trophic level incorporate the energy they receive into biomass; it is calculated using the following formula:
NPE=net consumer productivityassimilation×100
Net consumer productivity is the energy content available to the organisms of the next trophic level. Assimilation is the biomass (energy content generated per unit area) of the present trophic level after accounting for the energy lost due to incomplete ingestion of food, energy used for respiration, and energy lost as waste. Incomplete ingestion refers to the fact that some consumers eat only a part of their food. For example, when a lion kills an antelope, it will eat everything except the hide and bones. The lion is missing the energy-rich bone marrow inside the bone, so the lion does not make use of all the calories its prey could provide.
Thus, NPE measures how efficiently each trophic level uses and incorporates the energy from its food into biomass to fuel the next trophic level. In general, cold-blooded animals (ectotherms), such as invertebrates, fish, amphibians, and reptiles, use less of the energy they obtain for respiration and heat than warm-blooded animals (endotherms), such as birds and mammals. The extra heat generated in endotherms, although an advantage in terms of the activity of these organisms in colder environments, is a major disadvantage in terms of NPE. Therefore, many endotherms have to eat more often than ectotherms to get the energy they need for survival. In general, NPE for ectotherms is an order of magnitude (10x) higher than for endotherms. For example, the NPE for a caterpillar eating leaves has been measured at 18 percent, whereas the NPE for a squirrel eating acorns may be as low as 1.6 percent.
The inefficiency of energy use by warm-blooded animals has broad implications for the world’s food supply. It is widely accepted that the meat industry uses large amounts of crops to feed livestock, and because the NPE is low, much of the energy from animal feed is lost. For example, it costs about 1¢ to produce 1000 dietary calories (kcal) of corn or soybeans, but approximately $0.19 to produce a similar number of calories growing cattle for beef consumption. The same energy content of milk from cattle is also costly, at approximately $0.16 per 1000 kcal. Much of this difference is due to the low NPE of cattle. Thus, there has been a growing movement worldwide to promote the consumption of non-meat and non-dairy foods so that less energy is wasted feeding animals for the meat industry.
Modeling Ecosystems Energy Flow: Ecological Pyramids
The structure of ecosystems can be visualized with ecological pyramids, which were first described by the pioneering studies of Charles Elton in the 1920s. Ecological pyramids show the relative amounts of various parameters (such as the number of organisms, energy, and biomass) across trophic levels.
Pyramids of numbers can be either upright or inverted, depending on the ecosystem. As shown in Figure 26.2.3, typical grassland during the summer has a base of many plants and the numbers of organisms decrease at each trophic level. However, during the summer in a temperate forest, the base of the pyramid consists of few trees compared with the number of primary consumers, mostly insects. Because trees are large, they have great photosynthetic capability and dominate other plants in this ecosystem to obtain sunlight. Even in smaller numbers, primary producers in forests are still capable of supporting other trophic levels.
Another way to visualize ecosystem structure is with pyramids of biomass. This pyramid measures the amount of energy converted into living tissue at the different trophic levels. Using the Silver Springs ecosystem example, this data exhibits an upright biomass pyramid (Figure 26.2.3), whereas the pyramid from the English Channel example is inverted. The plants (primary producers) of the Silver Springs ecosystem make up a large percentage of the biomass found there. However, the phytoplankton in the English Channel example make up less biomass than the primary consumers, the zooplankton. As with inverted pyramids of numbers, this inverted pyramid is not due to a lack of productivity from the primary producers but results from the high turnover rate of the phytoplankton. The phytoplankton are consumed rapidly by the primary consumers, thus, minimizing their biomass at any particular point in time. However, phytoplankton reproduce quickly, thus they are able to support the rest of the ecosystem.
Pyramid ecosystem modeling can also be used to show energy flow through the trophic levels. Pyramids of energy are always upright, and an ecosystem without sufficient primary productivity cannot be supported. All types of ecological pyramids are useful for characterizing ecosystem structure. However, in the study of energy flow through the ecosystem, pyramids of energy are the most consistent and representative models of ecosystem structure.
Consequences of Food Webs: Biological Magnification
One of the most important environmental consequences of ecosystem dynamics is biomagnification. Biomagnification is the increasing concentration of persistent, toxic substances in organisms at each trophic level, from the primary producers to the apex consumers. Many substances have been shown to bioaccumulate, including classical studies with the pesticide dichlorodiphenyltrichloroethane (DDT), which was published in the 1960s bestseller, Silent Spring, by Rachel Carson. DDT was a commonly used pesticide before its dangers became known. In some aquatic ecosystems, organisms from each trophic level consumed many organisms of the lower level, which caused DDT to increase in birds (apex consumers) that ate fish. Thus, birds accumulated sufficient amounts of DDT to cause fragility in their eggshells. This effect increased egg breakage during nesting and was shown to have adverse effects on these bird populations. The use of DDT was banned in the United States in the 1970s.
Other substances that biomagnify are polychlorinated biphenyls (PCBs), which were used in coolant liquids in the United States until their use was banned in 1979, and heavy metals, such as mercury, lead, and cadmium. These substances were best studied in aquatic ecosystems, where fish species at different trophic levels accumulate toxic substances brought through the ecosystem by the primary producers. As illustrated in a study performed by the National Oceanic and Atmospheric Administration (NOAA) in the Saginaw Bay of Lake Huron (Figure 26.2.4), PCB concentrations increased from the ecosystem’s primary producers (phytoplankton) through the different trophic levels of fish species. The apex consumer (walleye) has more than four times the amount of PCBs compared to phytoplankton. Also, based on results from other studies, birds that eat these fish may have PCB levels at least one order of magnitude higher than those found in the lake fish.
Other concerns have been raised by the accumulation of heavy metals, such as mercury and cadmium, in certain types of seafood. The United States Environmental Protection Agency (EPA) recommends that pregnant women and young children should not consume any swordfish, shark, king mackerel, or tilefish because of their high mercury content. These individuals are advised to eat fish low in mercury: salmon, tilapia, shrimp, pollock, and catfish. Biomagnification is a good example of how ecosystem dynamics can affect our everyday lives, even influencing the food we eat.
Summary
Organisms in an ecosystem acquire energy in a variety of ways, which is transferred between trophic levels as the energy flows from the bottom to the top of the food web, with energy being lost at each transfer. The efficiency of these transfers is important for understanding the different behaviors and eating habits of warm-blooded versus cold-blooded animals. Modeling of ecosystem energy is best done with ecological pyramids of energy, although other ecological pyramids provide other vital information about ecosystem structure.
References
OpenStax, Biology. OpenStax CNX. June 26, 2020.
26.1: Ecology of Ecosystems
26.3: Biogeochemical Cycles |
12615 | https://askfilo.com/user-question-answers-mathematics/if-then-find-the-range-of-when-35313439363733 | If y=x^{2}-2 x-3 then find the range of y when. | Filo
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If y=x 2−2 x−3 then find the range of y when.
x∈R
x∈[0,3]
x∈[−2,0]
x&(−∞,−3]
x∈(0,∞)
x ε(4,10]
x∈[−4,4]
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Question Text If y=x 2−2 x−3 then find the range of y when.
Updated On Aug 31, 2023
Topic Sequence Series and Quadratic
Subject Mathematics
Class Class 11
Answer Type Video solution: 2
Upvotes 243
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12616 | https://math.arizona.edu/~gabitov/teaching/141/math_485/Midterm_Presentations/Elastic_Pedulum.pdf | Dynamics of the Elastic Pendulum Qisong Xiao; Shenghao Xia ; Corey Zammit; Nirantha Balagopal; Zijun Li Agenda • Introduction to the elastic pendulum problem • Derivations of the equations of motion • Real-life examples of an elastic pendulum • Trivial cases & equilibrium states • MATLAB models The Elastic Problem (Simple Harmonic Motion) • 𝐹 𝑛𝑒𝑡= 𝑚 𝑑2𝑥 𝑑𝑡2 = −𝑘𝑥 𝑑2𝑥 𝑑𝑡2 = − 𝑘 𝑚𝑥 • Solve this differential equation to find 𝑥𝑡= 𝑐1 cos 𝜔𝑡+ 𝑐2 sin 𝜔𝑡= 𝐴𝑐𝑜𝑠(𝜔𝑡−𝜑) • With velocity and acceleration 𝑣𝑡= −𝐴𝜔sin 𝜔𝑡+ 𝜑 𝑎𝑡= −𝐴𝜔2cos(𝜔𝑡+ 𝜑) • Total energy of the system 𝐸= 𝐾𝑡+ 𝑈𝑡 = 1 2 𝑚𝑣𝑡2 + 1 2 𝑘𝑥2 = 1 2 𝑘𝐴2 The Pendulum Problem (with some assumptions) • With position vector of point mass 𝑥= 𝑙𝑠𝑖𝑛𝜃 𝑖−𝑐𝑜𝑠𝜃 𝑗, define 𝑟such that 𝑥= 𝑙 𝑟and 𝜃= 𝑐𝑜𝑠𝜃 𝑖+ 𝑠𝑖𝑛𝜃 𝑗 • Find the first and second derivatives of the position vector: 𝑑 𝑥 𝑑𝑡= 𝑙𝑑𝜃 𝑑𝑡 𝜃 𝑑2 𝑥 𝑑𝑡2 = 𝑙𝑑2𝜃 𝑑𝑡2 𝜃−𝑙𝑑𝜃 𝑑𝑡 2 𝑟 • From Newton’s Law, (neglecting frictional force) 𝑚𝑑2 𝑥 𝑑𝑡2 = 𝐹 𝑔+ 𝐹 𝑡 The Pendulum Problem (with some assumptions) Defining force of gravity as 𝐹 𝑔= −𝑚𝑔 𝑗= 𝑚𝑔𝑐𝑜𝑠𝜃 𝑟− 𝑚𝑔𝑠𝑖𝑛𝜃 𝜃and tension of the string as 𝐹𝑡= −𝑇 𝑟: −𝑚𝑙𝑑𝜃 𝑑𝑡 2 = 𝑚𝑔𝑐𝑜𝑠𝜃−𝑇 𝑚𝑙𝑑2𝜃 𝑑𝑡2 = −𝑚𝑔𝑠𝑖𝑛𝜃 Define 𝜔0 = 𝑔/𝑙to find the solution: 𝑑2𝜃 𝑑𝑡2 = −𝑔 𝑙𝑠𝑖𝑛𝜃= −𝜔0 2𝑠𝑖𝑛𝜃 Derivation of Equations of Motion • m = pendulum mass • mspring = spring mass • l = unstreatched spring length • k = spring constant • g = acceleration due to gravity • Ft = pre-tension of spring • rs = static spring stretch, 𝑟 𝑠= 𝑚𝑔−𝐹 𝑡 𝑘 • rd = dynamic spring stretch • r = total spring stretch 𝑟 𝑠+ 𝑟 𝑑 Derivation of Equations of Motion -Polar Coordinates • r = r et • v = dr dt = r er + r θ eθ = vr er + vθ eθ • a = dv dt = r −r θ2 er + r θ + 2 r θ eθ + ar er + aθ eθ • vr magnitude change r direction change r θ • vθ magnitude change r θ + r θ direction change r θ2 Derivation of Equations of Motion -Rigid Body Kinematics x y z = cosθ sinθ 0 −sinθ cosθ 0 0 0 1 X Y Z i j k = cosθ sinθ 0 −sinθ cosθ 0 0 0 1 I J K Derivation of Equations of Motion -Rigid Body Kinematics Free Body Diagram After substitutions and evaluation: Derivation of Equations of Motion -Lagrange Equations Kinetic Energy Potential Energy Derivation of Equations of Motion -Lagrange Equations • Lagrange’s Equation, Nonlinear equations of motion Elastic pendulum in the real world Pendulum… but not elastic: Elastic… but not pendulum: Elastic pendulum in the real world – Spring Swinging Elastic pendulum in the real world -Bungee Jumping Trivial Cases • System not integrable • Initial condition without elastic potential • Only vertical oscillation • Initial condition with elastic potential Equilibrium States • Hook’s Law • Gravitational Force • At equilibrium • System at equilibrium Stable state Unstable at 0 ( ) e F k l l g F mg 0 0 ( ) e g F F F k l l mg 0 0 0 0 0 0 ( ) 0 x x y y l l k z z g z l m l 0 ( , , ) (0,0,2 ) x y z l l ( , , ) (0,0, ) x y z l Turning things into a handy system: 𝑥= −𝜔𝑧 2 𝑟−𝑙0 𝑟 𝑥 𝑦= −𝜔𝑧 2 𝑟−𝑙0 𝑟 𝑦 𝑧= −𝜔𝑧 2 𝑟−𝑙0 𝑟 𝑧−𝑔 Where 𝜔𝑧= 𝑘 𝑚 and 𝑟= 𝑥2 + 𝑦2 + 𝑧2 Some regimes make familiar shapes!
Initial Conditions: X = 1; Vx = 0; Y = 0; Vy = 0; Z = 1.1; Vz = 0; Parameters: w = 3; g = 9; l = 1; So motion stays in the XZ plane Positive Z is vertical.
Motion is shown to be relatively changeless over 50s What happens if we shake things up?
Initial Conditions X = 1.1; Vx = 0; Y = 0; Vy = 0; Z = 1.1; Vz = 0; Parameters: w = 3; g = 9; l = 1; Turning things up a bit… Initial Conditions: X = 1; Vx = 0; Y = 0; Vy = .2; Z = 1.1; Vz = 0; Parameters w = 1; g = 10; l = 1; totalTime = 200; stepsPerSec = 10; Let’s take a closer look at the same regime: This is the XZ 𝑿𝟎 plane Z-axis↑ X-Axis → Now look at the XY plane again.
Is it the swivel that is causing the pendulum to avoid the center?
Awesome they almost meet!
(Quasi-awesome) Δ𝑇= 308𝑠 𝑌↑ 𝑋→ ← Dr. Peter Lynch’s model: Initial Conditions: x0=0.01; xdot0=0.00; y0=0.00; ydot0=0.02; zprime0=0.1; zdot0=0.00; References (1/2) • Thanks to our mentor Joseph Gibney for getting us started on the MATLAB program and the derivations of equations of motion.
• Special thanks to Dr. Peter Lynch of the University College Dublin, Director of the UCD Meteorology & Climate Centre, for emailing his M-file and allowing us to include video of it’s display of the fast oscillations of the dynamic pendulum!
• Craig, Kevin: Spring Pendulum Dynamic System Investigation. Rensselaer Polytechnic Instititute.
• Fowles, Grant and George L. Cassiday (2005). Analytical Mechanics (7th ed.). Thomson Brooks/Cole.
• Holm, Darryl D. and Peter Lynch, 2002: Stepwise Precession of the Resonant Swinging Spring, SIAM Journal on Applied Dynamical Systems, 1, 44-64 • Lega, Joceline: Mathematical Modeling, Class Notes, MATH 485/585, (University of Arizona, 2013).
References (2/2) • Lynch, Peter, 2002: The Swinging Spring: a Simple Model for Atmospheric Balance, Proceedings of the Symposium on the Mathematics of Atmosphere-Ocean Dynamics, Isaac Newton Institute, June-December, 1996. Cambridge University Press • Lynch, Peter, and Conor Houghton, 2003: Pulsation and Precession of the Resonant Swinging Spring, Physica D Nonlinear Phenomena • Taylor, John R. (2005). Classical Mechanics. University Science Books • Thornton, Stephen T.; Marion, Jerry B. (2003). Classical Dynamics of Particles and Systems (5th ed.). Brooks Cole. • Vitt, A and G Gorelik, 1933: Oscillations of an Elastic Pendulum as an Example of the Oscillations of Two Parametrically Coupled Linear Systems. Translated by Lisa Shields, with an Introduction by Peter Lynch. Historical Note No. 3, Met Éireann, Dublin (1999) • Walker, Jearl (2011). Principles of Physics (9th ed.). Hoboken, N.J. : Wiley.
• Lynch, Peter, 2002:. Intl. J. Resonant Motions of the Three-dimensional Elastic Pendulum Nonlin. Mech., 37, 345-367. |
12617 | https://exceljet.net/formulas/count-numbers-with-leading-zeros | You are here
Count numbers with leading zeros
Related functions
Download Worksheet (17.5 KB)
Summary
To count numbers that contain leading zeros, you can use the SUMPRODUCT function with a simple logical expression. In the example shown, the formula in cell F5 is:
=SUMPRODUCT(--(code=E5))
where code is the named range B5:B16.
Generic formula
=SUMPRODUCT(--(range=value))
Explanation
In this example, the goal is to count numbers that contain leading zeros. In cell E5, we have the code "009875" and we want to count how many times this code appears in the range B5:B16. The challenge is that Excel can be finicky with leading zeros. Technically, the values in B5:B16 are text, as is the value in E5. However, sometimes text values that contain numbers are converted to numeric values as they go through Excel's calculation engine. When this happens, the leading zeros will be silently removed, which can cause an incorrect result. The article below explains the problem in more detail. For convenience, code (B5:B16) and qty (C5:C16) are named ranges.
COUNTIF with leading zeros
A common situation where leading zeros do not behave as expected is when functions like COUNTIF, COUNTIFS, SUMIF, SUMIFS, etc. are configured to use numbers with leading zeros. To demonstrate this problem, consider the formula below:
Here the COUNTIF function is set up to count values in B4:B8 that are equal to "01". We expect a result of 1, but COUNTIF returns 5.
=COUNTIF(B4:B8,"01") // returns 5
Somewhere in the calculation process, the leading zeros get dropped and all cells evaluate to 1. This is clearly not the result we want, and shows a limitation of the COUNTIF function. Similarly, if we apply COUNTIF to the worksheet shown above, we get the incorrect result of 4:
=COUNTIF(code,E5) // returns 4
The leading zeros in "009875" are stripped, and 9875 is counted 4 times, when the correct result for "009875"is 2.
Note: COUNTIF is in a group of 8 functions that share some particular quirks and limitations.
SUMPRODUCT solution
A simple solution to this problem is to use the SUMPRODUCT function like this:
=SUMPRODUCT(--(code=E5))
Working from the inside out, we are using the following expression as a logical test:
code=E5
Because code is the named range B5:B16, which contains 12 values, the expression yields 12 TRUE/FALSE results in an array like this:
{FALSE;FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;TRUE;FALSE;FALSE;FALSE}
The TRUE values in the array correspond to the cells in B5:B16 that contain "009875". You can see we have TRUE at the fourth cell (B8) and the ninth cell (B13).
Next, we use a double negative (--) to coerce the TRUE/FALSE values to 1s and 0s, which creates the following array:
{0;0;0;1;0;0;0;0;1;0;0;0}
This array is delivered directly to the SUMPRODUCT function, which sums the array and returns a final result:
=SUMPRODUCT({0;0;0;1;0;0;0;0;1;0;0;0}) // returns 2
This is an example of using Boolean algebra in Excel, and you will see many more advanced formulas use this technique. The nice thing about this approach is that it can be easily extended, as explained below.
Sum quantities
As you might have guessed, if you try to use the SUMIF function to sum the quantities associated with code "009875", the same problem will occur. The formula below returns 14, when the correct result is 7:
=SUMIF(code,E5,qty) // returns 14
The cause of the problem is the same: the leading zeros in "009875" get stripped during the SUMIF calculation, which causes "009875" to be grouped together with "9875", and SUMIF sums the quantities associated with both codes.
One of the nicest things about using SUMPRODUCT to perform conditional counts as we did above, is that we can easily extend the logic to perform conditional sums. In this case, all we need to do is multiply the counting logic by the named range qty (C5:C16) like this:
=SUMPRODUCT(--(code=E5)qty)
This is the formula used in cell G5 of the worksheet. Since we already know that the expression:
--(code=E5)
results in an array of 1s and 0s, we can simplify the formula like this:
=SUMPRODUCT({0;0;0;1;0;0;0;0;1;0;0;0}qty)
Then, evaluating quantities (qty), we get:
=SUMPRODUCT({0;0;0;1;0;0;0;0;1;0;0;0}{3;6;4;2;5;6;2;4;5;1;3;3})
After the two arrays are multiplied together we have:
=SUMPRODUCT({0;0;0;2;0;0;0;0;5;0;0;0}) // returns 7
Notice how the zeros in the first array "cancel out" the irrelevant quantities in the second array. In other words, the exact same logic we used to count code "009875" is used to sum quantities associated with code "009875". The final result from SUMPRODUCT is 7.
SUMPRODUCT is a workhorse function that can solve many tricky problems in Excel. See more examples here.
Note: technically the double negative (--) is not needed in the formula to sum quantities above because the math operation of multiplying the two arrays together will automatically coerce the TRUE and FALSE values in the first arrays with 1s and 0s. However, the double negative does no harm and makes the counting and summing formulas easier to compare.
Related functions
SUMPRODUCT Function
The Excel SUMPRODUCT function multiplies ranges or arrays together and returns the sum of products. This sounds boring, but SUMPRODUCT is an incredibly versatile function that can be used to count and sum like COUNTIFS or SUMIFS,...
COUNTIF Function
The Excel COUNTIF function returns the count of cells in a range that meet a single condition. The generic syntax is COUNTIF(range, criteria), where "range" contains the cells to count, and "criteria" is a condition that must be true for a cell to be counted. COUNTIF can be used to count...
SUMIF Function
The Excel SUMIF function returns the sum of cells that meet a single condition. Criteria can be applied to dates, numbers, and text. The SUMIF function supports logical operators (>,<,<>,=) and wildcards (,?) for partial matching....
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12619 | https://www.frontiersin.org/journals/oncology/articles/10.3389/fonc.2023.1308989/full | Frontiers | Epidemiology of thymomas and thymic carcinomas in the United States and Germany, 1999-2019
Frontiers in Oncology
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ORIGINAL RESEARCH article
Front. Oncol., 09 January 2024
Sec. Cancer Epidemiology and Prevention
Volume 13 - 2023 |
Epidemiology of thymomas and thymic carcinomas in the United States and Germany, 1999-2019
Tiemo Sven Gerber1Stephanie Strobl 1Alexander Marx 2Wilfried Roth 1Stefan Porubsky 1
1 Institute of Pathology, University Medical Center of the Johannes Gutenberg-University Mainz, Mainz, Germany
2 Institute of Pathology, University Medical Center Göttingen, Göttingen, Germany
Introduction: Mediastinal tumors, particularly non-neuroendocrine thymic epithelial tumors (TET) are relatively uncommon, posing challenges for extensive epidemiological studies. This study presents a comprehensive analysis of these tumors in the United States (US) and Germany (GER) from 1999 to 2019.
Methods: Patients aged 0-19 (n=478) and ≥20 years (n=17,459) diagnosed with malignant tumors of the anterior mediastinum were identified from the Surveillance, Epidemiology, and End Results registry (SEER) and the Zentrum für Krebsregisterdaten (ZfKD) databases.
Results: Among patients aged ≥20 years, TETs accounted for the most prevalent anterior mediastinal tumors (US/GER: 63%/64%), followed by lymphomas (14%/8%). For patients <20 years, predominant tumors included germ cell tumors (42%/14%), lymphomas (38%/53%), and TETs (10%/27%). The overall annual incidence of thymoma was 2.2/2.64 (US/GER) per million inhabitants and for thymic carcinomas 0.48/0.42. The male-to-female ratio was 1:1.09/1.03, and the mean age 59.48 ± 14.89/61.33 ± 13.94. Individuals with thymomas, but not thymic carcinomas, exhibited a 21%/29% significantly heightened risk of developing secondary malignancies compared to controls with non-thymic primary tumors.
Discussion: This study provides a comparative analysis of anterior mediastinal tumors, particularly TETs, in the US and GER over the past two decades. Furthermore, it highlights a significantly elevated incidence of secondary malignancies in thymoma patients.
1 Background
The anterior mediastinum is the origin of a variety of tumors such as thymic epithelial tumors (TETs), lymphomas, mediastinal germ cell tumors, mesenchymal neoplasms, and metastases (1, 2). These tumors have also a specific age distribution with TETs being the most prominent group in adults and relatively rare in children (3).
Formally, TETs consist of thymomas, thymic carcinomas, and thymic neuroendocrine tumors (NET), however, in the following manuscript, we will use TET referring only to thymomas and thymic carcinomas. The recognition of thymomas and thymic carcinomas as separate entities has been introduced in the second edition of the WHO Classification of Tumors in 1999 and the nomenclature of the major thymoma types has been maintained ever since, although the field has dynamically evolved with both increased molecular pathology, radiology, and more interdisciplinary tumor board emphasis (2, 4–7). Based on the morphology and architecture of the epithelial cells and their proportion to lymphocytes, thymomas are divided into types A, AB, B1, B2, and B3 and other rare types. Thymic carcinomas are pathologically similar to extrathymic carcinomas and show typically a squamous differentiation (8, 9).
The biological behavior of TETs differs significantly between the entities. Whereas thymic carcinomas show a clear malignant potential with often locally advanced or metastatic disease at the time of diagnosis, type B thymomas grow slowly and tend to invade locally. Type A and AB thymomas metastasize rarely and were, therefore, envisioned as benign in the absence of invasive growth. Consequently, these cases were not reported to cancer registries. Since 2015, all thymomas and thymic carcinomas have been recognized as malignant tumors and subjected to reporting (10). Next to the standard TNM classification published by the Union for International Cancer Control (UICC), the Masaoka-Koga system is in use that allows significantly better discrimination for the early stages as compared to the UICC system (8). However, since it is not recorded in cancer registries, a national evaluation taking the Masaoka-Koga system into account is difficult to achieve.
Due to the rarity of TETs, the data collection and analysis are challenging and many of the current studies rely on single-center experiences making the epidemiological data incomplete and of limited comparability. Therefore, the aim of our study was a comprehensive evaluation of cancer registry data between the years 1999 and 2019 in two industrialized countries, the United States (US) and Germany (GER), focusing on a) tumors of the anterior mediastinum in general, b) TETs, c) secondary tumors in TET patients, d) comparison of the data between the two countries, and e) in the time course.
2 Methods
2.1 Descriptive statistics
The National Cancer Institute’s Surveillance, Epidemiology, and End Results (SEER) data provide information on cancer statistics among the United States population. These data were obtained from the SEER Research Limited-Field Data, 22 Registries, Nov 2021 Sub (2000-2019; SEER 22) (11) using SEERStat software (version 8.4.0.1). The German data were retrieved from the Centre for Cancer Registry Data (Zentrum für Krebsregisterdaten, ZfKD). The data were based on the epidemiological state cancer registry data [Version Epi2021_3 (12)], searching for the International Classification of Diseases for Oncology (ICD-O-3) morphology codes 8580 to 8589, topography codes C379 and C381, as well as other tumors from those patients. The analysis was performed using Microsoft Access (version 2202) and MS Excel (version 2202). To retrieve data on tumors of the anterior mediastinum, we searched the SEER 22 and ZfKD data for the ICD-O-3 topography codes C370, C379, and C381. The tumors were grouped by differentiation. Non-solid entities (e.g., leukemias) and unspecified diagnoses (e.g. neoplasm) were excluded. Descriptive statistics on relative frequencies, male-to-female ratios, age, incidence rates, and survival time were calculated as the mean ± standard deviation for continuous variables and frequencies or percentages for categorical variables. Figures were assembled using Corel PaintShop Pro (version 22.2.0.8).
2.2 Incidence rate
Incidence rates were calculated per year, per million inhabitants, and directly age-standardized to the 2000 US standard population from the Census P25-1130 or the population count of 2010 in the German states from the Federal Statistics Office (Destatis), respectively. For Germany, due to incomplete data sets and, following the recommendations of the ZfKD in some federal states, only the data for Schleswig-Holstein, Hamburg, Bremen, Rheinland-Pfalz, Saarland, Brandenburg, Mecklenburg-Vorpommern, Sachsen, and Thüringen were evaluated. Standard errors and 95% confidence intervals (CI) were calculated using the Tiwari modification for CIs (13). Cases were selected by the ICD-O-3 (8580–8586) or the “Site and Morphology Site recode ICD-O-3/WHO 2008 (for SIRs)” to split thymus from other endocrine organs. In the analysis, significantly fewer patients were diagnosed with specific thymoma subtypes in 1999 and 2000, since the exact subtyping of thymoma only became widely used with the second edition of the WHO Classification of Tumors in 1999. Therefore, for the calculation of incidence rates of thymoma subtypes, we excluded patients diagnosed in 1999-2000. In addition to the 22 registries SEER and ZfKD data for the time frame between 1999 or 2000 and 2019, we obtained the SEER Research Data, 8 Registries, Nov 2021 Sub (1975-2019; SEER 8) for the time between 1975 and 2019 (14).
To identify the time points in which the incidence rate trend significantly changed, we conducted a logistic regression analysis using the Joinpoint Regression Program (version 4.9.1.0) (15). The settings were as follows: dependent variable: age-adjusted rate with uncorrelated standard error; independent variable: annual year of diagnosis, cohorts defined by sex; log transformation. The regression model fits linear trends to the log-transformed incidence rates, with the inclination of the trend changing at the inflection points (joinpoints). The software analyzes trend data and fits the simplest model. For the test of whether more joinpoints are statistically significant compared to no joinpoints, we used the permutation test with a significance level of .01 and 9999 permutations. The number of data points determines the default maximum possible number of joinpoints. The slope of the model represents the annual percent change (APC). The average of the overall observed period is measured in the average annual percent change (AAPC). A two-sided statistical significance t-test on a 5% significance level is used to determine if the APC and AAPC differ significantly from zero.
2.3 Survival analysis
For survival analysis of the SEER and ZfKD data, the inclusion criteria for stratification by subtype were: only malignant tumors, only with histological confirmation, only cases with the ICD-O-3 8581 to 8586, only first malignancy, and no other diagnosed malignancies. To avoid as many potential sources of error as possible, such as other malignant tumors or misclassifications, we used restrictive inclusion criteria. The inclusion criteria for stratification by tumor stage were: only malignant tumors, only with histological confirmation, only with known TNM, only cases with the ICD-O-3 8581 to 8586, only first malignancy, no other diagnosed malignancies, only valid TNM (only 8 edition), exclusion of cases before the release of the 8 th edition TNM (December 2016), and exclusion of perioperative mortality defined as death within the first month of diagnosis. The exclusion of cases predating the release of the 8 th edition TNM is attributed to the absence of a widely adopted and uniform TNM classification system for thymus tumors until that time, leading to the utilization of numerous disparate systems or the lack thereof (16). The tumor stages were classified using the Union for International Cancer Control (UICC) criteria of the 8 th edition. The analyzes were performed using SPSS (version 27.0.1.0). The log-rank test was performed to compare the estimated survival between the cohorts.
2.4 Consecutive malignancies
For the evaluation of consecutive malignancies, we identified both German and United States patients with TETs. The crude rate of the different consecutive malignancies of those patients between 2000 and 2019 was calculated. We determined the ratio between the rate of consecutive tumors and the respective number of primary tumors. As a control group, we chose all primary cancer patients in the SEER 22 cohort excluding epithelial tumors of the thymus. The rate of subsequent malignancies was reported as a ratio to the control group. We conducted a two-tailed Fisher`s exact test. Due to the comprehensive volume of data, the consecutive tumors were grouped according to lineage differentiation (i.e., for instance, lymphomas as hematopoietic neoplasms and both adenocarcinomas and squamous cell carcinomas as epithelial and so on). To avoid assessing erroneous multiple reports to the cancer registry, we excluded repeated reports of thymoma and thymic carcinoma cases. As no primary basal cell and squamous carcinomas of the skin are reported to the SEER registry, those tumors were excluded from the German cohort as well. Moreover, we calculated second tumors without considering other subsequent tumors.
3 Results
3.1 Anterior mediastinal tumors in the United States and Germany
Data on anterior mediastinal tumors obtained from the US (2000–2019) and GER (1999–2019) cancer registries were analyzed for the distribution of tumor entities in patients below 20 years of age and older (Figure 1). In patients younger than 20 years, the most frequent entities were: germ cell tumors (US: 42% and Germany: 14%), lymphomas (38% and 53%), and TETs (10% and 27%). This ratio changed in patients aged 20 years or more with TETs representing the by far most frequent entity (63% and 64%) followed by lymphomas (14%, 8%), carcinoma not otherwise specified (9%, 20%), germ cell tumors (7%, 2%), neuroendocrine tumors (5%, 5%) and sarcomas (2% and 1%). The notable disparities observed between the percentages of German and United States numbers can be attributed primarily to the relatively low number of overall patients in the German cohort of young patients, consisting of only 51 individuals, in contrast to the substantial sample size of 427 patients in the US cohort.
Figure 1
Figure 1 Tumors of patients younger than 20 years of the anterior mediastinum including thymus in the (A) United States (US; n=427) and (B) German (GER; n=51) cohort. Tumors of 20-year-old patients or older in the (C) US (n=12736) as well as (D) Germany (n=4723). In younger patients, most tumors are lymphomas and germ cell tumors, while in older patients the majority is comprised of non-neuroendocrine thymic epithelial tumors. Cases of nondescript carcinomas were considered a separate category (grey).
In the US, the inquiry identified 8,171 patients diagnosed with a TET with a male-to-female ratio of 1.09, the mean age of primary diagnosis 59.48 years (SD, 14.89), and the incidence rate of 2.68 per million inhabitants per year at risk. In Germany, 3,081 TET patients were found with a male-to-female ratio of 1.03, a mean age of 61.33 years (± 13.94), and an overall incidence rate of 3.06 per million inhabitants per year. The exact tumor subtype was reported in 5,979 (73.17%) and 2,905 cases (94.29%) in the United States and German data sets, respectively (Table 1). The differences between both countries regarding the frequency, male-to-female ratio, age, mean survival, and incidence rate were statistically not significant, except for the mean survival of thymic carcinoma patients, which was significantly better in Germany (Table 1).
Table 1
Table 1 Comparison of epidemiological features of thymomas and thymic carcinoma in Germany (GER) in 1999-2019 and the United States (US) in 2000-2019.
Most epithelial tumors of the thymus were diagnosed between 40 and 60 years of age, overall, with only minor variations between the sexes. The distribution of subtypes concerning the age at diagnosis also showed a consistent pattern (Figure 2). However, when conducting a subgroup analysis, we observed notable differences between the United States and Germany. Thymic carcinomas were diagnosed relatively frequently in the United States, while in Germany, AB thymomas were reported more frequently (Supplementary Figure 1). Additionally, in a comprehensive cohort analysis of patients with malignancies of the anterior mediastinum spanning from 1975 to 2019, it can be found that the overall incidence rate of Thymus tumors increased over time, primarily driven by the rise in carcinomas (Supplementary Figure 2).
Figure 2
Figure 2(A) Age distribution of thymic epithelial tumors, non-neuroendocrine (TETs) subtypes, and (B) age and sex distribution of all TETs taking the US (2000–2019) and German (1999–2019) data together.
3.2 Incidence analysis of thymic epithelial tumors subtypes
To reveal changes in the incidence of TETs during the observation period, an incidence analysis was performed. This analysis shows incidence rate trends and identifies joinpoints, at which the inclination of the trend changes. In the regression model for the evaluation of the age-adjusted incidence rates of TETs in the US, one such joinpoint was identified in 2007 (Figure 3). Overall, the incidence rates increased until 2007 and remained stable thereafter at 2.8 per million inhabitants per year. For both sexes, the average annual percent change (AAPC) was statistically significant on a 5% significance level [1.4%, 95% CI (0.3; 2.5), p =.01]. For female patients, the AAPC was 1.5% [95% CI (0.7; 2.3), p<.01]. For male patients the AAPC was lower with 0.6% [95% CI (-0.1; 1.3), p=.07]. Regarding tumor subtypes, AB thymomas showed a rise in incidence with an AAPC of 4.3% [95% CI (3.2; 5.5), p<.01]. This change occurred for both sexes (male 4.9%, female 3.9%, both p< .01). B2 thymomas showed an overall AAPC of 8.2% [95% CI (4.5; 11.9); p<.01]. This thymoma subtype increased more for females (AAPC 8.7% [95% CI (4.1; 13.5); p<.01] than for males (AAPC 5.9% [95% CI (3.6; 8.2); p<.01]. The incidence of thymic carcinomas also increased in the observed period (AAPC 1.6% [95% CI (0.3; 2.9); p =.02]. However, there were no sex differences, both showing an AAPC of 1.5%.
Figure 3
Figure 3(A, A’–G’) Age-adjusted incidence rates per million (y-axis) of thymomas not otherwise specified (NOS) and specific subtypes as well as thymic carcinomas (TCs) in the US from 2000-2019 (x-axis). Regression analysis of age-adjusted incidence rates per million for all non-neuroendocrine epithelial tumors of the thymus (A’) and each subtype separately (B’–G’). Morphology of different subtypes of epithelial tumors of the thymus: (B) A thymoma, (C) AB thymoma, (D) B1 thymoma, (E) B2 thymoma, (F) B3 thymoma, and (G) thymic carcinoma. (B’–G’) Separate regression analysis of age-adjusted incidence rates per million for each of the subtypes. (A’–G’) Bold font indicates that the annual percent change (APC) is significantly different from zero at the p =.05 level. (B–G) Hematoxylin and eosin (HE) staining. (B–E, G) 100x magnification. (F) 200x magnification.
3.3 Survival analysis and UICC stage of TETs
For survival analyzes, 1,869 (GER) and 4,156 (US) patients met the inclusion criteria for analysis by subtype and a total of 648 patients met the inclusion criteria for analysis by UICC stage. Regarding the UICC stages, AB thymoma most frequently showed low tumor stages followed by A thymomas to B1, B2, and B3 thymomas, whereas thymic carcinoma already reached stage IVB in more than 40% of cases and was detected in the first two stages in only about 15% (Figure 4). This was also reflected in the Kaplan-Meier curves, which differ significantly (p <.01). Surprisingly, UICC stage IIIB showed a significantly worse outcome than UICC stage IVA (p =.01). In the comparison between the United States and Germany, only thymic carcinoma patients diagnosed in GER show a significantly better prognosis (Supplementary Figure 3, p =.01). For the mean survival values, see Table 1.
Figure 4
Figure 4(A) Stage distribution of thymomas and thymic carcinomas in the United States (US) from 2018-2019 and Germany (GER) from 2017-2019 according to the Union for International Cancer Control (UICC) classification. Survival stratified (B) by TETs subtype in the US from 2000-2019 as well as GER from 1999-2019 and (C) by Union for International Cancer Control (UICC) stage of thymomas and thymic carcinomas in the US from 2018-2019 and GER from 2017-2019.
3.4 Occurrence of secondary malignancies in TET patients
In general, tumor patients show an increased incidence of secondary malignancies. We aimed to test the hypothesis that TET patients suffer from a disproportionally higher incidence of secondary tumors as compared to non-TET tumor patients (17). To this end, we compared the incidence of secondary malignant tumors in our thymoma and thymic carcinoma cohorts with a control group comprising 11,748,248 patients, consisting of all other primary cancer patients in the SEER 22 cohort excluding TET patients.
For the evaluation of sequential tumors, we identified 3806 and 2829 primary thymoma and 1166 and 376 primary thymic carcinoma patients in the US and German cohorts, respectively. In the thymoma cohort, 11.53% (US) and 13.18% (GER) suffered from a secondary malignancy.
Most secondary tumors in the US cohort were lung cancers (n=97, of which 70 were of epithelial origin, e.g., adenocarcinomas or squamous carcinomas), followed by adenocarcinomas of the prostate (n= 64), and thyroid carcinomas (n= 16). The most frequent subsequent tumors of thymoma patients in the GER cohort were lung cancers (n= 45, including 35 cases of epithelial differentiation), prostate cancer (n= 25), and breast -adeno-carcinomas (n= 7). In the thymic carcinoma cohort, secondary tumors emerged in 8.75% (US) and 10.90% (GER) patients, respectively. In the control group comprising 11,748,248 patients, 9.53% of patients had secondary cancers. As compared to the non-TET cohort, patients with thymomas were at a significantly higher risk of developing a secondary neoplasm in the US and Germany (Table 2). In contrast, in patients with thymic carcinoma, the risk of a secondary tumor did not differ from the non-TET cohort.
Table 2
Table 2 The relative risk of patients with thymomas and thymic carcinoma to develop consecutive malignancies in the United States (2000–2019) and Germany (1999-2019).
4 Discussion
The epidemiology of epithelial tumors of the thymus is not comprehensively documented due to their rarity and the fact that the current categorization was only introduced in 1999. Regarding survival data, the situation is even more difficult to assess, since only the latest edition of the UICC classification has introduced a uniform system for thymomas. In addition, previously published unicentric studies usually did not come up with sufficient case numbers to stratify survival data feasibly. This is complicated by the fact that neither the German nor the United States cancer registries collect data on Masaoka-Koga stages.
Our data show that in adults, the distribution of anterior mediastinal tumors did not show large differences between US and GER, with TETs representing the by far most frequent entity in this age group (63% and 64% in the US and Germany, respectively). In patients younger than twenty years, we observed germ cell tumors and lymphomas in 38%/53% (US/GER) and 42%/14%, respectively, whereas TETs accounted for only 10% (US) and 27% (GER) of tumors. As stated before, the notable disparities observed between the percentages of German and United States numbers can be attributed primarily to the relatively low number of overall patients in the German cohort. The relatively lower incidence of neurogenic tumors in our cohorts, compared to other studies, can be attributed to our exclusive focus on the anterior mediastinum, where such tumors are less prevalent (3, 18–20).
In our cohort, the overall incidence of TETs was 2.68 (US) and 3.06 (GER) per million inhabitants per year with only slight differences between the two countries in terms of subtypes and gender and age distribution. This is the first nationwide incidence study for Germany. For the US, the incidence is higher than the previously published data of 1.3 per million person-years considering thymomas between 1973 and 2006 (9). However, these incidence rates are limited by the fact that thymic carcinomas might be inconsistently considered, especially considering that it was only established as a separate entity in 2004 (21) and the fact that non-invasive thymomas were envisioned as benign tumors and excluded from reporting. This has gradually changed based on several publications to be finally defined in the fourth edition of the WHO Classification of Tumors of the Thymus so since then all thymomas have been classified as malignant (10).
Previous reports in the fifth edition of the WHO classification have suggested a higher occurrence of type A, AB, and B1 thymomas in females. However, our analysis of both the United States and the German cohort revealed that among the different thymoma subtypes, the incidences of AB, B1, and B2 thymomas were slightly higher in men, while the occurrence of A and B3 thymomas showed no significant difference between genders. These findings are mostly in line with the findings reported in several previous studies (22–25). The age distribution conforms closely to a normal distribution with sixty years as the mean. Only minor deviations were found in the subtype analysis. For example, type B1 and B2 thymomas were more frequent in younger patients than in older ones, while A thymomas and thymic carcinomas showed a tendency to occur more frequently in older patients. The incidence rates of TETs increased during the observed period. This development is certainly due to a certain extent to both improved radiological diagnostics and increased awareness of this tumor group (26). Considering the changes in incidence rates over the observed period, increasing trends were particularly evident for AB and B2 thymomas, with B2 thymomas particularly markedly increasing up to 2008.
Some reservations regarding the diagnosis of thymic carcinomas in cancer registry data need to be considered. First, there are many nondescript carcinomas in the anterior mediastinum classified, accounting for 9% (US) and 20% (GER) of all tumors reported. To a large extent, these tumors are most likely misclassified, since it can be assumed that no other epithelial tumors besides carcinomas of the thymus occur at this localization. The closest localization where carcinomas would be expected is the skin and lung, each of which has its ICD-O-3 code. Overall, the exact number of thymic carcinomas cannot be determined, although a higher number than reported here can be assumed. Consequently, the percentage of epithelial tumors of the thymus in adult patients could be as high as 72% to 84%. Second, these differences could preferentially affect a certain subgroup of patients, for example, patients with a higher TNM stage or lower differentiation. Consequently, the different survival times between the United States and German cohorts may be distorted. Despite these uncertainties, an unambiguous view of thymic carcinoma can be derived from the similarity of results for incidence and median survival between the data from Germany and the United States.
An increased incidence of secondary tumors has been reported in thymoma patients, often attributed to altered T-cell immune function or genetic predisposition (9). Nevertheless, a clear consensus regarding the specific tumor types that are more prevalent in thymoma patients remains elusive. Discrepancies in previous research findings have contributed to the need for further investigation to establish a better understanding of this association. For example, studies have described an association with hematological disorders (27), thyroid carcinomas (28), and gastrointestinal tract tumors (17, 24, 29), with the rate of increased occurrences varying significantly. In addition, the control groups in the aforementioned studies were of questionable size and comparability. Given the low number of thymoma patients, the subgroup of individuals with secondary tumors is inherently small. As a result, it is readily apparent that the subset of patients exhibiting secondary tumors is intrinsically diminutive, rendering statistical analysis challenging. To account for this circumstance, in our study, we grouped tumors by lineage differentiation, because it can be presumed that entities with similar lineages share comparable biological features. This approach mitigates the risk of statistical overvaluation resulting from individual rare secondary tumors by preventing them from causing undue influence due to their absence in the control group. The incidence of secondary malignancies in patients with thymomas and thymic carcinomas was compared with a control group of a large number of non-TET cancer patients in the same period. In both the German and US cohorts, thymomas predisposed significantly more for malignancies with epithelial, hematopoietic, and mesenchymal lineage differentiation. Neuroectodermal tumors did not occur more frequently in thymoma patients. These findings using large and matched control groups support previous studies indicating that thymoma patients are at a significantly higher risk of developing secondary malignancies compared to other tumor patients (17). In contrast, the incidence of secondary malignancies in patients with thymic carcinomas did not differ from the ones with non-TET primary tumors. These results allude to the hypothesis that the thymopoiesis typically occurring in thymomas but not in thymic carcinomas might be prone to generate dysfunctional T cells and thereby the decisive reason for a defective self-tolerance. This theory is supported, among other things, by the higher incidence of autoimmune diseases, as well as the increased likelihood in this patient group to develop immune-related adverse events during immune checkpoint blockade with Pembrolizumab (30). However, further research is needed to better understand the underlying mechanisms. In addition, these findings highlight the importance of closer monitoring of thymoma patients to detect and manage subsequent malignancies.
The present study has several limitations. First, it was a retrospective analysis of cancer registry data. This led to a particular limitation of available information on the respective patients, so that, for example, Masaoka-Koga stages were missing. In addition, the data are susceptible to misclassification mistakes due to a possibly limited experience of some physicians dealing with rare entities. Conversely, however, single-center studies are subject to even more significant limitations due to the rarity of the tumors and are increasingly influenced by individual investigators. Concerning the relative frequency of secondary tumors, an additional limitation is that only a control group from the United States but not from Germany was available. However, due to the comparable medical and demographic state of development of both countries, only minor deviations can be assumed. Concerning the elevated rate of secondary tumors in thymoma patients, it is essential to note certain limitations as well. The restricted overall tumor count poses challenges in developing a detailed subclassification based on therapy for statistical analysis. The precise nature of chemotherapy, immunotherapies (such as PD-L1 immune checkpoint inhibition), the dosage of radiotherapy, and the respective time intervals leading to the development of subsequent tumors must be considered to ascertain potential causality. In both thymoma and thymic cancers, curative intent resection remains the preferred treatment for resectable disease. Additionally, patients may undergo radiation with or without chemotherapy, contingent upon the Masaoka stage. In the case of stage II tumors, radiation may be contemplated for high-risk cases, while for stage III and IV tumors, radiation is usually part of a combination therapy (31, 32). For our cohorts, this implies that, due to the advanced stage of the tumor, a higher proportion of patients with thymic carcinoma likely underwent combination therapy involving radiation compared to patients with thymoma. Considering the carcinogenic effects of radiation, an augmented incidence of secondary tumors could reasonably be anticipated. Consequently, even though the survival rate of thymic carcinoma patients is significantly diminished, the inclination would be to expect a heightened frequency of secondary tumors in thymic carcinoma patients. However, the observed data does not support this expectation, thereby reinforcing the conclusion that the increased rate of secondary tumors in thymic patients is attributable to factors other than therapy.
In summary, our results demonstrate the age-specific tumor distribution of the tumors in the anterior mediastinum in a nationwide manner in the US and Germany over a time range of 20 years. The study identified stable incidence rates of TETs over time and revealed significant differences in survival outcomes among tumor subtypes and stages. Additionally, it highlighted an increased risk of secondary malignancies in thymoma patients compared to patients with non-TET primaries. These results contribute to our understanding of anterior mediastinal tumors, emphasizing the importance of tailored approaches based on tumor subtypes and the need for monitoring secondary malignancies in TET patients.
Data availability statement
The data analyzed in this study is subject to the following licenses/restrictions: The data that support the findings of this study are available from the Centre for Cancer Registry Data (Zentrum für Krebsregisterdaten, ZfKD) and Surveillance, Epidemiology, and End Results (SEER) Program (www.seer.cancer.gov). Restrictions apply to the availability of these data, which were used under license for this study. Requests to access these datasets should be directed to www.seer.cancer.gov and
Ethics statement
Ethical approval was not required for the study involving humans in accordance with the local legislation and institutional requirements. Written informed consent to participate in this study was not required from the participants or the participants’ legal guardians/next of kin in accordance with the national legislation and the institutional requirements.
Author contributions
TG: Conceptualization, Data curation, Formal analysis, Investigation, Methodology, Resources, Software, Validation, Visualization, Writing – original draft, Writing – review & editing. SS: Data curation, Methodology, Writing – review &editing. AM: Conceptualization, Validation, Writing – review & editing. WR: Supervision, Writing – review & editing. SP: Conceptualization, Data curation, Formal analysis, Validation, Writing – original draft, Writing – review & editing.
Funding
The author(s) declare financial support was received for the research, authorship, and/or publication of this article. The author (TG) received financial support from the Manfred-Stolte Foundation and is supported by the Clinician Scientist Fellowship “Else Kröner Research College: 2018_Kolleg.05”.
Acknowledgments
We thank Dr. Nina Buttmann-Schweiger and Dr. Klaus Kraywinkel of the Centre for Cancer Registry Data (Zentrum für Krebsregisterdaten, ZfKD) for their support.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
Supplementary material
The Supplementary Material for this article can be found online at:
Supplementary Figure 1 |(A) Age distribution of thymomas and thymic carcinomas in the (A) US from 2000-2019 and (B) Germany from 1999-2019. (B) Age and sex distribution of all epithelial tumors of the thymus in the (C) US from 2000-2019 and (D) Germany from 1999-2019.
Supplementary Figure 2 | Regression analysis of age-adjusted incidence rates per million between 1975 and 2019 of (A) all tumors of the thymus regardless of histology and (B) of all tumors of the thymus excluding epithelial tumors of the thymus, which are mostly comprised of tumors classified as squamous cell carcinomas (n=746), adenocarcinomas (n=502), and unspecified neoplasms (n=103).
Supplementary Figure 3 | Stage distribution of thymomas and thymic carcinomas in the (A) US from 2018-2019 and (B) Germany from 2017-2019. Survival stratified by subtype in the (C) US from 2000-2019 and (D) Germany from 1999-2019. (E) Significantly different survival of thymic carcinomas between the US and Germany.
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Keywords: thymic epithelial tumors, thymoma, thymic cancer, incidence rate, survival, epidemiology
Citation: Gerber TS, Strobl S, Marx A, Roth W and Porubsky S (2024) Epidemiology of thymomas and thymic carcinomas in the United States and Germany, 1999-2019. Front. Oncol. 13:1308989. doi: 10.3389/fonc.2023.1308989
Received: 07 October 2023; Accepted: 19 December 2023;
Published: 09 January 2024.
Edited by:
Sharon R. Pine, University of Colorado Anschutz Medical Campus, United States
Reviewed by:
Konstantin Shilo, The Ohio State University, United States
Margaret Ottaviano, G. Pascale National Cancer Institute Foundation (IRCCS), Italy
Copyright © 2024 Gerber, Strobl, Marx, Roth and Porubsky. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Stefan Porubsky, Stefan.Porubsky@unimedizin-mainz.de
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
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Edited by
Sharon R Pine Division of Medical Oncology, Department of Medicine, School of Medicine, University of Colorado Anschutz Medical Campus, United States
Reviewed by
K Shilo The Ohio State University, United States Margaret Ottaviano Unit of Melanoma, Cancer Immunotherapy and Development Therapeutics, G. Pascale National Cancer Institute Foundation (IRCCS), Italy
Table of contents
Abstract
1 Background
2 Methods
3 Results
4 Discussion
Data availability statement
Ethics statement
Author contributions
Funding
Acknowledgments
Conflict of interest
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Supplementary material
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Help with function question f(2 x+1)=2 f(x)+1 f(2 x+1)=2 f(x)+1 [closed]
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I've been working on this question for a few hours but I can't figure out the answer.
f(2 x+1)=2 f(x)+1 f(2 x+1)=2 f(x)+1 for all x x and f(0)=2 f(0)=2 what does f(3)f(3) equal?
algebra-precalculus
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edited Aug 27, 2018 at 2:29
Robert Soupe
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BLoby BlobBLoby Blob
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What is your question?user369816 –user369816 2018-08-27 02:14:35 +00:00 Commented Aug 27, 2018 at 2:14
edited. "its what does f(3) equal?" sorry about that.BLoby Blob –BLoby Blob 2018-08-27 02:16:06 +00:00 Commented Aug 27, 2018 at 2:16
2 We are told that f(0)=2 f(0)=2 and we are told that f(2⋅0+1)=2 f(0)+1 f(2⋅0+1)=2 f(0)+1 so... we now can know what f(1)f(1) is equal to. Continue.JMoravitz –JMoravitz 2018-08-27 02:16:52 +00:00 Commented Aug 27, 2018 at 2:16
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Well, f(3)=f(2⋅1+1)=2 f(1)+1 f(3)=f(2⋅1+1)=2 f(1)+1. Now, f(1)=f(2⋅0+1)=2 f(0)+1=5 f(1)=f(2⋅0+1)=2 f(0)+1=5. Thus f(3)=11 f(3)=11.
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answered Aug 27, 2018 at 2:18
user369816 user369816
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Thank you so much!BLoby Blob –BLoby Blob 2018-08-27 02:19:31 +00:00 Commented Aug 27, 2018 at 2:19
@BLobyBlob No problem. The key to these things it toying around with the equations for a little bit.user369816 –user369816 2018-08-27 02:20:36 +00:00 Commented Aug 27, 2018 at 2:20
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You have
f(2 x+1)=2 f(x)+1 f(2 x+1)=2 f(x)+1
Let
x=0 x=0
to get
f(1)=2 f(0)+1=2(2)+1=5 f(1)=2 f(0)+1=2(2)+1=5
Let x=1 x=1 to get
f(3)=2 f(1)+1=2(5)+1=11 f(3)=2 f(1)+1=2(5)+1=11
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answered Aug 27, 2018 at 2:19
Mohammad Riazi-KermaniMohammad Riazi-Kermani
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iteration speed of outside function is the same as inside function just they don't start with the same initial values F"Inside" starts with initial value of 0 and f"outside" starts with the initial value of 2 so you just solve for first order non homogenous recurrence relation of the form F(c+1)=2F(c)+1, with the seeds FI(0)=0, FO(0)=2. FI(t)=2^t-1, FO(t)=32^t-1, putting it in the form of parametric curve we get [FI(t),FO(t)], and solving for it as form of y(x). FI inverse of tI(x) = log base 2(x+1). 2^log base 2(x+1)=x+1, 3(x+1)-1=3x+2. so your final answer is 3x+2. checking 3(2x+1)+2=6x+3+2=6x+5. 2(3x+2)+1=6x+4+1=6x+5.
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edited Aug 27, 2018 at 5:29
answered Aug 27, 2018 at 5:22
Alireza RouhaniAlireza Rouhani
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2 I think it would be a good idea to format the math properly in your answer to make it more readable. Take a look at this page.Edmundo Martins –Edmundo Martins 2018-08-27 06:02:20 +00:00 Commented Aug 27, 2018 at 6:02
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12621 | https://www.penguin.co.uk/discover/articles/the-history-behind-the-lady-macbeth-coronavirus-meme | Published Time: 2020-03-12T00:00:00+00:00
'Out damned spot': the Lady Macbeth hand-washing scene that became a Coronavirus meme
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‘Out damned spot’: the Lady Macbeth hand-washing scene that became a Coronavirus meme
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‘Out damned spot’: the Lady Macbeth hand-washing scene that became a Coronavirus meme
Shakespeare's haunting soliloquy, shared widely this week in relation to COVID-19, has a rich history with popular culture.
Emma Smith 12 March 2020
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adapted by Fossilheads, original by World Health Organization; . Licence: CC BY-NC-SA 3.0 IGO
If you haven’t seen the handwashing instructions set to Lady Macbeth’s fractured soliloquy ‘Out damned spot’ you’ve not been visiting the right bathrooms. The poster adapts World Health Organisation’s guide to the use of soap and water, using Shakespeare’s words to establish the proper length of time to spend on sanitising.
In the context of the play Macbeth, those instructions are, of course, rather self-defeating. Lady Macbeth, sleepwalking in her chamber, ‘rubs her hands’ for ‘a quarter of an hour’, lamenting ‘what, will these hands ne’er be clean’. She can still smell blood: ‘All the perfumes of Arabia will not sweeten this little hand’. While soap does indeed incapacitate the COVID-19 virus, it cannot help mitigate the psychological aftermath of murder. Lady Macbeth’s involvement in the assassination of King Duncan echoes in her conscience. Her confident words to her nervily blood-stained husband – ‘A little water clears us of this deed’ – come back to haunt her. Lady Macbeth’s hand-washing is the sign of guilt. It speaks of a contamination that can never be washed away.
A young Judi Dench played the scene with ‘feverishly writhing hands'
This scene at the beginning of the play’s fifth act is Lady Macbeth’s swansong – as Verdi realised in his opera, giving her a gloriously controlled aria at this point. It is also a brilliant set piece for actors. Lady Macbeth must have originally been a role for the gifted young male actor whose presence in the King’s Men acting company made possible this play, and also Cleopatra, Coriolanus’s mother Volumnia, and John Webster’s heroine the Duchess of Malfi. These mature and compelling women required a special kind of talent, quite different from the androgynous women of the earlier comedies.
At the end of the eighteenth century the great Sarah Siddons made this her signature scene, bearing a ‘deathlike stare’ and wearing a flowing night-dress that reminded some reviewers of a shroud; two centuries later at Stratford-upon-Avon, a young Judi Dench played the scene with ‘feverishly writhing hands, only still when they are held for a long moment to the candle and scrutinised’. Roman Polanski’s movie version was notorious for Francesca Annis as a naked Lady Macbeth; Marian Coutillard played the role for Justin Kurzel’s film, and was fully awake, agonised, and alone in this scene.
Amy Shuard as Lady Macbeth, 1960. Getty.
If this scene is the highpoint of Lady Macbeth’s characterisation in the play, it has also had a vital afterlife entirely independent of the rest of the story. ‘Out damned spot’ has joined a handful of other Shakespearean lines – ‘to be or not to be’, ‘Romeo, Romeo, wherefore art thou Romeo’ - as instantly memorable and reapplicable. We pan past ‘Out Damned Spot Cleaners’ on the boulevard of Verona Beach in Baz Luhrmann’s Romeo + Juliet, while a nifty set-up with a dog called Spot brings the line into Gnomeo and Juliet. Lady Macbeth’s haunting scene reverberates from Agatha Christie to Chekhov, and from Brave New World to Family Guy.
It’s been, inevitably, commodified. Concealer pencils, erasers, tea towels, and above all, soap and soap products bear this logo. In 1945 an Aluminium Company of America advert told consumers that all Lady Macbeth needed was ‘some peroxide, cold water, and an electric washing machine…to change her destiny’. The American feminist artist Elizabeth Layton’s self-portrait as Lady Macbeth made spotlessness a domestic and a racial category, using bleach to try to conform to the patriarchal idea of housewife in which the idea of ‘whites only’ was about more than laundry. As the COVID-19 memes demonstrate, the iconography of obsessive hand-washing will always be associated with Lady Macbeth.
Emma Smith is the author of This Is Shakespeare.
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12622 | https://math.stackexchange.com/questions/2532231/writing-cosines-using-de-moivres-formula | complex numbers - writing Cosines using De Moivre's formula - Mathematics Stack Exchange
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writing Cosines using De Moivre's formula
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\begingroup
Given the question:
Use De Moivre’s formula to find a formula for \cos(3x) and \cos(4x) in terms of \cos(x) and \sin(x). Then use the identity \cos^2(x) + \sin^2(x) = 1 to express these formulas only in terms of \cos(x).
I started out by rewriting \cos(3x): \cos(3x)+i \sin(3x)=(\cos(x)+i \sin(x))^3
This could then be written into
\cos(3x) = \cos^3(3x)-3 \cos(x) \sin^2(x)
or
\sin(3x) = \cos^2(x) \sin(x)- \sin^3(x)
Then to use the identity I would substitute 1-\cos^2(x) for the \sin^2(x)
and in the second I would substitute \sin(x) for \sqrt{1 - cos(x)} right and would need to separate the \sin^3(x) into \sin^2(x) \sin(x) and substitute from there. I'm a lot less confident about the second equation substitution. Would this be the right way to go about doing this problem?
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edited Nov 22, 2017 at 11:59
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asked Nov 22, 2017 at 11:25
mathminutemaidmathminutemaid
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\endgroup
3
1 \begingroup So, you want to express \sin(3x) also in terms of \cos(x) only? Your original question only asks for \cos(3x) and \cos(4x) to be expressed in such a fashion. The method is right, anyway.\endgroup user279515 –user279515 2017-11-22 11:30:34 +00:00 Commented Nov 22, 2017 at 11:30
\begingroup I believe so since in the second part formulas was plural so it seemed to refer to both the formulas in terms of cos(x) and the sin(x) one. Thanks for the answer\endgroup mathminutemaid –mathminutemaid 2017-11-22 11:42:19 +00:00 Commented Nov 22, 2017 at 11:42
1 \begingroup You should not hope to express \sin(nx) in terms of cosines, ’cause the sine function is odd and any polynomial in cosine will be even.\endgroup Lubin –Lubin 2017-11-24 20:30:08 +00:00 Commented Nov 24, 2017 at 20:30
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\begingroup
You seem to have misunderstood what is meant by "formulas," based on your approach and comments. That said, the problem given to you is poorly phrased (perhaps due to a typo), so this is understandable. The problem should read as follows (with emphasis to make the adjustment clear):
Use DeMoivre's formula to find formulas for \cos(3x) and \cos(4x) in terms of \cos(x) and \sin(x). Then use the identity \cos^2(x)+\sin^2(x)=1 to express these formulas only in terms of \cos(x).
Your approach is just right for finding \cos(3x). Unfortunately, without more information, it is impossible to unambiguously define \sin(3x) in terms of \cos(x) only, since \sqrt{1-\cos^2(x)}=\sqrt{\sin^2(x)}=\left|\sin(x)\right|, so your substitution \sin(x)=\sqrt{1-\cos^2(x)} needn't be correct. Fortunately, you aren't being asked to do such a thing. Rather, it remains only to find a formula for \cos(4x) in terms of \cos(x) and \sin(x), then use \cos^2(x)+\sin^2(x)=1 to express the found formula only in terms of \cos(x).
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answered Nov 24, 2017 at 19:02
Cameron BuieCameron Buie
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\begingroup
{ \left( \cos { x } +i\sin { x } \right) }^{ 3 }=\cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \ \cos ^{ 3 }{ x } +3i\cos ^{ 2 }{ x\sin { x } -3\cos { x\sin ^{ 2 }{ x } -i\sin ^{ 3 }{ x } = } } \cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \ \ \cos { \left( 3x \right) =\cos ^{ 3 }{ x } -3\cos { x } \sin ^{ 2 }{ x } =\cos ^{ 3 }{ x } -3\cos { x } \left( 1-\cos ^{ 2 }{ x } \right) =4\cos ^{ 3 }{ x-3\cos { x } } } \ \sin { \left( 3x \right) =3\cos ^{ 2 }{ x } \sin { x } -\sin ^{ 3 }{ x } } =3\left( 1-\sin ^{ 2 }{ x } \right) \sin { x } -\sin ^{ 3 }{ x } =3\sin { x } -4\sin ^{ 3 }{ x } \
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answered Nov 22, 2017 at 11:34
haqnaturalhaqnatural
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\endgroup
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\begingroup I got the same thing for the cos(3x) written only in terms of cos(x) but it also wanted the sin(3x) one in terms of cos(x) only, and with that I'm not sure if I can take the square root of a negative to find what sin(x) is equal too. Thank you\endgroup mathminutemaid –mathminutemaid 2017-11-22 11:46:42 +00:00 Commented Nov 22, 2017 at 11:46
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12623 | https://sites.ecse.rpi.edu/~ssawyer/CircuitsSum2022_all/Labs/Unit1/Lab02.pdf | Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 1 LABORATORY 2: Transient circuits, RC, RL step responses, 2nd Order Circuits, Impedance, s-domain Circuits Material covered: • RC circuits • 1st order RC, RL Circuits • 2nd order RLC series circuits • 2nd order RLC parallel circuits • Thevenin circuits • S-domain analysis Part A: Transient Circuits RC Time constants: A time constant is the time it takes a circuit characteristic (Voltage for example) to change from one state to another state. In a simple RC circuit where the resistor and capacitor are in series, the RC time constant is defined as the time it takes the voltage across a capacitor to reach 63.2% of its final value when charging (or 36.8% of its initial value when discharging). It is assume a step function (Heavyside function) is applied as the source. The time constant is defined by the equation τ = RC where τ is the time constant in seconds R is the resistance in Ohms C is the capacitance in Farads The following figure illustrates the time constant for a square pulse when the capacitor is charging and discharging during the appropriate parts of the input signal. You will see a similar plot in the lab. Note the charge (63.2%) and discharge voltages (36.8%) after one time constant, respectively. Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 2 Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 3 Function Generator: In this lab, we will be looking at DC, sinusoidal, square wave and triangle wave functions. Your Proof of Skills Documentation is a good resource to learn how to make these transient signals for your lab. Signal types: DC source (constant) Square wave (pulse train) Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 4 Triangle wave (try to make this signal…add instructions to Proof of Skills for yoru board if not already there…) Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 5 LTSpice: To generate a square wave, use the voltage component. Use the following settings (do not include the information in parenthesis): Vinitial[V]: 0 (low voltage in volts) Von[V]: 2 (high voltage in volts) Tdelay[s]: 0 (time delay in seconds) Trise[s]: 0 (rise time in seconds) Tfall[s]: 0 (fall time in seconds) Ton[s]: variable (pulse width, 50% duty cycle, so should be half the period) Tperiod[s]: variable (as directed in the lab, 1/f, example a 1kHz wave has a 1ms period) Ncycles: variable (can leave blank, number of cycles) Plot quality/accuracy: When you set your simulation profile the Time Domain (Transient tab) analysis lets you set a few parameters. We will be looking at several periods of the signal. You should be using the following settings: Stop time: (sufficient time for several periods to be displayed) seconds Time to start saving data: 0 seconds Maximum timestep: (I usually set this so that there are ~1000 points per period) Plots: For transient analysis, we can place a Probe on the LTSpice circuit to obtain plots of voltage/current as a function of time. You can place that Probe at a node to obtain plots of voltage as a function of time. Note, this measurement is nodal and is measured relative to the indicated ground. If you are going to measure the voltage drop across a component that is not connected to ground, you can use differential probes, which can be set using the icon with the pair of probes. You place one probe on one side of the component and the other probe on the other side of the component. Cursors: Cursors are available in LTSpice as well. Right click on the Trace label at the top of the simulation screen. You can choose up to two cursors to use by clicking the pull down menu “Attached Cursor” and selecting 1st, 2nd, or 1st and 2nd. You can drag the cursor along your plot and it will follow the waveform. When you start the cursor, the data window should show up providing voltage levels, time, and frequency if using two cursors as well. More details with pictures are shown here. Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 6 A.1. RC Circuits Build the RC circuit shown in the top figures (They are the same circuit but the PSpice circuit on the right shows probe position). 1) On your instrumentation board, set your signal to a 1V amplitude (2V peak-to-peak) square wave with a DC offset of 1 Volts. Set the frequency to 1kHz initially. Implement your circuit with R1 = 1kΩ (1E3) and C1 = 0.1µF (0.1E-6). Connect Oscilloscope Channel 1 probes to the source and Oscilloscope Channel 2 probes across the capacitor. a. Reduce the frequency such that you can see the voltage across the capacitor reach a constant value (steady state) in each half period of the square wave (approaching the DC condition). You will need to adjust the Base (horizontal axis) of the Oscilloscope to see several periods. (An example is shown below, where red is the source voltage and green is the voltage across the capacitor.) Example of Voltage Output Waveform R1 1k C1 0.1E-6 V1 TD = 0 TF = 0 PW = 1E-3 PER = 2E-3 V1 = 0 TR = 0 V2 = 0.2 0 V V Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 7 b. Estimate the time constant. It will be easier if you use the cursors (described at the start of the lab). c. Compare your result with the expected time constant. (expected time constant is RC, resistance times capacitance, it will be explained in lecture) d. Implement your circuit in LTSpice and compare your results. 2) Repeat a-d. for R = 10kΩ and C = 0.01E-6F or any other RC combination. Part B: RC, RL Step Responses Overall notes: In your plots, you should compare input signals (sources) to outputs signals (component voltage/current). In Part B, you will investigate an RL parallel circuit. The derivation of the current across the inductor leads to the differential equation, s L L I i dt di R L = + For a step function source current, I1u(t), the solutions to the differential equation take the form ( ) 2 1 K e K t i L R t L + = − . This form is similar to what we saw for the RC circuit, with a slightly different time constant τ = L/R. I1 R1 L1 1 2 PART A: Proof of Concepts list A.1: RC Circuit-Prove that the time constant changes with different component values. Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 8 Unfortunately, there are two problems with investigating the above circuit. The first problem is that we don’t have access to a current probe. One practical solution is to add a small resistor in series with the inductor. You want the resistor to be small enough such that it does not significantly change the circuit and large enough that you can obtain an accurate voltage reading. Second, we don’t have a reliable current source. However, we can use source transformations to generate an equivalent circuit with a voltage source, as shown in the following circuit. R2 is the test resistor and should be much smaller than R1, (R2 << R1). In the experiment that follows, you could easily just determine the current by measuring the voltage across R1 without any need for R2. However, we will treat the experiment as representative of a more complex model and use a small R2 resistor. Note, you may want to prove to yourself that the source transformation circuit results in the same current response in L1. Instrumentation Board This lab investigates step responses. In Discovery Board, M1K and M2K, we use pulse streams and can’t easily make a step function source. To measure an equivalent I1 R1 L1 1 2 R2 L1 1 2 V1=(I1)(R1) R1 R2 Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 9 response using a pulse stream, we make sure that each half cycle is much longer than the time constant for the circuit, T/2 >> τ. By letting the circuit response reach DC steady state in each half cycle, the V=Vo half cycle is equivalent to a source turning on at t = 0 with zero initial conditions and the V = 0 half cycle is equivalent to a source turning off at t = T/2 with initial conditions determined at t = T/2-. Triggering: Oscilloscopes use a trigger to tell them when to start capturing data. For the Discovery Board, the trigger is available using the Source drop down list located near the top of the display, a bit to the right of center. (Please add to the Proof of Skills Documentation if you find how to do this for the M1K or M2K board!) Often, using the source in the circuit as a trigger is a good choice since that signal is reliable and known. In the experiments, you should connect Oscilloscope Channel 1 to the source to measure input voltages to the circuit and therefore set your Trigger Source to Channel 1 as well. We will want to use rising edge triggering, meaning we start capturing data when the slope of the signal is positive. Just to the right of the Source drop down list is a Cond. drop down list. Set that to Rising. Just below that drop box is the Level setting. This value tells the scope to start capturing data when you cross that voltage. You want to adjust the level somewhere between the maximum and minimum voltages of the channel input, which is again a good reason why you want to use a known signal like the source. With rising edge triggering and pulsed signals, setting the trigger Level slightly higher than low voltage of the pulse is a good choice. LTSpice We will use the voltage pulse component introduced above. B.1. RC Circuits Build the RC circuit shown in the top figure, with R1 = 6k, R2 = 3k, R3 = 2.7k and C1 = 0.4µF. Vs R1 R2 R3 C1 Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 10 1) Using the Discovery Board, set your signal to a 0.5 V Amplitude square wave with a DC offset of 0.5 V (the source should switch from 1 V to 0 V, repeating). Set the frequency such that you see the voltage across the capacitor reach DC steady state response for each half cycle of the square wave. Measure the source signal using Channel 1 of the Oscilloscope and the output signal (capacitor voltage in this case) with Channel 2 of the Oscilloscope. a. Analytically, determine the RC time constant for the above circuit. Finding the Thevenin equivalent circuit can help. (A similar circuit was discussed in class.) Make sure to include this calculation in the Proof of Concepts… b. Determine a differential expression for voltage across the capacitor. Solve the differential equation for a step function voltage source that turns on at t = 0 with a source voltage Vs = 1 V. Plot your result. c. In Discovery Board, obtain plots for the voltage (use differential probes) across the capacitor. Compare your results to the expression in part b. d. Estimate the RC time constant using the Discovery Board Oscilloscope e. Compare your result with the calculated value. f. Simulate the circuit in LTSpice using the voltage pulse component as your source and again compare your results. Include screen shots of your results in your Proof of Concept Report. There are multiple steps here. Arrange and label with headings in an easy to understand way for grading TAs! PART B: Proof of Concepts list B.1: RC Circuit: Prove the RC time constant using Thevenin calculations for 1 u(t) Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 11 Part C: 2nd Order Circuits Overall notes: Plots: In the above figure, an example of an underdamped current response is shown. The curve represents a step response decaying to zero. Using the plot, it is possible to determine the attenuation constant, α, and the oscillation constant, β. The exponential curve shown by the dark black line is defined by an amplitude, Io, and the attenuation constant, α. Both these terms can be obtained by identifying two points on the curve and using two equations/two unknowns to find the terms(for example: one point would be 0.9mA at 35µs). The sinusoidal peaks of the current response (any response) will always lie on the exponential curve. Similarly, the propagation constant, β, can be obtained from the plot by measuring the time between peaks (or minima) and using the relationship β=2π/T. Recall, α and β are determined from the differential equation, where the attenuation constant, α, is the real component of the quadratic solution of the s polynomial and the oscillation constant, jβ = jsqrt(ωo2 – α2), is the imaginary component of the quadratic solution of the s polynomial. Io exp(-αt) T = 2π/β ( ) ( ) ( ) [ ] 3 2 1 sin cos exp A t A t A at I o + + − β β Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 12 Overdamped case: A slightly similar discussion may be applied to the overdamped response curve. In this case, typically one of the exponential terms is much larger than the other term and will thus result in a faster decay. With this in mind, it is possible to estimate the smaller exponential term by looking at the region of the curve some time past the initial response. “Some time” depends on how strongly overdamped the circuit is and may be less than a time constant for a strongly overdamped circuit and ~4 time constants for a weakly overdamped circuit. C.1. RLC Series Circuit Build the RLC circuit shown in the above figure. The resistor is a 10k potentiometer. Note, one leg of the potentiometer is left floating (not used). 1) On the Discovery Board, set your signal to a 1Vp-p square wave with a DC offset of 0.5Volts. Set the frequency such that you see the voltage across the capacitor reach DC steady state response for each half cycle of the square wave. In the following, consider just one-half cycle of which, as we have seen, is equivalent to the step source circuit. a. Adjust the potentiometer so that you can clearly see an overdamped response. i. Measure the resistance setting of your potentiometer ii. Based on your measured resistance, determine an analytic (calculated) expression for the voltage across the capacitor with a step function source. Remember, for an overdamped circuit ( ) ( ) ( ) 3 2 2 1 1 exp exp A t s A t s A t Vc + + = iii. Simulate the same circuit in LTSpice. Use the resistance value you measured in place of Rpot (don’t use the potentiometer in your LTSpice circuit, floating nodes are bad in LTSpice). Vs L 22mH 1 2 C 0.1E-6 Rpot 10k Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 13 iv. Most likely, your Discovery Board curve is slightly different from the LTSpice and analytic results. In this case, we have ignored some resistances in the circuit. Any ideas on what they were? b. Adjust the potentiometer so that you clearly see an underdamped case. i. Again, compare your result to the analytic and LTSpice results. Remember, for an underdamped circuit ( ) ( ) ( ) ( ) [ ] 3 2 1 sin cos exp A t A t A t t Vc + + − = β β α Include screen shots of your results in your Proof of Concept Report. There are multiple steps here. Arrange in an easy to understand way for grading TAs! C.2. RLC Parallel Circuit Build a circuit equivalent to the RLC circuit shown in the above figure and measure the current through the capacitor. Recall, we don’t have a current source but we can build an equivalent circuit using source transformations (see Part B). Also, we use a small probe resistor to obtain a differential voltage measurement which is used to determine the current through the inductor. In your analysis, ignore the probe resistor and treat the circuit as a parallel RLC circuit. 2) On the Discovery Board, set your signal to a 1 V amplitude square wave with a DC offset of 0Volts (no offset). In effect, when considering one-half cycle, you will have nonzero initial conditions for the current through the inductor. a. Determine a reasonable period for the square wave in your instrumentation board (M1K, M2K, or Analog Discovery board) measurement interface and a reasonable run time for LTSpice. Your square wave should be sufficient that you see the steady state value in one half period, T/2. Measure the voltage across R2 to determine the current through the inductor. Based on the oscilloscope measurements, estimate the attenuation constant and oscillation constant (Refer to the Is R1 47k R2 47 L1 100mH 1 2 C1 0.1E-6 Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 14 first page of the lab). Ignore the 47Ω resistor in your analysis. You will check if that is reasonable in part d. In the instrumentation board (M1K, M2K, or Analog Discovery board) measurement interface, you will most likely see a DC offset. b. What type of response do you see (underdamped, overdamped or critically damped)? c. Using the LTSpice plot, estimate the attenuation constant and oscillation constant. How close is this estimate to your calculated values? d. In your LTSpice simulation, test the validity of ignoring R2 by running a simulation both with the resistor and without the resistor and comparing the results. Include screen shots of your results in your Proof of Concept Report. There are multiple steps here. Arrange in an easy to understand way for grading TAs! C.3. s-domain Analysis with Step Function Input Apply s-domain analysis to the RLC series in circuit in C.1. Initially, redraw the circuit using impedances (s-domain components). Include the initial value components which act like voltage sources in series with inductors and capacitors. Use symbolic notation in the circuit and the following analysis. Specific component values will be used later. 1. Draw your circuit. Circuit analysis is applied to obtain to determine voltage across or current through individual components. We will look at the voltage across inductor L1. In this case, a voltage divider is an appropriate choice. Recall, the voltage across the inductor in the s-domain includes the drop across the inductor and the drop across the initial value voltage source in the your circuit. 2. Write the transfer function that represents the voltage across inductor L1 when the initial conditions are zero (all the IL(0-) terms vanish, though in the Lab itself your circuit will have initial conditions). 3. We will use a source of the form Vs = 2-2u(t). (You may want to draw a sketch of this source.) In order to generate an equivalent source on the Discovery Board, set the function to 2Vpp (1V amplitude) with a 1V DC offset. To obtain Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 15 a clean signal, you should set one of the oscilloscope channels to the source, set the trigger to that channel and set the Trigger level around 0.15V. a. Components: L1 = 100mH, L2 = 100mH, R1 = 1kΩ, R2 = 1kΩ b. Based on the source, determine the initial conditions for the s-domain circuit L1i(0-) = L2i(0-) = 4. Using the circuit component values and initial conditions, determine the poles and zeroes of the expression on the previous page. zeroes: poles: a. Is the system underdamped, critically damped or overdamped? b. Which pole is the most significant in terms of transient behavior? c. What is an appropriate time scale when considering plots for the step response? Remember, the attenuation constant(s), α = 1/τ, provides a time scale for the attenuation of the exponential term and the period of oscillation, T=2π/β provides a time scale for harmonic oscillations if the system is underdamped. 5. Apply partial fraction expansion to the VL1(s) expression. 6. Apply the inverse Laplace transform to the above expression to obtain a time domain expression for your circuit. 7. Plot your function using any graphing tool (Excel/Matlab (for Proof of Skills…)/etc.) 8. Simulate the circuit in LTSpice and build the circuit. Compare the analytic plot, measured results from the Discovery Board, and simulation results from LTSpice. PART C: Proof of Concepts list C.1: RLC Series Circuit: Prove how adjusting the resistance changes the circuit from overdamped to underdamped. C.2: RLC Parallel Circuit: Prove how the attenuation constant and oscillation constant relates to an underdamped, overdamped, or critically damped circuit. (discuss why/whether you can ignore R2 used to calculate the current through the inductor. C.3 Using s-domain analysis (Laplace Transforms), prove that you can adjust the circuit from overdamped to underdamped. Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 16 Part D: Alpha Laboratories Applications D.1. Can a Circuit Tell Time, Keep Time, and Create Time?!? What are timing circuits? 1) Find three different timing circuits and describe the function of each component within it. (You can Google it!) 2) Demonstrate one of those via simulation and DESCRIBE its purpose in a larger circuit. (You may need another component to show it, but you do NOT have to physically build it). EXTRA CREDIT: Write in your metacognition journal (instructions and template in the link below, feel free to continue to edit a Google doc throughout the course to add entries). Docs/04_Deliverables/05_Circuits_Metacognition%20and%20Reflections.d ocx PART D: Proof of Concepts list D.1: Prove that your chosen timing circuit functions as you expected. Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 17 SUMMARY of Concepts Concept List that must be accounted for in your Proof of Concepts PART A: 1. RC Circuit-Prove that the time constant changes with different component values. PART B: 1. RC Circuit: Prove the RC time constant using Thevenin calculations for 1 u(t) PART C: 1. RLC Series Circuit: Prove how adjusting the resistance changes the circuit from overdamped to underdamped. 2. RLC Parallel Circuit: Prove how the attenuation constant and oscillation constant relates to an underdamped, overdamped, or critically damped circuit. (discuss why/whether you can ignore R2 used to calculate the current through the inductor. 3. Using s-domain analysis (Laplace Transforms), prove that you can adjust the circuit from overdamped to underdamped. PART D: Alpha Labs Applications 1. Prove that your chosen timing circuit functions as you expected. Alpha Laboratories ECSE-2010 Summer 2022 Written by J. Braunstein Modified by S. Sawyer Summer 2022: 7/19/2022 Rensselaer Polytechnic Institute Troy, New York, USA 18 Standards Based Assessment: You will be graded on the following Standards. Please ensure to achieve each standard. If you do not, you can resubmit to the missing standard to the end of the semester. CLEARLY mark the changes you make in you Proof of Concept submission by either Tracking Changes in Word or highlighting changes by writing comments in a different color and/or changing the color of the updated work. 1. I can change the time constant with different component values in an RC or RL circuit. 2. I can use Thevenin to simplify an RC circuit and find its time constant. 3. I can adjust the component values of an RLC series circuit to switch between underdamped and overdamped and analyze using both Diff. Eq. and Laplace Transforms. 4. I can adjust the component values of an RLC parallel circuit to switch between underdamped and overdamped and analyze it. 5. I can explain the use of a resistor in series with an inductor in a parallel circuit for measurement in a built parallel circuit. 6. I can design a timing circuit based on what is learned in this lab. 7. I can answer for myself “Is this right?” by comparing mathematical calculations to simulation and experimental results. 8. I can show plots and diagrams that are easy to read, scaled correctly and clearly labeled. 9. I can use consistent variable labels and component values in mathematical calculation, simulation and experimental results for easy comparison. 10. I can accurately answer conceptual questions found throughout the lab. |
12624 | https://4pda.to/forum/index.php?showtopic=753092 | Игры, портированные на Android с других платформ. - 4PDA
ТЕХНИКА)Смартфоны Новости, анонсы, рекомендацииНоутбуки Новости, анонсы, рекомендацииАудио Новости, анонсы, рекомендацииDevDB Каталог устройствТВ и Мониторы Новости, анонсы, рекомендацииБытовая техника Новости, анонсы, рекомендацииПК-железо Новости, анонсы, рекомендации
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ТЕХНИКА)Смартфоны Новости, анонсы, рекомендацииНоутбуки Новости, анонсы, рекомендацииАудио Новости, анонсы, рекомендацииDevDB Каталог устройствТВ и Мониторы Новости, анонсы, рекомендацииБытовая техника Новости, анонсы, рекомендацииПК-железо Новости, анонсы, рекомендации
ОБЗОРЫСмартфоны Выбираем, тестируем, рассказываемПланшеты Выбираем, тестируем, рассказываемУмные часы Выбираем, тестируем, рассказываемАксессуары Выбираем, тестируем, рассказываемНоутбуки Выбираем, тестируем, рассказываемАудио Выбираем, тестируем, рассказываем
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Игры, портированные на Android с других платформ. | Каталог игр
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Большая подборка портов с Java
ВНИМАНИЕ: В этой теме НЕ ЗАНИМАЮТСЯ портированием игр!!! Тема всего лишь каталог. Просьба не писать запросы "портируйте то, портируйте это".
Посты в духе: "А когда будет ххх?", "Может ли кто портировать ххх?" и оффтоп не относящийся к этой теме, будет удаляться.
Шаблон для оформления постов Перед выкладыванием ссылки в тему, просьба убедиться, что игра еще не присутствует в шапке! [URL=ссылка на тему с игрой/PM]Название игры[/URL] порт/околопорт
u Каталог игр
Новый список портированных игр
Список портированных игр:
0-9
1979 Revolution: Black Friday
2048 (Play Market)
48 Irons (48 утюгов)
60 Parsecs!
60 Seconds
60 Seconds! Reatomized
The 7th Guest: Remastered
A-B
A Boy and His Blob
AaaaaAAaaaAAAaaAAAAaAAAAA
The ABC Murders
Abuse
Ace Attorney - Dual Destinies
Ace Attorney Investigations - Miles Edgeworth
Ace Attorney - Spirit of Justice
Actraiser Renaissance (Google Play)
Actual Sunlight (GDrive)
Adam Wolfe: Dark Detective Mystery Game
After Burner Climax
Airline Tycoon Deluxe
Albion Online
Alien Breed
Alien Quarantine
Alien Shooter
ALite
ALONE IN SPACE: ESCAPE
Altered Beast (Play Market)
Amerzone - The Explorer's Legacy
Ancient Empire: Strike Back
Angband
Angelo and Deemon: One Hell of a Quest
ANIMUS: Revenant
Anodyne
Anomaly Korea
Anomaly Warzone Earth HD
Another World- 20th Anniversary
Ant Attack
AntVentor
Apollo Justice Ace Attorney
AquaPlane
Aquaria
Ara Fell: Enhanced Edition (Play Market)
Archmage
ARK: Survival Evolved
Assassin’s Creed 3
Assassin’s Creed: Altaпr’s Chronicles
ATOM RPG
Auralux
Baba Is You
Babel Rising 3D
Bad Hotel
Bad North: Jotunn Edition
Baldur's Gate Enhanced Edition
Baldur's Gate II
The Banner Saga
Banner Saga 2
The Bard's Tale
Bartender The Right Mix
Batman - The Telltale Series
Battle Chasers: Nightwar
The Battle for Wesnoth
Battle Worlds: Kronos
Bear With Me
Beatbuddy
Beat Cop
Beat Hazard Ultra
Beholder (Play Market)
Bendy and the Ink Machine
Beyond Oasis Classic (Play Market)
Book of Unwritten Tales 2
Bounce Original
Blackwell 1-3
BLAZING STAR
Blobby Volley 2
Blood Bowl
Bloodstained: Ritual of the Night
The Bluecoats - North vs South
Bomberman
Borderlands 2
Borderlands: The Pre-Sequel!
Botanicula
BoulderDash®-TheFullCollection
Bounce Original
Brawlhalla
Breach & Clear
Bridge Constructor
Bright Memory Mobile
Broadsword: Эпоха рыцарей
BrogueX
Broken Age
Broken Sword: Director's Cut
Broken Sword II Smoking Mirror
Broken Sword V (Сломанный Меч 5)
Brothers: a Tale of two Sons
Bud Spencer & Terence Hill - Slaps And Beans
Bully: Anniversary Edition
Bus Simulator 2015
C-D
Caapora Adventure - Ojibe's Revenge
Call of Myth
Canabalt HD
Candleman
Carmageddon
Car Mechanic Simulator 2016
Card Quest
Carp Fishing Simulator
Castle Agony
Castle of Illusion
Castlevania: Symphony of the Night
CastleStorm - Free to Siege
Cataclysm Dark Days Ahead
Cataclysm: Dark Days Ahead (Play Market)
The Cave
CHAOS;CHILD
CHAOS;HEAD NOAH
Chicken Invaders (Play Market)
Chicken Police – Paint it RED!
Chip and Dale
Chuchel
ChuChu Rocket!™
Chuck's Challenge 3D
Chronology (Play Market)
Civilization VI
Clannad
Cognition Episode 1-4
Cogs
Colin McRae Rally
Colonization (freecol)
Colossal Cave Adventure (Play Market)
The Coma Cutting Class
Combat Pillows (Pillow Fight)
Comix Zone (Play Market)
Command & Conquer 4: Tiberium Twilight
Commander Genius
The Conduit HD
Contrast
Combat In The Fortress
Company of Heroes
Conquest of Elysium 3
Corpse Party: BLOOD DRIVE
Costume Quest
Counter Strike
Counter Strike Deathmatch
CounterSpy™
Counter Terrorist Agency
Crashlands
Crawl \ Local Tiles \ Dungeon Crawl Stone Soup
Crayon Physics Deluxe
Crazy Taxi
Crimson Gray
Crimson Gray - Dusk and Dawn
Crimsonland HD
Crowntakers
Crying Suns
Crypt of the NecroDancer
Chrono Trigger
Chaos - The Battle of Wizards
Cubemen
Cubemen 2
Cultist Simulator
Cultures: 8th Wonder of the World
Cultures: Northland
CyberHive
Cyberlords - Arcology
Danganronpa
Danganronpa 2
Dangerous Dave
Dangerous Dave In The Haunted Mansion
Dark Mist
Darkestville Castle (Play Market)
Darkstone / Темный камень
Dead Cells
Dead Synchronicity: TCT
Death Coming
Death Rally Free
Death Road to Canada
Death Squared
Defender of the Crown
Deflektor Classic
Deltarune
Demon Slayer [Online]
Descent I (1994) (GitHub)
Descent II (1996) (GitHub)
Desktop Dungeons
Detention
DevilutionX: Diablo on Android
Diamond Rush Original
Didi
Dink Smallwood HD
Disgaea 1 Complete
DISTRAINT: Pocket Pixel Horror
Dog Mendonca
Doki Doki Literature Club (4PDA) | Doki-Doki: Literature Club (GDrive)
Don't Move
Don't Starve Pocket Edition
Don't Starve: Shipwrecked
DOOM
DOOM II
Doom (Prboom DOOM)
Doom 2D
Doom Alphar
Doom Touch (там же Heretic, Hexen, Strife)
Doom III
Doom 3: BFG Edition (официальный порт, PlayMarket, Nvidia Shield)
Doomdark's Revenge
Door Kickers
Door Kickers: Action Squad
Dokuro
Donut County
Double Dragon Trilogy
Double Switch
Downwell
Dracula 1: Resurrection (Full) (Play Market)
Dracula 4
Dracula 5: The Blood Legacy HD
DRAGON QUEST
DRAGON QUEST II
DRAGON QUEST III
Dragon Quest IV: Chapters of the Chosen
DRAGON QUEST V
Dragon Quest VIII
Dragon's Lair
Dragon`s Lair 2: Time Warp
Draw a Stickman: EPIC
Drawn: The Painted Tower
Driver San Francisco
Droid Rider
Duck Game
DuckTales: Remastered
Duke Nukem 3D
Dune 2
Dungeon of the Endless: Apogee
Dungeon Warfare 2
Duck Souls
Dynamite Jack
Dwarf Journey
E-F
Earthworm Joe
Earthworm Jim
EARTH WARS
EDGE Extended
Eisenhorn: Xenos
Egypt: Old Kingdom
Elite (3D Space Game)
Elona Mobile
EMERGENCY
Embracelet
Enigma
Empire TV Tycoon
The Escapee
The Escapists 2: Pocket Breakout
The Escapists
ESJ: Groove City
Ever 17: The Out of Infinity
Everlasting Summer
Evoland
Evoland 2
Eufloria HD
Expendable Rearmed
F1 2016
Fahrenheit: Indigo Prophecy
FAR: Lone Sails
Farm Frenzy 1
Farm Frenzy 3
Farm Jane (Ферма Джейн)
Farming Simulator 18
FATAL FURY SPECIAL
Fate of Kai
Fear & Hunger
Feist
Final Fantasy I
FINAL FANTASY II
FINAL FANTASY III
FINAL FANTASY IV
FINAL FANTASY V
FINAL FANTASY VI
FINAL FANTASY VII
FINAL FANTASY VIII Remastered
FINAL FANTASY IX for Android
FINAL FANTASY XIII
FINAL FANTASY XIII-2
FINAL FANTASY CRYSTAL CHRONICLES Remastered Edition
FireTaps
Five Nights at Freddy's
Five Nights at Freddy's 2
Five Nights at Freddy's 3
Five Nights at Freddy's 4
Flashback Mobile
FLASHOUT 3D
Flight of the Amazon Queen
Flight Control
Figment
Football Cards
Forager
Forgotton Anne
FOTONICA
FNaF 6: Pizzeria Simulator
Fran Bow [All Chapters]
The Franz Kafka Videogame
Freddy Fazbear’s Pizzeria Simulator
Freddi Fish 2: The Case of the Haunted Schoolhouse
Friday Night Funkin
Frogger (Play Market)
Frontline: The Longest Day
Frozen Bubble
Frozen Synapse
Frozen Synapse Prime
Full Pipe: Приключения
G-H
Gabriel Knight Sins of Fathers
Game of Thrones
Game Dev Tycoon
Gangstar: Crime City
Gangstar Rio: City of Saints
Gem Rush Strategy Board Game
Gemini Rue
Getting Over It
Giana Sisters
Gigalomania
Goat Simulator
Goat Simulator GoatZ
Goat Simulator MMO SImulator
Goat Simulator Waste of Space
Goat Simulator: PayDay
Godus
Gobliiins Trilogy
GoNNER (Play Market)
Golden Axe Classic (Play Market)
Gorogoa
Grand Theft Auto III
Grand Theft Auto: Chinatown Wars
Grand Theft Auto: Liberty City Stories
Grand Theft Auto: San Andreas
Порт GTA SA-MP на OS Android
Grand Theft Auto: Vice City
Graveyard Keeper
Greed Corp HD
Grid Autosport
Grim Fandango Remastered
Guilty Parade
Gumbelmon 3D (Play Market)
GUNS UP! Mobile
GWENT: The Witcher Card Game
Habbo
Hack, Slash, Loot
Haegemonia - Legions of Iron
Half-Life
Half Life I
Half Life II
Half-Life 2: Episode One
Half-Life 2: Episode Two
Hamlet / Гамлет
Hammurabi (Play Market)
Hanger
Hassel 1977
Hatoful Boyfriend
Healer's Quest: Pocket Wand
Heileen 2
Her Story
Hex of Steel
Hexen GLES
Hexen II
Hexen II Touch
Hero Siege: Pocket Edition
Heroes of Might and Magic II (Free Heroes 2)
Heroes of Might and Magic 3
Heroes of Might and Magic III
Hell Yeah! Pocket Inferno
Hello Neighbor
Hidden Secrets Nightmare (Тайна лотерейного билета)
High School: Hook Ups
History Great Battles Medieval
Homeworld
HotDog Bush
Hotline Miami
Hotline Miami 2: Wrong Number
House Flipper: Home Design, Renovation Games
House of Mist / Дом из тумана
Hue: A Pocket Adventure
Human Fall Flat
Huntdown[/url]
I-J
I am Bread
I Have No Mouth
I Was Rebuilt
lanternium
Icewind Dale Enhanced Edition
ICEY
Imperium Galactica 2
The Impossible Game
Indiana Jones and the Lost Puzzles
Incredible Mandy (Tap.io)
Incredipede
Infamous Machine
Inked
The Inner World
Iris and the Giant
Iron Snout
Ironclad Tactics
The Island: Castaway
Ittle Dew
Jade Empire: Special Edition
Jagged Alliance 2
Jazz Jackrabbit (OpenJazz)
Jetpack (Play Market)
Jet Set Radio
Joe Danger
K-L
Karateka Classic
Katawa Shoujo
Keep Talking and Nobody Explodes
Kerixep
Kid Chameleon (Play Market)
Kill It With Fire
KONA
King of Dragon Pass: Text Adventure RPG
Королевство Зерна. Kingdom of Grain
Kingdom: New Lands
Kingdom Two Crowns
Kingdom Rush
Kingsport Festival
Knights of Pen & Paper
Knights of Pen & Paper 2
Knock-Knock
Kung Fu Master (Play Market)
The Last Door: Collector's Edition
The Last Express
Last Horizon
Layton: Curious Village in HD
Layton’s Mystery Journey 1
Layton: Pandora's Box in HD
The Legend of Zelda: Mystery of Solarus DX
Legend Of Kunduns (Play Market)
Legion Gold
LEGO® Batman: Beyond Gotham
LEGO Batman: DC Super Heroes
LEGO Harry Potter: Years 1-4
LEGO Harry Potter: Years 5-7
LEGO® Jurassic World
LEGO® The Lord of The Rings
LEGO® Marvel Super Heroes
The LEGO Movie Video Game
LEGO® Star Wars: The Complete Saga
Leisure Suit Larry: Reloaded
Leisure Suit Larry - Wet Dreams Don't Dry
Levelhead
The Letter
Life is Strange
Life Is Strange: Before The Storm
Little Big Adventure
Little Inferno
LIMBO
Linelight
Loco Roco
Lock's Quest
Lode Runner
Lode Runner Classic
Lode Runner (ZX Spectrum) (Play Market)
The Lords of Midnight
Lost Horizon
Lost Horizon 2
Lost Kingdom of Samaria (Скрытая цель: затерянное Королевство Самарии)
LOVE – A Puzzle Box Filled with Stories
LovePlus (Google Play, онлайн отключен)
Luftrausers
Lume
Luxor HD
M-N
Machinarium
Machineers
Machinika Museum
Mad Games Tycoon
Majesty: Fantasy Kingdom Sim
Majesty: The Northern Expansion
Mahjong Artifacts® (Full)
ManaPlus
Marble Age: Remastered
Mario Forever
Max Payne Mobile
MechaNika
Mega-lo-mania (Gigalomania)
MEGA MAN MOBILE 1-6
Metal Gear Rising: Revengeance (Play Market)
Metal Slug
METAL SLUG 2
METAL SLUG 3
METAL SLUG X
Metroid 2
McPixel
Micetopia
Microsoft Solitaire Collection (Пасьянсы из ОС Windows)
Micro Pool (Micropool)
Might & Magic Clash of Heroes
Minicraft
Minecraft
Minecraft: Story Mode
Minetest Mobile
Mini Metro
Miragine War
Mitoza
MO: Astray
Moe Era
Moonlighter
Moncage
Monmusu Quest! / Monster Girl Quest!
MONOBOT
Monster Hunter Stories
Monster Loves You!
Monster Madness Grave Danger
Montezuma's Revenge! (Play Market)
Moon Patrol Run
The Mooseman
Morrowind (разработка порта движка)
Mortal Kombat 2
Mortal Kombat 3
Moto Racer 15th Anniversary
Monster Hunter Stories
Mount & Blade: Warband
Mount Your Friends
Multicraft
Mu Miracle / Mu Origin
Murderous Pursuits
My Time at Portia
Myst (realMyst)
Narcissu
Neighbours back From Hell
Neighbours from Hell: Season 1
Neigbhbours From Hell: Season 2
Neo Scavanger
Neon Chrome
Never Alone Kisima Ingitchuna
Never Alone: Ki Edition
Neverball ME
Neverending Nightmares
Neverwinter Nights: Enhanced Edition
NetHack
Nexomon: Extinction (Google Play)
NHL 2K
NightSky
Nihilumbra
Northgard (Play Market)
NOT A HERO
Nuclear Throne
O-P
Obama vs Romney
Octodad: Dadliest Catch
Oddworld: Munch's Oddysee
Oddworld: New 'n' Tasty
Oddworld: Stranger's Wrath
Одиссей Космос и его робот Квест
Oil Rush: морская 3D стратегия
OlliOlli2: Welcome to Olliwood
One Hour One Life (Play Market)
OpenLara (Tomb Raider)
Operate Now: Tonsil Surgery
Opsu! (Play Market)
Organ Trail: Director's Cut
Orwell
Oregon Trail HD
Osmos HD
Osu!
osu!Lazer
OlliOlli
OXENFREE
PAC-MAN Championship Edition
Party Hard
Parallax
Payback 2 - The Battle Sandbox
Peace, Death! (Play Market)
Pearl Puzzle
Pekka Kana 2 (Pekka the Rooster 2) (Play Market)
Phantasy Star II (Play Market)
Phantom Rose Scarlet
Pictoquest
Pigs Can't Fly
Pikachu Kawai
Planescape: Torment EE
Plants vs. Zombies
Prehistorik
Prince of Persia Classic
Prince of Persia Forgotten Sands
Prince of Persia Shadow&Flame
Prison Architect: Mobile
Pro Pilkki 2 Mobile
Project Highrise
Pix the Cat (Play Market)
Pocket MapleStory
Pocket Tanks
Police Sim 3D: London
Poly Bridge
Roman Empire
Ponpu
Portal
Portal Knights
Postal
The Powder Toy
Predynastic Egypt
Project IGI 2 Covert Strike
Project Stalker
PUSS!
Puzzle Quest 2
Q-R
Qbert: Rebooted
QB Planets
Quake
Quake GL
Quake I Touch
Quake II Touch
Quake 3 Touch
quake4droid
Quad Touch (Q1/Q2/Q3/Hexen2)
Quarterback Attack (Play Market)
Q.U.B.E.: Director's Cut
Quest of Dungeons
The Quest
R-Type
R-TYPE II
Race Into Space Pro
Race The Sun (Play Market)
Radio Commander
Raiden Fighter 2013 HD
Raiden Legacy
Rainy Day
Rage In Peace
Raging Loop
Ragnarok: AndRO
Ragnarok Online: Valkyrie Uprising
Railroad Ink Challenge
Randals Monday
RASPBERRY MASH
Rayman Classic
Realpolitiks Mobile
Record of Agarest War
Red Bull X-Fighters
Repton 1
ReGoth For Android
Republique
Resident Evil 4 Mobile (BioHazard 4 Mobile)
Resident Evil 5 (Play Market)
Resident Evil: Uprising
RESCUE: Heroes in Action
Retro City Rampage DX
Return To Castle Wolfenstein (RTCW) Touch
The Revenge of Shinobi Classic (Play Market)
Reventure
RE-VOLT Classic (Premium)
Ristar Classic (Play Market)
Rise of the Triad
Risk of Rain
River Raid
Road Not Taken
ROBLOX
Rochard
RollerCoaster Tycoon® Classic
RollerCoaster Tycoon® Classic
Roman Empire
Rome: Total War
Rome: Total War - Alexander
Run or Die (Play Market)
Run Sackboy! Run!
Runestone Keeper (Play Market)
S-T
Sacred Stones Mobile
Sanitarium
Sakura Angels
Sakura Dungeon
Sakura Fantasy ch 1
Sakura Santa (AniVisual)
Sakura Spirit
Sakura Shrine Girls
Saga Frontier Remastered (Play Market)
Samorost 1
Savant - Ascent
Saya No Uta / Песнь Сайи
Sclerosis: The Dark Descent [АМНЕЗИЯ]
SCP-087
Scribblenauts Remix
Scribblenauts Unlimited
Secret Files Tunguska
Secret Files Sam Peters
Secret of Mana
Serious Sam
The Settlers HD
Shadow Blade: Reload
Shadowrun: Dragonfall - DC
Shadowrun Returns
Shadowgate
Shadow Warrior (1997) (Play Market)
Sheltered
Shaq Fu: A Legend Reborn
Shovel Knight
The Silent Age
SimCity™ Deluxe
Simon The Sorcerer
Simon The Sorcerer 2
Simon Tatham's Puzzles
SIMULACRA
SIMULACRA 2
Sine Mora
Skool Daze Reskooled!
SKYHILL
Sky Force
Sky Force 2014
Skullgirls
Skulls of the Shogun
Stack Attack (как на Siemens A52)
Strikers 1945 (Play Market)
STRIKERS 1945-2
Slay
Slay the Spire
Slender: The Arrival
Slightly Magic
Small World 2
Solar 2
Sonic Advance
Sonic CD
Sonic The Hedgehog
Sonic The Hedgehog 2™
Sonic Jump
Sonic & SEGA All-Stars Racing™
Sonic Racing Transformed
Sophie's Curse: Horror Game (ApkPure)
SOUL CALIBUR
SoulWorker: Zero
SpaceChem Mobile
SPACECOM
Space Food Truck
Space Harrier II Classic (Play Market)
SPACE HULK
Spaceward Ho! (Play Market)
Sparklite
Spectral Souls
Spelunky Classic HD
Spider-Man: Ultimate Power (Человек-Паук: Невероятная Сила)
Splice: Tree of Life HD
SpongeBob SquarePants: Battle for Bikini Bottom
Spy Fox in Dry Cereal
Spy vs Spy
S.T.A.L.K.E.R mobile vol.2
Star Control II
Star Defender 4
Stardew Valley
Star Wars: The Clone Wars
Star Wars: Dark Forces (Amazon Appstore)
Star Wars: Jedi Academy Touch
Star Wars: Jedi Knight II Touch
Star Wars: Knights of the Old Republic™
STAR WARS™: KOTOR II
STAY
Stealth Bastard
Stealth Inc. 2: Game of Clones
STEINS;GATE
STEINS;GATE Linear Bounded Phenogram
Strife Touch
Sumotori Dreams
Supaplex
Super Auto Pets
Super Fancy Pants Adventure
Super Hexagon
Super Mario 64
Super Mario A
Super Meat Boy
Super Mega Baseball (Play Market)
Superbrothers: Sword & Sworcery EP
Super Crate Box
Surgeon Simulator
Swords & Soldiers
Syberia
Syberia 2
Syder Arcade HD
Symphony
T-Aiko!
Tales of the Mirror
Tales from the Borderlands
Tales of the Neon Sea
Tallowmere
Tank 1990 (Battle City)
Tank-O-Box
Tasty Planet
Tasty Planet: Back For Seconds
Tempest: Pirate Action RPG Premium (Play Market)
Tesla Force
The Alliance Alive HD Remastered (Play Market)
The First Tree
The Heroic Legend of Eagarlnia
The Last Door: Season 2
The Legend of Bum-Bo
The Longest Road on Earth
The Stanley Parable Full
The Talos Principle
The Tree
The Uncertain
The Wild Case
There Is No Game: Wrong Dimension
This Is Police
THUMPER
Templar Battleforce RPG
Terminal Velocity
Teslagrad
Tetrobot and Co.
The Dark Side of the Moon: An Interective Thriller
The Deer God
The Swords of Ditto
The Great Ace Attorney
THE LAST REMNANT Remastered
The Tiny Bang Story
Ticket to Ride
TileMap
Time Gal
Titan Souls
Titan Quest
Titan Quest: Legendary Edition
The Choice of Life: Middle Ages
Theme Hospital (Corsix-TH)
This Is The Police 2
This War of Mine
Third World: War of Kings (Третий Мир Война Королей)
Toki Tori
Tomb Raider 2013 (Shield)
Tomb Raider I
Tomb Raider II
Total War Battles: KINGDOM
Total War Battles: SHOGUN (Total War Battles)
To the Moon
Tower Bloxx: My City
Townsmen 6
Toy Odyssey: The Lost and Found
Touhou Project (Play Market)
TrackRacing Online
Trainz Simulator 3
Transport Tycoon Deluxe (OpenTTD)
Treasure Island
Trials of Mana
Tribble Troubles
Trine 2: Complete Story
Tropico
Through the Darkest of Times
Tsukihime / Лунная принцесса
Tux Rider - Android Edition
Tyrian (OpenTyrian)
Type: Rider
Typoman Mobile
U-V
UFO Alien Invasion
Undertale
Unit 404
Unmechanical
Unreal Tanks 3D HD
Unruly Heroes
Ultima 7 Exult
Ultima Online Client (Play Market)
Ultimate Chicken Horse (Play Market)
Ultimate Custom Night
Uplink
VA-11 Hall-A
Valiant Hearts: The Great War
VALKYRIE PROFILE: LENNETH
Virtua Tennis Challenge
Virtual Villagers 2
VoK
VVVVVV
W-X
Wagic
The Walking Dead: Michonne
The Walking Dead: Season One
The Walking Dead: Season Two
The Walking Dead: Season Three
Wall of Insanity
Warbands: Bushido
Warcraft II
Warcraft 3 на Android
Warfare Incorporated
Warhammer 40,000: Mechanicus
Warhammer 40,000: Regicide
War Thunder (Play Market)
Weird Worlds
Western Bar
WILL: A Wonderful World
Willy Jetman: Astromonkey's Revenge
Windward
Wings: Remastered Edition
Wintertale
The Witness (Play Market)
Higurashi no Naku Koro ni (When They Cry 1) / Когда плачут Цикады
Higurashi no Naku Koro ni Kai (When They Cry 2) / Когда плачут Цикады: Ответы
When They Cry to 3 / Когда плачут чайки
while True: learn()
White Day
WILL: A Wonderful World
Wizrogue - Labyrinth of Wizardry
The Wolf Among Us
WolfDroid
Wolfenstein 3D GLES
Wolfenstein 3D: Spear of Destiny
Wolfenstein 3D Touch
World of Goo
The World Ends With You
Worms
Worms II
Worms III
Worms IV
Wrath Of The Titans
xBill
Xevious
X-Com (OpenXcom)
XCOM®: Enemy Unknown
XCOM®: Enemy Within
XCOM 2 Collection
Xuanyuan Jian 3 Waizhuan: Tian zhi Hen
Y-Z
Yesterday
Yesterday Origins
Youtubers Life - Gaming
Ys Chronicles I
Ys Chronicles II
Z Origins - (Z The Game)
Z Steel Soldiers
Zen Bound 2
Zombie Farm / Зомби Ферма
Zombie Driver THD
Zombie Taxi
Zuma's Revenge!
А-Я
Армелло
Атлантида 3: Новый Мир
Братья Пилоты
Братья Пилоты 2
Братья Пилоты 3
Ведьмак. Истории: Кровная вражда
Как Достать Соседей
Кинг
Левиафан: Последний день Декады
Морской Бой
Петька и Василий Иванович (Play Market)
Петька и Василий Иванович 2 (Play Market)
Петька и Василий Иванович 3 (Play Market)
Приключения Поручика Ржевского (Play Market)
Поле Чудес
Путь Зоны (Play Market)
Сияющая Песня Сверхновой
Танчики 1990
Цветные Линии (Play Market)
ШтЫрлиц: операция “БЮСТ”
Список околопортов:
Alien Shooter 2 - The Legend
Assassin's Creed Identity
Assassin's Creed Pirates
The Amazing Spider-Man
The Amazing Spider-Man 2
Apex Legends Mobile
Battlefield: Bad Company 2
Beatmania IIDX ULTIMATE MOBILE (Google Play)
Black Desert Mobile
BlazBlueRr
CaesarIA
Call of Duty: Black Ops Zombies
Call of Duty®: Strike Team
Carnivores: Dinosaur Hunter HD
Carnivores: Ice Age
Civilization I/II (Android FreeCiv developer preview)
Civilization Revolution 2
Critical Strike Portable
Contra: Evolution
Dead Bunker 4
Dead by Daylight
Dead Space
Deus Ex: The Fall
Digger HD
Dizzy - Prince of the Yolkfolk
Double Dragon IV
Dungeon Hero RPG
Dungeon Keeper
eFootball PES 2021
Empire Deluxe Mobile Edition
FIFA 2010
FIFA 2012
FIFA 14 by EA SPORTS™
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Что такое порт/околопорт?
Для понимания:
Порт это не обрезанная версия, а полноценная, перенесенная на платформу Android.
Порты выходят значительно позже (спустя несколько лет) после релизов игр и изначально разрабами не задумываются, а выпускаются с целью собрать бабла с отжившей своё игры. Порты фанаты игр очень ждут.
Околопорт - это незначительная "вариация на тему", очень несущественно отличающаяся геймплеем.
Мультиплатформа - это когда разрабы заранее, еще до момента релиза игры или чуть позже, заявляют, для каких платформ сразу выйдет игра. Так же, мультиплатформа является игрой с кроссплеем с разными платформами, а не донатные помои с поверхностей маркета. К данному подразделу относятся также и порты браузерных игр (не всех, разумеется) доступные как на PC, так и на мобильных устройствах.
Какие портированные игры будут добавляться в шапку?
В данной теме будем собирать каталог игр, которые были портированы на Android с PC, PS, XBOX и других платформ.
Почему мой очень важный пост удалили?
Существует очень много факторов, в связи с которыми ваш пост был удалён. К примеру: Ваш пост чистой воды оффтоп; пост оформлен не по шаблону; игра уже есть в шапке; Ваш пост нарушает правила форума.
Список игр по жанрам. Актуально на 10.04.2020. | Исходник кода списка прикреплен к записи и каждый может его актуализировать по мере наполнения шапки. Ссылку на список по жанрам буду в шапке обновлять. Просьба к тому, кто будет актуализировать список по жанрам не забывать прикреплять текстовый файл с кодом записи для последующего наполнения.
Сообщение отредактировал Lеshij - 22.11.24, 08:12
Причина редактирования: Новый список портированных игр
20.06.16, 13:24| #2
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overgraf ᵥ
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Удобная тема была бы
Сообщение отредактировал overgraf - 20.06.16, 13:24
20.06.16, 13:28| #3
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John2013 ᵥ
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DOOM 3 на android
Counter strike
Half-Life[3D]
20.06.16, 13:28| #4
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mikika ᵥ
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overgraf @ 20.06.2016, 13:24
Удобная тема была бы
Согласен :) Еще несколько таких мнений и тема пойдет в закрепленные :)
Добавлено 20.06.2016, 13:32:
John2013 @ 20.06.2016, 13:28
DOOM 3 на android
Counter strike
Half-Life[3D]
Добавил.
20.06.16, 14:03| #5
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Oniksss ᵥ
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SW: KOTOR
20.06.16, 14:09| #6
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GioMobile ᵥ
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GTA: Liberty City Stories[3D]
Grand Theft Auto 3 [3D, G-sensor]
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Dead space [3D]
Joe Dever's Lone Wolf + Complete Edition[3D]
Crossfire mobile[3D, Online] (азиатский порт)
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Resident Evil 4 Mobile (BioHazard 4 Mobile) [3D]
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Skylanders Trap Team™
Все, что нашел на первых 4ех страницах раздела шутеров. Дальше искать влом. Порт - понятие относительное. Можно ли назвать портом WoT blitz или Mortal combat X?
Да и вообще я не особо понимаю смысл данной темы. По-вашему, если игра портирована, то это значит, что она хорошая?
Сообщение отредактировал GioMobile - 20.06.16, 14:11
20.06.16, 14:28| #7
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mikika ᵥ
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GioMobile @ 20.06.2016, 14:09
Можно ли назвать портом WoT blitz или Mortal combat X
Конечно нет. Порт это не обрезанная версия, а полноценная, перенесенная на платформу Android.
20.06.16, 15:33| #8
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Photon9 ᵥ
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dead space в таком случпе никак не порт
20.06.16, 15:45| #9
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SoulHunter22 ᵥ
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Самые лучшие игры это портированые с других платформ :) а не те что пишут под андроид :)
20.06.16, 15:52| #10
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mikika ᵥ
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Photon9 @ 20.06.2016, 15:33
dead space в таком случпе никак не порт
Перенесу в другой список. Спасибо.
20.06.16, 17:14| #11
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Montgolfier ᵥ
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Боже, идеальная тема, теперь и искать не надо. Благодарю.
20.06.16, 17:30| #12
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mikika ᵥ
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SonyFun12 @ 20.06.2016, 17:14
Боже, идеальная тема, теперь и искать не надо. Благодарю.
Если будет что добавить, не стесняйся сообщать :)
20.06.16, 18:08| #13
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Lord-Valington ᵥ
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Тема збс, нужно прикрепить ее
Dungeon Keeper не порт, Republique её же с андроид портировали на пк
Файнал фентези 9 и 3 можно добавить
RE-VOLT Classic (Premium) [3D]
Сообщение отредактировал Lord-Valington - 20.06.16, 19:09
20.06.16, 18:19| #14
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Synthetic Existence ᵥ
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Проекты с Java котируются?
20.06.16, 18:26| #15
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Wander2048 ᵥ
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The Sims 3 не полная, в околопорты её.
Hotline Miami 2: Wrong Number
Brothers: a Tale of two Sons[3D]
Rezident Evil 4 тоже околопорт.
Сообщение отредактировал Wander2048 - 20.06.16, 18:35
20.06.16, 18:37| #16
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Pavel004 ᵥ
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Еще бы с Symbian найти игры. Знаю, что были портированы sky force, final battle. Они вроде есть на форуме, но тяжело с телефона оформлять гиперссылки
20.06.16, 18:39| #17
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Wander2048 ᵥ
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Pavel004, не тяжело, просто введите ссылку и она автоматически определиться.
20.06.16, 18:50| #18
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axinek ᵥ
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интересно, а crashlands является портом с пк или наоборот, я вот не помню что вначале вышло
20.06.16, 18:51| #19
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duckes84 ᵥ
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Отличная тема в закрепленных была бы!
Хоть какой лучик света в темном царстве раздела "игры для андроид"
20.06.16, 18:57| #20
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xlotosx90 ᵥ
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Но ведь... Ведь minecraft - не порт..
20.06.16, 19:24| #21
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рапик ᵥ
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mikika,
Идея хороша)
А можешь в строке игры указывать ее жанр? Вроде будет удобнее)
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12625 | https://feldman.law/news/breach-of-fiduciary-duty-in-real-estate-transactions/ | Skip to content
Feldman & Feldman
Civil Litigation Law Firm
Breach of Fiduciary Duty in Real Estate Transactions
Posted on
A fiduciary duty is a legal obligation wherein one party is legally required to act in the best interests of another. The obligated party is known as the fiduciary and is entrusted with the care of the other party’s finances, property, or other valuables. When a breach of this legal duty occurs, it can result in serious repercussions for both parties. Understanding how a breach of fiduciary duty in real estate transactions can occur is essential to protecting one’s assets and finances.
When it comes to real estate, in the event a real estate professional is acting as an “agent,” they owe their client a fiduciary duty. Though commonly referred to as real estate agents, it is actually rare for real estate deals to involve real estate professionals acting within the legal definition of an agent.
Not All Real Estate Agents Are Bound by a Fiduciary Duty
If you’re thinking of buying a house, it’s easy to assume the real estate agent or agency you’re working with has your best interests at heart. Whether you’re buying or selling, it’s hard to imagine that these entities would only think of themselves, right?
The Consumer Federation of America found that “real estate agents often are not required by law to represent the interests of buyers or sellers.” This means some real estate agents’ clients can fall prey to poor transactions or self-dealing, as real estate agents are not legally obligated with a fiduciary duty to their clients.
Most states have enacted laws requiring a specific signed document to give a real estate professional the status of a legal agent. In Texas, the Texas Real Estate Commission (TREC) has specific rules in place governing a real estate broker or salesperson acting as an agent or fiduciary of another. Certain laws – like the Texas Real Estate License Act, the Inspector Act, the Residential Service Company Act, and the Timeshare Act – helped codify and solidify these rules.
How Can Real Estate Professionals Breach Their Fiduciary Duty?
Even with applicable laws and regulations in place, a breach of fiduciary duty can still occur in a real estate transaction. When a real estate professional is granted agent status, the resulting fiduciary duty requires them to be honest and candid with the buyer they are representing. The agent will spend the majority of the buying or selling process working independently of the client by negotiating with other parties, coordinating inspections, and reviewing documents. The real estate agent is ultimately required to act as a proxy for their client by making decisions for them and reporting information in an open and honest manner.
A breach of the real estate agent’s fiduciary duty to their client can occur when the agent decides to act in ways that benefit themself instead of in the best interest of their client.
Breaches of fiduciary duty in real estate transactions can occur when:
The agent receives secret profits or fees not disclosed to the client
The agent fails to inform a seller of other offers on the table after an offer has been accepted
The agent fails to advise a buyer of any material defects to the property
There is dual representation or facilitating a sale of a property by acting as the real estate agent for both the buyer and seller without either or both of the parties’ knowledge
The agent provides the client’s personal information to the other broker without the client’s permission
The agent declines or accepts an offer without the client’s approval
Each breach can have a serious financial impact on the client and can cause irreparable damage, including leaving a buyer stuck with a house full of defects or a client having their personal details disclosed to unknown individuals and/or entities.
In the event a breach of fiduciary duty in a real estate transaction occurs, there are several available legal remedies, including:
Rescission
When a real estate agent breaches their fiduciary duty, the client can ask the court to rescind the contract and restore them to its status before entering the agreement. This can involve the return of the sold property to the seller along with a refund of the purchaser’s money.
Forfeiture of Commission
A real estate agent is not entitled to payment if they breach their fiduciary duty. Because of this, an agent can be required to refund any compensation received.
Damages
If a breach of fiduciary duty by a real estate agent damages their client, the agent must compensate the client for those damages. In the event the agent of a seller failed to present a better offer than what was accepted, the agent would be required to reimburse the seller for the difference between the lower and higher offers.
Breach of Fiduciary Duty Attorneys
Failing to uphold a fiduciary duty can have serious legal ramifications. Any time fiduciary profits off of a real estate transaction through self-dealing or causes a loss to their client, they can be held legally accountable for their actions. The attorneys at Feldman & Feldman routinely handle allegations of breach of fiduciary duty. We understand these claims often involve delicate situations, so we will protect your interests while exploring all options for resolution. If you have been accused of breaching a fiduciary duty or if you believe your fiduciary’s actions constitute a breach, we can help.
Our commercial litigation attorneys understand the complexities and nuances of the fiduciary relationship, and we represent clients in pursuing and/or defending against these claims. Contact the real estate transactions attorneys at Feldman & Feldman today to discuss your needs. |
12626 | https://www.quora.com/How-can-you-solve-the-linear-Diophantine-equations-ax-by-c-and-ay-bx-c-with-integer-coefficients-and-variables-using-the-Euclidean-algorithm | Something went wrong. Wait a moment and try again.
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5
How can you solve the linear Diophantine equations ax + by = c and ay + bx = c with integer coefficients and variables using the Euclidean algorithm?
Sin Keong Tong
M.Sc. in Mathematics, University of New South Wales
·
Author has 283 answers and 361.1K answer views
·
2y
The normal process to use the Extended Euclid Algorithm (EEA) which is rather cumbersome. Presented below is an adaptation of Euclid Algorithm that does not involve backtracking or finding multiplicative inverse modulo.
[Extended Euclidean algorithm - Wikipedia
Method for computing the relation of two integers with their greatest common divisor In arithmetic and computer programming , the extended Euclidean algorithm is an extension to the Euclidean algorithm , and computes, in addition to the greatest common divisor (gcd) of integers a and b , also the coefficients of Bézout's identity , which are integers x and y such that a x + b y = gcd ( a , b ) . {\displaystyle ax+by=\gcd(a,b).} This is a certifying algorithm , because the gcd is the only number that can simultaneously satisfy this equation and divide the inputs. [ 1 ] It allows one to compute also, with almost no extra cost, the quotients of a and b by their greatest common divisor. Extended Euclidean algorithm also refers to a very similar algorithm for computing the polynomial greatest common divisor and the coefficients of Bézout's identity of two univariate polynomials . The extended Euclidean algorithm is particularly useful when a and b are coprime . With that provision, x is the modular multiplicative inverse of a modulo b , and y is the modular multiplicative inverse of b modulo a . Similarly, the polynomial extended Euclidean algorithm allows one to compute the multiplicative inverse in algebraic field extensions and, in particular in finite fields of non prime order. It follows that both extended Euclidean algorithms are widely used in cryptography . In particular, the computation of the modular multiplicative inverse is an essential step in the derivation of key-pairs in the RSA public-key encryption method. The standard Euclidean algorithm proceeds by a succession of Euclidean divisions whose quotients are not used. Only the remainders are kept. For the extended algorithm, the successive quotients are used. More precisely, the standard Euclidean algorithm with a and b as input, consists of computing a sequence q 1 , … , q k {\displaystyle q_{1},\ldots ,q_{k}} of quotients and a sequence r 0 , … , r k + 1 {\displaystyle r_{0},\ldots ,r_{k+1}} of remainders such that r 0 = a r 1 = b ⋮ r i + 1 = r i − 1 − q i r i and 0 ≤ r i + 1 < | r i | (this defines q i ) ⋮ {\displaystyle {\begin{aligned}r_{0}&=a\r_{1}&=b\&\,\,\,\vdots \r_{i+1}&=r_{i-1}-q_{i}r_{i}\quad {\text{and}}\quad 0\leq r_{i+1}<|r_{i}|\quad {\text{(this defines }}q_{i})\&\,\,\,\vdots \end{aligned}}} It is the main property of Euclidean division that the inequalities on the right define uniquely q i {\displaystyle q_{i}} and r i + 1 {\displaystyle r_{i+1}} from r i − 1 {\displaystyle r_{i-1}} and r i . {\displaystyle r_{i}.} The computation stops when one reaches a remainder r k + 1 {\displaystyle r_{k+1}} which is zero; the greatest common divisor is then the last non zero remainder r k . {\displaystyle r_{k}.} The extended Euclidean algorithm proceeds similarly, but adds two other sequences, as follows r 0 = a r 1 = b s 0 = 1 s 1 = 0 t 0 = 0 t 1 = 1 ⋮ ⋮ r i + 1 = r i − 1 − q i r i and 0 ≤ r i + 1 < | r i | (this defines q i ) s i + 1 = s i − 1 − q i s i t i + 1 = t i − 1 −]( "en.wikipedia.org")
Since the equations are symmetric in(x,y), the solution (x1,y1) in the first equation as (y1,x1) in the second equation. First divide the equation by hcf(a,b,c) so that it can be assume a,b,c are coprime. The following example demonstrates how to find the integer solutions for the system of equations.
143x+37y=11
37(3x+y)+32x=11
32(4
The normal process to use the Extended Euclid Algorithm (EEA) which is rather cumbersome. Presented below is an adaptation of Euclid Algorithm that does not involve backtracking or finding multiplicative inverse modulo.
[Extended Euclidean algorithm - Wikipedia
Method for computing the relation of two integers with their greatest common divisor In arithmetic and computer programming , the extended Euclidean algorithm is an extension to the Euclidean algorithm , and computes, in addition to the greatest common divisor (gcd) of integers a and b , also the coefficients of Bézout's identity , which are integers x and y such that a x + b y = gcd ( a , b ) . {\displaystyle ax+by=\gcd(a,b).} This is a certifying algorithm , because the gcd is the only number that can simultaneously satisfy this equation and divide the inputs. [ 1 ] It allows one to compute also, with almost no extra cost, the quotients of a and b by their greatest common divisor. Extended Euclidean algorithm also refers to a very similar algorithm for computing the polynomial greatest common divisor and the coefficients of Bézout's identity of two univariate polynomials . The extended Euclidean algorithm is particularly useful when a and b are coprime . With that provision, x is the modular multiplicative inverse of a modulo b , and y is the modular multiplicative inverse of b modulo a . Similarly, the polynomial extended Euclidean algorithm allows one to compute the multiplicative inverse in algebraic field extensions and, in particular in finite fields of non prime order. It follows that both extended Euclidean algorithms are widely used in cryptography . In particular, the computation of the modular multiplicative inverse is an essential step in the derivation of key-pairs in the RSA public-key encryption method. The standard Euclidean algorithm proceeds by a succession of Euclidean divisions whose quotients are not used. Only the remainders are kept. For the extended algorithm, the successive quotients are used. More precisely, the standard Euclidean algorithm with a and b as input, consists of computing a sequence q 1 , … , q k {\displaystyle q_{1},\ldots ,q_{k}} of quotients and a sequence r 0 , … , r k + 1 {\displaystyle r_{0},\ldots ,r_{k+1}} of remainders such that r 0 = a r 1 = b ⋮ r i + 1 = r i − 1 − q i r i and 0 ≤ r i + 1 < | r i | (this defines q i ) ⋮ {\displaystyle {\begin{aligned}r_{0}&=a\r_{1}&=b\&\,\,\,\vdots \r_{i+1}&=r_{i-1}-q_{i}r_{i}\quad {\text{and}}\quad 0\leq r_{i+1}<|r_{i}|\quad {\text{(this defines }}q_{i})\&\,\,\,\vdots \end{aligned}}} It is the main property of Euclidean division that the inequalities on the right define uniquely q i {\displaystyle q_{i}} and r i + 1 {\displaystyle r_{i+1}} from r i − 1 {\displaystyle r_{i-1}} and r i . {\displaystyle r_{i}.} The computation stops when one reaches a remainder r k + 1 {\displaystyle r_{k+1}} which is zero; the greatest common divisor is then the last non zero remainder r k . {\displaystyle r_{k}.} The extended Euclidean algorithm proceeds similarly, but adds two other sequences, as follows r 0 = a r 1 = b s 0 = 1 s 1 = 0 t 0 = 0 t 1 = 1 ⋮ ⋮ r i + 1 = r i − 1 − q i r i and 0 ≤ r i + 1 < | r i | (this defines q i ) s i + 1 = s i − 1 − q i s i t i + 1 = t i − 1 −
"en.wikipedia.org")
Since the equations are symmetric in(x,y), the solution (x1,y1) in the first equation as (y1,x1) in the second equation. First divide the equation by hcf(a,b,c) so that it can be assume a,b,c are coprime. The following example demonstrates how to find the integer solutions for the system of equations.
143x+37y=11
37(3x+y)+32x=11
32(4x+y)+5(3x+y)=11
5(27x+7y)+2(4x+y)=11(1)
By inspection, equation (1) is satisfied by
27x+7y=1
4x+y=3
Solving simultaneously yields x=20,y=−77
The general solution is given by (x,y)=(20+37k,−77+143k),k∈Z
Related questions
How do I solve a linear Diophantine equation using the Extended Euclid theorem?
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6
x
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Given a Diophantine equation with 1 = 1021x+51y, how can someone find more than 1 solution for x and y beside the one you get from the extended Euclidean algorithm? What are other pairs for (x,y)?
Leo Cutter
Studied Mathematics at Bishop Gorman High School (Graduated 2020)
·
Author has 356 answers and 4.1M answer views
·
6y
Related
What is the solution to this diophantine equation: ax+by+cz=d?
The question is asking for the general solution to the linear diophantine equation of 3 variables. I’ll admit that I don’t know the actual solution, but I come up with them as I write the answer. Thus, this answer you see here is probably different from the one in your textbook.
ax+by+cz=d
First, we need to determine if there even are integer solutions to the equation, so note that the solution only works when gcd(a,b,c)∣∣d. Now, onto the problem.
In order to solve the problem, I started by looking at the solution for 2 variables, which is given by:
ax+by=c,x=x1
The question is asking for the general solution to the linear diophantine equation of 3 variables. I’ll admit that I don’t know the actual solution, but I come up with them as I write the answer. Thus, this answer you see here is probably different from the one in your textbook.
ax+by+cz=d
First, we need to determine if there even are integer solutions to the equation, so note that the solution only works when gcd(a,b,c)∣∣d. Now, onto the problem.
In order to solve the problem, I started by looking at the solution for 2 variables, which is given by:
ax+by=c,x=x1+kv,y=y1−ku
Where k is the parameter, v=bgcd(a,b),u=agcd(a,b)
This is actually a pretty clever solution, because looking at the proof tells you instantly how it was come up with. Plugging the values of x and y back into the original equation, it all cancels nicely:
ax+by=c,x=x1+kv,y=y1−ku⟹a(x1+kv)+b(y1−ku)=c
Which reduces to
ax1+avk+by1−buk=c⟹ax1+by1+kabgcd(a,b)−kabgcd(a,b)=c
And of course it cancels all out to the original equation of ax1+by1=c. Since this is so, we’ll use this same trick to find the general solution for 3 variables. I will admit that it took me a while to come up with this, but I ended up finding the solutions to be:
ax+by+cz=d,x=x1+ck,y=y1+ck,z=z1−(a+b)k
I must say that I’m kind of proud of this because I tried it on a bunch of really large values for a,b,c and it held up. Anyways, the proof is the same for the 2-variables; simply substitute.
a(x1+ck)+b(y1+ck)+c(z−(a+b)k)=d
For the rest, it’s all a bunch of algebra, so I’ll probably leave to proof for you to verify.
Why did I show you the 2 variable equation? It’s to help you come up with your own solution for n-variables equations. Try to end up adding terms to the final answers but making sure that they’re canceled out by subtraction in another variable. If you’re adding 2k to the first variable, try to add −2k to another variable. When you think you have the correct solutions, substitute them back into the original equation and make sure you end up with the original equation again.
Anyways, I’ve probably been rambling on for too long, so good luck!
Bernard Blander
Studied at McGill University
·
8y
Related
How do I solve the linear diophantine equation
6
x
+
10
y
+
15
z
=
23
?
We are given the equation 6x+10y+15z=23 and want to find all integer solutions x,y,z. The equation is called linear because all the unknowns x,y,z appear with exponents equal to 1. Additionally, it is called a Diophantine equation because we are looking for integer solutions.
Does the equation look familiar? You may recognize it as the equation of a plane in 3-space R3. If our Diophantine equation has integer solutions x,y,z they will live (be found) on this plane and denoted by the 3-tuple (x,y,z).
Now we will explore the possible solutions. In order for this solution to be unde
We are given the equation 6x+10y+15z=23 and want to find all integer solutions x,y,z. The equation is called linear because all the unknowns x,y,z appear with exponents equal to 1. Additionally, it is called a Diophantine equation because we are looking for integer solutions.
Does the equation look familiar? You may recognize it as the equation of a plane in 3-space R3. If our Diophantine equation has integer solutions x,y,z they will live (be found) on this plane and denoted by the 3-tuple (x,y,z).
Now we will explore the possible solutions. In order for this solution to be understandable to as many people as possible the somewhat arcane concepts and operations of modular arithmetic shall not be explicitly used, so please bear with me.
6x+10y+15z=23(1)
1x+5x+5(2y+3z)=5(4)+3 after regrouping in multiples of 5,
x=5(4−x−2y−3z)+3=5k+3 with k∈Z
Doing the same to obtain y,
6x+10y+15z=23
3(2x+5z)+3(3y)+1y=3(7)+2 after regrouping in multiples of 3,
y=3(7−2x−3y−5z)+2=3m+2 with m∈Z
Lastly, to obtain z,
6x+10y+15z=23
2(3x+5y)+2(7z)+1z=2(11)+1 after regrouping in multiples of 2,
z=2(11−3x−5y−7z)+1
z=2n+1 with n∈Z
In summary, so far we have determined,
x=5k+3,y=3m+2,z=2n+1,∀k,m,n∈Z
We still have to determine the relationship among k,m, & n, integers. To that end, we substitute the expressions for x,y,z into 6x+10y+15z=23.
6(5k+3)+10(3m+2)+15(2n+1)=23
30(k+m+n)+53=23
30(k+m+n)=−30
k+m+n=−1 or k=−(1+m+n)
Therefore,
x=−2−5m−5n
y=2+3m
z=1+2n
(x,y,z)=(−2,2,1)+m(−5,3,0)+n(−5,0,2) where m,n are arbitrary integers. The integer solutions are an infinite subset of points on the infinite plane 6x+10y+15z=23 depicted below.
Marco Vona
Trinity Wrangler (Cambridge Maths Tripos 2021)
·
Author has 99 answers and 423.1K answer views
·
Updated 8y
Related
How do I solve the linear diophantine equation
3
x
+
6
y
+
5
z
=
7
?
Consider the equation
3x+6y+5z−7=0
In 3D space, this represents a plane with normal vector ˆv(3,6,5). Bearing this in mind, here’s what we are going to do:
Find a point on the plane with integer coordinates
Find two non-parallel vectors with integer components that lie on the plane
Express any other point on the plane with integer coordinates as a linear combination of these two vectors applied to our starting point
Step 1
Trying some values for x, y and z we quickly find the point P(−1,0,2) that lies in the plane and has integer coordinates.
Step
Consider the equation
3x+6y+5z−7=0
In 3D space, this represents a plane with normal vector ˆv(3,6,5). Bearing this in mind, here’s what we are going to do:
Find a point on the plane with integer coordinates
Find two non-parallel vectors with integer components that lie on the plane
Express any other point on the plane with integer coordinates as a linear combination of these two vectors applied to our starting point
Step 1
Trying some values for x, y and z we quickly find the point P(−1,0,2) that lies in the plane and has integer coordinates.
Step 2
The vector ^v(3,6,5) is normal to the plane; therefore, any vector parallel to the plane should be perpendicular to ^v. So, if ˆw lies in the plane, then ^v⋅^w=0. After some attempts (e.g for both vectors ^wa and ^wb that lie on the plane, set a component to 0 and choose the other two components in such a way as to make the scalar products ^wa⋅^v and ^wb⋅^v equal to 0) we quickly come up with ^wa(5,0,−3) and ^wb(0,5,−6).
Step 3
Let ^u(−1,0,2) be the positional vector of the first point we found. Any other point with integer coordinates (i.e its positional vector) can be expressed as
^u+k⋅^wa+h⋅^wa
Hence the solution
⎧⎨⎩x=−1+5ky=5hz=2−3k−6h
with k,h∈Z2.
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Related questions
How do you solve this Diophantine equation
73
x
+
85
y
=
7
?
Can linear Diophantine equations be solved without using Euclid's algorithm in any special cases?
What are the steps in solving linear diophantine equations with 3 variables?
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How do I solve for the simultaneous equations ax + by + c and bx + ay + c (with working out shown)?
Reuven Harmelin
Lecturer at Technion - Israel Institute of Technology (1982–present)
·
Author has 2.3K answers and 1.9M answer views
·
3y
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How do you use the Euclidean algorithm to find d=(a,b) and find a pair of integers x,y such that ax+by=d, a=85, b=41, and a=112,b=18?
I am not intending to solve this HW exercise for you, just to remind the Euclidean algorithm for finding d=gcd(a,b) for any pair of natural numbers a>b>1:
Denote
then perform division with a remainder:
If
and that is the end, with x=0, y=1.
If
and once again, if the last remainder is zero, then
that is:
and if the remainder is positive, you may continue that process with steps of the same pattern as above. As log as the remainders are not zero, this process would continue, and observe that the remainders are decreasing, as in two typical step we get:
and
Since the remainders form a strictly decreasing s
I am not intending to solve this HW exercise for you, just to remind the Euclidean algorithm for finding d=gcd(a,b) for any pair of natural numbers a>b>1:
Denote
then perform division with a remainder:
If
and that is the end, with x=0, y=1.
If
and once again, if the last remainder is zero, then
that is:
and if the remainder is positive, you may continue that process with steps of the same pattern as above. As log as the remainders are not zero, this process would continue, and observe that the remainders are decreasing, as in two typical step we get:
and
Since the remainders form a strictly decreasing sequence of natural number, it could not continue forever, that is, there must be some natural number n such that
and then, the last remainder which is not yet zero is the least common divisor:
so that is the end of the Euclidean Algorithm.
Now, we go backward as follows
and eventually it would yield the 2 integers x,y such that ax + by=d.
Deb P. Choudhury
Former Professor at University of Allahabad
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 10K answers and 8M answer views
·
5y
Related
How can we find the solutions of linear Diophantine ax + by = c?
The Diophantine equation ‘ax+by =c’ has solution if and only if d=GCD(a,b) is a divisor of c.
For finding a solution, in case c=d.c’, we first use the steps of the Euclidean Algorithm for finding d, to write d= k.a+m.b, where k and m are integers. Then we see that x=k.c’ and y=m.c’ satisfy the given equation.
The set of all the solutions is given by x= k.c’+t(b/d) and y = m.c’-t(a/d), where t can be any integer.
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Dean Rubine
Author has 10.6K answers and 23.6M answer views
·
6y
Related
How do I solve the linear diophantine equation
6
x
+
10
y
+
15
z
=
23
?
I like showing the continued fraction solution for finding particular solutions to two variable linear Diophantine equations. These are so simple it’s easier just to guess the appropriate particular solutions, which is what I’ll do below.
6x+10y+15z=23
We’re asking which lattice points are in this plane.
2(3x+5y)+15z=23
We can treat this as simultaneous Diophantine equations:
2u+15z=23
3x+5y=u
It’s pretty easy to guess a particular solution for 2u+15z=23
How about z=1,u=4
2(4)+15(1)=23
For integer k,
2(15k)+15(−2k)=0
Adding,
2(4+15k)+15(1−2k)=23
So the general solution to this
I like showing the continued fraction solution for finding particular solutions to two variable linear Diophantine equations. These are so simple it’s easier just to guess the appropriate particular solutions, which is what I’ll do below.
6x+10y+15z=23
We’re asking which lattice points are in this plane.
2(3x+5y)+15z=23
We can treat this as simultaneous Diophantine equations:
2u+15z=23
3x+5y=u
It’s pretty easy to guess a particular solution for 2u+15z=23
How about z=1,u=4
2(4)+15(1)=23
For integer k,
2(15k)+15(−2k)=0
Adding,
2(4+15k)+15(1−2k)=23
So the general solution to this equation is
u=4+15k
z=1−2k
The next equation is
3x+5y=u
We can just guess a particular solution to 3x+5y=1, how about x=2,y=−1
3(2)+5(−1)=1
3(2u)+5(−u)=u
3(5n)+5(−3n)=0
Adding,
3(2u+5n)+5(−u−3n)=u
So the general solution is, for any integers k and n,
x=2u+5n=8+30k+5n
y=−u−3n=−4−15k−3n
z=1−2k
which simultaneously indicates a plane and its lattice points in point-vector-vector form:
(x,y,z)=(8,−4,1)+k(30,−15,−2)+n(5,−3,0)
Dean Rubine
Author of: It's not just π; all of trig is wrong!
·
Upvoted by
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge · Author has 10.6K answers and 23.6M answer views
·
4y
Related
How can I find non-negative solutions to a diophantine equation of form Ax+By = C?
First we verify GCD(A,B) divides C; if it doesn’t there can’t be any solutions. Remove any common factors giving the equivalent equation
ax+by=c
Next it’s usually easy to just guess a particular solution, to ax+by=c or maybe to ax−by=±1, in which case we can scale by ±c for a solution (negating y):
ax0+by0=c
Of course I’m glossing over the interesting part here. Instead of guessing we can use Continued Fractions, which not-incidentally also would give us the GCD(A,B), so we could start there without reducing first.
I’ll give a brief description of how, because you asked. We expand A/
First we verify GCD(A,B) divides C; if it doesn’t there can’t be any solutions. Remove any common factors giving the equivalent equation
ax+by=c
Next it’s usually easy to just guess a particular solution, to ax+by=c or maybe to ax−by=±1, in which case we can scale by ±c for a solution (negating y):
ax0+by0=c
Of course I’m glossing over the interesting part here. Instead of guessing we can use Continued Fractions, which not-incidentally also would give us the GCD(A,B), so we could start there without reducing first.
I’ll give a brief description of how, because you asked. We expand A/B as a continued fraction using Euclid’s GCD algorithm.
AB=ab=a1+1a2+1a3+…+1an=[a1,a2,a3,…,an].
Then the penultimate convergent, pq=[a1,a2,…,an−1], gives (q,p), a particular solution to ax−by=±1.
Let’s do a quick example. Solve 30x−13y=±1. We have
3013=2+1134=2+13+14=[2,3,4].
So [2,3]=2+13=73 gives a particular solution,
30(3)−13(7)=−1✓
OK, at this point we have aq−bp=±1; let’s say aq−bp=−1 to be specific; the other case only differs in minor details. We scale by −c to get a particular solution to our original equation,
a(−cq)+b(cp)=c
The solution (x0,y0)=(−cq,cp) has one negative and one positive coordinate; we’re after the solutions that are two non-negative numbers. So we introduce integer parameter t and write:
a(bt)+b(−at)=0
Adding to our particular solution,
a(bt−cq)+b(cp−at)=c
gives our general solution, (x,y)=(bt−cq,cp−at). For non-negative solutions we solve simultaneous inequalities:
bt−cq≥0,cp−at≥0
We’ll assume a>0,b>0 — the details will be different if that’s not the case.
t≥cq/b,cp/a≥t
which gives our integers t that correspond to the non-negative solutions,
cq/b≤t≤cp/a
That’s all rather abstract; let’s do an example, the one we already started. Say our equation is
60x+26y=138
If we do the GCD algorithm for the continued fraction we’ll of course find GCD(60,26)=2 so we reduce this equation to
30x+13y=69
We found by the continued fraction method above
30(3)−13(7)=−1
Multiplying by −c=−69,
30(3(−69))+13(7(69))=69
so a particular solution (−207,483). Adding
30(13t)+13(−30t)=0
gives a general solution
(x,y)=(13t−207,483−30t)
For non-negative solutions we seek
13t≥207,483≥30t
t≥207/13≈15.9,483/30=16.1≥t
Clearly we get a single positive solution, given by t=16.
(x,y)=(13(16)−207,483−30(16))=(1,3)
This one was easy enough to guess, but the method is completely general.
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Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
·
Author has 3.9K answers and 4M answer views
·
4y
Related
Given a Diophantine equation with 1 = 1021x+51y, how can someone find more than 1 solution for x and y beside the one you get from the extended Euclidean algorithm? What are other pairs for (x,y)?
1021=51(20)+1
1021(1)+51(−20)=1
(x0,y0)=(1,−20)
For ax+by=c, where (x0,y0) is a solution, the general solution is:
x=x0+bgcd(a,b)k
y=y0−agcd(a,b)k
where k is any integer.
Here, we have a=1021, b=51 and their greatest common divisor isgcd(1021,51)=1 since they are relatively prime (51=(3)(17) and 1021 is prime). Therefore, the general solution is:
x=1+51k
y=−20−1021k
or
(1+51k,−20−1021k)
A plot looks like this:
1021=51(20)+1
1021(1)+51(−20)=1
(x0,y0)=(1,−20)
For ax+by=c, where (x0,y0) is a solution, the general solution is:
x=x0+bgcd(a,b)k
y=y0−agcd(a,b)k
where k is any integer.
Here, we have a=1021, b=51 and their greatest common divisor isgcd(1021,51)=1 since they are relatively prime (51=(3)(17) and 1021 is prime). Therefore, the general solution is:
x=1+51k
y=−20−1021k
or
(1+51k,−20−1021k)
A plot looks like this:
Amitabha Tripathi
have more than a working knowledge of Z
·
Author has 4.7K answers and 13.9M answer views
·
11mo
Related
How can one find all integer solutions for two variables in diophantine equations such as $ax+by=c$? Are there any software or websites that can assist with this problem?
The Diophantine equation ax+by=c has a solution if and only if g=gcd(a,b)∣c. So there’s nothing more to discuss when g∤c.
Now suppose g∣c. Then the solutions to ax+by=c and a1x+b1y=c1, where a1=a/g, b1=b/g, c1=c/g are identical. The advantage of reducing to the smaller set of coefficients is that gcd(a1,b1)=1, and you are guaranteed identical solutions to this reduced Diophantine equation.
Thus, we may as well consider the equation ax+by=c with gcd(a,b)=1. We know this admits a solution in integer pairs; so, in fact, {ax+by:x,y∈Z}=Z.
Suppose (x0,
The Diophantine equation ax+by=c has a solution if and only if g=gcd(a,b)∣c. So there’s nothing more to discuss when g∤c.
Now suppose g∣c. Then the solutions to ax+by=c and a1x+b1y=c1, where a1=a/g, b1=b/g, c1=c/g are identical. The advantage of reducing to the smaller set of coefficients is that gcd(a1,b1)=1, and you are guaranteed identical solutions to this reduced Diophantine equation.
Thus, we may as well consider the equation ax+by=c with gcd(a,b)=1. We know this admits a solution in integer pairs; so, in fact, {ax+by:x,y∈Z}=Z.
Suppose (x0,y0) is a solution to ax+by=c. Then if (x,y) is any solution to the same equation, we have ax+by=ax0+by0, or a(x−x0)=−b(y−y0). But then a∣b(y−y0), and so a∣(y−y0) because gcd(a,b)=1. If we write y−y0=at, t∈Z, and substitute in a(x−x0)=−b(y−y0), we get x−x0=−bt.
Conversely, ax+by=a(x0−bt)+b(y0+at)=ax0+by0=c.
Thus, we have
x=x0−bt,y=y0+at,t∈Z. (⋆)
We have shown that every solution can be expressed in terms of any one solution. There remains to discuss how to determine one solution to ax+by=c. Notice that any solution (x1,y1) to ax+by=1 gives rise to the solution (cx1,cy1) to ax+by=c. So we may focus on finding one solution to ax+by=1. (In fact, there is a unique solution with y∈{0,1,2,…,a−1} by (⋆).)
Perhaps the simplest method to find a solution is by repeated use of the Division Algorithm, which is captured by the Continued Fraction expansion of b/a; we may assume b>a. I will give an example to demonstrate this.
Let’s take a=23 and b=37. Then
3723=1+1423=1+12314=1+11+914=1+11+1149=1+11+11+195=1+11+11+11+154=1+11+11+11+11+14.
If we consider the partial sums in order, we get the sequence of rational numbers
11,21,32,53,85,3723.
The last fraction must equal b/a. Note the penultimate fraction satisfies (37⋅5)−(23⋅8)=1. In fact, the numerator in the difference between each successive pair of fractions alternates between +1 and −1. The numerator and denominator of the penultimate rational number in the continued fraction of b/a always gives a solution to the equation ax+by=1.
Robert Paxson
BSME in Mechanical Engineering, Lehigh University (Graduated 1983)
·
Author has 3.9K answers and 4M answer views
·
4y
Related
How do you linearize the equation y = axe^(bx)?
To linearize a function, we need a point about which to linearize.
Let’s assume that we want to linearize a given function y=f(x) about the point where x=c. Then, in general, we have:
y≈f(c)+f′(c)(x−c)
which is the linear truncation of the Taylor series expansion of the given function.
In this question, we have:
y=axebx
y′=aebx(bx+1)
Therefore, to linearize this function at x=c, we get:
y≈f(c)+f′(c)(x−c)
y≈acebc+(aebc(bc+1))(x−c)
y≈aebc(c+(bc+1)(x−c))
For example, if a=6 and b=−12, then the function is:
y=6xe−12x
and its linearization at c=1 i
To linearize a function, we need a point about which to linearize.
Let’s assume that we want to linearize a given function y=f(x) about the point where x=c. Then, in general, we have:
y≈f(c)+f′(c)(x−c)
which is the linear truncation of the Taylor series expansion of the given function.
In this question, we have:
y=axebx
y′=aebx(bx+1)
Therefore, to linearize this function at x=c, we get:
y≈f(c)+f′(c)(x−c)
y≈acebc+(aebc(bc+1))(x−c)
y≈aebc(c+(bc+1)(x−c))
For example, if a=6 and b=−12, then the function is:
y=6xe−12x
and its linearization at c=1 is:
y≈6e(−12)(1)(1+((−12)(1)+1))(x−1))
which simplifies to:
y≈1.819591979x+1.819591979
A plot of this function, in red, and its linearization at x=1, in green, looks like this:
Notice that the error in the linearization increases as we move away from the neighborhood around x=1.
Enrico Gregorio
Associate professor in Algebra
·
Author has 18.4K answers and 16M answer views
·
3y
Related
How do you use the Euclidean algorithm to find d=(a,b) and find a pair of integers x,y such that ax+by=d, a=85, b=41, and a=112,b=18?
Perform the Euclidean algorithm:
85=2⋅41+3
41=13⋅3+2
3=1⋅2+1
2=2⋅1
The last step leaves no remainder, so we know that 1 is the greatest common divisor. Why did I distinguished some numbers with boldface type? Because they’re to be considered as “variables” not to be touched during the reverse process. We start from the last remainder:
[math]\begin{align} \mathbf{1} &= \mathbf{3}-1\cdot\mathbf{2} \ &= \mathbf{3}-1(\mathbf{41}-13\cdot\mathbf{3}) = -1\cdot\mathbf{41}+14\cdot\mathbf{3} \ &= -1[/math]
Perform the Euclidean algorithm:
85=2⋅41+3
41=13⋅3+2
3=1⋅2+1
2=2⋅1
The last step leaves no remainder, so we know that 1 is the greatest common divisor. Why did I distinguished some numbers with boldface type? Because they’re to be considered as “variables” not to be touched during the reverse process. We start from the last remainder:
1=3−1⋅2=3−1(41−13⋅3)=−1⋅41+14⋅3=−1⋅41+14(85−2⋅41)=14⋅85−29⋅41
Anil Bapat
Lives in Mumbai, Maharashtra, India
·
Author has 2.8K answers and 3.8M answer views
·
Jan 17
Related
What are some methods for finding integer solutions to linear diophantine equations besides guess and check?
What are some methods for finding integer solutions to linear diophantine equations besides guess and check?
Besides the guess and check method, there are 2 other methods, as shown below:
Proper Analysis: If the problem could be analyzed properly then it often leads us to logical solutions
Brute Force: There are times when guess work doesn’t yield and even the analysis leads us nowhere, but a Brute Force really helps. Applying Brute Force is no mean task because huge calculations are often required to be performed over and over again. A Computer Program really comes handy, and I have come across
What are some methods for finding integer solutions to linear diophantine equations besides guess and check?
Besides the guess and check method, there are 2 other methods, as shown below:
Proper Analysis: If the problem could be analyzed properly then it often leads us to logical solutions
Brute Force: There are times when guess work doesn’t yield and even the analysis leads us nowhere, but a Brute Force really helps. Applying Brute Force is no mean task because huge calculations are often required to be performed over and over again. A Computer Program really comes handy, and I have come across certain solutions, which are hard to come by with guess Work or Proper Analysis.
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Mathematics Homework Ques...
Functions (mathematics)
Mathematical Sciences
Natural Logarithms
Mathematical Functions
Logarithmic Functions
5
What are the values of natural logarithms of 2 and 3?
Tarek Said
Masters of Science in Electrical Engineerinig, University of New South Wales (Graduated 2006)
·
3y
Why is the logarithm with base e called a natural logarithm?
Short answer
The short answer is that contrary to the popular belief, there is no clear reason as to why the logarithm with the base e is called the natural logarithm. (I know many people will disagree with me but bear with me.)
Brief history of the natural logarithm
The natural logarithm was discovered decades before the number e and the link between the two wasn’t recognized for more than 100 years!! This doesn’t make sense to our modern way of thinking since today we define the natural logarithm as a logarithm with the base e!
The natural logarithm was first discovered in 1647 through the work
Short answer
The short answer is that contrary to the popular belief, there is no clear reason as to why the logarithm with the base e is called the natural logarithm. (I know many people will disagree with me but bear with me.)
Brief history of the natural logarithm
The natural logarithm was discovered decades before the number e and the link between the two wasn’t recognized for more than 100 years!! This doesn’t make sense to our modern way of thinking since today we define the natural logarithm as a logarithm with the base e!
The natural logarithm was first discovered in 1647 through the work of Gregoire de Saint Vincent and his student Alfonso de Sarasa while studying the area under the square hyperbola (the shape we get from the equation y=1/x).
Saint Vincent realised that the relation between the distance and the area under the hyperbola is logarithmic, and de Sarasa wrote this relation explicitly as: area(a)=log(a). This logarithm was suitably called the hyperbolic logarithm however Saint Vincent and de Sarasa didn’t provide a way to compute it.
More than two decades later Nikolaus Mercator and Isaac Newton independently provided an infinite series to calculate the hyperbolic logarithm and Mercator called it the natural logarithm.
John Ellis Evans from Ohio state university summed up the three most common reasons it is called the natural logarithm as follows:
e is a number that arises frequently in nature
Natural logarithms have the simplest derivatives of all the systems of logarithms
In the calculation of logarithms to any base, logarithms to the base e are first calculated, then multiplied by a constant which depends on the system being calculated.
However none of these reasons were known at Mercator’s time and there must be another reason as to why he called it the natural logarithm.
In 1748, 101 years after the work of Saint Vincent and de Sarasa, Leonhard Euler calculated the base of the hyperbolic(natural) logarithm and found it to be what we call today the number e.
Here is a video I created about the history of the natural logarithm that hopefully gives more understanding on what it is:
Related questions
What is the value of 2 in logarithms?
What is the real value of natural logarithm of 3?
What are some examples of natural logarithms?
What is the logarithm of 0.2?
What is 3 times the natural logarithm of 5 equal to, and why?
Roy Østensen
Translator with fondness of numbers (mathematics)
·
Author has 1.1K answers and 1.2M answer views
·
11mo
According to Google, log 2 = 0.6932 (rounded), and log 3 = 1.09861228867 (again rounded off). I would advice you to ask Google such questions rather than put the question on Quora, since you then get the answer straight off.
George Ivey
Former Math Professor at Gallaudet University
·
Author has 23.7K answers and 2.6M answer views
·
11mo
The natural logarithm of 2 is the value of x so that ex=2.
It is approximately 0.69314718055994530941723212145818…
The natural logarithm of 3 is the value of x so that ex=3.
It is approximately 1.0986122886681096913952452369225,,,
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
Upvoted by
Michael Jørgensen
, PhD in mathematics and
Siddhant Grover
, MSc in Statistics & BSc Mathematics, Hindu College, University of Delhi (2022) · Author has 8.5K answers and 21.1M answer views
·
5y
Related
What is the value of
∫
1
0
x
7
−
1
log
x
d
x
, where log is the natural logarithm?
We want to determine the value of
∫10x7−1logxdx.
This is a classic example of an integral that is easier to evaluate by introducing a parameter. We consider the more general integral (with a>−1)
I(a)=∫10xa−1logxdx.
Why? See what happens when we differentiate both sides with respect to a; we can integrate under the integral sign:
\begin{align} I'(a) &= \displaystyle \frac{d}{da} \int_0^1 \frac{x^a-1}{\log{x}} \, dx\ &= \displaystyle \int_0^1 \frac{\partial}{\partial a} \frac{x^a-1}{\log{x}} \, dx\ &= \d
We want to determine the value of
∫10x7−1logxdx.
This is a classic example of an integral that is easier to evaluate by introducing a parameter. We consider the more general integral (with a>−1)
I(a)=∫10xa−1logxdx.
Why? See what happens when we differentiate both sides with respect to a; we can integrate under the integral sign:
I′(a)=dda∫10xa−1logxdx=∫10∂∂axa−1logxdx=∫10xalogx−0logxdx=∫10xadx.
This latter integral is easy to evaluate:
I′(a)=xa+1a+1∣∣∣10=1a+1.
From this, we can now find I(a) via integration:
I(a)=∫1a+1da=log(a+1)+C.
All that remains is to find the value of the constant C. To this end, it is easy to check that I(0)=0. Then, we obtain 0=log1+C and thus C=0.
Therefore, we have found that
I(a)=∫10xa−1logxdx=log(a+1).
Now, the original integral in question is an afterthought. Letting a=7 allows us to conclude that
∫10x7−1logxdx=log8.
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Related questions
What is 2 - 3log2 as a single logarithm?
What is the natural logarithm of 3 in terms of e?
How do you find the natural logarithm of ten to any base?
What is the natural logarithm of e?
What is 3 times the natural logarithm of the constant C? Why?
Daniel McLaury
Ph.D. Student in Mathematics at University of Illinois at Chicago
·
Upvoted by
Anurag Bishnoi
, Ph.D. Mathematics, Ghent University (2016) and
Omkar Girkar
, M.Sc. Mathematics, Indian Institute of Technology, Madras (2020) · Author has 3.3K answers and 15.8M answer views
·
12y
Related
Why do we use natural logarithms?
Obviously exponential functions of some sort are useful. For instance, the function f(x)=2x describes what happens when you have a population of bacteria that doubles every hour, and the function g(x)=2−x/t1/2 describes what happens when you have a substance decaying with a half-life t1/2. And obviously calculus is useful too. If we wanted to know how quickly a population growing or decaying exponentially is growing right now, we'd take the derivative.
Let's see what happens when we try to take the derivative of f(x)=2x:
f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) -
Obviously exponential functions of some sort are useful. For instance, the function f(x)=2x describes what happens when you have a population of bacteria that doubles every hour, and the function g(x)=2−x/t1/2 describes what happens when you have a substance decaying with a half-life t1/2. And obviously calculus is useful too. If we wanted to know how quickly a population growing or decaying exponentially is growing right now, we'd take the derivative.
Let's see what happens when we try to take the derivative of f(x)=2x:
f′(x)=limΔx→0f(x+Δx)−f(x)Δx
=limΔx→02x+Δx−2xΔx
=limΔx→02x(2Δx−1)Δx
=2xlimΔx→02Δx−1Δx
Here I've pulled the 2x out of the limit because it doesn't depend on Δx. Now, look at what's left inside the limit. It doesn't depend on x, so there are two possibilities: either the limit diverges, or it converges to some particular number. In the former case, this would say that f′(x) didn't exist, but that can't be right -- just look at the graph of 2x and tell me that it's not smooth:
(Image courtesy Wolfram|Alpha)
So this limit should converge to some number, which I'll call γ2 for the moment. Then we can rewrite this as
f′(x)=γ22x.
(I put the 2 there because we're looking at 2x) We can try getting an estimate here by plugging in, say, Δx=1/100, in which case we'd get
γ2≈21/100−11/100=0.6955…
So it doesn't look like we got lucky here -- γ2 doesn't look like any nice number we already know about.
Of course, if instead of looking at 2x we looked at h(x)=nx, the calculation would be basically the same. That is, we'd have h′(x)=γnnx, where
γn=limΔx→0nΔx+1Δx
So we'd discovered that any time you take a derivative of an exponential function, that derivative is the exponential function itself, times some mysterious factor γn. Just for the fun of it, let's approximate γ3. Clearly γ3 is going to be bigger than γ2, since changing the 2 to a 3 made all the corresponding terms in the limit bigger. And indeed, we see that
γ3≈31/100−11/100=1.0986…
This is interesting, though. γ2 was smaller than 1, and γ3 was bigger than 1, and we know that γn increases when we increase n. Under some relatively mild assumptions , there should be some number between 2 and 3, let's call it e, for which γe=1, i.e. for which ex is its own derivative.
And of course it turns out that there is such a number. Further, this demystifies the γn, because it's not hard to calculate that eγn=n. In other words, this says that γn is just the base-e logarithm of n.
Of course, there is really only one exponential function up to scaling, because, say, 2x=(3(log32))x=3(log32)x by basic exponent rules. So if we have a good handle on one particular exponential function, say ex, we can translate this to a full understanding of all exponential functions using basic calculus tools like the chain rule.
That is, using base e substantially simplifies our picture of the interplay between calculus and exponentiation, without losing any subtleties in the process.
Namely: that it makes sense to define nx for an arbitrary real number n by interpolating between the cases where n is rational, and that when you do this γn turns out to be a continuous function of n.
Michael Lamar
PhD in Applied Mathematics
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 3.7K answers and 17.5M answer views
·
7y
Related
Why is e the base of the natural logarithm?
The defining property of the natural logarithm (as opposed to some other logarithm) is that for x>0:
ln(x)=∫x1dtt
Furthermore, for a>0, we define the loga(x)=ln(x)ln(a).
So we see that:
loge(x)=ln(x) if and only if ln(e)=1
So your question is equivalent to, “Why is the natural log of e equal to one?”
So we must prove that:
∫e1dtt=1
But it turns out that e is often DEFINED to be the number for which
∫e1dtt=1
So if we choose this definition of e, we discover that there is nothing to prove.
Now
The defining property of the natural logarithm (as opposed to some other logarithm) is that for x>0:
ln(x)=∫x1dtt
Furthermore, for a>0, we define the loga(x)=ln(x)ln(a).
So we see that:
loge(x)=ln(x) if and only if ln(e)=1
So your question is equivalent to, “Why is the natural log of e equal to one?”
So we must prove that:
∫e1dtt=1
But it turns out that e is often DEFINED to be the number for which
∫e1dtt=1
So if we choose this definition of e, we discover that there is nothing to prove.
Now, if we choose some other definition for e like
e=limx→∞(1+1x)x
or
e=argmax(x√x) over any x>0
or
e=∞∑k=01k!
or any one of many other choices, we are left to show that our chosen definition of e implies ∫e1dtt=1. Depending on your chosen definition, that might be relatively simple to prove or it might be very challenging. But unless you tell me which definition you prefer, I don’t know what you would actually need to show to establish the truth of the claim. And since I don’t know which definition you want, I will just choose the one that is the simplest for this problem: e is the real number for which ∫e1dtt=1.
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Čîyaye Zagrose
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2y
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What is a natural logarithm?
The natural logarithm is the area in the image given below.
I think this is the most aesthetic, most wonderful definition of the natural logarithm.
The natural logarithm is represented by the symbols ln or lnx. So,
The answer was this short. However, i would like to extend this topic a little more.
So how do you calculate this natural logarithm?
First we need to accept that the natural logarithm is a function. This is now up to you. The area in light blue is not a regular geometric shape. If you want, you can get an approximation by dividing it into rectangles. It is almost im
The natural logarithm is the area in the image given below.
I think this is the most aesthetic, most wonderful definition of the natural logarithm.
The natural logarithm is represented by the symbols [math]\ln[/math] or [math]\ln x[/math]. So, [math]\boxed{S = \ln x}.[/math]
The answer was this short. However, i would like to extend this topic a little more.
So how do you calculate this natural logarithm?
First we need to accept that the natural logarithm is a function. This is now up to you. The area in light blue is not a regular geometric shape. If you want, you can get an approximation by dividing it into rectangles. It is almost impossible to calculate any value of the logarithm. Because to calculate this area, we have to draw infinitely small rectangles. However, this logarithmic value is a number between the upper value and the lower value. We can obtain approximations up to a certain level.
What is the natural logarithm of two( or [math]\ln3[/math])?
You would be insulting me if you went into the Wolftam Alpha calculator and told me the answer to the Android calculator or most scientific calculators. Read this question carefully. You are not asked for a numerical value in the question. It's asking you what natural logarithm three is.
Answer:
The natural logarithm three( [math]\ln3[/math]) is the area between 1 and 3 on the x axis of the [math]y=\frac{1}{x}[/math] curve.
I wrote the most basic definition of natural logarithm. This is the basic philosophy of the business and this is the essence of natural logarithm. It is just one area. And you won't regret believing this is a revolution. Alternatively, we can give the definition of the natural logarithm as follows: for all [math]x>0[/math],
[math]\boxed{\ln x = \int_{1}^{x} \frac{1}{t} \mathrm{d}t}.[/math]
If you notice, this alternative definition(or ln integral definition) derives from the basic definition. now this alternative definition is the integral version of the natural logarithm. This is very important for proving the properties of natural logarithm.
Can you prove that the natural logarithm is a function on this interval (for all [math]x>0[/math])?
(👆 Another question in connection with this question: Have you ever wondered why calculators give only one answer?)
Prove that [math]\ln(1)=0[/math].
Prove that for the positive real numbers a and b, [math]\ln (ab)=\ln (a)+\ln (b)[/math].
You can easily prove these questions by using the properties of the integral and with the help of your other information.
Actually, i think it would be more accurate to give you the definition of the natural logarithm in integral form as a theorem. But it's too late now. The basic definition of the natural logarithm is very useful for us to do numerical calculations, but the integral is useful to prove to us the properties of this logarithm.
Now let's do a numerical calculation. For this, let's choose the number 3. So let's do a numerical analysis of the natural logarithm of two:
For [math]x=3[/math], the shaded area gives us the natural logarithm of three. Now let's divide the interval [math][1,3][/math] into two equal parts:
As seen in the image, this interval is divided into two equal parts and let me calculate the lower area.
[math]\displaystyle{a+b= (2-1)\frac{1}{2}+(3-2)\frac{1}{3}}[/math]
[math]\displaystyle{=1/2+1/3=5/6}[/math]
. Now let's calculate the area named [math]S[/math] in the image. We already know this value, the answer is [math]\ln 3[/math].
[math]\displaystyle{S=\int_{1}^{3} \frac{1}{x} \mathrm{d}x= \ln 3}.[/math]
Now let's calculate the upper area.
[math]\displaystyle{A+B=(2-1)1+(3-2)\frac{1}{2}}[/math]
[math]\displaystyle{=1+1/2=3/2.}[/math]
Take a good look at the three graphic or images. So the following inequality can be written:
[math]\displaystyle{a+b<\ln 3 < A+B}[/math]
[math]\displaystyle{\Rightarrow \frac{5}{6} < \ln 3 < \frac{3}{2}}.[/math]
To make a better approximation, it is necessary to divide this interval into more parts. As we move towards infinity, the areas of all rectangles approach the desired value. This idea actually gave rise to the Riemann sum. The defined limit of this sum is called the Riemann integral. Anyway, we know that the logarithmic value of three is between [math]5/6[/math] and [math]3/2[/math].
Hence,
[math]\displaystyle{0,83 < \ln 3 < 1,50.}[/math]
The reason for choosing the area calculation of the rectangle here is to make the calculations simpler.
The natural logarithm function is a solution of the equation of [math]e^x=k[/math] (k real is number). However, in order to make this definition, we have to prove the existence of the number of [math]e[/math]. However, i don't think this definition is very useful in doing numerical calculations.
Luke Tang
Math Enthusiast and Coach
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Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979)
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12y
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Why do we use natural logarithms?
I'm going to assume you haven't taken Calculus yet, since most people understand this after they take the course. Logarithms of base "e" are very important in calculus, for two reasons.
First, the closely related exponential function [math]e^x[/math] is its own derivative. A derivative is a function which tells you what the instantaneous rate of change, or slope, of a curve is at any point. That means that the rate of change of [math]y=e^x[/math] (the slope of the line tangent to y ) is equal to the value of y. If you graph [math]y=e^x[/math], the slope of the line tangent to the curve y is equal to the y-value of the curve at that
I'm going to assume you haven't taken Calculus yet, since most people understand this after they take the course. Logarithms of base "e" are very important in calculus, for two reasons.
First, the closely related exponential function [math]e^x[/math] is its own derivative. A derivative is a function which tells you what the instantaneous rate of change, or slope, of a curve is at any point. That means that the rate of change of [math]y=e^x[/math] (the slope of the line tangent to y ) is equal to the value of y. If you graph [math]y=e^x[/math], the slope of the line tangent to the curve y is equal to the y-value of the curve at that point. However, if you find the derivative of another exponential curve, such as [math]y=2^x[/math], the derivative is [math]\frac{dy}{dx}=2^x \ln{2}[/math]. That means that if I want to find the slope of the line tangent to [math]y=2^x[/math] at [math]x=2[/math], the slope would be [math]m=2^2 \ln 2 = 4\ln2[/math]. In general, the derivative of an exponential function [math]y=b^x[/math] is [math]b^x \ln{b}[/math]
There are many things in Nature with a rate of change proportional to its own value, such as population (the more people, the faster the population grows). For these things in Nature, we use exponential functions to model them, since as I said above, the derivative of an exponential function is just itself times [math]\ln{b}[/math] If we want to find out how fast the population is growing at different points in time (again, meaning we want the derivative of the population function), the natural logarithm is bound to pop up in the derivative, just as it did in the case of [math]y=2^x[/math]
Second, the derivative of [math]y=\ln{x}[/math] is [math]\frac{dy}{dx}=1/x[/math].
So if we want to find the slope of the line tangent to the curve [math]y=\ln{x}[/math] at the point [math]x=2[/math], for example, its slope would just be 1/2. On the other hand, the derivative of [math]y=\log_2{x}[/math] is [math]\frac{dy}{dx}=\frac{1}{x\ln{2}}[/math]. If we wanted to find the slope of the line tangent to [math]y=\log_2{x}[/math] at [math]x=2[/math], it would be [math]\frac{dy}{dx}=\frac{1}{2\ln{2}}[/math]. The slope is more complicated, or less "natural" in this case than with [math]y=\ln{x}[/math]. Anyways, the slope of [math]y=\log_2{x}[/math]contains the natural logarithm! So you can't escape it.
For those two reasons (and a couple other misc. ones), the natural logarithm is used a lot in Calculus, so they teach it to people in Pre-Calculus and Algebra so that students are familiar with the notation before they get to Calculus.
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David Joyce
Professor Emeritus of Mathematics at Clark University
·
Upvoted by
Nuno Calaim
, MSc Applied Mathematics and
Gerhard Heinrichs
, Master Mathematics, Ludwig Maximilian University of Munich (1973) · Author has 9.9K answers and 68.4M answer views
·
11y
Related
Why is the logarithm with base e called a natural logarithm?
From the early 1600's there have been two kinds of logarithms—natural logarithms and common logarithms.
Common logarithms are base 10, denoted [math]\log_{10}.[/math] They were used for computing products, quotients, powers, and roots.
Natural logarithms are base [math]e,[/math] denoted [math]\log_e[/math] or [math]\ln.[/math] Although not so useful for computation, they're more useful for calculus since the natural logarithm is the integral of [math]1/x.[/math]
The function inverse to common logarithms is [math]10^x[/math] while the function inverts to natural logarithms is [math]e^x.[/math]
From Ernest William Hobson's John Napier and the invention of logarithms, 1614, page
From the early 1600's there have been two kinds of logarithms—natural logarithms and common logarithms.
Common logarithms are base 10, denoted [math]\log_{10}.[/math] They were used for computing products, quotients, powers, and roots.
Natural logarithms are base [math]e,[/math] denoted [math]\log_e[/math] or [math]\ln.[/math] Although not so useful for computation, they're more useful for calculus since the natural logarithm is the integral of [math]1/x.[/math]
The function inverse to common logarithms is [math]10^x[/math] while the function inverts to natural logarithms is [math]e^x.[/math]
From Ernest William Hobson's John Napier and the invention of logarithms, 1614, page 23.
When Napier first defined logarithms, he compared two quantities. One quantity grew with a rate of change proportional to time, and the other quantity grew with a rate of change proportional to its value. Note that he did this before calculus was invented. The conversion from the first quantity to the second is effected by the exponential function [math]e^x[/math] while the conversion from the second to the first is effected by the logarithmic function [math]\log_e x.[/math] In recognition of the natural description of these functions, they're called "natural".
Doug Hensley
Former retired at Texas A&M University (TAMU) (1977–2022)
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Author has 35.7K answers and 16.6M answer views
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2y
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How can the natural logarithm of two be calculated?
There’s a continued fraction expansion for it. The general expression involves a sequence of expressions [math]a_{2n}=\frac{n}{2(2n+1)},\quad a_{2n+1}=\frac{n+1}{2(2n+1)}[/math] and then
[math]\ln(z)=\cfrac{z}{1+\cfrac{a_1}{z+\cfrac{a_2}{z+\cfrac{a_3}{z+\ddots}}}}[/math] Just set z=1, truncate to taste, and do some arithmetic.
Convergence can be accelerated by observing that [math]2=(3/2)(4/3), \ln 2=\ln(3/2)+\ln(4/3)[/math] so that two calculations, one for z=1/2 and one for z=1/3, are needed, but each resulting piece of the calculation converges faster.
Similarly, series methods using the same trick will converge rapidly. Start with
There’s a continued fraction expansion for it. The general expression involves a sequence of expressions [math]a_{2n}=\frac{n}{2(2n+1)},\quad a_{2n+1}=\frac{n+1}{2(2n+1)}[/math] and then
[math]\ln(z)=\cfrac{z}{1+\cfrac{a_1}{z+\cfrac{a_2}{z+\cfrac{a_3}{z+\ddots}}}}[/math] Just set z=1, truncate to taste, and do some arithmetic.
Convergence can be accelerated by observing that [math]2=(3/2)(4/3), \ln 2=\ln(3/2)+\ln(4/3)[/math] so that two calculations, one for z=1/2 and one for z=1/3, are needed, but each resulting piece of the calculation converges faster.
Similarly, series methods using the same trick will converge rapidly. Start with the series expansion [math]\log(1+z)=z-z^2/2+z^3/3-z^4/4\cdots[/math] In this case it is essential to use the 3/2 and 4/3 trick or else convergence will be glacially slow. Naturally, the 3/2 and 4/3 trick can be modified to use a series of fractions that multiply to 2, each being just a little larger than 1. Thus, 3/2=(6/5)(5/4). You get the idea.
For standard functions, and the natural logarithm is about as standard as they come, there are routines based on ideas such as these but honed to perfection by generations of mathematicians and programmers. Ask Mathematica, running on a reasonably capable machine, for the numerical value of ln 2 to a thousand digits, and it’ll pop right up. Your screen probably doesn’t have room for a lot more, and you probably don’t need anywhere near that many.
It is a nice theorem that this natural log, and likewise for all positive integers after 1, is a transcendental number, meaning that it’s not (not exactly) the root of any polynomial with integer coefficients.
It’s a much simpler theorem that ln 2 is not rational. Let’s assume we already know that e is transcendental. In particular, e is not the nth root of an integer. Now, if ln 2 were rational, we would have [math]\ln 2=a/b\Leftrightarrow 2=e^{a/b}\Leftrightarrow 2^b=e^a[/math]. But as noted already, e not being the a-th root of an integer, [math]e^2\ne 2^b[/math].
So there won’t be any closed form expression with square roots and stuff that gives an exact value for the natural log of 2. We have to be content with knowing all sorts of fun facts about it, and in particular, knowing tricks for getting a truly ludicrous number of digits without too much trouble.
One last observation: in chemistry, logs come up. If you must do without calculators and you’re not handy with a slide rule, it doesn’t hurt to know that ln 2 is roughly 0.6931. Or if you want more digits (you don’t!) it goes on 0.693147180559945. Dang, I just dated myself big time.
Perspicacious cook of delicious existential thoughts
·
7y
Related
What is the value of log2?
As others have pointed out it depends on the value of the base,the output is the power the base should be raised to to get the number two or any other number for that matter.
But you could get this by simply googling the question,and is not the reason I am answering this.
An evergreen expert answer though;
Visualise math,always works.
As others have pointed out it depends on the value of the base,the output is the power the base should be raised to to get the number two or any other number for that matter.
But you could get this by simply googling the question,and is not the reason I am answering this.
An evergreen expert answer though;
Visualise math,always works.
Alexander Ng
Lives in California
·
7y
Related
What's the value of y if 3y=log2(3)?
The definition of a logarithm is “the number that, when base b is raised to it, will yield x”. Its relationship to exponents is akin to multiplication’s relationship to division, and addition to subtraction. That is, the expression [math]y = \log_b x[/math] is equivalent to the expression [math]b^y = x[/math] .
In this case, you are dealing with
[math]3y = \log_2 3[/math]
which can then be rearranged into
[math]2^{3y} = 3[/math]
which can then be simplified, because of exponent properties, to
[math]8^{y} = 3[/math]
As there’s no integer power that you can raise 8 to that gives you 3, y must be a decimal. If you solve for y in the original expression and plug that
The definition of a logarithm is “the number that, when base b is raised to it, will yield x”. Its relationship to exponents is akin to multiplication’s relationship to division, and addition to subtraction. That is, the expression [math]y = \log_b x[/math] is equivalent to the expression [math]b^y = x[/math] .
In this case, you are dealing with
[math]3y = \log_2 3[/math]
which can then be rearranged into
[math]2^{3y} = 3[/math]
which can then be simplified, because of exponent properties, to
[math]8^{y} = 3[/math]
As there’s no integer power that you can raise 8 to that gives you 3, y must be a decimal. If you solve for y in the original expression and plug that into a calculator, you’ll get the (correct) solution of 0.528, approximately.
There are also ways you can simplify this algebraically using log properties, giving you log(3)/log(8), which also approximates to 0.528.
Edit: Mathematical Notation with MathJax
Gopal Menon
B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010)
·
Author has 10.2K answers and 15.2M answer views
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7y
Related
What is 3 times the natural logarithm of 5 equal to, and why?
[math]3[/math]
times the natural logarithm of
[math]5[/math]
is equal to the natural logarithm of
[math]125[/math]
because
[math]a \log x = \log x^a.[/math]
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Home/ Calculus I/ Applications of Derivatives / L'Hospital's Rule and Indeterminate Forms
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Section 4.10 : L'Hospital's Rule and Indeterminate Forms
Back in the chapter on Limits we saw methods for dealing with the following limits.
lim x→4 x 2−16 x−4 lim x→∞4 x 2−5 x 1−3 x 2 lim x→4x 2−16 x−4 lim x→∞4 x 2−5 x 1−3 x 2
In the first limit if we plugged in x=4 x=4 we would get 0/0 and in the second limit if we “plugged” in infinity we would get ∞/−∞∞/−∞ (recall that as x x goes to infinity a polynomial will behave in the same fashion that its largest power behaves). Both of these are called indeterminate forms. In both of these cases there are competing interests or rules and it’s not clear which will win out.
In the case of 0/0 we typically think of a fraction that has a numerator of zero as being zero. However, we also tend to think of fractions in which the denominator is going to zero, in the limit, as infinity or might not exist at all. Likewise, we tend to think of a fraction in which the numerator and denominator are the same as one. So, which will win out? Or will neither win out and they all “cancel out” and the limit will reach some other value?
In the case of ∞/−∞∞/−∞ we have a similar set of problems. If the numerator of a fraction is going to infinity we tend to think of the whole fraction going to infinity. Also, if the denominator is going to infinity, in the limit, we tend to think of the fraction as going to zero. We also have the case of a fraction in which the numerator and denominator are the same (ignoring the minus sign) and so we might get -1. Again, it’s not clear which of these will win out, if any of them will win out.
With the second limit there is the further problem that infinity isn’t really a number and so we really shouldn’t even treat it like a number. Much of the time it simply won’t behave as we would expect it to if it was a number. To look a little more into this, check out the Types of Infinity section in the Extras chapter at the end of this document.
This is the problem with indeterminate forms. It’s just not clear what is happening in the limit. There are other types of indeterminate forms as well. Some other types are,
(0)(±∞)1∞0 0∞0∞−∞(0)(±∞)1∞0 0∞0∞−∞
These all have competing interests or rules that tell us what should happen and it’s just not clear which, if any, of the interests or rules will win out. The topic of this section is how to deal with these kinds of limits.
As already pointed out we do know how to deal with some kinds of indeterminate forms already. For the two limits above we work them as follows.
lim x→4 x 2−16 x−4=lim x→4(x+4)=8 lim x→4x 2−16 x−4=lim x→4(x+4)=8 lim x→∞4 x 2−5 x 1−3 x 2=lim x→∞4−5 x 1 x 2−3=−4 3 lim x→∞4 x 2−5 x 1−3 x 2=lim x→∞4−5 x 1 x 2−3=−4 3
In the first case we simply factored, canceled and took the limit and in the second case we factored out an x 2 x 2 from both the numerator and the denominator and took the limit. Notice as well that none of the competing interests or rules in these cases won out! That is often the case.
So, we can deal with some of these. However, what about the following two limits.
lim x→0 sin x x lim x→∞e x x 2 lim x→0sinx x lim x→∞e x x 2
This first is a 0/0 indeterminate form, but we can’t factor this one. The second is an ∞/∞∞/∞ indeterminate form, but we can’t just factor an x 2 x 2 out of the numerator. So, nothing that we’ve got in our bag of tricks will work with these two limits.
This is where the subject of this section comes into play.
L’Hospital’s Rule
Suppose that we have one of the following cases,
lim x→a f(x)g(x)=0 0 OR lim x→a f(x)g(x)=±∞±∞lim x→af(x)g(x)=0 0 OR lim x→af(x)g(x)=±∞±∞
where a a can be any real number, infinity or negative infinity. In these cases we have,
lim x→a f(x)g(x)=lim x→a f′(x)g′(x)lim x→af(x)g(x)=lim x→af′(x)g′(x)
So, L’Hospital’s Rule tells us that if we have an indeterminate form 0/0 or ∞/∞∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
Before proceeding with examples let me address the spelling of “L’Hospital”. The more modern spelling is “L’Hôpital”. However, when I first learned Calculus my teacher used the spelling that I use in these notes and the first text book that I taught Calculus out of also used the spelling that I use here.
Also, as noted on the Wikipedia page for L’Hospital's Rule,
“In the 17th and 18th centuries, the name was commonly spelled "l'Hospital", and he himself spelled his name that way. However, French spellings have been altered: the silent 's' has been removed and replaced with the circumflex over the preceding vowel. The former spelling is still used in English where there is no circumflex.”
So, the spelling that I’ve used here is an acceptable spelling of his name, albeit not the modern spelling, and because I’m used to spelling it as “L’Hospital” that is the spelling that I’m going to use in these notes.
Let’s work some examples.
Example 1 Example 1 Evaluate each of the following limits.
1. lim x→0 sin x x lim x→0sinx x
2. lim t→1 5 t 4−4 t 2−1 10−t−9 t 3 lim t→15 t 4−4 t 2−1 10−t−9 t 3
3. lim x→∞e x x 2 lim x→∞e x x 2
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a lim x→0 sin x x lim x→0sinx x Show Solution
So, we have already established that this is a 0/0 indeterminate form so let’s just apply L’Hospital’s Rule.
lim x→0 sin x x=lim x→0 cos x 1=1 1=1 lim x→0sinx x=lim x→0cosx 1=1 1=1
b lim t→1 5 t 4−4 t 2−1 10−t−9 t 3 lim t→15 t 4−4 t 2−1 10−t−9 t 3 Show Solution
In this case we also have a 0/0 indeterminate form and if we were really good at factoring we could factor the numerator and denominator, simplify and take the limit. However, that’s going to be more work than just using L’Hospital’s Rule.
lim t→1 5 t 4−4 t 2−1 10−t−9 t 3=lim t→1 20 t 3−8 t−1−27 t 2=20−8−1−27=−3 7 lim t→15 t 4−4 t 2−1 10−t−9 t 3=lim t→120 t 3−8 t−1−27 t 2=20−8−1−27=−3 7
c lim x→∞e x x 2 lim x→∞e x x 2 Show Solution
This was the other limit that we started off looking at and we know that it’s the indeterminate form ∞/∞∞/∞ so let’s apply L’Hospital’s Rule.
lim x→∞e x x 2=lim x→∞e x 2 x lim x→∞e x x 2=lim x→∞e x 2 x
Now we have a small problem. This new limit is also a ∞/∞∞/∞ indeterminate form. However, it’s not really a problem. We know how to deal with these kinds of limits. Just apply L’Hospital’s Rule.
lim x→∞e x x 2=lim x→∞e x 2 x=lim x→∞e x 2=∞lim x→∞e x x 2=lim x→∞e x 2 x=lim x→∞e x 2=∞
Sometimes we will need to apply L’Hospital’s Rule more than once.
L’Hospital’s Rule works great on the two indeterminate forms 0/0 and ±∞/±∞±∞/±∞. However, there are many more indeterminate forms out there as we saw earlier. Let’s take a look at some of those and see how we deal with those kinds of indeterminate forms.
We’ll start with the indeterminate form (0)(±∞)(0)(±∞).
Example 2 Evaluate the following limit. lim x→0+x ln x lim x→0+x lnx
Show Solution
Note that we really do need to do the right-hand limit here. We know that the natural logarithm is only defined for positive x x and so this is the only limit that makes any sense.
Now, in the limit, we get the indeterminate form (0)(−∞)(0)(−∞). L’Hospital’s Rule won’t work on products, it only works on quotients. However, we can turn this into a fraction if we rewrite things a little.
lim x→0+x ln x=lim x→0+ln x 1/x lim x→0+x lnx=lim x→0+lnx 1/x
The function is the same, just rewritten, and the limit is now in the form −∞/∞−∞/∞ and we can now use L’Hospital’s Rule.
lim x→0+x ln x=lim x→0+ln x 1/x=lim x→0+1/x−1/x 2 lim x→0+x lnx=lim x→0+lnx 1/x=lim x→0+1/x−1/x 2
Now, this is a mess, but it cleans up nicely.
lim x→0+x ln x=lim x→0+1/x−1/x 2=lim x→0+(−x)=0 lim x→0+x lnx=lim x→0+1/x−1/x 2=lim x→0+(−x)=0
In the previous example we used the fact that we can always write a product of functions as a quotient by doing one of the following.
f(x)g(x)=g(x)1/f(x)OR f(x)g(x)=f(x)1/g(x)f(x)g(x)=g(x)1/f(x)OR f(x)g(x)=f(x)1/g(x)
Using these two facts will allow us to turn any limit in the form (0)(±∞)(0)(±∞) into a limit in the form 0/0 or ±∞/±∞±∞/±∞. Which one of these two we get after doing the rewrite will depend upon which fact we used to do the rewrite. One of the rewrites will give 0/0 and the other will give ±∞/±∞±∞/±∞. It all depends on which function stays in the numerator and which gets moved down to the denominator.
Let’s take a look at another example.
Example 3 Evaluate the following limit. lim x→−∞x e x lim x→−∞x e x
Show Solution
So, it’s in the form (∞)(0)(∞)(0). This means that we’ll need to write it as a quotient. Moving the x x to the denominator worked in the previous example so let’s try that with this problem as well.
lim x→−∞x e x=lim x→−∞e x 1/x lim x→−∞x e x=lim x→−∞e x 1/x
Writing the product in this way gives us a product that has the form 0/0 in the limit. So, let’s use L’Hospital’s Rule on the quotient.
lim x→−∞x e x=lim x→−∞e x 1/x=lim x→−∞e x−1/x 2=lim x→−∞e x 2/x 3=lim x→−∞e x−6/x 4=⋯lim x→−∞x e x=lim x→−∞e x 1/x=lim x→−∞e x−1/x 2=lim x→−∞e x 2/x 3=lim x→−∞e x−6/x 4=⋯
Hummmm…. This doesn’t seem to be getting us anywhere. With each application of L’Hospital’s Rule we just end up with another 0/0 indeterminate form and in fact the derivatives seem to be getting worse and worse. Also note that if we simplified the quotient back into a product we would just end up with either (∞)(0)(∞)(0) or (−∞)(0)(−∞)(0) and so that won’t do us any good.
This does not mean however that the limit can’t be done. It just means that we moved the wrong function to the denominator. Let’s move the exponential function instead.
lim x→−∞x e x=lim x→−∞x 1/e x=lim x→−∞x e−x lim x→−∞x e x=lim x→−∞x 1/e x=lim x→−∞x e−x
Note that we used the fact that,
1 e x=e−x 1 e x=e−x
to simplify the quotient up a little. This will help us when it comes time to take some derivatives. The quotient is now an indeterminate form of −∞/∞−∞/∞ and using L’Hospital’s Rule gives,
lim x→−∞x e x=lim x→−∞x e−x=lim x→−∞1−e−x=0 lim x→−∞x e x=lim x→−∞x e−x=lim x→−∞1−e−x=0
So, when faced with a product (0)(±∞)(0)(±∞) we can turn it into a quotient that will allow us to use L’Hospital’s Rule. However, as we saw in the last example we need to be careful with how we do that on occasion. Sometimes we can use either quotient and in other cases only one will work.
Let’s now take a look at the indeterminate forms,
1∞0 0∞0 1∞0 0∞0
These can all be dealt with in the following way so we’ll just work one example.
Example 4 Evaluate the following limit. lim x→∞x 1 x lim x→∞x 1 x
Show Solution
In the limit this is the indeterminate form ∞0∞0. We’re actually going to spend most of this problem on a different limit. Let’s first define the following.
y=x 1 x y=x 1 x
Now, if we take the natural log of both sides we get,
ln(y)=ln(x 1 x)=1 x ln x=ln x x ln(y)=ln(x 1 x)=1 x lnx=lnx x
Let’s now take a look at the following limit.
lim x→∞ln(y)=lim x→∞ln x x=lim x→∞1/x 1=0 lim x→∞ln(y)=lim x→∞lnx x=lim x→∞1/x 1=0
This limit was just a L’Hospital’s Rule problem and we know how to do those. So, what did this have to do with our limit? Well first notice that,
e ln(y)=y e ln(y)=y
and so our limit could be written as,
lim x→∞x 1 x=lim x→∞y=lim x→∞e ln(y)lim x→∞x 1 x=lim x→∞y=lim x→∞e ln(y)
We can now use the limit above to finish this problem.
lim x→∞x 1 x=lim x→∞y=lim x→∞e ln(y)=e lim x→∞ln(y)=e 0=1 lim x→∞x 1 x=lim x→∞y=lim x→∞e ln(y)=e lim x→∞ln(y)=e 0=1
With L’Hospital’s Rule we are now able to take the limit of a wide variety of indeterminate forms that we were unable to deal with prior to this section.
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Geometric Problems by Homothety
Below is a list of problems either defined in terms of homothety or for which there is a solution exploiting the homothety transformation.
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Many more examples could be obtained by searching this site.
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12630 | https://www.sciencedirect.com/science/article/abs/pii/S1059056023003118 | Crowding out or crowding in? Reevaluating the effect of government spending on private economic activities - ScienceDirect
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International Review of Economics & Finance
Volume 89, Part A, January 2024, Pages 102-117
Crowding out or crowding in? Reevaluating the effect of government spending on private economic activities
Author links open overlay panel Joshua K.Park a, Xiangcai Meng b
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Abstract
The effect of government spending on private economic activities is an important topic that has been studied mainly in the time dimension, with relatively little research covering how government spending affects the private sector in the time-frequency space. With a continuous wavelet approach, this investigation contributes to the extant literature by characterizing the impact of government spending on private economic activities across frequencies and over time using a quarterly dataset from the Republic of Korea. The empirical results demonstrate that: First, private consumption changes could be largely captured by government spending fluctuations at high frequencies before 1970 but at only low frequencies between 1990 and 2010. Second, government spending crowds in private investment at the scale of 2–8 years between 1970 and 1985, but it crowds out private investment at the 8-16-year scale during the same period. Third, government spending increase leads to employment decrease at the 4-16-year scale between 1980 and 1990, but expansionary public spending results in employment increase at 2-4-year scale between 2000 and 2010. Our findings imply that policymakers should consider the heterogeneous effect of government spending on private economic activities when setting policies designed to rejuvenate the economy.
Introduction
As an important instrument of fiscal policy, government spending plays a crucial role in the macroeconomy. Evaluating the effect of government spending on the aggregate economy and examining how its impact evolves over time and across economies are of particular significance to the development and implementation of fiscal policy. While there is a plethora of theoretical and empirical studies that have examined the relationship between government spending and private sector behaviors (Afonso & Sousa, 2012; Argimon et al., 1997; Baxter & King, 1993; Blanchard & Johnson, 2013; Christiano & Eichenbaum, 1992; Cuestas et al., 2020; Furceri & Sousa, 2011; Mankiw, 2019; Mountford & Uhlig, 2009; Perotti, 2004; Wang, 2021), there is still no consensus even on whether government spending stimulates or constrains private economic activities (Afonso & Sousa, 2011; Şen and Kaya, 2014; Bahal et al., 2018; Funashima & Ohtsuka, 2019; Sie et al., 2021).
In estimating the effect of government spending on private economic activities, majority of the extant research uses conventional econometric methodologies including the structural vector autoregressive (SVAR) and error correction (SVECM) model (Perotti, 2004; Afonso & Sousa, 2011, 2012; Atabaev et al., 2018; Bahal et al., 2018; Cuestas et al., 2020; Mountford & Uhlig, 2009), the panel data approach (Argimon et al., 1997; Nieh & Ho, 2006; Furceri & Sousa, 2011; Funashima & Ohtsuka, 2019), and the dynamic stochastic general equilibrium (DSGE) model (Monacelli et al., 2010; Brückner & Pappa, 2012; Bermperoglou et al., 2017; Kuo & Miyamoto, 2019; Wang, 2021). Despite possessing recognized merits, these models can not characterize the frequency dimension of the effect of government spending on private sector behaviors, which is of high interest for researchers and policy makers (Crowley & Hudgins, 2015; Funashima, 2018).
Moreover, as heterogeneous economic participants usually have different targets over time and could behave simultaneously, while governments often have distinct targets that typically function at multiple timescales (Anguiar-Conraria et al., 2008; Crowley & Hudgins, 2017), the time series of government spending and private economic indicators are composed of components working at various frequencies (Aguiar-Conraria & Soares, 2011; Ramsey & Lampart, 1998). Hence, it is worthwhile to use the frequency dimension approach to examine how the effect of government spending on private economic activities varies in the time-frequency space, which would offer important new information to facilitate the development and evaluation of fiscal policy.
In the past few decades, the government spending and private economic activities experienced significant fluctuations in the Republic of Korea. An impressive feature is that the share of government spending in gross domestic product first decreased from 44% in Quarter 1, 1960 to 13% in Quarter 3, 2000, and then increased to 17% in Quarter 1, 2021, while the growth rate of government spending varied dramatically between −8.82% and 8.01%. At the same time, the growth rate of private consumption moved between −8.83% and 9.63%. Moreover, the growth rates of private investment and employment displayed much larger fluctuations, private investment growth varied largely between −22.14% and 28.68%, while employment growth varied from −24.84% to 27.56% with a standard deviation of 9.60%. These dramatic aggregate fluctuations in the Republic of Korea present a good opportunity to study the effect of government spending on private economic indicators, but not much research has appropriately addressed this issue in the Korea context.
This paper investigates the characteristics of the impact of government spending on private economic activities, i.e., private consumption, private investment, and employment, across frequencies and over time in Korea with a continuous wavelet partial coherency method. In particular, the novel partial approach will enable us to evaluate the effect of government spending on one specific private economic activity after controlling for the other variables. Moreover, the partial phase-differences could help us identify the lead-lag relationships in the time-frequency space, which would provide a more elaborate characterization about how government spending affects the private sector.
Our paper is closely related to the stream of literature that tries to examine the characteristics and policy implications of how government spending affects private economic activities (Afonso & Sousa, 2011, 2012; Andrade & Duarte, 2014; Atabaev et al., 2018; Bahal et al., 2018; Bermperoglou et al., 2017; Funashima & Ohtsuka, 2019; Kuo & Miyamoto, 2019; Perotti, 2004; Wang, 2021), however, this investigation extends the existing literature in several directions that constitute the contributions of this paper.
First, most of the extant literature evaluates the effect of government spending on private economic activities mainly in the time dimension while almost forgets the frequency domain (Ngeendepi and Phiri, 2021; Shankar & Trivedi, 2021; Sie et al., 2021; Cuestas et al., 2020; Omitogun, 2018; Ozerkek and Celik, 2010), however, it has been demonstrated that government spending and private economic indicators consist of components working at different timescales (Ramsey & Lampart, 1998; Anguiar-Conraria et al., 2008), ignoring the frequency dimension will result in inaccurate characterizations about the effect of government spending on private economic activities. To overcome this limitation, the wavelet coherency and partial coherency analysis are employed to estimate the impact of government spending on the private sector across frequencies and over time.
Second, in evaluating the crowding-out or crowding-in effect, most of the extant studies concentrate on examining the effect of government spending on private investment (e.g., Afonso & Sousa, 2011, 2012; Cuestas et al., 2020; Sen and Kaya, 2014; Shankar & Trivedi, 2021), and overlook to investigate the impact of public spending on other important private economic activities such as private consumption and employment. To address this issue, we investigate the effect of government spending on not only private investment but also private consumption and employment with a wavelet approach.
Third, very few studies employ discrete wavelet analysis and simulation method to study the optimal fiscal policy in the time-frequency space for the US and Euro area (Crowley & Hudgins, 2015, 2017, 2018). However, none of these researches has ever investigated the interaction between public spending and private economic activities across frequencies and over time. To deal with this limitation, we examine how government spending affects the private sector across frequencies and over time with a continuous wavelet partial coherency approach.
Last but not the least, many studies evaluate the crowding-out or crowding-in effect of government spending in different economies, for example, Funashima and Ohtsuka (2019) is about Japan, Cuestas et al. (2020) is on European economies, Bahal et al. (2018) is on India, Afonso and Sousa (2011) is about US, UK, Germany, and Italy, but there are relatively few studies about the Republic of Korea. To fill this gap, we employ a wavelet approach to analyze the time-frequency effect of government spending on private economic activities in Korea with a recently updated dataset from Quarter 01, 1960 to Quarter 01, 2021.
The remainder of this research is structured as follows. The literature review is presented in Section 2. The wavelet approach is discussed in Section 3. The data analysis is reported in Section 4. The effect of government spending on private consumption, private investment, and employment after controlling for the other variables across frequencies and over time is analyzed in Section 5. Section 6 concludes.
Section snippets
Literature review
The interaction between government spending and private economic activities is an important issue that has been intensively examined in the literature (Kydland & Prescott, 1982; Baxter & King, 1993; Christiano & Eichenbaum, 1992; Perotti, 2004; Mountford & Uhlig, 2009; Blanchard & Johnson, 2013; Mankiw, 2019; Furceri & Sousa, 2011, 2012, Atabaev et al., 2018; Kuo & Miyamoto, 2019; Cuestas et al., 2020; Wang, 2021). In this section, we briefly review the main existing theoretical and empirical
Empirical methodology
Fourier analysis can reveal the cyclical properties of time series in the frequency domain, but it cannot precisely study instantaneous interactions and identify structural changes. However, wavelet analysis could address these issues and investigate strongly non-stationary time series (Aguiar-Conraria et al., 2008; Aguiar-Conraria & Soares, 2011; Fan & Gençay, 2010; Gencay et al., 2002; Ramsey, 2002a, 2002b).
A wavelet is a real-valued square integrable function φ τ,s(t) derived from scaling a
Data analysis
We start from conducting basic data analysis to examine the statistical properties of the time series under consideration. Then we move to study the long-run relationship between government spending and fundamental private economic indicators through Johansen cointegration test. In addition, we analyze the characteristics of government spending volatilities using GARCH model.
The time-frequency effect of government spending on private activities
To elaborately examine the characteristics of the effect of government spending on the private economic
Conclusion
This paper evaluates the crowding-out and crowding-in effect of government spending on private economic activities in the time-frequency space through employing the continuous wavelet coherency approach with a quarterly dataset of the Republic of Korea from Quarter 01, 1960 to Quarter 01, 2021.
Through coherency analysis, we observe that government spending affects private consumption heterogeneously in the time-frequency space. One of the most impressive findings is that government spending
Author statement
We, Joshua K. Park and Xiangcai Meng, claim that we made equal contributions to the final version of this project.
Joshua K. Park: Conceptualization; Data curation; Formal analysis; Investigation; Validation; Visualization; Project administration; Supervision; Roles/Writing - original draft; Writing - review & editing.
Xiangcai Meng: Conceptualization; Data curation; Formal analysis; Investigation; Methodology; Software; Validation; Visualization; Roles/Writing - original draft; Writing - review
Acknowledgements
We would like to sincerely thank the two anonymous referees and the editor for their very constructive comments and extremely helpful suggestions.
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Analysis of the evolution path of new energy system under polymorphic uncertainty—A case study of China
2024, Energy Citation Excerpt :
On the one hand, due to significant initial investments and extended investment recovery periods, new energy in its nascent stages may impede economic growth. This hindrance can be divided into two aspects: Firstly, the implementation of fiscal subsidies, tax reductions, and other incentive measures that encroach upon government expenditures [22,23]; Secondly, the input on the production side is transferred to consumers via price mechanisms, thereby suppressing individual consumption . On the other hand, the development of new energy may stimulate the growth of related industrial sectors such as the energy internet and new types of energy storage , thereby benefiting employment and economic growth. Show abstract The exploration and promotion of new energy constitute a significant initiative for numerous countries in their pursuit of sustainable development and efforts to alleviate climate change. However, the influence of a multitude of uncertain factors on the evolutionary path of new energy is currently poorly studied. By establishing a dynamic evolution model for new energy under polymorphic uncertainty, this paper analyses the complex interaction between new energy, economic growth, and carbon trading over the period 2020–2060, using China as a case study. We employ numerical simulation and sensitivity analysis to investigate the evolutionary trajectories of the new energy system under various parameter value scenarios. We show that the evolutionary trajectories will undergo substantial fluctuations before 2030, subsequently stabilizing from 2030 onwards. The study highlights that an excessively rapid pace of new energy development is detrimental to new energy systems, whereas energy storage system deployment has the opposite effect. There exists a significant risk of the system collapsing as a result of slow new energy development, over-building energy storage facilities, or over-intervening in carbon trading. We also find that overdevelopment of energy storage system may have greater negative impacts on the economic system than underdevelopment.
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### Complementarity or Crowding Out: The Effects of Government-Led Philanthropic Development
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Calculating pH and pOH
This is part of the HSC Chemistry course under the topic Using Brønsted-Lowry Theory.
HSC Chemistry Syllabus
Calculate pH, pOH, hydrogen ion concentration([H+]) and hydroxide ion concentration ([OH–]) for a range of solutions (ACSCH102)
Calculations Involving pH and pOH in HSC Chemistry
pH and Acidity
Acids by various definitions are hydrogen ion (H+) producers (Arrhenius Theory) or proton donors (Brønsted-Lowry Theory)
Not all hydrogen containing molecules are acids. Molecules are only considered to be acids if their hydrogen ions can be ionised.
The acidity of a solution is measured by pH which stands for the potential of hydrogen. It is calculated using the formula
p H=−l o g 10[H+]p H=−l o g 10[H+]
Or the hydrogen ion concentration can be calculated using the formula
[H+]=10−p H[H+]=10−p H
pH is measured on a logarithmic scale which means that a change in value of 1 corresponds to a ten-fold change
pH = 1 corresponds to [H+] of 0.1 mol L–1
pH = 2 corresponds to [H+] of 0.01 mol L–1
Notice that a smaller pH corresponds to an increasing concentration of hydrogen ions.This means a smaller pH will mean a more acidic solution
pOH and Basicity
The basicity of a solution can be measured with pOH. Similar to pH, a change in value of 1 corresponds to a tenfold change in [OH–].
p O H=−l o g 10[O H−]p O H=−l o g 10[O H−]
Or the hydroxide concentration can be calculated using the formula
[O H−]=10−p O H[O H−]=10−p O H
Significant Figures in Calculations of pH and pOH
It is important to recognise that significant figures for hydrogen ion and hydroxide ion concentration, are expressed as decimal places when calculating pH and pOH
Example 1
Calculate the pH for a solution with hydrogen ion concentration of 0.0010 mol L–1
Solution:
Looking at our example the hydrogen ion concentration 0.0010 is given as 2 significant figures. This means that we will have to express our pH to 2 decimal places
To calculate pH, use the formula
p H=−l o g 10[H+]p H=−l o g 10[H+]
p H=−l o g 10[0.0010]=3 p H=−l o g 10[0.0010]=3
However remember that we must express our answer to 2 decimal places
∴p H=3.00∴p H=3.00
Example 2
Calculate the concentration of hydrogen ions in a solution of pH 4.35.
Solution:
This time we are going from pH to a hydrogen ion concentration. The pH has 2 decimal places so we need to give our answer in 2 significant figures.
To calculate the hydrogen ion concentration, we use the formula
[H+]=10−p H[H+]=10−p H
[H+]=10−4.35=4.466735922×10−5 m o l L−1[H+]=10−4.35=4.466735922×10−5 m o l L−1
However remember that we must express our answer to 2 significant figures
∴[H+]=4.5×10−5 m o l L−1∴[H+]=4.5×10−5 m o l L−1
Relationship between pH and pOH
To understand the quantitative relationship between pH and pOH, we must introduce the self-ionisation of water.
Self-ionisation of water is represented by the reaction:
2 H 2 O(l)⇌H 3 O+(a q)+O H−2 H 2 O(l)⇌H 3 O+(a q)+O H−
The equilibrium constant of this reaction K w=1.0×10−14 K w=1.0×10-14 at 25 ºC. This is extremely small which indicates that the equilibrium lies on the far-left side of the reaction, resulting in very small amounts of H 3 O+ and OH– equilibrium.
K w=[H 3 O+(a q)][O H−(a q)]=1.0×10−14 K w=[H 3 O+(a q)][O H−(a q)]=1.0×10−14
The relationship between pH and pOH can be derived by applying logarithmic function to both sides of the above equation:
log([H 3 O+][O H−])=log(10−14)log([H 3 O+][O H−])=log(10−14)
log[H 3 O+]+log[O H−]=−14 log[H 3 O+]+log[O H−]=−14
−log[H 3 O+]−log[O H−]=14−log[H 3 O+]−log[O H−]=14
p H+p O H=14 p H+p O H=14
Note that pH + pOH = 14 is only true at 25ºC because the value of K w K w changes with temperature.
BACK TO MODULE 6: ACID/BASE REACTIONS
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← Dividing Integers
Excel 2007 Charts and Graphs →
Integers Order of Operations
Posted onFebruary 25, 2012byPassy
In countries like Australia, we use the “BODMAS” or “BOMDAS” Rule for doing maths operations in the correct order.
In the USA, and some other countries, use the “PEMDAS” Rule for doing their maths operations.
“Operations” are any of the following: Brackets, Squares, Powers, Square Roots, Division, Multiplication, Addition, and Subtraction.
Operations Introduction
Consider the following question:
1 + 2 x 5
There are two possible ways of doing this question:
the right way, and the wrong way!
If you are having trouble thinking about which one is the correct answer, then think about a money sum.
If someone gave you one dollar plus two five dollar notes, then the total would be $11.
It would not be $15.
Now consider this next question:
2 + 3 x -4
Again there are two possible ways of doing this question:
the right way, and the wrong way.
We must follow the “Order of Operations” carefully in Mathematics to arrive at correct answers.
BODMAS Video
For countries like Australia that use BODMAS, the following five minute video covers how everything works.
[youtube
PEMDAS Video
For countries like the USA which use PEMDAS, the following ten minute Video explains all of the key concepts.
[youtube
Integers Examples
Example 1 : Multiplication and Addition
Example 2: Multiplication and Division
Example 3: Brackets Multiplication and Division
Example 4: Brackets Division and Multiplication
Example 5: Brackets Addition and Subtraction
BODMAS / Pemdas Summary
Once you have done a few practice questions, the following should become a lot more familiar.
BODMAS Song
Here is a catchy little tune to help you remember the BODMAS order of operations.
[youtube
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12633 | https://www.mdpi.com/2079-9292/11/15/2285 | Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers
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Table of Contents
Abstract
Introduction
Concept of Approximation by Residual Correction Method
Second Approximation by the Residual Correction Method for Trigonometric Functions
Second Approximation by the Residual Correction Method for Inverse Trigonometric Functions
Results
Discussion
Conclusions
Author Contributions
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Conflicts of Interest
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Open Access Article
Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers
by
Takashi Kusaka
Takashi Kusaka
SciProfilesScilitPreprints.orgGoogle Scholar
1,,† and
Takayuki Tanaka
Takayuki Tanaka
SciProfilesScilitPreprints.orgGoogle Scholar
2,†
1
Independent Researcher, Sapporo 063-0867, Japan
2
Graduate School of Information Science and Technology, Hokkaido University, Sapporo 060-0814, Japan
Author to whom correspondence should be addressed.
†
These authors contributed equally to this work.
Electronics2022, 11(15), 2285;
Submission received: 14 June 2022 / Revised: 11 July 2022 / Accepted: 19 July 2022 / Published: 22 July 2022
(This article belongs to the Section Computer Science & Engineering)
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Abstract
In modern computers, complicated signal processing is highly optimized with the use of compilers and high-speed processing using floating-point units (FPUs); therefore, programmers have little opportunity to care about each process. However, a highly accurate approximation can be processed in a small number of computation cycles, which may be useful when embedded in a field-programmable gate array (FPGA) or micro controller unit (MCU), or when performing many large-scale operations on a graphics processing unit (GPU). It is necessary to devise algorithms to obtain the desired calculated values without an accelerator or compiler assistance. The residual correction method (RCM) developed here can produce simple and accurate approximations of certain nonlinear functions with minimal multiply–add operations. In this study, we designed an algorithm for the approximate computation of trigonometric and inverse trigonometric functions, which are nonlinear elementary functions, to achieve their fast and accurate computation. A fast first approximation and a more accurate second approximation of each function were created using RCM with a less than 0.001 error using multiply–add operations only. This achievement is particularly useful for MCUs, which have a low power consumption but limited computational power, and the proposed approximations are candidate algorithms that can be used to stabilize the attitude control of robots and drones, which require real-time processing.
Keywords:
arctangent approximation; trigonometric functions approximation; elemental function approximation; fused multiply–add; algorithm design and analysis; computational cost reduction
1. Introduction
The approximation of elementary functions is a critical issue in science and engineering, including computer graphics, scientific computing, and signal processing [1,2,3,4]. Central processing units (CPUs) have evolved to have computation instructions for efficient processing, but electronic circuits with limited resources, such as field programmable gate arrays (FPGAs) and micro controller units (MCUs), may not have accelerators such as a floating-point unit (FPU) or digital signal processor (DSP). Therefore, techniques to avoid calculation with a large number of division and non-linear calculation cycles are important for large-scale calculations and embedded systems . Division and square root are frequently used, even though they are up to ten times slower than addition and multiplication, and this trend has not changed in the latest architectures [6,7].
In particular, MCUs have a low computational power with an instruction set limited to basic arithmetic and logic operations, but they are small in size, have a low power consumption, and are used in a wide variety of products, including smartphones, audio accessories, video game peripherals, and advanced medical equipment [8,9,10,11,12,13]. Furthermore, a small size and low power consumption features are required in areas such as IoT devices [14,15] and drone control [16,17,18,19].
Even in the rapidly developing field of machine learning, fast approximate computation is critical to solve problems such as short learning times and the real-time processing of inferences [20,21,22,23]. GPUs can process a large number of multiply–add operations in parallel, and by replacing nonlinear calculations used in machine learning with approximate calculations using multiply–add operations, it may be possible to achieve higher speeds and smaller models.
When the compiler optimization performance was low, implementation techniques to avoid division were very useful in the field of embedded electronic circuits . Even today, many MCUs do not have an extended instruction set, so such techniques are very important. For example, replacing division by a product of reciprocal constants and utilizing bit shifting for multiplying and dividing powers of two are well-known techniques . Computational tricks using the floating-point structure of the IEEE754 are effective, and useful techniques are known for obtaining fast approximations of exponential and logarithmic functions [21,26] and inverse square roots (reciprocal sqrt) [25,27,28,29,30,31,32]. In particular, the latter is a well-known algorithm called the fast inverse square root (FISR), which is a useful technique that can be used to find the reciprocal and square root.
As mentioned, it is important to use such techniques to obtain elementary functions at a high speed and low cost when considering integration into electronic circuits or use in machine learning with GPUs. Among elementary functions, the addition and multiplication in four fundamental arithmetic operations can generally be processed by CPUs at a high speed, and the division can be calculated by using the fast inverse square root described above. Moreover, exponential and logarithmic functions, which are nonlinear functions of elementary functions, can be obtained using IEEE754 calculation tricks. The remaining elementary functions include the computation of trigonometric and inverse trigonometric functions, for which, some methods have been proposed.
In general, approximations of trigonometric functions are often obtained using Taylor expansions, which are slow to converge and cannot be computed over a wide definition domain without the terms exceeding the seventh order. Inverse trigonometric functions are similarly computed using polynomial approximation , rational approximation [34,35,36,37], look up table (LUT), and coordinate rotation digital computer (CORDIC) algorithms [38,39,40], but each of them have disadvantages, such as requiring a large amount of memory and iterative calculations.
To solve this problem, we proposed an algorithm to approximate trigonometric functions (sin and cos) and inverse trigonometric functions (atan2) using only the multiply–add operation . In the proposed method, the approximation is achieved by using one addition and two multiplications. The proposed residual correction method (RCM) is an approximation that minimizes the error over the entire domain, rather than a locally precise approximation such as the Taylor expansion. The atan2 function created by the proposed method is faster than the previously used DSP-trick , and the approximation is obtained with fewer errors. The DSP-trick is a suitable algorithm for MCUs, but it must always perform division. Thus, it is a tightly coupled algorithm with division. In contrast, our proposed method can be executed with only multiply–add operations if the norm is known. Therefore, it does not necessarily require division. Hence, it is a loosely coupled algorithm with division. Since the norm of the measured values is known for most of the posture measurement sensors, the proposed method, skipping normalization, is suitable for embedded devices. If normalization is required, the algorithm can be integrated with any normalization method, such as fast inverse square root for MCUs or single instruction, multiple data (SIMD) instructions for advanced computing devices.
In this study, we extend our proposed method to compute even more accurate approximations of trigonometric and inverse trigonometric functions at a similar computational cost.
2. Concept of Approximation by Residual Correction Method
We have been developing wearable posture and fatigue measurement devices [43,44] and assistive suits [45,46,47] using small, low-power MCUs. For this purpose, it is necessary to measure human posture in real time from sensor data using only simple arithmetic and logical operations that are possible with MCUs. Therefore, in our previous research, we developed approximate formulas based on multiply–add operations of trigonometric and inverse trigonometric functions to realize posture calculation on MCUs .
We proposed the concept of designing approximate expressions utilizing the residual correction method using simple algebraic expressions with only multiply–add operations, without division operations and iterative convergence algorithms, to achieve efficient calculations for embedded systems.
In a previous study, as an example of approximation using the residual correction method, an approximation formula based on the multiply–add operation of trigonometric functions (sin and cos) and inverse trigonometric functions (atan2) was created.
The following procedure was used to design the approximate formula.
(1).
Linear approximation with fixed origin and domain endpoints:
𝑓(𝑥)≈𝑎 𝑥+𝑏;f(x)≈a x+b;
(2).
Evaluate residuals with approximate target function:
𝑒 𝑟 𝑒 𝑠=𝑎 𝑥+𝑏−𝑓(𝑥);e r e s=a x+b−f(x);
(3).
Design of residual correction function (RCF):
𝑟(𝑥)≈𝑒 𝑟 𝑒 𝑠;r(x)≈e r e s;
(4).
Approximate formula with residual correction:
𝑓 𝑎 𝑝 𝑝 𝑟 𝑜 𝑥(𝑥)=𝑎 𝑥+𝑏−𝑟(𝑥).f a p p r o x(x)=a x+b−r(x).
The concept of the residual correction method is to express the RCF using only multiply–add operations, which must be explored heuristically. Due to the restriction of using only multiply–add operations, the order of the candidate functions for the RCF is less than a third. The RCF was evaluated in terms of both computational cost and approximation error, and it was shown that, for the nonlinear functions mentioned above, the use of parabolic functions provides an efficient approximation formula that expresses each function using only multiply–add operations.
Although the above method can easily obtain a quadratic approximation formula for a nonlinear function, some applications require an even smaller error. In the next section, a method to reduce the approximated error while maintaining the computational cost by extending the approximation formula with residual correction is proposed.
3. Second Approximation by the Residual Correction Method for Trigonometric Functions
In a previous study, the approximations made by the residual correction method for trigonometric functions were as follows :
sin(𝜃)≈s(𝜃)=(2 𝜋)2 𝜃(𝜋−|𝜃|)sin(θ)≈s(θ)=2 π 2 θ(π−|θ|)
(1)
cos(𝜃)≈c(𝜃)=s(𝜋 2−|𝜃|)cos(θ)≈c(θ)=s π 2−|θ|
(2)
These are of the same form as the famous quadratic approximation for a sine function in [−𝜋,𝜋][−π,π], but the error is larger around 𝜃=𝜋 4 θ=π 4, as shown in Figure 1.
Figure 1. Summary of 1st approximations of sin-function and cos-function. (a) First approximation of sin and cos functions by RCM. (b) Comparison with unit circle.
Here, the above approximation formula is used as the first approximation by the residual correction method, and a method to further reduce the error by utilizing the residual correction method under the constraint of using only multiply–add operation is examined.
3.1. Exploring RCF Candidates
Here, we consider the sin-function and evaluate the residuals of the first approximation. The residuals of the first approximation have the shape of an odd function, as shown in Figure 2a. The following RCF candidates obtained by the multiply–add operation can be considered for this residual:
Figure 2. Residual error of 1st approximation of sin-function and candidates of its RCF. (a) Residual error of 1st approximation of sin function. (b) Comparison among designed RCFs.
(1).
RCF with signed quadratic function:
𝑟 𝑠 𝑞(𝜃)=𝛼 𝑠 𝑞 s(𝜃)(1−|s(𝜃)|);r s q(θ)=α s q s(θ)1−|s(θ)|;
(2).
RCF with cubic function:
𝑟 𝑐 𝑢 𝑏 𝑖 𝑐(𝜃)=𝛼 𝑐 𝑢 𝑏 𝑖 𝑐 s(𝜃)(1−s 2(𝜃));r c u b i c(θ)=α c u b i c s(θ)1−s 2(θ);
(3).
RCF with co-function:
𝑟 𝑐 𝑜 𝑓 𝑢 𝑛 𝑐(𝜃)=𝛼 𝑐 𝑜 𝑓 𝑢 𝑛 𝑐(s(𝜃)−sign(s(𝜃))1−𝑐 2(𝜃)−−−−−−−−√).r c o f u n c(θ)=α c o f u n c s(θ)−sign(s(θ))1−c 2(θ).
Although the square root appears in the RCF using the co-function, the built-in implementation will use the fast inverse square root.
Figure 2b shows the evaluation of the RCF candidate functions. Here, the coefficients of the RCF candidate functions are designed as 𝛼 𝑠 𝑞=0.224 α s q=0.224, 𝛼 𝑐 𝑢 𝑏 𝑖 𝑐=0.145 α c u b i c=0.145, and 𝛼 𝑐 𝑜 𝑓 𝑢 𝑛 𝑐=0.5 α c o f u n c=0.5 so that the residuals are well corrected. As can be seen from the graph, the signed quadratic function is closest to the residuals of the first approximation. In addition, the signed process is faster than the multiplication and fast inverse square root, indicating that the signed quadratic function is suitable for RCF.
3.2. Second Approximation of Trigonometric Functions
Using 𝑟 𝑠 𝑞(𝜃)r s q(θ) for RCF, the second approximation of the trigonometric function is
s 2(𝜃)=s(𝜃)−𝑟 𝑠 𝑞(𝜃)=s(𝜃)[(1−𝛼 𝑠 𝑞)−𝛼 𝑠 𝑞|s(𝜃)|].s 2(θ)=s(θ)−r s q(θ)=s(θ)(1−α s q)−α s q|s(θ)|.
(3)
A similar argument can be made for the second approximation of the cos function:
c 2(𝜃)=c(𝜃)[(1−𝛼 𝑠 𝑞)−𝛼 𝑠 𝑞|c(𝜃)|]c 2(θ)=c(θ)(1−α s q)−α s q|c(θ)|
(4)
The approximate shape of the final second approximation function is shown in Figure 3.
Figure 3. Summary of 2nd approximations of sin-function and cos-function. (a) Second approximation of sin and cos functions by RCM. (b) Comparison with unit circle.
To evaluate the performance of the second approximation, the final residuals and computation time are shown in Figure 4 and Table 1. The final residual is, at most, approximately 0.00092, which is sufficiently small when compared with the first approximation. The computation time was evaluated using an Intel Xeon E3 (2500 MHz) CPU. As a result, the processing time was 1.4 ns for the sine’s approximation and 2.04 ns for the cosine’s approximation, which is approximately twice as long as the first approximation, but more than 30 times faster than the built-in functions in math.h.
Figure 4. Residual error of 2nd approximation of sin-function.
Table 1. Calculation time evaluation of 2nd approximations of sin-function and cos-function (Average ± S.D.).
Comparing the first approximation, Equation (3), with the second approximation, Equation (4), we observe that they have the same parabolic function shape. The second approximation first keeps the results of the calculation of the first approximation in memory, and then uses the results to input them into a parabolic function of the same shape. The first approximation requires the computational cost of one parabolic function, and the second approximation requires the computational cost of two parabolic functions. Therefore, the computation cost is approximately doubled; however, the computation error is reduced to approximately 1/60.
4. Second Approximation by the Residual Correction Method for Inverse Trigonometric Functions
Approximate formulas based on the residual correction method for inverse trigonometric functions were also proposed in our previous study .
atan(𝑦/𝑥)≈at(𝑦,𝑥)=(𝜋 2−2 3 𝑥)𝑦 atan(y/x)≈at(y,x)=π 2−2 3 x y
(5)
Note that this equation is designed assuming that the norm is known, 𝑥 2+𝑦 2=1 x 2+y 2=1, and the domain of definition is [−𝜋/2,𝜋/2][−π/2,π/2]. In many cases, the norm is known for sensors that measure the directional cosine in embedded systems, but when the norm changes occasionally, it is assumed that the norm is normalized in advance using fast inverse square root. The computational complexity of this normalization will be discussed later. Next, we will consider a higher accuracy by the second approximation for inverse trigonometric functions and trigonometric functions.
4.1. Domain Expansion
Since the domain of Equation (5) is [−𝜋/2,𝜋/2][−π/2,π/2], it is necessary to extend the domain in order to treat it the same as atan 2(𝑦,𝑥)atan 2(y,x). Two methods are evaluated in this paper: one based on a conditional branch and the other based on an extension using the half-angle identity.
The method using the conditional branch changes the formula used by using the sign of the variable.
atan 2(𝑦,𝑥)≈at 2 𝑖 𝑓(𝑦,𝑥)=⎧⎩⎨at(𝑦,𝑥),𝜋−at(𝑦,−𝑥),−𝜋−at(𝑦,−𝑥),if 𝑥≥0 if 𝑥<0 and 𝑦≥0 otherwise.atan 2(y,x)≈at 2 i f(y,x)=at(y,x),if x≥0 π−at(y,−x),if x<0 and y≥0−π−at(y,−x),otherwise.
(6)
Another way to expand the domain of definition by a factor of two is to use the half-angle identity. This can be expressed in a single algebraic expression, eliminating the need to switch equations. The atan’s half-angle identity,
atan 2(𝑦,𝑥)=2 atan⎛⎝⎜⎜⎜⎜𝑦 𝑥 2+𝑦 2−−−−−−√+𝑥⎞⎠⎟⎟⎟⎟,atan 2(y,x)=2 atan y x 2+y 2+x,
(7)
together with the constraints of 𝑥 2+𝑦 2=1 x 2+y 2=1, can be organized into
atan 2(𝑦,𝑥)≈at 2 𝑒 𝑥(𝑦,𝑥)=(𝜋 2−−√1 1+𝑥−−−−√+2 3)𝑦 atan 2(y,x)≈at 2 e x(y,x)=π 2 1 1+x+2 3 y
(8)
=(𝜋 2−−√𝐺(1+𝑥)+2 3)𝑦.=π 2 G(1+x)+2 3 y.
(9)
Here, 1/·−−√1/· appears in the equation, and 𝐺(·)G(·) is used as the fast inverse square root. This allows us to extend the definition domain from [−𝜋/2,𝜋/2][−π/2,π/2] to [−𝜋,𝜋][−π,π], so that we can approximate atan2 with a single algebraic expression.
The results of each domain extension are shown in Figure 5a, and the residuals are shown in Figure 5b.
Figure 5. First approximation of atan2 and its domain expansion (plot using 𝑥=cos 𝜃,𝑦=sin 𝜃 x=cos θ,y=sin θ). (a) First approximation of atan2 and domain expansion. (b) Residual error in each expansion.
4.2. Exploring RCF Candidates
Next, we consider candidates for RCF, but, as the atan2 function is a two-variable function, it is difficult to search for functions by simple multiply–add operations, such as parabolic functions. Therefore, we consider using Newton’s method by referring to the steps for improving the accuracy of the fast inverse square root.
From the constraints on the trigonometric functions, the evaluation function 𝑓(𝜃)f(θ) and its derivative in Newton’s method by x, y, and 𝜃 θ are
𝑓(𝜃)=1 2[(𝑥−cos 𝜃)2+(𝑦−sin 𝜃)2]=1−𝑥 cos 𝜃−𝑦 sin 𝜃,f(θ)=1 2(x−cos θ)2+(y−sin θ)2=1−x cos θ−y sin θ,
(10)
𝑓′(𝜃)=𝑥 sin 𝜃−𝑦 cos 𝜃.f′(θ)=x sin θ−y cos θ.
(11)
Using this, the error converges to zero by repeating this equation
𝜃 𝑛+1=𝜃 𝑛−𝑓(𝜃 𝑛)𝑓′(𝜃 𝑛).θ n+1=θ n−f(θ n)f′(θ n).
(12)
With reference to this, the second term on the right-hand side, which is the correction term, can be used as a candidate for RCF.
Thus, it was found that the sin and cos functions are necessary for RCF to cancel the approximation error of atan2. From the evaluation in the previous chapter, the trigonometric functions in the math library are computationally expensive when implemented in programs. Therefore, we replace them with 𝗌 2(𝜃)s 2(θ) and 𝖼 2(𝜃)c 2(θ), which are highly accurate and fast approximate calculations.
Therefore, let the candidate functions of RCF be as follows:
𝑟 𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)=1−𝑥 c 2(𝜃)−𝑦 s 2(𝜃)𝑥 s 2(𝜃)−𝑦 c 2(𝜃)r a t a n 2(y,x)=1−x c 2(θ)−y s 2(θ)x s 2(θ)−y c 2(θ)
(13)
There are two problems with this RCF candidate. One is that it violates the design concept of the residual correction method because it involves division. The other is that the denominator may become zero due to approximation errors in the trigonometric functions. Therefore, 𝑟 𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)r a t a n 2(y,x) is not suitable for RCF. However, by analyzing the behavior of 𝑟 𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)r a t a n 2(y,x), a new RCF can be found. When there is no approximation error in the trigonometric function, the problem of the denominator being zero does not occur because the numerator also becomes zero at that time, and the effect of the correction term disappears. From this, the requirement for RCF is that the denominator of 𝑟 𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)r a t a n 2(y,x), 𝑓′(𝜃)f′(θ) is zero when it is also zero. Therefore, 𝑓′(𝜃)f′(θ) itself is a candidate for RCF here.
𝑟̃𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)=𝛼 𝑎 𝑡 𝑎 𝑛 2[𝑥 s 2(𝜃)−𝑦 c 2(𝜃)]r˜a t a n 2(y,x)=α a t a n 2 x s 2(θ)−y c 2(θ)
(14)
Although there are design degrees of freedom in the coefficients, the result of this optimization is approximately 1.0 as shown next, thus eliminating one multiplication.
𝛼 𝑎 𝑡 𝑎 𝑛 2=arg min 𝛼∫𝜋−𝜋|at 2∗(𝑦,𝑥)−𝑟̃𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)−𝜃|𝑑 𝜃≈1.0 α a t a n 2=arg min α∫−π π at 2∗(y,x)−r˜a t a n 2(y,x)−θ d θ≈1.0
(15)
The same result can be obtained using either at 2 𝑖 𝑓(𝑦,𝑥)at 2 i f(y,x) or at 2 𝑒 𝑥(𝑦,𝑥)at 2 e x(y,x) as at 2∗(𝑦,𝑥)at 2∗(y,x). A plot of 𝑟̃𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)r˜a t a n 2(y,x) is shown in Figure 6a. Here, we use the domain expansion at 2 𝑒 𝑥(𝑦,𝑥)at 2 e x(y,x) based on the half-angle identity. It can be seen that 𝑟̃𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)r˜a t a n 2(y,x), which can be computed by the multiply–add operation, agrees very well with the residual of at 2 𝑒 𝑥(𝑦,𝑥)at 2 e x(y,x) and is appropriate as RCF.
Figure 6. Residual error of atan2’s 1st approximation and consideration of its RCF candidates (plot using 𝑥=cos 𝜃,𝑦=sin 𝜃 x=cos θ,y=sin θ). (a) Evaluation of Newton’s method (NM) for RCF. (b) Applying desined RCF for each expansion.
Figure 6b shows the case where at 2 𝑖 𝑓(𝑦,𝑥)at 2 i f(y,x) is evaluated with the same RCF, and this RCF can always be applied if a good approximation is obtained in the first approximation. Therefore, other atan2 approximation methods, such as DSP-Trick, for example, are also RCFs that can be made highly accurate.
4.3. Second Approximation of Inverse Trigonometric Functions
Having obtained the RCF, the second approximation of atan2 can be computed using the second approximation of trigonometric functions as follows:
𝜃 1=at 2∗(𝑦,𝑥)θ 1=at 2∗(y,x)
(16)
at 2 2(𝑦,𝑥)=𝜃 1−𝑟̃𝑎 𝑡 𝑎 𝑛 2(𝑦,𝑥)=𝜃 1−𝑥 s 2(𝜃 1)+𝑦 c 2(𝜃 1).at 2 2(y,x)=θ 1−r˜a t a n 2(y,x)=θ 1−x s 2(θ 1)+y c 2(θ 1).
(17)
Figure 7 shows the residuals of at 2 𝑖 𝑓(𝑦,𝑥)at 2 i f(y,x) and at 2 𝑒 𝑥(𝑦,𝑥)at 2 e x(y,x). Both at 2 𝑖 𝑓(𝑦,𝑥)at 2 i f(y,x) and at 2 𝑒 𝑥(𝑦,𝑥)at 2 e x(y,x) can be used for at 2∗(𝑦,𝑥)at 2∗(y,x) in terms of error, but at 2 𝑖 𝑓(𝑦,𝑥)at 2 i f(y,x) should be used in terms of the computational cost.
Figure 7. Residual error of 2nd approximation of atan2 (plot using 𝑥=cos 𝜃,𝑦=sin 𝜃 x=cos θ,y=sin θ).
In addition, the constraint 𝑥 2+𝑦 2=1 x 2+y 2=1 was used in the approximate formula design up to this point, but normalization may be necessary in some applications. Our proposed atan2 approximation can be combined with any normalization algorithm. Recent CPUs can compute faster with the SIMD instruction set. However, for MCUs with only a simple instruction set, normalization is a major problem . Therefore, fast inverse square root, which can be executed using only multiply–add and logic operations, is a useful technique for MCUs. Any normalization algorithm or SSE instruction can be used in general; however, we assume an MCU and verify the use of fast inverse square root. The normalized second approximation is defined as at 2 2 𝑛(𝑦,𝑥)at 2 2 n(y,x) when the fast inverse square root is used for the variables. at 2 2 𝑛(𝑦,𝑥)at 2 2 n(y,x) can be calculated using the fast inverse square root 𝐺(·)G(·) as follows:
at 2 2 𝑛(𝑦,𝑥)=at 2 2(𝑛 𝑦,𝑛 𝑥),at 2 2 n(y,x)=at 2 2 n y,n x,
(18)
∵𝑛=𝐺(𝑥 2+𝑦 2)≈1 𝑥 2+𝑦 2−−−−−−√∵n=G(x 2+y 2)≈1 x 2+y 2
(19)
The error between the calculation results and the Math library is shown in Figure 8. The increase in computational complexity due to normalization is shown as at 2 2 𝑛 at 2 2 n in Table 2. Compared with at 2 2 at 2 2, it is two times greater, which indicates that a highly accurate approximation of at 2 2 at 2 2 can be performed with the same computational complexity as that of the fast inverse square root.
Figure 8. Atan2’s 2nd approximation with normalization using fast inverse square root(FISR) and its residual error. (a) Atan2 approx. over entire domain using FISR. (b) Error with atan2 in math library.
Table 2. Calculation time evaluation of 2nd approximations of atan2 (Average ± S.D.)
In this study, the original fast inverse square root is used for the evaluation implementation, but faster implementations have been studied, such as the algorithm , which is faster for FPGA.
5. Results
Second approximations of the trigonometric and inverse trigonometric functions have been developed, and the results are summarized here. The maximum error indicates the maximum value of the final residuals of the second approximation. The computational cost indicates the number of multiply–add operations used. Although an example of the computation time has already been shown, only the number of operations is shown here because it varies depending on the CPU performance and architecture.
For the second approximation of the trigonometric functions, Equations (3) and (4) are considered good approximations, and the error and computational cost are as shown in Table 3. The final error is less than 0.001 for the entire definition region, and the computational cost is very low.
Table 3. Error and computational cost of approximating trigonometric functions.
Equation (17) is considered a good second approximation of the inverse trigonometric function. The error and computational cost are shown in Table 4. The small approximation error is less than 0.001 rad over the entire definition region. The computational cost is a little higher because two second approximations of trigonometric functions are required, but the approximation formula is only a finite number of multiply–add operations.
Table 4. Error and computational cost of inverse trigonometric functions.
6. Discussion
In this study, we have designed a second approximation of RCF to achieve a higher accuracy in approximating trigonometric and inverse trigonometric functions.
6.1. Approximation of Trigonometric Function
As mentioned in Introduction, the Taylor series is slow to converge but can compute mathematically exact values of trigonometric functions. We compare the Taylor series and the proposed method in the domain of definition [−𝜋,𝜋][−π,π]. Taylor series are evaluated from the first to the ninth order. In addition, the computational cost is evaluated based on the definition formula and the following formula expansion to reduce the number of multiplications for embedding.
sin(𝑥)≈∑𝑛=0 𝑘(−1)𝑛 𝑥(2 𝑛−1)(2 𝑛−1)!=𝑥(1−𝑥 2(1 3!+𝑥 2(1 5!−𝑥 2(1 7!+⋯))))sin(x)≈∑n=0 k(−1)n x(2 n−1)(2 n−1)!=x 1−x 2 1 3!+x 2 1 5!−x 2 1 7!+⋯
(20)
The left-hand side of Equation (20) is the definition for the kth-order Taylor expansion and the right-hand side is the optimized number of multiplications by storing 𝑥 2 x 2 in memory. The error is evaluated numerically by Romberg integration, and the computational cost is compared to the number of multiplications.
Table 5 and Figure 9 show the comparison results. As mentioned above, the Taylor series is mathematically exact, but its convergence is slow, so a large error remains if a wide domain of definition is used. Although the proposed method cannot reduce the error to zero, it enables obtaining results with a reduced number of calculations and reduced error.
Figure 9. Comparison of maximum error and number of multiplications of sin function approximation between Taylor series and proposed method (log scale is used in vertical axis to compare error order).
Table 5. Comparison summary of sin function approximation between Taylor series and proposed method.
6.2. Approximation of Inverse Trigonometric Function
Next, we discuss the approximation of the inverse trigonometric function. Depending on the characteristics of the application, the desired accuracy can be used by selecting whether the first approximation is sufficient, second approximation is necessary, or whether normalization is necessary. Table 6 shows a comparison with other methods. Basically, all are trade-offs between accuracy and computational complexity, and should be used appropriately depending on the application. The approximate formula by the residual correction method is an algebraic formula and is extremely easy to implement as a program, so it can be one of the options for approximate calculation. While more exact calculation methods are suitable for obtaining exact values, such as in numerical simulations, multiply–add operations are very effective in compact systems such as MCUs, FPGAs, and large-scale machine learning operations. In such fields, the proposed method should be used to obtain highly accurate approximations with a finite number of multiply–add operations without using division.
Table 6. Comparison with other methods.
The following is an example of compilation and execution results on an actual MCU. The program memory occupancy of the proposed algorithm with an MCU grade is shown in Table 7. The famous Microchip PIC family and the XC8 compiler are considered in these examples. In the lowest grade model, the atan2 function alone consumes the entire program memory, and thus the program cannot be compiled. In contrast, the proposed method makes it possible to successfully implement the algorithm. Other grades of MCUs without FPU or DSP occupy the same amount of program memory. Since embedded systems usually require not only function calculations but also various processes, such as measurement and equipment control, the program memory of the MCU must be saved as much as possible in order to implement these processes. The second approximation of the proposed method also uses s 2 s 2 and c 2 c 2 in the calculation process; thus, trigonometric functions can be simultaneously used. A total of 7.7 KB is required to include them in the Math library, so the approximation is advantageous for performing complicated calculations.
Table 7. Comparison of occupied program memory compiled with actual MCUs.
Next, the calculation speed is measured using an actual MCU (PIC18F2580). The expetimental results are summarized in Table 8. The calculation speed is fast enough for the first approximation of atan2, and the speed of the second approximation is close to that of the Math library as the calculation volume increases. Therefore, if there is sufficient program memory, using the Math library is the most accurate way to calculate accurate values. However, for limited program memory, such as in low-grade MCUs, the first approximation can be used for speed and the second approximation can be used if accuracy is required. This increases the degree of freedom of implementation. As mentioned above, MCUs perform not only function calculations but also various other functions, such as measurement and equipment control, so function calculations should be as fast as possible.
Table 8. Experimental result on actual MCU (PIC18F2580@4MHz internal clock).
In this study, we developed highly accurate approximation formulas for sin and cos as trigonometric functions and atan2 as an inverse trigonometric function using the residual correction method. The tan, asin, and acos functions can be calculated from the fast inverse square root and the approximation formula proposed here using the trigonometric identities, respectively. Therefore, it is known that log and exp functions can be computed from the IEEE754 structure as the fast inverse square root, implying that, besides the four arithmetic operations, logarithm and exponential, trigonometric, and inverse trigonometric functions can also be computed quickly. This enables the computation of highly accurate approximations of all elementary functions by a finite number of multiply–add operations and bit shifts, when considered for embedded systems.
7. Conclusions
In this paper, we proposed a second approximation of the residual correction method that can approximate trigonometric and inverse trigonometric functions with a high accuracy using a small number of calculations. Using our proposed algorithm, nonlinear calculations that were previously impossible to implement can now be realized even on small, low-power computers, such as MCUs. It cannot reduce the error to completely zero; however, it can compute accurate approximations with low-cost computation. This can contribute to machine downsizing and cost reduction, and is a candidate algorithm for developing MCU-based devices that require real-time performance, such as drones and IoT devices.
To approximate trigonometric and inverse trigonometric functions with a high accuracy using finite multiply–add operations, a second approximation using the residual correction method was proposed. Trigonometric functions can be computed with a maximum error of less than 0.001 by two additions and four multiplications, and atan2 can be computed with a maximum error of less than 0.001 rad by seven additions and fourteen multiplications. For the formulation of the second approximation, we designed and evaluated a formula using floating-point arithmetic. In future research, we aim to optimize the formula for embedded systems by converting it to fixed-point arithmetic. Since the residual correction method is formulated using only the multiply–add operation without division, it is well suited to the fixed-point arithmetic, and further acceleration can be expected when considering implementation in embedded systems.
Author Contributions
Conceptualization, T.K.; investigation, T.K.; methodology, T.K.; supervision, T.T.; validation, T.K. and T.T.; writing—original draft, T.K.; writing—review and editing, T.T. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
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Figure 1. Summary of 1st approximations of sin-function and cos-function. (a) First approximation of sin and cos functions by RCM. (b) Comparison with unit circle.
Figure 2. Residual error of 1st approximation of sin-function and candidates of its RCF. (a) Residual error of 1st approximation of sin function. (b) Comparison among designed RCFs.
Figure 3. Summary of 2nd approximations of sin-function and cos-function. (a) Second approximation of sin and cos functions by RCM. (b) Comparison with unit circle.
Figure 4. Residual error of 2nd approximation of sin-function.
Figure 5. First approximation of atan2 and its domain expansion (plot using 𝑥=cos 𝜃,𝑦=sin 𝜃 x=cos θ,y=sin θ). (a) First approximation of atan2 and domain expansion. (b) Residual error in each expansion.
Figure 6. Residual error of atan2’s 1st approximation and consideration of its RCF candidates (plot using 𝑥=cos 𝜃,𝑦=sin 𝜃 x=cos θ,y=sin θ). (a) Evaluation of Newton’s method (NM) for RCF. (b) Applying desined RCF for each expansion.
Figure 7. Residual error of 2nd approximation of atan2 (plot using 𝑥=cos 𝜃,𝑦=sin 𝜃 x=cos θ,y=sin θ).
Figure 8. Atan2’s 2nd approximation with normalization using fast inverse square root(FISR) and its residual error. (a) Atan2 approx. over entire domain using FISR. (b) Error with atan2 in math library.
Figure 9. Comparison of maximum error and number of multiplications of sin function approximation between Taylor series and proposed method (log scale is used in vertical axis to compare error order).
Table 1. Calculation time evaluation of 2nd approximations of sin-function and cos-function (Average ± S.D.).
| | 1st Approx. | 2nd Approx. | Math Library (math.h) |
:---: :---: |
| Sine function | 0.54±0.16 ns | 1.4±0.2 ns | 68.28±0.52 ns |
| Cosine function | 1.22±0.13 ns | 2.04±0.39 ns | 69.1±0.90 ns |
Table 2. Calculation time evaluation of 2nd approximations of atan2 (Average ± S.D.)
| | 1st Approx. | 2nd Approx. at2 if/at2 ex | 2nd Approx. with FISR at2 nif/at2 nex | Math Library (math.h) |
:---: :---:
| Calculation time | 1.43±0.26 ns | 2.96±0.27 ns 4.76±0.26 ns | 7.3±0.19 ns 9.14±0.25 ns | 21.96±0.83 ns |
Table 3. Error and computational cost of approximating trigonometric functions.
| Sine Function | Maximum Error | Computational Cost |
:---:
| RCM 1st approximation | 5.6 × 10−2 | 1 addition 2 multiplications 1 absolute value |
| RCM 2nd approximation | 9.2 × 10−4 | 2 additions 4 multiplications 2 absolute values |
Table 4. Error and computational cost of inverse trigonometric functions.
| Atan2 Function | Maximum Error | Computational Cost |
:---:
| RCM 1st approximation | 4.2 × 10−2 rad | 1 addition 2 multiplications |
| RCM 2nd approximation | 9.2 × 10−4 rad | 7 additions 14 multiplications 4 absolute values |
Table 5. Comparison summary of sin function approximation between Taylor series and proposed method.
| | Taylor Series (Original/Optimized for Embedding) | Proposed Method |
:---:
| | 1st Order | 3rd Order | 5th Order | 7th Order | 9th Order | 1st Approx | 2nd Approx |
| Maximum abolute error | 3.14 | 2.03 | 5.24 × 10−1 | 7.52 × 10−2 | 6.93 × 10−3 | 5.6 × 10−2 | 9.2 × 10−4 |
| Mean absolute error | 5.84 | 2.25 | 4.23 ×10−1 | 4.80×10−2 4.80×10−2 | 3.66 ×10−3×10−3 | 1.9×10−1×10−1 | 3.3×10−3×10−3 |
| Number of multiplication | 0/0 | 3/3 | 8/4 | 15/5 | 24/6 | 2 | 4 |
Table 6. Comparison with other methods.
| Atan2 Approx. Methods | Accuracy | Division | Iteration | Memory |
:---: :---:
| Rational approx. | High | Necessary - |
| Tayler expansion | Depends on order -
| LUT | Depends on memory - | Large |
| CORDIC | Depends on iteration | Necessary | Necessary
| DSP-trick | Middle (<0.01) | Necessary - |
| RCM 1st approx. | Middle (<0.01) -
| RCM 2nd approx. | High (<0.001) -
: The proposed methods in this study.
Table 7. Comparison of occupied program memory compiled with actual MCUs.
| Grade | Example of MCU | Available Program Memory | Proposed Methods | Math Library (math.h) |
:---: :---:
| 𝐚𝐭 𝟐 𝒊 𝒇 at 2 i f 1 KB Required | 𝐚𝐭 𝟐 𝟐 at 2 2 (w/s2 and c2) 1.8 KB req’d | atan2 Only 3 KB req’d | atan2, sin, cos 7.7 KB req’d |
| Low | PIC12F1572 | 2 KB | A | AR | CC | CC |
| Middle | PIC16F1827 | 4 KB | A | A | AR | CC |
| Middle | PIC16F1769 | 8 KB | A | A | A | AR |
| High | PIC18F2580 | 16 KB | A | A | A | A |
A: available (less than 50% occupancy); AR: available with restriction; CC: cannot be compiled.
Table 8. Experimental result on actual MCU (PIC18F2580@4MHz internal clock).
| | 1st Approx of atan2 | 2nd Approx of atan2 | atan2 (math.h) |
:---: :---: |
| Average time of 100 iterations | 1.5 ms | 6.5 ms | 8.0 ms |
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Kusaka, T.; Tanaka, T. Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers. Electronics2022, 11, 2285.
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Kusaka T, Tanaka T. Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers. Electronics. 2022; 11(15):2285.
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Kusaka, Takashi, and Takayuki Tanaka. 2022. "Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers" Electronics 11, no. 15: 2285.
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Kusaka, T., & Tanaka, T. (2022). Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers. Electronics, 11(15), 2285.
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Kusaka, T.; Tanaka, T. Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers. Electronics2022, 11, 2285.
AMA Style
Kusaka T, Tanaka T. Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers. Electronics. 2022; 11(15):2285.
Chicago/Turabian Style
Kusaka, Takashi, and Takayuki Tanaka. 2022. "Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers" Electronics 11, no. 15: 2285.
APA Style
Kusaka, T., & Tanaka, T. (2022). Fast and Accurate Approximation Methods for Trigonometric and Arctangent Calculations for Low-Performance Computers. Electronics, 11(15), 2285.
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12634 | https://create.kahoot.it/details/percent-solve-word-problems/490f0f95-f5e8-4daf-8fa7-b3346907e645 | Percent: Solve Word Problems - Details - Kahoot!
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A solution is 12% bleach. There are 150 oz of solution. How many oz of bleach are in the solution?
A pair of shoes is $42. There is a 20% off sale. How much will be saved?
The class set a study goal. They are 75% to the goal with 450 hours studied. How many hours is their goal?
36 students speak three languages. If that is 12% of students, how many students are there?
There are 13 boys and 12 girls in choir. 3 boys and 4 girls are also in band. What percent of choir are in band too?
Jose bought a $400 bat with $250 coupon. What is the discount percent with the coupon?
Fruit punch contains 20% pineapple juice. How much punch can you make with 3 c of pineapple juice?
Joe correctly answered 96% of the test questions. There are 50 questions. How many are correct?
11 boys and 14 girls are in 1st period. What percent of 1st period are girls?
A game cost $524. It is on sale for 76% off. How much is taken off?
New cars lose 20% of value in first year. If value lost is $4,200 what is the original cost?
He got $45 off the price of the sweater. This is due to a 60% off sale. What was the original price?
There are 100 choices in the vending machine. 45 of them are candy. What percent are candy?
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12635 | https://www.youtube.com/watch?v=t4tcckF3vk4 | Solving A System Of Equations Word Problem
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Learn how to set up and solve 2 system of equations word problems. The method shown in this video is substitution.
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Intro this is mr masonette and what we're going to be doing in this tutorial is we're going to practice solving word problems involving systems of equations so let's go ahead and get started so this problem reads that the length of a rectangle is equal to three times its width write a system of equations that can be used to find the dimensions of the rectangle if the perimeter is 32 centimeters and then we have to figure out the length and the width of that rectangle so with systems of equations dealing with word problems sometimes it is difficult to figure out where to start well what i like to do is to determine what two variables we're going to use in order to solve so to figure out what variables we are going to use what we should do is figure out what it is we are trying to find well we know that we are dealing with the length and the width of a rectangle and that we have to find those two things so the two variables that we are going to use are going to be l for length and w for width now after you have declared your variables what we have to do is express what we know about the problem algebraically sometimes they might straight out give you an equation and other times you have to use a little bit of inference or prior knowledge to set up other equations so in this problem they are saying that the length of a rectangle is equal to or the same thing as three times its width and that's something that we can express algebraically so we can say that the length represented by l is equal to three times the width so right away we have one of our two equations in the system and now what we have to do is figure out what our second equation is going to be and the second equation is not that straightforward now it does say that the perimeter of this rectangle is 32 centimeters so we know that the perimeter is equal to 32 centimeters we cannot write p equals 32 for an equation though because that would give us a third variable and we can only have two variables in this situation so what we have to do is express perimeter using width and length now we do know that a rectangle has four sides and two of those sides are the length of the rectangle so we can say that two lengths plus two sides which are the width when added together will be equal to the given perimeter of 32 centimeters so this is our second equation in the system now notice our first equation has the length isolated already so we can say that length is the same thing as 3w which means we can substitute the length in the other equation with 3w so let's go ahead and rewrite this equation as 2 times 3w plus 2w is equal to 32. so all we did is we substituted the length in this equation for 3w because in the problem it is given that the length is the same thing as 3 multiplied by w now let's go ahead and simplify this equation a bit we're going to double 3w which is 6w plus 2w equals 32 and now we can go ahead and combine these two terms here 6w and 2w is of course 8 w and that is equal to 32 and now we're going to divide both sides by 8 to isolate this w and we have determined that w is equal to 32 divided by eight which is four so now we know that the width of our rectangle is equal to four all right now we know that the length is equivalent to 3 times the width and because the width is 4 the length is going to be 12. now we can be more formal though we can go ahead and rewrite that equation and substitute 4 in 4 w and that would give us the length is equal to 12. so the width of the rectangle is 4 and the length is equal to 12. all right let's go ahead and solve another example Solution all right this problem reads that tickets to a movie cost five dollars for adults and three dollars for students a group of friends purchased 18 tickets for 82 how many adult tickets and how many student tickets did they buy all right so just like i did with the first problem what i'm going to do is declare what our variable should be and that is going to be dictated by what you are looking for in the problem and we are looking for the number of adult tickets which i am going to use a to represent that amount and the number of student tickets which i'm going to use s to represent that amount so now that we have declared our variables let's go ahead and figure out what we know about the problem and express that algebraically one piece of information that sticks out about this problem is that the group of friends purchased a total of 18 tickets and we know that this is a total and that amount must be a combination of adult tickets and student tickets so we can express that algebraically as the number of adult tickets purchased plus the number of student tickets purchased is going to be a total or equal to 18 tickets in total now the other total that is given in the problem that we have to deal with is 82 we know that this is a total cost of all the tickets purchased so i know for sure that 82 can be written on this side of our equal sign and then we know that the amount spent for adults plus the amount spent on students is going to total 82 dollars now we know that it is five dollars for one adult ticket but we know that they paid for more than one adult ticket now if we knew the number of adult tickets we would just multiply that number by five and we can express that algebraically we can simply write five dollars multiplied by the number of adult tickets represented by a is going to give us the total amount spent on adult tickets and the same thing is true for student tickets we know that the number of student tickets represented by s would be multiplied by three dollars to get the total amount spent on student tickets and then when you add those two totals together that would give us a total of 82 dollars so now we have both equations in our system now to solve the system we can either use substitution or we can use elimination it just depends on what you're most comfortable with now in this situation i'm just going to use a little bit of substitution here what i'm going to do is take this first equation and isolate the a variable because if you use substitution one of the two equations has to have one variable that is isolated on one side of the equation so what i'm going to do is i'm going to take this s here and i'm going to write it on the other side of the equation which you're allowed to do as long as you write it as its inverse so i'm going to write minus s on this side so now we have the equation a is equal to 18 minus s now that we know what a is equal to 18 minus s in this case i'm going to substitute 18 minus s in for a to the other equation so let's go ahead and do just that i'm going to take 5 and substitute a with 18 minus s and then add 3s and set it equal to 82. all right now what we're going to do is distribute this 5 and 5 times 18 is 90 and 5 times negative s would be negative 5s plus 3s equals 82. all right now we can simplify the left side of our equation a bit more so let's bring down this 90 which is a constant and combine these two variable terms negative 5s and positive 3s is negative 2s and that is equal to 82. and now i'm going to take this positive 90 and write it as its inverse on the other side which is negative or minus 90. and that leaves us with negative 2s on the left and on the right this is negative 8 and to get rid of this coefficient we always divide coefficients by themselves to turn it into positive 1 which gives us s equals negative 8 divided by negative 2 which is positive 4. so the number of student tickets sold is four now we know that the number of adult tickets plus student tickets is equal to 18 because there were 18 tickets purchased all together and now that we know there are four student tickets we can determine the number of adult tickets because the total is 18. so a must be 14 because if we isolated this a here by subtracting 4 on the other side that would give us a equals 14. so now we know in the system that the number of student tickets is 4 and the number of adult tickets is equal to 14. Outro hey i just want to say thanks for checking out this map tutorial please don't forget to hit that subscription button and enable notifications so you can be informed as i upload new math tutorials to my math channel until next time this is shane masonette with masonette math [Music] you |
12636 | https://www.ahajournals.org/doi/pdf/10.1161/01.res.4.6.718 | Effect of Ganglionic Blocking Agents Upon Blood Flow and Resistance in the Superior Mesenteric Artery of the Dog
By JOSEPH H. TRAPOLD, P H . D .
With the technical assistance of Joan G. Sullivan, B.S.
The response of the mescnteric vascular circuit of anesthetized dogs to a group of ganglionic blocking agents has ljcen studied. The results demonstmte that these agent-s produce a decrease in blood pressure, blood flow and vasomotor activity of the mesenteric bed and either no change or a decrease in mesenteric resistance. When the perfusing pressure to the mescnteric artery was maintained at control levels an increase in flow with a concomitant decrease in mesenteric re-sistance followed the injection of the blocking agents. The present report presents experimental evidence that our preliminary report was in error. T HE ability of ganglionic blocking agents such as hexamcthonium, pentolinium and chlorisondamine to lower blood pressure in man and in laboratory animals is well known. Accompanying the fall in blood pressure induced by such agents there is an increase in blood flow to certain areas, 1 • 2 es-pecially in the extremities and usually a re-duction in cardiac output. 1 On the basis of the effectiveness of these agents in interrupting the transmission of impulses across the ganglia of the autonomic nervous system and the rela-tively weak activity of this group of drugs directly upon smooth muscle, it is felt that their cardiovascular effects are due primarily to a reduction in neurogenic control over the peripheral vascular system.' In spite of pe-ripheral vasodilatation in certain areas of the body following the administration of ganglionic blocking agents, total peripheral resistance frequently fails to change significantly from control values. 1 ' 4 Plummer and associates 2From the Department of Pharmacology, Louisiana Stato University, School of Medicine, New Orleans, Supported in part by a grant from the National Heart Institute (grant H-2030) of the National In-stitutes of Health, U. S. Public Health Service and by Ciba Pharmaceutical Products, Inc. A preliminary report of this work was presented at the Fortieth Annual Meeting of the Federation of American Societies for Experimental Biology, Atlan-tic City, N. J., April 1956. licceived for publication July 13, 1956. have reported that the injection of chlori-sondamine, a recently introduced ganglionic blocking agent, in anesthetized dogs produces a decrease in blood pressure without a sig-nificant change in total peripheral resistance. Concomitantly with this fall in blood pressure, resistance in the femoral artery was tempo-rarily decreased and a sustained significant increase in mesenteric resistance occurred. Em-ploying the same method as Plummer and associates, 2 we concluded in a previous re-port 6 that hexamethonium and pentolinium produced an increase in mesenteric resistance. Recently, however, we have obtained experi-mental evidence that the conclusion of our preliminary report was in error. The following report presents this evidence and indicates once again that false conclusions regarding the vasomotor effects of drugs can arise from the failure to appreciate technical errors which may be introduced by the use of flow recorders. In view of this error we have conducted aseries of experiments designed to evaluate more critically the effect of ganglionic blocking agents upon the mesenteric bed of the anes-thetized dog. METHODS
Animals. Fifty-seven mongrel dogs of both sexes weighing 12 to 17 Kg. were employed and were fasted for at least 12 hours prior to each acute experiment. Anesthesia was obtained with either
71S Circulation Rtttarch, Volume IV, Xovtmbrr 19S6 Downloaded from by on September 28, 2025
TRAPOLD 719
Inflow from carotid artery- •i Funnel with heating jacket (38 CJ Pressure recording Cranial mesenteric artery _Maisch metering pomp
Fia. 1. Schematic representation of arrangement employed to perfuse the mesenteric artery in .situ with the aid of a Maisch metering pump. barbital sodium, 250 mg./Kg. intravenously or Dial-urcthanc (Ciba) 0.5 ml./Kg. intravenously.
Recording. All events measured were recorded optically; respiration was recorded with the aid of a modified Anderson respirator}' manometer and blood pressure by Hamilton manometers via polyethylene catheters from the aortic arch and the outflow limb of the flow system. Mesenteric blood flow was measured with the aid of a Shipley-Wilson fotamcter. 8 The outflow tube of the rotamc-ter was connected via a glass cannula inserted into the superior mesenteric artery as close to the origin of this vessel from the abdominal aorta as possible. Inflow to the rotamcter was obtained from 1 of 3 sources, from the superior mesenteric artery via a glass cannula inserted cephalad to the outflow cannula (double cannulation method), from acarotid or femoral artery via a glass cannula of maximal tolerated diameter (single cannulation method), or from a Maisch variable-control metering pump as illustrated (fig. 1.) In this arrangement inflow to the pump was obtained from a carotid artery via a reservoir, the temperature of which was maintained at 38 C. The output of the Maisch metering pump was found to be constant against aconstant peripheral resistance. Clotting was prevented by the intravenous injection of 5 mg./Kg. of heparin sodium supple-mented by injections of 0.5 mg./Kg. at approxi-mately 20 min. intervals. As an additional pre-caution, the rotameter was rinsed with distilled water at 15 to 20 min. intervals during the period of the experiment. Frequent base-line checks were made. The system was calibrated before and after each experiment. Mesenteric resistance was calculated in PRU
= mm. Hg/vil./min.; using the blood pressure values obtained from the outflow end of the rotamc-ter.
Drugs. The following ganglionic blocking agents were employed in this study, chlorisondamine di-mcthochloridc (Ecolid-Ciba), pentolinium tartrate (Ansolyslen-Wycth), and hexamethonium chloride (Hexanieton-Burroughs Wellcome). All drugs were diluted in distilled water and injected intravenously. Doses were calculated as mg./Kg. of body weight unless otherwise stated. RESULTS The effects of the intravenous injections of chlorisondamine, pentolinium and hexameth-onium, upon canine mesenteric blood pressure, blood flow and calculated resistance are pre-sented in table 1. These results represent the averaged values and standard deviations ob-tained in 38 dogs.
Double Mesenteric Cannulation. The intra-venous injection of chlorisondamine, pento-linium or hexamethonium in 17 dogs produced a decrease in systemic and mesenteric pressure and mesenteric flow with a concomittant in-crease in mesenteric resistance. The response to hexamethonium differed from that to chlori-sondamine and pentolinium in that flow and pressure showed a partial return toward con-trol values within 1 hour. Bilateral renal liga-tion prior to the injection of the above blocking agents in 4 additional animals did not ob-servably alter the response of the mesenteric bed from the response of the other animals of this series. Acute bilateral adrenal ligation in 3 animals and the prior injection of 2.5 mg./Kg. of the adrenergic blocking agent phentolamine in 5 dogs greatly reduced, or prevented, the increase in mesenteric resistance induced by ganglionic blocking agents in this series. Re-positioning of the outflow cannula after in-jection of the ganglionic blocking agents fre-quently elevated blood flow toward control values. The most probable explanation of this Downloaded from by on September 28, 2025 720 MESEXTERLC RESISTANCE AFTER GANGLIONIC BLOCKADE TABLE \—The Effect of
Method
])oublc mesenteric cuniuiltition Single moscnteric cimiuilation Perfused in situ Agents C
V
H
c
PCPH
Ganylionic Blocking Agents Upon Blood Flow and Resistance in the Superior Mesenteric Artery
Dose mg./Kg. 0.3
0.3S 1.5 to 5.0 0.3 0.38 0.3 0.38 1.5 No. of dops S5465712Mesenteric pressure mm.Hg(S/D)t ± <r
Control
118±34.5 S7±27.6 92±28.6 65±28.9 127±19.9 OSdblS.l 105±44.7 6S±23.7 S6±7.4 69±6.0 124±26.4{ 93 \
S6±5.5J Post-Drug 70±21.1 47±18.5 57±10.4 33±11.2 70±16.7 50±13.4 57±10.5 30±5.6 42±6.5 25±8.7 114±20.9J 9H 68±1.0t Mesenteric blood flow ml./min. ± <rControl 21.5±17.1 21.7±9.6 9.5±8.4 23 3±12.4 45.3±33.6 43.3±9.6 47.3 54.5±28.0 Post-Drug 9.3±7.9 7.4±4.1 2.6±3.7 13.5±6.3 25.5±17.1 57.8±1S.5 51.2 57.7±32.0 Calculated mesenteric resistance PRU ± <rControl 8.73±7.5 5.12±3.S 20.36±9.7 4.65±1.9 2.45±1.14 2.90±0.54 1.97 2.68±1.68 Post-Drug 13.86±14.0 11.91±11.7 50.77±28.9 3.S2±1.46 1.76±0.S7 2.06±0.40 1.78 1.69±0.92 C = chlorisondumine dimethochloride P ™ pentolinium tartrate H = hexamethonium chloride observation is that the cannula was moved from its proper alignment in the vessel by an increased descent of the diaphgram during inspiration which occurred in most experiments following the injection of the blocking agents.
Single Mesenteric Cannidation. The intra-venous injection of either chlorisondamine or pentolinium in 11 dogs produced a decrease in systemic and mesenteric blood pressure with either no change or a slight decrease in mesen-teric resistance.
Difference in the Resistance of the "Single" ami "Double" Cannula Flow Systems. The difference between the resistance of the "single" and the "double" cannula flow systems per se was evaluated with whole blood maintained at 37 C. over a flow range of 5 to 65 ml./min. Inflow was obtained from and varied by a Maisch metering pump. The resistance of the rubber tubing through which the blood was returned via a reservoir to the pump and into which the blood flowed from the out-flow cannula was maintained as constant as pos-sible with a screw clamp. Inflow and outflow tS/D systolic pressure diastolic pressure | — Menn pressure pressure for both systems was measured with the aid of Hamilton optical manometers. The results of this evaluation demonstrated that the resistance of the double cannulation system was 13 to 20 per cent greater than the resistance of the single cannulation system at flow rates of 40 to 65 ml./min. and 40 to 50 per cent greater at flow rates of 5 to 20 ml./min.
Perfusion by "Maisch" Metering Pump. In view of the reduction of perfusion pressure and flow to the mesenteric bed of the intact animal following the administration of gan-glionic blocking agents, 10 experiments were conducted in which the perfusion pressure and flow was held constant by means of a Maisch metering pump. The effects of the intravenous administration of chlorisondamine, pentolinium and hexamethonium are presented in table 1. All of these agents caused a pro-nounced fall in systemic pressure, a slight fall in mesenteric perfusion pressure and a definite increase in mesenteric blood flow. Calculated mesenteric resistance decreased in each experi-Downloaded from by on September 28, 2025 TRAPOLD 721
PRU ...before G.BA. —after G.B.A.
140 120 100 80 60 40 20 0
Pressure in mm. Hg
Fio. 2. Top, effect of intravenous injection of ganglioriic blocking agents to ancsthotized dogs upon blood pressure-flow relutionship in the superior mesenteric artery. (Avcrago of 6 experiments.) Bot-tom, effect of ganglionic blocking agents upon blood pressure-calculated resistance relationship in the mesenteric artery. Flow was varied by aMaisch variable control metering pump. G.B.A. =• ganglionic blocking agent PRU —Mesenteric blood pressure (mm. Hg)/mesenteric blood now (ml./min.)
ment. The slight decrease following chlori-sondaminc was statistically significant (p <0.02 > 0.01).
Pressurc-Floiu Relationship Before and After Uic Injection of Ganglionic Blocking Agents.
The possibility that the decrease in mesenteric resistance was true only when the perfusion pressure to the mesenteric bed was high led to a study of the effect of ganglionic blocking agents upon the pressure-flow relationship in this vascular lied. Figure 2 illustrates the effect of 0.3 mg./Kg. chlorisondamine upon the pressure-flow relationship in the mesenteric bed of 6 dogs. Mechanically lowering the per-fusion pressure below 60 mm. Hg during the control period produced a progressive increase in mesenteric resistance. The injection of chlorisondamine shifted the pressure-flow curve to the right of the control curve and mechani-cally lowering the perfusion pressure to this area produced a progressive increase in mesen-teric resistance only when this pressure was reduced below 35 mm. of Hg. In 2 additional e x periments, 1 in which 0.25 mg./Kg. of hexamethonium was injected and the other in which 0.35 mg./Kg. of pentolinium was ad-ministered, essentially the same results were obtained. DISCUSSION The results of this study indicate that the intravenous administration of the ganglionic blocking agents chlorisondamine, pentolinium and hexamethonium to the anesthetized dog produces a decrease in calculated mesenteric resistance. The contradiction between the re-sult of this study and former studies 5' 6 is ap-parently due to the relatively high resistance of the flow system employed in earlier studies. In order to evaluate the effect of ganglionic blocking agents upon the vasomotor activity of the mesenteric bed we employed the method suggested by Green and associates 7 and de-termined the effect of these agents upon the pressure-flow relationship in the superior mesenteric artery. These studies demonstrated that the control pressure-flow relationship in this area is similar to that reported for the total systemic vascular bed of the anesthetized dog 8 and vascular areas such as the coronary system, 9 the pulmonary bed 10 and the skin. 7The observation that calculated resistance in the mesenteric bed varies inversely to flow and pressure particularly at pressures below 60 mm. Hg also has been reported for other vascular beds,'" 10 and agrees with the pressure-resistance relationship "uncorrected for yield pressure" reported by Selkurt 11 for the kidney of the dog. Furthermore, our studies demonstrated that the injection of ganglionic blocking agents consistently produced a shift of the mesenteric pressure-flow curve away from the pressure axis thereby indicating that these agents pro-duced a reduction in vasomotor activity 7 in the mesenteric bed. The progressive increase in mesenteric re-sistance which occurred in the control period when the perfusion pressure was mechanically reduced below 60 mm. Hg is consistent with a number of rheologic concepts 7' 12 ' "• ctc - but does not necessarily prove the correctness of any of these proposed explanations. However, for the purpose of discussion, we have chosen Downloaded from by on September 28, 2025 722 MIOSENTKRIC RESISTANCE AFTER GANGLIONIC BLOCKADE to refer to this phenomena as a manifestation of "critical closing pressure" as denned by Burton." After the administration of the ganglionic blocking agents of this study, the effects of "critical closing pressure" upon mesenteric resistance became apparent only after the perfusion pressure to this area had been mechanically reduced to 35 mm. Hg or less. The possibility that these results might have been due to back-pressure from an area of higher pressure, such as from the aorta via collateral circulation, was not supported by the absence of any consistent effect of aortic clamping upon the mesenteric pressure-flow relationship before or after the injection of ganglionic blocking agents. We feel, rather, that the agents tested decreased but did not abolish the "critical closing pressure" of the mesenteric bed by reducing the influence of the autonomic nervous system over this area. This continued existence of a "closing pressure" may be the primary cause of the former findings, 1- 6 pre-viously mentioned, wherein it was concluded that ganglionic blocking agents produce an in-crease rather than a decrease in mesenteric resistance. When the double cannula system is used, it is an effective device for reducing the mesenteric pressure below the "critical closing pressure," thereby leading to a physical increase in mesenteric resistance erroneously attributed to the action of ganglionic blocking agents. The antagonism of this increased re-sistance following either the injection of phentolamine or bilateral adrenal ligation is unexplained, but possibly might be clue to a further reduction of the "critical closing pres-sure" by these procedures. The continued presence of a "critical closing pressure" of blood vessels in even one vascular area, such as the mesenteric bed, after the administration of ganglionic blocking agents is of importance in relation to the effect of these agents upon total peripheral resistance. When the perfusion pressure falls below the "critical closing pres-sure" of such an area, the resistance of the area to blood flow will increase. On the other hand, if perfusion pressure is kept at the control level or increased to an area, such as the mesen-teric bed, a reduction in resistance occurs following ganglionic blockade. These findings are consistent with the ob-servations of Grob and associates, 4 that in patients whose cardiac output increased follow-ing hexamethonium, total peripheral resistance decreased whereas in patients whose output de-creased total peripheral resistance either was unchanged or increased. The effect of ganglionic blocking agents upon peripheral resistance can be readily explained by the following sequence of probable events: A reduction in ncurogenic influence on the arterial system leads to a reduction in "vaso-motor tone" and subsequently a decrease in peripheral resistance; a reduction in neurogenic influence on the venous system leads to areduction in "venomotor tone." Thus, if the pooling of venous blood and the subsequent decrease in venous return is not great enough to materially reduce cardiac output, a sustained decrease in peripheral resistance is obtained; if the reduction in venous return is sufficient to produce a reduction in cardiac output, ar-teriolar-capillary pressure falls below the "critical closing" level of various vascular beds hence total peripheral resistance either remains unchanged or increases. SUMMAHY Intravenous injection of the ganglionic blocking agents chlorisondamine dimetho-chloride, pentolinium tartrate and hexa-methonium chloride to intact, barbiturate anesthetized dogs resulted in a pronounced decrease in aortic pressure, mesenteric pressure and mesenteric blood flow with either no change or a slight decrease in calculated mesenteric resistance. The discrepancy between the results of this study and former studies in which it was re-ported that mesenteric resistance was increased by the administration of ganglionic blocking agents to the dog has been shown to be due to the relatively high resistance of the system used to measure flow in these earlier studies. This observation indicates once again that false conclusions regarding the vasomotor effect of drugs can arise from the failure to appreciate technical errors introduced by use of flow re-corders. Chlorisondamine chloride produced an in-crease in mesenteric blood flow with a con-comitant significant decrease in mesenteric re-Downloaded from by on September 28, 2025 TllAPOLD 723
Histance when the perfusion pressure to this area was held relatively constant with the aid of a Maisch metering pump. Following the injection of ganglionic block-ing agents the mesenteric pressure-flow curve which was demonstrated to be similar to that reported for other vascular areas was shifted away from the pressure axis indicating a de-crease in vasomotor activity. ACKNOWLEDGMENT We wish to express our sincere appreciation to Dr. Robert Heinle of Upjohn Company for the very generous supply of heparin used in this in-vestigation and to Dr. Isidore Cohn for the use of the Maisch metering pump. SuMMAJiio r.\ INTERLINGUA Le injection intravenose de agentes de blocagc ganglionic—dimethochlorido de chlo-risondamina, tartrato de pentolinium, e chlo-rido de hexamethonium—in canes intacte eancsthesiatc per barbituratos resultava in un pronunciate reduction del pression aortic, del pression mesenteric, e del fluxo sanguinee mesenteric, accompaniate per nulle alteration o per un leve reduction del calculate resistentia mesenteric. Le discrepantia inter le resultatos de iste studio e le reportos de previe studios (secundo le quales le resistentia esseva augmentate per le administration de agentes de blocage ganglionic a canes) se ha monstrate como de-bite al relativemente alte resistentia del sys-tenm que ille previe studios empleava pro mesurar le fluxo. Iste observation signala de novo que conclusiones erronee in re le effecto vasomotori de drogas pote resultar del non-rccognition del influentia technic de registra-tores dc fluxo. Chlorido de chlorisondamina produceva un augmento del fluxo sanguinee mesenteric e un concomitante reduction de grados significative in lc resistentia mesenteric quando le pression perfusional a iste area esseva mantenite a un nivello relativemente constanteper medio de un pumpa mesurante de Maisch. Post le injection del agentes de blocage ganglionic, le curva de pression-fluxo mesen-teric—que esseva demonstratemente simile al curva correspondente reportate pro altere areas vascular—manifestava un disviation ab le axe pressional. Isto indicava un reducite activitate vasomotori. REFERENCES 1 FREIS, E. D., ROSK, J. C , PARTENOPE, E. A., HIGGINS, T. F., KELLEY, R. T., SCHNAPER, H. W., AND JOHNSON, R. L.: The hemodynamic effects of hypotensive drugs in man. III. Hexamethonium. J. Clin. Invest. 32: 12S5, 1953. I PLUMMER, A. J., TRAPOLD, J. H., SCHNEIDER, J. A., MAXWELL, R. A., AND EARL, A. E.: Ganglionic blockade by a new bisquatcrnary series, including chlorisondamino dimethochlo-ride. J. Pharmacol. & Exper. Therap. 116:
172, 1955. J PATON, W. D. M.: The paralysis of autonomic ganglia, with special reference to the thera-peutic effects of ganglionic blocking agents. British Med. J. 1: 773, 1951.
G R O B , D., SCARBOROUGH, W. R., KATTUS, A. A.,
J R . , AND LANGFORD, H. G.: Further observa-tions on the effects of autonomic blocking agents in patients with hypertension. I I . Hemody-dynamic, ballistocardiographic and eleotro-cardiographic effects of hexamethonium and pentamethonium. Circulation 8: 352, 1953. 6 TRAPOLD, J. H . : The effect of chlorisondamino dimethochloridc, pentolinium turtrate, and hexamethonium upon mesenteric blood flow in the dog. J. Pharmacol. & Exper. Therap.
116: 5S, 1956. 8 SHIPLEY, R. E., AND WILSON, C : An improved recording rotameter. Proc. Soc. Exper. Biol. & M e d . 78:724, 1951. 7 G R E E N , H. D., L E W I S , R. N., NICKERSON, K. D., AND H E L L E R , A. L.: Blood flow, peripheral resistance and vascular tonus, with observations on the relationship between blood flow and cutaneous temperature. Am. J. Physiol. 141:
51S, 1944. 8 LEVY-, M. N., B R I N D , S. H., BRANDLIN, F. R., AND PHILLIPS, F. A.: The relationship between pressure and flow in the systemic circulation of the dog. Circulation Research 2: 372, 1954. • OSHER, W. J.: Pressure-flow relationship of the coronary system. Am. J. Physiol. 172: 403, 1953. 10 WILLIAMS, M. H., J R . : Relationship between pulmonary artery pressure and blood flow in the dog lung. Am. J. Physiol. 179: 243, 1954. II SELKURT, E. E.: The relation of renal blood flow to effective arterial pressure in the intact kidney of the clog. Am. J. Physiol. 147: 537, 194G. 15 LAMPORT, H . : In J. F. Fulton, A Textbook of Physiology, Philadelphia, W. B. Saunders Co., 1949, chapter 30. 13 BURTON, A. C : On the physical equilibrium of small blood vessels. Am. J. Physiol. 164: 319, 1951. Downloaded from by on September 28, 2025 |
12637 | https://math.stackexchange.com/questions/2532595/simplifying-absolute-value-expression-with-square-roots | inequality - Simplifying absolute value expression with square roots - Mathematics Stack Exchange
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Simplifying absolute value expression with square roots
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I'm attempting to simplify the expression |(|2–√+3–√|−|5–√−7–√|)||(|2+3|−|5−7|)|
Previously I've shown that if 0<a<b 0<a<b then a−−√<b√a<b
Thus we have 5–√−7–√<0 5−7<0 and therefore |5–√−7–√|=7–√−5–√|5−7|=7−5
This means |2–√+3–√|−|5–√−7–√|=2–√+3–√+5–√−7–√|2+3|−|5−7|=2+3+5−7
Now I've clearly simplified the expression, but I want to find out if it is possible to simplify further. As stated about 5–√−7–√<0 5−7<0, and 2–√+3–√>0 2+3>0. Intuitively I somehow see that this whole expression has to be greater than zero, but how do I prove it?
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inequality
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asked Nov 22, 2017 at 16:35
AfflonAfflon
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You seem to be asking two questions: (1) can this be simplified further, and (2) is the expression positive or negative?
No. There isn't really anything that you can do to make this expression simpler. Each radical expression has a different radicand, so there are no like terms to combine, and the radicals themselves are pretty simple.
One possible approach, among many, is as follows:
2–√+3–√+5–√−7–√≥0⟺2–√+3–√+5–√≥7–√⟺(2–√+3–√+5–√)2≥7.(x↦x 2 is increasing on[0,∞))2+3+5−7≥0⟺2+3+5≥7⟺(2+3+5)2≥7.(x↦x 2 is increasing on[0,∞)) Multiplying out the expression on the left, we get (2–√+3–√+5–√)2=2+6–√+10−−√+6–√+3+15−−√+10−−√+15−−√+5=10+2 6–√+2 10−−√+2 15−−√.(2+3+5)2=2+6+10+6+3+15+10+15+5=10+2 6+2 10+2 15. Therefore we have (2–√+3–√+5–√)2=10+2 6–√+2 10−−√+2 15−−√≥0≥10>7,(2+3+5)2=10+2 6+2 10+2 15⏟≥0≥10>7, from which it follows that 2–√+3–√+5–√−7–√>0.2+3+5−7>0.
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answered Nov 22, 2017 at 16:46
Xander Henderson♦Xander Henderson
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Yes you're right, I'm asking whether the expression is positive or negative, so that I know what to do with the absolute value signs. I really like your approach to this problem. First you see what it means that the expressions is greater than or equal to zero, and then you go about proving it. Thanks for your excellent input.Afflon –Afflon 2017-11-22 17:05:31 +00:00 Commented Nov 22, 2017 at 17:05
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When a>0,b>0 a>0,b>0, it is easy to see that a−−√+b√>a+b−−−−√a+b>a+b. Thus we have 2–√+5–√−7–√>0 2+5−7>0. So you can prove something stronger -
2–√+3–√+5–√−7–√>3–√+0=3–√.2+3+5−7>3+0=3.
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edited Nov 22, 2017 at 16:59
answered Nov 22, 2017 at 16:51
Abhiram NatarajanAbhiram Natarajan
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1 This is a great answer (I upvoted), though I am always suspicious when someone says that something is "easy" (one person's "easy" is another's "impossible"). For completeness: a−−√+b√>a+b−−−−√⟺(a−−√+b√)2>a+b.a+b>a+b⟺(a+b)2>a+b. But on the left, we have a+2 a b−−√+b a+2 a b+b, and 2 a b−−√>0 2 a b>0, from which the desired result follows.Xander Henderson –Xander Henderson♦ 2017-11-22 17:07:28 +00:00 Commented Nov 22, 2017 at 17:07
I totally agree with Xander. For me this inequality is not so obvious when looking at the general case, but we could for example consider the fact that 4–√+4–√=2+2=4 4+4=2+2=4 and 8–√=4+4−−−−√<9–√=3 8=4+4<9=3 and 3<4.3<4.Afflon –Afflon 2017-11-22 17:11:54 +00:00 Commented Nov 22, 2017 at 17:11
@Xander, Afflon, I agree that it is not obvious. I myself did what Xander did, i.e. square and then noting that 2 a b−−√>0 2 a b>0. Thanks, my bad!Abhiram Natarajan –Abhiram Natarajan 2017-11-22 17:19:23 +00:00 Commented Nov 22, 2017 at 17:19
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we have
2–√+3–√>0 2+3>0
and since
|a−b|=|b−a||a−b|=|b−a|
our term is
2–√+3–√+5–√−7–√2+3+5−7
and we have
2–√+3–√+5–√>7–√2+3+5>7
all Terms are positive and we get by squaring
2(6–√+15−−√+10−−√)>−3 2(6+15+10)>−3
which is true.
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answered Nov 22, 2017 at 16:45
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
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12638 | https://byjus.com/maths/relation-between-degree-and-radian/ | The amount of revolution made to get the terminal side from the initial side is called the measure of an angle. There are different units for measuring angles. The most commonly used units of measurement of an angle are degree measure and radian measure. The relation between degree and radian measures of angles is used to convert the given measurement of an angle from one unit to another unit. Before understanding the relation between degree measure and radian measure, let us recall the degree and radian measure.
Degree measure
If a rotation from the initial side to terminal side is (1/360)th of a revolution, then the angle is said to have a measure of one degree, written as 1°.
Radian measure
Angle subtended at the centre by an arc of length 1 unit in a unit circle is said to have a measure of 1 radian.
What is The Relation between Degree and Radian?
A complete angle of a circle is 360 degrees of 2π radians. This is the basis for converting the measures of angles from one unit to another. That means a circle subtends an angle whose radian measure is 2π and its degree measure is 360° at the centre. This can be written as:
2π radian = 360°
π radian = 180°
The above relationship allows us to formulate a radian measure in degree measure and a degree measure in radian measure. We know that the approximate value of pi (π) is 22/7. By substituting this value in the above relation, we get;
1 radian = 180°/π = 57° 16′ (approx)
Also,
1° = π/80 radian = 0.01746 radian (approx)
| |
| Read more: Degree measure Radian measure Degrees to Radians Calculator Radians and Degrees Calculator |
The relation between degree measures and radian measures of some standard angles can be observed from the below figure:
The most commonly used conversions from degrees to radians are tabulated below:
| | |
--- |
| Measure of an Angle in Degrees | Measure of an Angle in Radians |
| 0° | 0 |
| 30° | 30° × (π/180°) = π/6 = 0.524 Rad |
| 45° | 45° × (π/180°) = π/4 = 0.785 Rad |
| 60° | 60° × (π/180°) = π/3 = 1.047 Rad |
| 90° | 90° × (π/180°) = π/2 = 1.571 Rad |
| 120° | 120° × (π/180°) = 2π/3 = 2.094 Rad |
| 150° | 150° × (π/180°) = 5π/6 = 2.618 Rad |
| 180° | 180° × (π/180°) = π = 3.14 Rad |
| 210° | 210° × (π/180°) = 7π/6 = 3.665 Rad |
| 240° | 240° × (π/180°) = 4π/3 = 4.188 Rad |
| 270° | 270° × (π/180°) = 3π/2 = 4.713 Rad |
| 300° | 300° × (π/180°) = 5π/3 = 5.235 Rad |
| 330° | 330° × (π/180°) = 11π/6 = 5.764 Rad |
| 360° | 360° × (π/180°) = 2π = 6.283 Rad |
Relation Between Radian Degree and Minutes
The relation among radians, degrees and minutes can be defined from the following:
1 radian = 180°/π = 57° 16′ (approx)
Here,
1° = 60′ (1 degree = 60 minutes)
1′ = 60″ (1 minute = 60 seconds)
That means, one-sixtieth of a degree is called a minute and one-sixtieth of a minute is called a second.
Using this relationship, we can convert any measure of angle to other units of measuring angles.
Solved Examples
Example 1: Convert 135 degrees to radians.
Solution:
135° = 135° × (π/180°)
= 3π/4 rad
Or
= 2.355 rad (approx)
Example 2: Convert 6 radians into the degree measure.
Solution:
We know that,
1 radian = 180°/π
6 radian = 6 × (180°/π)
= (6 × 180°)/(22/7)
= (1080° × 7)/22
= 343 7/11 degrees
This can be further calculated as:
343 7/11 degrees = 343° + [(7 × 60)/11] minute
= 343° + 38′ + (2/11) minute
= 343° + 38′ + 10.9″
= 343° 38′ 11″ (approx)
Therefore, 6 radians = 343° 38′ 11″ (approx).
For more information on different units of measurement of angles and the relationship between them, visit byjus.com today!
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12640 | https://imat.entermedschool.com/t/equilibrium-constant-kc/5372 | Equilibrium Constant, Kc
I understand that changing the coefficients of the reaction affects Kc in a certain way;however, I did not know how to do the math part to find the solution. For example, in part ‘‘a’’ I know that I will write the Kc equation by doing the concentration of products over reactants and O2 will be to the power of 2, SO2 to the power of 4 / SO3^4, but I need help in finding how will this change affect the value of Kc and what should I do mathematically to change the number 278 in a correct way.
Hello,
I dont normally answer chemistry questions because I am traumatized from this subject, so please check if you have the answer key after my work.
These questions relate to manipulation of Kc, if you are interested in revising it. The basic and simple idea behind it is that they give you an initial reaction with its Kc then they ask you to derive the Kc for a number of similar other reactions which are all based on the initial one.
Here is the simple table showing how to manipulate Kc and solve all types of questions:
Basically, in part a, the reaction was doubled as we can infer from the question where all coefficients in the initial reaction became twice as initially. In the manipulation table, this will make Kc squared, so you square 278.
In part b, the reaction was reversed but the coefficients stayed the same, in this case we will inverse the original Kc, so make it 1/278
In part c, the reaction was reversed AND halved. so we take the square root AND make it 1/(whatever answer you get). i think you can also take the answer of part b and put it under square root.
If this was the answer and you found it helpful, consider marking it as the answer. Otherwise, I am not the best in chem but this is to the best of my abilities
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12641 | https://www.sciencedirect.com/science/article/abs/pii/S0096300312009034 | Skip to article
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Abstract
Introduction
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References (10)
Cited by (1)
Applied Mathematics and Computation
Volume 219, Issue 8, 15 December 2012, Pages 3461-3468
Parity symmetry with respect to both and requires periodicity with period : Connections between computer graphics, group theory and spectral methods for solving partial differential equations
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Abstract
A function is symmetric with respect to a point x = L if for all x and similarly is antisymmetric if . A function which is either symmetric or antisymmetric is said to be of definite parity with respect to L. The sines and cosines of a Fourier series have definite parity with respect to two points; all cosines are symmetric with respect to the origin while all sines are antisymmetric with respect to and for integral n are also symmetric with respect to while all other Fourier basis functions are antisymmetric with respect to the same point. Such symmetries can be exploited in numerical calculations; for example, computing the angular Mathieu functions using N basis functions can be split into four uncoupled eigenproblems each of dimension . It is natural to ask: Are there other classes of functions with similar symmetries? Using concepts from computer graphics, we prove that all functions which are symmetric with respect to two points separated by a distance L must be spatially periodic with period . We also prove that the only function which is of definite parity with respect to three distinct points must be a constant. These theorems define parity in the usual sense of a global property such that even parity with respect to the origin means for all . We construct counterexamples to both theorems that are functions with local parity, that is, symmetry which applies only for a finite interval in x.
Introduction
The theorems described here are elementary and do not even require calculus. Nevertheless, they lie in a nexus where computational geometry, group theory and spectral methods for partial differential equations combine to bear great fruit for crystallography, inorganic chemistry and quantum mechanics.
Eugene Wigner won his Nobel Prize largely for explaining the usefulness of group theory to physics both for simplifying numerical calculations and for classifying structure. If the wave function is approximated by a series of spectral basis functions, it is only necessary to retain those basis functions whose symmetry matches the symmetry of the solution. The coefficients of the basis functions whose symmetries are different from those of the crystal lattice or molecule will all be zero.
One of us (Boyd) devoted one chapter of his book on Chebyshev polynomial and Fourier spectral methods to the many uses of group symmetry, particularly parity, which is formally defined as follows:
Definition 1
A function is said to have definite parity with respect to for an arbitrary constant L if and only ifwhere either [even parity] or [odd parity]; in either case, is independent of x.
Although parity is a simple concept, it has great practical importance. By exploiting parity, the global weather forecasting models are able to reduce each associated Legendre summation or interpolation of one large matrixvector multiply to two matrixvector multiplies of half the size and half the combined total cost. Since such Legendre transforms are, even with parity exploitation, 50% of the running time of the model on the half billion dollar Japanese Earth Simulator computer , this simple trick is equivalent to buying two hundred and fifty million dollars of additional computer hardware. This is true even though atmospheric flow is not symmetric with respect to the equator so that the model must calculate both the symmetric and antisymmetric parts and sum them to make the forecast.
As explained in Chapter 7 of , the elements of a Fourier basis all have definite parity with respect to both and as classified in Table 1. This sometimes allows a Fourier numerical computation to be split into four smaller problems with a huge reduction in cost. In addition, double parity symmetry is often very useful for classification.Even parity with respect to means that the functions are symmetric with respect to that point, that is,or equivalently, for the shifted variable .
Similarly, antisymmetry with respect to implies that
The angular Mathieu functions share the fourfold double parity of the Fourier basis. These are the eigensolutions ofwhere is a parameter and is the eigenvalue. These arise when solving the Helmholtz equation in a domain bounded by an ellipse. The Mathieu functions are sufficiently important to merit an entire chapter in the old NBS Handbook of Mathematical Functions , a chapter in its online-and-book replacement, the NIST Handbook of Mathematical Functions , a 280-page book of tables , and monographs , . Recently, Shen and Wang have given Mathieu functions a new life as a spectral basis for ellipse-bounded domains; other recent applications include . Trefethens monograph and Boyds book use Mathieu functions to illustrate Fourier spectral methods , pp. 8889, and , pp. 443447.
Fig. 1 illustrates the lowest four Mathieu eigenfunctions, one of each class. The functions are all symmetric about while the are all of odd parity with respect to the origin; the odd and are all symmetric about while the other eigenfunctions are antisymmetric with respect to this same point.
Morse and Feshbach note wryly, We confine our formulas here to the minimum, though this minimum is not insignificant, owing to the unfortunate propensity of Mathieu formulas for coming in fours, being different for even and odd (about ) and also different depending on whether the functions are periodic in or , , vol. 2, p. 1408. The reason for the fourfold complexity is that each possible double parity class requires its own set of expressions.
In the next section, we ask: are there non-Fourier functions with two parity symmetries? The surprising answer is: No.
The proof, though elementary, is interesting because it exploits ideas from geometry and computer graphics.
In Section 3, we derive the consequences of having parity symmetry with respect to three different points (instead of two).
When parity is relaxed from global to merely local constraints, we show that double symmetry no longer guarantees periodicity. In Section 4, we analyze the relationship between a nonperiodic function, the sum of its Fourier series, and illustrate further this difference between local and global parity.
Section snippets
Implications of double parity
The following lemma is helpful for proving the theorem that follows:
Lemma 1 Alternative definition of parity
The parity conditionis equivalent to
Proof
By the change of variable becomes . ¡
In computer graphics, the usual strategy for creating a complex object is to first draw it in standard size and orientation centered on the origin. Next, apply a dilation to magnify or reduce the object. Lastly, use isometric transformations rotations, reflections and
Implications of parity with respect to three distinct points
Theorem 3 Triple parity implies constancy
Suppose, with L and M as constants and with not a rational number, that is a continuous function which has definite parity with respect to and :Then is a constant.
Proof
Theorem 2 implies bothandBecause of this periodicity, proving is constant on the interval immediately implies constancy for the whole real axis.
Let , and define . . To move this point back
Local and global parity
It may seem a paradox that functions with double parity are required to be periodic. After all, a non-periodic function can always be extended in a convergent Fourier series. The even cosines of the Fourier series would then seem to yield that part of the non-periodic function which is symmetric with respect to both and and so on. The flaw in the reasoning is that the Fourier series of a nonperiodic function converges to a periodic function.
For example, choosing yields the
Summary
The double and triple parity theorems lie in the intersection of three completely different areas of mathematics: computer graphics, group theory and spectral methods for the solution of partial differential equations. The practical value of parity is that it can greatly reduce the cost of numerical calculations even for problems where the solution lacks definite parity. It is therefore unfortunate that global double parity is possible only for periodic functions.
Acknowledgments
This work was supported by the National Science Foundation through Grants OCE 0451951 and ATM 0723440 and by the Undergraduate Research Opportunities Program (UROP) at the University of Michigan. We thank the reviewers for their comments.
References (10)
M. Abramowitz et al.
### Handbook of Mathematical Functions
(1965)
J.P. Boyd, Chebyshev and Fourier Spectral Methods, second ed., Dover, Mineola, New York, 2001, 665...
G. Chen et al.
### Visualization of special eigenmode shapes of a vibrating elliptical membrane
### SIAM Rev.
(1994)
A.N. Lowan and the Staff of the National Applied Mathematics Laboratories, National Bureau of Standards, Tables...
N.W. McLachlan
### Theory and Applications of Mathieu Functions
(1947)
There are more references available in the full text version of this article.
Cited by (1)
The Crane equation uuxx=2: The general explicit solution and a case study of Chebyshev polynomial series for functions with weak endpoint singularities
2017, Applied Mathematics and Computation
The boundary value problem appears in Cranes theory of laminar convection from a point source. We show that the solution is real only when . On this interval, denoting the constants of integration by A and s, the general solution is where the Crane function V is the parameter-free function and erfinv(z) is the inverse of the error function. V(x) is weakly singular at both endpoints; its Chebyshev polynomial coefficients an decrease proportionally to 1/n3. Exponential convergence can be restored by writing where the mapping is . Another option is singular basis functions. has a maximum pointwise error that is less 1/2000 of the maximum of the Crane function.
View full text
Copyright © 2012 Elsevier Inc. All rights reserved. |
12642 | https://math.stackexchange.com/questions/2090641/how-do-i-convert-an-arctangent-into-degrees-by-hand | Stack Exchange Network
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How do I convert an Arctangent into degrees by hand?
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Asked
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0
$\begingroup$
If I have $\arcsin(8/20)$, what angle is this? This is from Wikipedia and the answer is $21.8^{\circ}$. But I do not understand, by hand, how I can get the degrees.
This is from the Wikipedia page:
$8/20 = .4$
$.4$ is the ratio of rise to run, but I don't know what they did after that.
How did that get to $21.8$ degrees (or $.38$ radians)?
trigonometry
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edited Jun 12, 2020 at 10:38
CommunityBot
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asked Jan 9, 2017 at 18:03
johnnyjohnny
36722 gold badges55 silver badges1313 bronze badges
$\endgroup$
7
$\begingroup$ In general you cannot solve for the angle "by hand" unless it is a "nice" angle for which you have memorized its sine or tangent or whatever. For instance, $\arctan(1)=45^\circ$ or $\arctan(\sqrt{3})=60^\circ$. $\endgroup$
kccu
– kccu
2017-01-09 18:07:16 +00:00
Commented Jan 9, 2017 at 18:07
$\begingroup$ Arctangent rather than arcsine. They presumably used a calculator or some tables $\endgroup$
Henry
– Henry
2017-01-09 18:07:30 +00:00
Commented Jan 9, 2017 at 18:07
$\begingroup$ @kccu So they just used a calculator? Is it simply a massive set of steps do get 21.8? $\endgroup$
johnny
– johnny
2017-01-09 18:08:52 +00:00
Commented Jan 9, 2017 at 18:08
$\begingroup$ @Henry Thanks. I fixed it. $\endgroup$
johnny
– johnny
2017-01-09 18:09:21 +00:00
Commented Jan 9, 2017 at 18:09
1
$\begingroup$ @johnny It's not that it's a "massive set of steps," but rather there is no set of steps which would get you to the answer. Unless it is a "nice" value, you just have to use a calculator or a table. $\endgroup$
kccu
– kccu
2017-01-09 18:11:23 +00:00
Commented Jan 9, 2017 at 18:11
| Show 2 more comments
2 Answers 2
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2
$\begingroup$
$$ \arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+..... $$ The point to be noted here is that this series continues on without any end. This simply means that we cannot find the exact value of the $\arctan$ of any angle in radians. Hence, we approximate.
Calculators are able to find an approximated version of the $\arctan$ by using this series. They only consider a few hundred terms of this series.
If you consider more terms, you end up with an answer closer to the actual result.
Note that for small angles the second and the consecutive terms in the series become very small and can be neglected.
In the case of your question : $$ \arctan(0.4) \approx 0.4 - \frac{(0.4)^3}{3} + \frac{(0.4)^5}{5} \ \implies \arctan(0.4) \approx 0.38 $$
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answered Jan 9, 2017 at 18:42
Vishnu V.SVishnu V.S
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Generally, you don't. When $\sin x$ is algebraic $x$ is transcendental (with x represented in radians).
That means that the solution requires working with an infinite series.
One of the easier ones is.
$\arctan x = \sum_\limits{n=0}^{\infty} \frac {(-1)^n x^{2n+1})}{2n+1}$ (With the results in radians.)
You don't usually learn this until calculus. So, if you are in trig and this seems over your head, don't worry.
$\arcsin = \arctan \frac {0.4}{\sqrt {1-0.4^2}} = \arctan \frac {2}{\sqrt {21}}$
$\arctan \frac {2}{\sqrt {21}} \approx \frac {2}{\sqrt {21}} (1-\frac {4}{63}+\frac {16}{2205}\cdots)= 0.412$
$0.38$ radians is $\arctan 0.4 = \arcsin \frac 2{\sqrt{29}}$
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edited Jan 9, 2017 at 21:27
answered Jan 9, 2017 at 18:26
Doug MDoug M
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$\begingroup$ As stated above, the power series for $\tan^{-1}\quad$ yields an angle in radians. My understanding of your question is that you want to know how to get degrees out of radians. If that is your question, just multiply the radians by 180, then divide by $\pi.$ $\endgroup$
Senex Ægypti Parvi
– Senex Ægypti Parvi
2017-01-09 19:46:44 +00:00
Commented Jan 9, 2017 at 19:46
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12643 | https://www.fallbrookhealth.org/files/032c81ebc/Foundation+for+Senior+Care_Adult+Day+Care_The+Club_Application_FY22.23.pdf | FY 2022.2023 Fallbrook Regional Health District Community Health Contract Grant Application Organization Information Legal Name Fallbrook Healthcare Foundation DBA (if Applicable) Foundation for Senior Care Year Founded - use date of incorporation 1979 Program Name/Title Adult Day Care Brief Program Description Our Adult Day Program provides support, caregiving, and socialization for clients who are living with moderate to profound cognitive and medical impairments, as well as support for those who care for them. Caregiver mental and physical respite is also a significant driver for clients to enter our program.
Is this a new (pilot, recently developed) or established program?
Established Program Program Information - Type Ongoing How much funding was received for this program in the previous 2021.2022 CHC Grant cycle?
48168.90 Requested Amount 54214 Organization’s Mission Statement The Foundation for Senior Care’s mission is to provide programs and resources to Greater Fallbrook area seniors and disabled adults, enabling them to enhance their well-being and give them a more meaningful life. Since 2000, we have served seniors and disabled adults in the community through the provision of transportation, an adult day program, technology education, and through hands-on help and referrals for healthcare needs, food resources, housing, state and national entitlements, legal and financial assistance, abuse and neglect interventions, and so much more. The Fallbrook Adult Day Care Center, “The Club” fully supports our mission by providing resources for seniors and the disabled to have an enriched life, provide socialization and allow for respite support for those caregivers who work so hard caring for their family members. Indeed, most of the clients we serve have elderly caregivers, thus our program indirectly supports their health and wellbeing as well.
1 Organization’s Vision Statement Our vision is to be the go-to resource for seniors and the disabled in the communities we serve, acting as a trusted resource to promote health through advocacy and education, provide help and reliable referrals, and overcome obstacles, so that seniors and the disabled have what they need to age safely age in place with access to healthcare, a safe environment, food and housing security, transportation, and social outlets. Our vision, and all of our programs, are very much in alignment with the Fallbrook Regional Health District’s vision to offer and support services and programs that measurably improve physical and mental health, social engagement and increased life span and independence.
Agency Capability Our 501(c)3 agency was established in 1979. The programs we operate today have been successfully serving seniors in this community since the early 2000s, continuing to fill a gap in the health and wellbeing of local seniors, while providing affordable options for low-income families. In 2021, we provided direct services to approximately 1,100 clients through at least one of our interrelated programs.
Our Adult Day Care Program, affectionately known as “The Club” amongst our clients was established in 2000 when community and physician surveys revealed the significant need for caregiving for those with dementia and other cognitive impairments. While the majority of our clients do have some form of dementia, we serve many disabled adults including those with Parkinson’s and traumatic brain injuries. Over the past 5 years, up until the COVID-19 pandemic, our program was providing daytime caregiving for about 25 clients each month and providing critical respite care for 50-75 caregivers each month. Through our Care Van program, we even provide convenient rides to and from our Day Care program. Volunteers assist with activities and socialization in our program.
Because of the generous grants from FRHD, we have been able to offer scholarships for qualifying low-income District seniors and disabled adults, averaging more than 27 days per month of scholarship day care and respite for low-income District residents. That is approximately $28,200 of annual care that these residents directly receive through FRHD funding.
Agency Collaborations Our agency has established working relationships with many community and public offices. Our formal and informal partners include Fallbrook Regional Health District (COVID testing, vaccination and other health-related events), San Diego County Sheriff’s Office - You Are Not Alone (YANA), North County Fire Protection District, Adult Protective Services, Gary and Mary West PACE, North County Parkinson’s Support Group, Fallbrook Food Pantry, Meals on Wheels, In Home Support Services, Traveler’s Aide San Diego, Fallbrook Senior Center, Elder Law & Advocacy, San Diego County Aging and Independent Services, Age Well San Diego, Veteran’s Administration, area hospitals and skilled nursing facilities, many hospice agencies, most local area in-home caregiving agencies, independent and assisted living residential communities, and other medical offices and clinics. Additionally, our “Club” clients are often supported through other nonprofit partners such as the Fallbrook Food Pantry, REINS, and many of our community ancillary healthcare providers (physical therapy, urgent care centers), all of whom we regularly service with Care Van rides for clients.
Recently, we have also became Community Support Partners with Aetna and Health Net to provide housing support (through our Advocacy program) in 2022, with Day Program/Respite services partnership expected to start in 2023. 2 Although we have not yet seen new attendees from this partnership, we have collaborated with the North County Parkinson’s Support Group, now offering a few days of respite scholarship for Parkinson’s caregivers.
Target Population - Age Percent of program participants Children (infants to 12) Young Adults (13-17) Adults (18-60) 4 Seniors (60+) 96 We do not collect this data (indicate with 100%) Gender Percent of program participants Female 55 Male 45 Non-binary Unknown Income Level Percent of program participants Extremely Low-Income Limits, ceiling of $32,100 6 Very Low (50%) Income Limits, ceiling of $53,500 6 Low (80%) Income Limits, ceiling of $85,600 40 Higher Than Listed Limits 48 We do not collect this data (indicate with 100%) Target Population - Income Level Income categories are estimates. We do not gather income data for all of our clients, just for those who apply for scholarships. However, a qualitative review of our clients’ needs includes a high number of Medi-Cal and other low-income support programs. Hence, our target population breakdowns are estimates based on our assessment that over half of our agency's clients fall into HUD low to extremely low-income brackets. Projected number of residents that will directly benefit (participant/client) from this program.
75 3 Social Determinants of Health (SDOH) Program/Services Description - Social Determinants of Health Economic Stability (Employment, Food Insecurity, Housing Instability, Poverty) Social & Community Context (Civic Participation, Discrimination, Incarceration, Social Cohesion) Healthcare Access & Quality (Access to Health Care, Access to Primary Care, Health Literacy) Program/Services Description - FRHD Community Needs Assessment Mental Health (Social Support - Youth or Families) Health (Mobility) Health (Age Related Deficits) Social (Economic Security, Health Literacy, Family/Child Support, Legal/Advocacy) Statement of Need/Problem According to the American Association of Retired Persons (AARP), 87% of people age 65+ reported the desire to remain in their current homes and communities. But when healthcare needs become more complex, and seniors need assistance with the activities of daily living, there are few affordable options. The costs associated with caring for a loved one with dementia include increased medical care costs, decreased family income, and increased emergency medical care utilization for both patients and their caregivers. Consider the health, economic, and social impact of dementia alone. Nearly 100,000 San Diego County residents have been diagnosed with Alzheimer’s or another form of dementia – it is now the third leading cause of death in our County. Safe and trustworthy day programs that provide socialization and daily caregiving is critical. The negative impact of social isolation on people in good health has proven considerable - the impact on those with dementia or other serious illnesses has been devastating. And for those who are caring for them (nearly 250,000 County residents), such temporary respite for their own health and wellbeing is vital. According to the 2019 Alzheimer’s Disease Facts and Figures report, “Compared with caregivers of people without dementia, twice as many caregivers of those with dementia indicate substantial emotional, financial and physical difficulties…. Fifty-nine percent of family caregivers of people with Alzheimer’s or other dementias rated the emotional stress of caregiving as high or very high”. According to caregiver.org, more than 1 in 6 Americans who are providing caregiving must also work. In our Day Program, more than half our clients’ caregivers are also working but able to keep their loved one at home, with our help during the day. The respite our program provides for caregivers is desperately needed. Statement of Need/Problem - Others Adult Day Programs are uniquely suited to provide daytime care at an affordable rate. Unfortunately, there are very few such programs. We are one of two in North San Diego County, the only one north of Encinitas. While there are similar programs in Riverside and Hemet, there are no other similar 4 programs in the Temecula/Murrieta areas. We have families seeking our services from Menifee, Oceanside, and Escondido.
For $85 a day, our Adult Day program provides for all the Activities of Daily Living (ADLs) from 9am – 4pm weekdays, so that informal caregivers can tend to other demands. In-home care costs $27 - $38 per hour, and Seniors Centers do not provide caregiving. Our program provides the social interactions and active engagement that cannot be matched by in-home care. Additionally, through the FRHD grant, we are able to offer scholarships for 1-4 days a week for low-income residents.
Program/Services Description - Program Entry Doctor’s offices, Senior Centers, and other community agencies tell their clients about our program. We also conduct regular marketing and advertising. Most often, a family member will inquire about our program when they are exhausted and need the mental and physical break from caregiving.
As part of our intake process, we conduct an informal written assessment of the client and their caregiver(s) to determine if the level of care and behavioral conduct is a good fit for our program. Formal intake paperwork is then completed, including a physician’s report. We offer the first day free, as an opportunity for us to further assess the client’s needs and for the clients and their caregivers to see whether we meet their needs. After the first day, the client is then enrolled and scheduled, as often as desired. While our services do not require income verification, our records indicate that at least half of those we serve are low-income.
Our Day Care Administrator and staff keep the informal caregivers informed each day, providing updates on engagement, nutrition, movement, toileting, and other factors. Our Administrator often meets with family members to informally discuss caregiving or their own needs. Annually, we complete an Appraisal Needs and Services Plan for each client to determine the goals, changes in caregiving needs, and changes in health. We also conduct surveys regularly for new clients and semi-annually for all clients and caregivers to assess impact.
Program/Services Description - Program Activities Our Adult Day Program prevents isolation, depression, and undue cognitive and physical decline among seniors and disabled adults, as well as their informal caregivers (family, friends, neighbors). We provide a supportive, professionally staffed environment which attends to nutritional, daily living, and social needs of our clients within a group setting during the day, allowing them to continue to live at home rather than being institutionalized or paying high-cost home care. Social engagement is one of the primary reasons that families bring their loved ones to our program. We have seen, time and time again, how the social interactions at “The Club” engage and elevate dementia client’s interactions. Many times, we have seen non-verbal clients begin to sing along or start to talk again during an activity. Recently, we had a long-time non-verbal client who began to speak in her native tongue after one of our Caregivers spoke to her in a Chinese dialect. Through music and songs, games, physical exercise, arts and crafts, and other activities, clients’ minds and bodies are nourished and engaged.
Our program helps seniors living with complex health issues by promoting general health and wellness. For those clients living with dementia or cognitive disabilities, our program supports their well-being by providing a safe and caring environment for them, with staff who know how to effectively handle their needs. We provide nourishing snacks twice daily, physical assistance in transitioning from sitting to standing, toileting assistance, medication distribution, and other health and wellness support. We encourage physical movement daily, through tai-chi stretching, chair yoga, ball tosses, walks in our 5 garden, and other physical exercises. Fifty-eight percent of our clients’ caregivers reported that their loved one’s life has stabilized or improved since they began attending The Club.
Equally as important, the respite our program allows for informal caregivers makes a significant impact in their health and wellness. The Alzheimer’s Association reports that “Nearly 60 percent of Alzheimer’s and dementia caregivers rate the emotional stress of caregiving as high or very high; about 40 percent suffer from depression.” (Approximately 80-90% of our program attendees have a diagnosis of some form of dementia.) The phrase that we hear from nearly every family member who expressed an interest in our facility is, “I’m exhausted and need to find a better way to handle this.” In our own recent client survey, 86% of our client caregivers stated that our program helps reduce the stress of caregiving, and 72% said that they sleep better at night when their loved one attends our program. We support informal caregivers by giving them tips that we successfully use to overcome a challenge with some behavior. We provide information regarding County or other local classes that might be pertinent to their healthcare needs or expressed interests. Our Senior Care Advocates also help our Club clients’ families to overcome a variety of challenges and needs. In the coming weeks, we are seeking to begin a Dementia Support Group and other health-related classes relevant to caregivers and seniors.
Program Goal #1 Grow our Adult Day Program by 20% over last year, to address the specific physical and mental needs of cognitively or physically impaired seniors and disabled adults through socialization and enrichment activities, exercise, and addressing the activities of daily living. Provide scholarship opportunities for low-income district residents who could not otherwise afford the Day Care Program. Provide caregiving relief, support, and educational opportunities for our clients’ informal caregivers.
Program Objectives - Goal #1 Objective 1: During the FRHD grant period, provide 2,500 total days of day care for seniors and disabled persons, including those who have a diagnosis of Alzheimer’s, dementia, or related diseases, or are otherwise cognitively or physically impaired.
Objective 2: During the FRHD grant period, provide at least 360 days of day care scholarships for low-income district residents.
Objective 3: During the FRHD grant period, provide at least 4 educational and/or supportive services that will help informal caregivers in a manner that aims to address their expressed emotional, social, or physical health needs.
Objective 4: During the FRHD grant period, provide support through our Care Advocates for at least 10 of our client’s informal caregivers.
Program Outcomes/Measurables - Goal & Objectives #1 1) Track the number of day care days provided and the number of clients attending 2) Track the number of scholarship days provided and the number of scholarship clients attending 3) Measure the number of educational and/or support services we make available and the attendance at the events or support sessions 4) Measure the number of Day Care clients’ families that have received some form of documented 6 support from our Care Advocates team Anticipated Acknowledgment Anticipated Acknowledgment Social Media Postings Signage at Service Sites Print Materials to Service Recipients Website Display Other Anticipated Acknowledgment - The FRHD logo will be affixed to the sides of our Care Vans.
- We will include the FRHD logo and official sponsor designation in company-wide emails - Our organization brochure will feature the FRHD logo and official sponsor language - FRHD logo and sponsor status will appear on our website and in promotional event emails - FRHD logo and support will be highlighted at our annual fundraiser - We will promote District events every month via the following social media platforms: Facebook, Instagram.
- Our staff email signature lines will include the FRHD logo and grant supporter statement 7 Grantor Program/Project Amount Requested Angel Society Day Program/Plumbing repair and flooring $ 10,000.00 Legacy - Elizabeth Wilson Grant Day Program $ 8,000.00 The San Diego Foundation Age Friendly Communities 2021 for all 4 programs (combined): Advocacy Door Through Door Transportation Adult Day Program $ 30,000.00 SUBMITTED 202 Amount Awarded Date Submitted Status Month/Year of Funding $ 8,000.00 10/19/21 Approved Nov-21 $ 8,000.00 07/29/21 Approved Nov-21 $ - 8.12.21 Declined Oct-21 21 Board of Directors 2021 Officers Roger Shaver President rshaver@email.com o Pharmacist - U S Navy (Retired) o Past Pharmacy Director Menifee Valley Medical Center o Past Pharmacy Director Fallbrook Hospital o Life Member Military Officers Association of America o Life Member Veterans of Foreign Wars o Life Member and Past Commander Disabled American Veterans Chapter 95 Oceanside CA o Sea West Federal Credit Union (formerly on Board of Directors) o Menifee Valley Medical Center Foundation (formerly on Board of Directors) Mike McReynolds Past President mikem@mbarccarports.com o Founder of M Bar C Carports, Inc. o Served on the Board of Public Office, Vista Fire Protection District o Fraternal-Past Master, Culver City- Foshay Masonic Lodge #467 o Youth Leadership- Honorary American Degree, National Future Farmers of America Mark Haskell 1st Vice President haskclan@roadrunner.com o Member of Christ the King Lutheran Church o Served as an Officer/ Member of Orange County Fire Marshal, Metro Cities Fire Dispatch, Central Net Operations Authority Carlos Perez 2nd Vice President carlosperez@AQhomecare.com o Owner/Operator, Affordable & Quality Home Care Barbara Creech Secretary Bjcreech03@gmail.com o Operations Director (Retired) Herb Baker Treasurer herb@herbbaker.com o Operations and Controller (Retired) o Served on the Board of University of San Diego, College for Men o Fallbrook Rotary Club o Eucharistic Minister, San Rafael’s Catholic Church Other Board Members Cecilia Brown Cmbrownplus@yahoo.com o Medicare Broker Sarah Eckhardt Nordicprincess74@yahoo.com o Licensed Vocational Nurse o Owner/Operator, Angels Among Us, Vista CA Gail Jones gjonesjag@gmail.com o Owner of Alvarado Veterinary Hospital in Fallbrook. o Member of Fallbrook Rotary Club o Treasurer of St. John’s Episcopal Church Robert Pace Espirit1@roadrunner.com o MD, Orthopedic Surgeon (Retired) Lougene Williams lougenewilliams@sbcglobal.net o Senior Manufacturing Executive (Retired) o Served on Personnel Committee at church o SCORE Volunteer OMB No. 1545-0047 Form 990 Return of Organization Exempt From Income Tax 2020 Under section 501(c), 527, or 4947(a)(1) of the Internal Revenue Code (except private foundations) Open to Public G Do not enter social security numbers on this form as it may be made public.
Department of the Treasury Inspection Internal Revenue Service G Go to www.irs.gov/Form990 for instructions and the latest information.
A For the 2020 calendar year, or tax year beginning , 2020, and ending , 20 Employer identification number C D Check if applicable: B Address change Telephone number E Name change Initial return Final return/terminated $ Amended return Gross receipts G Is this a group return for subordinates?
H(a) Name and address of principal officer: F Application pending Yes No H(b) Are all subordinates included?
Yes No If "No," attach a list. See instructions H ( ) Tax-exempt status: 501(c)(3) 501(c) (insert no.) 4947(a)(1) or 527 I Group exemption number J Website: G H(c) G G Form of organization: Corporation Trust Association Other Year of formation: State of legal domicile: K L M Part I Summary Briefly describe the organization's mission or most significant activities: 1 if the organization discontinued its operations or disposed of more than 25% of its net assets.
Check this box G 2 Number of voting members of the governing body (Part VI, line 1a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 Number of independent voting members of the governing body (Part VI, line 1b). . . . . . . . . . . . . . . . . . . . . . . 4 4 Total number of individuals employed in calendar year 2020 (Part V, line 2a) . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 Total number of volunteers (estimate if necessary). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 6 Total unrelated business revenue from Part VIII, column (C), line 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7a 7a Net unrelated business taxable income from Form 990-T, Part I, line 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b 7b Prior Year Current Year Contributions and grants (Part VIII, line 1h). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Program service revenue (Part VIII, line 2g) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Investment income (Part VIII, column (A), lines 3, 4, and 7d) . . . . . . . . . . . . . . . . . . . . . . . . . 10 Other revenue (Part VIII, column (A), lines 5, 6d, 8c, 9c, 10c, and 11e). . . . . . . . . . . . . . . . 11 Total revenue ' add lines 8 through 11 (must equal Part VIII, column (A), line 12) . . . . . 12 Grants and similar amounts paid (Part IX, column (A), lines 1-3). . . . . . . . . . . . . . . . . . . . . . 13 Benefits paid to or for members (Part IX, column (A), line 4) . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Salaries, other compensation, employee benefits (Part IX, column (A), lines 5-10) . . . . . . 15 Professional fundraising fees (Part IX, column (A), line 11e). . . . . . . . . . . . . . . . . . . . . . . . . . 16a Total fundraising expenses (Part IX, column (D), line 25) G b Other expenses (Part IX, column (A), lines 11a-11d, 11f-24e). . . . . . . . . . . . . . . . . . . . . . . . . 17 Total expenses. Add lines 13-17 (must equal Part IX, column (A), line 25). . . . . . . . . . . . . 18 Revenue less expenses. Subtract line 18 from line 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 End of Year Beginning of Current Year Total assets (Part X, line 16) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Total liabilities (Part X, line 26) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Net assets or fund balances. Subtract line 21 from line 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Part II Signature Block Under penalties of perjury, I declare that I have examined this return, including accompanying schedules and statements, and to the best of my knowledge and belief, it is true, correct, and complete. Declaration of preparer (other than officer) is based on all information of which preparer has any knowledge.
A Signature of officer Date Sign Here A Type or print name and title Print/Type preparer's name Preparer's signature Date PTIN Check if self-employed Paid G Firm's name Preparer G Use Only Firm's EIN G Firm's address Phone no.
May the IRS discuss this return with the preparer shown above? See instructions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Yes No TEEA0101L 01/19/21 BAA For Paperwork Reduction Act Notice, see the separate instructions.
Form 990 (2020) Fallbrook Healthcare Foundation Inc.
PO Box 2155 Fallbrook, CA 92088 95-3389263 760-723-7570 X 1,241,667.
1,015,244.
60,574.
52,334.
1,302,241.
1,067,578.
179,711.
50,066.
1,005,428.
449,203.
378,187.
156,809.
164,810.
627,241.
292,394.
1,185,139.
499,269.
41,713.
97,105.
8,077.
8,846.
87,369.
80,384.
1,047,980.
312,934.
0.
0.
40 17 13 13 CA X 2,028,672.
President Roger Shaver X X Its purpose is to provide programs and resources enabling seniors to enhance their well-being and give them a more meaningful life. The geographic areas served include Fallbrook, Bonsall, Rainbow, DeLuz, southwest Temecula and areas of North County San Diego.
Phillip Howerzyl, CPA, CGMA P01363785 VanderSpek Howerzyl, CPAs 95-2770263 350 West Fifth Ave., Suite 300 (760)741-2659 Escondido, CA 92025 Same As C Above Phillip Howerzyl, CPA, CGMA Form 990 (2020) Page 2 Part III Statement of Program Service Accomplishments Check if Schedule O contains a response or note to any line in this Part III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Briefly describe the organization's mission: 1 Did the organization undertake any significant program services during the year which were not listed on the prior 2 Form 990 or 990-EZ?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Yes No If "Yes," describe these new services on Schedule O.
Did the organization cease conducting, or make significant changes in how it conducts, any program services?. . . . 3 Yes No If "Yes," describe these changes on Schedule O.
4 Describe the organization's program service accomplishments for each of its three largest program services, as measured by expenses.
Section 501(c)(3) and 501(c)(4) organizations are required to report the amount of grants and allocations to others, the total expenses, and revenue, if any, for each program service reported.
$ $ $ (Code: ) (Expenses including grants of ) (Revenue ) 4 a $ $ $ (Code: ) (Expenses including grants of ) (Revenue ) 4 b $ $ $ (Code: ) (Expenses including grants of ) (Revenue ) 4 c Other program services (Describe on Schedule O.) 4 d $ $ $ (Expenses including grants of ) (Revenue ) 4 e Total program service expenses G Form 990 (2020) TEEA0102L 10/07/20 BAA 549,974.
235,124.
216,660.
314,850.
217,892.
170,721.
408,895.
X X 95-3389263 Fallbrook Healthcare Foundation Inc.
X Assisting the elderly and their families with a varying degree of services.
Provide shuttle services for the elderly and their families to assure that they can make health care appointments.
See Schedule O Form 990 (2020) Page 3 Part IV Checklist of Required Schedules Yes No Is the organization described in section 501(c)(3) or 4947(a)(1) (other than a private foundation)? If 'Yes,' complete 1 Schedule A. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Is the organization required to complete Schedule B, Schedule of Contributors See instructions?
. . . . . . . . . . . . . . . . . . . . . . . 2 2 Did the organization engage in direct or indirect political campaign activities on behalf of or in opposition to candidates 3 for public office? If 'Yes,' complete Schedule C, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 4 Section 501(c)(3) organizations. Did the organization engage in lobbying activities, or have a section 501(h) election in effect during the tax year? If 'Yes,' complete Schedule C, Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Is the organization a section 501(c)(4), 501(c)(5), or 501(c)(6) organization that receives membership dues, 5 assessments, or similar amounts as defined in Revenue Procedure 98-19? If 'Yes,' complete Schedule C, Part III. . . . . . . 5 Did the organization maintain any donor advised funds or any similar funds or accounts for which donors have the right 6 to provide advice on the distribution or investment of amounts in such funds or accounts? If 'Yes,' complete Schedule D, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Did the organization receive or hold a conservation easement, including easements to preserve open space, the 7 environment, historic land areas, or historic structures? If 'Yes,' complete Schedule D, Part II . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Did the organization maintain collections of works of art, historical treasures, or other similar assets? If 'Yes,' 8 complete Schedule D, Part III. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Did the organization report an amount in Part X, line 21, for escrow or custodial account liability, serve as a custodian 9 for amounts not listed in Part X; or provide credit counseling, debt management, credit repair, or debt negotiation services? If 'Yes,' complete Schedule D, Part IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Did the organization, directly or through a related organization, hold assets in donor-restricted endowments 10 or in quasi endowments? If 'Yes,' complete Schedule D, Part V. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 If the organization's answer to any of the following questions is 'Yes', then complete Schedule D, Parts VI, VII, VIII, IX, 11 or X as applicable.
Did the organization report an amount for land, buildings, and equipment in Part X, line 10? If 'Yes,' complete Schedule a D, Part VI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11a Did the organization report an amount for investments ' other securities in Part X, line 12, that is 5% or more of its total b assets reported in Part X, line 16? If 'Yes,' complete Schedule D, Part VII. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11b Did the organization report an amount for investments ' program related in Part X, line 13, that is 5% or more of its total c assets reported in Part X, line 16? If 'Yes,' complete Schedule D, Part VIII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11c Did the organization report an amount for other assets in Part X, line 15, that is 5% or more of its total assets reported d in Part X, line 16? If 'Yes,' complete Schedule D, Part IX. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11d Did the organization report an amount for other liabilities in Part X, line 25? If 'Yes,' complete Schedule D, Part X. . . . . . e 11e Did the organization's separate or consolidated financial statements for the tax year include a footnote that addresses f the organization's liability for uncertain tax positions under FIN 48 (ASC 740)? If 'Yes,' complete Schedule D, Part X. . . . 11f Did the organization obtain separate, independent audited financial statements for the tax year? If 'Yes,' complete 12a Schedule D, Parts XI and XII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12a Was the organization included in consolidated, independent audited financial statements for the tax year? If 'Yes,' and b if the organization answered 'No' to line 12a, then completing Schedule D, Parts XI and XII is optional. . . . . . . . . . . . . . . . . 12b Is the organization a school described in section 170(b)(1)(A)(ii)? If 'Yes,' complete Schedule E . . . . . . . . . . . . . . . . . . . . . . . 13 13 Did the organization maintain an office, employees, or agents outside of the United States?. . . . . . . . . . . . . . . . . . . . . . . . . . . 14a 14a Did the organization have aggregate revenues or expenses of more than $10,000 from grantmaking, fundraising, b business, investment, and program service activities outside the United States, or aggregate foreign investments valued at $100,000 or more? If 'Yes,' complete Schedule F, Parts I and IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14b Did the organization report on Part IX, column (A), line 3, more than $5,000 of grants or other assistance to or for any 15 foreign organization? If 'Yes,' complete Schedule F, Parts II and IV. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Did the organization report on Part IX, column (A), line 3, more than $5,000 of aggregate grants or other assistance to 16 or for foreign individuals? If 'Yes,' complete Schedule F, Parts III and IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Did the organization report a total of more than $15,000 of expenses for professional fundraising services on Part IX, 17 column (A), lines 6 and 11e? If 'Yes,' complete Schedule G, Part I See instructions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Did the organization report more than $15,000 total of fundraising event gross income and contributions on Part VIII, 18 lines 1c and 8a? If 'Yes,' complete Schedule G, Part II. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Did the organization report more than $15,000 of gross income from gaming activities on Part VIII, line 9a? If 'Yes,' 19 complete Schedule G, Part III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 20a 20a Did the organization operate one or more hospital facilities? If 'Yes,' complete Schedule H . . . . . . . . . . . . . . . . . . . . . . . . . . . . If 'Yes' to line 20a, did the organization attach a copy of its audited financial statements to this return? . . . . . . . . . . . . . . . . b 20b Did the organization report more than $5,000 of grants or other assistance to any domestic organization or 21 domestic government on Part IX, column (A), line 1? If 'Yes,' complete Schedule I, Parts I and II. . . . . . . . . . . . . . . . . . . . . . 21 TEEA0103L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 X X X X X X X X X X X X X X X X X X X X X X X X X X X X Form 990 (2020) Page 4 Part IV Checklist of Required Schedules (continued) Yes No Did the organization report more than $5,000 of grants or other assistance to or for domestic individuals on Part IX, 22 column (A), line 2? If 'Yes,' complete Schedule I, Parts I and III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Did the organization answer 'Yes' to Part VII, Section A, line 3, 4, or 5 about compensation of the organization's current 23 and former officers, directors, trustees, key employees, and highest compensated employees? If 'Yes,' complete Schedule J . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Did the organization have a tax-exempt bond issue with an outstanding principal amount of more than $100,000 as of 24a the last day of the year, that was issued after December 31, 2002? If 'Yes,' answer lines 24b through 24d and complete Schedule K. If 'No, 'go to line 25a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24a Did the organization invest any proceeds of tax-exempt bonds beyond a temporary period exception?. . . . . . . . . . . . . . . . . . b 24b Did the organization maintain an escrow account other than a refunding escrow at any time during the year to defease c any tax-exempt bonds? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24c Did the organization act as an 'on behalf of' issuer for bonds outstanding at any time during the year? . . . . . . . . . . . . . . . . . d 24d 25a Section 501(c)(3), 501(c)(4), and 501(c)(29) organizations. Did the organization engage in an excess benefit 25a transaction with a disqualified person during the year? If 'Yes,' complete Schedule L, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . Is the organization aware that it engaged in an excess benefit transaction with a disqualified person in a prior year, and b that the transaction has not been reported on any of the organization's prior Forms 990 or 990-EZ? If 'Yes,' complete Schedule L, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25b Did the organization report any amount on Part X, line 5 or 22, for receivables from or payables to any current or 26 former officer, director, trustee, key employee, creator or founder, substantial contributor, or 35% controlled entity or family member of any of these persons? If 'Yes,' complete Schedule L, Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Did the organization provide a grant or other assistance to any current or former officer, director, trustee, key 27 employee, creator or founder, substantial contributor or employee thereof, a grant selection committee member, or to a 35% controlled entity (including an employee thereof) or family member of any of these 27 persons? If 'Yes,' complete Schedule L, Part III. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Was the organization a party to a business transaction with one of the following parties (see Schedule L, Part IV 28 instructions, for applicable filing thresholds, conditions, and exceptions): A current or former officer, director, trustee, key employee, creator or founder, or substantial contributor? If a 28a 'Yes,' complete Schedule L, Part IV. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A family member of any individual described in line 28a? If 'Yes,' complete Schedule L, Part IV . . . . . . . . . . . . . . . . . . . . . . . . b 28b A 35% controlled entity of one or more individuals and/or organizations described in lines 28a or 28b? If c Yes,' complete Schedule L, Part IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28c Did the organization receive more than $25,000 in non-cash contributions? If 'Yes,' complete Schedule M . . . . . . . . . . . . . . 29 29 Did the organization receive contributions of art, historical treasures, or other similar assets, or qualified conservation 30 contributions? If 'Yes,' complete Schedule M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Did the organization liquidate, terminate, or dissolve and cease operations? If 'Yes,' complete Schedule N, Part I. . . . . . . 31 31 Did the organization sell, exchange, dispose of, or transfer more than 25% of its net assets? If 'Yes,' complete 32 Schedule N, Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Did the organization own 100% of an entity disregarded as separate from the organization under Regulations sections 33 301.7701-2 and 301.7701-3? If 'Yes,' complete Schedule R, Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Was the organization related to any tax-exempt or taxable entity? If 'Yes,' complete Schedule R, Part II, III, or IV, 34 and Part V, line 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Did the organization have a controlled entity within the meaning of section 512(b)(13)?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35a 35a If 'Yes' to line 35a, did the organization receive any payment from or engage in any transaction with a controlled b entity within the meaning of section 512(b)(13)? If 'Yes,' complete Schedule R, Part V, line 2. . . . . . . . . . . . . . . . . . . . . . . . . . 35b 36 Section 501(c)(3) organizations. Did the organization make any transfers to an exempt non-charitable related 36 organization? If 'Yes,' complete Schedule R, Part V, line 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Did the organization conduct more than 5% of its activities through an entity that is not a related organization and that is 37 treated as a partnership for federal income tax purposes? If 'Yes,' complete Schedule R, Part VI . . . . . . . . . . . . . . . . . . . . . . 37 Did the organization complete Schedule O and provide explanations in Schedule O for Part VI, lines 11b and 19?
38 Note: All Form 990 filers are required to complete Schedule O. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Part V Statements Regarding Other IRS Filings and Tax Compliance Check if Schedule O contains a response or note to any line in this Part V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Yes No Enter the number reported in Box 3 of Form 1096. Enter -0- if not applicable. . . . . . . . . . . . . . 1 a 1 a Enter the number of Forms W-2G included in line 1a. Enter -0- if not applicable . . . . . . . . . . . . b 1 b Did the organization comply with backup withholding rules for reportable payments to vendors and reportable gaming c (gambling) winnings to prize winners?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 c TEEA0104L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 X X X X X X X X X X X X X X X X X X X X 7 0 X Form 990 (2020) Page 5 Part V Statements Regarding Other IRS Filings and Tax Compliance (continued) Yes No Enter the number of employees reported on Form W-3, Transmittal of Wage and Tax State-2 a ments, filed for the calendar year ending with or within the year covered by this return . . . . . 2 a If at least one is reported on line 2a, did the organization file all required federal employment tax returns? . . . . . . . . . . . . . b 2 b Note: If the sum of lines 1a and 2a is greater than 250, you may be required to e-file (see instructions) Did the organization have unrelated business gross income of $1,000 or more during the year?. . . . . . . . . . . . . . . . . . . . . . . . 3 a 3 a If 'Yes,' has it filed a Form 990-T for this year? If 'No' to line 3b, provide an explanation on Schedule O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b 3 b At any time during the calendar year, did the organization have an interest in, or a signature or other authority over, a 4 a financial account in a foreign country (such as a bank account, securities account, or other financial account)? . . . . . . . . . 4 a If 'Yes,' enter the name of the foreign countryG b See instructions for filing requirements for FinCEN Form 114, Report of Foreign Bank and Financial Accounts (FBAR).
Was the organization a party to a prohibited tax shelter transaction at any time during the tax year? . . . . . . . . . . . . . . . . . . . 5 a 5 a Did any taxable party notify the organization that it was or is a party to a prohibited tax shelter transaction?. . . . . . . . . . . . b 5 b If 'Yes,' to line 5a or 5b, did the organization file Form 8886-T?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c 5 c Does the organization have annual gross receipts that are normally greater than $100,000, and did the organization 6 a solicit any contributions that were not tax deductible as charitable contributions?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 a If 'Yes,' did the organization include with every solicitation an express statement that such contributions or gifts were b not tax deductible?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 b 7 Organizations that may receive deductible contributions under section 170(c).
Did the organization receive a payment in excess of $75 made partly as a contribution and partly for goods and a services provided to the payor?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 a If 'Yes,' did the organization notify the donor of the value of the goods or services provided? . . . . . . . . . . . . . . . . . . . . . . . . . . b 7 b Did the organization sell, exchange, or otherwise dispose of tangible personal property for which it was required to file c Form 8282? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 c If 'Yes,' indicate the number of Forms 8282 filed during the year. . . . . . . . . . . . . . . . . . . . . . . . . . d 7 d Did the organization receive any funds, directly or indirectly, to pay premiums on a personal benefit contract?. . . . . . . . . . e 7 e Did the organization, during the year, pay premiums, directly or indirectly, on a personal benefit contract?
. . . . . . . . . . . . . . f 7 f If the organization received a contribution of qualified intellectual property, did the organization file Form 8899 g as required?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 g If the organization received a contribution of cars, boats, airplanes, or other vehicles, did the organization file a h Form 1098-C? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 h 8 Sponsoring organizations maintaining donor advised funds. Did a donor advised fund maintained by the sponsoring organization have excess business holdings at any time during the year?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 9 Sponsoring organizations maintaining donor advised funds.
Did the sponsoring organization make any taxable distributions under section 4966? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a 9 a Did the sponsoring organization make a distribution to a donor, donor advisor, or related person?
. . . . . . . . . . . . . . . . . . . . . . b 9 b 10 Section 501(c)(7) organizations. Enter: Initiation fees and capital contributions included on Part VIII, line 12 . . . . . . . . . . . . . . . . . . . . . . a 10a Gross receipts, included on Form 990, Part VIII, line 12, for public use of club facilities. . . . . b 10b 11 Section 501(c)(12) organizations. Enter: Gross income from members or shareholders. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a 11a Gross income from other sources (Do not net amounts due or paid to other sources b against amounts due or received from them.). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11b 12a Section 4947(a)(1) non-exempt charitable trusts. Is the organization filing Form 990 in lieu of Form 1041?
. . . . . . . . . . . . . . 12a If 'Yes,' enter the amount of tax-exempt interest received or accrued during the year. . . . . . . b 12b 13 Section 501(c)(29) qualified nonprofit health insurance issuers.
Is the organization licensed to issue qualified health plans in more than one state? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a 13a Note: See the instructions for additional information the organization must report on Schedule O.
Enter the amount of reserves the organization is required to maintain by the states in b which the organization is licensed to issue qualified health plans. . . . . . . . . . . . . . . . . . . . . . . . . . 13b Enter the amount of reserves on hand . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c 13c Did the organization receive any payments for indoor tanning services during the tax year?. . . . . . . . . . . . . . . . . . . . . . . . . . . . 14a 14a If 'Yes,' has it filed a Form 720 to report these payments? If 'No,' provide an explanation on Schedule O. . . . . . . . . . . . . . . b 14b 15 Is the organization subject to the section 4960 tax on payment(s) of more than $1,000,000 in remuneration or 15 excess parachute payment(s) during the year?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . If 'Yes,' see instructions and file Form 4720, Schedule N.
16 Is the organization an educational institution subject to the section 4968 excise tax on net investment income?. . . . . . . . . 16 If 'Yes,' complete Form 4720, Schedule O.
TEEA0105L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 X X X X X X X X X X 17 X X X Form 990 (2020) Page 6 Part VI Governance, Management, and Disclosure For each 'Yes' response to lines 2 through 7b below, and for a 'No' response to line 8a, 8b, or 10b below, describe the circumstances, processes, or changes on Schedule O. See instructions.
Check if Schedule O contains a response or note to any line in this Part VI. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Section A. Governing Body and Management Yes No Enter the number of voting members of the governing body at the end of the tax year. . . . . . 1 a 1 a If there are material differences in voting rights among members of the governing body, or if the governing body delegated broad authority to an executive committee or similar committee, explain on Schedule O.
Enter the number of voting members included on line 1a, above, who are independent. . . . . b 1 b Did any officer, director, trustee, or key employee have a family relationship or a business relationship with any other 2 officer, director, trustee, or key employee? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Did the organization delegate control over management duties customarily performed by or under the direct supervision 3 of officers, directors, trustees, or key employees to a management company or other person?. . . . . . . . . . . . . . . . . . . . . . . . . 3 Did the organization make any significant changes to its governing documents 4 since the prior Form 990 was filed? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Did the organization become aware during the year of a significant diversion of the organization's assets?
. . . . . . . . . . . . . . 5 5 Did the organization have members or stockholders?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 6 Did the organization have members, stockholders, or other persons who had the power to elect or appoint one or more 7 a members of the governing body? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 a Are any governance decisions of the organization reserved to (or subject to approval by) members, b stockholders, or persons other than the governing body?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 b Did the organization contemporaneously document the meetings held or written actions undertaken during the year by 8 the following: The governing body?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a 8 a Each committee with authority to act on behalf of the governing body?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b 8 b Is there any officer, director, trustee, or key employee listed in Part VII, Section A, who cannot be reached at the 9 organization's mailing address? If 'Yes,' provide the names and addresses on Schedule O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Section B. Policies (This Section B requests information about policies not required by the Internal Revenue Code.) Yes No Did the organization have local chapters, branches, or affiliates?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10a 10a If 'Yes,' did the organization have written policies and procedures governing the activities of such chapters, affiliates, and branches to ensure their b operations are consistent with the organization's exempt purposes? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10b Has the organization provided a complete copy of this Form 990 to all members of its governing body before filing the form?. . . . . . . . . . . . . . . . . . . . . . 11a 11a Describe in Schedule O the process, if any, used by the organization to review this Form 990.
b Did the organization have a written conflict of interest policy? If 'No,' go to line 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12a 12a Were officers, directors, or trustees, and key employees required to disclose annually interests that could give rise b to conflicts? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12b Did the organization regularly and consistently monitor and enforce compliance with the policy? If 'Yes,' describe in c Schedule O how this was done . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12c Did the organization have a written whistleblower policy?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 13 Did the organization have a written document retention and destruction policy?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 14 Did the process for determining compensation of the following persons include a review and approval by independent 15 persons, comparability data, and contemporaneous substantiation of the deliberation and decision?
The organization's CEO, Executive Director, or top management official. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a 15a Other officers or key employees of the organization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b 15b If 'Yes' to line 15a or 15b, describe the process in Schedule O (see instructions).
Did the organization invest in, contribute assets to, or participate in a joint venture or similar arrangement with a 16a taxable entity during the year?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16a If 'Yes,' did the organization follow a written policy or procedure requiring the organization to evaluate its b participation in joint venture arrangements under applicable federal tax law, and take steps to safeguard the organization's exempt status with respect to such arrangements?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16b Section C. Disclosure List the states with which a copy of this Form 990 is required to be filed G 17 Section 6104 requires an organization to make its Forms 1023 (1024 or 1024-A, if applicable), 990, and 990-T (Section 501(c)(3)s only) 18 available for public inspection. Indicate how you made these available. Check all that apply.
Other (explain on Schedule O) Own website Another's website Upon request Describe on Schedule O whether (and if so, how) the organization made its governing documents, conflict of interest policy, and financial statements available to 19 the public during the tax year.
State the name, address, and telephone number of the person who possesses the organization's books and records G 20 TEEA0106L 10/07/20 BAA Form 990 (2020) 95-3389263 Fallbrook Healthcare Foundation Inc.
Perla Hurtado 135 S Mission Road Fallbrook CA 92028 (760) 723-7570 X X X X X X X X X X X X X X X X X X X X X X 13 13 X X CA See Schedule O See Schedule O See Schedule O See Schedule O See Sch. O Form 990 (2020) Page 7 Part VII Compensation of Officers, Directors, Trustees, Key Employees, Highest Compensated Employees, and Independent Contractors Check if Schedule O contains a response or note to any line in this Part VII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Section A. Officers, Directors, Trustees, Key Employees, and Highest Compensated Employees 1 a Complete this table for all persons required to be listed. Report compensation for the calendar year ending with or within the organization's tax year.
? List all of the organization's current officers, directors, trustees (whether individuals or organizations), regardless of amount of compensation. Enter -0- in columns (D), (E), and (F) if no compensation was paid.
? List all of the organization's current key employees, if any. See instructions for definition of 'key employee.' ? List the organization's five current highest compensated employees (other than an officer, director, trustee, or key employee) who received reportable compensation (Box 5 of Form W-2 and/or Box 7 of Form 1099-MISC) of more than $100,000 from the organization and any related organizations.
? List all of the organization's former officers, key employees, and highest compensated employees who received more than $100,000 of reportable compensation from the organization and any related organizations.
? List all of the organization's former directors or trustees that received, in the capacity as a former director or trustee of the organization, more than $10,000 of reportable compensation from the organization and any related organizations.
See instructions for the order in which to list the persons above.
Check this box if neither the organization nor any related organization compensated any current officer, director, or trustee.
(C) Position (do not check more (A) (D) (E) (F) (B) than one box, unless person Name and title Average Reportable Reportable is both an officer and a Estimated amount hours compensation from compensation from director/trustee) of other per the organization related organizations compensation from week (W-2/1099-MISC) (W-2/1099-MISC) the organization (list any and related hours for organizations related organiza-tions below dotted line) (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) TEEA0107L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 X Karen A Geuy 40 Interim Executive Director 0 X 77,251.
0.
0.
Keith Birkfield 40 Executive Director 0 X 48,224.
0.
0.
Barbara Creech 1 Director 0 X 0.
0.
0.
Jerry Kalman 1 2nd Vice Pres 0 X X 0.
0.
0.
Roger Shaver 2 President 0 X X 0.
0.
0.
Carlos Perez 2 Director 0 X 0.
0.
0.
Mike McReynolds 2 Past President 0 X X 0.
0.
0.
Cecilia Brown 2 Director 0 X 0.
0.
0.
Laurene Soper 1 Director 0 X 0.
0.
0.
Robert Pace 2 Director 0 X X 0.
0.
0.
Herb Baker 1 Treasurer 0 X X 0.
0.
0.
Laura Holck 1 Director 0 X 0.
0.
0.
Mark Haskell 1 Secretary 0 X X 0.
0.
0.
Gail Jones 1 Director 0 X 0.
0.
0.
Form 990 (2020) Page 8 Part VII Section A. Officers, Directors, Trustees, Key Employees, and Highest Compensated Employees (continued) (B) (C) Position (D) (E) (F) Average (do not check more than one (A) hours box, unless person is both an Reportable Reportable Name and title Estimated amount per officer and a director/trustee) compensation from compensation from of other week the organization related organizations compensation from (list any (W-2/1099-MISC) (W-2/1099-MISC) the organization hours and related for organizations related organiza - tions below dotted line) (15) (16) (17) (18) (19) (20) (21) (22) (23) (24) (25) G 1 b Subtotal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G c Total from continuation sheets to Part VII, Section A . . . . . . . . . . . . . . . . . . . . . . . G d Total (add lines 1b and 1c). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Total number of individuals (including but not limited to those listed above) who received more than $100,000 of reportable compensation 2 from the organization G Yes No 3 Did the organization list any former officer, director, trustee, key employee, or highest compensated employee 3 on line 1a? If 'Yes,' complete Schedule J for such individual. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 For any individual listed on line 1a, is the sum of reportable compensation and other compensation from the organization and related organizations greater than $150,000? If 'Yes,' complete Schedule J for 4 such individual . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Did any person listed on line 1a receive or accrue compensation from any unrelated organization or individual 5 for services rendered to the organization? If 'Yes,' complete Schedule J for such person . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Section B. Independent Contractors 1 Complete this table for your five highest compensated independent contractors that received more than $100,000 of compensation from the organization. Report compensation for the calendar year ending with or within the organization's tax year.
(A) (B) (C) Name and business address Description of services Compensation Total number of independent contractors (including but not limited to those listed above) who received more than 2 G $100,000 of compensation from the organization TEEA0108L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 0 X X X 0 0.
0.
125,475.
0.
0.
125,475.
0.
0.
0.
Lougene Williams 1 Director 0 X 0.
0.
0.
Form 990 (2020) Page 9 Part VIII Statement of Revenue Check if Schedule O contains a response or note to any line in this Part VIII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (A) (B) (C) (D) Total revenue Related or Unrelated Revenue exempt business excluded from tax function revenue under sections revenue 512-514 Federated campaigns . . . . . . . . . . 1 a 1 a Membership dues. . . . . . . . . . . . . b 1 b Fundraising events. . . . . . . . . . . . c 1 c Related organizations . . . . . . . . . d 1 d Government grants (contributions) . . . . . e 1 e All other contributions, gifts, grants, and f similar amounts not included above . . . . 1 f Noncash contributions included in g 1 g lines 1a-1f. . . . . . . . . . . . . . . . . . . . . . G h Total. Add lines 1a-1f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Business Code 2 a b c d e All other program service revenue. . . . f G g Total. Add lines 2a-2f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Investment income (including dividends, interest, and 3 G other similar amounts) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G Income from investment of tax-exempt bond proceeds 4 G Royalties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 (i) Real (ii) Personal Gross rents . . . . . . . . 6 a 6a Less: rental expenses b 6b Rental income or (loss) c 6c G Net rental income or (loss) . . . . . . . . . . . . . . . . . . . . . . . . . . . d (i) Securities (ii) Other Gross amount from 7 a sales of assets 7a other than inventory Less: cost or other basis b 7b and sales expenses Gain or (loss) . . . . . . . c 7c Net gain or (loss) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G d Gross income from fundraising events 8 a (not including $ of contributions reported on line 1c).
See Part IV, line 18 . . . . . . . . . . . . . 8a Less: direct expenses . . . . . . b 8b G Net income or (loss) from fundraising events . . . . . . . . . . c Gross income from gaming activities.
9 a See Part IV, line 19 . . . . . . . . . . . . . 9a Less: direct expenses . . . . . . b 9b G Net income or (loss) from gaming activities. . . . . . . . . . . c Gross sales of inventory, less . . . . . 10a returns and allowances. . . . . . . . . . 10a Less: cost of goods sold. . . . b 10b G Net income or (loss) from sales of inventory . . . . . . . . . . c Business Code 11a b c All other revenue . . . . . . . . . . . . . . . . . . . d G e Total. Add lines 11a-11d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G 12 Total revenue. See instructions . . . . . . . . . . . . . . . . . . . . . . TEEA0109L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 6,200.
1,041,780.
1,047,980.
87,369.
87,369.
87,369.
9,802.
9,802.
-1,725.
-1,725.
6,200.
41,713.
41,713.
1,185,139.
95,446.
0.
41,713.
Service fees 841,808.
843,533.
-1,725.
41,713.
Form 990 (2020) Page 10 Part IX Statement of Functional Expenses Section 501(c)(3) and 501(c)(4) organizations must complete all columns. All other organizations must complete column (A).
Check if Schedule O contains a response or note to any line in this Part IX. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (D) (A) (B) (C) Do not include amounts reported on lines Total expenses and Fundraising Program service Management 6b, 7b, 8b, 9b, and 10b of Part VIII.
expenses general expenses expenses Grants and other assistance to domestic 1 organizations and domestic governments.
See Part IV, line 21. . . . . . . . . . . . . . . . . . . . . . . . Grants and other assistance to domestic 2 individuals. See Part IV, line 22. . . . . . . . . . . . . Grants and other assistance to foreign 3 organizations, foreign governments, and for-eign individuals. See Part IV, lines 15 and 16 Benefits paid to or for members . . . . . . . . . . . . . 4 Compensation of current officers, directors, 5 trustees, and key employees . . . . . . . . . . . . . . . . Compensation not included above to 6 disqualified persons (as defined under section 4958(f)(1)) and persons described in section 4958(c)(3)(B) . . . . . . . . . . . . . . . . . . . . Other salaries and wages . . . . . . . . . . . . . . . . . . . 7 Pension plan accruals and contributions 8 (include section 401(k) and 403(b) employer contributions) . . . . . . . . . . . . . . . . . . . . Other employee benefits . . . . . . . . . . . . . . . . . . . 9 Payroll taxes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Fees for services (nonemployees): 11 Management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a Legal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b Accounting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c Lobbying. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . d Professional fundraising services. See Part IV, line 17. . . e Investment management fees . . . . . . . . . . . . . . . f g Other. (If line 11g amount exceeds 10% of line 25, column (A) amount, list line 11g expenses on Schedule O.). . . . . Advertising and promotion. . . . . . . . . . . . . . . . . . 12 Office expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Information technology. . . . . . . . . . . . . . . . . . . . . 14 Royalties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 Occupancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Travel. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Payments of travel or entertainment 18 expenses for any federal, state, or local public officials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Conferences, conventions, and meetings. . . . 19 Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Payments to affiliates. . . . . . . . . . . . . . . . . . . . . . 21 Depreciation, depletion, and amortization . . . . 22 Insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Other expenses. Itemize expenses not 24 covered above (List miscellaneous expenses on line 24e. If line 24e amount exceeds 10% of line 25, column (A) amount, list line 24e expenses on Schedule O.) . . . . . . . . . . . . . . . . . . a b c d All other expenses. . . . . . . . . . . . . . . . . . . . . . . . . e 25 Total functional expenses. Add lines 1 through 24e. . . . 26 Joint costs. Complete this line only if the organization reported in column (B) joint costs from a combined educational campaign and fundraising solicitation.
if following Check here G SOP 98-2 (ASC 958-720). . . . . . . . . . . . . . . . . . . BAA Form 990 (2020) TEEA0110L 10/07/20 Fallbrook Healthcare Foundation Inc.
95-3389263 125,475.
71,521.
38,897.
15,057.
0.
0.
0.
0.
436,247.
246,282.
144,214.
45,751.
14,717.
7,669.
5,513.
1,535.
50,802.
29,745.
16,073.
4,984.
13,869.
10,149.
3,220.
500.
24,282.
7,824.
2,918.
13,540.
24,472.
3,880.
2,777.
17,815.
8,088.
2,971.
4,440.
677.
18,252.
6,790.
11,016.
446.
16,545.
16,349.
196.
4,169.
4,169.
47,301.
47,301.
9,919.
7,823.
2,096.
161,524.
110,321.
-384.
51,587.
16,139.
11,065.
3,775.
1,299.
9,854.
6,313.
1,550.
1,991.
6,909.
4,104.
229.
2,576.
16,864.
7,168.
2,644.
7,052.
1,005,428.
549,974.
290,644.
164,810.
Events & programs Utilities Printing and Publications Bank / credit card fees Form 990 (2020) Page 11 Part X Balance Sheet Check if Schedule O contains a response or note to any line in this Part X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (A) (B) Beginning of year End of year Cash ' non-interest-bearing. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Savings and temporary cash investments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Pledges and grants receivable, net. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 Accounts receivable, net . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 4 Loans and other receivables from any current or former officer, director, 5 trustee, key employee, creator or founder, substantial contributor, or 35% controlled entity or family member of any of these persons . . . . . . . . . . . . . . . . . . . . . 5 Loans and other receivables from other disqualified persons (as defined under 6 6 section 4958(f)(1)), and persons described in section 4958(c)(3)(B) . . . . . . . . . . . . . . Notes and loans receivable, net. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 7 Inventories for sale or use. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 8 Prepaid expenses and deferred charges. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 9 Land, buildings, and equipment: cost or other basis.
10a Complete Part VI of Schedule D . . . . . . . . . . . . . . . . . . . . 10a Less: accumulated depreciation. . . . . . . . . . . . . . . . . . . . b 10b 10c 11 Investments ' publicly traded securities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 12 Investments ' other securities. See Part IV, line 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 13 Investments ' program-related. See Part IV, line 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 14 Intangible assets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 15 Other assets. See Part IV, line 11. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 16 16 Total assets. Add lines 1 through 15 (must equal line 33). . . . . . . . . . . . . . . . . . . . . . . Accounts payable and accrued expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 17 Grants payable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 18 Deferred revenue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 19 Tax-exempt bond liabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 20 Escrow or custodial account liability. Complete Part IV of Schedule D. . . . . . . . . . . 21 21 Loans and other payables to any current or former officer, director, trustee, 22 key employee, creator or founder, substantial contributor, or 35% controlled entity or family member of any of these persons . . . . . . . . . . . . . . . . . . . . . 22 Secured mortgages and notes payable to unrelated third parties . . . . . . . . . . . . . . . . 23 23 Unsecured notes and loans payable to unrelated third parties. . . . . . . . . . . . . . . . . . . 24 24 Other liabilities (including federal income tax, payables to related third parties, 25 and other liabilities not included on lines 17-24). Complete Part X of Schedule D. 25 26 Total liabilities. Add lines 17 through 25. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Organizations that follow FASB ASC 958, check here G and complete lines 27, 28, 32, and 33.
Net assets without donor restrictions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 27 Net assets with donor restrictions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 28 Organizations that do not follow FASB ASC 958, check here G and complete lines 29 through 33.
Capital stock or trust principal, or current funds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 29 Paid-in or capital surplus, or land, building, or equipment fund. . . . . . . . . . . . . . . . . . 30 30 Retained earnings, endowment, accumulated income, or other funds. . . . . . . . . . . . 31 31 Total net assets or fund balances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 32 Total liabilities and net assets/fund balances. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 33 TEEA0111L 10/07/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 76,067.
122,837.
1,200.
20,446.
850,570.
412,567.
485,304.
438,003.
505,007.
720,955.
1,067,578.
1,302,241.
22,994.
28,704.
29,340.
31,870.
52,334.
60,574.
X 933,241.
1,167,164.
82,003.
74,503.
1,015,244.
1,241,667.
1,067,578.
1,302,241.
Form 990 (2020) Page 12 Part XI Reconciliation of Net Assets Check if Schedule O contains a response or note to any line in this Part XI. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Total revenue (must equal Part VIII, column (A), line 12). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 Total expenses (must equal Part IX, column (A), line 25). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Revenue less expenses. Subtract line 2 from line 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3 Net assets or fund balances at beginning of year (must equal Part X, line 32, column (A)). . . . . . . . . . . . . . . . . . 4 4 Net unrealized gains (losses) on investments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 Donated services and use of facilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 6 Investment expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 7 Prior period adjustments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 8 Other changes in net assets or fund balances (explain on Schedule O). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 9 Net assets or fund balances at end of year. Combine lines 3 through 9 (must equal Part X, line 32, 10 column (B)) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Part XII Financial Statements and Reporting Check if Schedule O contains a response or note to any line in this Part XII . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Yes No Accounting method used to prepare the Form 990: Cash Accrual Other 1 If the organization changed its method of accounting from a prior year or checked 'Other,' explain in Schedule O.
Were the organization's financial statements compiled or reviewed by an independent accountant? . . . . . . . . . . . . . . . . . . . . 2 a 2 a If 'Yes,' check a box below to indicate whether the financial statements for the year were compiled or reviewed on a separate basis, consolidated basis, or both: Separate basis Consolidated basis Both consolidated and separate basis Were the organization's financial statements audited by an independent accountant?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b 2 b If 'Yes,' check a box below to indicate whether the financial statements for the year were audited on a separate basis, consolidated basis, or both: Separate basis Consolidated basis Both consolidated and separate basis c If 'Yes' to line 2a or 2b, does the organization have a committee that assumes responsibility for oversight of the audit, review, or compilation of its financial statements and selection of an independent accountant?
. . . . . . . . . . . . . . . . . . . . . . . . . 2 c If the organization changed either its oversight process or selection process during the tax year, explain on Schedule O.
As a result of a federal award, was the organization required to undergo an audit or audits as set forth in the Single 3 a Audit Act and OMB Circular A-133? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 a If 'Yes,' did the organization undergo the required audit or audits? If the organization did not undergo the required audit b or audits, explain why on Schedule O and describe any steps taken to undergo such audits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 b TEEA0112L 10/19/20 BAA Form 990 (2020) Fallbrook Healthcare Foundation Inc.
95-3389263 1,185,139.
1,005,428.
179,711.
1,015,244.
46,712.
0.
1,241,667.
X X X X X X Fallbrook Healthcare Foundation, Inc.
dba Foundation for Senior Care Statement of Financial Position December 31, 2020 See Notes to Financial Statements 3 Assets Current Assets Cash and cash equivalents 122,837 $ Investments 720,955 Prepaid expenses 20,446 Total current assets 864,238 Property and equipment, net 438,003 Total Assets 1,302,241 $ Liabilities and Net Assets Current Liabilities Accounts Payable and Accrued Expenses 1,885 $ Payroll liabilities 26,819 Current portion of notes payable 17,679 Total Current Liabilities 46,383 Other Liabilities Notes payable, net of current portion 15,142 Discount on non interest bearing notes (951) Total Liabilities 60,574 Net Assets Net Assets, without donor restrictions 1,167,164 Net Assets, with donor restrictions 74,503 Total Net Assets 1,241,667 Total Liabilities and Net Assets 1,302,241 $ Fallbrook Healthcare Foundation, Inc.
dba Foundation for Senior Care Statement of Activities For the Year Ended December 31, 2020 See Notes to Financial Statements 4 Without Donor With Donor Total Restriction Restriction Dec. 31, 2020 Support and revenues Donations - public support 454,130 $ -$ 454,130 $ Fees 87,369 87,369 Fundraising 47,913 47,913 Grants 587,650 587,650 Net investment income 54,789 54,789 Total support and revenues 1,231,851 - 1,231,851 Net assets released from restrictions Satisfaction of donor restrictions 7,500 (7,500) Expenses Programs 549,974 549,974 Fundraising 165,006 165,006 General and administrative 290,448 290,448 Total expenses 1,005,428 - 1,005,428 Change in net assets 233,923 (7,500) 226,423 Net assets, beginning 933,241 82,003 1,015,244 Net assets, ending 1,167,164 $ 74,503 $ 1,241,667 $ Fallbrook Healthcare Foundation, Inc.
dba Foundation for Senior Care Statement of Functional Expenses For the Year Ended December 31, 2020 See Notes to Financial Statements 5 General Program Fundraising Admin Totals Advertising 3,880 $ 17,815 $ 2,777 $ 24,472 $ Bank charges 4,104 2,576 229 6,909 Depreciation & amortization 47,301 47,301 Dues and subscriptions 5,333 188 5,521 Events/activities 110,321 51,587 (384) 161,524 Insurance 7,823 2,096 9,919 Interest expense 4,169 4,169 Licenses 4,420 389 4,809 Office expense & supplies 2,971 677 4,440 8,088 Outside Services 7,824 13,540 2,918 24,282 Professional fees 10,149 500 3,220 13,869 Publicity 6,313 1,991 1,550 9,854 Postage 556 749 594 1,899 Employee expenses 355,217 67,327 204,697 627,241 Repairs & maintenance 6,790 446 11,016 18,252 Taxes 75 36 41 152 Telephone 2,117 934 1,432 4,483 Travel & transport 16,349 196 16,545 Utilities 11,065 1,299 3,775 16,139 Totals for 2020 549,974 $ 165,006 $ 290,448 $ 1,005,428 $ Fallbrook Healthcare Foundation, Inc.
dba Foundation for Senior Care Statement of Cash Flows For the Year Ended December 31, 2020 See Notes to Financial Statements 6 Cash flows from operating activities: Change in net assets 226,423 $ Adjustments to reconcile changes in net assets to net cash used in operating activates Depreciation & amortization 47,301 Unrealized gain on investments (46,716) Increase in prepaid (19,246) Increase in accounts payable 1,008 Increase in payroll liabilities 23,123 Decrease accrued discount on non interest bearing notes (1,266) Net cash provided by operations 230,627 Cash flows from investing activities: Sale of investments 773,779 Purchase of investments (940,479) Net cash used by investing activities (166,700) Cash flows from financing activities: Proceeds from loans Payments on loans (17,157) Net cash used for financing activities (17,157) Net increase/(decrease) in cash 46,770 Cash balance at beginning of year 76,067 Cash balance at end of year 122,837 $ Interest expense 4,169 $ |
12644 | https://literacylearn.com/free-morphology-worksheets/ | Free Morphology Worksheets: Prefixes, Suffixes, Bases - Literacy Learn
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Free Morphology Worksheets: Prefixes, Suffixes, Bases
ByKatiePublished:May 7, 2024 September 16, 2025 Updated:September 16, 2025
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Get these FREE morphology worksheets to help students deepen their vocabulary knowledge by analyzing words by looking at morphemes. Get printable word cards with prefixes, suffixes, and bases PLUS a reusable word sum worksheet.
Get a new freebie every week!
Scroll to the bottom of the post to download the printables FREE!
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Why Teach Morphology?
How to Use Word-Part Clues
Free Word Building Worksheets
Recommended Resources
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Why Teach Morphology?
Morphology is a term that relates to thestudy of word structure and its meaning.
Morphemic analysis is the term for breaking down word parts to derive meaning. In order to do this, we must look at morphemes, or word part clues! These include bases (Greek/Latin/Ango-Saxon) and affixes (prefixes/suffixes).
The National Reading Panel (2000) identified morphemic analysis as an effective word-learning strategy. That means it should definitely be a part of all students’ literacy instruction!
Timothy Shanahan has a very helpful blog post entitled, What Morphology Instruction Should Look Like. In this post, he recommends beginning morphology instruction using words kids already know.
Then, using those word families to make connections between words to expand vocabulary knowledge.
For example, you can start with the base word ‘power.’ Then, you can add morphemes to change the word and begin to analyze how the meaning changes. Other words that are a part of this word family includes…
powerful
overpowering
superpower
empower
powerless
manpower
After you’ve spent lots of time discussing and teaching word parts, teachers should explicitly teach Greek bases and Latin bases.
How to Use Word-Part Clues
Leading literacy experts Timothy Risinski and Nancy Padak (2017) say that “[t]eachers at all grade levels and content areas can provide instruction in Greek and Latin roots to develop students’ vocabulary.”
This means that teachers must provide explicit instruction in word-part clues (morphemes). Kids must learn…
The meanings of word parts (prefix e- means out/away, Latin base rupt means to burst/break, suffix -s means 3rd person singular)
How to break apart words into word parts (erupted = e + rupt + s)
How to put words parts back together to form meaning (erupts means bursts or breaks out)
To summarize, kids need to look for base words, prefixes, and suffixes in words. Then, they must ask themselves… “What do each of those word parts mean? “
Then, they must put the meanings together to determine the meaning of the whole word.
🚀Looking for more? Read about the importance ofexploring morphemes with your students, and get our free worksheets to help! And check out how to use word chaining for morphology practice, too!
Free Word Building Worksheets
This free morphology resource is yours! Keep reading and download at the bottom of this page.
These free morphology printables are designed to help build your students’ knowledge of common morphemes. To help with your instruction, the resources include:
Reusable word sum worksheet
4 prefix cards ➡ dis-, un-, re-, il-/im-/in-/ir-
4 suffix cards ➡ -ed, -ing, -s/-es, -er/-or
4 Anglo-Saxon bases ➡ act, agree, like, count
💯This is the perfect resource to introduce students to structured word inquiry. Read our blog post to learn all about word sums, and then grab our free and customizable word matrix, too!
Using the Worksheets
This activity works best in pairs or a small group. An essential component of word inquiry is talking about the words and their meanings.
This type of activity helps to strengthen oral language while building vocabulary and deepening comprehension.
Printing Info: Print the word part cards double-sided. Then, cut the cards and laminate them for extended use. Print the word sum worksheet and place it in a reusable sleeve so kids can use it over and over again to build many words.
Get the expanded version of our popular Morphology Word Building Cards!
Looking for more?If you love this resource, grab the expanded version of our Morphology Word Cards.
It includes 156+ prefix, suffix, and base cards and includes tons of common Greek and Latin bases.
Recommended Resources
All About Phonemes, Graphemes, and Morphemes
3 Sounds of Suffix – ed Poster
Science of Reading Morphology Poster
Best OG Training Options
Science of Reading Professional Training
References:
Diamond, L. and L. Gutlohn (2019).Vocabulary handbook: core literacy library. Paul H. Brooks Publishing.
National Reading Panel (U.S.) & National Institute of Child Health and Human Development (U.S.). (2000).Report of the National Reading Panel: Teaching children to read : an evidence-based assessment of the scientific research literature on reading and its implications for reading instruction. U.S. Dept. of Health and Human Services, Public Health Service, National Institutes of Health, National Institute of Child Health and Human Development.
Rasinski, T. and N. Padak (2017). The roots of comprehension. ASCD (Advance and Elevate Learning).
Shanahan, T. (2018) What should morphology instruction look like? Shanahan on Literacy.
Download & Print
DOWNLOAD TERMS:All of our resources and printables are designed for personal use only in homes and classrooms. Each teacher must download his or her own copy. You may not:Save our files to a shared drive, reproduce our resources on the web, or make photocopies for anyone besides your own students. To share with others, please use the social share links provided or distribute thelink to the blog postso others can download their own copies. Your support in this allows us to keep making free resources for everyone! Please see ourCreative Creditspage for information about the licensed clipart we use. If you have any questions or concerns regarding our terms, pleaseemail us. Thank you!
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Katie Connersays: September 23, 2024 at 11:56 am Thanks! Reply
Katiesays: September 24, 2024 at 10:46 am You’re very welcome. Hope your students benefit from them.
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12645 | https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/tonoplast | Tonoplast - an overview | ScienceDirect Topics
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Tonoplast
In subject area:Biochemistry, Genetics and Molecular Biology
The tonoplast is the cytoplasmic membrane surrounding a vacuole, separating the vacuolar contents from the cell׳s cytoplasm.
From:Encyclopedia of Biological Chemistry (Third Edition), 2021
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On this page
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You might find these chapters and articles relevant to this topic.
Review article
Integrated approaches to cytoskeleton research
2007, Biochimica et Biophysica Acta (BBA) - Molecular Cell ResearchHannie S. van der Honing, ... Tijs Ketelaar
The tonoplast is the vacuolar membrane. Mature plant cells possess one or several large vacuoles, which can occupy over 90% of the total cell volume [3,4]. The cytoplasm fills up the rest of the cell interior, and surrounds the vacuole(s). The cytoplasmic organisation of plant cells varies with its developmental stage. Usually, a layer of cytoplasm is located in the cortical and perinuclear areas of the cell. These two areas of cytoplasm are interconnected by strands of cytoplasm that traverse the vacuole: the transvacuolar or cytoplasmic strands, bounded by the tonoplast (Fig. 1).
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Fig. 1. Differential interference contrast image of a tobacco bright yellow 2 suspension cell. Cytoplasmic strands (green) traverse the vacuole (yellow), and connect the cortical and perinuclear cytoplasm (cytoplasm is green; nucleus is dark red; nucleolus is light red). The cell wall (blue) encases the cell. Scale bar, 10 μm.
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Journal2007, Biochimica et Biophysica Acta (BBA) - Molecular Cell ResearchHannie S. van der Honing, ... Tijs Ketelaar
Review article
Mechanisms and models of the active transport of ions and the transformation of energy in intracellular compartments
2012, Progress in Biophysics and Molecular BiologyAlexey V. Melkikh, Vladimir D. Seleznev
3.5 A model for ion transport in vacuoles
Vacuoles are contained mainly in plant and protozoal cells. These organelles are typically round compartments contained within a thin envelope, the tonoplast. The tonoplast is a membrane composed of proteins and lipids that performs the following functions:
A.
Separates the contents of the cytoplasm of the vacuolar sap, the accumulation of which, for example, determines the yield of certain crops. The vacuole liquid contains minerals and organic matter (including carbohydrates, proteins, alkaloids, and tannins). Pigments can also be accumulated in these organelles.
B.
Provides the osmotic pressure in the cell (turgor of plants), which is particularly responsible for the water supply to the upper portions of plants.
C.
Supports basic characteristics (e.g., pH) at a stable level, despite possible abrupt changes in the concentrations of sodium and potassium in the external environment (salt stress).
To perform its functions, the vacuole membrane possesses a number of ion channels, pumps and exchangers (Stout and Griffing, 2001; Alberts et al., 2002; Grabe et al., 2000). The main transport systems in the vacuoles are shown in Fig.14. A vacuole contains two systems for proton transport (Sze et al., 1999; Drozdowicz and Rea, 2001; Hasegawa et al., 2000), generating Δ μ H e:
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Fig.14. The main transport systems in vacuole.
−
H+-ATP-ase, transporting protons into a vacuole (according to publications this V-ATPase is the basic for proton transport);
−
Pyrophosphatase, working at the expense of pyrophosphate and transporting protons into a vacuole.
Generated by active transport, Δ μ H e is used to pump the ions Na+, K+, and Ca 2+ inside the vacuoles. The role of these pumps is to act as exchangers (antiporters) of K+/H+, Ca 2+/H+ and Na+/H+.
Metabolites (e.g., sucrose) are also stocked due to Δ μ H e. In (Serrano and Rodriguez-Navarro, 2001), the following characteristics of vacuoles are reported, which were obtained experimentally and are summarized in Table 2.
Table 2. Characteristics of vacuoles.
| Parameter | Cytoplasm | Vacuole |
---
| pH | 7.6 | 5.5 |
| ΔpH | 2–3 | 1–2 |
| Potassium (мМ) | 100 | 150 |
| Potential (mV) | | −80–20 |
In addition, it is established (Yokoyama et al., 1998) that the differences in the concentrations of protons and the resting potential at the tonoplast are linearly dependent on the activity of V-ATPase. In (Yokoyama et al., 1998), ATP synthesis was observed in conditions when Δ μ H e became more than 10.
Let us now consider a theoretical model of the transport of substances across the membrane of the vacuole and the ability of this model to explain the experimental data.
The main source of vacuole non-equilibrium is V-ATPase. This enzyme has much in common with F-ATPase, which is responsible for ATP synthesis in the mitochondria (part 4). Similar to F-ATPase, V-ATPase has two subsystems: V 1, located under a tonoplast and having three sorption centers for ATP, and V 0, which is located in the membrane directly and has six “c” subunits with centers of proton sorption (Grabe et al., 2000). F 0, in contrast to V 0, has from 9 to 14 “c” subunits. According to the V-ATPase structure, during hydrolysis or synthesis, two protons are the share of one ATP molecule, as one revolution of the “c” cylinder subunits, 6 protons and 3 ATP molecules are consumed. The difference between F-ATPase and V-ATPase appears to be connected with their different functions. If F-ATPase is required for ATP synthesis with the shift of the reaction ATP⇔ADP+P from equilibrium Δ μ A=25, which demands Δ μ H e=25 3=8.3, V-ATPase is required to provide, at the expense of ATP hydrolysis with Δ μ A=20, an electrochemical potential difference of protons Δ μ H e from 10 to 4.
The equation of ATP hydrolysis on V-ATPase is written as in (2.40):
(7.1)J ATP=k H↓n F n D n P(n H i)2(exp(Δ μ A−2 Δ μ H e)−1).
In accordance with (7.1), if Δ μ H e is more than Δ μ A 2=10, JATP changes its sign. This shift corresponds to the fact that the corresponding reaction in (2.37) mainly proceeds from left to right and vice versa – if it proceeds from right to left, then ATP will be synthesized. This is what was recorded in (Yokoyama et al., 1988) at Δ μ H e>10.
In vivo in (Davies et al., 1994), the values Δ μ H e=e φ+ln n H i n H o=2+5.7=7.7 have been measured. This value corresponds to that Δ μ A−2 Δ μ H e=4.6. Such an excess of Δ μ A over 2 Δ μ H e is required to provide sufficient proton flow, which is particularly necessary for the accumulation of carbohydrates in the vacuole, or, in other words, to perform useful work. As shown in Section 2.4, to maximize the useful work, the difference Δ μ A−2 Δ μ H e is required to be equal to ∼3. This result agrees qualitatively with the experimental value of this quantity, which is∼4.6.
From the condition of J ATP=0 and based on (7.1), we easily obtain the following relationship of the concentrations of protons inside and outside of the vacuole:
(7.2)n H i=n H o exp(Δ μ A 2−e φ).
Such communication will characterize the vacuole in the rest state, when the process of the accumulation of carbohydrates is absent.
Let us consider other transport systems. We write the equation of transfer of sodium ions through the Na+/H+ antiporter, taking into account that a sodium ion is exchanged for one proton (2.31):
(7.3)J Na−H=A Na(n H i n Na o−n H o n Na i).
The second possible mechanism for the transport of sodium is passive and corresponds to the formula (2.12), which is written in simplified form:
(7.4)J Na p=B Na(n Na o exp(−e φ)−n Na i).
Potassium is transported by the same mechanism as sodium. Therefore, flow through the K+/H+ exchanger and passive flow of K+ can be written by analogy with (7.3, 7.4):
(7.5)J K−H=A K(n H i n K o−n H o n K i),
(7.6)J K p=B K(n K o exp(−e φ)−n K i).
Because no pumps for negative ions are found, we assume that these ions are transferred passively. This result corresponds to Nernst's law:
(7.7)n−i=n−o exp(e φ)
The overall system of equations of conservation of the major ions in vacuoles would be in the following form (see 3.3–3.5):
(7.8)V K∂n H i∂t=J ATP+J K−H+J Na−H+J P,
(7.9)V K∂n Na i∂t=J Na−H+J Na P,
(7.10)V K∂n K i∂t=J K−H+J K P,
(7.11)V K∂n−i∂t=J−,
(7.12)n N a i+n K i=n−i.
In these equations, J P represents proton flux through the pyrophosphatase. In the steady state, with an additional assumption that (7.8) the flux through the ATPase(7.1) is the main contributor, we obtain the following simplified system of equations:
(7.13)n H i=n H o exp(Δ μ A 2−e φ),
(7.14)J K=J K−H+J K P=A K(n H i n K o−n H o n K i)+B K(n K o exp(−e φ)−n K i)=0,
(7.15)J Na=J Na−H+J Na P=A Na(n H i n Na o−n H o n Na i)+B Na(n Na o exp(−e φ)−n Na i)=0,
(7.16)n−i=n−o exp(e φ),
(7.17)n Na i+n K i=n−i.
We now introduce the dimensionless variables that determine the ratio of the conductivities of the exchanger and the passive transport of K+ and Na+, respectively:
(7.18)α K=B K A K n H o H α Na=B Na A Na n H o.
Substituting (7.13) in (7.15) and (7.14) with (7.18), we obtain
(7.19)n K i n K o=exp(−e φ)(exp(Δ μ A 2)+α K)1+α K.
(7.20)n Na i n Na o=exp(−e φ)(exp(Δ μ A 2)+α Na)1+α Na.
Substituting (7.18), (7.19) and (7.16) into the condition of local neutrality (7.17), we obtain an equation for determining the resting potential:
(7.21)n K o n−o(exp(Δ μ A 2)+α K)1+α K+n Na o n−o(exp(Δ μ A 2)+α Na)1+α Na=exp(2 e φ).
Formula (7.21) satisfies the thermodynamic requirement that if Δ μ A=0, the potential φ=0.
Analyzing formula (7.21), we observe that it is very important from a practical point of view of the property. If we provide the closeness of the relationship of conductivity for Na+ and K+, i.e.,
α K=α Na,
we obtain an expression for the potential that does not depend on the concentrations of sodium and potassium in the cytoplasm:
(7.22)exp(Δ μ A 2)+α K 1+α K=exp(2 e φ),
This expression took into account the condition of local neutrality in the cytoplasm:
(7.23)n Na o+n K o=n−o.
This result is a manifestation of the properties of robustness (independence), as discussed in Section 3.2.
In the presence of experimental data, such as those shown in Table 2, when the time for the given values of n K o are determined, n K i∗ and e φ∗ based on (7.19) can easily restore the ratio of conductivity for potassium:
(7.24)α K=n K i∗n K o exp(−e φ∗)−exp(Δ μ A 2)1−n K i∗n K o exp(−e φ∗).
Substituting the values n K o, n K i∗ and e φ∗=2 from Table 2 into (7.24) we obtain:
α K≈2000.
Using this value, with the help of (7.22), we can identify a universal value for the potential that is independent of the concentration:
e φ≈1.2≈30 mV.
Thus, the resulting simplified system of equations (7.13)–(7.17) exhibits several well-known experimentally determined properties of the vacuole, such as the following:
1.
When ATPase enters the synthesis of ATP (7.1).
2.
The vacuole has a property of weak dependence of the potential on the sodium and potassium concentrations in the cytoplasm at similar values to the relations of the conduction of exchangers and the passive transport of sodium and potassium ions.
3.
The theoretical values of the resting potential of vacuoles in good agreement with experimental data.
A more complete and rigorous model of the vacuole, above all, requires a more complete account of all of the proton fluxes in Eq. (7.8), as well as the refinement of experimental data on organic matter and their incorporation into a theoretical model.
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Journal2012, Progress in Biophysics and Molecular BiologyAlexey V. Melkikh, Vladimir D. Seleznev
Review article
Multivesicular bodies: a mechanism to package lytic and storage functions in one organelle?
2002, Trends in Cell BiologyLiwen Jiang, ... John C Rogers
Models for the formation of internal membranes in multivesicular bodies (MVBs). (a) An incoming transport vesicle (red) carrying an integral membrane protein with a lumenal domain (blue). When the transport vesicle fuses with the endosome membrane (black), it releases its soluble cargo (green) into the lumen of the endosome. If invagination of the membrane at the site of fusion occurred, the resulting internal vesicle would have cytoplasm in its lumen (white) and the lumenal domain of the membrane protein would be on the vesicle exterior. (b) A variation in which the incoming vesicle is engulfed by an autophagic process. The engulfed vesicle then fuses with the surrounding membrane yielding an internal vesicle product identical to that in (a). (c) An internal vesicle interacts with the limiting membrane to give the structure on the right. Lines indicate individual leaflets of the lipid bilayers. This is a hemifusion intermediate. These very stable structures are well known from studies of fusion mediated by GPI-anchored influenza hemagglutinin[49–52] and of the fusion of artificial lipid bilayers [48,53,54]. The diameter of a hemifusion diaphragm ranges from 7 nm to 127 nm in cell–cell hemifusion but is on the order of microns in cell–bilayer hemifusion ; differences reflect effects of membrane tension and curvature as well as lipid composition. Here, we postulate a diameter larger than the incoming transport vesicle. The internal membranes are depicted in blue to indicate that their lipid composition will differ from that of the limiting membrane; in (d) and (e) for simplicity the individual bilayer leaflets are not defined. (d) The incoming cargo vesicle fuses with the hemifusion membrane and releases its soluble cargo into the internal compartment lumen. If the integral membrane cargo protein somehow moved past the hemifusion junction into the internal membrane, its lumenal domain would be inside the compartment. (e) Alternatively, the incoming transport vesicle might interact with the hemifusion site and be engulfed by autophagy, resulting in a vesicle within a vesicle within a vesicle.
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Journal2002, Trends in Cell BiologyLiwen Jiang, ... John C Rogers
Review article
Cell Biology
2007, Current Opinion in Plant BiologyKathryn S. Lilley, Paul Dupree
In studies that focus on organelles which are impossible to isolate to such a degree of purity, more stringent controls have to be taken to avoid mis-assignment of proteins to subcellular locations. The tonoplast has been subject of many plant subcellular localization studies, because of the importance of the tonoplast in plant cell metabolism and the enigmatic identity of many of the membrane transport proteins. Several vacuole proteomic studies have been attempted and yielded large datasets. A particularly successful recent example was the identification of HvSUT2 and AtSUT4 as tonoplast sucrose transporters after proteomic analysis of barley tonoplasts . The localization was supported by expression of GFP fusion proteins. However, in general vacuolar purification is difficult, and there is often some contamination within the datasets from other organelles, possibly because the vacuole has an autophagic function and therefore proteins from elsewhere within the cell will co-purify in the vacuolar lumen [37,38]. Of the proteins found in the vacuolar fraction of the dataset of Jaquinod et al, there is an overlap of 142 proteins with the LOPIT set. Only 22 proteins however are in agreement with Dunkley et al.'s vacuolar list, the remainder includes 90 which were classified by LOPIT not vacuolar in location. Attempts to further purify the organelle, for example using free flow zonal electrophoresis , may assist, but will not obviate the need for controls for the specific enrichment of the majority of the protein in the vacuoles.
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Journal2007, Current Opinion in Plant BiologyKathryn S. Lilley, Paul Dupree
Review article
Redox signalling in development and regeneration
2018, Seminars in Cell & Developmental BiologyFalco Krüger, Karin Schumacher
2 An old question: what is the origin of the tonoplast?
At the end of the 19th century it was de Vries who argued that vacuoles do not simply arise de novo when there is just enough water present in the cytoplasm to form distinct droplets and instead proposed that vacuoles develop from tonoplasts, special plastid-like precursors which subdivide to form new organelles if necessary. By the uptake of water the tonoplasts would inflate and become vacuoles . A few years later, Went proposed that all plant cells contain vacuoles and that they thus always originate from already existing vacuoles by division . However, this concept was challenged when Pfeffer showed that vacuoles can be experimentally induced de novo. Making use of the natural coloring of vacuoles in flower petals and the discovery of vital stains, especially Neutral red, it became possible to study also early stages of vacuole biogenesis. In the early 20th century the careful studies of Guilliermond as well as P.A. and P. Dangeard (father and son) confirmed vacuoles to be present already in the earliest meristematic cells and showed that they develop from small spherical, rod-like or tubular compartments into large central vacuoles present in differentiated cells in different plant species , reviewed in . Furthermore, they observed that vacuoles, especially in meristematic cells, are very dynamic organelles that constantly change their morphology, also fragment and fuse again, due to cytoplasmic streaming. They also noticed the transition of lytic vacuoles into protein storage vacuoles in maturing seeds and the reversed process during germination [11,24,23]. In fact, during this developmental stage lytic vacuoles and protein storage vacuoles do exist simultaneously . With the development of electron microscopy (EM) answering the central question from which donor compartment the tonoplast originates seemed in reach. However, EM studies by different groups came up with two seemingly contradictory answers.
Based on electron micrographs that showed stretches of rough ER transitioning into, or connecting, sphere-like enlargements reminiscent of lytic vacuoles, the ER was proposed to be the origin of the vacuole membrane . The similarity of the membrane texture of ER and tonoplast visualized by freeze-etching electron microscopy provided additional support for ER-derived biogenesis and a model suggesting cage-like enclosures of organelle-free cytoplasm by smooth ER that would then inflate to the central lytic vacuole was proposed [2,29]. Interestingly, an alternative model describes a very similar sequence of events leading to the formation of larger vacuoles; however, the trans-Golgi network (TGN) is postulated to be the membrane source . 3D high-voltage EM of mitotic cells revealed the presence of a complex tubular network and proximity of nascent provacuoles to nodes of the TGN combined with the fact that both compartments are acidified and contain acid hydrolases was taken as evidence that provacuoles are derived from the TGN . Interestingly, according to both models, autophagy is important in the process of vacuole formation as they predict that portions of cytosol that are entrapped by the vacuolar network need to be degraded. Clearly, the question if the TGN or the ER is the origin of the tonoplast membrane cannot be solved solely by electron microscopy but requires live cell imaging in a genetically tractable model organism.
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Journal2018, Seminars in Cell & Developmental BiologyFalco Krüger, Karin Schumacher
Review article
Phosphoinositides
2015, Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of LipidsMareike Heilmann, Ingo Heilmann
5 Tonoplast-associated PIs
Recently, the plant vacuole received attention as another cellular site containing a functional complement of PIs. The vacuole, surrounded by the tonoplast membrane, is an important orgenelle with functions in detoxification and storage as well as in the control of pH-homeostasis or the generation of turgor pressure for plant growth. The tonoplast has so far not been a focus of plant PI research. It has been proposed that the Arabidopsis PI 3-kinase AtVPS34 and its product PtdIns3P have functions in the control of endosomal membrane trafficking [10,11,105], which is consistent with the localization pattern of FYVE-based fluorescent reporters for PtdIns3P . AtVPS34 is an essential gene, limiting functional studies based on loss-of-function approaches. A number of studies have employed Wortmannin as a specific inhibitor of AtVPS34; however, as the inhibition is not strictly specific for PI 3-kinase~~s~~ the data should be reviewed with caution. The emerging picture suggests that the 3-phosphorylated PIs, PtdIns3P and PtdIns(3,5)P 2, are involved in trafficking to and from the vacuole and influence tonoplast function. Recently it has been reported that PtdIns(3,5)P 2 influences the morphology of the plant vacuole, which might have ramifications for the potential to build up turgor pressure . Arabidopsis mutants with defects in the formation of PtdIns(3,5)P 2 displayed aberrant stomatal closure, rationalized through a proposed effect of PtdIns(3,5)P 2 on a vacuolar proton-pyrophosphatase controlling vacuolar pH and, in consequence, vacuolar morphology and turgor . Similarly, an influence on vacuolar function has been attributed to Arabidopsis SAC-phosphatases, which are localized at the tonoplast, providing direct evidence for tonoplast-associated PIs . Arabidopsis mutants with defects in the expression of certain SAC-phosphatases displayed increased levels of PtdIns(3,5)P 2 and were characterized by altered vacuolar morphology . As there is so far no direct means to detect PtdIns(3,5)P 2 in living plant cells, and no biochemical analysis of tonoplast PIs has been reported as yet, the evidence for tonoplast-associated PI functions remains circumstantial.
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Journal2015, Biochimica et Biophysica Acta (BBA) - Molecular and Cell Biology of LipidsMareike Heilmann, Ingo Heilmann
Review article
Special Issue in memory of Professor Jeffrey B. Harborne
2003, PhytochemistryM.A.Susan Marles, ... Margaret Y Gruber
Tonoplast or ER localization of PA (11) biosynthesis has been known for some time, due to detailed histochemical studies in a variety of plant species (Parham and Kaustinen, 1977; Rao, 1988; Lees et al., 1995). However, genes specifying vacuolar localization were not cloned until recently. The ArabidopsisTT 12 gene has been shown to be involved with transport into the vacuoles of the seed coat epithelium (Debeaujon et al., 2001). Seeds of the tt 12 mutant appeared dull brown compared to wild type, and PA (11) precursors [reported as catechin (10a) and leucocyanidin (7b)] accumulated in the cytoplasm (Debeaujon et al., 2001). TT 12 was expressed early in seed development, but the transcript was undetectable later when the vacuoles were filled with PA (11) polymer (Debeaujon et al., 2001). The sequence of TT 12 appeared to encode a protein with twelve transmembrane regions and was identified as a putative secondary transporter protein (Debeaujon et al., 2001). Its transport mechanism is not yet known, but its sequence resembles a MATE protein family (Multidrug And Toxic compound Extrusion) (Debeaujon et al., 2001). The tt 12 mutant may well be the Arabidopsis analogue of ant 26 in barley. The availability of the collections of PA-specific mutations in Arabidopsis and barley should facilitate the isolation and characterization of additional genes that code for the condensation and vacuolar deposition steps of PA (11) in monocots and dicots. Fluorescence-labelled antibodies to recombinant proteins, increased attention to the exact structure of the polymer formed and differential substrate labelling experiments that include intact vacuoles will also enable a more accurate and detailed characterization of these events.
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Journal2003, PhytochemistryM.A.Susan Marles, ... Margaret Y Gruber
Review article
Plasmodesmata and their role in assimilate translocation
2022, Journal of Plant PhysiologyManuel Miras, ... Ji-Yun Kim
Greek (and New Latin) etymology (www.etymonline.com/)
protos: first
plásma: moldable substance, the liquid part of blood, etc., as distinguished from the corpuscles
hyalos: glass, clear alabaster, crystal lens used as a burning glass (as in hyaloplasma)
kytos: a hollow, receptacle, basket (as in cytoplasm)
desmá: band, bond, ligament (as in plasmodesmata)
apo-: word-forming element meaning "from, away from; after; in descent from" (apoplasm)
syn-; sym-: word-forming element meaning "together with, jointly; alike; at the same time’ (symplasm)
Box 1
Nomenclature: Plasm or Plast
There are various forms used for components of cells and for cellular domains. Most will likely agree that a protoplast is a plant cell from which the cell wall has been removed by cell wall digesting enzymes. Similarly, when cell walls of yeast cells are removed, we use the term spheroplast. The protoplast contains the protoplasm (protoplasma; 1839 Johannes Evangelista Purkinje; 1787–1869; gelatinous fluid in living tissue; Urschleim in German), also called cytoplasm. These terms are over 100 years old (von Hanstein, 1880). Thus, unitary entities such as a plasma membrane-encapsulated unit rend on a ‘t’, as in protoplast. Chloroplasts are entities ending on ‘t’. The cytoplasm consists of the cytosol, which had also been called hyaloplasm, plus the particles (named Kleinkörperchen by von Hanstein, 1880) or microsomata (microsomes, or better cell organelles and vesicles). PD connect the cytoplasm of most plant cells, creating a symplasm, as opposed to the continuous extracellular space (cell walls, xylem), which forms the apoplasm, in both cases not entities but both a plasm(a), a continuous liquid domain either inside or outside of plant cells. There is also some confusion regarding the term tonoplast. According to the definitions above, the vacuole is equivalent to the tonoplast, while its content would be a -plasm, i.e., the vacuolar protoplasm. However, the term tonoplast has, already early on, also been used for the membrane that encapsulates the vacuole (Weber, 1932).
Eduard Tangl is considered the first to recognize the importance of the structures underlying the cell-cell connections that we now call PD. Tangl had named them Protoplasmafortsätze (protoplasm extensions), Eduard Strasburger originally used the German term Plasmafäden (plasma threads). In 1901, he coined the German term Plasmodesmen (Strasburger, 1901). Based on the ethymology, a single one is called plasmodesma (sometimes plasmodesmos), plural plasmodesmata.
Note that much of the original literature was in German, nevertheless you will frequently find that the word symplast and apoplast are used, also historically.
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Journal2022, Journal of Plant PhysiologyManuel Miras, ... Ji-Yun Kim
Review article
Anion channels in higher plants: functional characterization, molecular structure and physiological role
2000, Biochimica et Biophysica Acta (BBA) - BiomembranesHélène Barbier-Brygoo, ... Christophe Maurel
3.5 Tonoplast anion channels as regulators of cell metabolism
With a storage capacity of more than 90% of the anions contained in the plant cell, the vacuole plays a central role in cell homeostasis and metabolism . In particular, the vacuolar membrane (or tonoplast) controls the availability in malate, nitrate and phosphate which can be metabolized in the cytoplasm and the plastids. The patch-clamp technique, which can be easily applied to isolated plant vacuoles, has permitted the identification of several distinct types of tonoplast anion channels. At low cytoplasmic Ca 2+ concentration, the tonoplast ion conductance is dominated by instantaneously activated fast-vacuolar (FV-type) channels, with a low selectivity for anions . In contrast, slow-vacuolar (SV-type) channels account for most of the ion conductances at high cytoplasmic Ca 2+ concentration. This channel is strongly voltage-dependent and was claimed to have similar selectivity for cations and anions [100,101], but this latter feature is still controversial . More recently, Dunlop and Phung provided some evidence in red beet for a phosphate permeable tonoplast channel that contributes to the slow vacuolar current. This set of data likely reflects the fact that several kinds of channels can contribute to the SV whole vacuolar current. In sugar beet vacuoles, Plant et al. described a chloride-activated inward anion channel, whose permeability sequence (chloride>malate>acetate>nitrate>phosphate) is shifted to (acetate>nitrate>phosphate>chloride>malate) by an increase in vacuolar chloride concentration. The activation of this channel by vacuolar chloride would then provide a pathway for the storage of nutrients, such as nitrate and phosphate in the vacuole, while the reduction in malate current would favor the use of malate in respiration and cytoplasmic pH control.
Malate is actually a ubiquitous anion which plays an important role in carbon metabolism and as a charge-balancing anion in the vacuole. In CAM plants, malate is produced as a result of dark CO 2 fixation and is stored in the vacuole, from which it can be remobilized during the following light period. Two anion channels, one favoring malate uptake and the other malate efflux have been identified by Iwasaki et al. on the tonoplast of Graptopetalum. These channels could provide a mechanism for homeostasis and diurnal rhythm in leaf cells of this CAM plant. In the tonoplast of mesophyll cells from Kalanchoe daigremontiana, another CAM plant, the membrane conductance is dominated by a vacuolar malate channel or VMAL which shows strong rectification, slow activation kinetics and lack of Ca 2+ dependence. In vacuoles from sugar beet and Arabidopsis cell suspensions, currents corresponding to malate 2− or succinate 2− entry in the vacuole have also been recorded [106,107]. All these studies clearly show that VMAL channels provide a major route for malate uptake into the vacuole of both C3 and CAM plants.
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Journal2000, Biochimica et Biophysica Acta (BBA) - BiomembranesHélène Barbier-Brygoo, ... Christophe Maurel
Review article
Vacuoles and prevacuolar compartments
2000, Current Opinion in Plant BiologyPaul C Bethke, Russell L Jones
Although there is agreement that plant cells might contain a prevacuolar compartment, the nature of this organelle remains unclear. Prevacuoles are defined as organelles that receive cargo from transport vesicles and subsequently deliver that cargo to the vacuole by fusion with the tonoplast. Alternatively, a prevacuole can be defined as an organelle that contains t-SNAREs and v-SNAREs, which bind to v-SNAREs on transport vesicles and t-SNAREs on vacuoles, respectively. To date, a prevacuolar compartment that fits either of these definitions has not been unequivocally identified .
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Journal2000, Current Opinion in Plant BiologyPaul C Bethke, Russell L Jones
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12646 | https://www.vumc.org/trauma-and-scc/sites/vumc.org.trauma-and-scc/files/public_files/Protocols/Substance%20Abuse%20Treatment%20Algorithm.pdf | TICU Substance Abuse Guidelines Alcohol Withdrawal: • For INTUBATED patients, if receiving a propofol or midazolam infusion, NO ADDITIONAL therapy required while on these infusions. • For patients who are not intubated or not receiving the above medications, DT prophylaxis may be initiated in patients who have 1) a history of delirium tremens or 2) a history of heavy alcohol use AND are demonstrating ≥ 2 of the following symptoms: - Nausea/vomiting - Tremor - Paroxysmal sweats and tachycardia (> 100 BPM) - Anxiety/agitation - Visual, tactile, or auditory disturbances - Clouded sensorium - Seizures • The above symptoms of withdrawal may present within 6-48 hrs after cessation of alcohol and may progress to DTs if untreated. • 5% of patients will develop DTs. This typically presents 48-72 hrs after the last drink, but has been reported up to 96 hrs later. • Symptoms of DTs include tachycardia, hypertension, fevers, increased respiratory rate/respiratory alkalosis, visual/auditory hallucinations, and marked agitation. These symptoms may last up to 5 days. The untreated mortality rate may be up to 15%, largely due to the risk of aspiration. As a result, the need for a secure airway should be discussed in patients experiencing DTs. Start olanzapine (Zyprexa) 5 mg PO/PT Q 6 hrs if dexmedetomidine is started. Dexmedetomidine continuation > 24 hrs requires attending approval. The addition of clonidine 0.1-0.3 mg PO/PT TID may also be used adjunctively and may facilitate transitioning off dexmedetomidine. Electrolyte disturbances are common in withdrawal. Potassium, magnesium, and phosphorus should be monitored daily. Consider the need for cardiac monitoring. Daily banana bag or oral vitamin supplementation x 3 days: 100 mg thiamine (prior to any dextrose), 1 mg folate, MVI, 2 gm magnesium (if banana bag) daily Diazepam 5-10 mg IV/PT/PO Q 8 hrs +/- Diazepam 5 mg IV Q 2 hrs prn severe symptoms x 72 hrs (If > 65 or known liver disease, consider lorazepam substitution) Unresponsive to at least diazepam 10 mg Q 8 hrs, consider an adjunctive agent: 1. Haloperidol 5 mg IV/IM Q 4 hrs prn severe agitation Symptoms controlled Continue current diazepam dose x 48 hrs, then decrease to 2.5 mg tid x 24 hrs Symptoms NOT controlled Treatment Algorithm Unresponsive to diazepam and prn haloperidol, consider: 2. Dexmedetomidine 0.2-1.5 mcg/kg/hr x 24 hrs Symptoms NOT controlled Opioid Withdrawal: • Opioid withdrawal is typically not life-threatening, in contrast to alcohol withdrawal. • Opioids may be detected on a urine drug screen. • Symptoms may include hypertension, tachycardia, vomiting, mydriasis, excessive lacrimation and salivation. • Symptoms may be alleviated by central alpha-2 adrenergic blockade: • Clonidine 0.1-0.3 mg PO/PT TID preferred • Clonidine 0.1-0.3 mg/24 hrs patch (TTS 1-3) TD Q week (may require 24 hrs enteral overlap) • Benzodiazepines are not required as they have no cross-reactivity with opioid receptor agonists. • Other substance abuse syndromes may be best treated by alleviating symptoms, according the sedation and analgesia protocol. • Patients with a history of regular benzodiazepine use may be restarted on their home medication or managed according to the alcohol withdrawal protocol. References: -Sarff M, Gold JA. Alcohol withdrawal syndromes in the intensive care unit. Crit Care Med. 2010; 38 (suppl.): S494-S501. -Mayo-Smith MF, Beecher LH, Fischer TL, et al. Management of alcohol withdrawal delirium. Arch Intern Med. 2004;164:1405-1412. -Mayo-Smith MF. Pharmacological management of alcohol withdrawal: a meta-analysis and evidence-based practice guideline. JAMA. 1997;278:144-151. Updated April 2011 Susan Hamblin, PharmD John Morris, MD Mickey Ott, MD |
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Applying the chain rule with the fundamental theorem of calculus 1
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I have the following problem in which I have to apply both the chain rule and the FTC 1. I got the right answer, but i'm confused about what's really going.
$$\frac{d}{dx} \int_1^{x^4} sec(t) \space dt $$
While trying to make sense of this, I have the following in mind.
If $g(x)=\int_a^{x}f(t)\space dt$
then $g(x)^{'}=f(x)$
And $(f(g(x)){'}$= $f^{'}(g(x)) \times g^{'}(x)$
and
$\frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx}$
Applying these to the this integral, i'm trying to break everything down to see what is what. So that for example I know which function is nested in which function.
I'm just looking for a clear and simple explanation of what's going on here.
calculus
integration
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chain-rule
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asked Oct 21, 2017 at 20:07
user140161user140161
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To avoid some of the confusion and to see how the Chain Rule applies here, use some other names of functions when stating the Chain Rule. What you wrote as the Chain Rule is perfectly correct and this is how it's stated in probably any textbook out there… but it makes you confused because in your question $g$ is the outside function, not inside, and $f$ is a totally different thing (not the outside function of the Chain Rule).
So let's do some renaming. If we call the outside function $g(x)$ and the inside function $h(x)$, then the very same Chain Rule will be written as $$\left[g(h(x))\right]'=g'(h(x))\cdot h'(x).$$ In your example: $$g(x)=\int_a^x f(t)\,dt, \quad \text{so} \quad g'(x)=f(x);$$ and $$h(x)=x^4, \quad \text{so} \quad h'(x)=4x^3.$$ The function $h(x)=x^4$ is the inside function here because it replaces $x$ in the expression for the outside function $g(x)$.
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edited Oct 21, 2017 at 22:50
answered Oct 21, 2017 at 20:37
zipirovichzipirovich
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$\begingroup$ zipirovich. Very nicely done!Everything to the point.+1 $\endgroup$
Peter Szilas
– Peter Szilas
2018-01-23 10:51:50 +00:00
Commented Jan 23, 2018 at 10:51
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12648 | https://www.youtube.com/watch?v=OIMvOMjBJro | Using vector to find the orthocentre of triangle
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12649 | https://en.wikipedia.org/wiki/Protein_kinase_A | Jump to content
Protein kinase A
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From Wikipedia, the free encyclopedia
Family of enzymes
Not to be confused with AMP-activated protein kinase or cyclin-dependent kinases.
| cAMP-dependent protein kinase (Protein kinase A) | |
--- |
| cAMP-dependent protein kinase heteroduodecamer, Sus scrofa | |
| Identifiers | |
| EC no. | 2.7.11.11 |
| CAS no. | 142008-29-5 |
| Alt. names | STK22, PKA, PKA C |
| Databases | |
| IntEnz | IntEnz view |
| BRENDA | BRENDA entry |
| ExPASy | NiceZyme view |
| KEGG | KEGG entry |
| MetaCyc | metabolic pathway |
| PRIAM | profile |
| PDB structures | RCSB PDB PDBe PDBsum |
| | Search | | --- | | PMC | articles | | PubMed | articles | | NCBI | proteins | | |
In cell biology, protein kinase A (PKA) is a family of serine-threonine kinases whose activity is dependent on cellular levels of cyclic AMP (cAMP). PKA is also known as cAMP-dependent protein kinase (EC 2.7.11.11). PKA has several functions in the cell, including regulation of glycogen, sugar, and lipid metabolism. It should not be confused with 5'-AMP-activated protein kinase (AMP-activated protein kinase).
History
[edit]
Protein kinase A, more precisely known as adenosine 3',5'-monophosphate (cyclic AMP)-dependent protein kinase, abbreviated to PKA, was discovered by chemists Edmond H. Fischer and Edwin G. Krebs in 1968. They won the Nobel Prize in Physiology or Medicine in 1992 for their work on phosphorylation and dephosphorylation and how it relates to PKA activity.
PKA is one of the most widely researched protein kinases, in part because of its uniqueness; out of 540 different protein kinase genes that make up the human kinome, only one other protein kinase, casein kinase 2, is known to exist in a physiological tetrameric complex, meaning it consists of four subunits.
The diversity of mammalian PKA subunits was realized after Dr. Stan McKnight and others identified four possible catalytic subunit genes and four regulatory subunit genes. In 1991, Susan Taylor and colleagues crystallized the PKA Cα subunit, which revealed the bi-lobe structure of the protein kinase core for the very first time, providing a blueprint for all the other protein kinases in a genome (the kinome).
Structure
[edit]
When inactive, the PKA apoenzyme exists as a tetramer which consists of two regulatory subunits and two catalytic subunits. The catalytic subunit contains the active site, a series of canonical residues found in protein kinases that bind and hydrolyse ATP, and a domain to bind the regulatory subunit. The regulatory subunit has domains to bind to cyclic AMP, a domain that interacts with catalytic subunit, and an auto inhibitory domain. There are two major forms of regulatory subunit; RI and RII.
Mammalian cells have at least two types of PKAs: type I is mainly in the cytosol, whereas type II is bound via its regulatory subunits and special anchoring proteins, described in the anchorage section, to the plasma membrane, nuclear membrane, mitochondrial outer membrane, and microtubules. In both types, once the catalytic subunits are freed and active, they can migrate into the nucleus (where they can phosphorylate transcription regulatory proteins), while the regulatory subunits remain in the cytoplasm.
The following human genes encode PKA subunits:
catalytic subunit – PRKACA, PRKACB, PRKACG
regulatory subunit type I - PRKAR1A, PRKAR1B
regulatory subunit type II - PRKAR2A, PRKAR2B
Mechanism
[edit]
Activation
[edit]
PKA is also commonly known as cAMP-dependent protein kinase, because it has traditionally been thought to be activated through release of the catalytic subunits when levels of the second messenger called cyclic adenosine monophosphate, or cAMP, rise in response to a variety of signals. However, recent studies evaluating the intact holoenzyme complexes, including regulatory AKAP-bound signalling complexes, have suggested that the local sub cellular activation of the catalytic activity of PKA might proceed without physical separation of the regulatory and catalytic components, especially at physiological concentrations of cAMP. In contrast, experimentally induced supra physiological concentrations of cAMP, meaning higher than normally observed in cells, are able to cause separation of the holoenzymes, and release of the catalytic subunits.
Extracellular hormones, such as glucagon and epinephrine, begin an intracellular signalling cascade that triggers protein kinase A activation by first binding to a G protein–coupled receptor (GPCR) on the target cell. When a GPCR is activated by its extracellular ligand, a conformational change is induced in the receptor that is transmitted to an attached intracellular heterotrimeric G protein complex by protein domain dynamics. The Gs alpha subunit of the stimulated G protein complex exchanges GDP for GTP in a reaction catalyzed by the GPCR and is released from the complex. The activated Gs alpha subunit binds to and activates an enzyme called adenylyl cyclase, which, in turn, catalyzes the conversion of ATP into cAMP, directly increasing the cAMP level. Four cAMP molecules are able to bind to the two regulatory subunits. This is done by two cAMP molecules binding to each of the two cAMP binding sites (CNB-B and CNB-A) which induces a conformational change in the regulatory subunits of PKA, causing the subunits to detach and unleash the two, now activated, catalytic subunits.
Once released from inhibitory regulatory subunit, the catalytic subunits can go on to phosphorylate a number of other proteins in the minimal substrate context Arg-Arg-X-Ser/Thr., although they are still subject to other layers of regulation, including modulation by the heat stable pseudosubstrate inhibitor of PKA, termed PKI.
Below is a list of the steps involved in PKA activation:
Cytosolic cAMP increases
Two cAMP molecules bind to each PKA regulatory subunit
The regulatory subunits move out of the active sites of the catalytic subunits and the R2C2 complex dissociates
The free catalytic subunits interact with proteins to phosphorylate Ser or Thr residues.
Catalysis
[edit]
The liberated catalytic subunits can then catalyze the transfer of ATP terminal phosphates to protein substrates at serine, or threonine residues. This phosphorylation usually results in a change in activity of the substrate. Since PKAs are present in a variety of cells and act on different substrates, PKA regulation and cAMP regulation are involved in many different pathways.
The mechanisms of further effects may be divided into direct protein phosphorylation and protein synthesis:
In direct protein phosphorylation, PKA directly either increases or decreases the activity of a protein.
In protein synthesis, PKA first directly activates CREB, which binds the cAMP response element (CRE), altering the transcription and therefore the synthesis of the protein. In general, this mechanism takes more time (hours to days).
Phosphorylation mechanism
[edit]
The Serine/Threonine residue of the substrate peptide is orientated in such a way that the hydroxyl group faces the gamma phosphate group of the bound ATP molecule. Both the substrate, ATP, and two Mg2+ ions form intensive contacts with the catalytic subunit of PKA. In the active conformation, the C helix packs against the N-terminal lobe and the Aspartate residue of the conserved DFG motif chelates the Mg2+ ions, assisting in positioning the ATP substrate. The triphosphate group of ATP points out of the adenosine pocket for the transfer of gamma-phosphate to the Serine/Threonine of the peptide substrate. There are several conserved residues, include Glutamate (E) 91 and Lysine (K) 72, that mediate the positioning of alpha- and beta-phosphate groups. The hydroxyl group of the peptide substrate's Serine/Threonine attacks the gamma phosphate group at the phosphorus via an SN2 nucleophilic reaction, which results in the transfer of the terminal phosphate to the peptide substrate and cleavage of the phosphodiester bond between the beta-phosphate and the gamma-phosphate groups. PKA acts as a model for understanding protein kinase biology, with the position of the conserved residues helping to distinguish the active protein kinase and inactive pseudokinase members of the human kinome.
Inactivation
[edit]
Downregulation of protein kinase A occurs by a feedback mechanism and uses a number of cAMP hydrolyzing phosphodiesterase (PDE) enzymes, which belong to the substrates activated by PKA. Phosphodiesterase quickly converts cAMP to AMP, thus reducing the amount of cAMP that can activate protein kinase A. PKA is also regulated by a complex series of phosphorylation events, which can include modification by autophosphorylation and phosphorylation by regulatory kinases, such as PDK1.
Thus, PKA is controlled, in part, by the levels of cAMP. Also, the catalytic subunit itself can be down-regulated by phosphorylation.
Anchorage
[edit]
The regulatory subunit dimer of PKA is important for localizing the kinase inside the cell. The dimerization and docking (D/D) domain of the dimer binds to the A-kinase binding (AKB) domain of A-kinase anchor protein (AKAP). The AKAPs localize PKA to various locations (e.g., plasma membrane, mitochondria, etc.) within the cell.
AKAPs bind many other signaling proteins, creating a very efficient signaling hub at a certain location within the cell. For example, an AKAP located near the nucleus of a heart muscle cell would bind both PKA and phosphodiesterase (hydrolyzes cAMP), which allows the cell to limit the productivity of PKA, since the catalytic subunit is activated once cAMP binds to the regulatory subunits.
Function
[edit]
PKA phosphorylates proteins that have the motif Arginine-Arginine-X-Serine exposed, in turn (de)activating the proteins. Many possible substrates of PKA exist; a list of such substrates is available and maintained by the NIH.
As protein expression varies from cell type to cell type, the proteins that are available for phosphorylation will depend upon the cell in which PKA is present. Thus, the effects of PKA activation vary with cell type:
Overview table
[edit]
| Cell type | Organ/system | Stimulators ligands → Gs-GPCRs or PDE inhibitors | Inhibitors ligands → Gi-GPCRs or PDE stimulators | Effects |
--- ---
| adipocyte | | epinephrine → β-adrenergic receptor glucagon → Glucagon receptor | | enhance lipolysis + stimulate lipase |
| myocyte (skeletal muscle) | muscular system | epinephrine → β-adrenergic receptor | | produce glucose + stimulate glycogenolysis - phosphorylate glycogen phosphorylase via phosphorylase kinase (activating it) - phosphorylate Acetyl-CoA carboxylase (inhibiting it) + inhibit glycogenesis - phosphorylate glycogen synthase (inhibiting it) + stimulate glycolysis - phosphorylate phosphofructokinase 2 (stimulating it, cardiomyocytes only) |
| myocyte (cardiac muscle) | cardiovascular | epinephrine → β-adrenergic receptor | | sequester Ca2+ in sarcoplasmic reticulum + phosphorylates phospholamban |
| myocyte (smooth muscle) | cardiovascular | β2 adrenergic agonists → β-2 adrenergic receptor histamine → Histamine H2 receptor prostacyclin → prostacyclin receptor Prostaglandin D2 → PGD2 receptor Prostaglandin E2 → PGE2 receptor VIP → VIP receptor L-Arginine → imidazoline and α2 receptor? (Gi-coupled) | muscarinic agonists, e.g. acetylcholine → muscarinic receptor M2 NPY → NPY receptor | Contributes to vasodilation (phosphorylates, and thereby inactivates, Myosin light-chain kinase) |
| hepatocyte | liver | epinephrine → β-adrenergic receptor glucagon → Glucagon receptor | | produce glucose + stimulate glycogenolysis - phosphorylate glycogen phosphorylase (activating it) - phosphorylate Acetyl-CoA carboxylase (inhibiting it) + inhibit glycogenesis - phosphorylate glycogen synthase (inhibiting it) + stimulate gluconeogenesis - phosphorylate fructose 2,6-bisphosphatase (stimulating it) + inhibit glycolysis - phosphorylate phosphofructokinase-2 (inactivating it) - phosphorylate fructose 2,6-bisphosphatase (stimulate it) - phosphorylate pyruvate kinase (inhibiting it) |
| neurons in nucleus accumbens | nervous system | dopamine → dopamine receptor | | Activate reward system |
| principal cells in kidney | kidney | Vasopressin → V2 receptor theophylline (PDE inhibitor) | | exocytosis of aquaporin 2 to apical membrane. synthesis of aquaporin 2 phosphorylation of aquaporin 2 (stimulating it) |
| Thick ascending limb cell | kidney | Vasopressin → V2 receptor | | stimulate Na-K-2Cl symporter (perhaps only minor effect) |
| Cortical collecting tubule cell | kidney | Vasopressin → V2 receptor | | stimulate Epithelial sodium channel (perhaps only minor effect) |
| Inner medullary collecting duct cell | kidney | Vasopressin → V2 receptor | | stimulate urea transporter 1 urea transporter 1 exocytosis |
| proximal convoluted tubule cell | kidney | PTH → PTH receptor 1 | | Inhibit NHE3 → ↓H+ secretion |
| juxtaglomerular cell | kidney | adrenergic agonists → β-receptor agonists → α2 receptor dopamine → dopamine receptor glucagon → glucagon receptor | | renin secretion |
In adipocytes and hepatocytes
[edit]
Epinephrine and glucagon affect the activity of protein kinase A by changing the levels of cAMP in a cell via the G-protein mechanism, using adenylate cyclase. Protein kinase A acts to phosphorylate many enzymes important in metabolism. For example, protein kinase A phosphorylates acetyl-CoA carboxylase and pyruvate dehydrogenase. Such covalent modification has an inhibitory effect on these enzymes, thus inhibiting lipogenesis and promoting net gluconeogenesis. Insulin, on the other hand, decreases the level of phosphorylation of these enzymes, which instead promotes lipogenesis. Recall that gluconeogenesis does not occur in myocytes.
In nucleus accumbens neurons
[edit]
PKA helps transfer/translate the dopamine signal into cells in the nucleus accumbens, which mediates reward, motivation, and task salience. The vast majority of reward perception involves neuronal activation in the nucleus accumbens, some examples of which include sex, recreational drugs, and food. Protein Kinase A signal transduction pathway helps in modulation of ethanol consumption and its sedative effects. A mouse study reports that mice with genetically reduced cAMP-PKA signalling results into less consumption of ethanol and are more sensitive to its sedative effects.
In skeletal muscle
[edit]
PKA is directed to specific sub-cellular locations after tethering to AKAPs. Ryanodine receptor (RyR) co-localizes with the muscle AKAP and RyR phosphorylation and efflux of Ca2+ is increased by localization of PKA at RyR by AKAPs.
In cardiac muscle
[edit]
In a cascade mediated by a GPCR known as β1 adrenoceptor, activated by catecholamines (notably norepinephrine), PKA gets activated and phosphorylates numerous targets, namely: L-type calcium channels, phospholamban, troponin I, myosin binding protein C, and potassium channels. This increases inotropy as well as lusitropy, increasing contraction force as well as enabling the muscles to relax faster.
In memory formation
[edit]
PKA has always been considered important in formation of a memory. In the fruit fly, reductions in expression activity of DCO (PKA catalytic subunit encoding gene) can cause severe learning disabilities, middle term memory and short term memory. Long term memory is dependent on the CREB transcription factor, regulated by PKA. A study done on drosophila reported that an increase in PKA activity can affect short term memory. However, a decrease in PKA activity by 24% inhibited learning abilities and a decrease by 16% affected both learning ability and memory retention. Formation of a normal memory is highly sensitive to PKA levels.
See also
[edit]
Protein kinase
Signal transduction
G protein-coupled receptor
Serine/threonine-specific protein kinase
Myosin light-chain kinase
cAMP-dependent pathway
References
[edit]
^ Jump up to: a b Turnham, Rigney E.; Scott, John D. (2016-02-15). "Protein kinase A catalytic subunit isoform PRKACA; History, function and physiology". Gene. 577 (2): 101–108. doi:10.1016/j.gene.2015.11.052. PMC 4713328. PMID 26687711.
^ Knighton, D. R.; Zheng, J. H.; Ten Eyck, L. F.; Xuong, N. H.; Taylor, S. S.; Sowadski, J. M. (1991-07-26). "Structure of a peptide inhibitor bound to the catalytic subunit of cyclic adenosine monophosphate-dependent protein kinase". Science. 253 (5018): 414–420. Bibcode:1991Sci...253..414K. doi:10.1126/science.1862343. ISSN 0036-8075. PMID 1862343.
^ Manning, G.; Whyte, D. B.; Martinez, R.; Hunter, T.; Sudarsanam, S. (2002-12-06). "The protein kinase complement of the human genome". Science. 298 (5600): 1912–1934. Bibcode:2002Sci...298.1912M. doi:10.1126/science.1075762. ISSN 1095-9203. PMID 12471243. S2CID 26554314.
^ Bauman AL, Scott JD (August 2002). "Kinase- and phosphatase-anchoring proteins: harnessing the dynamic duo". Nature Cell Biology. 4 (8): E203–6. doi:10.1038/ncb0802-e203. PMID 12149635. S2CID 1276537.
^ Alberts, Bruce (18 November 2014). Molecular biology of the cell (Sixth ed.). New York. p. 835. ISBN 978-0-8153-4432-2. OCLC 887605755.{{cite book}}: CS1 maint: location missing publisher (link)
^ Jump up to: a b Smith, FD; Esseltine, JL; Nygren, PJ; Veesler, D; Byrne, DP; Vonderach, M; Strashnov, I; Eyers, CE; Eyers, PA; Langeberg, LK; Scott, JD (2017). "Local protein kinase A action proceeds through intact holoenzymes". Science. 356 (6344): 1288–1293. Bibcode:2017Sci...356.1288S. doi:10.1126/science.aaj1669. PMC 5693252. PMID 28642438.
^ Jump up to: a b c Byrne, DP; Vonderach, M; Ferries, S; Brownridge, PJ; Eyers, CE; Eyers, PA (2016). "cAMP-dependent protein kinase (PKA) complexes probed by complementary differential scanning fluorimetry and ion mobility-mass spectrometry". Biochemical Journal. 473 (19): 3159–3175. doi:10.1042/bcj20160648. PMC 5095912. PMID 27444646.
^ Lodish; et al. (2016). "15.5". Molecular Cell Biology (8th ed.). W.H. Freeman and Company. p. 701. ISBN 978-1-4641-8339-3.
^ Voet, Voet & Pratt (2008). Fundamentals of Biochemistry, 3rd Edition. Wiley. Pg 432
^ Scott, JD; Glaccum, MB; Fischer, EH; Krebs, EG (1986). "Primary-structure requirements for inhibition by the heat-stable inhibitor of the cAMP-dependent protein kinase". PNAS. 83 (6): 1613–1616. Bibcode:1986PNAS...83.1613S. doi:10.1073/pnas.83.6.1613. PMC 323133. PMID 3456605.
^ "PKA Substrates". NIH.
^ Jump up to: a b c d e Rang HP (2003). Pharmacology. Edinburgh: Churchill Livingstone. ISBN 978-0-443-07145-4. Page 172
^ Rodriguez P, Kranias EG (December 2005). "Phospholamban: a key determinant of cardiac function and dysfunction". Archives des Maladies du Coeur et des Vaisseaux. 98 (12): 1239–43. PMID 16435604.
^ Jump up to: a b c d e Boron WF, Boulpaep EL (2005). Medical Physiology: A Cellular And Molecular Approach (Updated ed.). Philadelphia, Pa.: Elsevier Saunders. p. 842. ISBN 978-1-4160-2328-9.
^ Boron WF, Boulpaep EL (2005). Medical Physiology: A Cellular And Molecular Approaoch (Updated ed.). Philadelphia, Pa.: Elsevier Saunders. p. 844. ISBN 978-1-4160-2328-9.
^ Boron WF, Boulpaep EL (2005). Medical Physiology: A Cellular And Molecular Approach (Updated ed.). Philadelphia, Pa.: Elsevier Saunders. p. 852. ISBN 978-1-4160-2328-9.
^ Jump up to: a b c d Boron WF, Boulpaep EL (2005). Medical Physiology: A Cellular And Molecular Approach (Updated ed.). Philadelphia, Pa.: Elsevier Saunders. p. 867. ISBN 978-1-4160-2328-9.
^ Wand, Gary; Levine, Michael; Zweifel, Larry; Schwindinger, William; Abel, Ted (2001-07-15). "The cAMP–Protein Kinase A Signal Transduction Pathway Modulates Ethanol Consumption and Sedative Effects of Ethanol". Journal of Neuroscience. 21 (14): 5297–5303. doi:10.1523/JNEUROSCI.21-14-05297.2001. ISSN 0270-6474. PMC 6762861. PMID 11438605.
^ Ruehr, Mary L.; Russell, Mary A.; Ferguson, Donald G.; Bhat, Manju; Ma, Jianjie; Damron, Derek S.; Scott, John D.; Bond, Meredith (2003-07-04). "Targeting of Protein Kinase A by Muscle A Kinase-anchoring Protein (mAKAP) Regulates Phosphorylation and Function of the Skeletal Muscle Ryanodine Receptor". Journal of Biological Chemistry. 278 (27): 24831–24836. doi:10.1074/jbc.M213279200. ISSN 0021-9258. PMID 12709444.
^ Shah, Ajay M.; Solaro, R. John; Layland, Joanne (2005-04-01). "Regulation of cardiac contractile function by troponin I phosphorylation". Cardiovascular Research. 66 (1): 12–21. doi:10.1016/j.cardiores.2004.12.022. ISSN 0008-6363. PMID 15769444.
^ Boron, Walter F.; Boulpaep, Emile L. (2012). Medical physiology : a cellular and molecular approach. Boron, Walter F.,, Boulpaep, Emile L. (Updated second ed.). Philadelphia, PA. ISBN 9781437717532. OCLC 756281854.{{cite book}}: CS1 maint: location missing publisher (link)
^ Horiuchi, Junjiro; Yamazaki, Daisuke; Naganos, Shintaro; Aigaki, Toshiro; Saitoe, Minoru (2008-12-30). "Protein kinase A inhibits a consolidated form of memory in Drosophila". Proceedings of the National Academy of Sciences. 105 (52): 20976–20981. Bibcode:2008PNAS..10520976H. doi:10.1073/pnas.0810119105. ISSN 0027-8424. PMC 2634933. PMID 19075226.
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Cyclic+AMP-Dependent+Protein+Kinases at the U.S. National Library of Medicine Medical Subject Headings (MeSH)
Drosophila cAMP-dependent protein kinase 1 - The Interactive Fly
cAMP-dependent protein kinase: PDB Molecule of the Month
Overview of all the structural information available in the PDB for UniProt: P25321 (cAMP-dependent protein kinase catalytic subunit alpha) at the PDBe-KB.
Notes
[edit]
| v t e Intracellular signaling peptides and proteins | |
--- |
| MAP | see MAP kinase pathway |
| Calcium | Intracellular calcium-sensing proteins Calcineurin Ca2+/calmodulin-dependent protein kinase |
| G protein | | | | | | | | | | | | | | --- --- --- --- --- --- | | Heterotrimeric | | | | --- | | cAMP: | Heterotrimeric G protein + Gs/Gi Adenylate cyclase cAMP 3',5'-cyclic-AMP phosphodiesterase Protein kinase A | | cGMP: | Transducin Gustducin Guanylate cyclase cGMP 3',5'-cyclic-GMP phosphodiesterase Protein kinase G | | G alpha subunit Gα + GNAO1 + GNAI1 + GNAI2 + GNAI3 + GNAT1 + GNAT2 + GNAT3 + GNAZ + GNAS + GNAL + GNAQ + GNA11 + GNA12 + GNA13 + GNA14 + GNA15/GNA16 | | | G beta-gamma complex Gβ + GNB1 + GNB2 + GNB3 + GNB4 + GNB5 Gγ + GNGT1 + GNGT2 + GNG2 + GNG3 + GNG4 + GNG5 + GNG7 + GNG8 + GNG10 + GNG11 + GNG12 + GNG13 + BSCL2 | | | G protein-coupled receptor kinase AMP-activated protein kinase | | | | Monomeric | ARFs Rabs Ras + HRAS + KRAS + NRAS Rhos Arfs Ran Rhebs Raps RGKs | |
| Cyclin | Cyclin-dependent kinase inhibitor protein Cyclin-dependent kinase Cyclin |
| Lipid | Phosphoinositide phospholipase C Phospholipase C |
| Other protein kinase | | | | --- | | Serine/threonine: | Casein kinase + 1 + 2 eIF-2 kinase + EIF2AK3 Glycogen synthase kinase + GSK1 + GSK2 + GSK-3 + GSK3A + GSK3B IκB kinase + CHUK + IKK2 + IKBKG Interleukin-1 receptor associated kinase + IRAK1 + IRAK2 + IRAK3 + IRAK4 Lim kinase + LIMK1 + LIMK2 p21-activated kinases + PAK1 + PAK2 + PAK3 + PAK4 Rho-associated protein kinase + ROCK1 + ROCK2 Ribosomal s6 kinase + RPS6KA1 | | Tyrosine: | ZAP70 Focal adhesion protein-tyrosine kinase + PTK2 + PTK2B BTK | | Serine/threonine/tyrosine | Dual-specificity kinase + DYRK1A + DYRK1B + DYRK2 + DYRK3 + DYRK4 | | Arginine | Arginine kinase McsB | |
| Other protein phosphatase | | | | --- | | Serine/threonine: | Protein phosphatase 2 | | Tyrosine: | protein tyrosine phosphatase: Receptor-like protein tyrosine phosphatase Sh2 domain-containing protein tyrosine phosphatase | | both: | Dual-specificity phosphatase | |
| Apoptosis | see apoptosis signaling pathway |
| GTP-binding protein regulators | see GTP-binding protein regulators |
| Other | Activating transcription factor 6 Signal transducing adaptor protein I-kappa B protein Mucin-4 Olfactory marker protein Phosphatidylethanolamine binding protein EDARADD PRKCSH |
| see also deficiencies of intracellular signaling peptides and proteins | |
| v t e Kinases: Serine/threonine-specific protein kinases (EC 2.7.11-12) | |
--- |
| | Serine/threonine-specific protein kinases (EC 2.7.11.1-EC 2.7.11.20) | | --- | | | | | --- | | Non-specific serine/threonine protein kinases (EC 2.7.11.1) | LATS1 LATS2 MAST1 MAST2 STK38 STK38L CIT ROCK1 SGK SGK2 SGK3 Protein kinase B + AKT1 + AKT2 + AKT3 Ataxia telangiectasia mutated mTOR EIF-2 kinases + PKR + HRI + EIF2AK3 + EIF2AK4 Wee1 + WEE1 | | Pyruvate dehydrogenase kinase (EC 2.7.11.2) | PDK1 PDK2 PDK3 PDK4 | | Dephospho-(reductase kinase) kinase (EC 2.7.11.3) | AMP-activated protein kinase α + PRKAA1 + PRKAA2 β + PRKAB1 + PRKAB2 γ + PRKAG1 + PRKAG2 + PRKAG3 | | 3-methyl-2-oxobutanoate dehydrogenase (acetyl-transferring) kinase (EC 2.7.11.4) | BCKDK BCKDHA BCKDHB | | (isocitrate dehydrogenase (NADP+)) kinase (EC 2.7.11.5) | IDH2 IDH3A IDH3B IDH3G | | (tyrosine 3-monooxygenase) kinase (EC 2.7.11.6) | STK4 | | Myosin-heavy-chain kinase (EC 2.7.11.7) | Aurora kinase + Aurora A kinase + Aurora B kinase + Aurora C kinase | | Fas-activated serine/threonine kinase (EC 2.7.11.8) | FASTK STK10 | | Goodpasture-antigen-binding protein kinase (EC 2.7.11.9) | IκB kinase (EC 2.7.11.10) | CHUK IKK2 TBK1 IKBKE IKBKG IKBKAP | | cAMP-dependent protein kinase (EC 2.7.11.11) | Protein kinase A PRKACG PRKACB PRKACA PRKY | | cGMP-dependent protein kinase (EC 2.7.11.12) | Protein kinase G PRKG1 | | Protein kinase C (EC 2.7.11.13) | Protein kinase C Protein kinase Cζ PKC alpha PRKCB1 PRKCD PRKCE PRKCH PRKCG PRKCI PRKCQ Protein kinase N1 PKN2 PKN3 | | Rhodopsin kinase (EC 2.7.11.14) | Rhodopsin kinase | | Beta adrenergic receptor kinase (EC 2.7.11.15) | Beta adrenergic receptor kinase Beta adrenergic receptor kinase-2 | | G-protein coupled receptor kinases (EC 2.7.11.16) | GRK4 GRK5 GRK6 | | Ca2+/calmodulin-dependent (EC 2.7.11.17) | BRSK2 CAMK1 CAMK1A CAMK1B CAMK1D CAMK1G CAMK2 CAMK2A CAMK2B CAMK2D CAMK2G CAMK4 MLCK CASK CHEK1 CHEK2 DAPK1 DAPK2 DAPK3 STK11 MAPKAPK2 MAPKAPK3 MAPKAPK5 MARK1 MARK2 MARK3 MARK4 MELK MKNK1 MKNK2 NUAK1 NUAK2 OBSCN PASK PHKG1 PHKG2 PIM1 PIM2 PKD1 PRKD2 PRKD3 PSKH1 SNF1LK2 KIAA0999 STK40 SNF1LK SNRK SPEG TSSK2 Kalirin TRIB1 TRIB2 TRIB3 TRIO Titin DCLK1 | | Myosin light-chain kinase (EC 2.7.11.18) | MYLK MYLK2 MYLK3 MYLK4 | | Phosphorylase kinase (EC 2.7.11.19) | PHKA1 PHKA2 PHKB PHKG1 PHKG2 | | Elongation factor 2 kinase (EC 2.7.11.20) | EEF2K STK19 | | Polo kinase (EC 2.7.11.21) | PLK1 PLK2 PLK3 PLK4 | | | | |
| | Serine/threonine-specific protein kinases (EC 2.7.11.21-EC 2.7.11.30) | | --- | | | | | --- | | Polo kinase (EC 2.7.11.21) | PLK1 PLK2 PLK3 PLK4 | | Cyclin-dependent kinase (EC 2.7.11.22) | CDK1 CDK2 CDKL2 CDK3 CDK4 CDK5 CDKL5 CDK6 CDK7 CDK8 CDK9 CDK10 CDK12 CDC2L5 PCTK1 PCTK2 PCTK3 PFTK1 CDC2L1 | | (RNA-polymerase)-subunit kinase (EC 2.7.11.23) | RPS6KA5 RPS6KA4 P70S6 kinase P70-S6 Kinase 1 RPS6KB2 RPS6KA2 RPS6KA3 RPS6KA1 RPS6KC1 | | Mitogen-activated protein kinase (EC 2.7.11.24) | Extracellular signal-regulated + MAPK1 + MAPK3 + MAPK4 + MAPK6 + MAPK7 + MAPK12 + MAPK15 C-Jun N-terminal + MAPK8 + MAPK9 + MAPK10 P38 mitogen-activated protein + MAPK11 + MAPK13 + MAPK14 | | MAP3K (EC 2.7.11.25) | MAP kinase kinase kinases + MAP3K1 + MAP3K2 + MAP3K3 + MAP3K4 + MAP3K5 + MAP3K6 + MAP3K7 + MAP3K8 RAFs + ARAF + BRAF + KSR1 + KSR2 MLKs + MAP3K12 + MAP3K13 + MAP3K9 + MAP3K10 + MAP3K11 + MAP3K7 + ZAK CDC7 MAP3K14 | | Tau-protein kinase (EC 2.7.11.26) | TPK1 TTK GSK-3 | | (acetyl-CoA carboxylase) kinase (EC 2.7.11.27) | Tropomyosin kinase (EC 2.7.11.28) | Low-density-lipoprotein receptor kinase (EC 2.7.11.29) | Receptor protein serine/threonine kinase (EC 2.7.11.30) | Bone morphogenetic protein receptors + BMPR1 + BMPR1A + BMPR1B + BMPR2 ACVR1 ACVR1B ACVR1C ACVR2A ACVR2B ACVRL1 Anti-Müllerian hormone receptor | | | | |
| | Dual-specificity kinases (EC 2.7.12) | | --- | | | | | --- | | MAP2K | MAP2K1 MAP2K2 MAP2K3 MAP2K4 MAP2K5 MAP2K6 MAP2K7 | | | | |
| v t e | |
--- |
| Activity | Active site Binding site Catalytic triad Oxyanion hole Enzyme promiscuity Diffusion-limited enzyme Cofactor Enzyme catalysis |
| Regulation | Allosteric regulation Cooperativity Enzyme inhibitor Enzyme activator |
| Classification | EC number Enzyme superfamily Enzyme family List of enzymes |
| Kinetics | Enzyme kinetics Eadie–Hofstee diagram Hanes–Woolf plot Lineweaver–Burk plot Michaelis–Menten kinetics |
| Types | EC1 Oxidoreductases (list) EC2 Transferases (list) EC3 Hydrolases (list) EC4 Lyases (list) EC5 Isomerases (list) EC6 Ligases (list) EC7 Translocases (list) |
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Articles with short description
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12650 | https://pdfs.semanticscholar.org/6e33/889a5fc31c24d55ed61f291f577536e7ebf2.pdf | Journal of Artificial Intelligence Research 74 (2022) 917-955 Submitted 12/2021; published 06/2022 Path Counting for Grid-Based Navigation Rhys Goldstein rhys.goldstein@autodesk.com Kean Walmsley kean.walmsley@autodesk.com Jacobo Bibliowicz jacky.bibliowicz@autodesk.com Alexander Tessier alex.tessier@autodesk.com Autodesk Research, Toronto, ON, Canada Simon Breslav simon.breslav@trax.co Azam Khan azam.khan@trax.co Trax.Co, Toronto, ON, Canada Abstract Counting the number of shortest paths on a grid is a simple procedure with close ties to Pascal’s triangle. We show how path counting can be used to select relatively direct grid paths for AI-related applications involving navigation through spatial environments. Typ-ical implementations of Dijkstra’s algorithm and A prioritize grid moves in an arbitrary manner, producing paths which stray conspicuously far from line-of-sight trajectories. We find that by counting the number of paths which traverse each vertex, then selecting the vertices with the highest counts, one obtains a path that is reasonably direct in practice and can be improved by refining the grid resolution.
Central Dijkstra and Central A are introduced as the basic methods for computing these central grid paths. Theoretical analysis reveals that the proposed grid-based navigation approach is related to an existing grid-based visibility approach, and establishes that central grid paths converge on clear sightlines as the grid spacing approaches zero. A more general property, that central paths converge on direct paths, is formulated as a conjecture.
1. Introduction Computational methods for navigation are essential to many AI-related applications involv-ing spatial environments, particularly in the domains of video games (Botea et al., 2013), robotics (Noreen et al., 2016), and architectural design (Pelechano et al., 2008; Nagy et al., 2017). Some of these methods adhere to a grid-based approach in which (1) the spatial environment is represented by a regular grid of vertices; (2) the vertices are processed using some form of graph, tree, or array traversal algorithm; and (3) after processing each vertex, information is propagated to neighboring vertices. A typical grid-based navigation method will handle straight-line paths perfectly if they are oriented such that they pass directly through neighboring vertices, but not if they are oriented at any other angle. The purpose of this paper is to demonstrate a strictly grid-based approach for approximating straight path trajectories that head in arbitrary directions. The new approach is based on the simple and well-known procedure of counting the number of shortest paths on a grid.
The outcome of this work is a practical solution to a common problem in grid path planning. If there exists a grid path from some vertex A to another vertex B, there are often many shortest grid paths with the same length.
An example of a case involving multiple shortest grid paths is shown in Figure 1. Typical grid-based implementations of Dijkstra’s algorithm and A prioritize grid moves using arbitrary tie-breaking conventions, ©2022 AI Access Foundation. All rights reserved.
Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan (a) A relatively indirect shortest grid path (b) A more direct shortest grid path Figure 1: Two shortest grid paths (purple arrows) are shown alongside illustrations of corresponding smoothed paths (green curves). Whereas a simple rule for prioritizing moves might generate a grid path like the one in (a), the grid path in (b) is more direct and should yield a shorter smoothed path.
which tends to produce relatively indirect paths. For instance, a convention to lead with diagonal moves would produce the relatively indirect shortest grid path in Figure 1a. The problem is how to select more direct shortest grid paths like the one in Figure 1b. The first grid path strays conspicuously far from line-of-sight trajectories as it traverses obstacle-free regions of the environment. The second grid path adheres reasonably well to sightlines in open regions. Either grid path could be smoothed in a post-processing step, but the grid path in Figure 1b should yield a shorter smoothed path than the one in Figure 1a.
918 Path Counting for Grid-Based Navigation It is possible to select relatively direct grid paths by conducting numerous line-of-sight tests or performing other geometric calculations. However, we find that decent results can be achieved using a simple path counting technique. Our solution is to count the number of shortest grid paths which traverse each vertex en route from A to B, then select the vertices with the highest counts. The approach still requires a basic grid path planning method such as Dijkstra’s algorithm or A, as one must construct a directed acyclic graph of all shortest grid paths. Once this is done, the counting procedure can be applied. We refer to this approach as central grid path planning, and introduce Central Dijkstra and Central A as specific methods for computing central paths. We also present a theoretical analysis consisting of (1) observations on the relationship between path counting and an existing grid-based visibility approach; (2) an argument based on the central limit theorem that central grid paths converge on clear sightlines as the grid spacing approaches zero; and (3) a conjecture that central paths also converge on direct paths.
2. Background and Assumptions We begin by defining terms and reviewing the basic concepts and assumptions used through-out the paper. The review presents related work on paths, grids, grid paths, grid path planning, any-angle path planning, and grid-based visibility.
2.1 Paths A path is a directed curve which stretches from one point to another without intersecting itself or passing through obstacles. If one imagines a path as a string, then pulling on the string will generally cause the path to shorten until it is pulled taut. A taut path is a path that does not get shorter when one “pulls” on the ends (Oh & Leong, 2017). We introduce the concept of a direct path, which means that a sightline between any pair of points on the path must be part of the path itself. A direct path is always taut, but a taut path is not necessarily direct. There is also the well-known concept of a shortest path, a path with the minimum length of all possible paths between a given pair of endpoints. A shortest path is always direct, but a direct path is not necessarily among the shortest. Non-taut, taut, direct, and shortest paths are illustrated in Figure 2.
Applications involving navigation have diverse requirements. Nevertheless, we assume shortest paths are generally preferred over direct paths, direct paths over taut paths, and taut paths over non-taut paths. In addition, we assume paths which share the same topology as a shortest path are preferred over paths which do not. Two paths with the same endpoints share the same topology if the region between them is free of obstacles. Of the paths in Figure 2, only the first two have a common topology.
Paths may partially or fully reside on the boundaries of obstacles as long as there is always a traversable region on at least one side. A taut, direct, or shortest path can only contact obstacle boundaries at certain places: at the endpoints of the path, at convex corners encountered along the path, or alongside straight or convex surfaces of obstacles.
Everywhere else, such paths are perfectly straight. Paths may or may not be permitted to pass through single-point gaps, and the methods and associated theory presented in this paper should apply regardless of which convention is used. Examples of convex and concave surfaces and corners, and a single-point gap, are indicated in Figure 2.
919 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan Figure 2: An example featuring a (1) non-taut path, (2) taut path, (3) direct path, and (4) shortest path between points A and B. The dashed sightline connecting two points on the taut path illustrates why the path is not direct. Also shown are a (P) convex surface, (Q) concave surface, (R) convex corner, (S) concave corner, and (T) single-point gap.
While our focus is on grid-based methods, paths can also be computed on a continuous domain using analytic approaches. One such approach is the visibility graph, where (1) ob-stacles are represented by polygons, (2) corners of polygons and endpoints of paths become vertices in a graph, (3) sightlines between those vertices become edges in the graph, and (4) paths are computed using a graph-based shortest path algorithm. The visibility graph was introduced by Lozano-P´ erez and Wesley (1979), who also describe the useful technique of enlarging obstacles to enforce a minimum passage width. Analytic methods tend to give exact results if the analytic representation of the environment is exact. For example, the visibility graph gives exact shortest paths for environments with no curved surfaces.
2.2 Grids Conventional grid-based methods employ a regular grid of vertices, and restrict information flow such that the state of each vertex can directly influence only neighboring vertices. A neighborhood is a set of offsets that determine which vertices are considered neighbors. In 2D, commonly used neighborhoods include the 4- and 8-neighborhood on a rectangular grid and the 6-neighborhood on a triangular grid. Rivera et al. (2020) provide a detailed description of the broader set of 2k-neighborhoods on rectangular grids (k ≥2), and an analogous set of (3×2k)-neighborhoods could be described for triangular grids (k ≥1). We refer to these two sets of neighborhoods collectively as the standard 2D grid neighborhoods.
Examples of these neighborhoods appear in Figure 3.
A move is a vector from one vertex to any of its neighbors. The 4-neighborhood and the 6-neighborhood include only cardinal moves, which point to the nearest vertices along the primary axes of the grid.
We assume all cardinal moves have a length of one grid spacing. Each successive neighborhood of the same grid type is constructed by taking each pair of adjacent moves and inserting a new move between them. Each inserted move is the vector sum of the original two (Rivera et al., 2020). For example, each diagonal move in the 8-neighborhood is the vector sum of the two surrounding cardinal moves.
920 Path Counting for Grid-Based Navigation (a) 4-neighborhood (b) 6-neighborhood (c) 8-neighborhood (d) 12-neighborhood (e) 16-neighborhood Figure 3: Examples of standard neighborhoods for rectangular and triangular grids.
Unless otherwise stated, all grid path planning examples and illustrations in this paper use the 8-neighborhood. This should not be interpreted as an endorsement of 8-neighbor grids over the other options. The proposed methods and associated theory pertain to all of the standard 2D grid neighborhoods described above, and should generalize to certain 3D grids as well. Although grid-based navigation methods with larger neighborhoods tend to be more challenging to implement, they can be expected to yield shorter paths.
2.3 Grid Paths A grid path begins at a start vertex, then follows a sequence of moves to other not-yet visited vertices until reaching a destination vertex. Using a standard 2D grid neighborhood, and assuming no obstacles, the shortest grid paths between two vertices A and B are sequences involving at most two distinct bracketing moves. If the vector from A to B is aligned with a neighborhood move, then that move is the sole bracketing move. Otherwise, the bracketing moves are the pair of adjacent neighborhood moves that lie on either side of the A-B vector.
To get from [0, 0] to [6, 3] using the 8-neighborhood, for example, the bracketing moves are [1, 0] and [1, 1] and one needs three of each. More generally, if the vector from A to B is [x, y], and if the two bracketing moves are u = [ux, uy] and v = [vx, vy], then the shortest possible grid paths include exactly m moves in the u direction and k moves in the v direction, where m and k are given by the coordinate transformation below.
m k = ux vx uy vy −1 x y The transformation simplifies to the following.
m = vyx −vxy uxvy −vxuy k = −uyx + uxy uxvy −vxuy (1) 921 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan Adding the lengths of all moves in one of these sequences yields a well-known metric that we refer to as the standard 2D grid distance, and denote h(x, y).
h(x, y) = m q ux2 + uy2 + k q vx2 + vy2 (2) The standard grid distance reduces to what is known as the Manhattan distance for 4-neighbor grids or the octile distance for 8-neighbor grids.
Rivera et al. (2020) prove that h(x, y) is the minimum possible grid path length for all standard rectangular 2D grid neighborhoods, and provide algorithms which compute this metric for up to 64 neighbors.
Obstacles have the effect of disallowing moves between certain pairs of vertices. In the presence of obstacles, a shortest grid path between A and B may be longer than the standard grid distance, or there may be no grid paths at all between the two points. Given a set of obstacles defined on a continuous domain, a grid-based approximation of the environment can be constructed by overlaying a grid of vertices. Moves between two neighboring vertices are allowed if and only if there is a straight-line path between them.
Sometimes the obstacle geometry itself takes the form of a grid, either because a preexist-ing continuous representation of the environment was rasterized or because the environment was originally designed as a grid. Grid-based obstacle geometry can usually be regarded as an array of square, triangular, or hexagonal cells, where each cell is either blocked or unblocked. When a rectangular grid of vertices is overlaid on an environment of square cells, it is common practice to place the vertices on the centers of cells, as in Figure 4a, or on the corners of cells, as in Figure 4b. We assume that vertices on triangular grids would be placed on the centers of hexagonal cells or on the corners of triangular cells.
(a) Vertices placed on cell centers (b) Vertices placed on cell corners Figure 4: Vertex placement with grid-based obstacle geometry.
While theoretical and empirical studies have shown that placing vertices on square cell corners produces shorter paths (Bailey et al., 2015), the cell center convention may be more convenient for certain applications such as tile-based video games.
The grid-based methods proposed in this paper work with either convention, and in either case we treat grid-based obstacles the same as continuous obstacles. If a straight-line path exists between two neighboring vertices, moves between those vertices are allowed. An important consideration arises for grid-based obstacle geometry with single-point gaps, such as the gap between vertices P and Q in Figure 4a or at vertex R in Figure 4b. Paths through such gaps may be either permitted or prohibited.
922 Path Counting for Grid-Based Navigation 2.4 Grid Path Planning The classic approach to grid path planning involves setting up a grid-based representation of an environment, as described in Section 2.3, then applying Dijkstra’s graph-based shortest path algorithm (Dijkstra, 1959), or its A variant (Hart et al., 1968), to find one or more shortest paths.
Grid-based implementations of Dijkstra’s algorithm typically compute a hierarchy of paths between a source vertex and all accessible vertices in the environment. Every vertex is assigned an initial grid distance of infinity, except the source which has a grid distance of zero. The algorithm proceeds by expanding not-yet processed vertices in order of grid distance. When a vertex is expanded, finite grid distances are computed for its neighboring vertices making them eligible for expansion. If different grid distances are computed for the same vertex, the shortest is always selected. Once this search procedure is complete, shortest grid paths can be generated by retracing moves from any processed vertex to the source vertex. If needed, paths can be reversed so that they start at the source.
Grid-based implementations of A are similar, except that vertices are expanded in an order that more efficiently reaches a pre-selected goal vertex. With A, vertices are ordered by their grid distance from the source plus a lower-bound estimate of their distance to the goal. In this paper, we assume the lower-bound estimate is the standard 2D grid distance h(x, y) specified in (2), though other heuristics are sometimes used. If the goal is among the unprocessed vertices with the minimum combined distance, the search procedure is terminated and a shortest grid path is generated from the goal to the source. The path can be reversed if needed.
Typical implementations of Dijkstra’s algorithm or A employ some form of tie-breaking during the search procedure, thereby selecting a single solution from the multitude of short-est grid paths that usually exist between the source and goal. One tie-breaking convention is to generate what are known as canonical paths by prioritizing diagonal moves over car-dinal moves on an 8-neighbor grid (Sturtevant & Rabin, 2016). The shortest grid path in Figure 1a is an example of a canonical path, assuming vertex A is the source. The approach can be generalized to larger neighborhoods by prioritizing moves that are vector sums of the two surrounding moves (Rivera et al., 2020).
Canonical paths arise out of the work of Harabor and Grastien (2011), who demonstrate that prioritizing diagonal moves allows the A algorithm to jump over certain vertices dur-ing the search procedure. The result is a faster variant of A called Jump Point Search.
Sturtevant and Rabin (2016) use the same ordering of moves to propose the Canonical Di-jkstra and Canonical A methods. Canonical Dijkstra incorporates jumping into Dijkstra’s algorithm to reduce the number of vertex expansions. Canonical A avoids jumping, but demonstrates that canonical ordering alone can reduce the number of neighbors that need to be visited for each vertex expansion in A.
For implementations of Dijkstra’s algorithm or A that do not enforce a canonical order-ing, the selected shortest grid path is typically determined by other tie-breaking conventions.
These conventions may include a clockwise or counterclockwise ordering of neighborhood moves, a lexicographical ordering of vertex coordinates, or any tie-break rule for nodes in a heap-based priority queue. The resulting paths may not be as extreme as the example in Figure 1a, but they tend to be relatively indirect compared with the path in Figure 1b.
923 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan An alternative to the above tie-breaking conventions is to represent all possible shortest grid paths leading back to the source vertex. This adaptation of Dijkstra’s algorithm or A is largely a matter of recording, for every visited vertex, the set of predecessor vertices that are traversed by any shortest grid path en route to the source. The result is a directed acyclic graph of paths, rather than a hierarchy. For A, a slight change to the termination condition is also required. To capture all shortest paths, the search procedure must be continued until the goal vertex is not just one of the unprocessed vertices with the minimum combined distance, but rather the only unprocessed vertex with the minimum combined distance.
For long, meandering paths, this stricter stopping criterion should not cause a significant increase in the number of vertices that need to be expanded. Our work shows how the all-paths variants of Dijkstra’s algorithm and A can be used to select relatively direct grid paths, improving quality rather than speed.
2.5 Any-Angle Path Planning Any-angle path planning methods make use of a path planning grid and typically assume grid-based obstacle geometry, yet depart in some way from the usual constraints of grid-based approaches to achieve shorter, more direct paths. Field D departs from grid-based constraints by allowing paths to bend at points located between pairs of vertices (Ferguson & Stentz, 2006). Theta uses line-of-sight checks to connect grid vertices across distances far greater than the size of the neighborhood (Daniel et al., 2010). In the Anya method, the hierarchy that is constructed during the search procedure is not a hierarchy of vertices, but rather a hierarchy of line segments (Harabor et al., 2016). Whereas most any-angle path planning methods approximate shortest paths, Anya can be used to find exact shortest paths assuming the grid-based obstacle geometry is exact.
Some any-angle methods speed up the path planning process by performing a precompu-tation on the environment before any source or goal vertices are selected. Subgoal Graphs is an any-angle path planning method that uses a precomputation phase to identify key vertices, called subgoals, at the corners of grid-based obstacles (Uras et al., 2013). Paths are found on a graph of these subgoals, similar to the analytic visibility graph method, and then refined as needed. Block A uses a hierarchical approach, partitioning grid-based environments into blocks of, for example, 5-by-5 vertices (Yap et al., 2011). Shortest path distances are precomputed between the grid vertices on the boundaries of each block, al-lowing subsequent A searches to treat blocks rather than vertices as nodes. Block A is an any-angle method assuming an analytic or any-angle method is used to precompute distances within each block. If a strictly grid-based method is used for the precomputation, then the overall method could also be considered grid-based.
Any-angle path planning is often contrasted with the simple approach of using a classic grid path planning method, as in Section 2.4, then smoothing the resulting grid path in a post-processing step. Botea et al. (2004) describe what has become a well-known path smoothing technique employing a succession of line-of-sight checks. Beginning at a vertex on one end of a grid path, the procedure is to repeatedly delete the subsequent vertex as long as the resulting path does not go through an obstacle.
This greedy algorithm is usually successful at straightening and shortening paths, though it sometimes leaves conspicuous triangle-shaped detours unresolved. Han et al. (2020) propose an enhanced 924 Path Counting for Grid-Based Navigation “string-pulling” algorithm where line-of-sight checks are used not only to delete points, but also to insert new points on obstacle corners around which the path is “pulled” taut. Paths can also be smoothed using relaxation approaches that slightly shift the position of each point (Richardson & Olson, 2011), or interpolation approaches that replace piecewise linear intervals with polynomial curves, B´ ezier curves, or splines (Ravankar et al., 2018). Such local adjustments can be used to soften sharp turns or increase clearance from obstacles, but may be inefficient at straightening highly indirect grid paths. Regardless of the approach, smoothing a path does not change its topology.
2.6 Grid-Based Visibility Computational methods for visibility have several similarities to those developed for navi-gation. First, visibility methods are applied in many of the same domains, including video games (Cohen-Or et al., 2003), robotics (Morini et al., 2010), and architectural design (Turner et al., 2001; Nagy et al., 2017). They are also formulated in a similar way, with analytic or grid-based representations of obstacles in a 2D or 3D spatial environment. In a visibility context, obstacles are interpreted as optical barriers rather than travel barriers, and the paths of interest are generally straight-line paths called sightlines.
While visibility problems are most often tackled with analytic methods, as reviewed by Ghosh (2007), or raycasting, as described by Roth (1982), there are also grid-based methods that will prove closely related to our path counting technique for navigation. Although grid-based visibility methods appear to enjoy only limited use outside of academia, they offer similar advantages to grid path planning and other grid-based approaches. They are easy to implement, and allow the trade-offbetween speed and accuracy to be conveniently adjusted by varying the grid spacing.
A grid-based visibility method from the level set community was derived from a partial differential equation by Tsai et al. (2004) and further validated by Kao and Tsai (2008).
The method approximates the region visible from a source point. The original formulation assumes implicit obstacle geometry defined on a 4-neighbor grid, but here we adapt the method to use an explicit representation consistent with Section 2.3. We represent obstacles using a set of binary values Vr q, where Vr q = 1 means neighboring vertices q and r are mutually visible and Vr q = 0 means the sightline between them is blocked. The output of the method is a set of visibility scores ψx,y in the range 0 ≤ψx,y ≤1. If ψx,y ≥0.5, we classify [x, y] as visible. Assuming the source is at [0, 0] and the grid spacing is 1, the visibility scores for the first quadrant of a 4-neighbor grid are given by (3).
ψ0,0 = 1 ψx,0 = ψx−1,0Vx,0 x−1,0 x > 0 ψ0,y = ψ0,y−1V0,y 0,y−1 y > 0 ψx,y = xψx−1,yVx,y x−1,y + yψx,y−1Vx,y x,y−1 /(x + y) x, y > 0 (3) The calculation in (3) can be applied using a standard array traversal algorithm (e.g.
[x, y] = [0, 0], [1, 0], [2, 0], ..., [0, 1], [1, 1], [2, 1], ...). It can be extended to the other three 925 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan quadrants by negating the x coefficient, the y coefficient, or both. It can be generalized to 3D by adding the obvious z terms to the numerator and denominator of the quotient.
Although the original level set formulation assumes the 4-neighborhood, we can extend the method to employ any of the standard 2D grid neighborhoods in Section 2.2. Recall from Section 2.3 that every vertex [x, y] is bracketed by at most two moves u and v. Having identified these vectors, the coordinate transformation in (1) can be used to obtain the number of moves m and k in the respective directions. These move counts can then be substituted into the extended formula below. The effect of these steps is to transform the coordinate space such that u is mapped to [1, 0], v is mapped to [0, 1], and [x, y] is mapped to [m, k]. This linear transformation allows the validated 4-neighbor formula to be applied with [m, k] in place of [x, y].
ψ0,0 = 1 ψm,0 = ψm−1,0Vm,0 m−1,0 m > 0 ψ0,k = ψ0,k−1V0,k 0,k−1 k > 0 ψm,k = mψm−1,kVm,k m−1,k + kψm,k−1Vm,k m,k−1 /(m + k) m, k > 0 (4) Consider applying this extended method to the 8-neighborhood, specifically within the cone between the positive x axis and the x = y diagonal. Any vertex [x, y] in this region is bracketed by the moves [1, 0] and [1, 1]. Substituting these vectors into (1) yields m = x−y and k = y, and substituting these expressions into (4) yields the formula below. Similar equations could be derived for all 8 convex cones generated by pairs of adjacent moves.
ψx,y = (x −y)ψx−1,yVx,y x−1,y + yψx−1,y−1Vx,y x−1,y−1 /x 0 < y < x Formulating this method for all 3D neighborhoods would require a 3D coordinate trans-form analogous to the one in Section 2.3, but let us focus on the 26-neighborhood and consider just the convex cone generated by [1, 0, 0], [1, 1, 0], and [1, 1, 1].
The visibility scores for vertices in this region are computed as follows, and similar equations exist for all 48 tetrahedral regions.
ψx,y,z = (x −y)ψx−1,y,zVx,y,z x−1,y,z + (y −z)ψx−1,y−1,zVx,y,z x−1,y−1,z + zψx−1,y−1,z−1Vx,y,z x−1,y−1,z−1 /x 0 < z < y < x Another grid-based visibility method was proposed by Fisher-Gewirtzman et al. (2013) for the urban design community. Formulated specifically for 3D voxel grids using the 26-neighborhood, their method is similar to the 26-neighbor variation of the adapted level set method discussed above. However, instead of performing linear interpolations on sets of three neighboring vertices within 48 tetrahedral regions, Fisher-Gewirtzman et al. perform bi-linear interpolations on sets of four neighboring vertices within 24 pyramidal regions.
We refer to our adaptation of the level set visibility method as the linear grid-based visibility approach, and suggest that the 26-neighbor algorithm of Fisher-Gewirtzman et al.
be called the bi-linear grid-based visibility approach. The linear approach has a particularly strong relationship to our proposed use of path counting for grid-based navigation.
926 Path Counting for Grid-Based Navigation 3. Central Grid Path Planning We now demonstrate how path counting can be used to select relatively direct grid paths, and explain how the technique can be implemented as an extension to Dijkstra’s algorithm or A. An empirical study compares the proposed Central A method to A and Theta.
3.1 Path Planning with Counting Our path planning approach is inspired by Pascal’s triangle, the famous number pattern illustrated in Figure 5. In Pascal’s triangle, a 1 is placed at the apex and repeated along the two diverging sides. Every other number is generated by adding two preceding numbers.
Figure 5: The top several rows of Pascal’s triangle.
It is widely understood that the recursive procedure which generates Pascal’s triangle is also a path counting procedure. The apex of the triangle can be viewed as a source vertex on a standard 2D grid, and every integer is the number of shortest grid paths leading to that vertex from the source. For example, there are 70 shortest grid paths between the top of the triangle in Figure 5 and the vertex in the middle of the bottom row.
The relevance of Pascal’s triangle to our path planning problem is most easily seen on a grid with no obstacles. Consider the task of selecting a relatively direct shortest grid path from vertex A to vertex B on the obstacle-free grid in Figure 6a. Figure 6b shows all of the shortest grid paths, which collectively form a parallelogram. To obtain a path that heads through the middle of the parallelogram, a first attempt might be to label each vertex with the number of paths from A. As shown in Figure 6c, this yields a parallelogram-shaped slice of Pascal’s triangle. Prioritizing vertices with higher Pascal numbers would produce a path that heads relatively directly from A to the vertex labeled 70, and then proceeds toward B.
The goal is a grid path that heads from A to B as directly as possible.
The key to approximating a direct path is to perform the counting procedure from both ends. As shown in Figure 6d, the path counts from B form the same Pascal number pattern as the path counts from A, except flipped across both axes. The next step is to multiply the two opposing sets of path counts to produce another type of path count. By taking the number of paths from A to a vertex, then multiplying by the number of paths from the vertex to B, one ends up with the number of paths which traverse the vertex en route 927 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan (a) An obstacle-free grid (b) All shortest grid paths between A and B (c) The path counts from A (d) The path counts from B (e) The traversal counts (f) The central path Figure 6: The central path approach on a grid with no obstacles.
from A to B. We refer to this type of path count as a traversal count. Multiplying the path counts in Figure 6c and Figure 6d yields the traversal counts in Figure 6e. The final step is to generate a relatively direct grid path by starting at one end and selecting, for each move, the next vertex with the highest traversal count. Figure 6f shows the selected path for this obstacle-free example. We refer to such a path as a central grid path, or central path.
If obstacles are present in the environment, one cannot use Pascal’s triangle directly. Yet it is still possible to employ the recursive counting procedure on which the famous number pattern is based. To illustrate, consider the scenario in Figure 7a. The task is to identify a relatively direct shortest grid path that circumnavigates the travel barriers. First, one must find all shortest grid paths from A to B, as shown in Figure 7b. Next, path counts are computed from one end, as in Figure 7c, and then the other, as in Figure 7d. Similar to Pascal’s triangle, the path counting procedure begins by assigning 1 to the first vertex, then populating each succeeding vertex by adding the path counts of its predecessors. The remaining steps are the same with or without obstacles. Each vertex has two path counts, 928 Path Counting for Grid-Based Navigation (a) A grid with obstacles (b) All shortest grid paths between A and B (c) The path counts from A (d) The path counts from B (e) The traversal counts (f) The central paths Figure 7: The central path approach on a grid with obstacles.
one from A and one from B, and multiplying them yields the traversal counts in Figure 7e.
One then selects the highest traversal counts, starting at A and proceeding toward B.
As indicated by the fork near the top right of Figure 7f, there are actually two central paths in this example. The reason for the two possible solutions is that, as can be seen in Figure 7e, one must choose between two vertices that both have a traversal count of 54. If there are multiple central grid paths, an implementation can choose any one of them. In general, central paths account for a small subset of all shortest grid paths between a pair of vertices. The purpose of the central path approach is to obtain a relatively direct shortest grid path by ensuring the selected path is a member of this subset.
Figure 8 shows what a central grid path may look like in practice. The path, which was generated using the architectural space analysis tool described by Goldstein et al. (2020), remains reasonably close to line-of-sight trajectories as it traverses obstacle-free regions of the environment. It is a considerably more direct path than one would expect from a typical grid-based implementation of Dijkstra’s algorithm or A with no path counting.
929 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan Figure 8: A central path computed on a building floor plan using a grid spacing of 25 cm.
3.2 Central Dijkstra and Central A Conceptually, the central path approach is as simple as finding all shortest grid paths, counting the numbers of paths that traverse each vertex, and selecting the vertices with the highest traversal counts. Implementing the approach is also reasonably simple, though the steps involved deserve some discussion.
Due to the need to count all shortest grid paths, it makes sense to begin with an all-paths implementation of Dijkstra’s algorithm or A. Extending these classic algorithms with path counting yields the Central Dijkstra and Central A methods, which can be contrasted with the Canonical Dijkstra and Canonical A methods reviewed in Section 2.4.
Both the canonical and central approaches restrict the final output to a small subset of shortest grid paths. But whereas the canonical methods will likely yield somewhat shorter runtimes, the central methods offer more direct paths. The grid path in Figure 1a was previously identified as a canonical path. The grid path in Figure 1b is a central path.
Recall from Section 2.4 that an all-paths implementation of Dijkstra’s algorithm or A will record, for each visited vertex, the set of predecessor vertices that are traversed by any shortest grid path en route to the source. This directed acyclic graph of predecessors makes it easy to count paths from the goal vertex to the source, but not the other way around.
Fortunately, when counting paths from the goal to the source, one can simultaneously construct a directed acyclic graph of successors pointing in the opposite direction. These successors can then be used to count paths from the source vertex to the goal.
Algorithm 1 formalizes the above strategy for generating a central grid path using data from an all-paths Dijkstra or A search. Line 2 computes path counts from the goal vertex using predecessors from the all-paths search. Line 3 computes path counts from the source vertex using successors retrieved from the first path counting operation.
Lines 4 to 14 generate the central path by starting at the source and selecting, for each move, a successor 930 Path Counting for Grid-Based Navigation vertex with the highest traversal count. Each traversal count is calculated on line 10 as the product of the path counts from the source and from the goal.
Algorithm 1: Generate a central path using data from an all-paths search 1 function generate path(source, goal, predecessors, costs): 2 counts from goal, successors ←count paths(goal, predecessors, costs) 3 counts from source, ←count paths(source, successors, costs) 4 path ←list() 5 path.append(source) 6 while path.last() ̸= goal do 7 best successor ←none 8 highest count ←−∞ 9 foreach successor ∈successors[path.last()] do 10 count = counts from source[successor] × counts from goal[successor] 11 if count > highest count then 12 best successor ←successor 13 highest count ←count 14 path.append(best successor) 15 return path The costs argument in Algorithm 1 contains the grid distance from the source to each vertex, as computed by the initial all-paths search. For A, these travel costs are referred to as g-values to distinguish them from the heuristics, or h-values, used to estimate distances to the goal. We suggest supplying these costs to the function that counts paths, as done on lines 2 and 3. While travel cost information is not strictly necessary for path counting, it provides a convenient way to ensure vertices are processed in a viable order.
Algorithm 2 defines the function that starts at a given source vertex and counts shortest grid paths represented by the supplied successors. Recall from above that this function is first invoked with the goal and predecessors of the initial path planning problem, which causes paths to be counted in the opposite direction as the initial search. Lines 2 and 3 initialize the path counts to 0 for all vertices except the source, which gets a path count of 1.
Line 4 initializes all predecessors to the empty set. These predecessors may ultimately serve as successors in a subsequent invocation of the function. Lines 5 to 7 prepare a priority queue of unprocessed vertices, ordered using the travel costs from the initial all-paths search.
If the cost of the source vertex is 0, the traversal is assumed to be from source to goal, so vertices with lower costs are processed first; otherwise, the traversal is assumed to be from goal to source, so vertices with higher costs are processed first. Either way, the source vertex and its cost are pushed onto the queue. Lines 8 to 15 repeatedly process the highest priority vertex in the queue by adding its path count to those of its successors.
The time complexity of Algorithm 2 is O(N2log(N)), where N is the size of the spatial environment measured in grid spacings along one axis. The N2 factor is proportional to the number of vertices that must be processed, and the log(N) factor assumes that the pop operation on line 10 is performed on a typical heap-based priority queue. Since an all-paths 931 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan Algorithm 2: Count paths using shortest path successors and costs 1 function count paths(source, successors, costs): 2 counts ←array(costs.dimensions(), 0) 3 counts[source] ←1 4 predecessors ←array(costs.dimensions(), ∅) 5 prioritize lowest cost ←(costs[source] = 0) 6 queue ←priority queue(prioritize lowest cost) 7 queue.push(source, costs[source]) 8 while ¬ queue.empty() do 9 vertex ←queue.top() 10 queue.pop() 11 foreach successor ∈successors[vertex] do 12 if counts[successor] = 0 then 13 queue.push(successor, costs[successor]) 14 counts[successor] ←counts[successor] + counts[vertex] 15 predecessors[successor] ←predecessors[successor] ∪{vertex} 16 return counts, predecessors grid-based Dijkstra or A search procedure is also O(N2log(N)), introducing a subsequent path counting step does not increase the time complexity of the overall approach. In fact, since the counting procedure traverses a subset of the vertices visited during the initial search, intuition suggests that the additional runtime imposed by the counting step may be modest. This reasoning is explored experimentally in Section 3.3.
To ensure the central path methods are implemented efficiently and robustly, close atten-tion should be paid to the data types used for predecessors (and successors), grid distances, and path counts. On a rectangular grid, one can keep track of a vertex’s predecessors using an unsigned integer of one bit per neighborhood move. On an 8-neighbor grid, for example, recording all predecessors would require 8 bits, or one byte, per vertex. The successors would require another byte per vertex.
Since the exact length of a diagonal move is usually an irrational number, it may seem intuitive to represent grid distances using floating-point numbers. However, rounding errors characteristic of floating-point arithmetic will then produce small differences in the com-puted lengths of equally short paths. One can treat floating-point path lengths as equal if they differ by at most some small positive ϵ, but a simpler approach is to represent grid dis-tances using fixed-point numbers or integers. On an 8-neighbor grid, cardinal and diagonal moves could be prescribed integer lengths such as 5 and 7, or 408 and 577. It was known long ago that 577/408 ≈ √ 2 (Kraft & Washington, 2014). Alternatively, one could assign all cardinal moves an arbitrary integer length such as 1000, then round the lengths of all other moves to the nearest integer.
Although integers should be preferred over floating-point numbers for grid distances, the opposite is true for path counts. Because they grow exponentially with distance, path counts will quickly overflow if they are implemented using any standard integer data type. Even 932 Path Counting for Grid-Based Navigation with floats, overflow remains an issue. Path counts can exceed the maximum representable value of a standard 64-bit floating-point number, a value greater than 10308, after as few as 1030 moves. Note that restricting the width of an 8-neighbor grid to 1030 vertices or fewer, in both dimensions, does not necessarily avoid the problem; obstacles may cause shortest grid paths to meander around the environment and potentially surpass 10308 in number.
A solution to the overflow problem is to use 64-bit floating-point numbers to calculate and store logarithms of path counts rather than representing path counts directly. This technique requires a few changes to the above algorithms, though the changes can be en-capsulated by a custom data type if desired. On lines 2 and 12 of Algorithm 2, the default path count of 0 is replaced with the asymptotic limit of log(x) at x = 0, which is −∞. On line 3, the path count of the source vertex is replaced with log(1), which is 0. On line 14, the sum of two path counts a and b must be replaced with an operation that combines log(a) and log(b) to obtain log(a + b). No part of the operation can be vulnerable to overflow. As-suming all logarithms are base 2, a mathematical expression that meets the requirements is derived in (5). To ensure the subexpression 2log(b)−log(a) does not overflow, the two operands must be sorted such that log(a) ≥log(b).
log(a + b) = log(a(1 + b/a)) = log(a) + log(1 + b/a) = log(a) + log(1 + 2log(b/a)) = log(a) + log(1 + 2log(b)−log(a)) log(a) ≥log(b) (5) The operation on the last line of (5) suffices to compute log path counts from the goal and from the source. The next step is to obtain the traversal counts, which can also be handled using logarithms. Instead of multiplying a vertex’s path counts a and b to yield its traversal count ab, as indicated on line 10 of Algorithm 1, one adds the two log path counts log(a) and log(b) to produce the log traversal count log(ab).
log(ab) = log(a) + log(b) Since taking the log of a set of numbers preserves their ordering, central paths can be generated by selecting the vertices with the highest log traversal counts. It is not necessary to produce the traversal counts themselves. Due to rounding errors, a central path computed using logarithms may differ slightly from one based on exact path and traversal counts.
3.3 Empirical Comparison Intuition suggests that the central path approach could serve as a relatively simple upgrade to Dijkstra’s algorithm or A for applications demanding higher quality paths in exchange for modest increases in runtime. To investigate the extent to which the approach meets this expectation, at least for certain conditions, we retrieved a copy of the open source any-angle path planning C++ library by Uras and Koenig (2015), and extended it with a Central A implementation. We then collected statistics for the four basic methods below: • Regular A: The existing grid-based A implementation in which a shortest grid path is arbitrarily selected by way of various tie-breaking conventions.
933 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan • Smoothed A: The Regular A method followed by a smoothing operation that pro-duces a shorter path with the same topology. We used the smoothing function included in the library, an implementation of the greedy algorithm reviewed in Section 2.5 that deletes path vertices as dictated by a succession of line-of-sight checks.
• Smoothed Central A: The new Central A implementation followed by smoothing.
• Regular Theta: A basic any-angle path planning method included in the library.
To ensure an apples-to-apples comparison of runtimes, the all-paths A search within the new Central A implementation repurposes the existing data structures and Regular A code where possible. Aside from details such as the library’s use of floating-point numbers for distances, the subsequent path counting procedure closely reflects the algorithms and guidelines in Section 3.2. The A and Central A methods employ the standard grid distance heuristic reviewed in Section 2.3, which is also the octile distance since our study is limited to the 8-neighborhood.
The tests were conducted using benchmark sets from the repository of Sturtevant (2012).
Each set is a collection of maps containing 2D grid-based obstacle geometry, and each map has an associated list of path planning problems specifying start and end vertices.
We selected the following benchmarks: video game maps from Baldur’s Gate II, scaled to 512 × 512 grid cells; game maps from Dragon Age: Origins, ranging from 22 × 28 to 1260×1104; game maps from StarCraft, ranging from 384×384 to 1024×1024; and randomly generated maps with 10%, 20%, 30%, and 40% blocked cells, at 512×512 resolution. These are the same maps tested by Uras and Koenig (2015) and Harabor et al. (2016), though the problem sets were updated in 2018. The C++ library processes the maps by placing vertices on the corners of cells and permitting passage through single-point gaps.
Table 1 reports the average suboptimality of the paths produced by each method. Path-length suboptimality, the quality metric employed by Uras and Koenig (2015) and others, is the difference in length between the computed paths and the shortest possible paths ex-pressed as a percentage of the latter. Lower values indicate shorter, more desirable paths.
We calculate suboptimality scores by averaging path lengths across all problems associated with the same map, then calculating the suboptimality for each map, then averaging subop-timalities across all maps in the benchmark set. The shortest possible paths are computed using the library’s Anya implementation.
Suboptimality scores for Central A without smoothing are not shown, as they are identical to those of Regular A. Shortest grid paths produced by any method are necessarily equal in length, and must therefore be smoothed before they can be compared with one another using this metric.
The results in Table 1 support the expectation that one can obtain shorter paths by smoothing a central grid path than by smoothing the output of a typical grid-based A implementation. For the game maps, Smoothed Central A yielded suboptimality averages roughly ten times lower than those of Smoothed A, and only about 50% higher than those of Theta. The path planning errors that arise with such maps are discussed further below.
For the random maps, where the effectiveness of path counting is compromised by numerous single-cell passages and single-point gaps, Smoothed Central A yielded suboptimalities closer to the geometric average between those of Smoothed A and Theta.
934 Path Counting for Grid-Based Navigation Average Suboptimality (%) Benchmark # of # of Regular Smoothed Smoothed Regular Set Maps Problems A A Central A Theta Baldur’s Gate II 75 122600 4.8735 0.6686 0.0643 0.0421 Dragon Age: Origins 156 155620 4.6526 0.8935 0.1072 0.0642 StarCraft 75 211390 5.3018 1.1842 0.1091 0.0941 Random-10% 10 17970 4.5915 1.9175 0.4106 0.1446 Random-20% 10 19150 4.4357 2.2955 0.6336 0.2121 Random-30% 10 20780 4.4795 2.4990 0.7691 0.2421 Random-40% 10 36370 4.4051 2.1822 0.7092 0.2369 Table 1: Average path-length suboptimality for the four tested methods.
Average Runtime (ms) Average # of Expansions Benchmark Smoothed Smoothed Regular Smoothed Smoothed Regular Set A Central A Theta A Central A Theta BG II 3.07 4.71 9.25 13226 13868 11945 DAO 1.24 1.69 3.39 5667 5870 5734 SC 12.55 15.96 43.42 50706 51450 50046 R-10% 3.41 7.85 2.88 12518 14335 5812 R-20% 3.37 5.71 4.45 11768 12863 9951 R-30% 3.82 4.76 6.30 13081 13516 15058 R-40% 3.71 4.18 5.78 14358 14404 15658 Table 2: Average runtime and average number of expansions per search.
Table 2 reports the average runtime for generating a single path, as well as the number of expansions during the main search procedure of each method. The runtime and expansion statistics are calculated by averaging first across all problems associated with the same map, and then across all maps in the benchmark set. Runtimes for A and Central A without smoothing are not shown, but smoothing grid paths took only about 1% as long as generating them. Tests were run on a 2.7GHz Intel Core i7 laptop with 16GB of RAM.
The results in Table 2 mostly support the expectation that Central A imposes only modest increases in runtime compared with a typical implementation of A. For the game maps, Central A took roughly 30% to 50% longer than A as a result of (a) a small increase in the average number of expansions, as shown in the table; (b) a small increase in the time required to process each expansion; and (c) the additional step of counting paths. The additional expansions can be explained by the stricter termination condition associated with the all-paths A search, as described in Section 2.4. Compared with A, the central path approach introduced only a 13% increase in average runtime for random maps at 40% obstacle density; however, the additional counting step became progressively more significant at 30% density, 20% density, and 10% density where path counting was as costly as the initial search. Central A achieved shorter runtimes than Theta for all benchmark 935 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan sets except for the 10% and 20% random maps. Runtimes were roughly 2-3 times shorter for the game maps.
Several qualitative advantages and disadvantages of the central path approach are re-vealed by examining individual scenarios such as the one in Figure 9. For this problem, the A solutions in Figure 9a clearly exhibit the flaws that central paths are intended to address. First, typical A grid paths feature artifacts such as the pronounced terraces near label P, though these particular terraces are resolved by the smoothing operation. Second, it is relatively common for an arbitrarily selected grid path to pass an obstacle on a side that is unmistakably suboptimal, as seen near label Q. These topology errors cannot be resolved by smoothing. A third issue in the example, also near label Q, is that the greedy smooth-ing algorithm overshoots a corner of an obstacle and causes a conspicuous triangle-shaped detour to appear in the smoothed path.
(a) A (thin purple line) and Smoothed A (thick green line) (b) Central A (thin purple line) and Smoothed Central A (thick green line) (c) Theta Figure 9: A section of map “ost000a” from the Dragon Age: Origins benchmark set, dis-tributed by the Moving AI Lab with permission from BioWare. The paths are solutions to instance 800 (numbering from 0) of the 2018 problem set. The source is on the left.
936 Path Counting for Grid-Based Navigation The Central A grid path in Figure 9b avoids the artifacts near P and the topology error near Q. The greedy smoothing algorithm still overshoots obstacles, particularly near label R, but the relative directness of a central grid path limits the severity of these smoothing defects. Because central paths are nearly direct to begin with, they should accommodate relaxation- or interpolation-based smoothing techniques that avoid line-of-sight checks and are capable of producing more curved paths. Alternatively, applying the “string-pulling” algorithm of Han et al. (2020) to a central path should yield a smooth path that is taut, or nearly taut, and highly direct.
Although Theta outperforms 8-neighbor Central A in terms of average path length, we submit that defects in a basic Theta path may be more noticeable when they do occur.
One such defect appears in Figure 9c near label S, where an unfortunate confluence of factors leads Theta to skip certain line-of-sight checks that would have revealed a more direct trajectory. With its natural tendency to prioritize vertices near sightlines, the central path approach avoids such conspicuous errors. Central paths can be suboptimal, however, when there are two plausible path topologies and no sightline to clarify the better option.
A case in point occurs at label T, where the optimal topology found by Theta has no shortest grid path and is hence inaccessible to A and Central A. Our impression is that topology errors of this nature are less obvious to the human eye than defects involving a severe deviation from a sightline. Nevertheless, one might consider using a 16-neighbor Central A implementation for applications that demand even higher quality paths.
4. Theoretical Analysis A curious feature of the central path approach is that it tends to produce grid paths with nearly straight sections spanning arbitrary distances at any angle, yet the only vectors or line-of-sight tests required are between neighboring vertices. Here we pursue a theoretical understanding of why the approach works. We begin by observing a relationship between path counting and grid-based visibility. Next, we confirm that central grid paths converge on clear sightlines. Finally, we conjecture that central paths converge on direct paths.
4.1 Visibility by Counting The fact that central grid paths tend to approximate straight-line paths suggests that shortest grid paths are typically concentrated around sightlines. Intuitively, this implies that if a sightline from point A to point B is not blocked by an obstacle, then the number of shortest grid paths between A and B is likely close to the maximum possible number.
On the other hand, if the sightline is blocked, then the number of shortest grid paths is likely closer to zero. This idea can be used to develop a visibility method based on path counting. The visibility by counting method turns out to be mathematically equivalent to the linear grid-based visibility approach adapted from Tsai et al. (2004) in Section 2.6, but we present it here to shed light on the theoretical properties of central grid paths.
Recall from Section 2.3 that on an obstacle-free grid, shortest grid paths are sequences of at most two distinct bracketing moves. In Section 3.1, we observed that the number of such paths from a source vertex to any other vertex is the number at the corresponding position of Pascal’s triangle. As shown in Figure 10 below, counting the number of shortest grid paths heading outward from a source vertex on an λ-neighbor grid will generate λ 937 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan copies of Pascal’s triangle. The triangles will overlap by one vertex in directions for which there is only one shortest grid path to each vertex.
Figure 10: Vertices labeled with the number of shortest grid paths heading outward from a source vertex on a 6-neighbor grid (left) and an 8-neighbor grid (right). The numbers form overlapping copies of Pascal’s triangle, one of which is highlighted in each example.
The idea is now to compute the fraction of shortest grid paths heading outward from a source vertex that are not obstructed by obstacles. This fraction is the visibility score for each vertex. The region visible from the source is approximated as the set of vertices with at least 50% of the paths unobstructed.
To illustrate, consider the scenario in Figure 11a. The task is to evaluate sightlines from the source vertex A to all other vertices in the environment. Figure 11b shows all of the grid paths heading outward from A that are not obstructed by obstacles. Observe that unlike the counting approach proposed for navigation, the paths used for visibility do not wrap around obstacles but continue to advance from the source according to the bracketing moves.
Figure 11c shows the path counts for these outward grid paths. Because the paths always head outward from the source, the path counts can be computed using a standard array traversal (e.g. [x, y] = [0, 0], [1, 0], [2, 0], ..., [0, 1], [1, 1], [2, 1], ...). Figure 11d shows the path counts, or Pascal numbers, for all shortest grid paths ignoring any obstacles. Dividing the path counts in Figure 11c by those in Figure 11d yields the fractions, or visibility scores, in Figure 11e. Figure 11f highlights the vertices with visibility scores of at least 0.5, which happen to be exactly the set of vertices that are truly visible from A.
Like all grid-based visibility methods, it is possible for a vertex with a visibility score of 0.5 or higher to be occluded, perhaps only by a small obstacle. Similarly, it is possible for a vertex with a score below 0.5 to be visible, perhaps only through a small opening between obstacles. The accuracy of the method can be improved by choosing a larger neighborhood or by decreasing the grid spacing.
Similar to the central path approach, path counts computed for visibility can be stored and manipulated using logarithms to avoid overflow. Whereas the final counting step of the navigation approach involved adding two log path counts to obtain a log traversal count, here we must subtract two log path counts to obtain a log visibility score. If needed, it is always safe to exponentiate a log visibility score to obtain the fraction itself.
938 Path Counting for Grid-Based Navigation (a) A grid with obstacles (b) All outward grid paths from A (c) The path counts from A (d) The Pascal numbers from A (e) The visibility scores (f) The points visible from A Figure 11: The visibility by counting approach.
Another way to avoid overflow is to restrict the size of any visibility analysis to fewer than 1030 grid moves from the source. When represented using 64-bit floating-point numbers, path counts for visibility will never overflow on 4-neighbor grids smaller than 515 vertices in both dimensions, or 8-neighbor grids smaller than 1030 vertices. Note that we cannot make the same guarantee for the navigation approach, where paths meander around obstacles.
The apparent convergence of the approach can be observed in Figure 12, where each pixel is colored based on the visibility score of the associated vertex.
Fully obstructed vertices with scores of 0 are colored black, perfectly visible vertices with scores of 1 are colored white, and areas of uncertainty with scores between 0 and 1 are colored on a dark-to-light, cyan-colored spectrum.
The solution on the 501 × 501 grid exhibits a certain degree of uncertainty around the edges of the shadows. When the resolution is increased to 5001 × 5001 vertices, these uncertain regions are reduced to thin slivers.
939 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan (a) 501 × 501 vertices (b) 5001 × 5001 vertices Figure 12: Plots of the visibility scores produced by the path counting approach at two grid resolutions. The yellow boxes are obstacles and the source vertex is in the middle.
Observe that both solutions in Figure 12 feature two almost perfectly sharp shadow edges near the bottom center and bottom right of the field. This is due to the fact that two obstacles have been deliberately placed so that their corners line up with the source vertex in a cardinal or diagonal direction. In these exact directions there is one possible outward grid path, and in the vicinity of these locations there are not enough outward paths to produce the same tapering effect that occurs throughout most of the environment. This lack of isotropy is typical for grid-based visibility methods with explicit geometry, though view points and obstacles will not normally align as precisely as in Figure 12.
We now show that visibility by counting is mathematically equivalent to the linear grid-based visibility approach described in Section 2.6. The first step is to formally define the outward path counts, taking obstacles into account. We employ the notation established in Section 2.6 for any standard 2D grid neighborhood, where m and k are the numbers of bracketing moves required to get to any vertex from the source. In this case, however, m and k are not explicitly used in the calculations and need not be evaluated. The path count cm,k of each vertex is given by (6) below.
c0,0 = 1 cm,0 = cm−1,0Vm,0 m−1,0 m > 0 c0,k = c0,k−1V0,k 0,k−1 k > 0 cm,k = cm−1,kVm,k m−1,k + cm,k−1Vm,k m,k−1 m, k > 0 (6) 940 Path Counting for Grid-Based Navigation Recall from Section 2.6 that Vr q = 1 means there is a straight-line path between neigh-boring vertices q and r, and Vr q = 0 means the path is blocked by an obstacle. Ignoring all obstacles (Vr q = 1 everywhere), the same procedure generates the Pascal numbers Cm,k.
Dividing each path count cm,k by the corresponding Pascal number Cm,k produces the visibility score ψm,k.
ψm,k = cm,k Cm,k (7) Having listed the equations of the path counting approach, we now seek an alternative way of calculating the same scores. Instead of generating two sets of path counts and later dividing them, the division operations could be performed on the fly. This would allow one to store only the final visibility score for each vertex. Additionally, instead of computing the Pascal numbers Cm,k recursively, one could treat them as if they were calculated using the explicit formula in (8). The formula expresses the number of ways to reorder a sequence of m items of one type and k items of another, which is equivalent to the number of sequences of bracketing moves comprising shortest paths on an obstacle-free grid.
Cm,k = (m + k)!
m!k!
(8) To derive the alternative way of calculating ψm,k, we begin with (7), then substitute in the expression of (8), then incorporate the recursive step of (6), then simplify to remove all factorials. At the end of the analysis, we arrive at an alternative expression for ψm,k that is identical to the calculation in (4) from Section 2.6, the formula for the linear grid-based visibility method. Thus we find that, ignoring rounding errors, the linear and counting approaches will produce the same results.
ψm,k = cm,k Cm,k = m!k!
(m+k)!
cm,k = m!k!
(m+k)!
cm−1,kVm,k m−1,k + cm,k−1Vm,k m,k−1 = m!k!
(m+k)!
cm−1,kVm,k m−1,k + m!k!
(m+k)!
cm,k−1Vm,k m,k−1 = m m+k (m−1)!k!
(m+k−1)!
cm−1,kVm,k m−1,k + k m+k m!(k−1)!
(m+k−1)!
cm,k−1Vm,k m,k−1 = m m+k cm−1,k Cm−1,k Vm,k m−1,k + k m+k cm,k−1 Cm,k−1 Vm,k m,k−1 = mψm−1,kVm,k m−1,k + kψm,k−1Vm,k m,k−1 /(m + k) Although the two approaches are mathematically equivalent, it is worth noting that each involves a particular set of operations that may be more convenient for certain applications.
The linear approach requires a division operation to be performed for each vertex during the grid traversal. With the counting approach, the division operations are deferred to the end of the method, and can sometimes be avoided entirely. If one requires only the classification of each vertex as visible or not visible, then the final calculation cm,k/Cm,k ≥0.5 can be 941 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan replaced with 2cm,k ≥Cm,k. On the other hand, the counting approach requires that (a) two floating-point numbers are calculated and stored for each vertex, instead of just one; or (b) the Pascal numbers Cm,k are precomputed, then retrieved.
Aside from the minor practical trade-offs outlined above, the significance of visibility by counting is that it reveals a relationship between grid-based visibility and the proposed counting approach for grid-based navigation. Consider the scenarios in Figure 13. In the first scenario, a sightline between points A and B passes through a gap in a wall, clearing the obstacles by some small yet positive distance δ. It appears that the wall will block most of the possible shortest grid paths between A and B, thereby producing a visibility score less than 0.5. However, the work of Kao and Tsai (2008) suggests that the linear grid-based visibility score will converge on the correct result of 1 as the grid spacing approaches zero. It follows that the visibility by counting score will also converge on 1, and this in turn implies that nearly all possible shortest grid paths will go through the gap at a sufficiently fine grid resolution. In the second scenario, the sightline passes through the middle of a small circular obstacle of positive radius δ. Although the obstacle may appear to block only a small fraction of possible shortest grid paths, the fact that the visibility score converges on 0 assures us that nearly all of these paths will become blocked by the obstacle at a sufficient resolution. These insights about grid-based visibility support a theory that central grid paths converge on sightlines where possible and direct paths in general.
Figure 13: Two challenging scenarios in which grid-based visibility methods will nevertheless evaluate the sightlines correctly if the grid spacing is sufficiently small.
4.2 Convergence on Clear Sightlines The relationship between path counting and grid-based visibility suggests that shortest grid paths become increasingly concentrated around sightlines as the grid spacing approaches zero. We now provide an explanation for this effect by analyzing the distribution of path and traversal counts between endpoints A and B on an obstacle-free grid. We first relate these path counts to the binomial distribution, then approximate this distribution with a normal distribution, then appeal to the central limit theorem to argue that central grid paths converge on the A-B sightline as the grid resolution is increased. We then observe that the convergence property holds even if obstacles are introduced into the environment, provided these obstacles neither contact nor block the sightline.
Our analysis can be extended to all of the standard 2D grid neighborhoods using the transformation in (1), but for simplicity we consider an 8-neighbor grid with start vertex A 942 Path Counting for Grid-Based Navigation located at the origin and end vertex B at [N, K]. Given 0 < K < N, all shortest grid paths from A to B are sequences of the two bracketing moves u = [1, 0] and v = [1, 1]. We are interested in the relative proportion of these shortest grid paths that, after n moves, pass through some vertex [n, k]. This formulation of the problem is illustrated in Figure 14.
Figure 14: Endpoints A and B are separated by N moves, of which N −K are in direction u and K are in direction v. A point en route is reached after n moves from A, of which n−k are in direction u and k are in direction v. The illustration depicts an 8-neighbor grid with u = [1, 0] and v = [1, 1], but the analysis can be extended to other neighborhoods.
The number of ways to travel from A to [n, k] is n!
k!(n−k)! = 2nfn(k), where fn(k) is the probability mass function of a binomial distribution with success probability p = 1/2. Since we are interested in relative proportions, we drop the constant 2n and focus on the binomial distribution. The distribution is defined for 0 ≤k ≤n, but what matters for our analysis is that it accurately captures the relative proportions of path counts from A to any [n, k] within the parallelogram of shortest grid paths. For each n ∈{1, 2, 3, ..., N −1}, we now approximate the binomial distribution using a normal distribution with matching mean µA and variance σ2 A, as defined in (9). Depicted in Figure 15a, this normal distribution approximates the relative proportion of shortest grid paths that end at [n, k] after n moves forward from A.
µA = n 2 σ2 A = n 4 (9) Similarly in the other direction, the number of ways to travel from B to [n, k] is 2N−nf(N−n)(K −k). Dropping the constant 2N−n, we are left with a transformed bino-mial distribution that accurately captures the relative proportions of path counts from B to any [n, k] in the parallelogram. For each n ∈{1, 2, 3, ..., N −1}, we once again approximate the binomial distribution using a normal distribution. The transformed mean µB and vari-ance σ2 B are given in (10). Depicted in Figure 15b, this normal distribution approximates the relative proportion of shortest grid paths that end at [n, k] after N −n moves backward from B.
µB = K −N −n 2 σ2 B = N −n 4 (10) As indicated in Section 3.1, the number of ways to travel from A to B passing through [n, k] is the product of the number of paths from A to [n, k] and the number of paths from [n, k] to B. For each n ∈{1, 2, 3, ..., N −1}, we therefore approximate the proportion of 943 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan (a) Normal distribution for paths from A (b) Normal distribution for paths from B Figure 15: Visualizations of the normal distributions which approximate the proportion of shortest grid paths (a) forward from A and (b) backward from B. The red lines indicate where the distribution peak would be located for various n between 0 and N. The distri-butions extend into the region outside the parallelogram where there are no relevant paths.
However, since we are only interested in relative proportions within the parallelogram, the unused portions of the distributions do not impact the quality of the approximation.
A-B paths that traverse [n, k] using the pointwise product of the probability density func-tions of the two normal distributions described above. Given any two normal distributions N(µA, σ2 A) and N(µB, σ2 B), it is known that the pointwise product of their density func-tions is proportional to the density function of a new normal distribution N(µAB, σ2 AB), the parameters of which are defined as follows (Bromiley, 2014).
µAB = µAσ2 B + µBσ2 A σ2 A + σ2 B σ2 AB = σ2 Aσ2 B σ2 A + σ2 B (11) If we substitute the parameters of (9) and (10) into the formula for µAB in (11), most of the terms cancel and we are left with the following.
µAB = K N n (12) The result in (12) reveals that the mean value of the combined normal distribution, which is depicted in Figure 16, falls exactly on the straight line between vertex A at [0, 0] and vertex B at [N, K]. Since the density function of this distribution is approximately proportional to the traversal counts computed by the central path approach, we observe that the highest traversal counts should occur relatively close to the A-B sightline. This explains why central grid paths tend to adhere to line-of-sight trajectories, at least in the absence of obstacles. We further observe that, according to the central limit theorem, the 944 Path Counting for Grid-Based Navigation corresponding binomial and normal distributions converge as N →∞. It follows that if the grid resolution is successively refined, the proportion of paths traversing any region within the parallelogram will become increasingly consistent with the normal approximation. We therefore claim that a central path between two points in a 2D obstacle-free environment will converge on the sightline as the grid spacing approaches zero.
Figure 16: Visualization of the normal distribution which approximates the proportion of shortest grid paths between A and B. The red line indicates that the distribution peak coincides with the A-B sightline for all n between 0 and N.
We now consider the case of an obstacle somewhere within the parallelogram of shortest grid paths, but neither contacting nor blocking the A-B sightline. By cutting offsome of the paths on one side of the sightline, the obstacle may cause the highest traversal counts to shift toward the opposite side. Figure 17 depicts an extreme case of this effect, where an obstacle spanning the full length of the path is offset from the sightline by only some small positive distance δ. As illustrated, the obstacle will induce a bulge in the central path.
Figure 17: An illustration of an obstacle inducing a bulge in a central path.
945 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan To better understand the effect of obstacles on central paths, we now derive the stan-dard deviation σAB of the combined normal distribution. This parameter characterizes the extent to which all possible A-B grid paths are spread out on either side of the sightline.
Substituting the parameters of (9) and (10) into the formula for σ2 AB in (11), and taking the square root of both sides, we obtain the result in (13) below. The last step is based on the fact that n(N −n) reaches its maximum value at n = N/2.
σAB = s σ2 Aσ2 B σ2 A + σ2 B = 1 2 r n(N −n) N ≤ √ N 4 (13) Suppose now that the grid spacing is repeatedly halved by inserting new vertices among the existing vertices. After η such subdivisions, the grid resolution is increased by a factor of ρ = 2η. Point B is then found at [ρN, ρK], and all features of the map grow linearly with ρ when measured in grid spacings. In particular, there are now ρ times as many grid spacings between the clear sightline and some obstacle on one side. Yet the maximum standard deviation in (13) increases only from √ N/4 to √ρN/4 grid spacings, or by a factor of √ρ.
This implies that after a sufficient number of subdivisions, only an arbitrarily small faction of all possible A-B grid paths will deviate from the sightline to the extent that they are blocked by the obstacle. Thus in the limit, obstacles at any distance from the sightline have no impact on the resulting central path. This theory is tested in Figure 18, where a set of central paths can be seen converging on a clear sightline despite a nearby obstacle spanning its length.
Figure 18: A plot of 11 central paths (solid lines) between [0, 0] and [64δ, 32δ], where δ is the distance to an obstacle (dashed line) running parallel to the sightline (not rendered).
The grid spacing is 32δ for the first path, and is divided by 2 for each subsequent path.
The final path (closest to the sightline) has a grid spacing of δ/32. At resolutions for which there were multiple central paths, only the furthest from the sightline is plotted.
Since our theory pertains to the limiting behavior of central paths as the grid spacing approaches zero, we should also consider how obstacles may affect central paths in realistic scenarios where the grid spacing is fixed. The example in Figure 19 attempts to recreate the worst-case scenario in Figure 17, but using an actual architectural model with a reasonable 946 Path Counting for Grid-Based Navigation grid spacing of 25 cm. As expected, a wall that is not aligned with the grid induces a bulge in a central path running alongside it. Yet the bulge is arguably quite subtle. By contrast, a grid path produced by a typical Dijkstra or A implementation using the same endpoints might well pass through point P in the figure. The central path is significantly more direct even in this example, and in general the bulging effect tends to go unnoticed.
Figure 19: A central path traveling alongside a wall. Shown from different perspectives, the obstacle induces only a subtle bulge. Vertex P indicates how far a shortest grid path with the same endpoints can stray from the line-of-sight trajectory.
4.3 Convergence on Direct Paths The convergence theory in Section 4.2 pertains only to scenarios in which a sightline exists between the endpoints of a desired path. Even in those cases, the sightline must be clear of obstacles within some positive distance δ on both sides. We now propose a generalization of the theory that overcomes these limitations and applies when paths are forced to veer around obstacles. Recall from Section 2.1 that a direct path is one that must follow any sightline between any pair of points on the path. We conjecture that if the central path approach is applied to the general 2D path planning problem, then reapplied using successively shorter grid spacings, the resulting central paths will converge on a direct path. In essence, central paths are direct in the limit.
Since the notion of convergence can have multiple interpretations in the context of paths, we base our conjecture on a measure of path similarity devised by Fr´ echet (1906).
The Fr´ echet distance δF(s, t) between paths s and t can be understood as the minimum length of leash needed to connect a human traversing s to a dog traversing t (Alt & Godau, 1995). The human and the dog are allowed to independently vary their speeds, but are not allowed to go backwards. We say that a sequence of paths sn converges on path t if 947 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan δF(sn, t) →0 as n →∞. We also introduce a deviation-from-directness metric that accounts for the fact there can be multiple direct paths between a pair of endpoints. Defined in (14), the directness deviation δD(s) of path s is the minimum Fr´ echet distance between s and any path t ∈Ds, where Ds is the set of all direct paths with the same endpoints as s.
δD(s) = min t∈Ds(δF(s, t)) (14) We conjecture that central paths converge on direct paths regardless of whether obstacle geometry is continuous, as described in Section 2.1, or grid-based, as described in Section 2.3.
In either case, we assume that the endpoints A and B of the desired path are vertices on a standard rectangular or triangular 2D grid of spacing D, which is overlaid on the environment. To induce convergence, the grid spacing is repeatedly halved by inserting new vertices among the existing vertices. After η such subdivisions, the grid spacing is D/2η. The navigation grid at each resolution level is constructed as outlined in Section 2.3, where moves between neighboring vertices are allowed if and only if there is a straight-line path between them. We assume that for sufficiently large η, any topology that contains a direct path also contains at least one grid path.
The above sets up a path planning problem for each η ∈{0, 1, 2, 3, ...}. Applying the central path approach will yield a solution s ∈Cη, where Cη is the set of all possible central grid paths between A and B at resolution level η.
The upper bound on the directness deviation of a central path is therefore the maximum value of δD(s) over all of these possible solutions. The conjecture that central paths converge on direct paths can thus be formally stated as the property that this upper bound approaches zero as η →∞.
lim η→∞ max s∈Cη(δD(s)) = 0 (15) While a formal proof of (15) is a challenge for future research, we believe the conjecture holds for all 2D path planning scenarios that satisfy our assumptions. Our reasoning is that if there is a sightline between any two points P and Q on a central path, and if that sightline is not part of the path, then that portion of the central path should prove unstable as the grid spacing decreases. Specifically, we expect at least one of the following three effects to occur: • Centralization effect: If P and Q are on an obstacle-free stretch of the path, then the distribution of traversal counts along the entire stretch should increasingly conform with the normal approximation based on the central limit theorem; as a result, that stretch of the central path will converge on a sightline.
• Attraction effect: If P and Q are on a stretch of the path that veers around a convex corner, a convex curve, or in some cases a straight surface of an obstacle, then a concentration of sightlines on the inside of the turn will pull the central path closer to the obstacle.
• Shortcutting effect: If P and Q are mutually visible through an unutilized shortcut between obstacles, then an increase in the traversal counts along the sightline will cause the path to be rerouted through the shortcut at a sufficiently fine grid resolution.
948 Path Counting for Grid-Based Navigation The centralization and attraction effects cause central paths to converge on taut paths, whereas the shortcutting effect causes central paths to transition to topologies for which these taut paths are also direct paths. Note that central paths do not necessarily converge on shortest paths. Even at extremely fine grid resolutions, shortest grid paths and shortest smooth paths may have different topologies. Figure 2 shows a case where central paths on an 8-neighbor grid may converge on a direct path that is not a shortest path.
The remainder of this section focuses on the special cases illustrated in Figure 20, which serve to illuminate the nuances of the conjecture. We use the [n, k] notation of Section 4.2, where vertex A is at [0, 0] and vertex B is at [N, K]. We assume 0 < K < N.
(a) An obstacle spanning a sightline (b) Two obstacles creating a single-point gap Figure 20: Two scenarios in which obstacles contacting a sightline will block most of the possible shortest grid paths between A and B. The curves illustrate possible topologies for central paths.
We first consider the case in Figure 20a, where the A-B sightline is clear on one side and blocked on the other. As illustrated, the obstacle will induce a bulge in any central path from A to B. The scenario is similar to the ones discussed in Section 4.2, except in this case there is no clearance between the obstacle and the sightline. The question is whether central paths will still converge on the sightline as the grid spacing approaches zero.
It was argued in Sections 4.1 and 4.2 that the fraction of unobstructed shortest grid paths, or equivalently the grid-based visibility score, converges on 1 if the sightline is clear of obstacles within some positive distance δ on both sides. Similarly, the visibility score converges on 0 if any obstacle crosses the sightline. If a sightline is merely contacted by an obstacle, however, the theory in Sections 4.1 and 4.2 does not predict how the visibility scores will converge. Such scenarios must instead be examined on a case-by-case basis.
The grid-based visibility score for Figure 20a can be estimated by observing its relation-ship to the generalized ballot problem (Tak´ acs, 1962). In simplified terms, ballot problems are concerned with the probability ψ that the ratio of votes for two candidates never falls below some constant κ at any point during the ballot counting process. It is known that this probability can be interpreted as the fraction of shortest grid paths that never pass below a line L (Humphreys, 2010). If κ in the ballot problem is exactly the ratio of the total number of votes for each candidate, then L is the sightline in our scenario and ψ is the visibility score. Meng (2009) reports upper and lower bounds for ψ in the generalized ballot problem, and one can use the lower bound as an estimate ˆ ψN for the visibility score in our scenario.
ˆ ψN = 1 N (16) 949 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan The proof of (16) is based on a technique first applied to the ballot problem by Dvoretzky and Motzkin (1947). Consider any shortest grid path s from A to B, ignoring the obstacle.
If s passes below the sightline at any point, then select some vertex [n, k] on the path that is at the maximum distance below the sightline in the perpendicular direction. Now construct a cyclic permutation of s by (1) translating the entire path such that the point at [n, k] is moved to [0, 0]; (2) trimming offthe first n moves that are now between [−n, −k] and [0, 0]; and (3) inserting those same n moves in order onto the end of the path. It is easy to see that this cyclic permutation is also a shortest grid path from A to B, but one that never passes below the sightline. It follows that at least one of the N cyclic permutations of any possible shortest grid path s will not be blocked by the obstacle, and hence 1/N is a lower bound of ψ. Meng (2009) also provides an upper bound of ψ and an associated proof, and it can be shown that the upper bound is strictly less than double the lower bound for our conditions. It is therefore clear that if the sightline from the source to [N, K] is completely blocked on one side, the visibility score converges on 0.
Because the obstacle in Figure 20a blocks the vast majority of possible shortest grid paths at fine resolutions, we may not invoke the argument of Section 4.2 that the obstacle becomes increasingly irrelevant as the grid spacing approaches zero. Yet the conjecture in this section applies to all direct paths, including any sightline that is contacted but not blocked by an obstacle. As the grid spacing approaches zero, the attraction effect should steadily reduce the bulge illustrated in Figure 20a and cause central paths to converge on the sightline.
Finally, we consider the special case illustrated in Figure 20b. Here the previous example is augmented with a square obstacle of small but positive width δ, placed such that one of its corners creates a single-point gap at the midpoint of the sightline.
It was stated in Section 2.1 that paths may or may not be permitted to pass through single-point gaps, and that the theory presented in this paper should apply regardless. If passage through these gaps is prohibited, then all paths follow the outer topology illustrated in Figure 20b. The straight line from A to B is neither a path nor a sightline in that case, and consequently the only direct path is the taut path with a sharp bend at the outermost corner of the square obstacle. As the grid spacing is repeatedly halved, the attraction effect will cause the distance between successive central paths and this convex corner to converge on zero. At the same time, the centralization effect will cause these central paths to converge on the sightlines on either side.
If paths are permitted to traverse single-point gaps, then the straight line from A to B is both a direct path and a sightline. The taut path on the outside of the square is then disqualified from being a direct path, since it does not utilize the sightline. A technicality arises here as to whether the single-point gap is positioned such that any grid paths are able to go through. Due to its location, the gap in this scenario is accessible to grid-based solutions, and so the conjecture applies. The conjecture suggests that as the grid spacing is repeatedly halved, the number of shortest grid paths that pass through the gap will grow large compared with the number that go around the outside of the square obstacle. Thus even if central grid paths initially follow the outer topology, the shortcutting effect will eventually reroute them through the gap. The attraction effect will thereafter cause the central paths to converge on the sightline. We believe these effects will indeed occur, and submit that the generalized ballot theorem may be useful in proving this point.
950 Path Counting for Grid-Based Navigation 5. Discussion and Conclusions We introduced central grid path planning, a grid-based navigation approach in which Di-jkstra’s algorithm or A is extended with a path counting step to produce relatively direct shortest grid paths. The approach avoids line-of-sight checks between non-neighboring ver-tices, yet still approximates straight path trajectories of arbitrary length and direction. We also proposed the notion of a direct path, and argued that central paths converge on clear sightlines and likely converge on direct paths as the grid spacing approaches zero.
The central path approach represents a natural solution to the problem of selecting a relatively direct path from the multitude of shortest grid paths that typically exist between two endpoints. Even if a smooth path is ultimately desired, generating and then smoothing a central grid path will yield more optimal results on average than generating and then smoothing an arbitrarily selected shortest grid path. We submit that for a typical applica-tion involving navigation on a grid-based environment, a logical strategy for upgrading a conventional A implementation is as follows: • If A paths are noticeably indirect, and slightly longer runtimes can be tolerated, one can adopt Central A by switching to an all-paths search and computing logarithms of path counts from both ends. Empirical results suggest that Central A followed by smoothing could be an attractive alternative to the Theta any-angle method. For the game maps included in our experiment, Smoothed Central A achieved path lengths nearly as short as those of Theta with runtimes closer to A.
• If A paths are of acceptable quality, but runtimes are somewhat too long, one can adopt a canonical path method such as Canonical A or Jump Point Search by re-stricting vertex visits and expansions. An open question is whether a preliminary canonical path search could accelerate the computation of a central path.
• If A is found lacking in both speed and quality, one might consider any-angle methods like Subgoal Graphs or Block A that perform a precomputation on the environment to accelerate subsequent searches. Additional research would be needed to incorporate subgoals or blocks into a central path solver.
Future empirical studies could compare the central path approach to alternative path planning methods using a wider selection of maps, smoothing algorithms, and grid neighbor-hoods. The 16-neighborhood deserves particular attention, as it would improve the quality of the resulting central grid paths while also reducing the number of vertices for which log traversal counts need to be computed. Rather than using path length suboptimality as the sole measure of path quality, it is worth considering additional metrics to more fully capture the notion of relative directness. The directness deviation in Section 4.3 may be difficult to compute, but alternative metrics could include (a) the area or maximum deviation between the path being evaluated and the taut path with the same topology, or (b) the fraction of scenarios in which the topology of the evaluated path contains a shortest path. Whereas the conventional path length suboptimality metric requires shortest grid paths to be smoothed before they can be compared, these alternative metrics could be applied to grid paths and smooth paths alike.
951 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan Future theoretical work could include a formal proof of (15), the conjecture that central grid paths converge on direct paths provided the direct paths have topologies accessible to grid-based solutions. A first step might be to prove the conjecture for grid-based obstacle geometry. Another line of investigation would be to generalize the theory of central grid paths to 3D environments and higher dimensional spaces.
The finding that linear grid-based visibility scores can also be computed by path counting reveals that the existing visibility approach and the proposed navigation approach share a common theoretical basis. The relationship between the approaches is striking, as linear grid-based visibility was originally (a) based on implicit level set geometry instead of explicit geometry; (b) derived from differential equations instead of graph theory; (c) formulated using cardinal neighbors only instead of a variety of grid neighborhoods; (d) specified using normalized direction vectors instead of integer coefficients; and (e) validated using numerical analysis instead of probability theory. Differences in conventions may obscure a number of relationships between level set methods and methods derived from the concept of a grid path.
For example, it is common practice in the level set community to perform path planning by solving the Eikonal equation on a 4-neighbor grid. When the method is modified to use the 8-neighborhood (Danielsson & Lin, 2003), the resulting formula is equivalent to one of the interpolations performed by Field D.
We close with a few remarks about the use of the word “central” in naming the new grid-based navigation approach and the resulting paths. First, there is no intended connection between the central paths described here and those in linear programming and optimization (Nemirovski & Todd, 2008). The central paths in this work should therefore be referred to as “central grid paths” if the context is not clear. Our work is related, however, to the concept of centrality in graph theory (Borgatti & Everett, 2006), particularly betweenness centrality that quantifies the extent to which a vertex is between all others (Freeman, 1977).
Betweenness centrality is computed using traversal counts similar to those in this paper, except that the shortest paths are between all pairs of vertices in a graph. Our use of the word “central” was also motivated by the central limit theorem, which clarifies how the simple counting procedure of Pascal’s triangle can make straight lines, of any angle, appear on a grid.
Acknowledgments We sincerely thank the anonymous reviewers, whose insightful feedback and tireless at-tention to detail helped us significantly improve the quality and correctness of this paper.
Any remaining errors are of course our own. We also thank Nigel Morris, our colleague at Autodesk Research who introduced us to the level set method for computing visibility.
References Alt, H., & Godau, M. (1995).
Computing the Fr´ echet distance between two polygonal curves. International Journal of Computational Geometry & Applications, 5(1), 75– 91.
Bailey, J., Tovey, C., Uras, T., Koenig, S., & Nash, A. (2015). Path planning on grids: The 952 Path Counting for Grid-Based Navigation effect of vertex placement on path length. In Proceedings of the Artificial Intelligence and Interactive Digital Entertainment Conference (AIIDE), pp. 108–114.
Borgatti, S. P., & Everett, M. G. (2006). A graph-theoretic perspective on centrality. Social Networks, 28(4), 466–484.
Botea, A., Bouzy, B., Buro, M., Bauckhage, C., & Nau, D. (2013). Pathfinding in games. In Artificial and Computational Intelligence in Games, pp. 21–31. Dagstuhl Publishing.
Botea, A., M¨ uller, M., & Schaeffer, J. (2004). Near optimal hierarchical path-finding. Jour-nal of Game Development, 1, 7–28.
Bromiley, P. A. (2014). Products and convolutions of gaussian probability density functions.
Tech. rep., University of Manchester.
Cohen-Or, D., Chrysanthou, Y. L., & Silva, C. T.and Durand, F. (2003). A survey of visibil-ity for walkthrough applications. IEEE Transactions on Visualization and Computer Graphics, 9(3), 412–431.
Daniel, K., Nash, A., Koenig, S., & Felner, A. (2010). Theta: Any-angle path planning on grids. Journal of Artificial Intelligence Research, 39, 553–579.
Danielsson, P.-E., & Lin, Q. (2003). A modified fast marching method. In Proceedings of the Scandinavian Conference on Image Analysis, pp. 1154–1161.
Dijkstra, E. W. (1959). A note on two problems in connexion with graphs. Numerische Mathematik, 1, 269–271.
Dvoretzky, A., & Motzkin, T. (1947). A problem of arrangements. Duke Mathematical Journal, 14(2), 305–313.
Ferguson, D., & Stentz, A. (2006). Using interpolation to improve path planning: The Field D algorithm. Journal of Field Robotics, 23(2), 79–101.
Fisher-Gewirtzman, D., Shashkov, A., & Doytsher, Y. (2013).
Voxel based volumetric visibility analysis of urban environments. Survey Review, 45(333), 451–461.
Fr´ echet, M. (1906).
Sur quelques points du calcul fonctionnel.
Rendiconti del Circolo Matematico di Palermo, 22, 1–72.
Freeman, L. C. (1977). A set of measures of centrality based on betweenness. Sociometry, 40(1), 35–41.
Ghosh, S. K. (2007). Visibility Algorithms in the Plane. New York, NY: Cambridge Uni-versity Press.
Goldstein, R., Breslav, S., Walmsley, K., & Khan, A. (2020). SpaceAnalysis: A tool for pathfinding, visibility, and acoustics analyses in generative design workflows.
In Proceedings of the Symposium on Simulation for Architecture and Urban Design (SimAUD).
Han, J., Uras, T., & Koenig, S. (2020). Toward a string-pulling approach to path smoothing on grid graphs. In Proceedings of the Symposium on Combinatorial Search (SoCS), pp. 106–110.
Harabor, D., & Grastien, A. (2011). Online graph pruning for pathfinding on grid maps. In Proceedings of the AAAI Conference on Artificial Intelligence (AAAI), pp. 1114–1119.
953 Goldstein, Walmsley, Bibliowicz, Tessier, Breslav, & Khan Harabor, D., Grastien, A., ¨ Oz, D., & Aksakalli, V. (2016). Optimal any-angle pathfinding in practice. Journal of Artificial Intelligence Research, 56, 89–118.
Hart, P. E., Nilsson, N. J., & Raphael, B. (1968). A formal basis for the heuristic determina-tion of minimum cost paths. IEEE Transactions on Systems Science and Cybernetics, 4(2), 100–107.
Humphreys, K. (2010). A history and a survey of lattice path enumeration. Journal of Statistical Planning and Inference, 140(8), 2237–2254.
Kao, C.-Y., & Tsai, R. (2008). Properties of a level set algorithm for the visibility problems.
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Kraft, J. S., & Washington, L. C. (2014). Elementary Number Theory. Boca Raton, FL: CRC Press.
Lozano-P´ erez, T., & Wesley, M. A. (1979). An algorithm for planning collision-free paths among polyhedral obstacles. Communications of the ACM, 22(10), 560–570.
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Nagy, D., Lau, D., Locke, J., Stoddart, J., Villaggi, L., Wang, R., Zhao, D., & Benjamin, D.
(2017). Project Discover: An application of generative design for architectural space planning. In Proceedings of the Symposium on Simulation for Architecture and Urban Design (SimAUD).
Nemirovski, A. S., & Todd, M. J. (2008). Interior-point methods for optimization. Acta Numerica, 17, 191–234.
Noreen, I., Khan, A., & Habib, Z. (2016). Optimal path planning using RRT based ap-proaches: A survey and future directions. International Journal of Advanced Computer Science and Applications, 7(11), 97–107.
Oh, S., & Leong, H. W. (2017). Edge N-level sparse visibility graphs: Fast optimal any-angle pathfinding using hierarchical taut paths. In Proceedings of the International Symposium on Combinatorial Search (SoCS), pp. 64–72.
Pelechano, N., Allbeck, J. M., & Badler, N. I. (2008). Virtual Crowds: Methods, Simulation, and Control (Synthesis Lectures on Computer Graphics and Animation). Morgan & Claypool Publishers.
Ravankar, A., Ravankar, A. A., Kobayashi, Y., Hoshino, Y., & Peng, C.-C. (2018). Path smoothing techniques in robot navigation: State-of-the-art, current and future chal-lenges. Sensors, 18(9), 3170.
Richardson, A., & Olson, E. (2011). Iterative path optimization for practical robot planning.
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Rivera, N., Hern´ andez, C., Nicol´ as, H., & Baier, J. A. (2020). The 2k neighborhoods for grid path planning. Journal of Artificial Intelligence Research, 67, 81–113.
954 Path Counting for Grid-Based Navigation Roth, S. D. (1982). Ray casting for modeling solids. Computer Graphics and Image Pro-cessing, 18(2), 109–144.
Sturtevant, N. R. (2012). Benchmarks for grid-based pathfinding. Transactions on Com-putational Intelligence and AI in Games, 4(2), 144–148.
Sturtevant, N. R., & Rabin, S. (2016). Canonical orderings on grids. In Proceedings of the International Joint Conference on Artificial Intelligence (IJCAI), pp. 683–689.
Tak´ acs, L. (1962). A generalization of the ballot problem and its application in the theory of queues. Journal of the American Statistical Association, 57(298), 327–337.
Tsai, Y.-H. R., Cheng, L.-T., Osher, S., Burchard, P., & Sapiro, G. (2004). Visibility and its dynamics in a PDE based implicit framework. Journal of Computational Physics, 199(1), 260–290.
Turner, A., Doxa, M., & O’Sullivan, D. Penn, A. (2001). From isovists to visibility graphs: A methodology for the analysis of architectural space. Environment and Planning B: Planning and Design, 28(1), 103–121.
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In Proceedings of the Symposium on Combinatorial Search (SoCS), pp.
206–210, code available at:
Uras, T., Koenig, S., & Hernandez, C. (2013). Subgoal graphs for optimal pathfinding in eight-neighbor grids. In Proceedings of the International Conference on Automated Planning and Scheduling (ICAPS), pp. 224–232.
Yap, P., Burch, N., Holte, R., & Schaeffer, J. (2011). Block A: database-driven search with applications in any-angle path-planning. In Proceedings of the AAAI Conference on Artificial Intelligence (AAAI), pp. 120–125.
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12651 | https://s3da-design.com/the-importance-of-cubic-feet-calculations-in-construction-and-civil-ngineering/ | Importance of Cubic Feet Calculations in Construction and Civil Engineering
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The Importance of Cubic Feet Calculations in Construction and Civil Engineering
The cubic foot is an important unit to measure the volume of solid objects and plays an extensive role in the field of construction and civil engineering. It is used to calculate the volume of concrete needed for a foundation and the space for insulation in a house.
It helps to tell the internal space of the different objects such as the cylinder, container, room, and long vehicle that are used to transfer the goods from one city (or country) to another city (or country).
In this article, we will discuss the basics of cubic feet, the calculation involving cubic feet, and the Application of cubic feet. Furthermore, for a better understanding of cubic feet will solve different examples of cubic feet.
Definition of Cubic feet
A cubic foot is the unit of volume for a cube with boundaries that measure one foot in length. It represents the volume of a three-dimensional shape and is used to determine the capacity of an object.
One cubic foot is equal to 7.48 gallons or 28.32 liters. The abbreviation for cubic feet is ‘cu ft’ or ‘ft³’. Moreover, cubic feet can be easily converted to other units of measurement, helping communication across different industries. Using cubic feet for exact volume measurements can help save time, money, and resources in various applications.
Cubic Feet Formula
To calculate the volume in cubic feet you have to measure the length, width, and height of an object in feet. The formula for calculating cubic feet is:
Cubic feet (ft 3) = length x width x height
Cubic feet (ft 3) = L x W x H
Where
“L” is the length,
“W” is the width,
“H” is the height
For example, measure the cubic feet of a box, if the length of a box is 9 feet, the width is 15 feet, and the height is 1.25 feet,
Then, cubic feet = 9 x 15 x 1.25 = 168.75 ft3.
Conversion for cubic feet
The different conversions used from the different units to cubic feet.
Cubic Feet (ft³)Inches³LitersGallonsMeters³
1 ft³ 1728 0.04 7.48 0.03
10 ft³ 17280 0.35 74.81 0.28
How to Find Volume in Cubic Feet?
Here are a few simple steps to find cubic feet.
Measure the length, width, and height of the object in feet.
Calculate the volume by multiplying these measurements together:
Volume = Length × Width × Height
Alternatively, use Allmath’s cubic feet calculator for finding volume in cubic feet with steps to avoid time-consuming manual calculations.
This gives you the volume in cubic feet (ft³).
Example 1:
Evaluate the volume of the drawing room in cubic feet if a room with a width of 10 ft, a length is 15 ft, and a height is 19 ft.
Solution:
Step 1: Write the values of length, width, and height from the above data.
Length = 15 ft, width = 10 ft,
Height = 19 ft, v =?
Step 2:Put all values in the formula of cubic feet to find the volume of the room and simplify.
L = 15, W = 10, H = 19
Cubic feet (ft 3) = V = L x W x H
= 15 x 10 x 19
Cubic feet (ft3) = V = 2850 ft3
Example 2:
Find the volume of the basement room in cubic feet if its height is 36 ft, its length of 10 ft, and its width is 8 ft.
Solution:
Step 1: Write the values of length, width, and height from the above data.
Length = 10 ft, width = 8 ft
Height = 36 ft, v =?
Step 3:Put all values in the formula of cubic feet to find the volume of the room and simplify.
L = 10, W = 8, H = 36
Cubic feet (ft 3) = V = L x W x H
= 10 x 8 x 36
Cubic feet (ft3) = V = 2880 ft3
Application of Cubic Feet in Engineering and Construction
Cubic feet help to find the volume of the solid objects that ensure to determine the total cost to construct the object and determine the total quantity of material required to construct the object.
It is also useful in architecture and interior design to estimate the amount of materials required for any project and tell the total space construction project.
It plays an important role in the field of civil and electrical engineering designs to measure the volumes of different objects. Here we discuss some important uses of cubic feet in engineering and construction.
Material Estimation and Costing
Cubic foot measurements are useful to determine the amount of materials necessary for the completion of the projects. This helps to tell us the total concrete (foundations, slabs, beams), and total number of bricks required for the walls of the whole building.
Project Planning and Scheduling
By calculating the volume of materials, the required construction specialist and engineer can establish a project completion timeline. That can help us with the exact amount of concrete needed for the foundation and avoid delays caused due to excess materials or shortage of materials.
Structural Load Analysis and Design
Civil engineers with the help of cubic feet measure the weight and load-bearing capacity of structures by Calculating the volume of concrete or steel used in beams, columns, and slabs. It helps us to determine the overall weight of the structure that can specify load limits based on the area and volume of specific structural elements.
Quantifying Materials
Cubic feet help us for the estimation of the exact amount of materials needed for construction including concrete, bricks, lumber, and excavation materials. Measuring the accurate amount of material can avoid overspending due to excess materials and minimize material waste, which is very helpful for cost savings.
Wrap up:
In this article, we discussed the basic definition and application of cubic feet in the engineering or construction field. The measurements of the Cubic feet are helpful for accurate material estimation, cost control, structural design, and efficient construction site management.
A strong understanding of cubic feet calculations and determining the volume of the object is essential for anyone involved in construction and civil engineering, which helped us for the success and safety of projects.
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12652 | https://mathleaks.com/study/application_of_the_laws_of_sine_and_cosine | Published Time: 20210520T131738Z
Understanding the Law of Cosine and Sine Law in Applications
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arrow_back 15. Application of the Laws of Sine and Cosine
1. Congruence, Proof, and Constructions 16 Lessons
2. Similarity, Proof, and Trigonometry 15 Lessons
3. Extending to Three Dimensions 7 Lessons
4. Connecting Algebra and Geometry Through Coordinates 5 Lessons
5. Circles With and Without Coordinates 10 Lessons
6. Applications of Probability 12 Lessons
info_outline Introduction
arrow_right Theory 14 Slides
radio_button_unchecked Catch-Up and Review s. 1
radio_button_unchecked Which Law Should be Applied? s. 2
radio_button_unchecked Area of a Circle Circumscribed Around a Non-Right Triangle s. 3
radio_button_unchecked Solving Triangles Using the Law of Sines and the Law of Cosines s. 4
radio_button_unchecked Analyzing Different Triangles to Determine Which Law Can Be Used s. 5
radio_button_unchecked Calculating the Height of the Leaning Tower of Pisa s. 6
radio_button_unchecked Measuring Trees Using the Law of Sines s. 7
radio_button_unchecked Extending the Definition of Sine and Cosine to Obtuse Angles s. 8
radio_button_unchecked Calculating the Perimeter of a Piece of Land s. 9
radio_button_unchecked Law of Cosines in Soccer Practice s. 10
radio_button_unchecked Locating of a Smartphone Using Triangulation s. 11
radio_button_unchecked Finding the Altitude of a Helicopter s. 12
radio_button_unchecked Connection Between the Law of Sines and the Circumcircle of a Triangle s. 13
radio_button_unchecked Calculating the Area of a Circle Circumscribed Around a Non-Right Triangle s. 14
arrow_right Exercises level 1 4 Exercises
1.1
1.2
1.3
1.4
arrow_right Exercises level 2 2 Exercises
2.1
2.2
arrow_right Exercises level 3 2 Exercises
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Application of the Laws of Sine and Cosine expand_more
Chapter 2
15.
Application of the Laws of Sine and Cosine
Understanding mathematical principles and their applications can be intriguing. The lesson dives deep into how the law of cosine and law of sine play an essential role in solving real-world problems. These laws, especially when combined, can address intricate challenges that arise in fields like physics, engineering, and even everyday life. Recognizing their potential allows one to approach problems with a broader toolkit, facilitating more effective solutions. Whether you're trying to figure out the dimensions of a triangle in a construction project or understanding the relationships between angles and sides, these laws are indispensable.
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Application of the Laws of Sine and Cosine
Slide of 14
This lesson will explore the applications of the Law of Sines and the Law of Cosines to solve different types of triangles. Furthermore, real-life applications of these ideas will also be explored.
Catch-Up and Review
Here are a some recommended readings before getting started with this lesson.
Right Triangles
Trigonometric Ratios
Law of Sines
Law of Cosines
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Explore
Which Law Should be Applied?
Take a look at the following triangles. Think whether they can be solved by using the Law of Sines or the Law of Cosines.
0,0
How can this triangle be solved?
AAA
BBB
CCC
ccc
aaa
bbb
It is correct because....
It is not correct because....
Next Triangle
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Challenge
Area of a Circle Circumscribed Around a Non-Right Triangle
The following figure shows a circle circumscribed around a non-right triangle. Notice that none of the triangle's sides correspond to the diameter of the circle. What is the area of the circle?
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Discussion
Solving Triangles Using the Law of Sines and the Law of Cosines
Trigonometric ratios are often used to solve right triangles, but they cannot be used to solve non-right, or oblique, triangles. For these triangles, the Law of Sines and the Law of Cosines are particularly useful because they can be used to solve any triangle, right or oblique. To solve a triangle using the Law of Sines or the Law of Cosines, three pieces of information must be known.
| Case | Given Information | Law | Strategy |
--- --- |
| 1 | Two angles and a side length | Law of Sines | The Triangle Angle Sum Theorem can be used to find the missing angle measure. Then the Law of Sines can be used to find the unknown side lengths. |
| 2 | Two side lengths and a non-included angle | Law of Sines | The Law of Sines can be used to solve for one of the unknown angle measures. Then the Triangle Angle Sum Theorem can be used to find the third angle measure. Finally, the Law of Sines can be applied one more time to find the unknown side length. |
| 3 | Three side lengths | Law of Cosines | The Law of Cosines can be used to find any of the unknown angle measures. Then, either the Law of Sines or the Law of Cosines can be used to find another missing angle measure. Finally, the Triangle Angle Sum Theorem can be used to find the third angle measure. |
| 4 | Two side lengths and their included angle | Law of Cosines | The Law of Cosines can be used to find the missing side length. Then, either the Law of Cosines or the Law of Sines can be used used to find a missing angle measure. Finally, the Triangle Angle Sum Theorem can be used to find the last angle measure. |
In the following applet, a triangle and three of its parts are shown. Analyze the given information to decide which law can be used to solve the triangle.
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Pop Quiz
Analyzing Different Triangles to Determine Which Law Can Be Used
0,0
Which law should be used?
AAA
BBB
CCC
ccc
aaa
bbb
Correct! This is Case 1\text{Correct! This is Case } 1Correct! This is Case 1
Incorrect, this is Case 1\text{Incorrect, this is Case } 1Incorrect, this is Case 1
Correct! This is Case 2\text{Correct! This is Case } 2Correct! This is Case 2
Incorrect, this is Case 2\text{Incorrect, this is Case } 2Incorrect, this is Case 2
Correct! This is Case 3\text{Correct! This is Case } 3Correct! This is Case 3
Incorrect, this is Case 3\text{Incorrect, this is Case } 3Incorrect, this is Case 3
Correct! This is Case 4\text{Correct! This is Case } 4Correct! This is Case 4
Incorrect, this is Case 4\text{Incorrect, this is Case } 4Incorrect, this is Case 4
Law of Sines
Law of Cosines
Next Triangle
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Example
Calculating the Height of the Leaning Tower of Pisa
Zain is vacationing in Italy. They were in Pisa to see the famous Leaning Tower when a question came across their mind. What would the tower's height be if it was not a leaning tower? Zain's distance to the tower is 80 meters and they can measure an angle of elevation of 37∘ to the tower's top. Furthermore, the guidebook states that the tower's inclination is about 4∘.
Remembering a Geometry lesson, they realize that the situation can be modeled using a non-right triangle. Help Zain calculate the height of the upright tower. Write the answer rounded to one decimal place.
7
8
9
4
5
6
1
2
3
0
meters
Check Answer
Hint
The inclination angle and its adjacent angle are complementary angles.
Solution
The situation can be modeled using a non-right triangle.
The tower's inclination angle and ∠B are complementary angles, so the sum of their measures adds to 90∘. The measure of ∠B can now be calculated.
4∘+m∠B\=90∘⇕m∠B\=86∘
Now that two angles and their included side are known, it is possible to use the Law of Sines. To do so, the measure of ∠A must be found. This will be done by using the Triangle Angle Sum Theorem.
m∠A+86∘+37∘\=180∘⇕m∠A\=57∘
Finally, the Law of Sines will be used to write a proportion. This equation will be solved for c, the height of the upright tower.
sinAa\=sinCc
SubstituteValuesSubstitute values
sin57∘80\=sin37∘c
▼
Solve for c
MultEqnLHS⋅sin37∘\=RHS⋅sin37∘
sin57∘80(sin37∘)\=c
UseCalcUse a calculator
57.406571…\=c
RearrangeEqnRearrange equation
c\=57.406571…
RoundDecRound to 1 decimal place(s)
c≈57.4
Therefore, the height of the Leaning Pisa Tower would be around 57.4 meters if it was standing straight.
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Pop Quiz
Measuring Trees Using the Law of Sines
Magdalena is doing some research in the forest for a biology project. She has a device that allows her to measure angles. She can also measure the distance from a tree to her device. To help Magdalena complete her research project, find the lengths of different trees, rounded to one decimal place.
0,0
12.3 m12.3\text{ m}12.3 m
20∘20^\circ20∘
43∘43^\circ43∘
Correct answer: 9.1\text{Correct answer: }9.1Correct answer: 9.1
Length:
Check the Answer
Show Answer
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Discussion
Extending the Definition of Sine and Cosine to Obtuse Angles
As mentioned before, the Law of Sines and the Law of Cosines are valid for all types of triangles, including both right and non-right triangles. However, the definitions of the sine and cosine of an angle are given in terms of the ratios of a right triangle's sides.
Therefore, these definitions do not seem to be compatible with obtuse angles. Nevertheless, they can be extended to deal with obtuse angles by considering the following identities.
Rule
Sine and Cosine of Supplementary Angles
The sine of supplementary angles are equal. Conversely, the cosine of supplementary angles are opposite values.
sin(180∘−θ)cos(180∘−θ)\=sinθ\=-cosθ
The following graph verifies both identities for different angles.
0,0
72∘72^\circ72∘
108∘108^\circ108∘
sin72∘\=0.951\sin{\color{darkorange}72^\circ}=0.951sin72∘\=0.951
sin108∘\=0.951\sin{\color{green}108^\circ}=0.951sin108∘\=0.951
cos72∘\=0.309\cos{\color{darkorange}72^\circ}=0.309cos72∘\=0.309
cos108∘\=-0.309\cos{\color{green}108^\circ}=\color{red}\text{-}0.309cos108∘\=-0.309
These identities can also be used when the angle is given in radians.
sin(π−θ)cos(π−θ)\=sinθ\=-cosθ
As shown, these identities allow the definition of the value of the sine and cosine of obtuse angles and the application the Law of Sines and the Law of Cosines to any triangle.
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Example
Calculating the Perimeter of a Piece of Land
Ignacio's grandparent wants to construct a fence for a quadrilateral piece of land. To find the perimeter of the land, he starts measuring its sides using an old trundle wheel. Unfortunately, after measuring just two sides, the trundle wheel breaks.
Ignacio wants to help his grandparent and, in an attempt to simplify the problem, he divides the land into two triangles. Then, by using a compass, he is able to measure the angles of these triangles.
Help Ignacio find the perimeter of the piece of land, rounded to nearest integer.
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Hint
Since two side lengths and the measure of their included angle are known in △ABC, the Law of Cosines can be used to solve for the missing side length.
Solution
Note that two side lengths and all the angle measurements are known in △ABC.
In particular, since two side lengths and the measure of their included angle are known, the Law of Cosines can be used to solve for the missing side length.
b2\=a2+c2−2accosB
SubstituteValuesSubstitute values
b2\=4002+5602−2(400)(560)cos80∘
▼
Solve for b
CalcPowCalculate power
b2\=160000+313600−2(400)(560)cos80∘
AddTermsAdd terms
b2\=473600−2(400)(560)cos80∘
MultiplyMultiply
b2\=473600−448000cos80∘
SqrtEqnLHS\=RHS
b\=473600−448000cos80∘
UseCalcUse a calculator
b\=629.130842…
RoundIntRound to nearest integer
b≈629
When solving the above equation, only the principal root was considered because side lengths are always positive. Therefore, the length of AC is about 629 meters.
Now, △ACD will be considered.
Since one side length and all the angle measures are known, the missing side lengths can be found by using the Law of Sines. The length of DC will be calculated.
sinAa\=sinDd
SubstituteValuesSubstitute values
sin21∘a\=sin120∘629
▼
Solve for a
MultEqnLHS⋅sin21∘\=RHS⋅sin21∘
a\=sin120∘629(sin21∘)
UseCalcUse a calculator
a\=260.285020…
RoundIntRound to nearest integer
a≈260
Therefore, the length of DC is about 260 meters. Next, the length of AD can be calculated by following the same procedure.
sinCc\=sinDd
SubstituteValuesSubstitute values
sin39∘c\=sin120∘629
▼
Solve for c
MultEqnLHS⋅sin39∘\=RHS⋅sin39∘
c\=sin120∘629(sin39∘)
UseCalcUse a calculator
c\=457.079577…
RoundIntRound to nearest integer
c≈457
Now, all the sides of the piece of land are known.
Finally, the perimeter will be calculated by adding all the side lengths.
P\=457 m+560 m+400 m+260 m⇕P\=1677 m
The perimeter of the piece of land is about 1677 meters.
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Pop Quiz
Law of Cosines in Soccer Practice
Kriz is setting up for a free shots on an empty goal. When considering their distance to both goal posts, they realize that the Law of Cosines can be used to calculate the top measure of the angle in which they must kick the ball in order to score. They are practicing with a standard 7.3 meter net. Help Kriz calculate this angle and score the goal! Write the answer rounded to one decimal place.
0,0
7.37.37.3
Wawawiwa!
11.511.511.5
15.515.515.5
Correct answer: 26.4\text{Correct answer: }26.4Correct answer: 26.4
Angle:
Check the Answer
Show Answer
New Question
External credits: @kdekiara
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Example
Locating of a Smartphone Using Triangulation
A burglar robbed a store and took the cashier's smartphone. In an attempt to outsmart the police, the burglar turned off the phone's GPS. Once at their secret location, the burglar felt safe and made a call to plan their next move. However, the smartphone signal was detected by two nearby towers, estimating their distance to the phone in use.
0,0
d1 ≈4.5 blocksd_1 \ \approx 4.5 \text{ blocks}d1 ≈4.5 blocks
d2 ≈2.8 blocksd_2 \ \approx 2.8 \text{ blocks}d2 ≈2.8 blocks
Tower 1\text{Tower }1Tower 1
Tower 2\text{Tower }2Tower 2
Reset\text{Reset}Reset
With the distance from the towers to the location of the smartphone and knowing that the towers are 6 blocks apart, the police think they have enough information to locate the burglar. Help the police find the burglar's location! Write the horizontal and vertical distance in blocks with respect to the phone tower at the left. Round the answers to the nearest integer.
Horizontal distance
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Vertical distance
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Hint
The situation can be modeled using a triangle with three known side lengths.
Solution
The distance from the tower at the left to the smartphone is about 4.5 blocks. The distance from the tower at the right to the phone is about 2.8 blocks. The towers are 6 blocks apart. Therefore, the situation can be modeled using a triangle with three known sides lengths.
Even if no angle is known at first, knowing the three side lengths allows the use of the Law of Cosines to solve for any angle. In this case, it is convenient to solve for ∠A, which corresponds to the vertex represented by the left tower.
a2\=b2+c2−2bccosA
SubstituteValuesSubstitute values
2.82\=62+4.52−2(6)(4.5)cosA
▼
Solve for A
CalcPowCalculate power
7.84\=36+20.25−2(6)(4.5)cosA
AddTermsAdd terms
7.84\=56.25−2(6)(4.5)cosA
MultiplyMultiply
7.84\=56.25−54cosA
SubEqnLHS−56.25\=RHS−56.25
-48.41\=-54cosA
DivEqnLHS/(-54)\=RHS/(-54)
0.896481…\=cosA
RearrangeEqnRearrange equation
cosA\=0.896481…
cos-1(LHS)\=cos-1(RHS)
A\=cos-10.896481…
UseCalcUse a calculator
A\=26.300639…∘
RoundIntRound to nearest integer
A≈26∘
Now that the measure of ∠A is known, the horizontal and vertical distance from the left tower to the smartphone, H and V, respectively, can be calculated. To do this, a right triangle formed by the left tower and the phone will be considered. Then, trigonometric ratios will be used.
The vertical distance to the smartphone V is the opposite leg to the angle whose measure is 26∘. Therefore, the length of the leg can be found by using the sine ratio.
sinθ\=HypotenuseOpposite Leg
SubstituteValuesSubstitute values
sin26∘\=4.5V
▼
Solve for V
MultEqnLHS⋅4.5\=RHS⋅4.5
(4.5)sin26∘\=V
UseCalcUse a calculator
1.972670…\=V
RoundIntRound to nearest integer
2≈V
RearrangeEqnRearrange equation
V≈2
Similarly, the horizontal distance H is the adjacent leg to the angle whose measure is 26∘. Therefore, the cosine ratio can be used to find the value of H.
cosθ\=HypotenuseAdjacent Leg
SubstituteValuesSubstitute values
cos26∘\=4.5H
▼
Solve for H
MultEqnLHS⋅4.5\=RHS⋅4.5
(4.5)cos26∘\=H
UseCalcUse a calculator
4.044573…\=H
RoundIntRound to nearest integer
4≈H
RearrangeEqnRearrange equation
H≈4
The horizontal distance from the left tower to the burglar's location is about 4 blocks, and the vertical distance is about 2 blocks.
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Example
Finding the Altitude of a Helicopter
Two radar stations located 30 kilometers apart detect a passing helicopter. The first station measures an angle of elevation to the helicopter of 40∘, while the second station measures an angle of elevation of 45∘. At what altitude, rounded to one decimal place, is the helicopter flying?
0,0
30 km30 \text{ km}30 km
Station 1\text{Station } 1Station 1
Station 2\text{Station } 2Station 2
40∘40^{\circ}40∘
45∘45^{\circ}45∘
Altitude \=
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Check Answer
Hint
The missing angle can be found by using the Triangle Angle Sum Theorem.
Solution
Because two angles and the included side are known, this problem can be approached by using the Law of Sines to find the distance between the helicopter and one of the radar stations. Once one of the missing side lengths is determined, trigonometric ratios can be used to find the altitude, or height, of the helicopter. The first thing to do is to find the unknown angle.
To do this, the Triangle Angle Sum Theorem can be used. The measure of ∠A will now be found.
m∠A+40∘+45∘\=180∘⇕m∠A\=95∘
Now that the opposite angle to the known side length a has been found, the Law of Sines can be used to write a proportion and solve for any of the missing sides. For instance, b can be calculated.
sinAa\=sinBb
SubstituteValuesSubstitute values
sin95∘30\=sin45∘b
▼
Solve for b
MultEqnLHS⋅sin45∘\=RHS⋅sin45∘
sin95∘30(sin45∘)\=b
UseCalcUse a calculator
21.294234…\=b
RearrangeEqnRearrange equation
b\=21.294234…
RoundDecRound to 1 decimal place(s)
b≈21.3
The distance from Station 1 to the helicopter is about 21.3 kilometers. This distance is the hypotenuse of the right triangle formed by considering the helicopter and Station 1 as vertices. Furthermore, in this right triangle, the opposite leg to the angle that measures 40∘ represents the helicopter's altitude.
Finally, the sine ratio can be used to find the altitude of the helicopter.
sinθ\=hypotenuseopposite leg
SubstituteValuesSubstitute values
sin40∘\=21.3h
▼
Solve for h
MultEqnLHS⋅21.3\=RHS⋅21.3
sin40∘(21.3)\=h
UseCalcUse a calculator
13.691376…\=h
RearrangeEqnRearrange equation
h\=13.691376…
RoundDecRound to 1 decimal place(s)
h≈13.7
The helicopter is flying at an altitude of about 13.7 kilometers.
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Discussion
Connection Between the Law of Sines and the Circumcircle of a Triangle
The Law of Sines states that for any triangle, the ratio of the sine of an angle to the length of its opposite side is constant. However, this is not just any constant. In fact it has an important geometrical interpretation.
Rule
Extended Form of the Law of Sines
The following goes for any triangle. The diameter of a triangle's circumcircle is equal to the ratio of a side length to the sine of its opposite angle.
Now, consider the above figure and let D be the diameter of the circle. With the given information, the following equation holds true.
sinAa\=sinBb\=sinCc\=D
Proof
The proof can be completed in two parts. Part I shows how the sides of a triangle are proportional to the sines of the opposite angles. Part II shows this proportion is equal to the diameter of the circle circumscribed around the triangle.
Part I: Proportional Sides
Consider a triangle ABC with side lengths a, b, and c, and angle measures A, B, and C.
By the Law of Sines, the ratio of the side length of the triangle to the sine of the opposite angle is the same for all sides.
sinAa\=sinBb\=sinCc
Part II: Diameter of the Circumscribed Circle
Draw the circumcircle of the triangle ABC with its circumcenter O.
By the Inscribed Angle Theorem, the measure of A is half of the measure of the intercepted arc BC.
Now, draw the central angle with the intercepted arc BC. Recall that a central angle and its intercepted arc have the same measure. Therefore, the measure of BC is equal to the measure of m∠COB.
By the Substitution Property of Equality, the measure of ∠A is half of the measure of ∠COB.
⎩⎪⎨⎪⎧m∠A\=21mBCmBC\=m∠COB⇓m∠A\=21m∠COB
Let m∠A be α. Then m∠COB becomes 2α.
Note that OB and OC are both equal to the radius R. Hence, △COB is an isosceles triangle with two congruent sides OB and OC.
Now, focus on △COB. Draw its altitude from the vertex angle O. Since △COB is an isosceles triangle, the altitude bisects both the vertex angle and the opposite side.
The sine of α — the ratio of the opposite side to the hypotenuse — can be written using the right triangle CIO.
sinα\=R2a\=2Ra
Rearrange sinα by substituting α\=A.
sinα\=2Ra
Substituteα\=A
sinA\=2Ra
▼
Solve for 2R
MultEqnLHS⋅2R\=RHS⋅2R
2RsinA\=a
DivEqnLHS/sinA\=RHS/sinA
2R\=sinAa
RearrangeEqnRearrange equation
sinAa\=2R
Since R represents the radius, 2R is equal to the diameter D of the circle.
sinAa\=2R⇒sinAa\=D
Conclusion
Combining the results of Part I and Part II, the extended form of the Law of Sines can be obtained.
⎩⎪⎪⎪⎨⎪⎪⎪⎧sinAa\=sinBb\=sinCcsinAa\=D⇓sinAa\=sinBb\=sinCc\=D
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Closure
Calculating the Area of a Circle Circumscribed Around a Non-Right Triangle
The challenge presented at the beginning of this lesson can be solved by using a combination of the Law of Sines, the Law of Cosines, and the extended form of the Law of Sines.
The question here was to find the area of the circle. This will be answered by first finding the ratio of the triangle's side lengths to the sine of their opposite angles.
a What is the ratio of a side length to the sine of its opposite angle? Write the answer rounded to one decimal place.
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b What is the area of the circumscribed circle? Write the answer rounded to one decimal place.
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Hint
a Two side lengths and the measure of the included angle are known.
b The area of a circle is A\=πr2, where r is the radius.
Solution
Since two side lengths and the included angle are known, the Law of Cosines can be used to find the missing side length. Because the missing side is opposite to the known angle, finding the side length will allow to calculate the desired ratio.
The Law of Cosines will be used to find the missing side a.
a2\=b2+c2−2bccosA
SubstituteValuesSubstitute values
a2\=4.92+4.62−2(4.9)(4.6)cos44∘
▼
Solve for a
SqrtEqnLHS\=RHS
a\=4.92+4.62−2(4.9)(4.6)cos44∘
CalcPowCalculate power
a\=24.01+21.16−2(4.9)(4.6)cos44∘
AddTermsAdd terms
a\=45.17−2(4.9)(4.6)cos44∘
MultiplyMultiply
a\=45.17−45.08cos44∘
UseCalcUse a calculator
a\=3.569616…
RoundDecRound to 1 decimal place(s)
a≈3.6
When solving the above equation, only the principal root was considered. This is because a side length is always positive. Therefore, the missing side length is about 3.6 units.
Now that the measure of ∠A and the length of its opposite side a are known, the desired ratio can be calculated.
sinAa
SubstituteIIa\=3.6, A\=44∘
sin44∘3.6
▼
Evaluate
UseCalcUse a calculator
5.182403…
RoundDecRound to 1 decimal place(s)
5.2
The ratio of a side length to the sine of its opposite angle is about 5.2. Therefore, by the extended form of the Law of Sines, the length of the circle's diameter is about 5.2. Recall that the radius of a circle is half its diameter.
Radius:25.2\=2.6 units
With this information, the area of the circle can be calculated.
A\=πr2
Substituter\=2.6
A\=π(2.6)2
▼
Simplify right-hand side
CalcPowCalculate power
A\=π(6.76)
CommutativePropMultCommutative Property of Multiplication
A\=6.76π
UseCalcUse a calculator
A\=21.237166…
RoundDecRound to 1 decimal place(s)
A≈21.2
The area of the circumscribed circle is about 21.2 square units.
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Level 1Level 2Level 3
Application of the Laws of Sine and Cosine
Exercises
Exercise Hint & Answer Solution more_vert
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A satellite orbiting the Earth emits a signal to calculate the distance between itself and Earth. It found that the distances between itself and two cities are 398km and 361 km, respectively, and that the angle between them is 5∘. Find the distance between the cities. Round the answer to the nearest kilometer.
Distance ≈
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Because two sides lengths and the included angle measure are known, we can apply the Law of Cosines to solve for the missing side length representing the distance between the cities.
Let's use the Law of Cosines to solve for the distance.
When solving the equation, only the principal root was considered because side lengths are always positive. The distance between the two cities is about 50 kilometers.
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Exercise Hint & Answer Solution more_vert
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A security camera has been installed in one of the corners in a room. The diagram shown below indicates the area that is visible in the camera's recording. What is the angle of the camera's field of vision? Write the answer rounded to the nearest degree.
≈
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Examining the diagram, we see that the field of vision can be modeled using a non-right triangle where all side lengths are known.
Because of this, the Law of Cosines can be used to find m∠ A, which corresponds to the desired field of vision angle.
Therefore, the field of vision angle is about 26^(∘).
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Exercise Hint & Answer Solution more_vert
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A solar panel was installed on the roof of a house and set to an angle that optimizes the energy produced. The length of the panel is 1.96 meters and the length of the leg supporting it is 0.68 meters. If the support leg makes an angle of 117∘ with the roof, what is the optimal inclination angle of the solar panel? Write the answer rounded to the nearest degree.
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The situation can be modeled by using a non-right triangle.
Because two side lengths and an angle measure are known, it is possible to apply the Law of Sines to solve for the required angle measure. Let's use the Law of Sines to set up a proportion and solve for m∠ A.
The optimal angle for the solar panel is about 18 ^(∘).
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Two cellphone towers detect a smartphone in use. One tower finds that the smartphone is about 650 meters away at an angle of 41∘. The second tower is at an approximate distance of 534 meters from the smartphone at an angle of 53∘. Find the distance between the cellphone towers. Write the answer rounded to the nearest meter.
≈
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The situation can be modeled by a non-right triangle.
The missing side length could be found by using the Law of Sines if the measure of ∠ C were known, so let's start by finding m∠ C. This can be done by using the Interior Angles Theorem. 41 ^(∘) + 53 ^(∘) + m∠ C = 180 ^(∘) ⇕ m∠ C = 86 ^(∘) Now that m∠ C is known, we can use the Law of Sines to write a proportion and solve for the distance between the towers c.
Therefore, the distance between the two towers is about 812 meters.
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Application of the Laws of Sine and Cosine
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12653 | https://www.wolframalpha.com/input/?i=area+between+y%3Dx+and+y%3Dx%5E2&lk=4 | area between y=x and y=x^2 - Wolfram|Alpha
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12654 | https://math-gpt.org/problems/question/a-use-the-quadratic-equation-to-find-the-zeros-of-the-function-b-average-t | a. Use the quadratic equation to find the zeros of the funct | MathGPT
a. Use the quadratic equation to find the zeros of the function.b. Average the zeros to find the x -coordinate of the v...
Question:
a. Use the quadratic equation to find the zeros of the function.b. Average the zeros to find the x -coordinate of the vertex of the parabola.y=x^2+2 x-8
Answer:
a. Finding the zeros using the quadratic equation:
Given the quadratic function , we find its zeros by setting :
To find the zeros of the function, we use the quadratic formula:
where , , and . Substituting these values into the for...
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mula gives:
Simplifying inside the square root:
Thus, the zeros are:
and
b. Averaging the zeros to find the -coordinate of the vertex:
To find the -coordinate of the vertex, we average the zeros:
Therefore, the zeros of the function are and , and the -coordinate of the vertex of the parabola is .
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12655 | https://www.gauthmath.com/solution/1835861333793889/Evaluate-frac-1-7-2- | Solved: Evaluate (frac 1(-7)^-2). [Math]
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Evaluate (frac 1(-7)^-2).
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Answer
The answer is 49
Explanation
Rewrite the expression using the reciprocal property of exponents
The expression is $$\frac{1}{(-7)^{-2}}$$(−7)−2 1. Using the property $$\frac{1}{a^{-n}} = a^n$$a−n 1=a n, we can rewrite the expression as $$(-7)^{2}$$(−7)2
2. Evaluate the exponent
$$(-7)^{2} = (-7) \times (-7) = 49$$(−7)2=(−7)×(−7)=49
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12656 | https://www.youtube.com/watch?v=QW4-GArDWW0 | Finding a Sinusoidal Equation Given a Maximum and Minimum
Mario's Math Tutoring
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Posted: 13 Jun 2016
Learn how to write the equation of a sinusoidal graph given the maximum and minimum points. We go through an example and discuss some helpful formulas in this free math video tutorial by Mario's Math Tutoring.
Timestamps:
00:00 Intro
0:32 General format for sine and cosine equation
0:53 Example 1
2:08 Formula to find amplitude
2:42 How to find the period from the max and min and the b value
3:18 Phase Shift or Horizontal shift
3:36 Formula for finding the vertical shift
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Intro Uh the basic equation for s and cosine the general equation is this one here. A which is the amplitude b which is related to the period through this formula. 2 pi / b okay period is p h and k. h is the horizontal shift. K is the vertical shift. Okay. And that's pretty much it. Let's take a look at uh this example here so you can kind of understand this a little bit better. First thing I like to do is draw a graph. So pi over4a 5. Okay. That's going to be the maximum. So I'm just going to put that right here. So 1 2 3 4 5. Okay. So it's pi over4 General format for sine and cosine equation and I'm just going to label that pi over 4 comma 5. And then the minimum over here is at pi / 2. Okay. Which is right here pi / 2. And that's at negative 1. So that's going to be right basically there. So pi / 2,1. And our graph's going to look something like this. It's going to go down like Example 1 that and back like that. Okay? So you understand? So it has that Sshaped you know graph that s cosine syosoidal look right that snake type type look okay so the next thing we want to do now is we want to decide okay we're going to do a sign graph or a cosine graph well for these ones I think sometimes it's easier just to do a cosine graph because you're given the maximum and normally we graph cosine it starts at the maximum okay it goes down and comes back up and that's one period so let's go ahead and start with the cosine graph I'll show you how to find the sign graph as well and we'll uh take a look at that so the First thing is to find the the amplitude. Now to find the amplitude, we're trying to find the height of these waves, right? But you want to measure from the midline. So the way to do that is you take the maximum value minus the minimum value divided by two and then you take the absolute value. So we're going to take 5 -1. Okay? So amplitude equals 5 -1 / 2. Take the absolute value. So that's 6 / 2, which is three. And so, so far we've got uh let's see, we've got y = 3 cosine. Okay, now we want to find the period. One thing I want you to uh to notice is that when they give us the maximum and the minimum, see that horizontal distance from p /4 to Formula to find amplitude p /2. If we subtract, that gives us p /4. But that's only half the period. The other half here I drew in. And uh so what you want to do is you want to take that and double it. So p /4 doubled gives us p /2. So that means our b value is p / 2. Okay, which means uh I'm sorry, our period is p /2. So to solve for our b value, we've got p / 2 = 2 over b. And if we cross multiply, you get 4 = pi b. And if you divide by pi, you can see that b equals 4. So we How to find the period from the max and min and the b value have four. And then we want to take into account the, you know, the phase shift. So the left and right shift. Normally this point would be uh starting over here. Okay, right at the y ais. So you can see it's being shifted right pi over4. So this is going to be x - /4. And if you want to find the vertical shift, okay, up or down, you can use this formula for k max plus the min / 2. So I'm going to take the y values 5 + 1. Okay, so 5 +1 / 2. So that comes out to Phase Shift or Horizontal shift two. So that's our K value. Okay. So that means that this is going to be + two. And I'm just going to draw this midline in here. |
12657 | https://en.wikipedia.org/wiki/Sign_convention | Sign convention - Wikipedia
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1 RelativityToggle Relativity subsection
1.1 Metric signature
1.2 Curvature
2 Other sign conventions
3 See also
4 References
Sign convention
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From Wikipedia, the free encyclopedia
Agreed-upon meaning of a physical quantity being positive or negative
In physics, a sign convention is a choice of the physical significance of signs (plus or minus) for a set of quantities, in a case where the choice of sign is arbitrary. "Arbitrary" here means that the same physical system can be correctly described using different choices for the signs, as long as one set of definitions is used consistently. The choices made may differ between authors. Disagreement about sign conventions is a frequent source of confusion, frustration, misunderstandings, and even outright errors in scientific work. In general, a sign convention is a special case of a choice of coordinate system for the case of one dimension.
Sometimes, the term "sign convention" is used more broadly to include factors of the imaginary uniti and 2 π, rather than just choices of sign.
Relativity
[edit]
Metric signature
[edit]
In relativity, the metric signature can be either (+,−,−,−) or (−,+,+,+). (Throughout this article, the signs of the eigenvalues of the metric are displayed in the order that presents the timelike component first, followed by the spacelike components). A similar convention is used in higher-dimensional relativistic theories; that is, (+,−,−,−,...) or (−,+,+,+,...). A choice of signature is associated with a variety of names, physics discipline, and notable graduate-level textbooks:
Comparison of metric signatures in general relativity| Metric signature | (+,−,−,−) | (−,+,+,+) |
---
| Spacetime interval convention | timelike, τ 2=x μ x μ{\displaystyle \tau ^{2}=x^{\mu }x_{\mu }} | spacelike, τ 2=−x μ x μ{\displaystyle \tau ^{2}=-x^{\mu }x_{\mu }} |
| Subject area primarily using convention | Particle physics and Relativity | Relativity |
| Corresponding metric tensor | (1 0 0 0 0−1 0 0 0 0−1 0 0 0 0−1){\textstyle {\begin{pmatrix}1&0&0&0\0&-1&0&0\0&0&-1&0\0&0&0&-1\end{pmatrix}}} | (−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1){\textstyle {\begin{pmatrix}-1&0&0&0\0&1&0&0\0&0&1&0\0&0&0&1\end{pmatrix}}} |
| Mass–four momentum relationship | m 2=p μ p μ{\textstyle m^{2}=p^{\mu }p_{\mu }} | m 2=−p μ p μ{\textstyle m^{2}=-p^{\mu }p_{\mu }} |
| Common names of convention | West coast convention "Mostly minuses" Landau–Lifshitz sign convention | East coast convention "Mostly pluses" Pauli convention |
| Graduate textbooks using convention | Landau & Lifshitz The Mathematical Theory of Black Holes (Subrahmanyan Chandrasekhar) Gravitation: an introduction to current research (L. Witten) Introducing Einstein's relativity (Ray D'Inverno) General relativity (Michael P. Hobson, George Efstathiou& Anthony N. Lasenby) | Gravitation (Misner, Thorne, and Wheeler) Spacetime and Geometry: An Introduction to General Relativity (Sean M. Carroll) General Relativity (Wald) (Wald changes signature to the timelike convention for Chapter 13 only) |
Curvature
[edit]
The Ricci tensor is defined as the contraction of the Riemann tensor. Some authors use the contraction R a b=R c a c b{\displaystyle R_{ab}\,=R^{c}{}{acb}}, whereas others use the alternative R a b=R c a b c{\displaystyle R{ab}\,=R^{c}{}_{abc}}. Due to the symmetries of the Riemann tensor, these two definitions differ by a minus sign.
In fact, the second definition of the Ricci tensor is R a b=R a c b c{\displaystyle R_{ab}\,={R_{acb}}^{c}}. The sign of the Ricci tensor does not change, because the two sign conventions concern the sign of the Riemann tensor. The second definition just compensates the sign, and it works together with the second definition of the Riemann tensor (see e.g. Barrett O'Neill's Semi-riemannian geometry).
Other sign conventions
[edit]
The sign choice for time in frames of reference and proper time: + for future and − for past is universally accepted.
The choice of ±{\displaystyle \pm } in the Dirac equation.
The sign of the electric charge, field strength tensorF a b{\displaystyle \,F_{ab}} in gauge theories and classical electrodynamics.
Time dependence of a positive-frequency wave (see, e.g., the electromagnetic wave equation):
e−i ω t{\displaystyle \,e^{-i\omega t}} (mainly used by physicists)
e+j ω t{\displaystyle \,e^{+j\omega t}} (mainly used by engineers)
The sign for the imaginary part of permittivity (in fact dictated by the choice of sign for time-dependence).
The signs of distances and radii of curvature of optical surfaces in optics.
The sign of work in the first law of thermodynamics.
The sign of the weight of a tensor density, such as the weight of the determinant of the covariant metric tensor.
The active and passive sign convention of current, voltage and power in electrical engineering.
A sign convention used for curved mirrors assigns a positive focal length to concave mirrors and a negative focal length to convex mirrors.
It is often considered good form (since it reduces ambiguity and the risk of misunderstanding) to state explicitly which sign convention is to be used at the beginning of each book or article.
See also
[edit]
Orientation (vector space)
Symmetry (physics)
Gauge theory
Negative logic
References
[edit]
Charles Misner; Kip S Thorne&John Archibald Wheeler (1973). Gravitation. San Francisco: W. H. Freeman. p.cover. ISBN0-7167-0344-0.{{cite book}}: CS1 maint: multiple names: authors list (link)
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12658 | https://www.haverford.edu/sites/default/files/Office/Writing-Center/How-to-Write-a-Close-Reading.pdf | Close Reading for English Literature Assignments What is a close reading? A close reading is a very in-depth, careful analysis of a short text. This text can be a passage selected from a novel, a poem, an image, a short story, etc. The analysis looks carefully at what is happening in the short text, but isn’t necessarily isolated from references outside the text. For example, a close reading of a passage of a novel can invoke or refer to the novel more broadly, but focuses its analysis and thesis on just a small section. Crucially, the thesis of a close reading must argue why and how this reading is important in a context beyond the text itself. Here’s how to get started: I. Literal reading: First, read to understand on a literal level what is going on in your passage: who, what, when, where, why, how? ● List characters, setting attributes, motivations within the text. Create a simple plot summary for yourself to ensure you understand what is happening, what the setting is, who is involved, and especially why it’s happening. What are the sources of conflict? Then, start analyzing… Use a pencil to write and mark up the passage, if possible! II. Figures of speech: Does the passage contain significant metaphors, similes, allegory, personification, ellipsis, alliteration, etc.? ● Depending on the length of the text you’re looking at, these literary devices might be broad and stretch out an entire length of a book, or they may be tucked away in a single line of text. For now, look at ones contained in the language of a single passage. ● Consult the list at It offers definitions and examples for many literary devices. Some of the most common to begin with are metaphor, simile, allegory, alliteration, repetition, allusion, archetype, and imagery. ● Make physical annotations on the text you are reading (if possible) to mark the literary devices in use. As you read closely, think about what the effect of each device is and why the author might have made the choice to use this device in this particular way and place. III. Grammatical structure: Look at syntax or word arrangement, grammar, parallel structures/grammatical repetition, punctuation, length and structure of the sentence or line, ambiguous pronouns, word choice, the overabundance of nouns, verbs, adjectives, etc. ● The exact way something is written out affects the way we interpret it even if we don’t notice. Your job is to understand why and how this is happening. Some things to pay attention to are: ○ What strikes you as strange or interesting about the way the passage is written? Is it different or similar to the structure and tone of other texts you’ve read? If it’s located within a longer text, how does it compare to other portions of the text? ○ Are there words that have multiple meanings? Do these meanings, in the context of the passage, change the way you interpret the significance? ○ Is there an absence or overabundance of any particular grammatical structure? For instance, if there is no punctuation in a passage, you might be inclined to read it very rapidly without stopping or slowing down. If there are lots of adjectives, you might see something materialize visually more clearly than in other passages. How does this affect your experience of reading and what you take away? IV. Images and Themes: Words within a passage can evoke previous scenes, images or ideas that the text has already presented. You can often build a strong argument by analyzing a repeating image in a text. ● Think about what in the passage is repeated, or alludes to something that is built in other places of the text. Repetitive elements can be words, storylines, images, ideas, forms, or structure that shows up more than once. ○ What are “structure,” “form,” and “language” in literature? Good question. Check out this link for examples of how they operate in poetry: ● What is the significance of these repetitions? ○ A theme in literature is an organizing principle that the text will explore in various ways. Some examples might be, “war,” “love and loss,” “masculinity,” “crossing borders,” “race,” “mortality,” “intergenerational wisdom,” etc. As you can see, themes night be either broad categories or specific concepts played out in your text. Regardless of the breadth of the theme, it is important to develop an understanding of how the theme operates in your specific passage/text, and how this theme relates to other literary devices you’ve analyzed. ○ How do the repetitive elements build a theme throughout the text that ties its various parts together? ● Start thinking about context: once you have developed some ideas about theme, contextualize them with the work. What is the text trying to say about the issues which it explores? How is this done? Why does the author make the choices that they make? V. Context (IF ANALYZING A PASSAGE IN A BROADER TEXT) When you analyze a passage you are temporarily taking it out of context. Make sure that you can put it back into context. That is, how does this passage connect with the rest of the work? ● How is this passage different from or characteristic of the rest of the text? Are the ideas you’re extracting from it relevant and true more broadly across the rest of the text? When constructing a thesis, make sure you locate your ideas as part of something larger than just your passage. VI. Putting it all together: After you have some ideas about the literal meaning, form, figures of speech, themes, images and context, develop an idea of what your passage communicates. ● You do not (and should not) need to incorporate every single literary device that you’ve analysed into your thesis. Instead, focus on one or a few that communicate something you can organize around a single idea. Hold on to your other thoughts, because they might be useful to include later in your paper. ● Useful questions to think about are: ○ Why did the author choose to use literary devices the way they did? ○ What effect does reading the passage this way have? ○ What does the passage communicate about the broader world? ○ Why is this particular analysis of the passage important? VII: What does a good thesis statement look like? ● A good close-reading thesis statement should be clear, concise, argumentative (but still provable) and specific. ● Make sure your thesis refers to the specific devices/themes/concepts in the text that you will be analyzing in your paper, but also expresses what your reading of them is and why it’s important. ● Talk to your professor or make a WC appointment if you have trouble crafting your thesis statement, since this is the organizing principle for your entire paper. Here is Prof. Asali Solomon’s “Quick and dirty” thesis test to see if you have an argumentative thesis, taken from Suzanne Keen: -Will everyone agree with it? Then it’s too obvious or something we already figured out in class. Keep refining. -Will everybody say I’m crazy? Then it’s controversial, which is good, but unconvincing. Keep refining. -Can a reasonable person disagree with it? This is what you want. -Does it explore complex relationships between aspects of the text or text(s)? This is also what you want. Pointers for writing your paper: 1. Try writing your introduction last. Begin with your thesis, write your paragraphs, and then your conclusion. By the time you get to the conclusion, you’ll have a better idea of what you need to introduce. You may even realize that your thesis changed over the course of writing your paper. Don’t be afraid to change your thesis statement to reflect what you’re actually arguing. 2. Check your topic sentences. Each paragraph should open with a “mini-thesis” about what you’re going to argue in that paragraph. Often, what you initially write as the concluding sentence of the paragraph makes the strongest topic sentence since it is a culmination of all your thoughts from writing. If this is the case, use it as your topic sentence! 3. Do a reverse outline. Highlight just your thesis statement, topic sentences, and concluding sentences of every paragraph. Then read them in order. Does the argument make sense? Does it flow in a logical direction? Does it actually reflect what you’re discussing in each paragraph? 4. Stay grounded in the text. In a literary essay specifically, you don’t want to make any claims you can’t back up with textual evidence. If you’re arguing something that isn’t already proven in earlier evidence or analysis, find a place in the text where you can support your claim. 5. Be specific. Try to avoid making generalizing statements in your writing. Literature is complex, nuanced, and thought provoking. Your writing should reflect this by making specific statements that state what you actually mean to say about the text. For example, instead of saying, “Everyone spends the summer on vacation,” you might write: “Because most of Amanda’s peers had extravagant vacations planned for the summer, she felt particularly unexcited about getting a job at the new pizzeria.” Compiled by Diana Varenik ’22 in consultation with Prof. Asali Solomon of the Haverford English Department, who in turn has drawn on materials created by Prof. Suzanne Keen (Hamilton College) and Prof. Elizabeth Abel (UC Berkeley) |
12659 | https://math.stackexchange.com/questions/3793023/congruent-sets-of-an-arithmetic-sequence-and-a-geometric-sequence | Skip to main content
Congruent sets of an arithmetic sequence and a geometric sequence
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Suppose we have a a,d, and q such that a≠0,d≠0. Then, let M={a,a+d,a+2d} and N={a,aq,aq2}. Given that M=N, find the value of q.
(A) 12
(B) 13
(C) −14
(D) −12
(E) −2
I immediately thought about setting a+d=aq and a+2d=aq2. I then proceeded to do aq2−aq=d, and substitute in for d, which gave me a+(aq2−aq)=aq. Simplifying then gave me aq2−2aq+a=0, and dividing by a gave me q2−2q+1=0, which should mean that q=1. However, that's not an answer choice. What should I do instead?
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edited Aug 16, 2020 at 17:49
Michael Rozenberg
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We have three cases.
2aq=a+aq2
or
q=1,
which is impossible because d≠0.
2aq2=a+aq
or
2q2−q−1=0,
which gives
q=−12.
3. 2a=aq+aq2.
Can you end it now?
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answered Aug 16, 2020 at 16:43
Michael RozenbergMichael Rozenberg
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Ah, I see how to do it now. Thank you!
– questionasker
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Aug 16, 2020 at 16:45
@questionasker You are welcome!
– Michael Rozenberg
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Aug 16, 2020 at 16:46
Actually the third case is immediately impossible because a and a must be equal in the two sets and hence a must be the smallest or the largest element in the geometric set as well.
– cr001
Commented
Aug 16, 2020 at 16:49
@cr001 The third case gives q=−2. No? So d=3a.
– Michael Rozenberg
Commented
Aug 16, 2020 at 16:51
q=−2 does not match the first set and must be discarded. In (a,a+d,a+2d), a must be either the largest or the smallest. It cannot be the average of the other two.
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Surely q=1 would mean that d=0, which is precluded by the conditions of your problem.
Have you considered that M=N means only that the elements are the same, but does not imply anything about in which ordering. What you have proven is that the ordering you've attempted doesn't work.
Let's try adifferent ordering. Due to d≠0, M and N both consist of three different numbers, and a is common in both. Thus, you have only two possibilities for the ordering. One is the one you've examined (a+d=aq,a+2d=aq2), and the other is: a+d=aq2,a+2d=aq. Now, applying the same logic as you did, but to this ordering, you get:
a(2q2−q−1)=0
which, after cancelling a (a≠0) and solving for q gives you the solutions q=1 or q=−1/2. The first solution is forbidden, so we are left with q=−1/2 (answer D).
One example is 4,1,−2 (arithmetic progression) vs. 4,−2,1 (geometric progression).
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answered Aug 16, 2020 at 16:54
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One could also work with the difference between terms, rather than eliminating it. The two possible orderings of the elements in the set are
{a , aq = a+d , aq2 = a+2d} or {a , aq = a+2d , aq2 = a+d} . [Initially, it "feels like" the second arrangement should be unreasonable...]
For the first arrangement, subtracting the first term from the second and third produces a⋅(q−1) = d , a⋅(q2−1) = 2d . Taking a≠0 (otherwise, set N would only contain zeroes), dividing the latter equation here by the former yields
q2 − 1q − 1 = q + 1 = 2dd = 2 ⇒ q = 1 ⇒ q − 1 = da = 0 .
This just gives us the "constant" sequence of terms you were concerned about (and which the stated conditions exclude). But in fact, q=1 isn't even permissible for this ratio. So it turns out that the first arrangement of elements is the "incorrect one".
The seemingly unreasonable ordering leads to
a⋅(q2−1) = d , a⋅(q−1) = 2d
⇒ q2 − 1q − 1 = q + 1 = d2d = 12 ⇒ q = −12
⇒ q2 − 1 = −34 = da ⇒ d = −34⋅a .
We note that this leaves a unspecified, so there are an infinite number of such sequences possible. The elements of the sets are thus
{ a , a − 34a = 14a = aq2 , a − 2⋅34a = −12a = aq } .
[I also had an argument in which we could avoid the question of sequence ordering by looking at the sum of the elements, which gives us
a⋅(1 + q + q2) = a + (a+d) + (a+2d) = 3a + 3d
⇒ q2 + q − (2 + 3⋅da) = 0
⇒ q = −12 ± 1 + 4⋅(2 + 3⋅da)−−−−−−−−−−−−−−−−−√2 = −12 ± 9+(12⋅da)−−−−−−−−−−−√2
= −12 ± 3⋅ 1+(4d3a)−−−−−−−−√2 .
Setting the discriminant equal to zero gives us q = −12 and 4⋅d = −3⋅a as above, but I didn't really see a satisfying explanation for doing so, other than "tidiness" of the result.]
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12660 | https://www.geeksforgeeks.org/dsa/count-n-digit-numbers-whose-adjacent-digits-have-equal-gcd/ | Count N-digit numbers whose adjacent digits have equal GCD
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Given a positive integer N, the task is to find the number of all N-digit numbers whose adjacent digits have equal Greatest Common Divisor(GCD).
Examples:
Input: N = 2
Output: 90
Explanation:
All 2-digit numbers satisfy the condition as there is only one pair of digits and there are 90 2-digit numbers.
Input: N = 3
Output: 457
Naive Approach: The simplest approach to solve the given problem is to generate all possible N-digit numbers and count those numbers whose adjacent digits have equal GCD. After checking for all the numbers, print the value of the count as the resultant total count of numbers.
Time Complexity: O(N 10N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using Dynamic Programming because the above problem has Overlapping subproblems and an Optimal substructure. The subproblems can be stored in dp[][][] table memoization where dp[index][prev][gcd] stores the answer from the indexth position till the end, where prev is used to store the previous digit of the number and gcd is the GCD between existing adjacent digits in the number. Follow the steps below to solve the problem:
Initialize a global multidimensional array dp with all values as -1 that stores the result of each recursive call.
Define a recursive function, say countOfNumbers(index, prev, gcd, N) and performing the following steps:
If the value of the index as (N + 1), then return 1 as a valid N-digit number has been formed.
If the result of the state dp[index][prev][gcd] is already computed, return this value dp[index][prev][gcd].
If the current index is 1, then any digit from [1- 9] can be placed.
If the current index is greater than 1, any digit from [0-9] can be placed.
If the index is greater than 2, a digit can be placed if the gcd of the current digit and the previous digit is equal to the GCD of already existing adjacent digits.
After making a valid placement of digits, recursively call the countOfNumbers function for (index + 1).
Return the sum of all possible valid placements of digits as the answer.
Print the value returned by the function countOfNumbers(1, 0, 0, N) as the result.
Below is the implementation of the above approach:
C++
````
// C++ program for the above approach
include
using namespace std;
int dp;
// Recursive function to find the count
// of all the N-digit numbers whose
// adjacent digits having equal GCD
int countOfNumbers(int index, int prev,
int gcd, int N)
{
// If index is N+1
if (index == N + 1)
return 1;
int& val = dp[index][prev][gcd];
// If the state has already
// been computed
if (val != -1)
return val;
// Stores the total count of all
// N-digit numbers
val = 0;
// If index = 1, any digit from
// [1-9] can be placed.
// If N = 0, 0 can be placed as well
if (index == 1) {
for (int digit = (N == 1 ? 0 : 1);
digit <= 9;
++digit) {
// Update the value val
val += countOfNumbers(
index + 1,
digit, gcd, N);
}
}
// If index is 2, then any digit
// from [0-9] can be placed
else if (index == 2) {
for (int digit = 0;
digit <= 9; ++digit) {
val += countOfNumbers(
index + 1, digit,
__gcd(prev, digit), N);
}
}
// Otherwise any digit from [0-9] can
// be placed if the GCD of current
// and previous digit is gcd
else {
for (int digit = 0;
digit <= 9; ++digit) {
// Check if GCD of current
// and previous digit is gcd
if (__gcd(digit, prev) == gcd) {
val += countOfNumbers(
index + 1, digit, gcd, N);
}
}
}
// Return the total count
return val;
}
// Function to find the count of all
// the N-digit numbers whose adjacent
// digits having equal GCD
int countNumbers(int N)
{
// Initialize dp array with -1
memset(dp, -1, sizeof dp);
// Function Call to find the
// resultant count
return countOfNumbers(1, 0, 0, N);
}
// Driver Code
int main()
{
int N = 2;
cout << countNumbers(N);
return 0;
}
````
// C++ program for the above approach
// C++ program for the above approach
```
include #include
```
using namespace std; using namespace std
int dp; int dp 100 10 10
// Recursive function to find the count
// Recursive function to find the count
// of all the N-digit numbers whose
// of all the N-digit numbers whose
// adjacent digits having equal GCD
// adjacent digits having equal GCD
int countOfNumbers(int index, int prev, int countOfNumbers int index int prev
int gcd, int N) int gcd int N
{
// If index is N+1
// If index is N+1
if (index == N + 1) if index == N + 1
return 1; return 1
int& val = dp[index][prev][gcd]; int& val = dp index prev gcd
// If the state has already
// If the state has already
// been computed
// been computed
if (val != -1) if val!= - 1
return val; return val
// Stores the total count of all
// Stores the total count of all
// N-digit numbers
// N-digit numbers
val = 0; val = 0
// If index = 1, any digit from
// If index = 1, any digit from
// [1-9] can be placed.
// [1-9] can be placed.
// If N = 0, 0 can be placed as well
// If N = 0, 0 can be placed as well
if (index == 1) {if index == 1
for (int digit = (N == 1 ? 0 : 1); for int digit = N == 1? 0 1
digit <= 9; digit<= 9
++digit) {++ digit
// Update the value val
// Update the value val
val += countOfNumbers(val += countOfNumbers
index + 1, index + 1
digit, gcd, N); digit gcd N
}
}
// If index is 2, then any digit
// If index is 2, then any digit
// from [0-9] can be placed
// from [0-9] can be placed
else if (index == 2) {else if index == 2
for (int digit = 0; for int digit = 0
digit <= 9; ++digit) {digit<= 9 ++ digit
val += countOfNumbers(val += countOfNumbers
index + 1, digit, index + 1 digit
__gcd(prev, digit), N); __gcd prev digit N
}
}
// Otherwise any digit from [0-9] can
// Otherwise any digit from [0-9] can
// be placed if the GCD of current
// be placed if the GCD of current
// and previous digit is gcd
// and previous digit is gcd
else {else
for (int digit = 0; for int digit = 0
digit <= 9; ++digit) {digit<= 9 ++ digit
// Check if GCD of current
// Check if GCD of current
// and previous digit is gcd
// and previous digit is gcd
if (__gcd(digit, prev) == gcd) {if __gcd digit prev == gcd
val += countOfNumbers(val += countOfNumbers
index + 1, digit, gcd, N); index + 1 digit gcd N
}
}
}
// Return the total count
// Return the total count
return val; return val
}
// Function to find the count of all
// Function to find the count of all
// the N-digit numbers whose adjacent
// the N-digit numbers whose adjacent
// digits having equal GCD
// digits having equal GCD
int countNumbers(int N) int countNumbers int N
{
// Initialize dp array with -1
// Initialize dp array with -1
memset(dp, -1, sizeof dp); memset dp - 1 sizeof dp
// Function Call to find the
// Function Call to find the
// resultant count
// resultant count
return countOfNumbers(1, 0, 0, N); return countOfNumbers 1 0 0 N
}
// Driver Code
// Driver Code
int main() int main
{
int N = 2; int N = 2
cout << countNumbers(N); cout<< countNumbers N
return 0; return 0
}
Java
````
// Java program for the above approach
import java.util.;
class GFG
{
static int[][][] dp = new int;
static int _gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return _gcd(a-b, b);
return _gcd(a, b-a);
}
// Recursive function to find the count
// of all the N-digit numbers whose
// adjacent digits having equal GCD
static int countOfNumbers(int index, int prev,
int gcd, int N)
{
// If index is N+1
if (index == N + 1)
return 1;
int val = dp[index][prev][gcd];
// If the state has already
// been computed
if (val != -1)
return val;
// Stores the total count of all
// N-digit numbers
val = 0;
// If index = 1, any digit from
// [1-9] can be placed.
// If N = 0, 0 can be placed as well
if (index == 1) {
for (int digit = (N == 1 ? 0 : 1);
digit <= 9;
++digit) {
// Update the value val
val += countOfNumbers(
index + 1,
digit, gcd, N);
}
}
// If index is 2, then any digit
// from [0-9] can be placed
else if (index == 2) {
for (int digit = 0;
digit <= 9; ++digit) {
val += countOfNumbers(
index + 1, digit,
_gcd(prev, digit), N);
}
}
// Otherwise any digit from [0-9] can
// be placed if the GCD of current
// and previous digit is gcd
else {
for (int digit = 0;
digit <= 9; ++digit) {
// Check if GCD of current
// and previous digit is gcd
if (_gcd(digit, prev) == gcd) {
val += countOfNumbers(
index + 1, digit, gcd, N);
}
}
}
// Return the total count
return val;
}
// Function to find the count of all
// the N-digit numbers whose adjacent
// digits having equal GCD
static int countNumbers(int N)
{
int i, j, k;
// Initialize dp array with -1
for(i = 0; i < 100; i++)
{
for(j = 0; j < 10; j++)
{
for(k = 0; k < 10; k++)
{
dp[i][j][k] = -1;
}
}
}
// Function Call to find the
// resultant count
return countOfNumbers(1, 0, 0, N);
}
public static void main(String[] args)
{
int N = 2;
System.out.println(countNumbers(N));
}
}
// This code is contributed by target_2
````
Python3
````
Python3 program for the above approach
dp = [[[-1 for i in range(10)]
for j in range(10)]
for k in range(100)]
from math import gcd
Recursive function to find the count
of all the N-digit numbers whose
adjacent digits having equal GCD
def countOfNumbers(index, prev, gcd1, N):
# If index is N+1
if (index == N + 1):
return 1
val = dp[index][prev][gcd1]
# If the state has already
# been computed
if (val != -1):
return val
# Stores the total count of all
# N-digit numbers
val = 0
# If index = 1, any digit from
# [1-9] can be placed.
# If N = 0, 0 can be placed as well
if (index == 1):
digit = 0 if N == 1 else 1
while(digit <= 9):
# Update the value val
val += countOfNumbers(index + 1,
digit, gcd1, N)
digit += 1
# If index is 2, then any digit
# from [0-9] can be placed
elif (index == 2):
for digit in range(10):
val += countOfNumbers(index + 1, digit,
gcd(prev, digit), N)
# Otherwise any digit from [0-9] can
# be placed if the GCD of current
# and previous digit is gcd
else:
for digit in range(10):
# Check if GCD of current
# and previous digit is gcd
if (gcd(digit, prev) == gcd):
val += countOfNumbers(index + 1, digit,
gcd1, N)
# Return the total count
return val
Function to find the count of all
the N-digit numbers whose adjacent
digits having equal GCD
def countNumbers(N):
# Function Call to find the
# resultant count
return countOfNumbers(1, 0, 0, N)
Driver Code
if name == 'main':
N = 2
print(countNumbers(N))
This code is contributed by ipg2016107
````
C#
````
// C# program for the above approach
using System;
public class GFG
{
static int[,,] dp = new int[100, 10, 10];
static int _gcd(int a, int b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return _gcd(a-b, b);
return _gcd(a, b-a);
}
// Recursive function to find the count
// of all the N-digit numbers whose
// adjacent digits having equal GCD
static int countOfNumbers(int index, int prev,
int gcd, int N)
{
// If index is N+1
if (index == N + 1)
return 1;
int val = dp[index,prev,gcd];
// If the state has already
// been computed
if (val != -1)
return val;
// Stores the total count of all
// N-digit numbers
val = 0;
// If index = 1, any digit from
// [1-9] can be placed.
// If N = 0, 0 can be placed as well
if (index == 1) {
for (int digit = (N == 1 ? 0 : 1);
digit <= 9;
++digit) {
// Update the value val
val += countOfNumbers(
index + 1,
digit, gcd, N);
}
}
// If index is 2, then any digit
// from [0-9] can be placed
else if (index == 2) {
for (int digit = 0;
digit <= 9; ++digit) {
val += countOfNumbers(
index + 1, digit,
_gcd(prev, digit), N);
}
}
// Otherwise any digit from [0-9] can
// be placed if the GCD of current
// and previous digit is gcd
else {
for (int digit = 0;
digit <= 9; ++digit) {
// Check if GCD of current
// and previous digit is gcd
if (_gcd(digit, prev) == gcd) {
val += countOfNumbers(
index + 1, digit, gcd, N);
}
}
}
// Return the total count
return val;
}
// Function to find the count of all
// the N-digit numbers whose adjacent
// digits having equal GCD
static int countNumbers(int N)
{
int i, j, k;
// Initialize dp array with -1
for(i = 0; i < 100; i++)
{
for(j = 0; j < 10; j++)
{
for(k = 0; k < 10; k++)
{
dp[i,j,k] = -1;
}
}
}
// Function Call to find the
// resultant count
return countOfNumbers(1, 0, 0, N);
}
// Driver code
static public void Main ()
{
int N = 2;
Console.Write(countNumbers(N));
}
}
// This code is contributed by shubham singh
````
JavaScript
````
// Javascript program for the above approach
let dp = new Array(100).fill(0).map(() => new Array(10).fill(0).map(() => new Array(10).fill(-1)));
// Recursive function to find the count
// of all the N-digit numbers whose
// adjacent digits having equal GCD
function countOfNumbers(index, prev, gcd, N)
{
// If index is N+1
if (index == N + 1)
return 1;
let val = dp[index][prev][gcd];
// If the state has already
// been computed
if (val != -1)
return val;
// Stores the total count of all
// N-digit numbers
val = 0;
// If index = 1, any digit from
// [1-9] can be placed.
// If N = 0, 0 can be placed as well
if (index == 1) {
for (let digit = (N == 1 ? 0 : 1);
digit <= 9;
++digit) {
// Update the value val
val += countOfNumbers(
index + 1,
digit, gcd, N);
}
}
// If index is 2, then any digit
// from [0-9] can be placed
else if (index == 2) {
for (let digit = 0; digit <= 9; ++digit) {
val += countOfNumbers(
index + 1, digit,
__gcd(prev, digit), N);
}
}
// Otherwise any digit from [0-9] can
// be placed if the GCD of current
// and previous digit is gcd
else {
for (let digit = 0; digit <= 9; ++digit) {
// Check if GCD of current
// and previous digit is gcd
if (__gcd(digit, prev) == gcd) {
val += countOfNumbers(
index + 1, digit, gcd, N);
}
}
}
// Return the total count
return val;
}
// Function to find the count of all
// the N-digit numbers whose adjacent
// digits having equal GCD
function countNumbers(N)
{
// Function Call to find the
// resultant count
return countOfNumbers(1, 0, 0, N);
}
function __gcd(a, b)
{
if (b == 0)
return a;
return __gcd(b, a % b);
}
let N = 2;
document.write(countNumbers(N));
// This code is contributed by gfgking.
````
Output:
90
Time Complexity: O(N 1000)
Auxiliary Space: O(N 10 10)
S
shreyasshetty788
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12661 | https://pmc.ncbi.nlm.nih.gov/articles/PMC6340385/ | Red and orange flags for secondary headaches in clinical practice: SNNOOP10 list - PMC
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Neurology
. 2019 Jan 15;92(3):134–144. doi: 10.1212/WNL.0000000000006697
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Red and orange flags for secondary headaches in clinical practice
SNNOOP10 list
Thien Phu Do
Thien Phu Do, MD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Thien Phu Do
1, Angelique Remmers
Angelique Remmers, MD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Angelique Remmers
1, Henrik Winther Schytz
Henrik Winther Schytz, MD, PhD, DMSc
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Henrik Winther Schytz
1, Christoph Schankin
Christoph Schankin, MD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Christoph Schankin
1, Sarah E Nelson
Sarah E Nelson, MD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Sarah E Nelson
1, Mark Obermann
Mark Obermann, MD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Mark Obermann
1, Jakob Møller Hansen
Jakob Møller Hansen, MD, PhD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
Find articles by Jakob Møller Hansen
1, Alexandra J Sinclair
Alexandra J Sinclair, MD, PhD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
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1, Andreas R Gantenbein
Andreas R Gantenbein, MD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
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1, Guus G Schoonman
Guus G Schoonman, MD, PhD
1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
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1 From the Headache Diagnostic Laboratory (T.P.D., H.W.S.), Danish Headache Center and Department of Neurology (J.M.H.), Rigshospitalet-Glostrup, Faculty of Health Sciences, University of Copenhagen, Glostrup, Denmark; Department of Neurology (A.R., G.G.S.), Elisabeth-TweeSteden Hospital, Tilburg, the Netherlands; Department of Neurology (C.S.), Inselspital, Bern University Hospital, University of Bern, Switzerland; Department of Neurology and Anesthesiology/Critical Care Medicine (S.E.N.), Johns Hopkins University, Baltimore, MD; Center for Neurology (M.O.), Asklepios Hospitals Schildautal, Seesen; Department of Neurology (M.O.), University Hospital Essen, University of Duisburg-Essen, Germany; Neurometabolism (A.J.S.), Institute of Metabolism and Systems Research, College of Medical and Dental Sciences, University of Birmingham, UK; and Neurorehabilitation (A.R.G.), RehaClinic Bad Zurzach and University of Zürich, Switzerland.
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Correspondence Dr. Schoonman g.schoonman@etz.nl
Go to Neurology.org/N for full disclosures. Funding information and disclosures deemed relevant by the authors, if any, are provided at the end of the article.
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Corresponding author.
Received 2018 Apr 15; Accepted 2018 Aug 27.
© 2018 American Academy of Neurology
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PMCID: PMC6340385 PMID: 30587518
Abstract
A minority of headache patients have a secondary headache disorder. The medical literature presents and promotes red flags to increase the likelihood of identifying a secondary etiology. In this review, we aim to discuss the incidence and prevalence of secondary headaches as well as the data on sensitivity, specificity, and predictive value of red flags for secondary headaches. We review the following red flags: (1) systemic symptoms including fever; (2) neoplasm history; (3) neurologic deficit (including decreased consciousness); (4) sudden or abrupt onset; (5) older age (onset after 65 years); (6) pattern change or recent onset of new headache; (7) positional headache; (8) precipitated by sneezing, coughing, or exercise; (9) papilledema; (10) progressive headache and atypical presentations; (11) pregnancy or puerperium; (12) painful eye with autonomic features; (13) posttraumatic onset of headache; (14) pathology of the immune system such as HIV; (15) painkiller overuse or new drug at onset of headache. Using the systematic SNNOOP10 list to screen new headache patients will presumably increase the likelihood of detecting a secondary cause. The lack of prospective epidemiologic studies on red flags and the low incidence of many secondary headaches leave many questions unanswered and call for large prospective studies. A validated screening tool could reduce unneeded neuroimaging and costs.
Introduction
A minority of headache patients have a secondary headache disorder. There are many possible laboratory tests available for a suspected secondary headache.1 However, running a standard test battery including brain imaging and blood tests in every headache patient is costly2 and entails the risk of false-positive test results and incidental findings.3 Furthermore, it is not possible to detect every secondary headache with standard neuroimaging or laboratory tests.
The medical literature promotes red flags to direct the clinician to initiate a workup plan.4–12 The absence of red flags may suggest that no workup is needed. Clinical experience and large case series of patients with a specific secondary headache form the basis for many red flags. This only allows for the sensitivity of red flags to be known. The specificity and predictive value of a red flag are also necessary to determine whether further testing is necessary.
In this review, we aim to discuss the incidence and prevalence of secondary headaches as well as the data on sensitivity, specificity, and predictive value of red flags for secondary headaches.
Epidemiology of secondary headaches
Neurologists worldwide estimate that 18% of patients with a headache have a secondary headache disorder.13 The International Classification of Headache Disorders 3 (ICHD-3) provides a list of 8 categories and 46 subcategories of potential secondary causes.14
In a Norwegian population study, the 1-year prevalence of chronic secondary headaches was 2.1% and the majority of these patients had medication overuse headache (MOH).15
Regarding studies conducted at tertiary treatment sites, an epidemiologic study identified a secondary cause in 12.9% of their headache patients.16 The most common cause was MOH, accounting for 7.4%. Other secondary causes were identified in ≤1.5% of patients. Another study at a tertiary treatment site also reported similar findings, with one-fifth of the headache patients having a secondary cause identified.17
There appears to be a large risk of selection bias in the above-mentioned studies as one-third of patients with chronic secondary headaches do not have their headache investigated.18
The chance of encountering a secondary headache disorder is high in the emergency department, where headache is often the primary neurologic complaint.19–21
The studies show a high prevalence of secondary headaches. The prevalence variates depending on the setting, i.e., primary care, emergency department, or tertiary headache centers. This must be kept in mind when applying knowledge from the literature into clinical practice.
The evidence for using red and orange flags
We define red flags as symptoms or other pieces of information that encourage testing or observation of the patient. We define orange flags as information that is only alarming when it occurs with other orange or red flags.
A secondary headache disorder is more often suspected than detected. Many headache patients undergo neuroimaging without detecting anything of clinical relevance. A prospective study conducted over 5 years at a tertiary headache service included 3,655 nonacute headache patients.22 The authors used red flags23 to determine whether a patient should undergo imaging or not. After screening, 530 (14.5%) patients underwent imaging. Only 11 had an abnormality (2.1% of patients scanned and 0.3% of the total study population).22 A retrospective study reviewed 360 patients referred for MRI evaluation of a chronic or a recurrent headache.24 Only 0.7% of the patients had relevant findings on MRI.24 In a review of 328 patients with a nonfocal headache, only 1.5% had clinically relevant results.2 A review of 402 chronic headache cases identified abnormalities in 3.7%.25
These studies reveal that even with the use of red flags, most scans still return no findings. Thus, there is a need for assessing the predictive value of red flags.
In 2003, the mnemonic SNOOP (systemic symptoms/signs and disease, neurologic symptoms or signs, onset sudden or onset after the age of 40 years, and change of headache pattern) was proposed as a red flag detection tool for secondary headaches.7 National and regional guidelines8–11 have since provided more items to screen for potential secondary causes leading to the current SNNOOP10 (table 1). We review these items in the following paragraphs.
Table 1.
SNNOOP10 list of red and orange flags
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Systemic symptoms including fever
The combination of headache and fever prompts the clinician to rule out a systemic or neurologic infection or vasculitis, rheumatic disease, or any other inflammatory disease. Neuroinfections include bacterial meningitis, viral meningitis, encephalitis, and brain abscesses. Bacterial meningitis has an annual incidence ranging 0.7–1.38 per 100,000,26,27 with higher incidences reported in specific countries.28 Viral meningitis has an annual incidence of 4.7–7.6 per 100,000 in adults,29,30 higher in children aged <14 years at 27.8 per 100,00030 and infants <1 year old at 219 per 100,000.30 Acute encephalitis has an annual incidence of 6.43–7.4 per 100,000.31 Brain abscesses are rare in developed countries, with an annual incidence of 1.3 per 100,000.32
Fever accounted for 4.8% of all emergency department consultations in the United States in 2007.33 Only a minority of patients with fever have a neurologic infection.34 In a cohort of 213 patients with fever, pneumonia accounted for 27.2% and urinary tract infections for 21.1%; one patient, 0.5%, was diagnosed with meningitis.34
In bacterial meningitis (table 2), the triad of fever, neck stiffness, and decreased consciousness was present in 21.4% to two-thirds of confirmed episodes.28,35–37 Headache was present in 51.7%–87%,35,37,38 fever in 77%–97%,28,35–38 and neck stiffness in 50%–88% of the cases.28,35–38
Table 2.
The frequency of headache, fever, neck stiffness, and decreased consciousness in bacterial meningitis, viral meningitis, encephalitis, and brain abscesses
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Regarding viral meningitis, the triad of fever, neck stiffness, and decreased consciousness never occurred in a 10-year cohort of viral meningitis cases.37 Headache was present in 72.2%, fever in 61.1%, and neck stiffness in 50% of the cases. In a pediatric emergency department, 5.2% of cases with headache as the primary complaint had viral meningitis.39 In a review of patients who underwent lumbar puncture for investigation of putative meningitis and headache with fever as main complaints, 15% had viral meningitis.40
Encephalitis is most commonly caused by herpes simplex virus. In a study of 113 patients with herpes simplex encephalitis, headache was present in 79% and fever in 87% of the episodes.41
In brain abscesses, the frequency of headache at presentation ranges from 49% to 81%, while the frequency of fever ranges from 29% to 57%.42–45 Isolated fever was recorded in 29% of 21 episodes of brain abscesses caused by streptococcus pneumonia.44
Based on the above-mentioned studies, concurrent fever and headache has a relatively high sensitivity for neuroinfections. The specificity cannot be determined from these studies, but presumably it is low as these 2 symptoms occur in a variety of infections. The presentation of the triad of fever, neck stiffness, and decreased consciousness is variable and a lack of this does not rule out a neuroinfection. Headache with systemic symptoms is categorized as a red flag or orange flag in the case of isolated fever.
Key points
Headache with fever is primarily alarming when accompanied by relevant symptoms (e.g., neck stiffness, decreased consciousness, and neurologic deficit).
However, the presentation of a triad of fever, neck stiffness, and decreased consciousness occurs in far from all episodes of neuroinfections.
Neoplasm in history
The worldwide annual incidence of brain tumors is estimated to be 3.4 per 100,000 with a 5-year prevalence of 6.6 per 100,000.46
The risk of finding a brain tumor in a headache patient without a history of any neoplasm is low. It is less than 0.1%,22,47 of which the majority presents after the age of 50 years.47 Lung cancer, breast cancer, and malignant melanoma have the highest risk of intracranial metastasis.48
The frequency of brain metastasis was 24% in a review of patients with breast cancer.49 Headache was the most common symptom and was present in 35% of the cases. In a study of oncology patients, there were intracranial metastases in 32% of 68 patients presenting with a newly developed headache.5 Other risk factors were emesis, headache duration of ≤10 weeks, and atypical headache pattern. In a similar study, 54% of 54 oncology patients with newly developed headache had brain metastasis.6 The authors identified 4 independent predictors: pulsating quality and moderate to severe intensity, emesis, gait instability, and extensor plantar response.
A high proportion of patients with cancer with new onset of a headache have an intracranial metastasis especially if they have a cancer type prone to metastasis to the brain48 or one of the above-mentioned clinical predictors.5,6 Thus, every oncology patient with newly developed headache should undergo MRI. Consequently, neoplasm in history is categorized as a red flag.
Key points
A newly developed headache in a patient with neoplasm is highly suspect for an intracranial metastasis.
Relevant accompanying symptoms include emesis, headache duration ≤10 weeks, atypical headache pattern, pulsating quality and moderate to severe intensity, gait instability, and extensor plantar response.
Neurologic deficit (including decreased consciousness)
Neurologic deficits presenting with headache may have several causes. The most common cause is presumably migraine with aura (causing reversible deficits) followed by intracranial hemorrhage and ischemic stroke. Primary headache disorders with neurologic deficits50–52 are often known by the patient and do not cause alarm. Other causes are infections, abscesses, tumors, and others. The annual incidence of stroke is approximately 300–500 per 100,000.53
In a review of headache in ischemic cerebrovascular disease, headache presented in 16%–65% of patients with TIA and nondisabling stroke.54 Headache occurred in 3%–44% depending on stroke subtype in studies including both TIA and ischemic stroke. Headache occurred more often due to ischemia in the posterior than in the anterior circulation. In a prospective study of 240 patients with acute stroke, headache occurred in 64.5% of the cases with hemorrhagic stroke and 32% of the cases with ischemic stroke.55 In the hemorrhagic stroke group, headache was present in all patients with subarachnoid hemorrhages and in 58% of intraparenchymal hemorrhages. Headache occurred in 59% of patients with a stroke caused by vertebrobasilar artery disease and 26% of patients with stroke caused by anterior circulation disease. In patients with cortical stroke, 56.5% had a headache, whereas only 26.5% of those with subcortical stroke reported a headache. Ten percent had a sentinel headache before the occurrence of the neurologic deficits. A study of headache at stroke onset included 2,196 patients with ischemic stroke or TIA.56 A quarter of the patients presented with headache. Headache at stroke was associated with female sex, history of migraine, younger age, cerebellar stroke, and low blood pressure values on admission. This distribution is similar to a study of 238 patients.57 This study also found that headache severity is not proportional to the size of ischemic stroke lesion. Another study of 876 patients reports equivalent findings.58
As headache with a neurologic deficit has a high sensitivity for stroke, this is categorized as a red flag. Unquestionably, neurologic deficits should always raise serious concern regardless if a headache is present or not.
Key points
Headache occurs in one-fourth of episodes of acute stroke with a higher frequency in hemorrhagic than in ischemic stroke.
The severity of headache is not related to the size of the lesion.
Onset of headache is sudden or abrupt (thunderclap headache)
The ICHD-3 defines thunderclap headache (TCH) as a high-intensity headache of abrupt onset.14 The ICHD-3 defines abrupt onset as reaching maximum intensity in <1 minute. Besides the ICHD-3, there is no clear consensus of what sudden/abrupt onset constitutes, varying from <1 minute to 1 hour.14,59 It should be kept in mind that not all acute headaches are thunderclap headaches as both the temporal aspect and high intensity need to be present. The annual incidence of acute severe headache was 43 per 100,000 in a Swedish cohort.60 It is associated with serious intracranial disorders of vascular origin, in particular subarachnoid hemorrhage (SAH). Less common secondary causes are cerebral venous or sinus thrombosis and other stroke subtypes, intracranial hypotension, reversible cerebral vasoconstriction syndrome, and others.61
A large multicenter cohort study conducted at tertiary care emergency departments included a total of 2,131 patients with acute headache peaking within 1 hour and no neurologic deficits.59 SAH was identified in 6.2%. The combination of age older than 40 years, neck pain or stiffness, loss of consciousness, or onset during exertion resulted in a sensitivity of 98.5% and specificity of 27.5% for SAH. The addition of thunderclap headache and limited neck flexion on examination increased sensitivity to 100% with a reduction of specificity to 15.3%. The positive predictive value was 7.2% and the negative predictive value was 100%. A recent review on sudden and severe headache reported a cerebrovascular cause in 27% of the cases, an infectious origin in 7%, and nonvascular origin in 5%.62 A prospective study identified SAH in 25% of 148 episodes of TCH.63 It was the only symptom in half of the cases. The term sentinel headache defines a headache attack before the rupture of an aneurysm.64 The incidence of a sentinel headache before SAH was 10%–43% in a systematic review.65
In 3 prospective studies,59,60,63 6.2%–25% of patients presenting with an acute headache had SAH. In addition, 12% of the included patients had another serious neurologic condition (e.g., meningoencephalitis).63 Thus, TCH is categorized as a red flag.
Key points
TCH can be the only initial symptom of SAH.
A high sensitivity for SAH can be achieved with a combination of TCH and age older than 40 years, neck pain or stiffness, loss of consciousness, onset during exertion, or limited neck flexion on examination.
TCH also occurs in other neurologic conditions.
Older age (onset after 65 years)
A prospective study from a tertiary headache center showed a higher frequency of secondary headache in patients aged >65 years compared to a younger population (11.2% vs 8.0%).66 Infection was the most common cause of secondary headache, accounting for 29.4% of the episodes. A prospective study included 262 headache patients ≥65 years old.67 The authors identified a secondary cause in 16%. This is in line with findings from a population-based study.68 A study reviewed the records of 193 patients ≥65 years old with headache as their initial and primary symptom.69 While the likelihood of a patient seeking care at a general hospital for headache decreased with age, the risk of a serious underlying cause increased 10-fold at age ≥65 years compared to a younger population. A serious secondary cause (e.g., stroke, temporal arteritis, and others) was identified in 15% of the elderly patients vs 1.6% in a group of patients <65 years old.
Older age is categorized as a red flag; the odds are higher for identifying a secondary cause of alarming etiology in this age group.
Key point
Headache patients ≥65 years old have a higher frequency of serious secondary cause than younger patients.
Pattern change or recent onset of new headache
It is unknown how often pattern change in headache occurs. Presumably, the odds of encountering a secondary etiology are higher in patients with pattern change or recent onset (<3 months) compared to patients with chronic symptoms (>3 months).
Twelve of 30 (40%) patients diagnosed with cerebral venous thrombosis had headache as the only symptom.70 The diagnosis was delayed in the group with isolated headache (9 ± 6.7 days vs 4.5 ± 4.2 days). A prospective study investigated the incidence of isolated headache in patients with intracranial tumors.71 Fifteen of 183 (8.2%) patients had recent onset of headache as their only symptom. Recent onset of headache was an independent predictor for intracranial metastasis in oncology patients.5 A prospective population-based study investigated 100 adult patients with recent onset of a new headache or pattern change of an existing headache.72 There were intracranial lesions in 21% of the patients and normal neurologic examinations in 62% of these patients. The mean duration of headache was shorter in the secondary headache group (2.9 vs 8.2 months).
While there are a limited amount of data regarding this red flag, it has been suggested that recent change of pattern or a newly developed headache (<3 months) can be the only signs of a serious underlying etiology. These are accordingly red flags.
Key points
A recent change of pattern or a newly developed headache can be the only signs of a serious underlying etiology. A correct diagnosis is often delayed in these cases.
Positional headache
Headache occurring immediately or within seconds of assuming an upright position and resolving quickly after lying horizontally is suggestive for low CSF pressure.50 It is commonly recognized in patients who have recently undergone a lumbar puncture or neurosurgical operation. Positional headaches occur spontaneously, with an annual incidence of 5 per 100,000.73
Spontaneous intracranial hypotension differs widely in onset and presentation. In a case series of 5 patients, the onset varied from sudden to gradually over 3 months.74 Four of the patients had orthostatic headache as the presenting complaint. Common accompanying symptoms are nausea, visual disturbances, neck stiffness, vertigo, tinnitus, and cognitive abnormalities.74–76 Spontaneous intracranial hypotension is commonly caused by CSF leaks at the spinal level.76 The orthostatic features may disappear in the chronic patient.76
Many migraine patients report an increase of their headache during physical activity,77 which may mimic positional headache, but is not expected to resolve after lying horizontally.
Positional headache is a red flag as it is a common symptom of intracranial hypotension. The symptoms of intracranial hypotension can be masked by other headache disorders.
Key points
Positional headache is the trademark of intracranial hypotension and the most common cause is CSF leak at the spinal level.
The orthostatic features may diminish over time and they can be masked by other headache disorders.
Precipitated by sneezing, coughing (Valsalva), or exercise
Secondary cough headache is associated with Chiari malformation type 1 (a herniation of the cerebellar tonsils) in 40%.78 In an MRI study of 2,000 healthy volunteers aged ≥45 years, 0.9% had a Chiari malformation.3 Other studies estimate the prevalence of Chiari malformation at 0.24%–3.6% of the general population.79 The prevalence of symptomatic Chiari malformation is different from the prevalence of image-defined Chiari malformation.79 Only one-third of detected cases were symptomatic at the time of diagnosis of MRI-positive Chiari malformations.80 Besides Chiari malformations, posterior fossa lesions account for approximately 15% of cases with cough headache.78 Other less common causes were arachnoid cysts,81 dermoid tumors,81 meningiomas,81 os odontoideum,81 subdural hematoma,83 brain metastases,82 acute sphenoid sinusitis,82 CSF pressure related,83−86 infection,87 hydrocephalus,88 and vascular diseases.89−91 Other triggers for headache in patients with primary cough headache could be laughing,81,92 lifting heavy objects,82,92 postural changes,81,92 and defecation.81,82
Cough headache is categorized as a red flag as it can be a sign of serious underlying pathology such as Chiari malformations and posterior fossa lesions.
Key points
Cough headache can be a sign of serious underlying pathology.
Papilledema
Papilledema was a predictor of intracranial abnormalities in an MRI study.4 Papilledema should thus always lead to further investigation. However, in a prospective study of 34 pediatric patients initially suspected of having papilledema, the majority had pseudo-papilledema or a normal variant.93 Only 2 patients had real papilledema. This type of misinterpretation has also been described in a case report of an adult patient who sustained a head injury after tripping and was suspected to have raised intracranial pressure.94 An ophthalmologist later identified bilateral optic nerve head drusen. Even with these studies in mind, one should still be wary of a potential intracranial tumor. In a retrospective study of 74 pediatric patients diagnosed with primary brain tumors, 38% presented with papilledema and 31% had a triad of headache, vomiting, and papilledema.95
As there is a high prevalence of patients with papilledema who have serious pathology, papilledema is categorized as a red flag.
Key point
A high prevalence of patients with papilledema have a serious underlying pathology.
Progressive headache and atypical presentations
New or recent onset of daily or continuous headache should be alarming, especially if the pain or associated symptoms are progressive. In a series of 17 cases of cerebral venous thrombosis with headache as the only presenting symptom, 65% had a progressive course of their headache.96
An atypical presentation of headache is associated with increased likelihood of abnormality. A study used MRI to evaluate 402 patients for the primary complaint of chronic headache.25 Relevant abnormalities were identified in 15 patients (3.7%). Two-thirds of the patients retrospectively presented with an atypical headache pattern defined as not fulfilling the criteria for any of the primary disorders.
Progressive headache and atypical headache presentation are not commonly described and defined in detail in the literature but given that they may be associated with serious underlying pathology they are thus considered to be red flags.
Key point
Progressive headache and atypical headache presentation can be the only signs of serious underlying pathology.
Pregnancy or puerperium
The risk of secondary headache is higher during pregnancy and puerperium due to physiologic changes (e.g., hypercoagulability, hormonal changes, or intervention, such as epidural anesthesia).97
In a review of de novo headache during pregnancy, the incidence of new onset of headache during pregnancy is estimated to be at 5% of all pregnant women.97 The prevalence of secondary causes for headache is believed to be higher during the third trimester. In a retrospective study of 140 pregnant women presenting to acute care with headache, a secondary cause was identified in one-third.98 The majority presented during the third trimester (56.4%). The authors identified additional risk factors that should prompt a more thorough investigation: the absence of a headache history and the presence of seizures, hypertension, or fever. The most common secondary causes were hypertensive disorders, accounting for approximately half of the cases of secondary headache, followed by pituitary adenoma or apoplexy (20.1%). A secondary cause was identified in 27 of 63 (42.8%) consecutive cases of pregnant women with new onset of headache.99 A nationwide cohort investigating the incidence of stroke during pregnancy found an incidence of 1.5 per 100,000 women.100 This occurred predominantly during the third trimester and puerperium (85% of the cases). A multicenter prospective observational study found that 17% of the cases of venous stroke in women occurred either during pregnancy or puerperium.101 Immediate postpartum headache affects 40% of women.102 Accidental post–dural puncture headache accounts for the majority in those receiving epidural or spinal anesthesia.102,103
One should also keep non-neurologic disorders in mind when a pregnant woman presents with headache.97 Hypertension seems to be the most common, with potential severe consequences such as preeclampsia, but other non-neurologic pathology such as hypothyroidism, anemia, electrolyte imbalance, and diabetes should also be suspected.97,98
As headache during pregnancy can be related to severe pathology, it is categorized as a red flag. Further concern should be raised if there is a presence of additional risk factors.
Key points
Headache during pregnancy and puerperium has a higher risk of severe pathology due to physiologic changes and interventions.
Other risk factors are absence of headache history, occurring during third trimester, seizures, hypertension, and fever.
Painful eye with autonomic features
Eye problems constitute 2%–3% of all primary care consultations.104,105 Orbital pain with redness of the eye occurs also in primary headache disorders and is typical for cluster headache and other trigeminal autonomic cephalalgias (TAC).106 The 1-year prevalence of cluster headache is 53 per 100,000 persons and the lifetime prevalence is 124 per 100,000 persons.107 The annual incidence ranges between 2 and 10 per 100,000.108 Population studies on the epidemiology of TAC are sparse. One study found that 5.3% of all headache referrals in a general neurology setting had a form of TAC.109 The proportion of TAC that is secondary is unknown. Thus, it is recommended that patients presenting with TAC or similar clinical features should always undergo MRI1,110 as they can be secondary to pathology in the posterior fossa, pituitary region, or cavernous sinus.110−113 While atypical presentations are more often associated with secondary findings,110−113 even typical presentations of cluster headache can be due to a structural lesion.114 There are many other diagnoses to consider, both of neurologic and ophthalmic etiology, such as glaucoma, inflammation, and corneal disorders.
Even though secondary etiology is rare, painful eye with autonomic features is categorized as a red flag because even typical presentations can be due to a structural lesion.
Key point
Patients with presentations of painful eye with autonomic features should undergo neuroimaging as it can be due to a structural lesion.
Posttraumatic onset of headache
The annual incidence of traumatic brain injury (TBI) is 235–556 per 100,000 114 and accounts for 4.8% of all emergency department visits in the United States.115 The prevalence of people living with long-term disabilities due to TBI is estimated to be 1.1%–1.7% of the US population.116 The main causes of TBI are traffic accidents, assaults, and falls.117
In a prospective study of 212 patients with mild TBI, the cumulative incidence of headache 1 year after injury was 91%.118 The majority met the criteria for migraine and probable migraine. A smaller proportion was found in a Lithuanian study of patients with concussion.119 Only 11.5% reported headache after 3 months, and after 1 year, the prevalence was similar to controls. The authors suggest that the difference might be related to sociodemographic difference such as a lack of a third-party insurance scheme. A prospective observational study of acute and chronic posttraumatic headache after TBI included 628 patients with momentary loss of consciousness.120 At 2 weeks, 51% had acute posttraumatic headache. At 3 months, 23% had developed chronic posttraumatic headache. Risk factors for acute posttraumatic headache were female sex, younger age, headache present in the emergency department, and CT scan abnormalities. Risk factors for chronic development were female sex and headache present at the emergency department. Patients with posttraumatic headache more often reported anxiety and depression after trauma.
The ICHD-3 guidelines state that posttraumatic headache must develop within 7 days of trauma.14 A 7-day interval yields a higher specificity at the cost of sensitivity.14 To our knowledge, there are no studies on the long-term outcome of patients with persistent headache attributed to traumatic injury to the head. Most experts try to treat this secondary headache based on the phenotype, e.g., if the headache is migraine-like, it is treated with prophylactic drugs with proven efficacy in migraine. However, no clinical trials have investigated which prophylactic or acute medication is effective in posttraumatic headache.
Depending on the time factor, this is categorized as either an orange or red flag. If the duration of the headache is chronic, it is an orange flag. If headache occurs directly in relation to the trauma, it is a red flag.
Key point
Posttraumatic headache is an unspecific marker depending sociodemographic factors. Nonetheless, headache related to trauma should always be explored.
Pathology of the immune system such as HIV
Approximately 1% of the global adult population is infected with HIV.121 The majority of these cases are in sub-Saharan Africa, where it is estimated that 5.2% of the population is infected.121 About 2.7 million people are infected with HIV every year worldwide.121
Headache is the most common pain problem in patients with HIV,122,123 affecting up to 34%–61% of patients.124 The incidence of opportunistic infections increases if the CD4 count is ≤ 200 cells/μL.124 The most common etiologies of CNS lesions in advanced HIV are cerebral toxoplasmosis, primary CNS lymphoma, and progressive multifocal leukoencephalopathy.125 Besides infections, one should be wary of inappropriate immune reactions such as aseptic meningitis.125
Pathology of the immune system is a red flag due to risk of opportunistic infections.
Key points
Headache is the most common pain problem in patients with HIV.
The risk of severe pathology is dependent on the degree of immunosuppression.
Painkiller overuse (MOH) or new drug at onset of headache
The prevalence of MOH among the adult population is 0.5%–7.2% and is the most common secondary headache disorder.16,17,126 Several studies have shown that withdrawal of medication can revert chronic headache to episodic headache in up to 70% of patients.127−129 Furthermore, abrupt withdrawal or tapering down of overused medication is recommended according to the European Federation of Neurologic Societies.130 Yet the approach to treatment and the underlying pathophysiology behind MOH are still highly debated among headache experts. It is more common in women aged 40–50 years.126 While MOH is not a fatal condition, it affects a large proportion of headache patients and should thus always be screened for.
The introduction of any drug can be associated with new onset of headache. In particular, nitric oxide donors, phosphodiesterase inhibitors, carbon monoxide intoxication, histamine intoxication from food sources, cocaine use, exogenous hormones, and acute pressor agents can cause secondary headaches.50 The phenotype of some medication-induced headaches can be identical to primary headache disorders,131 especially migraine, requiring an exact history of the temporal relationship between the onset of headache and medication intake. Furthermore, one should be aware of headache related to substance withdrawal, for instance due to caffeine, opioids, and estrogen.50
Medication headache is categorized as a red flag and it is critical to consider the temporal aspect. We emphasize that it is important to screen for MOH as it is a relatively straightforward treatable secondary headache, even though the course of the headache may be relatively harmless. On the other hand, onset of headache due to a new drug can be a sign of incompatibility with the given drug.
Key points
Medication overuse is the most common cause of secondary headache.
Onset of headache due to a new drug can be a sign of incompatibility with the given drug.
Discussion
An approximation of the sensitivity of a red flag is often possible while the specificity is more difficult to assess as most studies on the occurrence of clinical symptoms are based on retrospective studies of patients with a known secondary headache disorder.
New headache patients should be screened using the list to increase the likelihood of detecting a secondary cause. A combination of red flags might increase the chances of predicting the underlying etiology of a secondary headache. Much remains unclear since there is a lack of prospective epidemiologic studies. Furthermore, some red flags such as pattern change are poorly elucidated. Large-scale studies are necessary given the low incidence of many secondary causes. A validated screening tool for secondary headaches using red flags would be helpful, like the assessment for mild TBI, to minimize the amount of unnecessary testing and patient anxiety. Finally, a validated screening tool will lead to increased awareness of secondary headaches.
Glossary
ICHD-3
International Classification of Headache Disorders 3
MOH
medication overuse headache
SAH
subarachnoid hemorrhage
TAC
trigeminal autonomic cephalalgia
TBI
traumatic brain injury
TCH
thunderclap headache
Author contributions
T.P. Do: drafting the original manuscript. A. Remmers: drafting the original manuscript. S. Henrik Winther: conceptualization and revising the manuscript for intellectual content. C. Schankin: conceptualization and revising the manuscript for intellectual content. S.E. Nelson: conceptualization and revising the manuscript for intellectual content. M. Obermann: conceptualization and revising the manuscript for intellectual content. J. Møller Hansen: conceptualization and revising the manuscript for intellectual content. A.J. Sinclair: conceptualization and revising the manuscript for intellectual content. A.R. Gantenbein: conceptualization and revising the manuscript for intellectual content. G.G. Schoonman: conceptualization and revising the manuscript for intellectual content.
Study funding
Thien Phu Do was funded by a grant from Candys Foundation. Christoph Schankin was funded by grants from German Migraine and Headache Society, Eye on Vision Foundation, and Baasch-Medicus-Stiftung.
Disclosure
T. Do, A. Remmers, H. Schytz, C. Schankin, and S. Nelson report no disclosures relevant to the manuscript. M. Obermann has received scientific support, travel support, and/or honoraria from Biogen Idec, Novartis, Sanofi/Genzyme, Pfizer, Teva, Lilly, Schwarz, and Heel; and received research grants from Allergan, Electrocore, Heel, and the German Ministry for Education and Research (BMBF). J. Hansen, A. Sinclair, A. Gantenbein, and G. Schoonman report no disclosures relevant to the manuscript. Go to Neurology.org/N for full disclosures.
Publication history
Received by Neurology April 15, 2018. Accepted in final form August 27, 2018.
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Abstract
Introduction
Epidemiology of secondary headaches
The evidence for using red and orange flags
Discussion
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12662 | https://duncan.math.sc.edu/s23/math241/exams/mt1_solns.pdf | Solutions Midterm Examination 1 Problem 1.
Given vectors a = ⟨2, 5, −1⟩and b = ⟨2, −1, 1⟩, express the vector 5a −b in component form.
Solution: 5a −b = ⟨10, 25, −5⟩−⟨2, −1, 1⟩= ⟨8, 26, −6⟩.
Problem 2.
A vector v has initial point (1, 2, 4) and terminal point (0, 2, 2). Find the unit vector in the direction of v. Express the answer in component form.
Solution: First, we find v = (0, 2, 2) −(1, 2, 4) = ⟨−1, 0, −2⟩. Then ∥v∥= p (−1)2 + (0)2 + (−2)2 = √ 5.
The unit vector is thus v ∥v∥= −1 √ 5, 0, −2 √ 5 .
Problem 3.
Find the component form of the two-dimensional vector u where ∥u∥= 3 and the angle the vector makes with the positive direction of the x-axis is 30◦in a counter-clockwise direction.
Solution: u = ⟨∥u∥cos(θ), ∥u∥sin(θ)⟩= ⟨3 cos(30◦), 3 sin(30◦)⟩= 3 √ 3 2 , 3 2 + .
Problem 4.
Given vectors a = i + k, b = 2i −j, and c = j + k, compute (a · b)c.
Solution: (a · b)c = (⟨1, 0, 1⟩· ⟨2, −1, 0⟩)⟨0, 1, 1⟩= 2⟨0, 1, 1⟩= ⟨0, 2, 2⟩.
Problem 5.
Given vectors a = ⟨1, 2, 3⟩and b = ⟨0, 2, 1⟩, compute a × b.
Solution: a × b = ⟨−4, −1, 2⟩.
Problem 6.
Find a vector c that is orthogonal to both a = ⟨1, 1, 1⟩and b = ⟨1, 2, 0⟩.
Solution: One way to construct such a c is to use the cross product. Thus c = a × b = ⟨−2, 1, 1⟩.
Problem 7.
Determine the vector projection proju v of the vector v = ⟨2, 1, 1⟩onto the vector u = ⟨1, 1, 1⟩.
Solution: proju v = u · v ∥u∥2u = 4 3⟨1, 1, 1⟩= 4 3, 4 3, 4 3 Problem 8.
Find a normal vector n to the plane with equation x −2y + 3z −60 = 0.
Solution: n = ⟨1, −2, 3⟩ Problem 9.
Let L be the line passing through the point P = (0, 1, 1) with direction v = ⟨3, 2, 1⟩. Find symmetric equations of the line L.
Solution: x 3 = y −1 2 = z −1 1 .
Problem 10.
Find the general equation of the plane passing through P = (0, 0, 1), Q = (2, 3, 0), and R = (1, 0, −1).
MATH 241-H01 Spring 2023 1 of 3 Solutions Midterm Examination 1 Solution: We find a normal vector n to the plane via the cross product n = − → PQ × − → PR = ⟨2, 3, −1⟩× ⟨1, 0, 2⟩= ⟨−6, 3, −3⟩.
Now our plane satisfies −6x + 3y −3z = d for some d. Since P lies on the plane −6(0) + 3(0) −3(1) = d.
Thus d = −3. Thus the equation of our plane is −6x + 3y −3z = −3 or 2x −y + z = 1.
Problem 11.
Find the distance from the point P = (4, 1, 1) to the plane with equation 2x −y + z = 0.
Solution: A normal vector to the plane is n = ⟨2, −1, 1⟩. Note that Q = (0, 0, 0) lies on the plane. Thus, we want to compute the scalar component of the projection of − → QP = ⟨4, 1, 1⟩onto n. This is given by |− → QP · n| ∥n∥ = 8 √ 6 = 4 √ 6 3 .
Problem 12.
Determine whether the line with parametric equations x = 1 + t, y = 1 −t, z = 2 + t, t ∈R intersects the plane with equation x + 2y + z −10 = 0. If it does intersect, find the point of intersection.
Solution: Substituting the parametrization into the equation for the plane we obtain 0 = x + 2y + z −10 = (1 + t) + 2(1 −t) + (2 + t) −10 = −5.
Now 0 = −5 has no solution. Thus the line does not intersect the plane.
Alternatively: The vector v = ⟨1, −1, 1⟩is pointing in the direction of the line. The plane has normal vector n = ⟨1, 2, 1⟩. Since v · n = 0, we conclude that the line is orthogonal to the normal vector. Thus the line is parallel to the plane. Either the line is contained in the plane or they do not intersect. The point (1, 1, 2) is on the line, but is not on the plane so they do not intersect.
Problem 13.
Evaluate lim t→1 t2 −1 t −1 , ln(t), e−t .
Solution: We find lim t→1 t2 −1 t −1 = lim t→1 (t + 1)(t −1) t −1 = 2 and know ln(1) = 0. Thus, the limit is ⟨2, 0, e−1⟩.
Problem 14.
Find the tangent vector to r(t) = ⟨t2, ln(t), et⟩at t = 1.
Solution: First, r′(t) = ⟨2t, 1 t , et⟩. Then r′(1) = ⟨2, 1, e⟩.
Problem 15.
Evaluate the integral Z t2i + cos(t)j −etk dt.
Solution: We find t3 3 i + sin(t)j −etk + C for some constant vector C.
Problem 16.
Determine the length of the parametric curve given by x = 3t2, y = 2t3 where 0 ≤t ≤1.
MATH 241-H01 Spring 2023 2 of 3 Solutions Midterm Examination 1 Solution: The arc length formula gives us s = Z 1 0 sdx dt 2 + dy dt 2 dt = Z 1 0 q (6t)2 + (6t2)2 dt = Z 1 0 √ 36t2 + 36t4 dt = Z 1 0 6t √ 1 + t2 dt Use the substitution u = 1 + t2, du = 2t dt: = Z 2 1 3√u du = 2u 3 2 2 1 = 4 √ 2 −2 Problem 17.
Find the principal unit tangent vector and the principal unit normal vector for the curve r(t) = ⟨sin(t), cos(t), 1 −3t⟩ at t = 0.
Solution: For the principal unit tangent vector, we first compute r′(t) = ⟨cos(t), −sin(t), −3⟩ and ∥r′(t)∥= p cos2(t) + sin2(t) + 32 = √ 10. Thus, T(t) = r′(t) ∥r′(t)∥= cos(t) √ 10 , −sin(t) √ 10 , −3 √ 10 .
For the principal unit normal vector, we compute T′(t) = −sin(t) √ 10 , −cos(t) √ 10 , 0 and ∥T′(t)∥= 1 √ 10. Thus N(t) = T′(t) ∥T′(t)∥= ⟨−sin(t), −cos(t), 0⟩.
Plugging in t = 0, we obtain T(0) = 1 √ 10, 0, −3 √ 10 , and N(0) = ⟨0, −1, 0⟩.
MATH 241-H01 Spring 2023 3 of 3 |
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Maternal Serum Screening
Last Literature Review: March 2021 Last Update:
3000145
Maternal Serum Screen, First Trimester, hCG, PAPP-A, NT 3000145
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
First-trimester screening test for T21 and T18
Does not include alpha fetoprotein (AFP) for ONTD screening
Requires nuchal translucency (NT) measurement performed by an ultrasonographer certified by the Fetal Medicine Foundation (FMF) or the Nuchal Translucency Quality Review (NTQR)
3000146
Maternal Screening, Sequential, Specimen #1, hCG, PAPP-A, NT 3000146
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
First-trimester screening test for T21 and T18
Requires NT measurement performed by an ultrasonographer certified by the FMF or NTQR
Risks provided in both first and second trimesters
3000148
Maternal Screening, Sequential, Specimen #2, Alpha Fetoprotein, hCG, Estriol, and Inhibin A 3000148
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
Second-trimester screening test for T21, T18, and ONTD
Requires a previously submitted first-trimester specimen, Maternal Screening, Sequential, Specimen #1, hCG, PAPP-A, NT (3000146)
Requires NT measurement performed by an ultrasonographer certified by the FMF or NTQR
Risks provided in both first and second trimesters
3000147
Maternal Serum Screening, Integrated, Specimen #1, PAPP-A, NT 3000147
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
First-trimester screening test for T21, T18, and ONTD
Risks determined using a combination of first- and second-trimester serum markers, with or without first-trimester NT measurement
Risks provided after testing is completed for second-trimester specimen, Maternal Serum Screening, Integrated, Specimen #2, Alpha Fetoprotein, hCG, Estriol, and Inhibin A (3000149)
3000149
Maternal Serum Screening, Integrated, Specimen #2, Alpha Fetoprotein, hCG, Estriol, and Inhibin A 3000149
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
Second-trimester screening test for T21, T18, and ONTD
Requires a previously submitted first-trimester specimen, Maternal Serum Screening, Integrated, Specimen #1, PAPP-A, NT (3000147)
Risks are determined after second-trimester specimen is received, using a combination of first- and second-trimester serum markers with or without first-trimester NT measurement
3000143
Maternal Serum Screen, Alpha Fetoprotein, hCG, Estriol, and Inhibin A (Quad) 3000143
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
Second-trimester screening test for T21, T18, and ONTD
3000144
Maternal Serum Screen, Alpha Fetoprotein 3000144
Method
Quantitative Chemiluminescent Immunoassay (CLIA)
Second-trimester screening test for ONTD
The American College of Obstetricians and Gynecologists (ACOG), American College of Medical Genetics and Genomics (ACMG), and Society for Maternal-Fetal Medicine (SMFM) recommend offering both screening and diagnostic testing for chromosomal abnormalities and neural tube defects (NTD) to all pregnant women.,, Screening options include maternal serum screening (MSS), cell-free DNA (cfDNA) screening, and ultrasound. Testing is optional; women may decline screening, as well as prenatal diagnosis. High-risk results merit prompt, appropriate follow-up with critical clinical decisions based on diagnostic rather than screening test results. Refer to the ARUP Consult Prenatal Testing for Chromosomal Abnormalities and Neural Tube Defects topic for additional details.
Disease Overview
Incidence
Open neural tube defects (ONTD): 1/1,400 pregnancies
Trisomy 21 (T21): 1/660 births
Trisomy 18 (T18): 1/3,300 births
Background
ONTD: pretest risk is independent of maternal age.
Most common ONTDs include:
Spina bifida: variable presentation which includes some degree of paralysis of lower limbs, loss of bowel and bladder control, ventriculomegaly
Anencephaly: incompatible with life
T21: pretest risk increases with maternal age.
Caused by an extra chromosome 21 in all cells
Clinical features include hypotonia, characteristic facial features, developmental delays/intellectual disability, and short stature
T18: pretest risk increases with maternal age.
Caused by an extra chromosome 18 in all cells
Clinical features include intrauterine growth restriction, multiple congenital anomalies, and intellectual disability
High risk for pre- and postnatal mortality
Test Description
MSS uses biochemical markers present in maternal blood to identify pregnancies with a higher risk for ONTDs, T21, and T18. Some of the panel tests require NT measurements obtained by certified sonographers to be provided to the laboratory. Gestational age windows for test components are specific. Please refer to the ARUP First and Second Trimester Screening Options table for more information.
Test Interpretation
Results
NOTE: The cutoff values were selected based on a ≤5% false-positive rate.
| Disorder(s)a | Result | Posttest Risk Cutoff |
---
| Maternal Serum Screen, First Trimester Only (3000145) |
| Maternal Serum Screen, Sequential (3000146 [first trimester] and 3000148 [second trimester]) |
| Maternal Serum Screen, Integrated (3000147 [first trimester] and 3000149 [second trimester]) |
| Maternal Serum Screen, Quad (3000143) |
| Maternal Serum Screen, Alpha Fetoprotein (3000144) |
| First Trimester |
| T21 | Screen positive | ≥1/230 |
| Screen negative | <1/230 |
| T18 | Screen positive | ≥1/100 |
| Screen negative | <1/100 |
| First Trimester |
| T21 T18 | Screen positive | ≥1/25 |
| Screen negative | <1/25 |
| Second Trimester |
| T21 | Screen positive | ≥1/110 |
| Screen negative | <1/110 |
| T18 | Screen positive | ≥1/100 |
| Screen negative | <1/100 |
| ONTDsb | Screen positive | ≥1/250 and/or AFP ≥2.5 MoM |
| Screen negative | <1/250 and AFP <2.5 MoM |
| Second Trimester |
| T21 | Screen positive | ≥1/110 |
| Screen negative | <1/110 |
| T18 | Screen positive | ≥1/100 |
| Screen negative | <1/100 |
| ONTDs | Screen positive | ≥1/250 and/or AFP ≥2.5 MoM |
| Screen negative | <1/250 and AFP <2.5 MoM |
| Second Trimester |
| T21 | Screen positive | ≥1/150 |
| Screen negative | <1/150 |
| T18 | Screen positive | ≥1/100 |
| Screen negative | <1/100 |
| ONTDsb | Screen positive | ≥1/250 and/or AFP ≥2.5 MoM |
| Screen negative | <1/250 and AFP <2.5 MoM |
| Second Trimester |
| ONTDsb | Screen positive | ≥1/250 and/or AFP ≥2.5 MoM |
| Screen negative | <1/250 and AFP <2.5 MoM |
| aOther measurements that may indicate areas of increased risk include: uE3 of ≤0.14 MoM: congenital steroid sulfatase deficiency or Smith-Lemli-Optiz syndrome hGC of ≥3.5 MoM: poor fetal outcome bCutoffs for ONTDs vary as follows: Diabetic: ≥1/250 and/or AFP ≥1.90 MoM Twins: ≥1/103 and/or AFP ≥4.50 MoM Twin and diabetic: ≥1/103 and/or AFP ≥2.94 MoM MoM, multiple of median |
Limitations
For test specific sensitivity, see Supplemental Resources.
False positives may occur with incorrect gestational age, multiple gestation pregnancies, fetal demise, placental abnormalities, fetal ventral wall defects, fetal conditions not targeted by MSS, or due to other fetal and maternal biological factors.
References
32976375
Screening for fetal chromosomal abnormalities: ACOG Practice Bulletin Summary, No. 226. Obstet Gynecol. 2020;136(4):859-867.
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Driscoll DA, Gross SJ, Professional Practice Guidelines Committee. Screening for fetal aneuploidy and neural tube defects. Genet Med. 2009;11(11):818-821.
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Gregg AR, Skotko BG, Benkendorf JL, et al. Noninvasive prenatal screening for fetal aneuploidy, 2016 update: a position statement of the American College of Medical Genetics and Genomics. Genet Med. 2016;18(10):1056-1065.
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Palomaki GE, Bupp C, Gregg AR, et al. Laboratory screening and diagnosis of open neural tube defects, 2019 revision: a technical standard of the American College of Medical Genetics and Genomics (ACMG). Genet Med. 2020;22(3):462-474.
Smith’s Recognizable Patterns of Human Malformation
Jones KL, Jones MC, Del Campo, M. Smith’s Recognizable Patterns of Human Malformation. 7th ed. Elsevier Saunders; 2013:7-13.
Related Information
Prenatal Testing for Chromosomal Abnormalities and Neural Tube Defects
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Semi-algebraic Sets
Part of the book series:
Ergebnisse der Mathematik und ihrer Grenzgebiete / A Series of Modern Surveys in Mathematics ((MATHE3,volume 36))
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Abstract
This chapter deals with semi-algebraic sets over a real closed field R. These are the sets defined by a boolean combination of polynomial equations and inequalities. This class of sets has a remarkable property: stability under projection. Several applications of this basic property are investigated. The study of semi-algebraic sets is based mainly on the ∜slicingℝ technique, which makes it possible to decompose them into a finite number of subsets semi-algebraically homeomorphic to open hypercubes. Using this decomposition, we show that a semi-algebraic set has a finite number of semi-algebraically connected components. The notions of connectedness and compactness over a real closed field, other than ℝ, require some care. Nevertheless, closed and bounded semi-algebraic subsets of R
n preserve several of the properties known in the case R = ℝ. They are proved using the curve-selection lemma. All this is the subject of the first five sections of this chapter. In Section 6, we study continuous semi-algebraic functions and we show Łojasiewicz’s inequality. Section 7 deals with the separation of disjoint closed semi-algebraic sets. Section 8 introduces the notion of dimension for a semi-algebraic set and establishes its expected properties. Finally, the last section contains essentially an implicit function theorem in the semi-algebraic framework (this result is well known over R but it is also useful over real closed fields other than ℝ).
Throughout this chapter R is a fixed real closed field.
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Mathematisch Instituut, Vrije Universiteit, De Boelelaan 1081, NL-1081 HV, Amsterdam, The Netherlands
Jacek Bochnak & Marie-Françoise Roy
Institut Mathématique de Rennes, Université de Rennes 1, Campus de Beaulieu, F-35042, Rennes cedex, France
Michel Coste
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Bochnak, J., Coste, M., Roy, MF. (1998). Semi-algebraic Sets.
In: Real Algebraic Geometry. Ergebnisse der Mathematik und ihrer Grenzgebiete / A Series of Modern Surveys in Mathematics, vol 36. Springer, Berlin, Heidelberg.
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Right vs Rightways: Meaning And Differences
Home » Grammar » Word Usage
Are you confused about the difference between “right” and “rightways”? You’re not alone. These two words can be easily mixed up, but in reality, they have distinct meanings that are important to understand. In this article, we will explore the differences between “right” and “rightways” and when to use each one.
Let’s start with the basics. “Right” is a common English word that has a few different meanings. It can refer to something that is true or correct, as in “that answer is right.” It can also refer to a direction, as in “turn right at the next intersection.” Finally, “right” can also be used to indicate permission or authority, as in “you have the right to remain silent.”
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On the other hand, “rightways” is not a commonly used word in English. It is actually a combination of two words: “right” and “ways.” Together, they form a compound word that means “the correct or proper way to do something.” Essentially, “rightways” is a more specific version of “right” that refers to the correct method or process for accomplishing a task.
Now that we have a better understanding of what each word means, let’s explore the differences between them in more detail.
Define Right
Right is a term that refers to something that is just, fair, or morally correct. It is a concept that is often used in discussions about ethics, law, and human rights. When something is considered “right,” it means that it is in accordance with what is expected or required by society, or what is deemed acceptable by a particular group or individual.
For example, in a legal context, the term “right” is often used to refer to a person’s entitlement to certain privileges or protections under the law. This might include the right to free speech, the right to a fair trial, or the right to own property. In a broader sense, the term “right” can also refer to more abstract concepts, such as the right to happiness, the right to equality, or the right to life.
Define Rightways
Rightways, on the other hand, is a term that is less commonly used and may be unfamiliar to many people. It refers to the idea of doing something in the right or proper way, as opposed to simply doing what is considered “right.” In other words, rightways is about the process of achieving a desired outcome, rather than just the outcome itself.
For example, if someone is trying to lose weight, they might be focused on eating the “right” foods or following a particular diet. However, this alone may not be enough to achieve their goal. To truly succeed, they may need to focus on doing things the “rightway,” such as setting achievable goals, tracking their progress, and making sustainable lifestyle changes.
Similarly, in a business context, the concept of rightways might involve focusing on the process of achieving success, rather than just the end result. This could mean prioritizing ethical business practices, investing in employee training and development, or taking a long-term approach to growth and sustainability.
How To Properly Use The Words In A Sentence
Using the right word in a sentence can make all the difference in conveying the intended meaning. In this section, we will discuss how to properly use the words “right” and “rightways” in a sentence.
How To Use “Right” In A Sentence
The word “right” has multiple meanings and can be used as an adjective, adverb, noun, or verb. It can mean correct, just, appropriate, or the opposite of left. Here are some examples of how to use “right” in a sentence:
Adjective: She gave the right answer.
Adverb: He drove straight down the right lane.
Noun: Everyone has the right to freedom of speech.
Verb: You need to right the wrongs you have done.
It is important to note that “right” can also be used in idiomatic expressions such as “right as rain” or “right off the bat”. These expressions have a figurative meaning and should not be taken literally.
How To Use “Rightways” In A Sentence
The word “rightways” is not a commonly used word in the English language. It is often used in informal speech and may be considered nonstandard. However, it can be used to mean “the correct or proper way”. Here is an example of how to use “rightways” in a sentence:
Adverb: He always does things rightways.
It is important to note that “rightways” is not a substitute for the more commonly used word “correctly”. It is best to use “correctly” in formal writing or speech.
More Examples Of Right & Rightways Used In Sentences
In order to better understand the distinction between “right” and “rightways”, let’s take a look at some examples of how these words can be used in sentences.
Examples Of Using “Right” In A Sentence
It’s not right to judge someone based on their appearance.
Everyone has the right to freedom of speech.
The company made the right decision by investing in new technology.
She was right about the outcome of the election.
It’s only right to apologize when you’ve made a mistake.
The teacher marked my answer as right on the test.
He has the right to remain silent.
It’s not right to take credit for someone else’s work.
She made the right choice by pursuing her passion.
Everyone has the right to a fair trial.
Examples Of Using “Rightways” In A Sentence
The rightways to approach a difficult conversation is with empathy and understanding.
There are rightways and wrongways to handle a conflict in the workplace.
She learned the rightways to care for her plants from her grandmother.
There are rightways and wrongways to train a dog.
Following the rightways to prepare a meal can make all the difference in the taste.
He showed me the rightways to lift weights without injuring myself.
There are rightways and wrongways to negotiate a business deal.
She followed the rightways to study for the exam and aced it.
Knowing the rightways to budget your money can help you save for the future.
There are rightways and wrongways to approach a difficult customer.
Common Mistakes To Avoid
When it comes to using the words “right” and “rightways” interchangeably, there are a few common mistakes that people tend to make. These mistakes can lead to confusion and misunderstandings, so it’s important to be aware of them and learn how to avoid them in the future.
Using “Right” When You Mean “Rightways”
One of the most common mistakes is using “right” when you actually mean “rightways.” While these two words may seem interchangeable, they have different meanings. “Right” refers to something that is correct or true, while “rightways” refers to the correct or best way of doing something.
For example, if you were to say “I did it right,” you are saying that you did something correctly. However, if you were to say “I did it the rightways,” you are saying that you did it in the best way possible.
To avoid this mistake, it’s important to pay attention to the context in which you are using these words. If you are referring to the correct way of doing something, use “rightways” instead of “right.”
Using “Rightways” When You Mean “Right”
On the flip side, another common mistake is using “rightways” when you actually mean “right.” This mistake is less common, but it can still cause confusion and misunderstandings.
For example, if you were to say “I know the rightways answer,” you are saying that you know the best way to answer a question. However, if you were to say “I know the right answer,” you are saying that you know the correct answer to a question.
To avoid this mistake, it’s important to be clear about what you are trying to say. If you are referring to the correct answer to a question, use “right” instead of “rightways.”
Offering Tips To Avoid These Mistakes
To avoid these common mistakes, it’s important to be aware of the differences between “right” and “rightways.” Here are a few tips to help you avoid these mistakes in the future:
Pay attention to the context in which you are using these words.
If you are referring to the correct way of doing something, use “rightways” instead of “right.”
If you are referring to the correct answer to a question, use “right” instead of “rightways.”
Take your time when speaking or writing to ensure that you are using the correct word.
Context Matters
When it comes to choosing between “right” and “rightways,” context matters. While both terms may seem interchangeable, they can have different connotations depending on the situation in which they are used.
Examples Of Different Contexts
Let’s take a look at some examples of different contexts and how the choice between “right” and “rightways” might change:
| Context | Choice between “Right” and “Rightways” |
--- |
| Legal | In a legal context, the choice between “right” and “rightways” can be crucial. “Right” may refer to something that is legally correct, while “rightways” may refer to the correct procedure or process to follow in order to achieve a legal outcome. |
| Ethical | When it comes to ethical considerations, the choice between “right” and “rightways” can be complex. “Right” may refer to something that is morally correct, while “rightways” may refer to the correct process or procedure to follow in order to achieve a morally acceptable outcome. |
| Business | In a business context, the choice between “right” and “rightways” may depend on the company’s values and goals. “Right” may refer to something that is in line with the company’s values, while “rightways” may refer to the most effective or efficient process or procedure to achieve the company’s goals. |
As you can see, the choice between “right” and “rightways” can vary depending on the context in which they are used. It’s important to consider the nuances of each term and how they may be perceived in different situations.
Exceptions To The Rules
While the proper use of “right” and “rightways” is generally straightforward, there are some exceptions to the rules that should be noted. In certain situations, the traditional usage of these words may not apply, and it is important to understand these exceptions in order to communicate effectively.
1. Colloquial Speech
In casual conversation or informal writing, it is not uncommon for people to use “right” and “rightways” interchangeably, regardless of their traditional meanings. This can lead to confusion or ambiguity in certain situations, but it is generally accepted in casual settings.
For example, someone might say “I think you’re rightways” to mean “I think you’re correct,” even though “rightways” is not a recognized word in standard English usage.
2. Regional Dialects
Regional dialects and variations in language usage can also affect the proper use of “right” and “rightways.” In some areas, for example, “right” may be used more frequently than “correct,” while in other areas, the opposite may be true.
Additionally, some dialects may use “rightways” to mean “in the proper manner,” even though this usage is not common in standard English. For example, someone might say “You need to do it rightways” to mean “You need to do it correctly.”
3. Technical Jargon
In certain technical fields, such as engineering or computer programming, “right” and “rightways” may have specialized meanings that differ from their traditional usage. In these cases, it is important to understand the specific jargon of the field in order to use the words correctly.
For example, in the context of computer programming, “right” may refer to a specific type of code or syntax that is considered optimal or efficient, while “rightways” may refer to a particular methodology or approach to programming.
4. Idiomatic Expressions
Finally, there are some idiomatic expressions that use “right” or “rightways” in non-standard ways. These expressions can be confusing or difficult to understand for non-native speakers or those unfamiliar with the idioms.
For example, the expression “right as rain” means “completely correct or accurate,” while the expression “all right” can mean “acceptable” or “satisfactory,” even though it does not refer to correctness or accuracy in the traditional sense.
Overall, while the rules for using “right” and “rightways” are generally straightforward, it is important to be aware of these exceptions in order to communicate effectively in a variety of situations. By understanding the nuances of language usage, we can avoid confusion and ensure that our messages are clear and accurate.
Practice Exercises
Now that we have established the difference between “right” and “rightways,” it’s time to put our knowledge into practice. Here are some exercises to help you improve your understanding and use of these words in sentences:
Exercise 1: Fill In The Blank
Choose the correct word to fill in the blank in the following sentences:
It’s not just about being _, it’s about doing things the _ way.
She always tries to do the _ thing, even if it’s not the _ way.
You have the _ to do what’s _, even if it’s not the easiest way.
Answer Key:
right, rightways
right, right
right, rightways
Exercise 2: Rewrite The Sentence
Rewrite the following sentences to use “right” or “rightways” correctly:
He always does things the wrong way.
She thinks she’s always right.
There’s no right or wrong way to do it.
Answer Key:
He always does things the wrongways.
She thinks she’s always right (no change needed).
There’s no rightways or wrong ways to do it.
By practicing these exercises, you can improve your understanding and use of “right” and “rightways.” Remember, it’s not just about being right, it’s about doing things the rightways.
Conclusion
In conclusion, the use of the words “right” and “rightways” can be confusing for many people. As we have seen in this article, “right” is a commonly used word that can have different meanings depending on the context. On the other hand, “rightways” is a less common word that refers to the correct or proper way of doing something.
It is important to understand the subtle differences between these two words and use them appropriately in our writing and speech. Using the wrong word can lead to miscommunication and confusion.
Key takeaways from this article include:
The word “right” can have different meanings depending on the context in which it is used.
“Rightways” refers to the correct or proper way of doing something.
It is important to use these words appropriately to avoid confusion.
As we continue to learn and grow in our understanding of grammar and language use, it is important to stay curious and open to new knowledge. By doing so, we can improve our communication skills and become more effective communicators.
Shawn Manaher
Shawn Manaher is the founder and CEO of The Content Authority. He’s one part content manager, one part writing ninja organizer, and two parts leader of top content creators. You don’t even want to know what he calls pancakes.
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The 30 Dermatomes Explained and Located
Medically reviewed by Susan W. Lee, DO — Written by Jill Seladi-Schulman, Ph.D. — Updated on July 31, 2024
A dermatome is an area of skin supplied by a single spinal nerve. There are 31 pairs of spinal nerves, forming nerve roots that branch from your spinal cord, but only 30 dermatomes.
Your spinal nerves help to relay sensory, motor, and autonomic information between the rest of your body and your central nervous system (CNS).
So why are dermatomes important? How many are there? And where can they be found? Continue reading as we answer these questions about dermatomes and more.
Dermatomes in context
A single spinal nerve supplies sensation to certain areas of your skin, including each of your dermatomes.
Your spinal nerves
Spinal nerves are part of your peripheral nervous system (PNS). Your PNS works to connect the rest of your body with your CNS, which is made up of your brain and spinal cord.
You have 31 pairs of spinal nerves. They form nerve roots that branch from your spinal cord. Spinal nerves are named and grouped by the region of the spine that they’re associated with.
The five groups of spinal nerves are:
Cervical nerves: There are eight pairs of these cervical nerves, numbered C1 through C8. They originate from your neck.
Thoracic nerves: You have 12 pairs of thoracic nerves, numbered T1 through T12. They originate in the part of your spine that makes up your torso.
Lumbar nerves: There are five pairs of lumbar spinal nerves, designated L1 through L5. They come from the part of your spine that makes up your lower back.
Sacral nerves: Like the lumbar spinal nerves, you have five pairs of sacral spinal nerves. They’re associated with your sacrum, one of the bones in your pelvis.
Coccygeal nerves: You have only a single pair of coccygeal spinal nerves, which originate from the area of your coccyx, or tailbone.
Your dermatomes
Each of your dermatomes is associated with a single spinal nerve. These nerves transmit sensations, such as pain, from a specific area of your skin to your CNS.
Your body has 30 dermatomes, one less than the number of spinal nerves. This is because the C1 spinal nerve typically doesn’t have a sensory root. As a result, dermatomes begin with spinal nerve C2.
Dermatomes have a segmented distributionTrusted Source throughout your body. The exact dermatome pattern can actually vary from person to person. Some overlap between neighboring dermatomes may also occur.
Because spinal nerves exit the spine laterally, dermatomes associated with the torso and core are distributed horizontally. When viewed on a body map, they appear like stacked discs.
The dermatome pattern in the limbs is slightly different. This is due to the shape of the limbs as compared with the rest of the body. In general, dermatomes associated with your limbs run vertically along the long axis of the limbs, such as down your leg.
Where is each dermatome located?
Your dermatomes are numbered based on which spinal nerve they correspond to. Below, we’ll outline each dermatomeTrusted Source and the area of the body that it’s associated with.
Remember that the exact area that a dermatome may cover can vary by individual. Some overlap is also possible. As such, consider the outline below to be a general guide.
Cervical spinal nerves
C2: lower jaw, back of the head
C3: upper neck, back of the head
C4: lower neck, upper shoulders
C5: area of the collarbones, upper shoulders, lateral forearm
C6: shoulders, outside of arm, thumb
C7: upper back, back of arm, pointer and middle finger
C8: upper back, inside of arm, ring and little finger
Thoracic spinal nerves
T1: upper chest and back, armpit, front of arm
T2: upper chest and back
T3: upper chest and back
T4: upper chest (area of nipples) and back
T5: mid-chest and back
T6: mid-chest and back
T7: mid-chest and back
T8: upper abdomen and mid-back
T9: upper abdomen and mid-back
T10: abdomen (area of belly button) and mid-back
T11: abdomen and mid-back
T12: lower abdomen and mid-back
Lumbar spinal nerves
L1: lower back, hips, groin
L2: lower back, front and inside of thigh
L3: lower back, front and inside of thigh
L4: lower back, front of thigh and calf, area of knee, inside of ankle
L5: lower back, front and outside of calf, top and bottom of foot, first four toes
Sacral spinal nerves
S1: lower back, back of thigh, back and inside of calf, last toe
S2: buttocks, genitals, back of thigh and calf
S3: buttocks, genitals
S4: buttocks
S5: buttocks
Coccygeal spinal nerves
buttocks, area of tailbone
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Dermatomes diagram
Why are dermatomes important?
Dermatomes are important because they can help to assess and diagnose a variety of conditions. For instance, symptoms that occur along a specific dermatome may indicate a problem with a specific nerve root in the spine.
Examples of this includeTrusted Source:
Radiculopathies: refers to conditions in which a nerve root in the spine is compressed or pinched. Pain from radiculopathies can follow one or more dermatomes. One form of radiculopathy is sciatica. Symptoms can include:
pain
weakness
tingling sensations
Shingles: Shingles is a reactivation of the varicella zoster (chickenpox) virus that lies dormant in the nerve roots of your body. Symptoms of shingles, such as pain and a rash, occur along dermatomes associated with the affected nerve root.
Frequently asked questions
What does dermatome pain feel like?
Depending on the condition that’s affecting your dermatomes, the pain you feel may be different.
Commonly associated with dermatome conditions that may affect more than one dermatome are symptoms like:
tingling sensations
itching
burning
weakness
What’s the difference between dermatomes and myotomes?
While dermatomes are areas of the skin a spinal nerve supplies sensation to, myotomesTrusted Source are the groups of skeletal muscles a spinal nerve supplies sensation to.
Myotomes are also connected to more than one spinal nerve, so there is more overlap. This is unlike dermatomes, which are associated with a single spinal nerve.
The takeaway
Dermatomes are areas of skin connected to a single spinal nerve. There are 31 spinal nerves and 30 dermatomes. The exact area that each dermatome covers can be different from person to person.
Spinal nerves help relay information from other parts of the body to the central nervous system. As such, each dermatome transmits sensory details from a particular area of skin back to the brain.
Dermatomes can be helpful in evaluating and diagnosing conditions affecting the spine or nerve roots. Experiencing symptoms along a specific dermatome can help inform doctors about which area of the spine may be affected.
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How we reviewed this article:
Healthline has strict sourcing guidelines and relies on peer-reviewed studies, academic research institutions, and medical journals and associations. We only use quality, credible sources to ensure content accuracy and integrity. You can learn more about how we ensure our content is accurate and current by reading our editorial policy.
12.6E: Dermatomes. Anatomy and physiology. (2019).
12.6F: Function and physiology of the spinal nerves. Anatomy and physiology. (2019).
Guo L. (2022). Dermatomes.
Kaiser JT, et al. (2019). Neuroanatomy, spinal nerves. StatPearls.
Whitman PA, et al. (2023). Anatomy, skin, dermatomes. StatPearls.
Our experts continually monitor the health and wellness space, and we update our articles when new information becomes available.
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Foundational Knowledge
Understanding Numerical Aperture & Image Resolution
5 June 2024 · 5 min read
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Abstract
This foundational knowledge article explores the effects of the numerical aperture (NA) of an objective lens on the resolution of images produced by a microscope. It explains the diffraction pattern produced by an objective lens and how increasing the NA results in higher resolution images. The tutorial demonstrates the changes in image structure as the NA is adjusted.
Key Learnings:
The numerical aperture (NA) of an objective lens affects the resolution of images produced by a microscope.
Increasing the NA results in higher resolution images.
The light from a point-like structure of a sample passing through the objective pupil is imaged as a three-dimensional diffraction pattern in the intermediate image. The cross section through this three-dimensional diffraction pattern visible in the intermediate image plane consists of a central disc surrounded by bright and dark higher order diffraction rings which together form the Airy disc
How the Numerical Aperture Affects the Airy Disk with its Specific Diffraction Pattern
The image formed by a perfect, aberration-free objective lens at the intermediate image plane of a microscope is a diffraction pattern with a very specific intensity distribution. This tutorial explores the effects of the objective´s numerical aperture (NA) on the diffraction pattern and the resolution of a microscope. The three-dimensional representation of the diffraction pattern is the Point-Spread-Function (PSF) which, in a coma- and/or astigmatism-free system, is symmetrically periodic both along the optical axis, and radially across the image plane. This diffraction pattern can be sectioned in the focal plane to produce a two-dimensional diffraction pattern, having a bright circular disk surrounded by an alternating series of bright and dark higher-order diffraction rings whose intensity decreases with distance from the central disk, the so-called Airy disk. Under visual microscopical observation, only two or three of the circular luminous rings are usually visible in the intermediate image plane.
Tutorial Guide
The tutorial starts with a pattern of Airy disks appearing in the focal plane of the microscope and the point-spread function / three dimensional of a corresponding, single Airy disk pattern shown on the right. To operate the tutorial, use the Numerical Aperture slider to change the objective´s numerical aperture and the resolution of the Airy patterns. The left position of the slider shows the pattern at the lowest objective numerical aperture (= 0.20), and the right position illustrates the highest degree of resolution (numerical aperture = 1.30). As the slider is moved from left to right, the objective’s numerical aperture increases and the complex Airy pattern, as visible in the image, results in a progressively increased resolution of image detail. Correspondingly, the central peak and higher-order diffraction rings in the three-dimensional Airy pattern drawing grow smaller in diameter.
How to Control Image Resolution with Numerical Aperture
The resolving power of an objective determines the size of the formed Airy diffraction pattern: The radius of the central disk is determined by the combined numerical apertures of the objective and condenser. When condenser and objective have equivalent numerical apertures or the objective acts also as the condenser like in an inverted fluorescence microscope, the Airy pattern radius from the central peak to the first minimum is given by the equation:
r(Airy) = 1.22λ/2NA(Obj)(1)
r(Airy) is the Airy radius, λ is the wavelength of the illuminating light, and NA(Obj) is the objective´s numerical aperture (objective aperture = condenser aperture). The numerical aperture depends on the aperture angle of the illumination entering the objective aperture, as well as the refractive index of the imaging medium:
NA(Obj) = n(sin(θ))(2)
θ is the objective’s angular aperture and n is the refractive index of the medium (air, water, or oil) between the objective and the specimen.
The image resolution (D) is defined by this equation and hence corresponds to the Airy radius:
D = 0.61λ/NA(3)
Resolution is clearly influenced by the objective’s numerical aperture. Note that lower values of D indicate higher resolution. In the tutorial, the Numerical Aperture slider is used to control how the image structure evolves as the objective’s numerical aperture is increased. At the lowest numerical aperture value (0.20), the image details visible in the microscope are poorly defined and surrounded by diffraction fringes. As the slider is moved to higher numerical aperture values (0.50-0.80), the structural outline of the image becomes sharper and higher-order diffraction rings begin to emerge. At the highest numerical apertures (1.00-1.30), the diffraction disks are resolved individually as discrete luminous points surrounded by alternating series of bright and dark higher-order diffraction rings of decreasing intensity.
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12668 | http://www.urotoday.com/recent-abstracts/urologic-oncology/testicular-cancer/138748-effect-of-testosterone-replacement-therapy-on-quality-of-life-and-sexual-function-in-testicular-cancer-survivors-with-mild-leydig-cell-insufficiency-results-from-a-randomized-double-blind-trial-beyond-the-abstract.html | Effect of Testosterone Replacement Therapy on Quality of Life and Sexual Function in Testicular Cancer Survivors with Mild Leydig Cell Insufficiency: Results from a Randomized Double-Blind Trial - Beyond the Abstract
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Testicular Cancer
Effect of Testosterone Replacement Therapy on Quality of Life and Sexual Function in Testicular Cancer Survivors with Mild Leydig Cell Insufficiency: Results from a Randomized Double-Blind Trial - Beyond the Abstract
August 9, 2022 Testicular cancer (TC) is one of the most curable cancers nowadays, with a 5-year survival rate of 97,3% in Europe,1 but several long-term complications have been related to the treatment.2 Approximately 50% of patients have disease confined to the testicle and will be cured with unilateral orchiectomy alone, while the remaining 50% will need additional cisplatin-based chemotherapy or abdominal radiotherapy, because of disease spread.3
The pituitary gland excretes luteinizing hormone (LH) and follicle stimulating hormone (FSH) in response to gonadotropin releasing hormone (GnRH) from the hypothalamus. LH stimulates the production of testosterone from Leydig cells in the testicles.4
Post orchiectomy, a great proportion of TC survivors develop mild Leydig cell insufficiency, which means they are in a compensated state with elevated LH and borderline low levels of s-testosterone.5 Furthermore, the toxic effects of radiotherapy and chemotherapy will reduce the Leydig cell function and contribute to a reduction in testosterone production too,6 with the risk appearing to be highest in the most heavily treated patients.7
In non-cancerous populations, testosterone deficiency, defined as total testosterone below the normal range, is associated with a higher incidence of fatigue,8 depression,9 and impaired sexual function 10 as well as several other illnesses.11-13 Testosterone replacement therapy (TRT) has shown a positive effect on these conditions.14,15 To our knowledge, only one other study has investigated this question 16 why it’s still not clear if TRT has a similar effect in TC survivors with mild Leydig cell insufficiency.
This study is a randomized double blind clinical trial, investigating the effect of testosterone replacement therapy in 69 TC survivors over a 12-month period.
The primary aim was to evaluate the effect on metabolic health, which has been reported in a previous publication.17 In this article, we present the secondary findings on patient reported anxiety and depression, overall quality of life, fatigue, and sexual function among TC survivors with. There was no improvement in the patient reported outcomes, routine testosterone substitution in this patient group is not recommended, but the article contains recommendations for further investigation of this subject.
Written by: Emma Grunwald Højer, MD, Department of Oncology, Copenhagen University Hospital, Rigshospitalet, Copenhagen, Denmark
References:
Verdecchia A, Francisci S, Brenner H, et al. Recent cancer survival in Europe: a 2000-02 period analysis of EUROCARE-4 data. Lancet Oncol. 2007;8(9):784-796. doi:10.1016/S1470-2045(07)70246-2
Fung C, Fossa SD. Long-term Morbidity of Testicular Cancer Treatment. Published online 2015:420318.
Krege S, Beyer J, Souchon R, et al. European Consensus Conference on Diagnosis and Treatment of Germ Cell Cancer: A Report of the Second Meeting of the European Germ Cell Cancer Consensus group (EGCCCG): Part I. Eur Urol. 2008;53(3):478-496. doi:10.1016/j.eururo.2007.12.024
Oldenburg J. Hypogonadism and fertility issues following primary treatment for testicular cancer. Urol Oncol. 2015;33(9):407-412. doi:10.1016/j.urolonc.2015.01.014
Bandak M, Aksglaede L, Juul A, Rørth M, Daugaard G. The pituitary-Leydig cell axis before and after orchiectomy in patients with stage I testicular cancer. Eur J cancer. 47(17):2585-2591. doi:10.1016/j.ejca.2011.05.026
Pühse G, Secker A, Kemper S, Hertle L, Kliesch S. Testosterone deficiency in testicular germ-cell cancer patients is not influenced by oncological treatment. Int J Androl. 2011;34(5 PART 2):351-357. doi:10.1111/j.1365-2605.2010.01123.x
Bandak M, Jørgensen N, Juul A, et al. Testosterone deficiency in testicular cancer survivors - a systematic review and meta-analysis. Andrology. 2016;4(3):382-388. doi:10.1111/andr.12177
Mullhall, J. P., MD; Trost, L. W., MD; Branigan, R. E, MD; Kurtz, E. G., MD; Redmon, J. B., MD; Chiles, K. A., MD MSc; Lightner, D. J., MD; Miner, M. M., MD; Murad M. H. MD, MPH; Nelson C. J., PhD; Platz, E. A., ScD, MPH; Ramanathan, L. V., PhD; Lewis, R. M. AUA Guideline TRT. Am Urol Assoc Clin Guidel. 2018;(February).
Zitzmann M. Testosterone and the brain. In: Aging Male. Vol 9. ; 2006:195-199. doi:10.1080/13685530601040679
Nazareth I, Lewin J, King M. Sexual dysfunction after treatment for testicular cancer: a systematic review. J Psychosom Res. 2001;51(6):735-743.
de Haas EC, Oosting SF, Lefrandt JD, Wolffenbuttel BHR, Sleijfer DT, Gietema JA. The metabolic syndrome in cancer survivors. Lancet Oncol. 2010;11(2):193-203. doi:10.1016/S1470-2045(09)70287-6
Howell, S. J. Shalet S. Testosterone deficiency and replacement. Horm Res. 2001;56(17/OCT.):43.
Zitzmann M, Nieschlag E. Hormone substitution in male hypogonadism. Mol Cell Endocrinol. 2000;161(1):73-88. doi:
Ponce OJ, Spencer-Bonilla G, Alvarez-Villalobos N, et al. The Efficacy and Adverse Events of Testosterone Replacement Therapy in Hypogonadal Men: A Systematic Review and Meta-Analysis of Randomized, Placebo-Controlled Trials. J Clin Endocrinol Metab. 2018;103(5):1745-1754. doi:10.1210/jc.2018-00404
Bassil N, Alkaade S, Morley JE. Therapeutics and Clinical Risk Management The benefits and risks of testosterone replacement therapy: a review. Ther Clin Risk Manag. Published online 2009:5-427. www.dovepress.com
Walsh JS, Marshall H, Smith IL, et al. Testosterone replacement in young male cancer survivors: A 6-month double-blind randomised placebo-controlled trial. PLoS Med. 2019;16(11):1-18. doi:10.1371/journal.pmed.1002960
Michael Kreiberg, 1 Niels Jørgensen, 2, 3 Anders Juul, 2, 3 Jakob Lauritsen, 1 Peter Sandor Oturai, 4 Jørn Wulff Helge, 5 Jesper Frank Christensen, 6 Lise Aksglæde, 2 Tim Schauer, 6 Thomas Wagner, 1 Josephine Rosenvilde, 1 Christian Dehlendorff 7 Gedske Daugaard1 & Mikkel Bandak1. A randomized double-blind study of testosterone replacement therapy or placebo in testicular cancer survivors with mild Leydig cell insufficiency (Einstein-intervention). Published online 2021.
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12669 | https://www.tutorchase.com/answers/ib/chemistry/how-can-interpolation-and-extrapolation-be-used-on-a-graph | How can interpolation and extrapolation be used on a graph?
Interpolation and extrapolation are used on a graph to estimate values within or beyond the given data range, respectively.
Interpolation is a method used to find a value between two known values on a graph. It involves creating a line or curve that best fits the data points and then using this line or curve to predict values within the range of the data. For example, if you have data points for the boiling points of a substance at various pressures, you can use interpolation to estimate the boiling point at a pressure that falls between two of your data points. This is particularly useful in chemistry when you want to predict the behaviour of a substance under conditions that you haven't directly measured.
Extrapolation, on the other hand, is used to estimate values outside the range of the known data. This involves extending the line or curve that you've created through interpolation beyond the range of your data. For instance, if you have data for the reaction rate of a chemical reaction at various temperatures, you can use extrapolation to predict the reaction rate at a temperature higher or lower than those you have data for. However, extrapolation should be used with caution as it assumes that the trend observed within the data will continue outside the range of the data, which may not always be the case.
Both interpolation and extrapolation are powerful tools in data analysis, allowing scientists to make predictions and draw conclusions from their data. However, it's important to remember that these are estimates and may not be 100% accurate. The accuracy of interpolation and extrapolation depends on the quality of the data and the appropriateness of the line or curve used to fit the data. Therefore, it's crucial to carefully consider these factors when using these methods in your own data analysis.
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12670 | https://www.ck12.org/book/cbse_maths_book_class_11/section/3.3/ | Trigonometric Functions | CK-12 Foundation
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HomeAlgebraFlexBooksCK-12 CBSE Maths Class 11Ch33. Trigonometric Functions
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3.3 Trigonometric Functions
Difficulty Level: basic | Created by: CK-12
Last Modified: Feb 18, 2016
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You've been working hard in your math class, and are getting to be quite the expert on trigonometric functions. Then one day your friend, who is a year ahead of you in school, approaches you.
“So, you're doing pretty well in math? And you're good with trigonometric functions?” he asks with a smile.
“Yes,” you reply confidently. “I am.”
“Alright, then what's the sine of 150∘?” he asks.
"What? That doesn't make sense. No right triangle has an angle like that, so there's no way to define that function!” you say.
Your friend laughs. “As it turns out, it is quite possible to have trigonometric functions of angles greater than 90∘.”
Is your friend just playing a joke on you, or does he mean it? Can you actually calculate sin 150∘?
At the conclusion of this Concept, you'll be able to answer this question.
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Guidance
Just as it is possible to define the six trigonometric functions for angles in right triangles, we can also define the same functions in terms of angles of rotation. Consider an angle in standard position, whose terminal side intersects a circle of radius r. We can think of the radius as the hypotenuse of a right triangle:
[Figure 1]
The point (x,y) where the terminal side of the angle intersects the circle tells us the lengths of the two legs of the triangle. Now, we can define the trigonometric functions in terms of x,y, and r:
cos θ=x r sin θ=y r tan θ=y x sec θ=r x cosec θ=r y cot θ=x y
And, we can extend these functions to include non-acute angles.
Consider an angle in standard position, such that the point (x,y) on the terminal side of the angle is a point on a circle with radius 1 unit.
[Figure 2]
This circle is called the unit circle. With r=1, we can define the trigonometric functions in the unit circle:
cos θ=x r=x 1=x sin θ=y r=y 1=y tan θ=y x sec θ=r x=1 x cosec θ=r y=1 y cot θ=x y
Notice that in the unit circle, the sine and cosine of an angle are the y and x coordinates of the point on the terminal side of the angle. Now we can find the values of the trigonometric functions of any angle of rotation, even the quadrantal angles, which are not angles in triangles.
[Figure 3]
We can use the figure above to determine values of the trigonometric functions for the quadrantal angles.A quadrantalangle is an angle that has its terminal side on one of the four lines of axis:positive “x”, negative “x”, positive “y”, or negative “y”.For example,sin 90∘=y=1.
Example 1
The point (-3, 4) is a point on the terminal side of an angle in standard position. Determine the values of the six trigonometric functions of the angle.
Solution:
Notice that the angle is more than 90 degrees, and that the terminal side of the angle lies in the second quadrant. This will influence the signs of the trigonometric functions.
[Figure 4]
cos θ=−3 5 sin θ=4 5 tan θ=4−3 sec θ=5−3 cosec θ=5 4 cot θ=−3 4
Notice that the value of r depends on the coordinates of the given point. You can always find the value of r using the Pythagoras Theorem. However, often we look at angles in a circle with radius 1. As you can see, doing this allows us to simplify the definitions of the trigonometric functions.
Example 2
Use the unit circle above to find the value of cos 90∘
Solution:
cos 90∘=0
The ordered pair for this angle is (0, 1). The cosine value is the x coordinate, 0.
Example 3
Use the unit circle above to find the value of cot 180∘
Solution:
cot 180∘ is undefined
The ordered pair for this angle is (-1, 0). The ratio x y is −1 0, which is undefined.
Guided Practice
Use this figure:
[Figure 5]
to answer the following questions.
Find cos θ on the circle above.
Find cot θ on the circle above.
Find cosec θ on the circle above.
Solutions:
We can see from the “x” and “y” axes that the “x” coordinate is −3–√2, the “y” coordinate is 1 2, and the hypotenuse has a length of 1. This means that the cosine function is:
cos θ=b a s e h y p o t e n u s e=−3√2 1=−3–√2
We know that cot θ=1 tan θ=1 p e r p e n d i c u l a r b a s e=b a s e p e r p e n d i c u l a r. The adjacent side to θ in the circle is −3–√2 and the opposite side is 1 2. Therefore,
cot θ=−3√2 1 2=−3–√
We know that cosec θ=1 sin θ=1 p e r p e n d i c u l a r h y p o t e n u s e=h y p o t e n u s e p e r p e n d i c u l a r. The opposite side to θ in the circle is 1 2 and the hypotenuse is 1. Therefore,
cosec θ=h y p o t e n u s e p e r p e n d i c u l a r=1 1 2=2
Concept Problem Solution
Since you now know that it is possible to apply trigonometric functions to angles greater than 90∘, you can calculate the sin 150∘. The easiest way to do this without difficulty is to consider that an angle of 150∘ is in the same position as 30∘, except it's in the second quadrant. This means that it has the same “x” and “y” values as 30∘, except that the “x” value is negative.
Therefore,
sin 150∘=1 2
Recall special right triangles from Geometry. In a(30∘−60∘−90∘)triangle, the sides are in the ratio 1:3–√:2.
In an isosceles triangle(45∘−45∘−90∘), the congruent sides and the hypotenuse are in the ratio 1:1:2–√.
[Figure 6]
In a(30∘−60∘−90∘)triangle, the sides are in the ratio 1:3–√:2.
[Figure 7]
Now let’s make the hypotenuse equal to 1 in each of the triangles so we’ll be able to put them inside the unit circle. Using the appropriate ratios, the new side lengths are:
[Figure 8]
[Figure 9]
Using these triangles, we can evaluate sine, cosine and tangent for each of the angle measures.
sin 45∘=2–√2 sin 60∘=3–√2 sin 30∘=1 2 cos 45∘=2–√2 cos 60∘=1 2 cos 30∘=3–√2 tan 45∘=1 tan 60∘=3√2 1 2=3–√tan 30∘=1 2 3√2=3–√2
These triangles can now fit inside the unit circle.
[Figure 10]
Putting together the trigonometric ratios and the coordinates of the points on the circle, which represent the lengths of the legs of the triangles,(Δ x,Δ y), we can see that each point is actually(cos θ,sin θ), where θ is the reference angle. For example,sin 60∘=3–√2 is the y– coordinate of the point on the unit circle in the triangle with reference angle 60∘. By reflecting these triangles across the axes and finding the points on the axes, we can find the trigonometric ratios of all multiples of 0∘,30∘and 45∘(or 0,π 6,π 4 radians).
[Figure 11]
Example 4
Find sin 3 π 2.
Solution:
Find 3 π 2 on the unit circle and the corresponding point is (0,−1). Since each point on the unit circle is (cos θ,sin θ)., sin 3 π 2=−1
Example 5
Find tan 7 π 6.
Solution:
This time we need to look at the ratio sin θ cos θ. We can use the unit circle to find sin 7 π 6=−1 2 and cos 7 π 6=−3–√2. Now, tan 7 π 6=−1 2−3√2=1 3–√=3–√3.
Another way to approach these exact value problems is to use the reference angles and the special right triangles. The benefit of this method is that there is no need to memorize the entire unit circle. If you memorize the special right triangles, can determine reference angles and know where the ratios are positive and negative you can put the pieces together to get the ratios. Looking at the unit circle above, we see that all of the ratios are positive in Quadrant I, sine is the only positive ratio in Quadrant II, tangent is the only positive ratio in Quadrant III and cosine is the only positive ratio in Quadrant IV.
Keeping this diagram in mind will help you remember where cosine, sine and tangent are positive and negative. You can also use the pneumonic device - All Students Take Calculus, or ASTC, to recall which is positive (all the others would be negative) in which quadrant.
The coordinates on the vertices will help you determine the ratios for the multiples of 90∘ or π 2.
[Figure 12]
Example 6
Find the exact values for the following trigonometric functions.
a) cos 120∘
b) sin 5 π 3
c) tan 7 π 2
Solution:
a) First, we need to determine in which quadrant the angles lies. Since 120∘ is between 90∘ and 180∘ it will lie in Quadrant II. Next, find the reference angle. Since we are in QII, we will subtract from 180∘ to get 60∘. We can use the reference angle to find the ratio, cos 60∘=1 2. Since we are in QII where only sine is positive, cos 120∘=cos(180∘−60∘)=−cos 60∘=−1 2.
b) This time we will need to work in terms of radians but the process is the same. The angle 5 π 3 lies in QIV and the reference angle is π 3. This means that our ratio will be negative. Since sin π 3=3–√2,sin 5 π 3=−3–√2.
c) The angle 7 π 2 represents more than one entire revolution and it is equivalent to 2 π+3 π 2. Since our angle is a multiple of π 2 we are looking at an angle on an axis. In this case, the point is (0,−1). Because tan θ=sin θ cos θ,tan 7 π 2=−1 0, which is undefined. Thus, tan 7 π 2 is undefined.
Guided Practice
Evaluate the following:
cos 7 π 3
tan 9 π 2
sin 405∘
tan 11 π 6
cos 2 π 3
Answers:
7 π 3 has a reference angle of π 3 in QI. cos π 3=1 2 and since cosine is positive in QI, cos 7 π 3=1 2.
9 π 2 is coterminal to π 2 which has coordinates (0, 1). So tan 9 π 2=sin 9 π 2 cos 9 π 2=1 0 which is undefined.
405∘ has a reference angle of 45∘ in QI. sin 45∘=2–√2 and since sine is positive in QI, sin 405∘=2–√2.
11 π 6 is coterminal to π 6 in QIV. tan π 6=3–√3 and since tangent is negative in QIV, tan 11 π 6=−3–√3.
2 π 3 is coterminal to π 3 in QII. cos π 3=1 2 and since cosine is negative in QII, cos 2 π 3=1 2.
Explore More
Find the values of the six trigonometric functions for each angle below.
1.0∘
2.90∘
3.180∘
4.270∘
Find the sine of an angle that goes through the point(2–√2,2–√2).
Find the cosine of an angle that goes through the point(2–√2,2–√2).
Find the tangent of an angle that goes through the point(2–√2,2–√2).
Find the secant of an angle that goes through the point(−3–√2,1 2).
Find the cotangent of an angle that goes through the point(−3–√2,−1 2).
Find the cosecant of an angle that goes through the point(3–√2,1 2).
Find the sine of an angle that goes through the point(1 2,−3–√2).
Find the cosine of an angle that goes through the point(−3–√2,1 2).
The sine of an angle in the first quadrant is 0.25. What is the cosine of this angle?
The cosine of an angle in the first quadrant is 0.8. What is the sine of this angle?
The sine of an angle in the first quadrant is 0.15. What is the cosine of this angle?
Find the exact values for the following trigonometric functions.
sin 3 π 4
cos 3 π 2
tan 300∘
sin 150∘
cos 4 π 3
tan π
cos(−15 π 4)
sin 225∘
tan 7 π 6
sin 315∘
cos 450∘
sin(−7 π 2)
cos 17 π 6
tan 270∘
sin(−210∘)
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Description
This maths text was created using CK-12 resources to be seed content for a complete Maths Class 11 course for CBSE students.
Learning Objectives
Difficulty Level
Basic
Subjects
mathematics,Algebra,Geometry
Concept Nodes
Standards Correlations
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Language
English
Date Created
Feb 18, 2016
Last Modified
Feb 18, 2016
Vocabulary
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12671 | http://repository.ias.ac.in/72500/1/72500.pdf | J. Astrophys. Astr. (1997) 18, 87–90 Inverse Compton Scattering – Revisited T. Padmanabhan Inter University Centre for Astronomy and Astrophysics, Post bag 4, Ganeshkhind, Pune 411 007, India; email: paddy@iucaa.ernet.in Received 1996 October 28; accepted 1996 November 30 Abstract. The inverse Compton scattering of high energy electrons by photons is discussed and a simple derivation of the total power radiated is presented. The derivation is completely classical and exhibits clearly why similar formulas are applicable in the case of inverse compton scattering and synchroton radiation. Key words: Radiation mechanisms. An important radiative process, used in several astrophysical contexts to generate high energy photons, is the inverse compton scattering. In this process, relativistic electrons transfer part of their kinetic energy to low energy photons thereby creating high energy photons. Conventional text book derivation (Rybicki & Lightman 1979) of this process uses the photon picture, kinematics of electron-photon scattering and a judicious choice of Lorentz transformations to arrive at the final result: the net energy transferred per second from electron to photons is (1) where σT = (8π/3) (e2/mc2)2 is the Thompson scattering cross section, Urad is the energy density of the radiation field, v is the speed of the electron and γ = (1 – v2/c2)–1/2 The conventional derivation raises some interesting questions regarding the nature of this process. To begin with, the final answer has no h dependence, suggesting that the result has nothing to do with the quantum nature of the radiation. In other words, there must exist a purely classical derivation of the total power radiated in inverse compton process. The second interesting feature regarding (1) is the striking similarity between this formula and the one describing the power radiated by an isotropic distribution of electrons in a synchroton process. Several textbooks have emphasized the fact that the net power radiated by relativistic electrons, moving in a constant magnetic field B, can be obtained by replacing Urad by UΒ = (Β2/8π). In the case of synchroton emission, it is virtually impossible to provide an interpretation in terms of photons and the derivation is completely classical. The above two observations suggest that there must exist an alternative derivation of (1) based on simple classical considerations. In this note, we shall provide such a derivation. Consider an electron moving with a four velocity ui through a region containing electromagnetic radiation. Let the stress tensor corresponding to the radiation be 87 – 88 Τ. Padmanabhan Tab. The electron, accelerated by the electromagnetic field of the radiation, will radiate energy and – consequently – will feel a drag (four) force gi. If we can find gi then the rate of emission of energy can be determined from the component g0. It turns out that gi can be determined quite easily from the following considerations. In the rest frame of the electron, the spatial components of the drag force can be expressed as σTTµ 0 where Tµ 0 is the momentum flux of the radiation in the rest frame of the electron with µ = 1, 2, 3 (Landau & Lifshitz 1979). The spatial components of the four vector fi = σTTkuk will have this form in the rest frame of the electron. We can, of course, add to f i any vector of the form σ TAui without altering this conclusion, since the spatial components of the latter vector will vanish in the rest frame. Thus, we expect gi to have the form gi = σT (Tkuk + Aui) where A is yet to be determined. We can fix A by using the requirement that, for any four force, g ui = 0; this gives A = –Tabuaub. Hence we find that the drag four force acting on an electron, moving through a region containing electromagnetic radiation stress tensor Tab , must have the form (2) As far as the author knows, this result has not been stated explicitly in the literature. A direct derivation of the above formula from the expression for radiation reaction is given in the appendix. This result remains valid whenever Fik does not change rapidly in the region at which the particle is moving. If that is not the case, one can still derive an expression for gi but it will involve derivatives of Fik. We shall now use the above formula to obtain the rate of energy emission in the inverse Compton process and some related cases. In the case of an electron moving through a radiation bath, we have Tab =Urad dia(l, 1/3,1/3,1/3) and ui = (γ ,γ v). Hence (3) From these results, we immediately find that gi — (γf·v,γf) with (4) Since the rate of energy emission by inverse Compton scattering is –f·ν we immediately obtain the result (1). The simplicity of the above derivation, compared to the conventional analysis, is noteworthy. In the case of a charged particle moving in a magnetic field (taken to be along the z-axis) we can perform a similar analysis. In this case we have Tk= UΒ dia(1, –1, –1, 1). Simple calculation gives (5) where vz = v cos α with α being the pitch angle. This is the standard formula for energy emitted in synchroton radiation by a single charge. For a system of charged particles emitting synchroton radiation it is usual to assume 〈vz〉 = v2/3 or – equivalently – 〈sin2α〉 = 2/3. In this case, we find that g0 = – (4/3)γ(UΒ σT γ2v2). – – i i i i 2 Inverse Compton Scattering – Revisited 89 The rate of emission of energy in the synchroton process is therefore (6) The correspondence between this result and (1) arises due to two facts. (i) The structure of Tab for a constant magnetic field and (ii) the assumption of isotropic distribution of velocities for the electrons allowing 〈vz〉= (v2 /3). Finally, the equation (2) can also be used to estimate the radiative force on a charged fluid embedded in a slightly anisotropic radiation field. If the charged particles in the fluid are moving with non relativistic velocities, then equation (2) approximates to (7) We take the slightly perturbed radiation field to have an energy momentum tensor of the form (8) where δTb has a non zero flux Jµ =δTµ 0 .In this case we easily find that gµ is given by (9) The first term represents the “push” exerted by the radiation flux and the second term is the drag arising from the inverse Compton effect. APPENDIX Consider a particle with charge q and mass m moving in an electromagnetic field Fik which is constant in space and time. The radiation reaction force acting on the particle is given by (10) When Fik is a constant we have (11) Substituting these expressions in (10) and rearranging the terms we get (12) Using the definition of Tab we can write FilFkl as (13) 2 a 90 Τ. Padmanabhan Now we can express gi in terms of Tab alone. Note that (14) since the term involving F2 = FabFab cancels out. Therefore, (15) with σT = (8π/3) (q2/mc2)2. This relation expresses the radiation reaction in terms of the energy density of electromagnetic field. When Fik is not a constant, one picks up an additional term on the right hand side of the form (16) This additional term is ignorable when Fik is constant or when it is due to electromagnetic radiation with 〈Fik〉 = 0. References Landau, L. D., Lifshitz, E. M. 1979, Classical Theory of Fields (New York: Pergamon), page 219; equation 78.8. Rybicki, G. B., Lightman, A. P.1979, Radiative Processes in Astrophysics (New York: Wiley). |
12672 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4069649/ | A Case of Acquired Idiopathic Pterygium Inversum Unguis - PMC
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Ann Dermatol
. 2014 Jun 12;26(3):374–376. doi: 10.5021/ad.2014.26.3.374
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A Case of Acquired Idiopathic Pterygium Inversum Unguis
Ji Hye Baek
Ji Hye Baek
1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
Find articles by Ji Hye Baek
1, Hei Sung Kim
Hei Sung Kim
1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
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1, Young Min Park
Young Min Park
1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
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1, Hyung Ok Kim
Hyung Ok Kim
1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
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1, Baik Kee Cho
Baik Kee Cho
1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
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1, Jun Young Lee
Jun Young Lee
1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
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1,✉
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1 Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, Seoul, Korea.
✉
Corresponding author: Jun Young Lee, Department of Dermatology, Seoul St. Mary's Hospital, College of Medicine, The Catholic University of Korea, 222 Banpo-daero, Seocho-gu, Seoul 137-701, Korea. Tel: 82-2-2258-6222, Fax: 82-2-594-3255, jylee@catholic.ac.kr
✉
Corresponding author.
Received 2011 Jul 27; Revised 2011 Nov 25; Accepted 2011 Dec 20; Issue date 2014 Jun.
Copyright © 2014 The Korean Dermatological Association and The Korean Society for Investigative Dermatology
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC4069649 PMID: 24966638
Abstract
Pterygium inversum unguis (PIU) is a rare nail abnormality in which the distal nail bed adheres to the ventral surface of the nail plate, with obliteration of the distal groove. Because of the rarity of this condition, its exact origin is unknown. This disorder can be either congenital or acquired, with or without a family history. The acquired forms may be idiopathic or secondary to systemic connective tissue diseases or other causes such as stroke, neurofibromatosis, leprosy, or the use of nail fortifiers. We present an unusual case of acquired idiopathic PIU of the 10 fingernails in a 22-year-old man.
Keywords: Malformed nails, Pterygium inversum unguis
INTRODUCTION
Pterygium inversum unguis (PIU) is rare nail abnormality characterized by the adherence of the distal portion of the nail bed to the ventral surface of the nail plate, resulting in a subungual extension of the hyponychium and obliteration of the distal groove1,2. The term 'pterygium inversum unguis' was first described in 1973 by Caputo and Prandi3, and its cause is not well understood. This rare condition may be congenital or acquired. The acquired forms, which may be idiopathic or secondary to systemic connective tissue diseases or other causes, have been reported in a few patients. Herein, we report a rare case of PIU of all 10 fingernails, which is the first report of its kind in the Korean dermatologic literature.
CASE REPORT
A 22-year-old healthy man presented with aberrant nails on all 10 fingers, a condition that he had since childhood. He had relatively long fingernails because he did not clip his nails very often owing to pain, discomfort, and easy bleeding that resulted from nail clipping. He had no other symptoms, history of systemic disease, previous trauma, nail disease, or a family history of a similar abnormality. Physical examination of the fingers showed overgrowth of the hyponychium attached to the ventral surface of the free edge of the distal part of the nail plate, and marked subungual thickening owing to keratosis, with loss of the nail grooves of all 10 fingers (Fig. 1). His toenails showed a normal appearance. A punch biopsy specimen from the center of the right index fingernail did not show any specific findings on histopathological examination. We could not conduct further evaluation or administer treatment because the patient was staying abroad.
Fig. 1.
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Fingers showing adherence of the distal portion of the nail bed to the ventral surface of the nail plate with an abnormal elevated onychodermal band.
DISCUSSION
The first case of PIU was described in 1973 in a woman who developed ventral pterygium on multiple fingers of both hands without any apparent causes3. Initially, the authors summarized this abnormality to be an acquired, non-familial, gradually developing, symmetrical, pathologic entity that did not alter the nail plate or periungual soft tissues.
From the time of the first description, 37 cases have been reported in the English literature (Table 1)1,2,3,4,5,6,7,8,9,10,11,12,13. Among the 37 reported cases of PIU, 34 were acquired and 4 were congenital; 26 patients were women and 12 were men. Three patients had a positive familial history. In 12 cases, there were no associated abnormalities, whereas associations with different diseases were reported in the other 25 cases. In most patients, PIU was associated with connective tissue diseases such as systemic sclerosis or lupus erythematosus5,8. Other postulated associations were acrylate allergy, acrylic nail uses among women11, use of nail fortifiers12, stroke2,7, neurofibromatosis5, and leprosy13. Most patients showed PIU lesions on both hands, on either all or some of the fingernails. The toenails are rarely affected, with only 3 reported cases of PIU involving the toes1,7,9. The patient in this case was a 22-year-old man with extra skin at the base of all 10 of his fingernails since childhood. This case is an acquired idiopathic form of PIU affecting all the 10 fingernails without involvement of the toenails.
Table 1.
Review of published cases
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F: female, M: male, A: acquired, C: congenital, FN: fingernails, NM: not mentioned, TN: toenails, BL: bilateral, UL: unilateral, F/Hx: family history, +: positive, .: negative, SLE: systemic lupus erythematosus, DM: dermatomyositis, SS: systemic sclerosis.
The cause of PIU is unknown. Congenital PIU is thought to be caused by an early defect in the development of the fetal groove and ridge. This abnormal embryonic onychogenesis causes mechanical stretching of the hyponychium, which consequently undergoes distal migration through its attachment to the growing nail plate4. Caputo and Prandi3 suggested that idiopathic forms of PIU might be due to the distal extension of the zone of the nail bed that normally contributes to the formation of the nail plate. Patterson5 reported that secondary PIU is the consequence of abnormal distal circulation. Ischemia and altered blood flow result in recurrent ulceration and scarring, which leads to hyperkeratosis of the stratum corneum and pterygium formation. This was hypothesized as the possible pathologic mechanism for pterygium formation in patients with leprosy13 or stroke2,7. However, all these pathophysiologic hypotheses are speculative.
Three cases of PIU have been histopathologically investigated1,2,10. Histological examination revealed a marked hyperkeratotic stratum corneum that extended and firmly attached to the undersurface of the nail plate with normal nail bed vasculature. Vadmal et al.2 concluded that vascular and neurosensory disturbances might have contributed to reactive hyperkeratosis in their patient with stroke. Recently, Oiso et al.10 observed a marked, eosinophilic, keratinized substance with nucleated corneocytes attached to the distal and ventral nail plate, and a whorled keratinized substance in the horny layer of the fingertip. They suggested that the keratinized structure was derived from the nail isthmus, and that PIU was possibly caused by aberrantly regulated keratinization in the isthmus.
The management of PIU is not well defined; different treatments including the use of keratolytics or topical steroids as well as surgical excision with electrocautery are reported to be ineffective4,7,9. The most effective strategy is the treatment of the underlying cause of PIU11. Patients with PIU need to be evaluated to rule out associated rheumatologic diseases, and require long term follow-up examinations to detect the development of connective tissue diseases such as scleroderma. In our case, unfortunately, the patient could not receive any treatment or further long term follow-up because he was living abroad. Future investigations involving a large number of cases will result in a more accurate classification as well as clarification of the pathogenesis of PIU, and the determination of an effective form of therapy.
References
1.Nogita T, Yamashita H, Kawashima M, Hidano A. Pterygium inversum unguis. J Am Acad Dermatol. 1991;24:787–788. doi: 10.1016/s0190-9622(08)80372-6. [DOI] [PubMed] [Google Scholar]
2.Vadmal M, Reyter I, Oshtory S, Hensley B, Woodley DT. Pterygium inversum unguis associated with stroke. J Am Acad Dermatol. 2005;53:501–503. doi: 10.1016/j.jaad.2005.04.057. [DOI] [PubMed] [Google Scholar]
3.Caputo R, Prandi G. Pterygium inversum unguis. Arch Dermatol. 1973;108:817–818. [PubMed] [Google Scholar]
4.Odom RB, Stein KM, Maibach HI. Congenital, painful, aberrant hyponychium. Arch Dermatol. 1974;110:89–90. [PubMed] [Google Scholar]
5.Patterson JW. Pterygium inversum unguis-like changes in scleroderma. Report of four cases. Arch Dermatol. 1977;113:1429–1430. [PubMed] [Google Scholar]
6.Catterall MD, White JE. Pterygium inversum unguis. Clin Exp Dermatol. 1978;3:437–438. doi: 10.1111/j.1365-2230.1978.tb01523.x. [DOI] [PubMed] [Google Scholar]
7.Morimoto SS, Gurevitch AW. Unilateral pterygium inversum unguis. Int J Dermatol. 1988;27:491–494. doi: 10.1111/j.1365-4362.1988.tb00928.x. [DOI] [PubMed] [Google Scholar]
8.Caputo R, Cappio F, Rigoni C, Scarabelli G, Toffolo P, Spinelli G, et al. Pterygium inversum unguis. Report of 19 cases and review of the literature. Arch Dermatol. 1993;129:1307–1309. doi: 10.1001/archderm.129.10.1307. [DOI] [PubMed] [Google Scholar]
9.Balma A, Pope E. Acquired idiopathic pterygium inversum unguis. Clin Pediatr (Phila) 2010;49:394–395. doi: 10.1177/0009922809346731. [DOI] [PubMed] [Google Scholar]
10.Oiso N, Narita T, Tsuruta D, Kawara S, Kawada A. Pterygium inversum unguis: aberrantly regulated keratinization in the nail isthmus. Clin Exp Dermatol. 2009;34:e514–e515. doi: 10.1111/j.1365-2230.2009.03601.x. [DOI] [PubMed] [Google Scholar]
11.Paley K, English JC, 3rd, Zirwas MJ. Pterygium inversum unguis secondary to acrylate allergy. J Am Acad Dermatol. 2008;58:S53–S54. doi: 10.1016/j.jaad.2006.05.040. [DOI] [PubMed] [Google Scholar]
12.Daly BM, Johnson M. Pterygium inversum unguis due to nail fortifier. Contact Dermatitis. 1986;15:256–257. doi: 10.1111/j.1600-0536.1986.tb01358.x. [DOI] [PubMed] [Google Scholar]
13.Patki AH. Pterygium inversum unguis in a patient with leprosy. Arch Dermatol. 1990;126:1110. doi: 10.1001/archderm.126.8.1110b. [DOI] [PubMed] [Google Scholar]
Articles from Annals of Dermatology are provided here courtesy of Korean Dermatological Association and Korean Society for Investigative Dermatology
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12673 | https://learnenglish.britishcouncil.org/grammar/english-grammar-reference/ability | Ability | LearnEnglishSkip to main content
Level: beginner
We use can and can't to talk about someone's skill or general abilities:
She can speak several languages.
He can swim like a fish.
They can't dance very well.
We use can and can't to talk about the ability to do something at a specific time in the present or future:
I can see you.
Help! I can't breathe.
We use could and couldn't to talk about the past:
She could speak several languages.
I couldn't see you.
Ability: can and could 1
Ability: can and could 2
Level: intermediate
We use could have to say that someone had the ability or opportunity to do something, but did not do it:
She could have learned Swahili, but she didn't want to.
I could have danced all night. [but I didn’t]
Ability: could have 1
Ability: could have 2
Resource skill
Resource type
Grammar reference
What's the different between he can speak several languages and she could speak several languages ?
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Hello AHMED-22,
In this context, which is describing ability, 'can' is the present form and describes his ability now. 'Could' is the past form and describes his ability in the past. We would probably use 'could' if the person is no longer alive, for example, or has forgotten the languages for some reason.
Peter
The LearnEnglish Team
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Hello sir ,
Can we replace could to can in any situation,
For example we replace it in impossibility when we use sentence like cannot have got lost .
Can we do same thing in ability with sentence like she could have learned English , we say she can have learned English .
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Hi g-ssan,
No, sometimes their meanings are different. For ability, can refers to the present and could refers to the past. Also, to talk about past opportunities, you can say could have but can have is ungrammatical.
I hope that helps.
Jonathan
The LearnEnglish Team
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Hello sir ,
What about couldn’t have in ability what’s it mean ?
for example : ( I couldn’t have danced all night but i want to ).
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Hello g-ssan,
I'm afraid that sentence doesn't really make sense because 'couldn't have' refers to the past, but 'I want to' refers to the present.
If you meant to say 'I couldn't have danced all night, but I wanted to', I'm afraid this doesn't work either. When we're speaking about ability, 'could have' has this specific meaning of having the opportunity to do something but not doing it. It's not used in a negative form in this way, i.e. 'couldn't have' does not mean you didn't have the opportunity to do something but did it anyway. This doesn't make sense because you can't do something that you don't have the opportunity to do.
I hope this helps you make sense of it.
All the best,
Kirk
The LearnEnglish Team
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Can we use "ability" to describe things that non-animate things can or can't do? I found some examples like "generalization ability of AI" but I couldn't find a solid source.
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Hi omer3939,
Yes! Here are some examples I found.
What impresses more about this car is its handling ability.
The city has thrived on its ability to sell.
The machine has a superior cutting ability.
But overall, it seems more common to use this word to refer to human (or animate) abilities.
Best wishes,
Jonathan
The LearnEnglish Team
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Hi,
In this sentence "She could get up at 9 Am, but She did not clean her face"Why we use -could-instead of
-was able to- ?In this situation getting up is special ability,isn't it?
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Hello Yigido,
That sentence sounds wrong to me. Perhaps it would make sense in its context, but looking at it now, I can't imagine how it is correct.
All the best,
Kirk
The LearnEnglish Team
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Never mind
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hi how to learn tofel
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Hello abdirahman mohamed yousuf,
The British Council focuses on the IELTS exam rather than TOEFL. We have plenty of resources for the former:
~
For TOEFL, I suggest you try the homepage of the exam, where you should be able to find more information:
~
Peter
The LearnEnglish Team
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Hello Team!
I want to ask a question.
I saw this sentence in a technical book.But I don't understand the meaning and duty of "to" used after shall modal verb.Here is the sentence:
"Unless otherwise specified,all codes,standarts and recommended practises herein shall be to latest editions,addendums and suplements issued before March 17, 2017"
Is this sentence grammatically true?And the meaning of this sentence unclear for me.
Could you please explain?
Thank you!
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Hello Goktung123,
The meaning of the sentence is that all of the specified codes etc are relevant to the most recent editions (etc), not to earlier ones.
~
The sentence is not completely grammatical. There are spelling mistakes and grammar mistakes in it and I would not like to try to explain something which may also be an error. This is why we tend not to provide explanations here of language from unknown sources, but rather focus on explaining the material on our own pages.
~
Peter
The LearnEnglish Team
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Hello Peter!
Thank you for your explanation.Can we say that "to" in that sentence has spelling mistake?I thought that it would be "the".Can we say that?
Thank you so much again!
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Hello Goktug123
'to' makes sense in this sentence; as Peter says, it shows which editions the codes are relevant to. If you changed it to 'the', the sentence would become even more difficult to understand.
All the best
Kirk
The LearnEnglish Team
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Could you please help me?
What is the difference between "didn't have to do" and "needn't have done"?
We didn’t have to run to the museum because it was already closed when we got there.
We needn’t have run to the museum because it was already closed when we got there.
Thank you. I appreciate your helping me.
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Hello Ahmed Imam,
When we use needn't have it means we did something and it was not necessary.
When we usedidn't need to it is not clear if we did something or not.
For example:
I didn't need to go to work. [we don't know if I went to work or not]
I needn't have gone to work. [I went to work and it was not necessary]
Peter
The LearnEnglish Team
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Do we use 'shall' instead of 'should'? For instance we talk about the result of drizzling that it causes mess on streets. We know that streets are in bad condition so there is a mess on the streets due to their bad condition itself, not because of the drizzling. So can we say: Therefore, drizzling shall not be called a distress. Our streets shall be called a distress.
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Hello Zeeshan Siddiqii
I would recommend saying 'we shouldn't call our streets a mess' or something similar here. 'should' works better because you are describing the best thing to talk about the streets in this situation. Note that 'distress' isn't really appropriate in this context in standard British English -- I think 'mess', the word you used earlier, works better here.
All the best
Kirk
The LearnEnglish Team
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I'm really confused and I need your help with these modals.
You ……… have the car inspected next week. The registration expires soon.
(must - have to - need to)
Are all correct in this context?
Thank you.
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Hello Ahmed Imam,
All three options are grammatically correct.
I think the third option (need to) is the best, but the second (have to) is also possible. The first option (must) does not seem a natural choice in any normal context.
Peter
The LearnEnglish Team
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sorry, I don't understand the point related to "must". Is "must" correct here or not.
Thank you.
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Hello Ahmed Imam,
In the context you provided, we would not use must.
Please note that we generally do not comment on questions from other sources. We're happy to answer questions about our own material or about the language generally, but we don't check exercises or questions from textbooks or similar sources.
Peter
The LearnEnglish Team
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I'm sorry for disturbing you but I just try to improve my English. I am a teacher of English in Egypt and I sometimes face some exercises in our outside books which really confuse me. You are a reliable and trusted source so I hope you still receive my notes. Thank you so much.
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12674 | http://faculty.up.edu/wootton/discrete/section2.2.pdf | Section 2.2: Introduction to the Logic of Quantified Statements In this section, we shall continue to examine some of the fundamentals of predicate calculus. Specifically, we shall look at the negations of quantified statements, consider conditional statements in more detail (and the variants of conditional statements), and look at other related statements.
1. Negations of Quantified Statements When we defined quantified statements, we already considered their negations, so we shall briefly summarize them and then look at some examples.
Theorem 1.1. The negation of a statement ∀x ∈D, P(x) is equivalent to ∃x ∈D such that ∼P(x).
Theorem 1.2. The negation of a statement ∃x ∈D, P(x) is equivalent to ∀x ∈D, ∼P(x).
In summary, the negation of an existential statement is a uni-versal statement and the negation of a universal statement is an existential statement. We illustrate with some examples.
Example 1.3. Negate the following statements informally: (i) “All cars are red” The negation would be “There exists a car which is not red” or, “there are cars which are not red” (ii) “Some birds cannot fly” The negation would be “all birds can fly”.
Example 1.4. Write the following statements formally and then negate them: (i) “Some math classes are not fun” Let D be the set of all math classes and P := “is fun”. Then formally, this statement is: ∃x ∈D, ∼P(x).
The formal negation of this statement is ∀x ∈D, ∼∼P(x) 1 2 which is equivalent to ∀x ∈D, P(x).
(ii) “All math classes are fun” Let D be the set of all math classes and P := “is fun”. Then formally, this statement is: ∀x ∈D, P(x).
The formal negation of this statement is ∃x ∈D, ∼P(x).
We can use the rules for negating existential and universal statements together with De Morgans laws and our knowledge of the negations of the other basic propositional connectives to negate much more difficult statements. We illustrate with an example.
Example 1.5. Negate the following statements: (i) ∀x, P(x) ∧Q(x) We have ∼(∀x, P(x) ∧Q(x)) ≡∃x, ∼(P(x) ∧Q(x)) ≡∃x, ∼P(x)∨∼Q(x) (ii) ∀x, P(x) ∧Q(x) →R(x) We have ∼(∀x, P(x) ∧Q(x) →R(x)) ≡∃x, ∼(P(x) ∧Q(x) →R(x)) ≡∃x, (P(x) ∧Q(x))∧∼R(x) since we are negating a conditional statement.
2. Universal Conditional Statements As with the regular conditional statement, there are a number of differ-ences between universal conditional statements and regular universal statements, so we consider them separately here.
2.1. Negating a Universal Conditional Statement. In the last example we considered, we negated a universal conditional statement.
Since it is a fairly important technique in math and logic, we formalize our answer.
Theorem 2.1. The negation of the universal conditional statement ∀x, P(x) →Q(x) is the existential statement ∃x, P(x)∧∼Q(x).
We illustrate with an example.
Example 2.2. Negate the statement “all red cars are fast” The negation of this statement will be “there exists a car which is red and which is not fast”.
3 2.2. Related Conditional Statements. Just as with the regular conditional statement, there are a few closely related statements to the universal conditional statement.
Definition 2.3. Consider the universal conditional statement ∀x ∈D, P(x) →Q(x) (i) The contrapositive of this statement is ∀x ∈D, ∼Q(x) →∼P(x) (ii) The converse of this statement is ∀x ∈D, Q(x) →P(x) (iii) The inverse of this statement is ∀x ∈D, ∼P(x) →∼Q(x) In general, the contrapositive of a universal conditional statement is equivalent to a statement and the inverse and converse of a universal conditional statement are equivalent. However, as statement and its converse are not logically equivalent.
We illustrate with some examples.
Example 2.4. Consider the statement “if an integer x does not equal 0 or 1, then x2 > x”. Write the statement as a universal conditional statement and state the contrapositive, converse and inverse.
Universal conditional statement: “∀integers x, if x ̸= 0 and x ̸= 1 then x2 > x”.
Contrapositive: “∀integers x, if x2 ⩽x then either x = 0 or x = 1”.
Converse: “∀integers x, if x2 > x then x ̸= 0 and x ̸= 1”.
Inverse: “∀integers x, if either x = 0 or x = 1 then x2 ⩽x”.
Notice that the converse and inverse are certainly not logically equiva-lent to the original statement since the original statement is false (take x = 1/2 as a counter example) and the converse is true.
2.3. Vacuous Truth Of Universal Statements. Just as with the regular conditional statement, a universal conditional statement can be vacuously true or true by default. Specifically, we have the following: Result 2.5. Consider the universal conditional statement ∀x ∈D, P(x) →Q(x).
If P(x) is false for all x ∈D, then the universal conditional statement is true. In such circumstances, we say it is true by default or vacuously true.
4 This seems like a silly statement to make, but we need predicate logic to be consistent with propositional logic. Moreover, when we consider examples, it is more clear why it seems reasonable.
Example 2.6. Consider the statement “∀integers x, if x is odd and even, then x is made of cheese.” Clearly this is absurd, there is no way that an integer can be made of cheese, but if we look at a little closer, we shall see this is not the case.
To show this statement is false, we would need to show its negation is true i.e “∃an integer x, such that x is odd and even, and x is not made of cheese.” However, notice that there are no even and odd integers i.e.
if “P := is odd and even”, then P(x) is false for every integer x. Thus this existential expression can never be true (since there are no even and odd integers). It follows that the original statement must be true (since its negation was false) 3. Necessary and Sufficient Conditions, Only if The definitions we introduced for “necessary”, “sufficient” and “only if” can also be extended to universal statements.
Definition 3.1.
(i) The statement “∀x, r(x) is a sufficient condi-tion for s(x)” means “∀x, r(x) →s(x)”.
(ii) The statement “∀x, r(x) is a necessary condition for s(x)” means “∀x, ∼r(x) →∼s(x)” or equivalently “∀x, s(x) → r(x)”.
(iii) The statement “∀x, r(x) only if s(x)” means “∀x, ∼s(x) →∼ r(x)” or equivalently “∀x, r(x) →s(x)”.
We illustrate with some examples.
Example 3.2. Rewrite the following statements in formal logic (the domain is the set of all complex numbers denoted C) and in informal language.
(i) x2 < 0 only if x is an imaginary number Let P(x) := x2 < 0 and Q :=“is an imaginary number”.
Then this formally translates to “∀x ∈C, ∼Q(x) →∼P(x)” or “∀complex numbers, if x is not imaginary, then x2 ⩾0” (ii) x2 < 0 is a sufficient condition for x to be a imaginary number Let P(x) := x2 < 0 and Q :=“is an imaginary number”.
Then this formally translates to “∀x ∈C, P(x) →Q(x)” or “∀complex numbers, if x2 < 0, then x is imaginary” (iii) x2 < 0 is a necessary condition for x to be a imaginary number Let P(x) := x2 < 0 and Q :=“is an imaginary number”.
Then this formally translates to “∀x ∈C, ∼P(x) →∼Q(x)” or “∀complex numbers, if x2 ⩾0, then x is not imaginary 5 Homework (i) From the book, pages 95-97: Questions: 2, 3, 4, 8, 9, 13, 15, 19, 25, 28, 32, 37, 38, 42, 43 |
12675 | https://liavas.net/courses/calc3/files/Line_arc_length.pdf | Calculus 3 Lia Vas Line Integrals with Respect to Arc Length Suppose that C is a curve in xy-plane given by the equations x = x(t) and y = y(t) on the interval a ≤t ≤b. Recall that the length element ds is given by ds = q (x′(t))2 + (y′(t))2 dt.
Let z = f(x, y) be a surface. The line inte-gral of C with respect to arc length is Z C f(x, y)ds = Z b a f(x(t), y(t)) q (x′(t))2 + (y′(t))2 dt This integral represents the area of the surface between the curve C in the xy-plane and the pro-jection of the curve C on the surface z = f(x, y).
This area is represented by the “curtain” in or-ange in the figure on the right.
Three dimensional curves. Suppose that C is a curve in space given by the equations x = x(t), y = y(t), and z = z(t) on the interval a ≤t ≤b. Recall that the length element ds is given by ds = |⃗ r(t)| = q (x′(t))2 + (y′(t))2 + (z′(t))2 dt Let f(x, y, z) be a function. The line integral of C with respect to arc length is Z C f(x, y, z) ds = Z b a f(x(t), y(t), z(t)) q (x′(t))2 + (y′(t))2 + (z′(t))2 dt Note that the quotient ds dt = q (x′(t))2 + (y′(t))2 + (z′(t))2 is always positive because the right side of the equation is always positive. Thus, if the lower t-value is used as the lower bound of the integral and the larger t-value as the upper, that ensures that dt is positive and, hence, ds is positive also.
Three applications. (1) The total length L of a curve C parametrized by ⃗ r(t) = ⟨x(t), y(t), z(t)⟩ can be found by integrating ds from the beginning to the end of C.
L = Z C ds = Z b a q (x′(t))2 + (y′(t))2 + (z′(t))2 dt (2) If C is the trajectory of an object and it is parametrized by ⃗ r(t) = ⟨x(t), y(t), z(t)⟩, the quotient ds dt = q (x′(t))2 + (y′(t))2 + (z′(t))2 computes the speed of the object at time t. Thus, the integral R b a ds computes the total distance traveled from the time t = a to the time t = b.
1 (3) If a wire C in space has the density ρ(x, y, z), then the mass m and the center of mass (x, y, z) are given by m = Z C ρ(x, y, z) ds x = 1 m Z C x ρ(x, y, z) ds y = 1 m Z C y ρ(x, y, z) ds z = 1 m Z C z ρ(x, y, z) ds.
Practice problems. Evaluate the following line integrals.
1.
R C y ds, C : x = t2, y = t, 0 ≤t ≤2 2.
R C x y4 ds, C is the right half of the circle x2 + y2 = 16 3.
R C x y3 ds, C : x = 4 sin t, y = 4 cos t, z = 3t, 0 ≤t ≤π/2 4.
R C x eyz ds, C is the line segment from (0,0,0) to (1, 2, 3) 5. Find the mass and the center of mass of a wire in the shape of the curve C with the given density function ρ.
a) C is the right half of the circle x2 + y2 = 4 and the density function is a constant k.
b) C is the helix x = 2 sin t, y = 2 cos t, z = 3t, 0 ≤t ≤2π, and the density function is a constant k.
Solutions.
1. x′ = 2t, y′ = 1, so ds is √ 4t2 + 1.
R C y ds = R 2 0 t √ 4t2 + 1dt = 173/2−1 12 = 5.76.
2. The circle x2 + y2 = 16 parametrizes as x = 4 cos t, y = 4 sin t. So ds = √ 16 sin2 t + 16 cos2 t = √ 16 = 4. Since we are integrating over its right half, −π 2 ≤t ≤π 2.
Z C x y4 ds = Z π/2 −π/2 4 cos t 44 sin4 t 4dt = 46sin5 t 5 π/2 −π/2 = 8192 5 = 1638.4.
3. x′ = 4 cos t, y′ = −4 sin t, z′ = 3 ⇒ds = √ 16 cos2 t + 16 sin2 t + 9 = √ 25 = 5.
R C x y3 ds = R π/2 0 4 sin t 43 cos3 t 5dt = (5)44 −cos4 t 4 π/2 0 = (5)43 = 320.
4. The curve C is a line passing (0, 0, 0) in the direction of the vector − → PQ = (1, 2, 3) −(0, 0, 0) = (1, 2, 3). So, an equation of this line is x = 0 + 1t = t, y = 0 + 2t = 2t and z = 0 + 3t = 3t. So, x′ = 1, y′ = 2, z′ = 3, and ds = √1 + 4 + 9 = √ 14. The t = 0 is the t-value that corresponds to (0,0,0) and t = 1 is the value that corresponds to (1,2,3). So, the bounds are 0 ≤t ≤1. The integral is R C x eyz ds = R 1 0 te2t 3t√ 14dt = √ 14 R 1 0 te6t2dt = √ 14 1 12e6t2 1 0 = √ 14 e6−1 12 = 125.48.
5. a) The right half of the circle x2 + y2 = 4 parametrizes as x = 2 cos t, y = 2 sin t with −π 2 ≤t ≤π 2. So ds = √ 4 sin2 t + 4 cos2 t = √ 4 = 2. The mass is m = R C kds = R π/2 −π/2 2kdt = 2kπ.
The x-coordinate is x = 1 2kπ R C kxds = 1 2kπ R π/2 −π/2 2k 2 cos tdt = 8k 2kπ = 4 π. The y-coordinate is y = 1 2kπ R C kyds = 1 2kπ R π/2 −π/2 2k 2 sin tdt = 0.
b) ds = √ 4 cos2 t + 4 sin2 t + 9 = √4 + 9 = √ 13. The mass is m = R C kds = R 2π 0 k √ 13dt = 2kπ √ 13. The x-coordinate is x = 1 2kπ √ 13 R C kxds = 1 2kπ √ 13 R 2π 0 k 2 cos t √ 13dt = 0. Similarly y = 0. The z-coordinate is z = 1 2kπ √ 13 R C kzds = 1 2kπ √ 13 R 2π 0 k 3t √ 13dt = 6kπ2√ 13 2kπ √ 13 = 3π.
2 |
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Schur's Inequality
Schur's Inequality is an inequality that holds for positive numbers. It is named for Issai Schur.
Contents
1 Theorem
1.1 Common Cases
1.2 Proof
1.3 Generalized Form
2 References
3 See Also
Theorem
Schur's inequality states that for all non-negative and :
The four equality cases occur when or when two of are equal and the third is .
Common Cases
The case yields the well-known inequality:
When , an equivalent form is:
Proof
Without loss of Generality, let . Note that . Clearly, , and . Thus, . However, , and thus the proof is complete.
Generalized Form
It has been shown by Valentin Vornicu that a more general form of Schur's Inequality exists. Consider , where , and either or . Let , and let be either convex or monotonic. Then,
The standard form of Schur's is the case of this inequality where .
References
Mildorf, Thomas; Olympiad Inequalities; January 20, 2006;
Vornicu, Valentin; Olimpiada de Matematica... de la provocare la experienta; GIL Publishing House; Zalau, Romania.
See Also
Olympiad Mathematics
Inequalities
Number Theory
Retrieved from "
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Inequalities
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12677 | https://pmc.ncbi.nlm.nih.gov/articles/PMC1867141/ | Focusing on the Glomerular Slit Diaphragm: Podocin Enters the Picture - PMC
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Am J Pathol
. 2002 Jan;160(1):3–5. doi: 10.1016/S0002-9440(10)64341-6
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Focusing on the Glomerular Slit Diaphragm
Podocin Enters the Picture
Jeffrey H Miner
Jeffrey H Miner
1 From the Department of Medicine, Renal Division and Department of Cell Biology and Physiology, Washington University School of Medicine, St. Louis, Missouri
Find articles by Jeffrey H Miner
1
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1 From the Department of Medicine, Renal Division and Department of Cell Biology and Physiology, Washington University School of Medicine, St. Louis, Missouri
Accepted 2001 Oct 31.
Copyright © 2002, American Society for Investigative Pathology
PMC Copyright notice
PMCID: PMC1867141 PMID: 11786391
The nature of the glomerular filtration barrier is currently one of the most intensely studied and exciting problems in nephrology research, primarily because of the relatively recent discoveries that two novel and two previously known genes are intimately involved in glomerular filtration. The two novel genes, NPHS1 and NPHS2 (encoding nephrin and podocin, respectively) were identified by positional cloning. NPHS1 is mutated in congenital nephrotic syndrome of the Finnish type,1 and NPHS2 is mutated in steroid resistant nephrotic syndrome.2 The other two genes, previously characterized but not known to be involved in filtration, are ACTN4, encoding α-actinin-4, and Cd2ap, encoding CD2-associated protein (CD2AP). ACTN4 was found to be mutated in several families exhibiting inherited focal segmental glomerulosclerosis,3 and knockout mice lacking CD2AP exhibit congenital nephrotic syndrome.4 Several recent reviews have addressed the potential roles of these molecules in glomerular filtration.5-10
All four of these genes have been shown to be expressed by podocytes, the specialized epithelial cells that lie atop the glomerular basement membrane (GBM) in the urinary space. Podocytes make up the final cell layer across which the glomerular ultrafiltrate must pass before flowing down the tubular portions of the nephron toward the ureter and bladder. Podocytes elaborate long, regularly spaced, interdigitated foot processes that enwrap the glomerular capillaries. Near the GBM, the foot processes are connected by a thin structure termed the glomerular slit diaphragm. The existence of the slit has been known at an ultrastructural level for decades,11 but until recently its molecular composition and role in glomerular filtration had been a mystery.
Several groups have shown by immunoelectron microscopy (IEM) that nephrin is found at the slit diaphragm.12-15 Nephrin is a transmembrane protein with a substantial extracellular domain, and both humans and mice lacking nephrin are born without typical slit diaphragms and exhibit massive proteinuria.1,16 Together, these data suggest that nephrin is an actual component of the slit and lend support to the hypothesis that the slit has a crucial role in glomerular filtration. The localization of CD2AP and α-actinin-4 proteins has also been investigated by immunohistological methods. CD2AP is an adapter molecule17 that localizes to podocytes and is capable of binding to the cytoplasmic tail of nephrin, perhaps linking the slit diaphragm to the podocyte cytoskeleton and providing stability to what might otherwise be a fragile structure.4 Our more recent studies demonstrate by IEM that CD2AP is present at the slit and binds to nephrin via its carboxyl terminal domain.18 Other IEM studies are consistent with these results but suggest that CD2AP is more widely distributed in the foot process,19 in agreement with our original immunofluorescence data.4 α-actinin-4 is a member of a family of proteins that cross-link and anchor actin filaments, and it was shown to be associated with the actin cytoskeleton in podocyte foot processes.3,20 The fact that mutations in ACTN4 cause glomerulosclerosis suggests that α-actinin-4 plays a crucial role in maintaining the structure of podocyte foot processes and/or the slit diaphragm via the actin cytoskeleton.3
But what about the localization of podocin, the fourth member of this genetically defined quartet? The article by Corinne Antignac and colleagues21 published in this issue of The American Journal of Pathology clearly confirms that podocin, like nephrin, is present in the podocyte plasma membrane in the area of the slit diaphragm. The authors made several antisera that specifically recognize podocin and showed by both high-resolution confocal microscopy and IEM that podocin is found in foot processes and concentrated at the slit diaphragms. Thus, this entire quartet of molecules localizes to podocyte foot processes, with nephrin, CD2AP, and podocin being present at slit diaphragms. The fact that mutations in any of the four genes causes proteinuria and renal failure indicates that podocytes, their foot processes, and the intervening slits are likely the most critical components of the glomerular filtration barrier.
With regard to other molecules at the slit, it has been known for a long time that ZO-1, a protein normally associated with tight junctions, is localized to the cytoplasmic face of the slit.22 This led to the conclusion that the slit diaphragm is a modified tight junction. However, it was shown by IEM that P-cadherin localizes to the extracellular portion of the slit diaphragm and co-localizes with ZO-1 on the cytoplasmic face of the slit, suggesting that the slit has features of adherens junctions, which can also contain ZO-1.23 However, the importance of P-cadherin at the slit remains to be demonstrated. The facts that 1) knockout mice lacking P-cadherin are viable and fertile with no reported kidney defects,24 and 2) humans with a mutation in CDH3, the gene encoding P-cadherin, have hair and retinal but not kidney defects25 suggests that any role P-cadherin plays in glomerular filtration is either subtle or at least partially redundant. Indeed, another member of the cadherin superfamily, FAT, has also been localized to slit diaphragms.26 However, a role for FAT in kidney function has not yet been addressed. Finally, Neph1, the gene encoding NEPH1, a protein with weak homology to nephrin, is transcribed by podocytes, and knockout mice lacking NEPH1 develop proteinuria. Most Neph1−/− mice die between 1 and 12 days after birth, but a few live up to 8 weeks of age.27 It will be very interesting to determine the subcellular localization of NEPH1 and the mechanisms leading to the onset of proteinuria. Unlike Nphs1, Neph1 is also expressed by mesangial, proximal tubular, and collecting duct cells (and in many other tissues besides kidney).27 Thus, NEPH1 may have functions in the kidney quite distinct from those attributed to nephrin.
The notion that the glomerular slit diaphragm merely serves as a filter to keep albumin and other plasma proteins from entering the urinary space is being challenged by evidence that slit-associated molecules are involved in signaling events. For example, Huber and colleagues have shown that transfection of nephrin into 293T cells initiates activation of AP-1 transcriptional activity and of stress-activated p38 and c-Jun N-terminal protein kinases.28 Interestingly, co-transfection of podocin synergistically increased nephrin-initiated AP-1 activation,28 suggesting that nephrin and podocin may cooperate at the slit to mediate signaling. Furthermore, two groups have shown that nephrin is associated with lipid rafts, specialized cholesterol-rich membrane domains associated with signaling molecules and signaling events.29,30 Podocin is also associated with lipid rafts and interacts directly with nephrin and CD2AP at the slit diaphragm.31
What of the forgotten role of the GBM in glomerular filtration? With all of the focus on podocytes and glomerular slit diaphragms, less attention is being paid to the GBM. Although the GBM may not be the major size-selective filter, its negative charge—imparted by the presence of heparan sulfate proteoglycans such as agrin—is likely important in establishing a charge-selective barrier to negatively charged plasma proteins. In addition, the unique molecular structure of the GBM seems to be crucial for maintaining podocyte homeostasis. Mutations that alter the composition of the GBM and that result in filtration defects may affect podocytes directly and filtration indirectly. The GBM is normally composed of laminin-11 (α5β2γ1), the collagen α3, α4, and α5(IV) chains, agrin, and nidogen, as well as other matrix molecules.32 Knockout mice lacking the laminin β2 chain have no laminin-11 in their GBMs and develop massive proteinuria at 7 days of age.33 Laminin-11 may therefore be necessary for proper podocyte adhesion and for maintenance of foot process architecture. Similarly, mice, humans, and dogs with collagen IV gene mutations exhibit Alport syndrome, which is characterized by later onset filtration defects.34,35 The GBM becomes thickened and split, and novel laminin chains, such as α2 and β1, accumulate within it.36,37 These ectopic laminins may provide atypical signals to the overlying podocytes that disrupt their behavior, leading to foot process effacement and proteinuria. Indeed, reduced levels of laminin α2 in Alport GBM was associated with a reduction in podocyte damage.36
In conclusion, it is clear that the glomerular slit diaphragm is a major, if not the most important, component of the kidney’s ultrafiltration barrier. The finding that podocin is concentrated there reinforces this notion, because mutations in NPHS2, the gene encoding podocin, cause nephrotic syndrome and glomerulosclerosis. Determining the function of podocin will be challenging, given that podocin is a member of the stomatin family of proteins, about which little is known. Nevertheless, the generation of knockout mice lacking podocin, which is presumably underway, will allow for those lines of experimentation that are not possible with human patients, but that will hopefully lead to a better understanding of podocin and the structure and function of the glomerular slit diaphragm.
Table 1.
Proteins that Have Been Localized to the Glomerular Slit Diaphragm
| Protein | Protein class | References |
:---:
| ZO-1 | PDZ domain-containing | 22 |
| Nephrin | Ig superfamily | 12–15 |
| P-cadherin | Classical cadherin | 23 |
| FAT | Cadherin superfamily | 26 |
| CD2AP | SH3 domain-containing adapter | 18, 19 |
| Podocin | Stomatin | 21, 31 |
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Footnotes
Address reprint requests to Dr. Jeffrey H. Miner, Renal Division, Box 8126, Washington University School of Medicine, 660 S. Euclid Ave., St. Louis, MO 63110. E-mail: minerj@pcg.wustl.edu.
References
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12678 | https://connectedmath.msu.edu/the-math/development-of-the-math-strands/number-operations-rates-and-ratio.aspx | Toggle Accessibility Tools
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MICHIGAN STATE UNIVERSITY
College of Natural Science
Connected Mathematics Project
Number, Operations, Rates and Ratio
Every branch of mathematics and nearly every application of mathematics uses numbers and operations in essential ways for reasoning and problem-solving tasks. The applications in Connected Mathematics (CMP) involve whole numbers, integers, fractions, decimals, percents, ratios, and irrational numbers. The overarching goal in the CMP3 Number and Operations strand is to extend student understanding and skill in the use of numbers and operations to represent and reason about quantitative information.
There are ten Units in the CMP3 Number, Operations, and Proportional Reasoning strand. These Units aim to:
Extend knowledge of whole numbers, fractions, and decimals, develop the ability to know which representation and which operation to choose, and develop fluency in operations with those numbers;
Develop concepts of ratio, rate, and proportion, particularly equivalence of ratios, and apply to real-world problems;
Develop understanding of the meaning of percents and facility with applications;
Introduce negative, irrational, and complex numbers and the structural properties of each number system.
The learning progressions that work toward each of those broad goals are described in the following sections.
Extending Whole Numbers, Fractions and Decimals
Proficiency in the use of common fractions depends on understanding the multiplicative structure of whole numbers. This includes understanding the concepts of factor, multiple, prime number, greatest common factor, and least common multiple. Prime Time begins the CMP3 curriculum by developing ideas of elementary number theory while engaging students in games and problem-solving tasks that set the tone for a classroom that operates as a community of learners in which all participants are active participants. Prime Time also begins the Algebra and Functions strand of the curriculum by developing student understanding of arithmetic expressions, the Distributive Property, and the Order of Operations. The review and extension of fraction and decimal number understanding and skill is focused in the three Grade 6 Units, Comparing Bits and Pieces, Let’s Be Rational and Decimal Ops. In theseUnits, students develop number sense and problem solving habits of mind by addressing four key questions:
What kind of numbers will accurately represent the quantities involved?
What operations on those numbers will provide answers to the core questions posed?
What is a good estimate for the result of those operations?
What exact result is produced by applying a standard computational algorithm for the operations?
The name change from Bits and Pieces II (CMP2) to Let’s be Rational, Bits and Pieces I to Comparing Bits and Pieces, and Bits and Pieces III to Decimal Ops signals an extension of students’ understandings about fractions and decimals. Negative numbers are introduced and students order these numbers on the number line, observing that, for example, - ¾ and ¾ are the same distance from zero but on opposite sides of zero. A consequence of the addition of by-hand computations with multidigit numbers in Decimal Operations is that students can see that any fraction a/b can be written as a decimal, which will either repeat or terminate, by computing a ÷ b.
Let's Be Rational and Decimal Ops Units focused on fraction and decimal operations also begin the work on basic ideas of algebra by exploring fact families that connect pairs of related operations, equivalent equations, and equivalent expressions.
Example
The multiplication/division fact family relating 4, 6, and 24 includes:
4 x 6 = 24
6 x 4 = 24
4 = 24 ÷ 6
6 = 24 ÷ 4
Students work on computational skills and simultaneously develop skill in solving simple equations. Use of fact-family reasoning enables solution of one-step equations. For example, to solve 4x = 24, students reason that x = 24 ÷ 4, or to solve 6% of x = $4.80, students reason that x = $4.80 ÷ 0.06.
The Distributive Property of Multiplication Over Addition and Subtraction, introduced in Prime Time allows writing of arithmetic calculations in a variety of equivalent forms.
Example
4(5 + 1) = 4(5) + 4(1).
The Distributive Property is used in the Algebra and Functions strand in support of the overarching goal of writing and interpreting equivalent expressions. As the integer, rational, real, and complex number systems are developed in succeeding Units, students are repeatedly asked whether such very useful structural properties of operations do or should continue to apply in the new contexts.
Ratio, Rates, Scale Factors and Proportions
Problem 1.1 in Comparing Bits and Pieces presents some fundraising goals and then asks students whether some claims about each fundraising goal are true. The aim of Problem 1.1 is to focus student attention on the common mathematical challenge of comparing quantities accurately and fairly. It also highlights the difference between comparison by subtraction (or addition) and comparison by division (or multiplication).
Problem 1.1 introduces the proportional reasoning concepts of ratio, rate, and percent that are then developed in subsequent Problems of the Unit. The Problem serves a second important function of helping teachers to determine the breadth and depth of students’ prior knowledge of ratio, rate, and percent. Problem 1.1 also illustrates our guiding principle that students will grasp new concepts most readily if they encounter them in familiar contexts to which they can first apply their informal sense-making and then abstract underlying common mathematical structures.
Subsequent problems in Comparing Bits and Pieces develop the concept of ratio and rate more explicitly, based on the idea that a ratio of a to b means every a of one quantity is related to b of a second quantity. Equivalence of ratios is developed by reasoning about what makes sense in meaningful contexts and then equivalence of ratios is connected to equivalence of fractions. The ways that working with ratios is different from working with fractions are also highlighted.
Example
Suppose there are 10 boys and 15 girls in one class and 12 boys and 9 girls in another class. The ratio of boys to girls in the two classes combined is 22 to 24.
From this informal, sense-making, intuitive beginning, the concepts of ratio and rate develop in breadth and depth across the Grade 6 and Grade 7 CMP3 courses. Unit rates and rate tables are introduced in the second Investigation of Comparing Bits and Pieces and revisited in an algebraic context in Variables and Patterns near the end of Grade 6. In Grade 7, Stretching and Shrinking develops thinking about ratios in terms of geometric similarity and scale factors, and Comparing and Scaling brings ratios, rates, scale factors, and rate tables together to develop strategies for solving proportions. These fundamental proportional reasoning ideas are then extended and applied to work on linear functions, rates of change, and slope of graphs in the algebra Units Moving Straight Ahead in Grade 7 and Thinking with Mathematical Models at the start of Grade 8.
The Number strand and the Algebra and Functions strand both make use of the concepts of rate and proportionality.
Throughout this development of proportional reasoning concepts and skills, students are continually asked two key questions.
When does it make sense to compare quantities by ratio or rate?
How can ratios or rates be expressed in equivalent forms to answer questions that involve proportions?
Building on the concepts of equivalence and scaling, Comparing and Scaling enables students to develop reasonable algorithms for solving proportions, underpinned by sound conceptual understanding.
Percents
In almost every practical quantitative reasoning task, percents are the most common tool for framing and resolving questions that call for comparison of quantities. They offer standardized language and procedures for describing the relationship between two quantities from both additive and multiplicative perspectives.
Additive Perspective Examples
If an item originally priced at $15 is on sale for $9, the price reduction is $6. Also, $6 is 40% of the original price, or the sale price is 60% of the original price.
If a city’s population increases from 25,000 to 30,000, the increase is 5,000 which is 20% of the original population. The new population is 120% of the original population.
While each of these examples represent additive reasoning, the increase or decrease can be expressed as a percent.
Multiplicative Perspective Examples
What is 30% of 90?0.30 x 90 = n
36 is what percent of 90?p x 90 = 36
45 is 25% of what number?45 = 0.25 x n
The prior work with fact families relating multiplication and division pays rich dividends in learning how to deal with percent problems. Of course, percents appear throughout the geometry, data, probability, and algebra Units of each grade.
Negative Numbers
Traditional introductions to negative numbers focus on integers-positive and negative whole numbers and 0. The development in CMP3 takes a somewhat different path. In compliance with CCSSM requirements, we begin in the sixth-grade Unit Comparing Bits and Pieces by extending the rational number line to include negative numbers. The development there is limited to location of positive numbers and their opposites and the concept of absolute value. Later in Grade 6, a problem in Variables and Patterns extends the coordinate plane to all four graphing quadrants, using negative numbers informally with the assumption that sixth-grade students will almost certainly have had some encounters with them.
The heart of the development of negative numbers is the Grade 7 Unit Accentuate the Negative. We promote sense-making through games with wins and losses and other story contexts, number line models, and chip models to build on student intuitions and derive standard rules for operating with negative numbers. In the spirit of connected mathematics, we mix tasks about negative rational numbers with those limited to integer values.
In addition to using informal reasoning and models for numbers and operations, we again connect with the fact-family idea in the derivations of algorithms.
Example
After establishing rules for multiplication (product of two negative numbers is the subtle case), rules for the division follow:
(-30) ÷ 5 = -6 because -30 = 5(-6)(-30) ÷ (-5) = 6 because -30 = (-5)(6)
The final Investigation of Accentuate the Negative takes a retrospective view of the family of number systems—whole numbers, integers, and rationals—and highlights the basic structural properties that are common to all, especially the distributive property. Since Accentuate the Negative occurs relatively early in Grade 7, negative numbers are available for applications in other Units, most notably in Moving Straight Ahead about linear functions.
Irrational and Complex Numbers
The development of standard number systems and operations in CMP3 is completed in two Units of the Grade 8 curriculum-Looking for Pythagoras and Function Junction. Looking for Pythagoras makes the standard connection between square roots and diagonals of squares or hypotenuses of right triangles. It deals with decimal representation of rational (repeating) and irrational (non repeating) numbers, and, per CCSSM specifications, introduces cube roots as well. The number π pops up earlier, in Grade 7, when circumference and area of circles is tackled in Filling and Wrapping.
Complex numbers have traditionally been a topic of high school Algebra II courses, but the CCSSM syllabus for Algebra I calls for introduction of complex numbers. Thus, in our effort to offer CMP3 materials that support a full introductory algebra course, we have included complex numbers in Function Junction. The development there is mathematically fairly standard, continuing the pattern of extending the number system to include meaningful numbers that provide solutions for equations not solvable in preceding simpler number systems.
Example
3x = 2 is not solvable in whole numbers, but it is solvable in rational numbers;
3 + x = 2 is not solvable in positive numbers, but it is solvable in integers;
x2 = 2 is not solvable in rational numbers, but it is solvable in real numbers;
x2 = - 1 is not solvable in real numbers, but it is solvable in complex numbers.
Conceptual Knowledge and Procedural Skills
The specific Number and Operations Units of CMP3 develop all the concepts and procedural skills specified in the CCSSM with a consistent focus on meaningful derivations of ideas, techniques, and applications. When students complete all Units in the strand and the important connections in other content strands, they should be well prepared conceptually and technically:
To represent quantities with appropriate numerical forms, to identify operations that will answer questions of interest;
To estimate results of planned operations, to perform algorithms to produce exact computational results; and
To interpret those results in the contexts from which questions arose.
They should have sound understanding of the key structural properties of number systems that allow and guide the more general reasoning about quantity using algebra. |
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Proof of the Polynomial Remainder Theorem
Synopsis: Written below is a brief description of the polynomial remainder theorem. The theorem has a wide range of applications spanning from Algebra to Number Theory. This depicts how important the polynomial remainder theorem truly is, and why it must be taught in all courses and is a great tool.
The remainder theorem states when a polynomial denoted as is divided by for some value of , whether real or unreal, the remainder of Written below is the proof of the polynomial remainder theorem.
All polynomials can be written in the form , where is the divisor of the function/polynomial , is the quotient. amd is the remainder. Because the or the and the fact that degrees must be whole numbers( and the positive numbers), the , and so to speak, is a constant, which we will denote as .
Knowing this, we can write
We have hereby proven when the quantity is divided into a polynomial of any degree, the value of , where b is the remainder. The remainder must be a constant because .
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12680 | https://www.physics.rutgers.edu/~eandrei/389/xrays/xrays%2017.pdf | Powder X-ray 1 X-RAY DIFFRACTION in POWDERS PURPOSE: To learn x-ray powder-pattern diffraction techniques, to verify Vegard's law for copper-nickel alloys, to determine the nickel content in some American and Canadian "nickels", and to identify an unknown sample. APPARATUS: Rigaku Miniflex 30 kV x-ray diffraction spectrometer, with water cooling circulator, x-ray power supply and gun, proportional counter detector, safety interlocks; fluorescent paddle; known and unknown samples. THEORY: An excellent, very readable discussion of all of the aspects of x-ray diffraction that you will need for this experiment can be found in Cullity1 “Elements of X-ray Diffraction”. References to specific sections have been given throughout this write-up. Appendices 5 and 10, which contain useful parameters and information, are attached in the appendix to this write-up. 1. Crystal Structure and Vegard’s Law: Solids can be classified as being either amorphous or crystalline. In crystalline materials the ions occupy specific locations in a regular lattice. The simplest such lattice is a simple cube with ions on each of the corners of the cube. However, in all materials other crystalline structures are preferred because they have a lower total energy. The most common simple structures are the face-centered-cubic (fcc) and the body-centered cubic (bcc). The fcc structure consists of a cube of ions with six additional ions located at the centers of each of the six faces of the cube. Aluminum, copper, and nickel are metals with the fcc structure. The bcc structure consists of a cube of ions with one additional ion located at the center of the cube. Tungsten, molybdenum and iron take on the bcc structure. For both the fcc and bcc structures the distance between the ions on the corners of the cube is called the lattice constant a. In this experiment you will use x-ray diffraction to determine the lattice constant of a number of different samples. The size of the lattice constant depends on the size of the ions. In metals the outermost electrons become unbound and wander freely through the lattice (making them good Powder X-ray 2 conductors). This lowers the energy of the crystal and creates an attractive metallic bond that holds the metal together. But when the (positive) ions come so close to each other, the Coulomb repulsion between the ions sets in and prevents the crystal from further compressing. The equilibrium value for the lattice constant is then determined by the balance between the attractive metallic bond and the Coulomb repulsion of the ions. This distance varies from metal to metal. In this experiment you will study how the lattice constant changes for some alloys of copper in nickel. Copper is totally soluble in nickel. That is, you can form a copper-nickel alloy with any composition: CuxCuNi1-x , where x is the number fraction of Cu ions, 0 x 1. You will study how the lattice constant changes in CuxNi1-x alloys as you vary x. For copper a = 3.6148 Å (one Angstrom = 1 Å = 1x10-10 m), while for nickel, a = 3.5239 Å. If one makes the simplistic assumption that the ions of an alloy behave like hard spheres packed as close together as possible, then Vegard’s law tells us that the alloy lattice constant should vary linearly with composition: aalloy = x aCu + (1-x) aNi . (1) Note that x is defined as the number (not weight) fraction of Cu atoms. Vegard’s law is found to be a fairly good approximation in many alloy systems. Nickel is a magnetic element with a Curie point of Tc = 368o C. Copper is nonmagnetic. In CuxNi1-x alloys the Curie point is found to vary linearly with concentration: Tc = 368 -1170x (oC), (2) which implies samples with more than about 30% copper are nonmagnetic at room temperature. 2. X-ray Diffraction. In 1912 W.L. Bragg proposed a simple way of understanding x-ray diffraction by crystalline materials. He pointed out that, for any crystal, one can draw a set of equidistant parallel planes that pass through all of the atoms in the crystal, and that there are many different sets of such "Bragg" planes. Figure 1 shows a few of these planes on a simple cubic crystal. The planes are labeled with a set of integers (hkl), called the Miller Powder X-ray 3 indices, which identify the reciprocals of the fractional intercepts which the plane of interest makes with the crystal axes. (See Cullity section 2-6 for a more complete discussion.) Thus in Fig. 1, where the z-axis is out of the page, we see that the plane labeled d120 intercepts the cube shown as a dotted box in the figure at x = 1, y = 1/2, and z =. So the Miller indices are (1, 2, 0). For a cubic crystal the Miller indices also correspond to the x, y, and z components of a vector perpendicular to the plane. Notice that the distance dhkl between adjacent Bragg planes becomes smaller as the Miller indices increase. For cubic crystals this distance is: dhkl = a 1 h2 + k2 + l2 . (3) BRAGG PLANES IN A CUBIC SOLID Figure 1 When a beam of x-rays strikes one of the ions in the crystal, the x-rays are diffracted Powder X-ray 4 in all directions. In general these diffracted waves from different ions will be, on average, out of phase and cancel out. However, Bragg showed that, for an x-ray of wavelength and angle of incidence with respect to a Bragg plane (not the normal to the plane), scattered waves from the various ions lying in a single Bragg plane will be coherent if the angle of reflection equals the angle of incidence, Fig. 2. Scattering in this direction from successive planes a distance dhkl apart will also be coherent, and will interfere constructively if n= 2 dhkl sin (4) where n is integer. This is the Bragg law of diffraction. Thus the diffracted beam makes an angle with the Bragg plane and the angle between the incident beam and the diffracted beam is 2. Figure 2 Two different x-ray diffraction techniques are frequently used to study the structure of solids. The Laue method (see Cullity, Section 3.6) uses a single crystal sample (as distinguished from a powdered sample), so the incident beam makes the same angle with all parts of the sample and the angle of incidence with each set of Bragg planes is fixed. The beam used for the Laue method contains a continuous distribution of x-ray wavelengths. The diffraction pattern consists of a set of spots, each corresponding to a different set of (h k l) values, which are formed by those x-rays from the beam which have the appropriate wavelength for the angle of incidence, Eq. (4). The Laue method is a very Powder X-ray 5 powerful crystallographic tool, but the spectra are somewhat hard to interpret. In this experiment you will use the powder method, which is easier to interpret and is capable of high accuracy, especially for determining the spacing of atoms in a solid. In the powder method, monochromatic x-rays are used and the sample is very finely powdered. As can be seen from Eq. (4) for a given dhkl, since is fixed, there are only a few values of (sometimes only one, since sin ≤ 1) that will give a diffracted beam. But since the sample is powdered, the beam will encounter small crystallites oriented at random so that there will always be some that are oriented properly for diffraction. The diffracted beams corresponding to each dhkl are recorded by rotating a detector around the sample so that the angle 2between the incident beam and the detector sweeps from 0 o to 180o while the sample itself rotates from 0o to 90o. It is not strictly necessary to have a powdered sample in order to use the powder method. If a metal sample is rapidly cooled, it will form many small crystallites whose x-ray spectra will mimic that of a powdered sample. However, the orientation of these crystallites is frequently not completely random, especially if the sample has been mechanically worked (rolled flat or stretched, etc.), and the relative intensity of the x-ray lines will vary from that of a powder sample. In extreme cases some lines may even be missing. For such samples an accessory that spins the sample is sometimes used to give reproducible intensities for the diffracted lines. 3. Interpretation of Spectra. The interpretation of the powder pattern spectrum for cubic crystals is very straightforward. From the measured value for and the known value for (1.5418 Å for the x-ray tube you will use), you can use Bragg's law to calculate the spacing of the Bragg planes dhkl. Then, since the lattice constant a must be the same for all the lines, a set of Miller indices can be assigned to each line giving the same value of a for each line. (This is called indexing the spectrum.). This task is easy, since the order of the reflections is determined by the value of s 1 / h 2 k 2 l 2 . Then, in order of increasing 2, the first few reflections for a simple cubic crystal will be (100), (110), (111), (200), (210), (220). A complication arises for cubic crystals of high symmetry such as fcc Powder X-ray 6 or bcc where total destructive interference occurs for x-rays diffracted from some planes. For example, in an fcc crystal (see Section 2-7 in Cullity) such as copper or nickel, only those reflections are seen for planes for which (h k l) are all odd or all even (counting zero as even). Then the first few reflections for a fcc crystal are (111), (200), (220), ... . A complete list of allowed reflections for various crystals is given in Appendix 10 of Cullity. If the crystal structure is unknown, the indexing is carried out by trial-and-error until a consistent set of values is obtained. For the x-ray apparatus you will be using there can be an error of several tenths of a degree in the measured angle, which can affect how well your data agrees with Vegard’s law. To reduce this error, the powdered alloy samples are mixed with powdered silicon. You will use the silicon x-ray diffraction lines as a reference to correct for systematic errors in The crystal structure of silicon is the same as that of diamond. Blakemore3 states: "Close packing is not the primary consideration in the diamond lattice. This (e.g., diamond) is the structure adopted by solids for which the existence of four symmetrically placed valence bonds overrides any other consideration, which is the situation in silicon, germanium, and gray tin (20 C), as well as diamond. Diamond has the face-centered cubic (fcc) translational symmetry. We can view diamond as the result of two interpenetrating fcc lattices displaced from each other by one quarter of the cube diagonal distance. Since for diamond the two lattices are of the same type of atom, each atom ends up with four tetrahedrally arranged nearest neighbors just like itself and 12 next-nearest neighbors." It is interesting to note that some of the diffraction lines seen in the fcc structure are missing in silicon due to destructive interference between the two interpenetrating fcc lattices. Thus the silicon diffraction pattern is even more sparse than the fcc pattern. (See Cullity, Appendix 10). The intensity of the spectrum varies widely from line to line. This variation is due to a number of different factors (see Cullity, Sections 4-12 and 4-13), which make it difficult to use intensity to estimate the relative amount of different crystal phases in a mixed phase sample. 4. Generation of X-rays An x-ray tube generates x-rays by slamming a beam of electrons Powder X-ray 7 that has been accelerated across a large potential difference (30 kV for the Rigaku Miniflex) into a water-cooled copper target. The spectrum has a continuous distribution of bremsstrahlung ("braking radiation") wavelengths, with a cutoff at short wavelengths corresponding to the maximum electron kinetic energy (30 keV for the Rigaku Miniflex). The shape of the spectrum is independent of the material used as the target. Superimposed on this continuous spectrum are narrow, much more intense lines called the characteristic spectrum, whose wavelengths are characteristic of the material used in the target, independent of the exciting electron energy. For this experiment we need a monochromatic x-ray beam and will be using one of these characteristic spectral lines for copper. Tungsten or molybdenum targets are used if shorter wavelength x-rays are needed. These lines arise when an incoming electron strikes an atom and ejects one of its orbiting electrons. One of the remaining, less tightly bound atomic electrons then cascades down to the unoccupied orbit and gives up the energy difference as an x-ray characteristic of the element. In a multi-electron atom the shells of electrons are labeled with the letters K, L, M, with the K shell being closest to the nucleus (most tightly bound). When a K-shell electron is ejected and an L-shell electron "drops down" into the empty orbit, the emitted x-ray is labeled K. When an M-shell electron drops into the K-shell, a K x-ray is emitted, etc. In the apparatus you will use, the x-ray tube has a copper target and the K line is most intense compared to the other characteristic lines or the continuous spectrum. The K line actually consists of two closely spaced lines, K1 = 0.15406 nanometers (8.047 keV) and K2 = 0.15444 nm (8.027 keV) , but for most purposes it is sufficient to use an average wavelength weighted (K1:K2) by the relative intensities 2:1 (av K0.15418 nm). In some samples you might see a splitting of the x-ray spectrum due to these two source lines (especially for large values of 2). A thin nickel foil filter is placed in front of the proportional counter (gas) detector to eliminate most of the higher energy copper K line. Nickel is almost transparent to copper Kradiation but highly absorptive to K. EQUIPMENT: The apparatus you will use is a Rigaku 30 kV Miniflex powder pattern x-ray spectrometer which is extremely easy and safe to use. You will be issued a radiation Powder X-ray 8 badge. You must wear the radiation badge whenever you are in room 133 and the x-ray machine is in operation. 1. Safety devices: Improperly controlled X-rays are dangerous and can give serious burns. The machine is well shielded with a series of safety devices and interlocks to insure your safety. It is important that you do not attempt to defeat these safety measures in any way. There is a red light mounted on top of the machine that is lit whenever the machine is generating x-rays, to remind you of possible hazards. The door to the machine is made of a special plastic that is highly absorbent to x-rays, even though it is transparent to visible light. When the door is opened the machine will continue to generate x-rays and the light will remain on. However, a shutter automatically covers the source so that it is safe for you to reach into the machine to make adjustments. If for some reason the shutter should fail to cover the source, there is an additional interlock which automatically shuts off the machine. (The shutter moves across the source whenever the handle on the door is turned by even few degrees. Therefore it is important, when recording a spectrum, to be sure that the handle is completely closed. Otherwise your x-ray beam may be partially or totally blocked.) 2. Spectrometer operation: Before you begin using the machine make sure you and your partners have your radiation badges on. Sign in on the USER LOG located on the top of the machine. a. Mount the sample in the machine by slipping the holder into the two spring clips provided on the machine. b. Adjust the detector to the desired starting value for 2. The mechanism which moves the detector is called a goniometer. It turns the sample through an angle with respect to the x-ray beam while at the same time moving the detector through 2 Note that it always sweeps in the direction of decreasing 2. There are two sweep speeds, 2o/minute and 0.5o/minute. The slower sweep speed is for resolving overlapping lines or accurately determining angles. The sweep speed can be adjusted Powder X-ray 9 with the knob located next to a small piece of metal that reads “o/min.” Push the knob all the way in for 0.5o/minute, and pull it all the way out for 2o/minute. To adjust the detector push the chrome metal lever (located inside the machine near the center at the bottom) counter-clockwise, to release the gears that move the detector. Move the detector near the desired angle and release the lever to re-engage the gears. To fine-adjust the detector position, move the sweep speed knob to the center position and turn the degree knob to move the detector. Set the detector at the desired position with the degree knob exactly on zero. Pull the sweep rate knob out to 2o/minute. c. Close the door to the x-ray machine. Make sure it closes completely! d. Turn on the switch for the cooling water circulator. It is located on the side of the table holding the machine. [Do not run it longer than necessary since the noise can annoy students doing the speed of light experiment.] e. Turn on the main power switch for the machine and the high voltage power supply switch for the detector. Press the x-ray On switch to begin generating x-rays. The red light on top of the machine should go on. To turn off the x-rays use the red switch beside the On switch. f. Set the count switch to Count and the count range to a convenient setting, such as 2000 cps. g. Start the Logger Pro program and make sure that the Universal Lab Interface is turned on. Select “COM1” as the port to scan. Go to “File” ”New.” Then go to “Setup” “Data Collection.” In the Sampling tab, you can adjust the experiment length and the number of samples per second. To run the experiment, press the “Collect” button and set the goniometer switch to “Start.” PROCEDURE: 1. Place the fluorescent paddle in the sample holder (Figure 3) . The Powder X-ray 10 paddle absorbs x-rays and emits visible light, so that you can observe the location of the x-ray beam. Notice how the shape and location of the beam changes for several widely different values of 2. You have to be sure that you place your samples in the sample holder so that the beam strikes them as 2 is swept. 2. Record the x-ray spectrum for silicon, sweeping 2 from 70o to 25o. Silicon has a diamond structure with a lattice constant of 5.4309 Å (see Appendix 5 of Cullity). Calculate the theoretical x-ray spectrum, using this value, and compare with your data. You can expect a discrepancy of several tenths of a degree due to goniometer misalignment. To take this into account, silicon is frequently mixed with other samples to serve as a reference whose spectrum is exactly known. In your report present the data in a table giving for each line the Miller indices, the measured and theoretical angles, and the approximate observed intensity. The only purpose of the intensity measurement is to record whether the lines are weak or strong. 3. For either a copper or a nickel sample, measure as many lines as you can over the entire range of 2 from 0o to 160o and compare with theory using the lattice constant listed in Appendix 5 of Cullity. You doubtless found sweeping over a wide range of angles to be very tedious, even at 2o/min. Instead of making a wide sweep, predict the location of these Figure 3. Fluorescent paddle. Left panel – Paddle resting on sample box. Right panel – when exposed to the X-ray beam the paddle emits green light highlighting the location of the bam. Powder X-ray 11 Figure 4. Definition of half width points. lines and then search for them with a small sweep. In your report present the data in a table giving for each line the Miller indices, the measured and theoretical angles, and the approximate observed intensity. 4. Carefully compare the observed spectra of copper and nickel for the two lowest index lines. You will see that the difference in 2 between copper and nickel is fairly small (less than 1.5o). To get good results, use Si(220) as a reference assuming you know it’s location exactly. Measure the angles as precisely as possible. In order to do this use the 0.5o/min sweep. This will give a spread out spectrum that will be easier to measure accurately. To locate the position of the line by locating the maximum value. It is more accurate to measure the width of the line at several positions to locate the half width points, Fig. 3. From your data for these two lines determine the lattice constant for the two metals and estimate your error. Compare with the accepted value given in Cullity. You may assume the concentrations given for the alloys are accurate to ± 2%. 5. Verify Vegard's law for Cu-Ni alloys. You need measure only the Ni(Cu) (1 1 1) and the Si (2 2 0) lines, which can be observed by sweeping from 2= 48oto 43o. Use the slow sweep in order to get as accurate a determination of 2as possible and follow the procedure outlined in step 4. Present your data as a graph of lattice constant versus copper concentration. Use you data from step 4 for the lattice constants of pure copper and Powder X-ray 12 nickel; these are good data points too. Make a least square fit to a straight line. On your graph also show the theoretical Vegard’s law, Eq. (1), using the accepted lattice constants for Cu and Ni. Are any of the alloys magnetic? Check with a small, permanent magnet. Do your results agree with Eq. (2)? 6. Determine the nickel concentrations in a U.S. nickel, and in both a 1979 and a 1984 Canadian nickel. Estimate your error. Are any of these nickels magnetic? 7. Record the x-ray spectrum of one of the unknown samples and use Appendix 5 in Cullity to identify it. Be careful; it’s easy to miss! Note that there may be very weak or missing lines in these non powdered samples and, even worse, extra lines not associated with the unknown. If you have trouble indexing the unknown, try dropping one or two of the weakest lines; they may be spurious. Also note that both a simple cubic lattice and a body-centered cubic lattice can give a 1:2:3:4:5 set of d2 ratios but there are no simple cubic metals! If you find several metals seem to be possible, try determining the density and whether they are magnetic to help you discriminate between them. APPENDIX: Since you will need to frequently refer to them, the Appendices from Cullity are included in the supplementary material REFERENCES: 1. B.D. Cullity: Elements of X-ray Diffraction, Addison-Wesley, Reading, Mass. 1978 2. D.W. Preston & E.R. Dietz: The Art of Experimental Physics, Wiley, 1991. Section 10, X-ray and Microwave Diffraction by Periodic Structures, Bragg Spectroscopy, pp 180-1952. 3. J. S. Blakemore, Solid State Physics; Saunders, 1969, pp 41,42 |
12681 | https://math-salamanders.s3-us-west-1.amazonaws.com/Fractions/Converting-Decimals-to-Fractions-Worksheet/converting-decimals-to-fractions-1.pdf | Name Date CONVERTING DECIMALS TO FRACTIONS 1 Convert these decimals to fractions. • Leave your fraction answers as a decimal fraction with the denominator as a power of 10, you do not need to simplify your fraction. • If the decimal is greater than one, leave your answer as a mixed fraction. 1) 0.6 = 13) 0.34 = 25) 1.8 = 2) 0.3 = 14) 0.2 = 26) 3.4 = 3) 0.1 = 15) 0.81 = 27) 2.63 = 4) 0.9 = 16) 0.7 = 28) 9.6 = 5) 0.8 = 17) 0.325 = 29) 4.39 = 6) 0.27 = 18) 0.289 = 30) 7.28 = 7) 0.58 = 19) 0.452 = 31) 0.375 = 8) 0.81 = 20) 0.013 = 32) 6.41 = 9) 0.43 = 21) 0.67 = 33) 1.372 = 10) 0.92 = 22) 0.29 = 34) 5.391 = 11) 0.78 = 23) 0.316 = 35) 8.29 = 12) 0.55 = 24) 0.527 = 36) 11.83 = Name Date CONVERTING DECIMALS TO FRACTIONS 1 ANSWERS Convert these decimals to fractions. • Leave your fraction answers with the denominator as a power of 10, you do not need to simplify your fraction. • If the decimal is greater than one, leave your answer as a mixed fraction. 1) 0.6 = 6 10 13) 0.34 = 34 100 25) 1.8 = 1 8 10 2) 0.3 = 3 10 14) 0.2 = 2 10 26) 3.4 = 3 4 10 3) 0.1 = 1 10 15) 0.81 = 81 100 27) 2.63 = 2 63 100 4) 0.9 = 9 10 16) 0.7 = 7 10 28) 9.6 = 9 6 10 5) 0.8 = 8 10 17) 0.325 = 325 1000 29) 4.39 = 4 39 100 6) 0.27 = 27 100 18) 0.289 = 289 1000 30) 7.28 = 7 28 100 7) 0.58 = 58 100 19) 0.452 = 452 1000 31) 0.375 = 375 1000 8) 0.81 = 81 100 20) 0.013 = 13 1000 32) 6.41 = 6 41 100 9) 0.43 = 43 100 21) 0.67 = 67 100 33) 1.372 = 1 372 1000 10) 0.92 = 92 100 22) 0.29 = 29 100 34) 5.391 = 5 391 1000 11) 0.78 = 78 100 23) 0.316 = 316 1000 35) 8.29 = 8 29 100 12) 0.55 = 55 100 24) 0.527 = 527 1000 36) 11.83 = 1 1 83 100 |
12682 | https://www.facebook.com/groups/HEMAAlliance/posts/2477579955601050/ | HEMA United | This week on Sword STEM: back to tournament rulesets | Facebook
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Tournament rulesets: non-zero-sum concepts explained
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Sean Franklin
· May 17, 2018 ·
This week on Sword STEM: back to tournament rulesets. And why you should be thinking of Non Zero-Sum concepts when you design them.
If that sounds too abstract and or intimidating, don't worry. I explained it using pie.
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Non Zero-Sum Concepts in Tournament Rulesets – Sword STEM
Non Zero-Sum Concepts in Tournament Rulesets May 17, 2018 Sean Franklin Rule Sets 0 The more I think about tournament rules, the more it becomes apparent to me that a zero-sum solution to scoring will never get us the results we want. But I’m going to guess that all “zero-sum” means to you is...
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Ashanti Ziths
Flower point this year did modified FNY rules, and the only thing counted for progressing in the tournament was the amount of lives you had left.
I feel like that worked well.
7y
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Vincent Le Chevalier
Nice, thought provoking article as usual.
I'm not sure I entirely agree that a game with a winner and a loser has to be zero-sum. There are ways to reflect the cleanness of the fight in the score as well. An idea I had, a bit similar to David Reddy's …
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doesnt read synopsis. Gets disappointed this isn a pie recipe
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12683 | https://www.whitman.edu/mathematics/calculus_online/section09.06.html | Home » Applications of Integration » Center of Mass
9.6 Center of Mass
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Introduction
1 Analytic Geometry
1. Lines
2. Distance Between Two Points; Circles
3. Functions
4. Shifts and Dilations
2 Instantaneous Rate of Change: The Derivative
1. The slope of a function
2. An example
3. Limits
4. The Derivative Function
5. Properties of Functions
3 Rules for Finding Derivatives
1. The Power Rule
2. Linearity of the Derivative
3. The Product Rule
4. The Quotient Rule
5. The Chain Rule
4 Transcendental Functions
1. Trigonometric Functions
2. The Derivative of $\sin x$
3. A hard limit
4. The Derivative of $\sin x$, continued
5. Derivatives of the Trigonometric Functions
6. Exponential and Logarithmic functions
7. Derivatives of the exponential and logarithmic functions
8. Implicit Differentiation
9. Inverse Trigonometric Functions
10. Limits revisited
11. Hyperbolic Functions
5 Curve Sketching
1. Maxima and Minima
2. The first derivative test
3. The second derivative test
4. Concavity and inflection points
5. Asymptotes and Other Things to Look For
6 Applications of the Derivative
1. Optimization
2. Related Rates
3. Newton's Method
4. Linear Approximations
5. The Mean Value Theorem
7 Integration
1. Two examples
2. The Fundamental Theorem of Calculus
3. Some Properties of Integrals
8 Techniques of Integration
1. Substitution
2. Powers of sine and cosine
3. Trigonometric Substitutions
4. Integration by Parts
5. Rational Functions
6. Numerical Integration
7. Additional exercises
9 Applications of Integration
1. Area between curves
2. Distance, Velocity, Acceleration
3. Volume
4. Average value of a function
5. Work
6. Center of Mass
7. Kinetic energy; improper integrals
8. Probability
9. Arc Length
10. Surface Area
10 Polar Coordinates, Parametric Equations
1. Polar Coordinates
2. Slopes in polar coordinates
3. Areas in polar coordinates
4. Parametric Equations
5. Calculus with Parametric Equations
11 Sequences and Series
1. Sequences
2. Series
3. The Integral Test
4. Alternating Series
5. Comparison Tests
6. Absolute Convergence
7. The Ratio and Root Tests
8. Power Series
9. Calculus with Power Series
10. Taylor Series
11. Taylor's Theorem
12. Additional exercises
12 Three Dimensions
1. The Coordinate System
2. Vectors
3. The Dot Product
4. The Cross Product
5. Lines and Planes
6. Other Coordinate Systems
13 Vector Functions
1. Space Curves
2. Calculus with vector functions
3. Arc length and curvature
4. Motion along a curve
14 Partial Differentiation
1. Functions of Several Variables
2. Limits and Continuity
3. Partial Differentiation
4. The Chain Rule
5. Directional Derivatives
6. Higher order derivatives
7. Maxima and minima
8. Lagrange Multipliers
15 Multiple Integration
1. Volume and Average Height
2. Double Integrals in Cylindrical Coordinates
3. Moment and Center of Mass
4. Surface Area
5. Triple Integrals
6. Cylindrical and Spherical Coordinates
7. Change of Variables
16 Vector Calculus
1. Vector Fields
2. Line Integrals
3. The Fundamental Theorem of Line Integrals
4. Green's Theorem
5. Divergence and Curl
6. Vector Functions for Surfaces
7. Surface Integrals
8. Stokes's Theorem
9. The Divergence Theorem
17 Differential Equations
1. First Order Differential Equations
2. First Order Homogeneous Linear Equations
3. First Order Linear Equations
4. Approximation
5. Second Order Homogeneous Equations
6. Second Order Linear Equations
7. Second Order Linear Equations, take two
18 Useful formulas
19 Introduction to Sage
1. Basics
2. Differentiation
3. Integration
Suppose a beam is 10 meters long, and that there are three weights on the beam: a 10 kilogram weight 3 meters from the left end, a 5 kilogram weight 6 meters from the left end, and a 4 kilogram weight 8 meters from the left end. Where should a fulcrum be placed so that the beam balances? Let's assign a scale to the beam, from 0 at the left end to 10 at the right, so that we can denote locations on the beam simply as $x$ coordinates; the weights are at $x=3$, $x=6$, and $x=8$, as in figure 9.6.1.
Figure 9.6.1. A beam with three masses.
Suppose to begin with that the fulcrum is placed at $x=5$. What will happen? Each weight applies a force to the beam that tends to rotate it around the fulcrum; this effect is measured by a quantity called torque, proportional to the mass times the distance from the fulcrum. Of course, weights on different sides of the fulcrum rotate the beam in opposite directions. We can distinguish this by using a signed distance in the formula for torque. So with the fulcrum at 5, the torques induced by the three weights will be proportional to $(3-5)10=-20$, $(6-5)5=5$, and $(8-5)4=12$. For the beam to balance, the sum of the torques must be zero; since the sum is $-20+5+12=-3$, the beam rotates counter-clockwise, and to get the beam to balance we need to move the fulcrum to the left. To calculate exactly where the fulcrum should be, we let $\ds \bar x$ denote the location of the fulcrum when the beam is in balance. The total torque on the beam is then $\ds (3-\bar x)10+(6-\bar x)5+(8-\bar x)4=92-19\bar x$. Since the beam balances at $\ds \bar x$ it must be that $\ds 92-19\bar x=0$ or $\ds \bar x=92/19\approx 4.84$, that is, the fulcrum should be placed at $x=92/19$ to balance the beam.
Now suppose that we have a beam with varying density—some portions of the beam contain more mass than other portions of the same size. We want to figure out where to put the fulcrum so that the beam balances.
Figure 9.6.2. A solid beam.
Example 9.6.1 Suppose the beam is 10 meters long and that the density is $1+x$ kilograms per meter at location $x$ on the beam. To approximate the solution, we can think of the beam as a sequence of weights "on'' a beam. For example, we can think of the portion of the beam between $x=0$ and $x=1$ as a weight sitting at $x=0$, the portion between $x=1$ and $x=2$ as a weight sitting at $x=1$, and so on, as indicated in figure 9.6.2. We then approximate the mass of the weights by assuming that each portion of the beam has constant density. So the mass of the first weight is approximately $\ds m_0=(1+0)1=1$ kilograms, namely, $(1+0)$ kilograms per meter times 1 meter. The second weight is $\ds m_1=(1+1)1=2$ kilograms, and so on to the tenth weight with $\ds m_9=(1+9)1=10$ kilograms. So in this case the total torque is $$ (0-\bar x)m_0+(1-\bar x)m_1+\cdots+(9-\bar x)m_9= (0-\bar x)1+(1-\bar x)2+\cdots+(9-\bar x)10. $$ If we set this to zero and solve for $\ds \bar x$ we get $\ds \bar x=6$. In general, if we divide the beam into $n$ portions, the mass of weight number $i$ will be $\ds m_i=(1+x_i)(x_{i+1}-x_i)=(1+x_i)\Delta x$ and the torque induced by weight number $i$ will be $\ds (x_i-\bar x)m_i=(x_i-\bar x)(1+x_i)\Delta x$. The total torque is then $$\eqalign{ (x_0-\bar x)(1+x_0)\Delta x&+(x_1-\bar x)(1+x_1)\Delta x+\cdots+ (x_{n-1}-\bar x)(1+x_{n-1})\Delta x\cr &=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x- \sum_{i=0}^{n-1}\bar x(1+x_i)\Delta x\cr &=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x- \bar x\sum_{i=0}^{n-1}(1+x_i)\Delta x.\cr }$$ If we set this equal to zero and solve for $\ds \bar x$ we get an approximation to the balance point of the beam: $$\eqalign{ 0&=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x- \bar x\sum_{i=0}^{n-1}(1+x_i)\Delta x\cr \bar x\sum_{i=0}^{n-1}(1+x_i)\Delta x&=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x\cr \bar x&={\ds\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x\over \ds\sum_{i=0}^{n-1}(1+x_i)\Delta x}.\cr }$$ The denominator of this fraction has a very familiar interpretation. Consider one term of the sum in the denominator: $\ds (1+x_i)\Delta x$. This is the density near $\ds x_i$ times a short length, $\Delta x$, which in other words is approximately the mass of the beam between $\ds x_i$ and $\ds x_{i+1}$. When we add these up we get approximately the mass of the beam.
Now each of the sums in the fraction has the right form to turn into an integral, which in turn gives us the exact value of $\ds \bar x$: $$ \bar x={\ds\int_0^{10} x(1+x)\,dx\over\ds\int_{0}^{10}(1+x)\,dx}. $$ The numerator of this fraction is called the moment of the system around zero: $$\int_0^{10} x(1+x)\,dx=\int_0^{10} x+x^2\,dx={1150\over 3},$$ and the denominator is the mass of the beam: $$\int_0^{10} (1+x)\,dx=60,$$ and the balance point, officially called the center of mass, is $$\bar x={1150\over 3}{1\over 60} = {115\over 18}\approx 6.39.$$ $\square$
It should be apparent that there was nothing special about the density function $\sigma(x)=1+x$ or the length of the beam, or even that the left end of the beam is at the origin. In general, if the density of the beam is $\sigma(x)$ and the beam covers the interval $[a,b]$, the moment of the beam around zero is $$M_0=\int_a^b x\sigma(x)\,dx$$ and the total mass of the beam is $$M=\int_a^b \sigma(x)\,dx$$ and the center of mass is at $$\bar x={M_0\over M}.$$
Example 9.6.2 Suppose a beam lies on the $x$-axis between 20 and 30, and has density function $\sigma(x)=x-19$. Find the center of mass. This is the same as the previous example except that the beam has been moved. Note that the density at the left end is $20-19=1$ and at the right end is $30-19=11$, as before. Hence the center of mass must be at approximately $20+6.39=26.39$. Let's see how the calculation works out. $$\eqalign{ M_0&=\int_{20}^{30} x(x-19)\,dx=\int_{20}^{30} x^2-19x\,dx= \left.{x^3\over3}-{19x^2\over2}\right|_{20}^{30}={4750\over3}\cr M&=\int_{20}^{30} x-19\,dx=\left.{x^2\over2}-19x\right|_{20}^{30}=60\cr {M_0\over M}&={4750\over3}{1\over60}={475\over18}\approx 26.39.\cr }$$ $\square$
Example 9.6.3 Suppose a flat plate of uniform density has the shape contained by $\ds y=x^2$, $y=1$, and $x=0$, in the first quadrant. Find the center of mass. (Since the density is constant, the center of mass depends only on the shape of the plate, not the density, or in other words, this is a purely geometric quantity. In such a case the center of mass is called the centroid.)
Figure 9.6.3. Center of mass for a two dimensional plate.
This is a two dimensional problem, but it can be solved as if it were two one dimensional problems: we need to find the $x$ and $y$ coordinates of the center of mass, $\ds \bar x$ and $\ds \bar y$, and fortunately we can do these independently. Imagine looking at the plate edge on, from below the $x$-axis. The plate will appear to be a beam, and the mass of a short section of the "beam'', say between $\ds x_i$ and $\ds x_{i+1}$, is the mass of a strip of the plate between $\ds x_i$ and $\ds x_{i+1}$. See figure 9.6.3 showing the plate from above and as it appears edge on. Since the plate has uniform density we may as well assume that $\sigma=1$. Then the mass of the plate between $\ds x_i$ and $\ds x_{i+1}$ is approximately $\ds m_i=\sigma(1-x_i^2)\Delta x=(1-x_i^2)\Delta x$. Now we can compute the moment around the $y$-axis: $$M_y=\int_0^1 x(1-x^2)\,dx={1\over4}$$ and the total mass $$M=\int_0^1 (1-x^2)\,dx={2\over3}$$ and finally $$\bar x = {1\over4}{3\over2}={3\over8}.$$ Next we do the same thing to find $\ds \bar y$. The mass of the plate between $\ds y_i$ and $\ds y_{i+1}$ is approximately $\ds n_i=\sqrt{y}\Delta y$, so $$M_x=\int_0^1 y\sqrt{y}\,dy={2\over5}$$ and $$\bar y={2\over5}{3\over2}={3\over5},$$ since the total mass $M$ is the same. The center of mass is shown in figure 9.6.3. $\square$
Example 9.6.4 Find the center of mass of a thin, uniform plate whose shape is the region between $y=\cos x$ and the $x$-axis between $x=-\pi/2$ and $x=\pi/2$. It is clear that $\ds \bar x=0$, but for practice let's compute it anyway. We will need the total mass, so we compute it first: $$ M=\int_{-\pi/2}^{\pi/2} \cos x\,dx=\sin x\Big|_{-\pi/2}^{\pi/2}=2. $$ The moment around the $y$-axis is $$ M_y=\int_{-\pi/2}^{\pi/2} x\cos x\,dx= \cos x+x\sin x\Big|_{-\pi/2}^{\pi/2}=0 $$ and the moment around the $x$-axis is $$ M_x=\int_{0}^{1} y\cdot2\arccos y\,dy= \left.y^2\arccos y-{y\sqrt{1-y^2}\over2}+{\arcsin y\over 2} \right|_{0}^{1}={\pi\over4}. $$ Thus $$\bar x={0\over2},\quad \bar y={\pi\over8}\approx 0.393.$$ $\square$
Exercises 9.6
You can use Sage to help check your work.
Ex 9.6.1 A beam 10 meters long has density $\ds \sigma(x)=x^2$ at distance $x$ from the left end of the beam. Find the center of mass $\ds \bar x$. (answer)
Ex 9.6.2 A beam 10 meters long has density $\sigma(x)=\sin(\pi x/10)$ at distance $x$ from the left end of the beam. Find the center of mass $\ds \bar x$. (answer)
Ex 9.6.3 A beam 4 meters long has density $\ds \sigma(x)=x^3$ at distance $x$ from the left end of the beam. Find the center of mass $\ds \bar x$. (answer)
Ex 9.6.4 Verify that $\ds\int 2x\arccos x\,dx= x^2\arccos x-{x\sqrt{1-x^2}\over2}+{\arcsin x\over 2}+C$.
Ex 9.6.5 A thin plate lies in the region between $\ds y=x^2$ and the $x$-axis between $x=1$ and $x=2$. Find the centroid. (answer)
Ex 9.6.6 A thin plate fills the upper half of the unit circle $\ds x^2+y^2=1$. Find the centroid. (answer)
Ex 9.6.7 A thin plate lies in the region contained by $y=x$ and $\ds y=x^2$. Find the centroid. (answer)
Ex 9.6.8 A thin plate lies in the region contained by $\ds y=4-x^2$ and the $x$-axis. Find the centroid. (answer)
Ex 9.6.9 A thin plate lies in the region contained by $\ds y=x^{1/3}$ and the $x$-axis between $x=0$ and $x=1$. Find the centroid. (answer)
Ex 9.6.10 A thin plate lies in the region contained by $\ds \sqrt{x}+\sqrt{y}=1$ and the axes in the first quadrant. Find the centroid. (answer)
Ex 9.6.11 A thin plate lies in the region between the circle $\ds x^2+y^2=4$ and the circle $\ds x^2+y^2=1$, above the $x$-axis. Find the centroid. (answer)
Ex 9.6.12 A thin plate lies in the region between the circle $\ds x^2+y^2=4$ and the circle $\ds x^2+y^2=1$ in the first quadrant. Find the centroid. (answer)
Ex 9.6.13 A thin plate lies in the region between the circle $\ds x^2+y^2=25$ and the circle $\ds x^2+y^2=16$ above the $x$-axis. Find the centroid. (answer) |
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Acute pyelonephritis
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Acute pyelonephritis
Renal and urinary system
Congenital disorders
Bladder exstrophyHorseshoe kidneyHydronephrosisHypospadias and epispadiasPotter sequenceRenal agenesis
Infectious, immunologic, and inflammatory disorders
Alport syndromeGoodpasture syndromeIgA nephropathy (NORD)Lupus nephritisPoststreptococcal glomerulonephritisRapidly progressive glomerulonephritisAmyloidosisDiabetic nephropathyFocal segmental glomerulosclerosis (NORD)Lupus nephritisMembranoproliferative glomerulonephritisMembranous nephropathyMinimal change diseaseAcute tubular necrosisRenal papillary necrosisAcute pyelonephritisChronic pyelonephritisLower urinary tract infection
Metabolic and regulatory disorders
Acute tubular necrosisPostrenal azotemiaPrerenal azotemiaRenal azotemiaChronic kidney diseaseKidney stonesRenal tubular acidosis
Neoplasms
AngiomyolipomaMedullary cystic kidney diseaseMedullary sponge kidneyMulticystic dysplastic kidneyPolycystic kidney diseaseBeckwith-Wiedemann syndromeNephroblastoma (Wilms tumor)Non-urothelial bladder cancersRenal cell carcinomaTransitional cell carcinomaWAGR syndrome
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Neurogenic bladderPosterior urethral valvesUrinary incontinenceVesicoureteral reflux
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Renal artery stenosisRenal cortical necrosis
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Metabolic acidosisMetabolic alkalosisRespiratory acidosisRespiratory alkalosisHypercalcemiaHyperkalemiaHypermagnesemiaHypernatremiaHyperphosphatemiaHypocalcemiaHypokalemiaHypomagnesemiaHyponatremiaHypophosphatemia
Renal and urinary system pathology review
Congenital renal disorders: Pathology reviewNephritic syndromes: Pathology reviewNephrotic syndromes: Pathology reviewUrinary tract infections: Pathology reviewKidney stones: Pathology reviewRenal failure: Pathology reviewRenal tubular acidosis: Pathology reviewRenal tubular defects: Pathology reviewRenal and urinary tract masses: Pathology reviewUrinary incontinence: Pathology reviewAcid-base disturbances: Pathology reviewElectrolyte disturbances: Pathology review
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A 67-year-old woman is currently postoperative day 5 after having an open cholecystectomy. She reports worsening suprapubic pain and malaise. Her vital signs are 37 °C (98.6°F), pulse is 98/min, respirations are 14/min, blood pressure is 137/64 mmHg, and oxygen saturation is 99% on room air. Physical examination shows suprapubic tenderness on palpation as well as cloudy urine in her Foley catheter. Which of the following pathogens is most likely causative of this patient’s clinical presentation?
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Rishi Desai, MD, MPH
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Tanner Marshall, MS
With acute pyelonephritis, pyelo- means pelvis, and -neph- refers to the kidney, so in this case it’s the renal pelvis, which is the funnel-like structure of the kidney that drains urine into the ureter, and -itis means inflammation. So acute pyelonephritis describes an inflamed kidney that develops relatively quickly, usually as a result of a bacterial infection.
Now a urinary tract infection, or UTI, is any infection of the urinary tract, which includes the upper portion of the tract—the kidneys and the ureters, and the lower portion of the tract—the bladder and urethra. So acute pyelonephritis is a type of upper urinary tract infection.
Acute pyelonephritis is most often caused by ascending infection, meaning bacteria start by colonizing the urethra and bladder, which would be a lower urinary tract infections, and make their way up the ureters and kidney, therefore upper UTI shares a lot of the same risk factors as lower UTI, things like female sex, sexual intercourse, indwelling catheters, diabetes mellitus, and urinary tract obstruction.
One major factor that increases the risk of an upper UTI from a lower UTI spreading upward is vesicoureteral reflux, or VUR, which is where urine is allowed to move backward up the urinary tract, which can happen if the vesicoureteral orifice fails. The vesicoureteral orifice is the one-way valve that allows urine to flow from each ureter into the bladder, but not in the reverse direction.
VUR can be the result of a primary congenital defect or it can be caused by bladder outlet obstruction, which increases pressure in the bladder and distorts the valve.
As kind of a double-whammy, obstruction also leads to urinary stasis, where urine stands still, which makes it easier for bacteria to adhere and colonize the urinary tract.
So, for ascending infections that cause acute pyelonephritis, the most common organisms are E coli, Proteus species, and Enterobacter species, all of which are commonly found in the bowel flora.
Now, it’s also possible that kidneys get infected via hematogenous infection, or spread through the bloodstream, although this is a lot less common.
Usually pyelonephritis from hematogenous spread is a consequence of septicemia or bacteremia—which is bacteria in the blood, as well as infective endocarditis, an infection of the inner layer of the heart. In these situations, the most common organisms are Staphylococcus species and again E. coli.
Summary
Acute pyelonephritis is a sudden, severe infection of the kidney. The infection may be caused by bacteria that travel up from the bladder or bacteria circulating in the blood (bacteremia). Symptoms of acute pyelonephritis can include fever, chills, nausea, vomiting, and pain in the lower back and sides. If left untreated, acute pyelonephritis can lead to serious complications like sepsis (a potentially life-threatening condition caused by infection).
Sources
"Robbins Basic Pathology" Elsevier (2017)
"Harrison's Principles of Internal Medicine, Twentieth Edition (Vol.1 & Vol.2)" McGraw-Hill Education / Medical (2018)
"Pathophysiology of Disease: An Introduction to Clinical Medicine 8E" McGraw-Hill Education / Medical (2018)
"CURRENT Medical Diagnosis and Treatment 2020" McGraw-Hill Education / Medical (2019)
"International Clinical Practice Guidelines for the Treatment of Acute Uncomplicated Cystitis and Pyelonephritis in Women: A 2010 Update by the Infectious Diseases Society of America and the European Society for Microbiology and Infectious Diseases" Clinical Infectious Diseases (2011)
"Renal ultrasound and DMSA screening for high-grade vesicoureteral reflux" Pediatrics International (2016)
"EAU Guidelines on Vesicoureteral Reflux in Children" European Urology (2012)
"International Clinical Practice Guidelines for the Treatment of Acute Uncomplicated Cystitis and Pyelonephritis in Women: A 2010 Update by the Infectious Diseases Society of America and the European Society for Microbiology and Infectious Diseases" Clinical Infectious Diseases (2011)
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12685 | https://www.ck12.org/flexi/math-grade-8/function-notation/ | Function Notation | Flexi Homework help & answers | CK-12 Foundation
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Math Grade 8
Function Notation
Function Notation
Identifying FunctionsFunctions in Tables, Graphs, and Equations
Concept Summary:
A function may be named with any letter you wish, but the most common are f,g,and h.
The x term is replaced by the input value that you want to determine the output value for.
The function statement f(x)=x−5 is read as "The function f,performed on the value x,is equal to x−5."
Ask your own question
Translate each of the following statements into an equation, using x as the variable: (b) One fifth of a number is 5 less than that number.
Which of the following equations can be formed starting with x=0 ?
Which of the following is an equation?
Which of the following is an equation?
How to tell if an equation is a function?
How do you find the equation of a function from points?
Find f if f''(t) = t^2 + 1/t^2, t > 0, f(2) = 3, f'(1) = 2.
### Some Special Cases f(x)=x 2 What is f(x)+f(x)+f(x)?∼3 f(x) Complete: 3 f(x)=3 x 2 Complete: Evaluate 3 f(2)= Done.x2less-by-mathgreaterless-by-pgreater-lesspgreaterwhat-is-lessmathgreaterf(x)plusf(x)plusf(x)-quad-sim-3-f(x)less-by-mathgreaterless-by-pgreat-28d26fc/ "Some Special Cases
@$\begin{align}f(x)=x^{2}\end{align}@$
What is @$\begin{align}f(x)+f(x)+f(x) ? \quad \sim 3 f(x)\end{align}@$
Complete:
@$\begin{align}3 f(x)= 3 x^{2}\end{align}@$
Complete:
Evaluate @$\begin{align}3 f(2)=\end{align}@$
Done.
")
Using the following equations, find f(g(x)): f(x)=−2 x g(x)=x 2+3 f(g(x))=[?]x 2+:
@$\begin{align}
\begin{array}{c}
f(x)=-2 x \
g(x)=x^{2}+3 \
f(g(x))=[?] x^{2}+
\end{array}
\end{align}@$")
Using the following equations, find f(g(x)): f(x)=3 x+2 g(x)=6 x−5 f(g(x))=[?]x+:
@$\begin{align}
\begin{array}{r}
f(x)=3 x+2 \
g(x)=6 x-5 \
f(g(x))=[?] x+
\end{array}
\end{align}@$")
Using the following equations, find f(g(x)): f(x)=3 x+2 g(x)=6 x−5 f(g(x))=[?]x+
Find f(g(-2)) and g(f(1)) for the pair of functions. f(x)=x 2−2 g(x)=x+2 f(g(−2))= g(f(1))=.
@$\begin{align}
\begin{array}{l}
f(x)=x^{2}-2 \
g(x)=x+2 \
\end{array}
\end{align}@$
@$\begin{align}f(g(-2))=\end{align}@$
@$\begin{align}g(f(1))=\end{align}@$
")
For the function f(x) = 5x + 3, find the matching value for x when f(x) = -7.
What is (f+g)(x)? f(x)=3x^2–7x g(x)= – 3x^2. Write your answer as a polynomial or a rational function in simplest form.
F(x) = 3 – 2x g(x) = 2x + 3 h(x) = 2x (a) (i) Find f(–3). ............................................... (ii) Find gf(–3). ............................................... (b) Find f–1(x). f–1(x) = .............................................. (c) Find x when gg(x) = 7 . (d) Find x when h–1(x) = 5. x = ..............................................
The function f is defined by f(t) = (t-2)(t+3)^2. If f(t-1)=0, what is one possible value of t?
Consider the function f(x) = 8x + 6. Step 2 of 2: Find the value of f(a − 1).
Let f(x)=x^2+6 and g(x)=x^2-2x. Find (f⋅g)(x).
Evaluate f(x + 2) for the function f(x) = x^2 − 3x.
Find f(x) and g(x) such that h(x) = (fog)(x). h(x) = (9x + 3)^7.
Find f(g(-2)) if f(x) = 4x + 5 and g(x) = x^2 - 1.
The value of the function f(x) = 3x^2 − 2x for x = 4 is?
Find f(g(−2)). Given: f(x)=2 x 2+3 x−4 and g(x)=2 x−1.
Complete the table: f(s) = 9s^2. s f(s) –3 –1 1 3.
Find g(f(3)). f(x) = 2x2 + 3x - 4, g(x) = 2x - 1.
If f(x)=x^{2} and g(x)=x-1, find f(g(x)).
Solve for f: -3(f + -3) = 5.4
For the function f(x), find all values of a for which f(a) is the indicated value. f(x)=12 x−36 x, and f(a)=6.
Find f(3) for this function. f(x)=3 x 2+6 x−2 After you enter your answer press GO. ____ GO..
GO.")
Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers. f(x) = x^2 + 6, g(x) = x + 7.
Given the function g(x) = 5 - x^{2}, evaluate g(x+h)−g(x)h, h ≠0.
Let f(x)=1 x 4+5 and g(x)=1 x 2+1. What is the value of f(1-g(0))?
Find (g ∘ f)(x) and (f ∘ g)(x) for the given functions f and g. f(x) = 2x + 5, g(x) = 4x − 7.
Find (g ∘ f)(x) and (f ∘ g)(x) for the given functions f and g. f(x) = 3x + 4, g(x) = 5x − 8.
Find (g ∘ f)(x) and (f ∘ g)(x) for the given functions f and g. f(x) = 2x + 5, g(x) = 3x - 7.
Use the pair of functions to find f(g(x)) and g(f(x)). f(x) = x^2 + 6, g(x) = x + 7
Which equation written in two variables is equivalent to this function? f(x) = -2(x – 4)
Given the function f(x) = 8 - 3x: evaluate f(-2), solve f(x) = -1.
Given the function f(x) = x^2 - 3x, evaluate f(5) and solve f(x) = 4.
Use the following function rule to find f(16). f(x) = 8 + 32/x.
Find f(x) and g(x) such that h(x) = (fog)(x). h(x) 6/(x-3)^2.
If f(x) = −2 + 8, what is f(−1.8)?
Find a function f(x) such that f'(x) = 3x^2-8, and f(0) = -2.
Find a function f(x) such that f'(x) = -6x^2+4, and f(0) = 4.
Find f′(a) for the function f(t)=5t+1/(t+4).
For the function f(x) =−10x−8, find the slope of the tangent line at x = -6.
Let 𝑓(𝑥) = 20. Find the derivative of the function, denoted by 𝑓′(𝑥).
Let f(x) be a polynomial function such that f(2)=3,f′(2)=0 and f′′(2)=−4 . Classify the point (2,3).
Evaluate the function at the given value of the independent variable. Simplify the results. f(x) = 2x^2. Find f(x + Δx) − f(x) Δx (Δx ≠ 0).
Miss Cormier is going to the Taylor Swift concert in November of this year. She wants to make friendship bracelets to pass out to other concert goers as is tradition of the Eras Tour. She needs to put together a kit on Amazon in order to have enough materials to make them all. Kits cost $75.00 plus $15.00 per set of Taylor-themed beads. a. Use the information given to create an equation for the cost of the friendship bracelet kits. Be sure to include what each variable represents.
The equation P = -300t + 2,300, can be used to find the value, P, of a notebook computer at the end of t years. a. What is the value of the notebook computer at the end of three years? At the end of three years, the notebook will be worth $ . b. When is the notebook computer worth $800? At the end of years the notebook will be worth $800.
H(t)=2t+6.h(20) = The height of the balloon seconds after it is released is feet. b. Explain how to use the function to find the height of the balloon 2 minutes after it is released. 2 minutes is seconds, so calculate the value of the function by substituting t= in the equation. The height of the balloon 2 minutes after it is released is feet. c. What is the height of the balloon just before it is released? How do you know? feet t = before the balloon is relea
The function f(x)=3x+10 models the yearly membership of an online movie rental club, where x is the number of movies rented. (a) In one year, Rachel rented 10 movies. How much was Rachel's membership? (b) In one year, Toby rented 30 movies. How much more than Rachel did Toby pay in one year?
Michael received $275 more from relatives than friends. He received $4275 dollars in all, how much money did he receive from relatives and how much money did he receive from friends?
Evaluate e 4+2 f−3 when e=12 and f=1 2.
Find a function f(x) such that f'(x) = 9e^x+x, and f(0) = -5.
Miss Cormier is going to the Taylor Swift concert in November of this year. She wants to make friendship bracelets to pass out to other concert goers as is tradition of the Eras Tour. She needs to put together a kit on Amazon in order to have enough materials to make them all. Kits cost $75.00 plus $15.00 per set of Taylor-themed beads. a. Use the information given to create an equation for the cost of the friendship bracelet kits. Be sure to include what each variable represents.
What is the value of x in the equation f(x) = x /2 + 5 , if the value of f(x) = 8?
Let f(x) be a polynomial function such that f(−2)=4, f′(−2)=0, and f′′(−2)=4. Classify the point (−2,4).
The equation P = -300t + 2,300, can be used to find the value, P, of a notebook computer at the end of t years. a. What is the value of the notebook computer at the end of three years? At the end of three years, the notebook will be worth $ . b. When is the notebook computer worth $800? At the end of years the notebook will be worth $800.
Given the following function, determine the difference quotient, f(x+h)−f(x)h . f(x)=5x−2
Given the following function, determine the difference quotient, f(x+h)−f(x)h . f(x)=−8x−8
For f(x) = (x + 20) / x, find (a) f(x + h) and (b) (f(x + h) - f(x)) / h.
Let f(x) be a polynomial function such that f(2)=−4, f′(2)=0, and f′′(2)=5. Classify the point (2,−4).
Is it true that the notebook computer will be worth $1,300 after four years? How much did the notebook computer originally cost?
If z = x^{5} + y^{5}, then what is the value of x{∂z}{∂x} + y{∂z}{∂y}? A) 0 B) 5 C) z D) 5z-0-b)-5-c)-z-d)-5z/ "If z = x^{5} + y^{5}, then what is the value of x{∂z}{∂x} + y{∂z}{∂y}? A) 0 B) 5 C) z D) 5z")
H(t)=2t+6.h(20) = The height of the balloon seconds after it is released is feet. b. Explain how to use the function to find the height of the balloon 2 minutes after it is released. 2 minutes is seconds, so calculate the value of the function by substituting t= in the equation. The height of the balloon 2 minutes after it is released is feet. c. What is the height of the balloon just before it is released? How do you know? feet t = before the balloon is relea
Write the equation of the linear function that satisfies f(4)=35 and f(7)=56. What is f(x)?
High Tech Graphics bought a fax machine for $750. The fax machine depreciates at a rate of $25 a month. Find a function F that can be used to determine the value of the fax machine t months after purchase. Complete the model below.
F(x)=12-3x determine f(x+h)-f(x)
The function f(x)=3x+10 models the yearly membership of an online movie rental club, where x is the number of movies rented. (a) In one year, Rachel rented 10 movies. How much was Rachel's membership? (b) In one year, Toby rented 30 movies. How much more than Rachel did Toby pay in one year?
Let f equals {(6, negative 7), (negative 6, 6), (3, negative 4)}. Find f(6).
Differentiate f(x)=9e^V(x).
For the function f(x)=7x^2-8, determine f(x+h)-f(x)/h. Is it correct?
The revenue function for a student council for selling T-shirts is given by R(n)=8n, and the total cost function is given by C(n)=210+5n where n represents the number of T-shirts produced. Determine the profit function P(n).
What does the "x" operator do in relational algebra?
Define function notation.
How to find the equation of a piecewise function?
Which set of ordered pairs represents a function?
How do you find the equation of a piecewise function?
What is the definition of function notation?
What is the sign of the leading coefficient of a function?
What is an identity function equation?
How is function notation written?
What is the equation of a quadratic function?
What is the purpose of function notation?
What is functional notation?
How do you write function notation?
How do you do function notation?
What is function notation?
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12686 | https://instituteofeducation.ie/wp-content/uploads/2022/10/1-CIRCLE-Notes.pdf | © H DORGAN IOE LC Maths HL 1 Name: MATHEMATICS Hilary Dorgan Leaving Certificate Higher Level 2021 - 2022 C I R C L E N O T E S 1 © H DORGAN IOE LC Maths HL 2 CIRCLE Strand 2, Paper 2 NOTES x2+ y2 = 25 tangent radius (– 4, –3) © H DORGAN IOE LC Maths HL 3 Syllabus, What You Must Know Page 4 Equation of a Circle Page 5 Junior Cert Circle Theorems Page 7 Tangents Page 8 Problems with 𝑔, 𝑓 and 𝑐 Page 10 Touching Circles and Common Tangent Page 13 Points to Remember Page 14 CONTENTS © H DORGAN IOE LC Maths HL 4 CIRCLE SYLLABUS • Recognise that (𝑥−ℎ)2 + (𝑦−𝑘)2 = 𝑟2 represents the relationship between the 𝑥 and 𝑦 coordinates of points on a circle of centre (ℎ, 𝑘) and radius 𝑟. (OL) • Solve problems involving a line and a circle with centre (0, 0). (OL) • Recognize that 𝑥2 + 𝑦2 + 2𝑔𝑥+ 2𝑓𝑦+ 𝑐= 0 represents the relationship between the 𝑥 and 𝑦 coordinates of points on a circle of centre (−𝑔, −𝑓) and radius 𝑟 where 𝑟= √𝑔2 + 𝑓2 −𝑐. (HL) • Solve problems involving a line and a circle. (HL) WHAT YOU MUST KNOW Recognize the equation of a circle in various forms (e.g. degree 2, cannot have an 𝑥𝑦 term, coefficients of 𝑥2 and 𝑦2equal) Familiarity with all the formulae for coordinate geometry of the Straight Line Find the centre and radius of a circle given its equation in any form Find the equation of a circle given its centre and radius Find the equation of a circle given the two end points of a diameter Check whether a particular point is inside, outside or on a given circle Familiarity with Junior Cert theorems involving circles Find the point of intersection between a line and a circle Prove that a given line is a tangent to a circle Find the equation of a tangent of a circle given the equation of the circle and their point of intersection Find the equation of a circle given its centre and given the equation of a tangent to the circle Find the equations of the two tangents that can be drawn from a given point outside a given circle Solve problems involving 𝑔 , 𝑓 and 𝑐. © H DORGAN IOE LC Maths HL 5 Equation of a Circle What is a Circle: A circle is a set of points that are equidistant from a common point c (h, k) called the centre. The fixed distance r from the centre to any point on the circle is called the radius. The standard equation of a circle with centre c (h, k) and radius r is: (x – h)2 + (y – k)2 = r2. c (h, k) If the centre of the circle is at the origin, c (0, 0), this equation becomes x2 + y2 = r2. r Example 1: Find the equation of a circle whose centre is at (3, – 7) and has a radius of 2. Solution 1: Given (ℎ , 𝑘 ) = (3 , – 7) and 𝑟 = 2 ⇒ required circle equation: (x – 3)2 + (y – (– 7))2 = 22 ⇒ required equation: (x – 3)2 + (y + 7)2 = 4 Example 2: Find the equation of a circle that has a diameter with the endpoints given by the points 𝒂(– 3, 4) and 𝒃(5, 10). Solution 2: Whenever we want the equation of a circle, we look always for two pieces of information (i) What is the centre of the circle? (ii) What is the radius of the circle? The centre of this circle, c, is the midpoint of the line segment or diameter ab. c = ( −3 + 5 2 , 4 + 10 2 ) = (1, 7) [Using the mid-point formula] The radius of this circle is half the length of the diameter between a and b. r = 1 2 √(5 + 3)2 + (10 −4)2 = 1 2 √64 + 36 = 1 2 × 10 = 5 Using standard equation (x – h)2 + (y – k)2 = r2 ⇒ (x – 1)2 + (y – 7)2 = 52 ⇒ (x – 1)2 + (y – 7)2 = 25 The equation of a circle is also very often given in the form: x2 + y2 + 2gx + 2fy + c = 0. The centre (𝑜) of this circle is: ( – g , – f ) and the length of the radius of this circle is: 𝑜 [Both formulae on page 19 of Tables] 𝒓= √𝒈𝟐+ 𝒇𝟐−𝒄 Note: The equation of a circle should always have the following properties: (i) it is of degree 𝟐, (ii) it has no 𝒙𝒚 term (iii) the coefficients of 𝒙2 and 𝒚2 are equal © H DORGAN IOE LC Maths HL 6 Example 3: Find the centre and radius of the circle with equation 𝑥2 + 𝑦2 – 4𝑥 – 6𝑦 + 9 = 0. Solution 3: We will do this question by two different methods. [This will verify the previous formula given.] Method 1: This method involves re-writing the given equation in more standard form with all x terms together and all terms with y terms together (using brackets). (𝑥2 – 4𝑥 ) + (𝑦2 – 6𝑦 ) + 9 = 0 We now complete the square within each bracket and compensate by adding 13 to other side also: (𝑥2 – 4𝑥 + 4) + (𝑦2 – 6𝑦 + 9) + 9 = 0 + 13 This gives us : (𝑥2 – 4𝑥 + 4) + (𝑦2 – 6𝑦 + 9) = 4 [Adding – 9 to both sides] Now we simplify and write in standard form: (𝑥 – 2)2 + ( 𝑦 – 3)2 = 4 OR (𝑥 – 2)2 + ( 𝑦 – 3)2 = 22 Comparing this equation with the standard equation: centre of circle at 𝒄 (𝒉, 𝒌) = 𝒄 (𝟐 , 𝟑) and radius of circle = 𝒓 = 2. Method 2: This method involves using formulae (on previous page). Equation of circle: 𝑥2 + 𝑦2 – 4𝑥 – 6𝑦 + 9 = 0 implies that 𝒈= – 2, 𝒇= – 3 and 𝒄 = 9. Thus, the centre of this circle is: ( – 𝒈 , – 𝒇 ) = (𝟐 , 𝟑 ) and the radius of this circle is: √𝒈𝟐 + 𝒇𝟐 − 𝒄 = √(−2)2 + (−3)2 − 9 = √4 + 9 −9 = √4 = 2. Example 4: Is the point q (3 , 4) inside, outside or on the circle with equation (𝒙 + 𝟔)2 + ( 𝒚 – 𝟏)2 = 𝟖𝟓 Solution 4: Find the distance from the centre of the circle to q and compare this with the length of the radius. Method 1: centre 𝒄 = (– 𝒈, – 𝒇 ) = (– 6 , 1) ; radius 𝒓 = √85 = 9.230 distance from c to q = √(−6 −3)2 + (1 −4)2 = √81 + 9 = √90 = 9.487 Since the distance from c to q is greater than the radius, point q is outside the circle. Method 2: Substitute the point (3 , 4) in for 𝒙 and 𝒚 in the LHS of the circle equation (𝒙 + 𝟔)2 + ( 𝒚 – 𝟏)2 = 𝟖𝟓. LHS = (3 + 6)2 + (4 – 1)2 = 81 + 9 = 90 since LHS > RHS, point (3, 4) is outside the circle. Note: Point (3, 3) is on the circle (𝒙 + 𝟔)2 + (𝒚 – 𝟏)2 = 𝟖𝟓, since LHS = 81 + 4 = 85 = RHS Point (3, 2) is inside the circle (𝒙 + 𝟔)2 + (𝒚 – 𝟏)2 = 𝟖𝟓, since LHS = 81 + 1 = 82 < RHS © H DORGAN IOE LC Maths HL 7 Some important Junior Cert Theorems Angles standing on the. C D A B same arc of a circle are equal |𝑎𝑛𝑔𝑙𝑒 𝑨| = |𝑎𝑛𝑔𝑙𝑒 𝑩| Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So 𝑪= 𝑫 = 𝟗𝟎o. A tangent to a circle forms a right angle with the circle's The angle formed at the centre of a circle by lines radius, at the point of contact of the tangent. originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. 𝑨 = 𝟐𝑩 𝑩 𝑨 𝑴 𝐴 𝐵 𝑸 𝐶 𝑷 𝐷 A cyclic quadrilateral is a four-sided figure inside a circle, A radius which is perpendicular to a chord with each vertex of the quadrilateral touching the of a circle will always bisect that chord. circumference of the circle. The opposite angles of such i.e. |𝑃𝑄| = |𝑄𝑀| a quadrilateral add up to 180o. i.e. 𝑨 + 𝑪 = 180o and 𝑩 + 𝑫 = 180o © H DORGAN IOE LC Maths HL 8 Tangents Tangents are lines that touch a circle at one point only. As can be seen in the figure across, the tangent line is always at right angles to the radius at the point of contact. Also, the perpendicular distance from the centre of a circle to any tangent is equal to the length of the radius. Example 5: By finding the point of intersection, show that the line 𝟕𝒙 – 𝟗𝒚 + 𝟏𝟑𝟎 = 𝟎 is a tangent to the circle 𝒙2 + 𝒚2 = 𝟏𝟑𝟎. Solution 5: If 7𝑥−9𝑦+ 130 = 0, then 𝑥= 9𝑦−130 7 ⇒ ( 9𝑦−130 7 )2 + 𝑦2 = 130 [Using 𝑥2 + 𝑦2 = 130] ⇒ (9𝑦−130)2 + 49 𝑦2 = 6370 [Multiplying across by 49] ⇒ 81 𝑦2 − 2340𝑦 + 16900 + 49𝑦2 = 6370 [Expanding LHS] ⇒ 130 𝑦2 − 2340𝑦 + 10530 = 0 [Gathering like terms] ⇒ 𝑦2 − 18𝑦 + 81 = 0 [Dividing across by 130] ⇒ (𝑦−9)(𝑦−9) = 0 [Factorising] ⇒ y = 9 ⇒ 𝑥= 9𝑦−130 7 = −7 [Solving] ⇒ There is just a single point of intersection (– 7, 9) ⇒ 𝟕𝒙 – 𝟗𝒚 + 𝟏𝟑𝟎 = 𝟎 is a tangent to the circle 𝒙2 + 𝒚2 = 𝟏𝟑𝟎. Note: If there are two solutions for the intersection of a line and a circle, then the line cuts the circle in two places. If there are no solutions for the intersection of a line and a circle, then the line does NOT meet the circle at all. Example 6: Find the equation of the circle that has its centre at (3, 5) and has a tangent whose equation is given by 𝑥 + 𝑦 = 2. Solution 6: Perpendicular distance from (𝟑, 𝟓) to 𝒙 + 𝒚 = 𝟐 is given by 𝑑 = | 𝑎 𝑥1 +𝑏𝑦1 +𝑐 √𝑎2 + 𝑏2 | . [Note minus sign here] d = | (𝟏) (𝟑) + (𝟏) (𝟓) +(−𝟐) √𝟏𝟐 + 𝟏𝟐 | d = r = | 𝟔 √𝟐| = 𝟑√𝟐 = length of radius ∴ equation of circle: (𝒙 – 𝟑)2 + (𝒚 – 𝟓)2 = (𝟑√𝟐)𝟐 = 𝟏𝟖 Note: Solution is drawn in graphical form in the diagram above. This is NOT required by the question. a = 1 b = 1 c = –2 x1 = 3 y1 = 5 © H DORGAN IOE LC Maths HL 9 Example 7: Find the equation of the tangent to the circle (𝒙 – 𝟓)2 + (𝒚 + 𝟒)2 = 𝟐𝟓 at the point (1, – 1). Solution7: Slope of radius, R, = 𝑦2− 𝑦1 𝑥2− 𝑥1 = −4 + 1 5 − 1 = −3 4 (1, – 1) R ⇒ slope of tangent, T, = 4 3 [because T ⊥𝑅] (5, – 4) ⇒ equation of tangent, T, = 𝑦− 𝑦1 = 𝑚 ( 𝑥− 𝑥1 ) T ⇒ eqn of T : 𝑦 + 1 = 4 3 ( 𝑥− 1 ) ⇒ 3 (𝑦 + 1) = 4( 𝑥−1) ⇒ 3𝑦+ 3 = 4𝑥−4 ⇒ T : 4𝑥−3𝑦−7 = 0 Example 8: Find the equations of the two tangents from the point (3, 5) to the circle 𝒙2 +𝒚2 +𝟐𝒙 – 𝟒𝒚 – 𝟒= 𝟎. Solution 8: Equation of any line containing the point (3, 5) is: 𝑦−5 = 𝑚( 𝑥− 3 ) ⇒ 𝑚𝑥−𝑦+ 5 −3𝑚= 0 Centre of circle = (– 1, 2) Radius of circle = √12 + (−2)2 −(−4) = √9 = 3 Distance from centre= (– 1, 2) to tangent 𝑚𝑥−𝑦+ 5 −3𝑚= 0 is equal to length of radius = 3. We will use the perpendicular distance from a point to a line formula: 𝑑 = | 𝑎 𝑥1 +𝑏𝑦1 +𝑐 √𝑎2 + 𝑏2 | d = | (𝒎) (−𝟏) + (−𝟏) (𝟐) +(𝟓−𝟑𝒎) √𝒎𝟐 +(− 𝟏)𝟐 | (3, 5) ⇒ d = | −𝒎 −𝟐 + 𝟓 −𝟑𝒎 √𝒎𝟐 + 𝟏 | ⇒ 3 = | − 𝟒𝒎 + 𝟑 √𝒎𝟐 + 𝟏 | ⇒ 𝟗 (𝒎𝟐+ 𝟏) = (𝟑−𝟒𝒎)𝟐 [Cross multiplying and squaring] r ⇒ 𝟗𝒎𝟐 + 𝟗 = 𝟗 − 𝟐𝟒𝒎 + 𝟏𝟔𝒎𝟐 (–1, 2) ⇒ 0 = 7𝒎𝟐 − 𝟐𝟒 𝒎 ⇒ 𝒎(𝟕𝒎−𝟐𝟒) = 𝟎 ⇒ 𝒎 = 𝟎 𝒐𝒓 𝒎 = 𝟐𝟒 𝟕 If m = 0: eqn of tangent: − 𝒚+ 𝟓= 𝟎 𝒐𝒓 𝒚= 𝟓 If m = 𝟐𝟒 𝟕: eqn of tangent: 𝟐𝟒 𝟕𝒙 – 𝒚+ 𝟓−𝟑 ( 𝟐𝟒 𝟕) = 𝟎 𝒐𝒓 𝟐𝟒𝒙 – 𝟕𝒚 + 𝟑𝟓 − 𝟕𝟐 = 𝟎 ∴ 𝒚 = 𝟓 𝒐𝒓 𝟐𝟒𝒙 – 𝟕𝒚 − 𝟑𝟕 = 𝟎 a = m b = –1 c = 5 – 3m x1 = – 1 y1 = 2 © H DORGAN IOE LC Maths HL 10 Proof of the Tangent Theorem Prove that the tangent to the circle 𝒙2 + 𝒚2 = 𝒓2 at the point (𝒙1, 𝒚1) is given by 𝒙𝒙1 + 𝒚𝒚1 = 𝒓2. Proof: (𝒙1, 𝒚1) Let R be the radius of the circle, joining the centre (𝟎, 𝟎) to (𝒙1, 𝒚1). R T Let T be the tangent of the circle, whose equation we are trying to find. Slope of radius, R, = 𝑦2− 𝑦1 𝑥2− 𝑥1 = 𝑦1 − 0 𝑥1 − 0 = 𝑦1 𝑥1 (𝟎, 𝟎) ⇒ slope of tangent, T, = − 𝑥1 𝑦1 Equation of tangent, T, = 𝑦− 𝑦1 = 𝑚 ( 𝑥− 𝑥1 ) ⇒ equation of T : 𝑦− 𝑦1 = − 𝑥1 𝑦1 ( 𝑥− 𝑥1 ) ⇒ 𝑦𝑦1 − 𝑦1 2 = − 𝑥𝑥1 + 𝑥1 2 ⇒ 𝑥𝑥1 + 𝑦𝑦1 = 𝑥1 2 + 𝑦1 2 But, (𝒙1, 𝒚1) is a point on the circle 𝒙2 + 𝒚2 = 𝒓2 ⇒ 𝑥1 2 + 𝑦1 2 = 𝑟2 Hence, the equation of the tangent, T, is given by: 𝑥𝑥1 + 𝑦𝑦1 = 𝑟2. Problems involving g, f and c When asked to find the equation of a circle which obeys certain conditions, it is usually better to write the circle in the form 𝒙2 + 𝒚2 + 𝟐𝒈𝒙 + 𝟐𝒇𝒚 + 𝒄 = 𝟎. Then apply the various conditions given to find the values of 𝑔, 𝑓 and 𝑐. Some special cases that apply (you should try to justify each of them yourself, one is done for you below) (i) If (0, 0) is on the circle, then c = 0 (ii) If the centre of the circle is on the x-axis, then 𝒇 = 𝟎 (iii) If the centre of the circle is on the y-axis, then 𝒈 = 𝟎 (iv) If the x-axis is a tangent to the circle, then 𝒓 = |𝒇| ⇒ 𝒄 = 𝒈𝟐 (see diagram below) (v) If the y-axis is a tangent to the circle, then 𝒓 = |𝒈| ⇒ 𝒄 = 𝒇𝟐 (vi) If both axes are tangents to the circle, then 𝒓 = |𝒈| = |𝒇| ⇒ 𝒄 = 𝒈𝟐= 𝒇𝟐 (vii) If the centre of the circle lies on the line 𝒂𝒙 + 𝒃𝒚 + 𝒄 = 𝟎, then 𝑎(−𝑔) + 𝑏(−𝑓) + 𝑐= 0 ⇒ 𝑎𝑔 + 𝑏𝑓 − 𝑐 = 0 (viii) If (𝒙1, 𝒚1) is a point on the circle 𝒙2 + 𝒚2 + 𝟐𝒈𝒙 + 𝟐𝒇𝒚 + 𝒄 = 𝟎, then (𝒙1)2 + (𝒚1)2 + 𝟐𝒈(𝒙1) + 𝟐𝒇(𝒚1) + 𝒄 = 𝟎. © H DORGAN IOE LC Maths HL 11 𝑦−𝑎𝑥𝑖𝑠 K In this diagram, the x-axis is a tangent to the circle K, 𝒙2 + 𝒚2 + 𝟐𝒈𝒙 + 𝟐𝒇𝒚 + 𝒄 = 0. (– g, – f) From the diagram, it is clear that the radius of the circle, r = |𝒇|. r From our formula, we know that the radius of the circle, r = √𝑔2 + 𝑓2 − 𝑐 Therefore, |𝒇| = √𝑔2 + 𝑓2 − 𝑐 ⇒ 𝑓2 = 𝑔2 + 𝑓2 −𝑐 𝑥−𝑎𝑥𝑖𝑠 ⇒ 0 = 𝑔2 −𝑐 ⇒ 𝑐 = 𝑔2 Example 9: Find the equation of the circle such that the three points 𝑷(0 , 4), 𝑸(3 , 5) and 𝑹(7 , 3) are on the circle. Solution 9: Let equation of required circle, K, be: x2 + y2 + 2gx + 2fy + c = 0 (0, 4) ε K ⇒ 02 + 42 + 2g(0) + 2f(4) + c = 0 Q ⇒ 8f + c = – 16 … eqn (i) P R (3, 5) ε K ⇒ 32 + 52 + 2g(3) + 2f(5) + c = 0 ⇒ 6g + 10f + c = – 34 … eqn (ii) (7, 3) ε K ⇒ 72 + 32 + 2g(7) + 2f(3) + c = 0 ⇒ 14g + 6f + c = – 58 … eqn (iii) 7 × (eqn (ii)) : 42g + 70f + 7c = – 238 3 × (eqn (iii)): 42g + 18f + 3c = – 174 Subtracting: 52f + 4c = – 64 … eqn (iv) 4 × (eqn (i)) : 32f + 4c = – 64 ⇒ 20f = 0 ⇒ f = 0 ⇒ c = – 16 If f = 0, and c = – 16, then eqn (ii): 6g = – 34 – 10f – c 6g = – 34 – (10)(0) – (– 16) = – 18 ⇒ g = – 3 ∴ Required equation: x2 + y2 – 6x – 16 = 0 or (x – 3)2 + y2 = 25 © H DORGAN IOE LC Maths HL 12 Example 10: Find the equation of the circle such that the two points (1 , 0), and (0 , 2) are on the circle and the centre of the circle lies on the line, D: x + 3y – 11 = 0. Solution 10: Let equation of required circle, K, be: x2 + y2 + 2gx + 2fy + c = 0 (1, 0) ε K ⇒ 12 + 02 + 2g(1) + 2f(0) + c = 0 [Substituting given point into circle equation] ⇒ 2g + c = – 1 … eqn (i) (0, 2) ε K ⇒ 02 + 22 + 2g(0) + 2f(2) + c = 0 [Substituting given point into circle equation] ⇒ 4f + c = –4 … eqn (ii) (– g, – f) ε D ⇒ – g – 3f – 11 = 0 [Centre of circle is (– g, – f) lies on line x + 3y – 11 = 0] ⇒ g + 3f = – 11 … eqn (iii) (eqn (i)) : 2g + c = – 1 (eqn (ii)): 4f + c = – 4 Subtracting: 2g – 4f = 3 … eqn (iv) But, 2 × (eqn (iii)): 2g + 6f = – 22 ⇒ – 10f = 25 ⇒ f = – 2.5 Using eqn (iii): g = – 11 – (3)(– 2.5) ⇒ g = – 3.5 If f = – 2.5 and g = – 3.5, then eqn (i): c = – 1 – (2)(– 3.5) = 6 ∴ Required equation: x2 + y2 – 7x – 5y + 6 = 0 or (x – 𝟕 𝟐)2 + (y – 𝟓 𝟐) 2 = 𝟐𝟓 𝟐 . © H DORGAN IOE LC Maths HL 13 Touching Circles, Common Tangent If c1 and c2 are the centres of two circles, K1 and K2, and r1 and r2 are the radii of these two circles then, r1 r2 these circles: c1 c2 |𝑐1 𝑐2| = 𝑟 1 + 𝑟2 touching externally touch externally iff |𝑐1 𝑐2| = 𝑟 1 + 𝑟2 K1 K2 and touch internally iff |𝑐1 𝑐2| = 𝑟 1 − 𝑟2 K1 K2 |𝑐1 𝑐2| = 𝑟 1 − 𝑟2 c1 c2 touching internally r1 r2 Example 11: Prove that the two circles K1: x2 + y2 – 6x – 8y + 24 = 0 and K2: x2 + y2 = 16 touch externally and find their point of contact. Solution 11: c1 = (3, 4) ; c2 = (0, 0) r1 = √32 + 42 −24 = 1 and r2 = 4. |𝑐1 𝑐2| = √32 + 42 = 5 𝑟 1 + 𝑟2 = 1 + 4 = 5 Therefore, as |𝑐1 𝑐2| = 𝑟 1 + 𝑟2, the circles touch externally. Note: K1 – K2 = (x2 + y2 – 6x – 8y + 24) – (x2 + y2 – 16) = 0 ⇒ – 6x – 8y + 40 = 0 ⇒ 3x + 4y – 20 = 0 [This is the equation of the common tangent. It is got by subtracting the two circle equations. This method will give you the common chord when two circles cut each other in two places.] Now find where 3x + 4y – 20 = 0 meets, say, K2, x2 + y2 = 16 Using simultaneous equations, we find the point of intersection by substituting 𝑥= 20−4𝑦 3 into x2 + y2 = 16 and finding a single solution, i.e. 𝑦= 16 5 , 𝑥= 12 5 Therefore, point of intersection or contact: ( 12 5 , 16 5 ). © H DORGAN IOE LC Maths HL 14 Example 12: (i) C is the circle x2 + y2 – 8x + 4y – 5 = 0. A circle K touches C internally and goes through g, the centre of C. If 3x – 4y + 5 = 0 is the tangent common to both circles, find the equation of K. (ii) Circle H is the image of circle K under the central symmetry in g. Find the equation of H and the equation of the common tangent to H and C. Solution 12: (i) Equation of C: x2 + y2 – 8x + 4y – 5 = 0 ⇒ centre of C = g = (𝟒, 𝟐) M ⇒ radius of C = √−42 + 22 −(−5) = √25 = 5 u C ⇒ radius of K = 1 2 (5) = 5 2 To find s: T ∩ C = s q H T: 𝟑𝒙−𝟒𝒚+ 𝟓= 𝟎 ⇒ 𝒙 = 𝟒𝒚−𝟓 𝟑 C: 𝑥2 + 𝑦2 −8𝑥+ 4𝑦−5 = 0 g(4, –2) ( 4𝑦−5 3 ) 2 + 𝑦2 −8( 4𝑦−5 3 ) + 4𝑦−5 = 0 ⇒ (4𝑦−5)2 + 9𝑦2 − 24(4𝑦−5) + 36𝑦−45 = 0 T ⇒ 16𝑦2 −40𝑦+ 25 + 9𝑦2 − 96𝑦+ 120 + 36𝑦−45 = 0 K p ⇒ 25𝑦2 −100𝑦+ 100 = 0 ⇒ 𝑦2 −4𝑦+ 4 = 0 s ⇒ (𝑦−2)2 = 0 ⇒ 𝑦= 2 ⇒ 𝒙 = 𝟒𝒚−𝟓 𝟑 = 𝟒(𝟐)−𝟓 𝟑 = 1 ⇒ point of intersection, s = (1, 2) To find equation of circle K Centre of K = p = mid-point of [𝑟𝑠] = ( 4+1 2 , −2+2 2 ) = ( 5 2 , 0) ; radius of K = 5 2 Equation of K: (𝑥− 5 2 )2 + (𝑦−0)2 = ( 5 2)2 ⇒ 𝑥2 −5𝑥+ 25 4 + 𝑦2 = 25 4 ⇒ 𝑥2 + 𝑦2 −5𝑥= 0 (ii) To find equation of circle H H is the image of K under the central symmetry in g. ∴ radius of H = radius of K = 5 2 ∴ q is the image of p under the central symmetry in g. p ( 5 2 , 0) → g (4, –2) → q ( 𝟏𝟏 𝟐, –4) Equation of H: (𝑥− 11 2 )2 + (𝑦+ 4)2 = ( 5 2)2 ⇒ 𝑥2 −11𝑥+ 121 4 + 𝑦2 + 8𝑦+ 16 = 25 4 ⇒ 𝑥2 + 𝑦2 −11𝑥+ 8𝑦+ 40 = 0 To find equation of line M (see diagram) Line M is parallel to line T and contains the point u (see diagram). ∴ Slope of M = slope of T = 3 4 . u is the image of g under the central symmetry in q. g (4, –2) → q ( 𝟏𝟏 𝟐, –4) → u (7, – 6) ∴ equation of M : 𝑦− 𝑦1 = 𝑚 ( 𝑥− 𝑥1 ) ⇒ 𝑦 + 6 = 3 4 ( 𝑥− 7 ) ⇒ 4 (𝑦 + 6) = 3( 𝑥−7) ⇒ 4𝑦+ 24 = 3𝑥−21 ⇒ 3𝑥−4𝑦−45 = 0 Note: The common tangent, M can also be found using: M = C – H [Intersection of circles C and H] ∴ equation of M = (x2 + y2 – 8x + 4y – 5) – (x2 + y2 – 11x + 8y + 40) = 0 ∴ equation of M = 3x – 4y – 45 = 0 © H DORGAN IOE LC Maths HL 15 POINTS TO REMEMBER 1) Check the formulae available on page 19 and page 9 of your Mathematical Tables and Formulae booklet. 2) Drawing a quick sketch/diagram can often be a great help with problems on the circle. For accurate circles you have to make the gaps on each axis the same. 3) Realise that you may need a lot of your formulae from “the straight line” here (page 18 Tables). 4) Length of arc of circle, 𝒍= 𝒓 𝜽 () and area of sector 𝑨= 𝟏 𝟐 𝒓𝟐 𝜽 () [angle, 𝜃, is given in radians] (Also, 𝑙= 𝜃 360 (2𝜋𝑟) is used when the angle, 𝜃, is given in degrees.) (Also, the equation 𝐴= 𝜃 360 (𝜋𝑟2) is used when the angle, 𝜃, is given in degrees.) 5) C: 𝒙2 + 𝒚2 = 𝒓2 () represents a circle of centre (0, 0) and radius 𝒓. 6) (𝒙 – 𝒉)2 + (𝒚 – 𝒌)2 = 𝒓2 () represents a circle of centre (𝒉, 𝒌) and radius 𝒓. 7) The equation of a circle should be of degree 𝟐, cannot have an 𝒙𝒚 term and coefficients of 𝑥2 and 𝑦2 must be equal. 8) When the equation of a circle is given in the form: 𝒙2 + 𝒚2 + 𝟐𝒈𝒙 + 𝟐𝒇𝒚 + c = 𝟎 () the centre of the circle is: (– g, – f) () and the radius is given by: 𝒓= √𝒈𝟐+ 𝒇𝟐−𝒄 (). 9) A point is inside a circle if its distance from the centre of the circle is less than the radius of the circle. 10) A point is on a circle if its distance from the centre of the circle is equal to the radius of the circle. 11) A point is outside a circle if its distance from the centre of the circle is greater than the radius of the circle. 12) Angles standing on the same arc of a circle are equal. 13) The angle at a circle standing on a diameter is a right angle. 14) A tangent to a circle is perpendicular to the radius drawn to the point of contact. 15) The angle at the centre of a circle is twice the angle at the circle, which is standing on the same arc. 16) In a cyclic quadrilateral, opposite angles add up to 180o. 17) A radius which is perpendicular to a chord of a circle bisects that chord. 18) A line drawn perpendicularly from the centre of a circle on to any chord will bisect that chord. 19) The tangent to the circle 𝒙2 + 𝒚2 = 𝒓2 at the point (𝒙1, 𝒚1) is given by 𝒙𝒙1 + 𝒚𝒚1 = 𝒓2 (not needed). © H DORGAN IOE LC Maths HL 16 20) The perpendicular distance from the centre of a circle to any tangent of that circle is equal to the radius. 21) The slope of a tangent will be the negative inverse of the slope of the radius to the circle at the point of contact. 22) When finding the equation of two tangents from a given point outside a given circle, let m be the required slope of the tangent, find the equation of the tangent in terms of m, and use the fact that the perpendicular distance from the centre of the circle to the tangent is equal to the radius. 23) When solving equations using the equation 𝒙2 + 𝒚2 + 𝟐𝒈𝒙 + 𝟐𝒇𝒚 + 𝒄 = 𝟎, it is useful to know that (i) if (0, 0) is on the circle, then c = 0 (ii) if the centre of the circle is on the x-axis, then f = 0; if the centre of the circle is on the y-axis, then g = 0 (iii) if the x-axis is a tangent to the circle, then r = |𝒇| ⇒𝒄 = 𝒈𝟐 (iv) if the y-axis is a tangent to the circle, then r = |𝒈| ⇒𝒄= 𝒇𝟐 (v) if both axes are tangents to the circle, then r = |𝒈| = |𝒇| ⇒ 𝒄 = 𝒈𝟐= 𝒇𝟐 (vi) if the centre of the circle lies on the line a𝒙 + b𝒚 + c = 0, then 𝑎(−𝑔) + 𝑏(−𝑓) + 𝑐= 0 ⇒𝑎𝑔+ 𝑏𝑓−𝑐= 0. 24) When circles touch externally, |𝒄𝟏 𝒄𝟐| = 𝒓𝟏 + 𝒓𝟐 . () 25) When circles touch internally, |𝒄𝟏 𝒄𝟐| = 𝒓𝟏− 𝒓𝟐, () where r1 is the radius of the bigger circle. © H DORGAN IOE LC Maths HL 17 NOTES |
12687 | https://www.youtube.com/watch?v=O7PgogMB-Rg | Differential Manometers Example Problems - Pressure Difference Measurement - Fluid Mechanics
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Posted: 26 Aug 2020
Differential Manometers Example Problems - Pressure Difference Measurement - Fluid Mechanics - Hydraulics - Civil Engineering
In this video, we work through two example problems, calculating the pressure difference using a closed differential manometer and an inverted, closed differential manometer.
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Introduction welcome everyone to this video by learn civil engineering where we will be working through a couple of example problems on the use of differential manometers to measure pressure differences within fluids if you haven't already seen the last video on the use of standard manometers to measure pressure in a single pipeline go and check it out as it covers the fundamental theory that we'll be using to solve these problems today Example Problem 1 for the first problem then we have a differential manometer used to find the difference in pressure between two separate pipelines two separate pipelines each convey a gas of density rho a differential manometer containing a monometric liquid of density rho m is used to measure the pressure difference p a minus pb between the point a in pipeline one and the point b in pipeline two this is done by connecting one end of the manometer to pipeline one at point a and then the other end to pipeline two at point b as shown in the diagram the pressure difference pa minus pb can be determined by measuring the heights h1 and h2 show that pa minus pb is equal to rho m minus rho times by h1 minus h2 times by g and hence find the difference in pressure when rho equals 1.2 kilograms per meter cubed rho m equals 800 kilograms per meter cubed h1 equals 58 millimeters and h2 equals 176 millimeters pause the video here if you'd like to attempt the problem before we work through it firstly we must identify an elevation which we will denote x at which the pressure px can be related to the pressures p a and p b so we will take x to be the elevation of the lower interface of the monometric liquid in the right hand side of the manometer therefore in the right hand side we have px is equal to pb plus rho g h2 and in the left hand side we have px is equal to pa plus rho g h1 plus rho m g times by h2 minus h1 equating these two equations gives pa minus pb is equal to rho gh2 minus rho gh1 minus rho mg times by h2 minus h1 and then simplifying this we can show that pa minus pb is equal to rho m minus rho times by h1 minus h2 times by g as required and then to finish off the question substituting rho is equal to 1.2 kilograms per meter cubed rho m is equal to 800 kilograms per meter cubed h1 equals 0.058 meters and h2 equals 0.176 meters this gives pa minus pb is equal to 800 minus 1.2 times by 9.81 times by 0.058 minus 0.176 which is equal to minus 925 newtons per meter squared so the pressure of the gas in pipeline 2 at point b is 925 newtons per meter squared greater than the pressure at point a in pipeline 1. Example Problem 2 moving on to the next question we have an inverted differential manometer where the monometric liquid lies on top of the pipeline fluid as the density of the monometric liquid is now less than that of the pipeline fluid a pipeline conveys a liquid of specific weight gamma an inverted differential manometer is used to measure the pressure difference pa minus pb between two points a and b located either side of a construction which is just a point in the pipeline where the diameter is reduced the monometric liquid has specific weight gamma m and here the manometer is inverted and so the pipeline liquid lies below the monometric liquid inside the tube hence as described before in this case we must have gamma m is less than gamma or rho m is less than rho the pressure difference pa minus pb is determined by measuring h1 h2 and h3 show that pa minus pb is equal to gamma times h1 minus h2 plus gamma m times by h2 minus h3 and then find the pressure difference when gamma is equal to 9800 newtons per meter cubed gamma m is equal to 0.8 gamma h1 is equal to 23.1 centimeters h2 equals 10.8 centimeters and h3 equals 24.3 centimeters again you can pause the video here if you would like to attempt the problem before we work through it firstly we must identify an elevation which we will donate x at which the pressure px can be related to the pressures pa and pb taking x to be the elevation of the upper interface of the monometric liquid in the left hand side of the manometer hence in the left hand side we have pa is equal to px plus gamma h1 and in the right hand side we have pb is equal to px plus gamma m times by h3 minus h2 plus gamma h2 then eliminating px results in pa minus pb equals gamma h1 minus gamma m times h3 minus h2 minus gamma h2 and by simplifying this like we did for the previous question we can show that pa minus pb is equal to gamma times by h1 minus h2 plus gamma m times by h2 minus h3 as required and then to finish off substituting gamma equals 9810 newtons per meter cubed gamma m equals 7848 newtons per meter cubed h1 equals 0.231 meters h2 equals 0.108 meters and h3 equals 0.243 meters gives pa minus pb is equal to 9810 times by 0.231 minus 0.108 plus 7848 times by 0.108 minus 0.243 is equal to 147 newtons per meter squared so the pressure of the gas at point b in the pipeline is 147 newtons per meter squared less than the pressure at point a in the pipeline well done if you solve both of these problems and if you didn't manage to i hope after working through them together you now have a better understanding of them this has been a video by learned civil engineering if you have found this video useful at all please leave a like and subscribe to the channel to show your support if you do have any remaining questions or would like me to cover a specific topic please leave them in the comment section below and i'll try to respond as soon as possible thank you for watching |
12688 | https://math.stackexchange.com/questions/2708719/vector-equation-of-cylinder-given-radius-around-specified-line | linear algebra - Vector equation of cylinder given radius around specified line - Mathematics Stack Exchange
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Vector equation of cylinder given radius around specified line
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I’m a little bit stumped on the following problem.
Write down a vector equation for the cylinder in 3-dimensional space of radius r around the line described by
{(t,2 t,3 t):t∈R}{(t,2 t,3 t):t∈R}
Could anyone point out what I should do? I’m thinking about finding end-points using parametrics—something like
x=r cos θ,y=r cos θ,z=z x=r cosθ,y=r cosθ,z=z
But I’m not sure. Thanks! (:
linear-algebra
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asked Mar 26, 2018 at 12:40
MickeyMickey
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Hint: compute the distance between a point and the line. The distance for points in the cylinder must be r r Exodd –Exodd 2018-03-26 12:47:50 +00:00 Commented Mar 26, 2018 at 12:47
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As a comment from @Exodd suggests, you can establish a coordinate equation by computing the distance from the line, this distance is r r for all points on the cylinder.
A vectorial approach is as follows. Let u⃗=(1,2,3)u→=(1,2,3) be the direction of the cylinder and let v⃗v→ be a unitary vector, perpendicular to u u. Then every vector z⃗z→ pointing a point on the cylinder can be relied to the origin by the equations
z⃗=t u⃗+r v⃗z→=t u→+r v→
u⃗⋅v⃗=0,||v⃗||=1 u→⋅v→=0,||v→||=1
Note also that ⟨u⃗⟩⊥⟨u→⟩⊥ is the plane generated by (−2,1,0),(−3,0,1)(−2,1,0),(−3,0,1). If you give an orthonormal base instead, you can write in a single equation what you want.
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edited Mar 26, 2018 at 23:29
answered Mar 26, 2018 at 13:04
Tal-BotvinnikTal-Botvinnik
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Thank you! This makes a lot of sense (:Mickey –Mickey 2018-03-26 15:38:32 +00:00 Commented Mar 26, 2018 at 15:38
Thanks, corrected Tal-Botvinnik –Tal-Botvinnik 2018-03-26 23:30:10 +00:00 Commented Mar 26, 2018 at 23:30
@Mickey, I corrected the answer given amd's comment. You can find an orthonormal base and write as I did before. And please accept an answer to avoid going into the unanswered section Tal-Botvinnik –Tal-Botvinnik 2018-03-26 23:33:41 +00:00 Commented Mar 26, 2018 at 23:33
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You’re starting off in the right direction by thinking about tilting the cylinder x 2+y 2=r 2 x 2+y 2=r 2 so that its axis lines up with the direction vector (1,2,3)(1,2,3) of the given line. A vector parameterization of this equation is
r cos t(1,0,0)+r sin t(0,1,0)+s(0,0,1)(1)(1)r cost(1,0,0)+r sint(0,1,0)+s(0,0,1)
so one might expect the tilted version to also have the form
r cos t u+r sin t v+s w.(2)(2)r cost u+r sint v+s w.
Obviously we can take w=(1,2,3)w=(1,2,3) (we don’t care about the change of scale in this direction), but what to use for the other two? For a fixed s s, we want (2) to represent a cross-section of the cylinder taken perpendicular to its axis, so u u and v v must be orthogonal to w w. In order to avoid distortions and get a circle of the correct size as this cross-section, they must be unit vectors. So, u u can be any convenient unit vector that’s orthogonal to w w and v=1∥w∥w×u v=1‖w‖w×u also satisfies the criteria. These vectors form an orthogonal basis aligned with the cylinder.
For any nonzero vector (a,b,c)(a,b,c), at least two of (0,c,−b)(0,c,−b), (−c,0,a)(−c,0,a) and (b,−a,0)(b,−a,0), which are all orthogonal to (a,b,c)(a,b,c), are also nonzero. For our w w, the choices are (0,3,−2)(0,3,−2), (−3,0,1)(−3,0,1) and (2,−1,0)(2,−1,0). None of these result in particularly nice-looking unit vectors, so let’s just take u=1 10√(−3,0,1)u=1 10(−3,0,1), giving v=1 35√(1,−5,3)v=1 35(1,−5,3). A possible vector parameterization of the cylinder is therefore
1 10−−√r cos t(−3,0,1)+1 35−−√r sin t(1,−5,3)+s(1,2,3).(3)(3)1 10 r cost(−3,0,1)+1 35 r sint(1,−5,3)+s(1,2,3).
As a sanity check, an implicit Cartesian equation for the cylinder can be obtained in a similar fashion. Reflecting in the angle bisector of the z z-axis and the given line will tilt the reference cylinder’s axis in the right way. We can use a reflection because of the cylinder’s symmetry, and I find a reflection more convenient than a rotation for this because the reflection matrix is both symmetric and is its own inverse, so I don’t have to remember which combination of inverse and transpose is required to transform the conic. As well, I don’t have to make any decisions about other degrees of freedom as I would for a rotation. Leaving out the details of this computation, the resulting equation is
13 x 2+10 y 2+5 z 2−4 x y−3 x z−6 y z=14 r 2.13 x 2+10 y 2+5 z 2−4 x y−3 x z−6 y z=14 r 2.
The parameterization obtained above does indeed satisfy this equation.
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answered Mar 31, 2018 at 0:16
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12689 | https://www.mathcelebrity.com/search.php?q=tan%2850%29 | Crop Image
tan(50)
Answer
↓Steps Explained:↓
↓Steps Explained:↓
Calculate tan(50)
tan is found using Opposite/Adjacent
Determine quadrant:
Determine quadrant:
Since 0 ≤ 50 ≤ 90 degrees
it is in Quadrant I
sin, cos and tan are positive.
Determine angle type:
Determine angle type:
50 < 90°, so it is acute
tan(50) = 1.1917535901808
Write tan(50) in terms of cot
Write tan(50) in terms of cot
Since 50° is less than 90...
We can express this as a cofunction
tan(50) = cot(90 - 50)
tan(50) = cot(40)
Special Angle Values
Special Angle Values
θ° | θrad | sin(θ) | cos(θ) | tan(θ) | csc(θ) | sec(θ) | cot(θ) || 0° | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
| 30° | π/6 | 1/2 | √3/2 | √3/3 | 2 | 2√3/3 | √3 |
| 45° | π/4 | √2/2 | √2/2 | 1 | √2 | √2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 | 2√3/3 | 2 | √3/3 |
| 90° | π/2 | 1 | 0 | N/A | 1 | 0 | N/A |
| 120° | 2π/3 | √3/2 | -1/2 | -√3 | 2√3/3 | -2 | -√3/3 |
| 135° | 3π/4 | √2/2 | -√2/2 | -1 | √2 | -√2 | -1 |
| 150° | 5π/6 | 1/2 | -√3/2 | -√3/3 | 2 | -2√3/3 | -√3 |
| 180° | π | 0 | -1 | 0 | 0 | -1 | N/A |
| 210° | 7π/6 | -1/2 | -√3/2 | √3/3 | -2 | -2√3/3 | √3 |
| 225° | 5π/4 | -√2/2 | -√2/2 | 1 | -√2 | -√2 | 1 |
| 240° | 4π/3 | -√3/2 | -1/2 | √3 | -2√3/3 | -2 | √3/3 |
| 270° | 3π/2 | -1 | 0 | N/A | -1 | 0 | N/A |
| 300° | 5π/3 | -√3/2 | 1/2 | -√3 | -2√3/3 | 2 | -√3/3 |
| 315° | 7π/4 | -√2/2 | √2/2 | -1 | -√2 | √2 | -1 |
| 330° | 11π/6 | -1/2 | √3/2 | -√3/3 | -2 | 2√3/3 | -√3 |
θ° | θrad | sin(θ) | cos(θ) | tan(θ) | csc(θ) | sec(θ) | cot(θ) || 0° | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
| 30° | π/6 | 1/2 | √3/2 | √3/3 | 2 | 2√3/3 | √3 |
| 45° | π/4 | √2/2 | √2/2 | 1 | √2 | √2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 | 2√3/3 | 2 | √3/3 |
| 90° | π/2 | 1 | 0 | N/A | 1 | 0 | N/A |
| 120° | 2π/3 | √3/2 | -1/2 | -√3 | 2√3/3 | -2 | -√3/3 |
| 135° | 3π/4 | √2/2 | -√2/2 | -1 | √2 | -√2 | -1 |
| 150° | 5π/6 | 1/2 | -√3/2 | -√3/3 | 2 | -2√3/3 | -√3 |
| 180° | π | 0 | -1 | 0 | 0 | -1 | N/A |
| 210° | 7π/6 | -1/2 | -√3/2 | √3/3 | -2 | -2√3/3 | √3 |
| 225° | 5π/4 | -√2/2 | -√2/2 | 1 | -√2 | -√2 | 1 |
| 240° | 4π/3 | -√3/2 | -1/2 | √3 | -2√3/3 | -2 | √3/3 |
| 270° | 3π/2 | -1 | 0 | N/A | -1 | 0 | N/A |
| 300° | 5π/3 | -√3/2 | 1/2 | -√3 | -2√3/3 | 2 | -√3/3 |
| 315° | 7π/4 | -√2/2 | √2/2 | -1 | -√2 | √2 | -1 |
| 330° | 11π/6 | -1/2 | √3/2 | -√3/3 | -2 | 2√3/3 | -√3 |
Show Unit Circle
Show Unit Circle
Final Answer
Final Answer |
12690 | https://quillbot.com/blog/rhetoric/rhetorical-devices/ | Rhetorical Devices | Examples, Definition & List
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Rhetoric
Rhetorical Devices | Examples, Definition & List
Rhetorical Devices | Examples, Definition & List
Published on September 27, 2024 by Kassiani Nikolopoulou, MSc Revised on September 7, 2025
Rhetorical devices are linguistic tools used by speakers and writers to make their arguments more compelling. These devices involve playing with sentence structure, sound, or meaning to evoke a particular reaction from the audience and ultimately persuade them. We often encounter rhetorical devices in public speaking and written communication.
Rhetorical devices example
Her smile was like sunshine on a cold winter day. [simile]
All birds lay eggs; a penguin is a bird; therefore a penguin lays eggs. [syllogism]
The project fell apart faster than the wings of Icarus as he flew too close to the sun. [allusion]
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Table of contents
What are rhetorical devices?
Rhetorical devices examples
Rhetorical devices list
Frequently asked questions about rhetorical devices
What are rhetorical devices?
Rhetorical devices are techniques used to engage and persuade an audience. These devices help shape structure, tone, and communication style, making the message more impactful.
Rhetoric, the art of persuasion, has been studied since ancient times, particularly by philosophers like Aristotle, who identified key strategies for effective argumentation, such as the rhetorical triangle. Public speaking developed in many places worldwide; however, much of what we know about it came to English speakers from the ancient Greeks. For this reason, most rhetorical devices have names that come from Greek.
Rhetorical devices are some of the methods that writers and speakers can use to apply the three rhetorical strategies or modes of persuasion—ethos, pathos, and logos—in their speeches or written works. These devices help craft arguments that are credible (ethos), emotionally engaging (pathos), and logically sound (logos).
For example, a speaker might use asimileto appeal to pathos, creating emotional connection. Similarly, syllogisms and enthymemes strengthen logos, whileallusions to respected sources can bolster ethos. However, there are other strategies and methods that can also be used to make these appeals, such as using facts and figures or nonverbal communication.
Note
Rhetorical devices as a category overlaps partially with literary devices. For example, metaphor, personification, and euphemism (and many others) can all be used as either rhetorical devices or literary devices. The difference is in their purpose:
Literary devices are used to add artistry or interest to a piece of writing.
Rhetorical devices are used to persuade an audience or evoke emotion.
Rhetorical devices examples
Some rhetorical devices rely on the repetition of sounds, while others rearrange the order or meaning of words.
Anadiplosis
Anadiplosis is the repetition of the final words of a sentence or line at the beginning of the next. It is often used to link a series of ideas in a memorable way.
Anadiplosis example:Gladiator(2000)
“They call for you: The general who became aslave; theslave who became a gladiator; the gladiatorwho defied an Emperor.”
Anaphora
Anaphora is the repetition of a word or phrase at the beginning of successive phrases, sentences, or verses, for emphasis.
Anaphora example: “I Have a Dream” speech by Dr. Martin Luther King Jr.
Go back to Mississippi, go back to Alabama, go back to South Carolina,
go back to Georgia, go back to Louisiana, go back to the slums and ghettos of our northern cities, knowing that somehow this situation can and will be changed.
Assonance
Assonanceis the repetition of similar vowel sounds in nearby words. The repeated sound usually occurs in the middle of the word. Assonance helps to create a pleasant rhythm between words.
Assonance example: “The Bells” by Edgar Allan Poe
Hear the mellow wedding bells,
Golden bells!
Chiasmus
Chiasmus involves reversing the order of grammatical structures or concepts in a phrase, creating an A-B-B-A pattern, without necessarily repeating the same words.
Chiasmus example: Paradise Lost by John Milton
“Divine compassion visibly appeared
Love without end, and without measure grace”
Here, “love” and “grace” reflect related concepts, while “without end” and “without measure” both convey the idea of infinity.
Hyperbole
Hyperbole is the use of exaggerated language to create emphasis or add humor to a statement. Hyperbolic statements effectively communicate the intensity of a situation, emotion, or characteristic.
Hyperbole examples
Once you try our pizza, you will never want to eat anything else.
I am so tired I couldsleep for a week.
The show was so good, we were dying of laughter.
Hypophora
Hypophora is a rhetorical device in which a speaker or writer poses a question and immediately provides an answer.
Hypophora example:Palm Sunday: An Autobiographical Collageby Kurt Vonnegut
“What should young people do with their lives today? Many things, obviously. But the most daring thing is to create stable communities in which the terrible disease of loneliness can be cured.”
Litotes
Litotes is a form of understatementin which a sentiment is expressed by denying its opposite, such as “not bad” to mean “good.” This rhetorical device often expresses irony or softens a statement.
Litotes examples
This wasn’t half bad. [to indicate it was good]
It wasn’tthe worst dish I’ve had. [implying it was good]
The report is not without its errors. [to indicate that there are mistakes]
Metaphor
A metaphor is an implicit comparison between two unrelated things, where one is described as if it were the other to highlight a shared characteristic or quality.
Metaphor examples
Today’s meetingwas a total nightmare.
Our nation’s economyis the engine that drives our success.
Lifeis a journey, so enjoy each step along the way.
Metonymy
Metonymy is the substitution of the name of one thing for that of another to which the former is closely related. Metonymy helps create concrete and vivid images in place of generalities.
Metonymy examples
Wall Streetreacted sharply to the news. [the U.S. financial and banking sector]
The press has been covering the story all week. [the news media]
We need to get more boots on the ground for this operation. [soldiers]
Onomatopoeia
Onomatopoeiais the use of words that mimic the sound of the thing they describe, such as animal or machine sounds.
Onomatopoeia example
The dripof the faucet kept me awake all night long.
The bees buzzed around the flowers.
The thunderrumbled in the distance.
Oxymoron
An oxymoron is the combination of two apparently contradictory terms to produce an expression that conveys a new complex meaning. Because oxymorons seem absurd at first, they invite readers or members of an audience to pause and think about their meaning.
Oxymoron examples
The teacher’s questions were met withdeafening silence.
Our affordable luxurycollection gives you the best of both worlds.
He tried to act naturally while giving his presentation, but his nerves got the best of him.
Personification
Personification is giving human attributes to non-human animals, inanimate objects, or abstract concepts.
Personification example
The leaves danced joyfully in the breeze.
The sun smiled down on the children playing in the park.
My laptop does not want to cooperate today.
Pleonasm
Pleonasm is the use of more words than necessary to express an idea. While typically considered a stylistic fault, pleonasm can be used deliberately for emphasis.
Pleonasm examples
I saw it with my owneyes.
It is a true fact that he was there.
She made a final conclusion.
Polysyndeton
Polysyndeton is the use of multiple conjunctions like “and,” “or,” and “but” in close succession within a sentence, particularly where they could have been omitted.
Polysyndeton example
We laughed andcried andscreamed and shouted all at once.
The repeated use of “and” is not necessary here, but it emphasizes the intensity of the emotions.
Rhetorical devices list
Here is a list of some common rhetorical devices with definitions and examples.
| Rhetorical device | Definition | Example |
---
| Anadiplosis | Repetition of the final words of a sentence or line at the beginning of the next. | “When I give, I give myself” —”Song of Myself” by Walt Whitman |
| Anaphora | Repetition of a word or phrase at the beginning of successive sentences. | “[…] we shall fight on the beaches, we shall fighton the landing grounds, we shall fightin the fields and in the streets […] —Winston Churchill |
| Assonance | Repetition of similar vowel sounds in nearby words. | Keep your eyes on the prize. |
| Chiasmus | Reversal of the structure of phrases or clauses in a mirrored arrangement (A-B-B-A). | “Who dotes [A], yet doubts [B]; suspects [B], yet strongly loves [A]” —Othello by Shakespeare |
| Hyperbole | Exaggeration used for emphasis or effect. | Ittook us ages to find the place. |
| Hypophora | A question followed directly by an answer. | What makes a good leader? A good leader inspires and motivates others … |
| Litotes | Using double negatives or negation to affirm a positive meaning. | It wasnot too shabby. |
| Metaphor | A direct comparison between two unlike things without using “like” or “as.” | The newswas music to my ears. |
| Metonymy | Replacing the name of an object or concept with a word closely related to the original. | That is a nice ride you have |
| Onomatopoeia | A word that imitates the sound of the thing it represents. | The chickens clucked as they dug the dirt for worms. |
| Oxymoron | Placing contradictory or opposite words together for effect. | Parting is such sweet sorrow. |
| Personification | Attribution of human characteristics or qualities to non-human entities or abstract concepts. | My battery died. |
| Pleonasm | Using redundant words to express an idea, often for effect. | Thanks to our advanced planning, we delivered the project on time. |
| Polysyndeton | Deliberate use of multiple conjunctions in close succession to create an effect. | Over the weekend, we went to the parkand the movies and the swimming pool. |
Frequently asked questions about rhetorical devices
What is the difference between literary devices and rhetorical devices?
Literary devices andrhetorical devices are closely related, but they differ in their purpose and use:
Literary devices encompass a broad category of artistic techniques that writers use to enhance their writing, adding interest and depth. For example, devices like irony or metaphor and narrative styles like first-person point of view fall under this category.
Rhetorical devices, on the other hand, are techniques used to persuade and evoke emotion. Examples include metonymy, appeal to emotion, and understatement. Although these can be used as literary devices, we often encounter them in various forms of communication, such as speeches, advertisements, and debates where their primary role is to influence decision-making.
In short, literary devices are used to enrich a narrative, whereas rhetorical devices are more focused on persuasion and emotional appeal. However, literary and rhetorical devices sometimes overlap (e.g., irony).
What are some rhetorical devices examples?
The following are some common rhetorical devices with examples:
Metaphor: an implicit comparison between two unlike things (e.g., “He is a night owl. He prefers working at night.”)
Hyperbole: An exaggerated statement, not to be taken literally (e.g., “I was bored to tears during the lecture.”)
Synecdoche: A part is used to represent a whole or vice versa (e.g., “I got a new set of wheels.”)
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Kassiani Nikolopoulou, MSc
Kassiani has an academic background in Communication, Bioeconomy and Circular Economy. As a former journalist she enjoys turning complex information into easily accessible articles to help others.
Rhetoric
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Rhetorical triangle
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12691 | https://www.math.utoronto.ca/mccann/199/ass6.pdf | Cutting space with planes IDIS 303 Oct 24 2006 cutplane 1 2-D lines regions 0 1 1 2 2 4 3 7 4 11 5 16 6 22 3-D planes regions 0 1 1 2 2 4 3 8 4 15 3-D planes regions 0 1 1 2 2 4 3 8 4 15 Suppose I cut space with a plane. How many regions do I get? The answer is 2. Suppose I now use 2 planes––how many regions? Well, it depends––if the planes are parallel, there will be 3, otherwise 4. To remove this ambiguity, let's agree that our planes should be randomly chosen, or "in general position", so that coincidences do not occur. Then the answer is 4 regions. Now try 3 planes, again in general posi-tion. The answer is 8 regions––indeed, we can think of the three planes x=0, y=0 and z=0 in the Cartesian coordinate system. The regions which these planes create are called "octants" because there are eight of them. It's a bit harder to visualize 4 planes, and the question touches off a murmur of debate in the classroom. Can all 8 of the existing regions be cut into by the new plane? The answer is no, but perhaps the argument is not so easy to pin down. But 7 of them can certainly be sliced. One nice construction is to take the three coordinate planes as above, and then to slice everything with the plane that cuts the three axes at a dis-tance +1 from the origin. This cuts a tetrahedron out of the positive oc-tant, and cuts through all the other octants except the "negative" one–– x, y and z all negative. So 7 of the 8 octants are "cut in half," giving 15 regions. The question is, what is the pattern? That is, how does the table at the right continue? I give the class a few moments to play with 5 planes, but things are very hard to visualize. Can the new plane pass through one of the four exist-ing intersection points? No, that would be a coincidence. With 4 planes, one of the 15 regions is bounded, so it’s different from the oth-ers. Does it make a difference whether the fifth plane goes through the bounded region? Donno—that’s a good question. One way of coping with a hard problem is to solve an easier one first. In this case, if we go down to two dimensions we get an easier problem because pictures can be more easily drawn. So our base space is now a plane, and the objects we cut it with are—lines! So let's ask how many regions are created in a plane by cutting it with lines. Again we want the lines to be in general position, so we don't have any parallel lines or concurrencies. It is not hard to produce the following table. The case of 5 lines is illustrated at the right. Now what's the pattern here? The class is quick to notice the interesting difference pattern––successive differences in the right column of the 2-D table are the integers, 1,2,3,4,5, and 6. By this reckoning the next number should be 29, and a careful picture will verify this. But the im-portant task of course is to find a way to “see” why this pattern is ex-pected. Alas, the class is not really interested in that question—at least not right away. What they clamour to do is to look at the difference pattern in the MAT 199 Aha! ... Assignment #6 Due on Crowdmark 2 pm Thurs Mar 4, 2021 3-D table. This turns out to be the sequence 1,2,4, and 7. Now that's really interesting: these are the first four numbers of the 2-D table! Dare we conjecture that this pattern continues––that therefore the next num-ber of the 3-D table should be 15+11=26? Do 5 planes cut space up into 26 regions? To answer this question, we really do need some new understanding. Why should the 2-D table provide the set of differences for the 3-D ta-ble? 2-D lines regions 0 1 1 2 2 4 3 7 4 11 5 16 6 22 A powerful idea is to go down one dimension. Let's look again at the simple difference pattern for the 2-D table—where does that nice set of 1,2,3,4… really come from? What really are those differences? For example, let’s focus on the transition between 5 and 6 lines in the 2-D table. That causes the number of regions to change from 16 to 22, a difference of 6. Now where do those 6 new regions really come from? Well, if we study the picture, those 6 regions are created by the sixth line slicing through 6 of the old regions and cutting them “in half.” But why 6? Is it because it’s the 6th line? Not really. If we stare hard at the picture, we see it is because the 6Pth line is divided into 6 segments by its intersections with the 5 existing lines, and each of these segments represents a cutting of one of the existing regions in half! So what that 6 really is, is the number of segments created on a line by 5 points. The reason I want to say it this way, is because now we see that this is really the same problem as before but one more di-mension down. This is really the problem of how many regions are cre-ated on a line if it is “cut” by n points. Of course! If the 2-D problem somehow provides the differences for the 3-D table, we might expect the 1-D problem to provide the differences for the 2-D table. So let’s formulate the 1-D problem. What do we cut 1-dimensional space with?––well with points! If we place n points on a line the number of regions (which are really intervals now) we get is n+1. The table is at the right. And what we have noticed above is the reason that this table provides the differences for the 2-D table. For example, the difference between 16 and 22 in the 2-D table is 6, but exactly what is that 6? (because there are lots of 6’s around here)—it’s the 6 in the “regions” column of the 1-D table which sits in the n=5 row. That’s how to think of that 6. 1-D points regions 0 1 1 2 2 3 3 4 4 5 5 6 6 7 1-D points regions 0 1 1 2 2 3 3 4 4 5 5 6 6 7 2-D lines regions 0 1 1 2 2 4 3 7 4 11 5 16 6 22 cutplane 2 Now this should show us the way to argue that the differences in the 3-D table are provided by the 2-D table. Let's see if we can run the whole argument through one dimension up. Take, for example, the addition of the fifth plane. We want to know how many new regions it creates. Well, this plane will intersect all the 4 previous planes (no parallelism) and these intersections will etch a pattern of 4 lines on the plane. These 4 lines must cut the plane into 11 regions (from the 2-D table) and each of these regions really represents a "cutting in half" (by the 5th plane) of each of the original regions in 3-space. So the 5th plane should create 11 new regions in space––that "11" being the n=4 entry of the 2-D table. So to get the n=5 entry of the 3-D table from the n=4 entry, we need to add the n=4 entry of the 2-D table. There it is! 3-D planes regions 0 1 1 2 2 4 3 8 4 15 5 26 6 42 It all works out quite beautifully. The 1-D table provides the differences for the 2-D table, and the 2-D table provides the differences for the 3-D table. With this we deduce that 5 planes will indeed cut space into 26 regions, and we can keep going and find the answer for 6 planes, and 7 planes, etc. This is a superb example of the power of pattern recognition, analogical reasoning, and persis-tence in getting at the heart of the matter. This was a hard problem! But we have arrived at a convincing analysis. I stand regarding the blackboard with awe. The students are silent. They can see how the tables can now be generated. Is it really that easy? It's important to emphasize that along with this computational scheme, we also have an elegant proof that it works––a way to argue with absolute certainty that this scheme will calculate the correct entries. For example, let’s go back to our small uncertainties with four planes in 3-space. We know that three planes will determine 8 regions and we now have a nice argument that the 4th plane can't cut through all 8 of the existing regions. Because if it did, we'd contradict the 2-D table–– we'd have a plane with three lines on it defining 8 regions. And in the same way the 2-space table must be right because we know we can trust the 1-space table. 4-D hyper-planes regions 0 1 1 2 2 4 3 8 4 16 5 31 6 57 Okay, what's next? Another silence from the class, but this one expec-tant. A voice from the back catches on: What about 4-space? Right! Suppose I slice 4-space with 6 hyperplanes: how many regions? Well it would seem to be an almost impossible problem, but the argu-ments behind our results are quite general and would apply to the transi-tion from 3 dimensions to four. Using the entries of the 3-D table as differences, we get the 4-D table at the right. With 6 hyperplanes, we get 57 regions. cutplane 3 This lovely problem was used by the famous mathematician and teacher George Polya in his movie "Let's Teach Guessing". Problems 1. Find a formula for the nth entry of the 2-D and 3-D tables. 2. Suppose I draw 20 circles in the plane, all passing through the origin, but no two tan-gent at the origin. Also, except for the origin, no three circles pass through a common point. How many regions are created in the plane? 3. Take a sphere and draw on it a great circle (a great circle is a circle whose centre is the centre of the sphere). There are two regions created. Here, I am referring to regions on the surface of the sphere. Now draw another great circle: there are four regions. Now draw a third, not passing through the points of intersection of the first two. How many re-gions? Here's the general question: How many regions are created by n great circles, no three concurrent, drawn on the surface of the sphere? cutplane 4 NOT TO BE HANDED IN: Read Sections 7.1-7.2 of Heart' 4th ed. (= sections 6.1-6.2 of 3rd ed.) Do section 7.2 #14 of 4th ed (= 6.2 #14 of 3rd ed.) Do section 1.1 story #6A tight weave' BE PREPARED, AS USUAL, TO BE QUIZZED ON THIS MATERIAL This handout reproduced from `In process' with permission of its author, Peter Taylor of Queen's University. TO BE HANDED IN: #1, 2, 3 above. |
12692 | https://www.sciencedirect.com/science/article/abs/pii/S1349007924000756 | Skip to article
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Abstract
Introduction
Section snippets
References (83)
Cited by (6)
Journal of Oral Biosciences
Volume 66, Issue 2, June 2024, Pages 272-280
Revolutionizing the diagnosis of irreversible pulpitis Current strategies and future directions
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Highlights
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Determining pre-treatment pulp inflammation is crucial amid resurgence of vital pulp therapy.
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In clinical practice, ambiguous data renders diagnosis of pulpal status challenging, compromising success of treatment.
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Pulp tissue secretes multitude biological products to counter external adverse stimuli termed biomarkers of pulpitis.
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Molecular changes precede macroscopic andmicroscopic changes, providing platforms for early diagnosis, improving clinical outcomes.
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Molecular diagnosis of the biomolecular profile can help invent non-invasive tools for adoption to clinical practice.
Abstract
Background
Pulpitis primarily arises from the pulp space infection by oral microbiota. Vital pulp therapy is a minimally invasive approach that relies on assessing the severity of pulpal inflammation to facilitate repair. However, the current evaluation methods prescribed by the American Association of Endodontics are subjective, leading to ambiguity in assessment. Therefore, this review aims to explore molecular strategies for evaluating the severity of pulpal inflammation to accurately predict the success of pulp vitality preservation in clinical settings.
Methodology
This review was conducted by searching relevant keywords, such as irreversible pulpitis, pulpitis biomarkers, molecular diagnosis, inflammation, and genomic strategies, in databases such as PubMed, Web of Science, and Scopus to address the subjective nature of diagnosis. The data included in this review were collected up to April 2023. The literature search revealed well-documented limitations in clinically assessing the pulp inflammatory. Molecular approaches that aid in clinical differentiation between irreversible and reversible pulpitis may potentially enhance favorable outcomes in vital pulp therapy. Non-invasive diagnostic methods for pulpal assessment would also be valuable for determining whether the inflamed pulp is reversible, irreversible, or necrotic.
Conclusion
The present review examines the various molecular diagnostic approaches that have revolutionized the medical field and are considered the most promising empirical methodologies for the proactive detection of pulpal diseases. It also provides comprehensive insights into the current diagnostic methods, associated challenges, next-generation strategies, and future directions for diagnosing the severity of pulp inflammation.
Introduction
Dental pulp is a form of connective tissue that is contained within the pulp cavity and encased in hard dentin and enamel. It plays a fundamental role in providing nutrition, innervation, and immunocompetency to the teeth. However, owing to the lack of collateral circulation, any breach in dental hard tissue may stimulate various pathological changes in the pulp, compromising its healing . Pulpitis is an inflammatory state of the pulp tissue, with microbial infection speculated to be its primary genesis. The invasiveness of traditional dental treatments, concerns about overtreatment, and the restorative cycle have led the profession to advocate for minimally invasive biologically based therapies. The likelihood of pulp exposure has decreased because of the switch from nonselective to selective removal of caries . Vital pulp therapy (VPT) techniques such as partial pulpotomy and total pulpotomy have seen an upsurge with an increase in contemporary strategies for the management of caries-exposed pulp . The importance of these developments to the profession of endodontics cannot be emphasized sufficiently because preserving pulpal health and preventing apical periodontitis are fundamental tenets of the specialty .
According to Nyvad et al., caries is a biofilm-activated disease fueled by a source of fermentable carbohydrates . The ensuing ecological shift produces a cariogenic and acidogenic niche, which leads to the degradation of the hard tissues of teeth and development of caries . Although pulpitis exists throughout the progression of the carious process, the inflammatory response does not become severe unless the demineralized infected carious dentin is close to the pulp with the probability of bacteria entering the pulp. According to Mjör et al., pulp tissue in animal models involving healthy dentin has an inherent capability to recuperate if the microbial challenge is eliminated and the tooth is effectively healed . The primary, secondary, and tertiary dentinogenesis, immune response, and mechanoreceptors of the pulp are maintained by conserving pulpal health.As a therapeutic alternative to pulpectomy and root canal therapy, vital pulp therapy procedures are quicker, less technique-sensitive, and less invasive, while minimizing side effects, such as discoloration, fracture, or lingering periapical irritation. Nevertheless, it should be emphasized that VPT for deep carious lesions are frequently "biologically demanding, frequently necessitating magnification and competence to treat due to advancing microbial contamination. Therefore, the key objectives of deep caries management by vital pulp therapy procedures are to prevent caries progression, control bacterial contamination, stimulate pulpal repair, stimulate the formation of tertiary dentin, and seal the cavity to establish a profound barrier for preservation of the tooth from a long-term perspective, rendering it functional and free of symptoms with a healthy pulp. Therefore, even in cases in which the pulp is not exposed, an aseptic process must be followed.
Various diagnostic investigations have been conducted to ascertain the existence of distinct ailments and the most suitable course of therapy. Histopathological evaluation is considered the most effective method for identifying pulpal and periapical inflammation . Modern endodontic diagnostic methods, however, employ a combination of case history, clinical examination, and radiographic evaluation to establish varied grades of pulpal and periapical inflammation because it is not possible to histologically examine pulpal and periapical inflammation. Clinical evaluation involves an array of strategies, including visual inspection, evaluation of pulp sensitivity to thermal (hot and cold) and/or electrical stimulation, and determination of the degree of pain during percussion and palpation tests. Diagnosis of pulpal status, if reversibly or permanently inflamed, or necrosis of the pulp including periapical conditions such as apical periodontitis, periapical abscess, or condensing osteitis, is based on these findings . Clinical procedures for pulpal diagnosis have evolved remarkably in the last century . However, research has indicated a low association between these tests and an accurate histological scenario of the pulp [12,13]. Even with a combination of diagnostic tests, a recent systematic review revealed inadequate evidence to support the reliability of clinical diagnostic tests in determining the inflammatory condition of pulp tissue. The authors concluded that further research is needed to explore biological markers of pulpal inflammation and to improve existing endodontic diagnostic techniques. Extracted cell cultures studied in vitro have been used to investigate the molecular regulatory pathways of pulpal inflammation [14,15]. However, it is difficult to directly apply these reductionist experimental findings to therapeutic situations. In contrast to isolated in vitro models, the presence of immune cells, protease inhibitors, and other modifying molecules in vivo may elicit noticeably different inflammatory responses, and as a result, distinct clinical outcomes. The number of studies investigating potential tissue inflammation biomarkers in clinical samples is increasing. Identifying biomarkers has enormous clinical implications for the diagnosis and staging of pulpal diseases. Consequently, the need for additional meticulous investigations and structured critiques of existing reports appears self-evident. Hence, the aim of this paper is to review currently available published data regarding pulpal biomarkers derived from clinically defined normal or inflamed tissues and to decipher the possible future directions towards achieving the goal of vital pulp therapy procedures as a paradigm shift towards contemporary minimally invasive techniques.
Section snippets
Current understanding of pulpitis - pathological & molecular changes
Pulpitis is a tightly orchestrated series of vascular and cellular events mediated by molecular factors that characterize the inflammatory response in dental pulp tissues . Pulpitis is most commonly caused by an opportunistic infection by commensal oral microbes that enter the pulp space through channels such as dental caries, cracks, exposed dentinal tubules, or the apical foramen . Specific groups of cells that live in the dental pulp, such as endothelial cells, dendritic cells, and
Current methods of classification of pulpitis
Pulpal diagnostic classification systems categorize the underlying disease states based on patient signs, symptoms, and detailed clinical examination findings. Modern prospective studies on the treatment of deep caries primarily use radiographs to characterize lesion depth to recruit participants, classifying decay as deep or extremely deep. Detailed radiographic descriptions are prominently featured in the experimental protocols. However, inadequate preoperative pulpal manifestation
Current methods of diagnosis
As stated by Schweitzer JL(2009), the diagnosis of endodontic origins corresponds to solving a jigsaw puzzle, wherein the diagnosis is made from several pieces of information. To arrive at a probablediagnosis, clinicians must methodically compile all essential data. Clinicians should already have a tentative but rational diagnosis in mind while taking the patient's medical and dental histories, especially if a chief complaint is present, a comprehensive evaluation of the periodontium, and
Challenges faced in the current methods of diagnosis
Accurate diagnosis is crucial for assessing the status of the dental pulp in carious teeth or alternative forms of injury. Vitality or necrosis of the pulp is critical in this regard. Periods of spontaneous, radiating discomfort that linger after the trigger is removed are signs of irreversible damages. It is also critical to ascertain whether the pulp tissue is reversibly or irreversibly inflamed, particularly if the tissue has been severely or traumatically exposed. Current approaches have
Next generation strategies
Molecular diagnosis refers to the use of molecular biology techniques to analyze the genome and proteome in medical evaluation. These cutting-edge methods have had a big impact on clinical practice, especially in fields like oncology, pharmacogenomics, infectious illnesses, and prenatal genetic testing. Biomarker expressions function as reliable markers for physiological well-being, pathological processes, and therapeutic responses in a variety of assessment scenarios . Numerous host
Future directions
Discovering a revolutionary method for assessing the current inflammatory state of dental pulp would be highly beneficial, as pulp necrosis is a common outcome of coronal restoration in presumed non-inflamed vital teeth. The risk of overtreatment, such as performing unnecessary pulpectomies on teeth that could have been partially preserved, raises concerns . Therefore, it is vital for endodontic diagnosis to concentrate on assessing the degree of microbial infection or the host tissue's
Ethical approval
Ethical approval was not required for the present review article.
Contribution
Raksha Bhat: Literature search, figures, data collection, Formal analysis, data interpretation, writing, Shishir Shetty: Data analysis, Data interpretation, Writing, Proofreading, Praveen Rai: Data interpretation, writing, Proofreading., Ballamoole Krishna Kumar: Study design, writing, Proofreading. Preethesh Shetty: Formal analysis, data interpretation, writing
Declaration of competing interest
Revolutionizing the diagnosis of Irreversible Pulpitis Current Strategies and Future Directions, All authors are required to disclose any COI within the period of 12 months prior to the submission of any manuscript in the subject matter of which any company, entity, or organization has an interest. - Nil, The corresponding author is required to complete this form with information from all the authors listed in the manuscript.
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Cited by (6)
Nano-apatite with Doxycycline for pulp capping: A potential strategy to reduce inflammation and promote pulp healing
2025, Journal of Drug Delivery Science and Technology
The aim of the study is to develop a specialist dental capping agent that is particularly formulated to target pulp inflammation. The study included the synthesis of nano-apatite using Lepidium sativum (nHALS) that was loaded with Doxycycline (nHALS@DXC) utilizing an in vitro model of pulpitis with an O-antigen (O-A). The nanoscale HALS were analysed using scanning electron microscopy (SEM) and Transmission electron microscopy (TEM) to assess their odontogenic properties. The diameter of the nHALS nanoparticles ranged from 20 to 33 nm in width and 80170 nm in length. Doxycycline has been identified as an appropriate loading material for medicinal applications. While the effect of nHALS@DXC on the odontogenic competence of HDPSCs was not superior to that of nHALS alone, it exhibited a more significant anti-inflammatory effect. The findings indicate that nHALS@DXC has the capacity to serve as an effective agent for pulp capping, perhaps surpassing nHALS in terms of reducing inflammation and activating dental pup cells to enhance pulp repair after injury.
### Oral biosciences: The annual review 2024
2025, Journal of Oral Biosciences
The Journal of Oral Biosciences is committed to advancing and disseminating fundamental knowledge across all areas of oral biosciences. This editorial review features review articles covering diverse topics, including the mandible, tooth remineralization, dental pulpitis, dental implants, mesenchymal stem cells, microbiota, facial pain, stomatitis, odontogenic tumors, oral submucous fibrosis, insights on orofacial pain, tissue engineering, melatonin, and regenerative medicine.
This editorial review focuses on forensic anthropology, calcium sucrose phosphate, pulp biomarkers, zirconia, mesenchymal stem cells, microflora, stomatitis, ameloblastoma, areca nut, orofacial pain, collagen, melatonin, and tooth regeneration.
The review articles featured in the Journal of Oral Biosciences have significantly contributed to expanding readers knowledge across various domains of oral biosciences. The current editorial review discusses the key findings and significance of these review articles.
### From pulp to cementum: 3D visualization of soft and hard dental tissues using different ex vivo nano-CT contrast-enhancement techniques
2025, International Endodontic Journal
### Dual-functional injectable hydrogels as antimicrobial and angiogenic therapeutics for dental pulp regeneration
2025, Journal of Materials Chemistry B
### Current Insights into the Roles of LncRNAs and CircRNAs in Pulpitis: A Narrative Review
2024, International Journal of Molecular Sciences
### Peri-implant medication-related osteonecrosis of the jaw mimicking endodontic disease in a cancer patient: A case report
2024, Australian Endodontic Journal
View full text
© 2024 Japanese Association for Oral Biology. Published by Elsevier B.V. All rights reserved. |
12693 | https://www.youtube.com/watch?v=NgWav3Fj_kU | OpenStax Precalculus Ch 4.5 Properties of Log Functions # 2
Kwai Chan
4770 subscribers
Description
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Posted: 26 Mar 2021
Transcript:
hello pre-calculus student this video is for chapter 4.5 properties of log functions so we can going to continue with using the properties of log function um to you know work on log expressions so in this particular examples i'm going to use the law of rules to expand a single log into multiple individual logs so when we say individual logs will be just log x log y log z like nothing so all these x square and dividing and multiplying and dividing you know root have to become a coefficient so let's review the properties that we have before so summaries of long properties are if i have the last page if i have log and inside long m and n multiply i can split them into add if i have m divided by n i can split them into subtraction if i have an exponent i can move the exponent out as a coefficient so in the last example we combine them into one now we are expanding one both of them you will need to know for getting into calculus so this is important skill so log no base means log base 10 so i'm just going to keep on writing that so first of all is to split the multiply and divide don't worry about the exponent yet so this is a multiply so it becomes add right so be log x squared plus log root cube root of y minus log z to the 4. so this is not still not individual log so i'm going to move the exponent out 2 log x plus y to the one third right cube root cube root of y is right to the power one third so one third can move outside as long y and then minus this four can move outside log c and that's individual log and that's how we uh this you need to get to this point and uh what about this one it want to expand this individual log this is a big fourth root of a over c b cubed and there's two things in the denominator so first of all change this two to the one fourth right so this will be long a over c b cubed to the one fourth so first of all you're going to move this 1 4 then you might worry okay what about this 3 here there's an exponent but we'll do that later so it's 1 4 log a over c b cube now one form we're going to stay outside and you're going to split this into three things a and c and b cube right so belong a minus log c and i would want you this is minus again log b cubed why because i am still dividing this b cube is in the denominator so i do subtract okay so and then i do 1 4 clean up those exponents long a has nothing to clean up log c has nothing to clean up this 3 can move outside minus 3 log b so i will be working in individual logs so in some example in some uh answer they might want you to do the distributed properties one-fourth long a minus one-fourth log c minus three-fourth log b but uh this is also good enough okay um c expand this log of y square y plus 3 cubed so again ignore the exponent look at the multiply and divide so this is multiply there's no divide and you don't split right so become ln y squared plus ln y plus 3 cubed now you're going to tackle the exponent move them as coefficient 2 l and y plus 3 ln y plus 3 that's it right no more cannot split this y and 3. last one you have a square root of x squared minus 1 over x now this square within this square cannot cancel because you have any other things inside so i'm going to split them into the definition first so that would be equal to ln square root of x squared minus 1 minus ln x right just think of split them as a fraction by subtraction now what can i simplify this i can simplify this as one half because x squared minus one square root is x squared minus one one half right that is equal to the whole thing x squared minus 1 to the power one half but you can you can simplify this in okay you can't do that so that would be one half of ln x squared minus one and you can split x squared minus 1 because and then minus ln x because there is not a multiplication so you can you can split them anymore that would be the best you can expand them individual log um so and this and then the third part of four point five is change of base formula for log so in our scientific calculator we um have log 10 and log and log base e which is ln right in our scientific calculator but if someone asks you to do log base 2 or log base 5 you don't have the button in the calculator to do it so the change of base formula is log base b m is equal to log base m log 10 m over log 10 b basically the base becomes in the denominator or you can do ln m over l and b the proof of it is based on the log properties okay the log properties you can talk about this in class example one if you want to evaluate log base 5 36 um i can rewrite it as log base 10 36 over log base 10 5 in the calculator it becomes log log 36 over log 5 that would be 2.2266 can use ln also it's exactly the same so ln 36 over ln5 the same exact number two two six six right so this is called change of phase formula so you can evaluate any log you want uh summaries of the log property so uh if you have long a plus b can you split them as a plus b no you can also just can use with them as long a minus and multiply no if it's log a plus b inside that is the simplest one log a plus b is not equal to log a plus log b because log a plus log b can be simplified with log a times b now a plus b outside cannot be added inside also if you subtract just b you cannot move it inside or outside but log a times b can be simplified as long a plus log b right um and then log a a times b is not log a times log b only this is true okay inside is multiply you can split it as n and then if you have a times log b you can move it as an exponent or if you have a number outside of the log you can move it in as an exponent but you cannot just move it in as a times b and if you have log b divided by a this means one over a you become a root um because you are basically um it's as if you're doing 1 over a right or 1 over a means b to the 1 over a is a root and um it is not equal to dividing inside okay if you have a log b over a is not you cannot move the divide by a inside it's actually a root so this one is true so this is true these are not true if you do subtract uh subtract 2a you can combine this you can combine them as dividing but you can move them as a minus b if you have a negative log a it is not equal to moving in but it's actually 1 over a because that that means you can move the negative 1 as a negative 1 equal to 1 over a okay a to the power negative 1 is 1 over a now log zero is undefined because anything in the log has to be greater than zero not negative is undefined but log one is zero so these are some of the properties of log and we'll do 4.6 which is solving exponential and log function in the next video |
12694 | https://www.biologic.net/topics/internal-resistance-series-part-i-what-is-internal-resistance-in-a-battery/ | Internal Resistance series. Part I: What is internal resistance in a battery? - BioLogic Learning Center
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Internal Resistance series. Part I: What is internal resistance in a battery? - BioLogic Learning Center
homechevron_rightTopicschevron_right Internal Resistance series. Part I: What is internal resistance in a battery?
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Topic5 min read
Internal Resistance series. Part I: What is internal resistance in a battery?
Latest updated: November 15, 2024
Part I: What is internal resistance in a battery?
Introduction
The field of battery and energy storage continues to grow exponentially with the development of consumer electronics and electric vehicles, among other key technologies, As a result, laboratories and industry are constantly looking to find new ways of improving battery performance and gaining competitive advantage, notably via reliability and longer battery life. They also need to understand and quantify battery degradation during its cycle life. One of the key parameters affecting those challenges is battery internal resistance. This series of 3 articles will help you to understand what internal resistance is and how it can be measured.
Batteries cannot be considered to be perfect devices, and neither are they pure voltage sources. One of the imperfections of the battery is the voltage drop and the loss of electromotive force $emf$ that occurs when we start drawing current from the battery.
This loss can be characterized by a quantity called internal apparent resistance ($R_\mathrm{int}$) in $\Omega$ or apparent resistance ($\mathrm{R_{app}}$). Internal apparent resistance is a generic term that does not refer to any specific resistance in the battery. In practical terms, it is a generic term that refers to the battery’s natural predisposition to hinder current flow.
Figure 1 shows the evolution of the battery voltage $U$ as it delivers a current $I_0$ with $I_0>0$. Its voltage is instantly decreased by a value $R_\mathrm{\Omega}I_0$ (see the ohmic resistance definition below) and then by a value $R_\mathrm{int}I_0$ that varies with time.
Figure 1: Battery electrode voltage $U$ as a function of time when it delivers a current $I_0$. $E$ represents the batterie $emf$.
The International Electrotechnical Commission (IEC) gives the following definition of internal apparent resistance: Quotient of change of the voltage of a battery by the corresponding change in discharge current under specified conditions.
To introduce the effects of internal resistance on the battery, we can draw an analogy with two ski races. In Figure 2, two skiers simultaneously descend a slope. On the left (Fig. 2a.), a downhill skier heads straight down while, on the right (Fig. 2b.), a mogul skier will encounter some obstacles/bumps and therefore cannot go as fast as the downhill skier.
Figure 2: Internal resistance skier analogy: a) the downhill skier; b) the mogul skier.
The analogy of electrochemistry and battery cycling relates to the following:
The skier’s speed corresponds to the current that passes through the system, with $I_1<I_2$.
The height difference can then be compared to the voltage $U$ that decreases during this action.
The bumps represent the internal resistance $R_\mathrm{int}$.
The more bumps there are, the more difficult it will be for the skier to gain speed! We can extend this analogy to batteries: the ability to deliver a high current or power can be reduced because of internal resistance.
Which phenomena contribute to internal resistance?
Internal resistance can be described as a dipole that follows Ohm’s law and that is an approximative combination of the ohmic drop resistance of all the components of the battery, the charge transfer resistance, and the diffusion resistance of the battery’s electrodes.
A battery can be represented, as shown in Figure 2, as an electromotive force, $E$, in series with a resistor $R_\mathrm{int}$ i.e., the internal apparent resistance.
Figure 3: Simplified electrical model of a battery.
Please note: the “perfect” battery would be represented without the resistance $R_\mathrm{int}$.
Throughout the lifespan of the battery, the internal resistance increases because of the aging of components, electrolyte, etc.
What are the differences between internal resistance and ohmic resistance?
Ohmic resistance, $R_\Omega$, also referred to as high-frequency resistance, $R_\mathrm{HF}$, is defined by the electric resistance of electronic and ionic conductor materials such as connectors, contacts, electrodes, electrolytes, etc. It is also represented by a dipole that follows Ohm’s law.
Internal resistance is a more general quantity including phenomena relevant to ohmic resistance.
What are the consequences of internal resistance on the battery?
Internal resistance can have a significant impact on the battery’s performance, durability, and safety.
As already shown in Figure 1, the most direct effect of internal resistance on batteries when a current flows, is the voltage drop due to the presence of this resistance. Figure 4 shows a battery in a circuit with a resistance $R$, delivering a current $I=V/R$ with $I>0$ and where $V$ is the voltage at the resistance terminals, which is equal to the voltage at the battery terminals (Fig. 4).
As established previously, $R_\mathrm{int}$ can be approximated by a dipole that follows Ohm’s law.
$V=E – R_\mathrm{int} I $
Thus, the battery voltage drop due to internal resistance is $IR_\mathrm{int}$.
Figure 4: Circuit with a battery discharging in a resistance $\pmb{R}$.
Note that if the circuit is opened, using a switch, no current flows through the circuit so the battery does not experience any voltage drop.
The effect of internal resistance can differ from one battery chemistry to the other. Using the example of the lithium-ion battery, an increase in internal resistance results in a loss of lithium inventory, which in turn, reduces the battery’s capacity.
Internal resistance is also a safety issue. Indeed, according to the Joule effect, the fact that the battery hinders the current’s passage will generate thermal heating. If the temperature becomes too high, it will eventually lead to a thermal runaway and be harmful to the system and everything around it.
To learn more about how to measure internal resistance, please visit the 2nd article in this series: “How to determine the internal resistance of a battery”
For more specific information about Alternating and Direct Current Internal Resistance techniques (ACIR and DCIR), please see the 3rd article in the series.
References
Quantities, Units and Symbols in Physical Chemistry, International Unions of Pure and Applied Chemistry (IUPAC) Commission on physicochemical symbols, terminology and units, 2nd Edition, Blackwell Science Ltd Eds., Oxford (1993).
Internal resistance Battery Aging ACIR DCIR
Further reading...
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12695 | https://www.youtube.com/watch?v=4ee5UDN59zg | Orthogonality | Chapter 3 Applied Linear Algebra
Sabetta Talks Math
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Posted: 30 Aug 2021
In the third video in this series on Applied Linear Algebra, we'll start by generalizing the concept of length to arbitrary vector spaces using inner products and norms. This introduces us to orthogonality -- when two vectors have an inner product equal to zero. We are then able to construct spaces and subspaces that are spanned by orthonormal bases, using the Gram-Schmidt algorithm. Matrices, also, can be orthogonal. Orthogonal matrices give rise to unitary transformations of objects and are used frequently in video games and other numerical applications
Music "Everything" by Vi Hart
Transcript:
Intro hello and welcome everyone to the third video in our unit on linear algebra in Applied Linear Algebra this video we'll be generalizing how to measure distances in ukian Space by looking at inner products and Norms then we'll explore what orthogonality means for sets of vectors in Vector spaces and then for Matrix along the way we'll discover some algorithms for creating orthogonal sets at orthogonal matrices inner products are the generalization of the ukian dot Inner Product product to Vector spaces an inner product space is a vector space with a binary operation denoted V comma W in Brackets called an inner product that must satisfy three conditions the first is linearity the inner product of a V with W is equal to a the inner product of V with W the inner product is also distributive over vector addition so the inner product of U plus V with W is equal to the inner product of U with W plus the inner product of V with W the second property is symmetry the inner product of V with W is equal to the inner product of w with v the third property is positivity the inner product of V with itself is always greater than zero unless V is is the zero Vector in which case it is zero inner product spaces have a norm Associated to their inner product denoted by the magnitude of V which is the square root of the inner product of V with itself the ukian dot product is an example of an inner product here we're looking at the dotproduct of V1 with vs2 this gives us 2 2 + 1 0 is equal to four orthogonality is another consequence of Orthogonality the inner product two elements of an inner product space are called orthogonal if their inner product is zero for example using the dot product again the inner product of V2 and V3 is zero so they are orthogonal we can also use the inner product to define a unit Vector equal to V / the magnitude of V this new Vector is parallel to V and has Norm equal to 1 for example we can divide V1 by its magnitude theare OT of 5 to get a unit Vector parallel to V 1 we defined one type of Norm the magnitude of V but Norms too can be generalized a norm on a vector space assigns a real number confusingly also denoted as the magnitude of V to every Vector in the vector space subject to three properties firstly positivity the norm of V is always greater than or equal to zero and it's only equal to zero when V is the zero Vector secondly homogeneity the norm of a scalar a time a vector v is equal to the absolute value of a the norm of v and thirdly the triangle inequality where the norm of V + W is less than or equal to the norm of V plus the norm of w every inner product can be used as a way to measure distances on a vector space however many Norms that are useful don't come from inner products for example the lp Norm the lp Norm of V is equal to the P root of the sum on I of the I element of V to the P power three interesting LP Norms are the ukian norm or the L2 Norm which is the familiar magnitude of a vector in ukian space for example this Vector has length equal to theare < TK of 5 the taxi cab or Manhattan Norm is the L1 Norm it measures distances when you're restricted to move only left and right or up and down down it's equal to the sum of the absolute value of the components of V our Vector has taxi cab Norm equal to 3 the last is the L Infinity Norm which is the maximum of the set of the absolute values of the components of V our Vector has L Infinity Norm of two each Norm has a unit sphere defined for it which is the set of all vectors that have Norm equal to one our ukian Norm has a unit sphere that is just the unit Circle any pair whose absolute values add to one forms the unit Sphere for the L1 Norm this looks like a diamond with coordinates 1 0 0 1 - 1 0 and 0 -1 the unit Sphere for the L Infinity Norm also looks a bit curious it is a square of side length two centered at the origin see if you can convince yourself why this makes sense now that we have understood the Orthognal Bases concepts of inner product and orthogonality we can use to understand properties of vectors in Vector spaces and subspaces one very useful concept is that of a basis the set s of vectors in the vector space V is called a basis of V if the elements of s are linearly independent and every Vector in V can be written as a linear combination of elements in s thus every basis is a linearly independent spanning Set n dimensional Vector spaces have bases consisting of n linearly independent vectors every Vector space that we know of has a basis however the proof of this relies on Zorn's Lemma or the aium of choice conversely if every Vector space can be proven to have a bases then the axium of choice is true so assuming that the axium of choice is true a basis B of an n-dimensional inner product space V is called orthogonal if the inner product of any pair of vectors in B is equal to zero let's look at this example of an orthogonal basis for R3 V1 is equal to 1 1 -1 V2 is equal to -1 2 1 and V3 is equal to 3 03 we can easily check that it is an orthogonal basis given an orthogonal basis any Vector can be written as a linear combination of its elements where the coefficient of the I Vector is given by the inner product of X with VI divided by the magnitude of VI i^ 2 you can think of the inner product as projecting the vector X onto the vector VI for example we can write x = 1 - 31 as a linear combination of our basis vectors V1 through V3 the component of X that's parallel to V1 is -1 -1 1 the component of X that's parallel to V2 is 1 - 2 1 and parallel to V3 is 1 0 1 when we add those vectors together we recover our original Vector X additionally we call an orthogonal basis orthonormal if the magnitude of each of its elements is equal to one we can get this by dividing any vector by its magnitude next we'll have a look at the Gram-Schmidt Orthogonalization gram Schmid orthogonalization process this is an algorithm that creates an orthonormal basis for a vector space or Subspace imagine we start with a linearly independent set of vectors V1 through VK in an N dimensional inner product space the gram schmit process generates a new set of K orthonormal vectors that spans the same Subspace as the original set this takes advantage of the same projection operator we just used to decompose vectors into components of an orthogonal basis the projection of V onto U is given by the inner product of U with v divided by the inner product of U with itself the grit algorithm is as follows pick one vector from the original basis to be the first element of the orthogonal basis and call it U1 next we Define U2 as V2 minus the projection of V2 on U1 basically we're finding the component of V2 that is parallel to U1 and subtracting it from V2 then the next Vector in our orthogonal basis u3 is equal to V3 minus the projection of V3 onto U1 minus the projection of V3 onto U2 and we repeat this process until we get to the K Vector UK K is equal to VK minus the sum on I from I = 1 to K of the projection of VK on onto the new orthogonal basis Vector UI then we divide each of the vectors in our orthogonal basis by its magnitude to get an orthonormal basis for our Vector space in addition to sets of vectors being orthogonal there is a notion of Matrix orthogonality too an orthogonal Orthogonal Matrices Matrix is a square Matrix whose rows and columns are orthonormal vectors this can be written as Q transpose Q is equal to QQ transpose is equal to the Identity or equivalently Q transpose is equal to Q inverse orthogonal matrices are a special kind of linear transformation called a unitary transformation unitary Transformations are transformations that preserve the inner product here are a few examples of orthogonal 2x2 matrices and the effect of their linear transformation on the set of shapes first is the identity this of course leaves the shapes unchanged secondly we have a rotation Matrix which rotates our objects about the origin by some angle Theta and lastly we have a reflection Matrix which reflects our shapes about the x-axis we've been thinking about systems of linear equations and solving them using matrices in the first video in this series we used gaus Jordan elimination to solve systems of equations however there are several Matrix decompositions that give us more information about a system one of those is the QR decomposition the QR our decomposition is the decomposition of any Square Matrix a into the product of an orthogonal Matrix q and an upper triangular Matrix R you can think of this as a unitary operation on a particular basis for a one algorithm for QR decomposition uses the gram Schmid process using the basis of column vectors for a we can write Q as a matrix of unit vectors coming from the Gramm orthogonalization of a the upper right triangle Matrix R is then a matrix of projections of the components of The Columns of a onto the basis vectors of Q we will return to this technique in the next video where we explore data fitting and linear regression [Music] |
12696 | https://www.kcci.com/article/kaelin-meinders-charged-motorcycle-crash-killed-santiago-rhone/68037387 | Iowa crime: Man charged for motorcycle crash that killed teen
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Iowa man charged after motorcycle crash that killed Des Moines teen
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IN AN IMMIGRATION CENTER IN TEXAS, AND WE HAVE BREAKING NEWS HERE AT HOME. TWO, A MAN IS BEING CHARGED IN CONNECTION TO A CRASH THAT KILLED A TEENAGE BOY. THE POLK COUNTY SHERIFF’S OFFICE SAYS THAT 23 YEAR OLD KAITLYN MEINDERS FROM ALBIA, WAS DRIVING THE MOTORCYCLE THAT HIT AND KILLED 15 YEAR OLD SANTIAGO ROAN OVER THE WEEKEND. MEINDERS IS NOW CHARGED WITH SEVERAL OFFENSES THAT INCLUDE RECKLESS DRIVING, DRIVING WHILE BARD IS A OFFENDER TO MORE CHARGES. MORE CHARGES COULD BE PENDING. THOSE WHO KNEW ROAN SAY THAT HE HAD THREE LOVES. FAMILY, FRIENDS AND BASEBALL. COMMUNITY MEMBERS WILL GATHER TONIGHT AT THE BASEBALL FIELD TO REMEMBER THE 15 YE
Updated: 10:33 AM CDT Sep 25, 2025
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Iowa man charged after motorcycle crash that killed Des Moines teen
Updated: 10:33 AM CDT Sep 25, 2025
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An Iowa man is now facing charges for his role in a motorcycle crash that resulted in a teenager's death. Kaelin Meinders, 23, of Albia, was the operator of a motorcycle Saturday night when the motorcycle crashed into a pedestrian in the 4900 block of NE 29th Street. The pedestrian was identified as Santiago "Santi" Rhone, 15, a Des Moines East High School sophomore.Rhone was transported to a nearby hospital but died from his injuries.The Polk County Sheriff's Office on Wednesday announced Meinders has been charged with multiple offenses. Meinders has been charged with reckless driving; driving while barred (habitual offender); and no insurance — causing an accident.Authorities are still investigating the crash. Meinders was hospitalized following the wreck. Sheriff's department asks for help The Polk County Sheriff's Office is encouraging anyone with information or video from the crash to reach out to them.You can submit information at this email: Pcso.detectives@polkcountyiowa.gov.» Subscribe to KCCI's YouTube page» Download the free KCCI app to get updates on the go: Apple | Google PlayWATCH: Des Moines East High mourns loss of sophomore hit by motorcycle
DES MOINES, Iowa —
An Iowa man is now facing charges for his role in a motorcycle crash that resulted in a teenager's death.
Kaelin Meinders, 23, of Albia, was the operator of a motorcycle Saturday night when the motorcycle crashed into a pedestrian in the 4900 block of NE 29th Street. The pedestrian was identified as Santiago "Santi" Rhone, 15, a Des Moines East High School sophomore.
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Rhone was transported to a nearby hospital but died from his injuries.
The Polk County Sheriff's Office on Wednesday announced Meinders has been charged with multiple offenses. Meinders has been charged with reckless driving; driving while barred (habitual offender); and no insurance — causing an accident.
Authorities are still investigating the crash. Meinders was hospitalized following the wreck.
Sheriff's department asks for help
The Polk County Sheriff's Office is encouraging anyone with information or video from the crash to reach out to them.
You can submit information at this email: Pcso.detectives@polkcountyiowa.gov.
»Subscribe to KCCI's YouTube page
» Download the free KCCI app to get updates on the go:Apple | Google Play
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12697 | https://simple.wikipedia.org/wiki/Hyperbolic_functions | Hyperbolic functions - Simple English Wikipedia, the free encyclopedia
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Contents
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1 DefinitionsToggle Definitions subsection
1.1 Exponential definitions
2 Useful relations
3 Comparison with circular functions
4 Relationship to the exponential function
5 Related pages
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7 Other websites
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From Simple English Wikipedia, the free encyclopedia
The English used in this article or section may not be easy for everybody to understand. You can help Wikipedia by reading Wikipedia:How to write Simple English pages, then simplifying the article.(April 2020)
In mathematics, hyperbolic functions are analogues of the ordinary trigonometric functions, but defined for the unit hyperbola rather than on the unit circle: just as the points (cos t, sin t) form a circle with a unit radius, the points (cosh t, sinh t) form the right half of the hyperbola.
Hyperbolic functions occur in the calculations of angles and distances in hyperbolic geometry. They also occur in the solutions of many linear differential equations, cubic equations, and Laplace's equation in Cartesian coordinates. Laplace's equations are important in many areas of physics.
The basic hyperbolic functions are:
hyperbolic sine "sinh" (/ˈ s ɪ ŋ,ˈ s ɪ n tʃ,ˈ ʃ aɪ n/),
hyperbolic cosine "cosh" (/ˈ k ɒ ʃ,ˈ k oʊ ʃ/),
from which are derived:
hyperbolic tangent "tanh" (/ˈ t æ ŋ,ˈ t æ n tʃ,ˈ θ æ n/),
hyperbolic cosecant "csch" or "cosech" (/ˈ k oʊ s ɛ tʃ,ˈ k oʊ ʃ ɛ k/)
hyperbolic secant "sech" (/ˈ s ɛ tʃ,ˈ ʃ ɛ k/),
hyperbolic cotangent "coth" (/ˈ k ɒ θ,ˈ k oʊ θ/),
corresponding to the derived trigonometric functions.
The inverse hyperbolic functions are:
area hyperbolic sine "arsinh" (also denoted "sinh−1", "asinh" or sometimes "arcsinh")
area hyperbolic cosine "arcosh" (also denoted "cosh−1", "acosh" or sometimes "arccosh"
and so on.
A ray through the unit hyperbolax 2 − y 2 = 1 in the point (cosh a, sinh a), where a is twice the area between the ray, the hyperbola, and the x-axis. For points on the hyperbola below the x-axis, the area is considered negative (see animated version with comparison with the trigonometric (circular) functions).
The hyperbolic functions take a real argument called a hyperbolic angle. The size of a hyperbolic angle is twice the area of its hyperbolic sector. The hyperbolic functions may be defined in terms of the legs of a right triangle covering this sector.
Hyperbolic functions were introduced in the 1760s independently by Vincenzo Riccati and Johann Heinrich Lambert.
Definitions
[change | change source]
sinh, cosh and tanh
csch, sech and coth
There are various equivalent ways to define the hyperbolic functions.
Exponential definitions
[change | change source]
sinh x is half the difference of e x and e−x
cosh x is the average of e x and e−x
In terms of the exponential function:
Hyperbolic sine: the odd part of the exponential function, that is sinhx=e x−e−x 2=e 2 x−1 2 e x=1−e−2 x 2 e−x.{\displaystyle \sinh x={\frac {e^{x}-e^{-x}}{2}}={\frac {e^{2x}-1}{2e^{x}}}={\frac {1-e^{-2x}}{2e^{-x}}}.}
Hyperbolic cosine: the even part of the exponential function, that is coshx=e x+e−x 2=e 2 x+1 2 e x=1+e−2 x 2 e−x.{\displaystyle \cosh x={\frac {e^{x}+e^{-x}}{2}}={\frac {e^{2x}+1}{2e^{x}}}={\frac {1+e^{-2x}}{2e^{-x}}}.}
Hyperbolic tangent: tanhx=sinhx coshx=e x−e−x e x+e−x=e 2 x−1 e 2 x+1{\displaystyle \tanh x={\frac {\sinh x}{\cosh x}}={\frac {e^{x}-e^{-x}}{e^{x}+e^{-x}}}={\frac {e^{2x}-1}{e^{2x}+1}}}
Hyperbolic cotangent: for x ≠ 0, cothx=coshx sinhx=e x+e−x e x−e−x=e 2 x+1 e 2 x−1{\displaystyle \coth x={\frac {\cosh x}{\sinh x}}={\frac {e^{x}+e^{-x}}{e^{x}-e^{-x}}}={\frac {e^{2x}+1}{e^{2x}-1}}}
Hyperbolic secant: sechx=1 coshx=2 e x+e−x=2 e x e 2 x+1{\displaystyle \operatorname {sech} x={\frac {1}{\cosh x}}={\frac {2}{e^{x}+e^{-x}}}={\frac {2e^{x}}{e^{2x}+1}}}
Hyperbolic cosecant: for x ≠ 0, cschx=1 sinhx=2 e x−e−x=2 e x e 2 x−1{\displaystyle \operatorname {csch} x={\frac {1}{\sinh x}}={\frac {2}{e^{x}-e^{-x}}}={\frac {2e^{x}}{e^{2x}-1}}}
Useful relations
[change | change source]
The hyperbolic functions satisfy many identities, all of them similar in form to the trigonometric identities.
Odd and even functions:
sinh(−x)=−sinhx cosh(−x)=coshx{\displaystyle {\begin{aligned}\sinh(-x)&=-\sinh x\\cosh(-x)&=\cosh x\end{aligned}}}
Hence:
tanh(−x)=−tanhx coth(−x)=−cothx sech(−x)=sechx csch(−x)=−cschx{\displaystyle {\begin{aligned}\tanh(-x)&=-\tanh x\\coth(-x)&=-\coth x\\operatorname {sech} (-x)&=\operatorname {sech} x\\operatorname {csch} (-x)&=-\operatorname {csch} x\end{aligned}}}
Thus, cosh x and sech x are even functions; the others are odd functions.
arsechx=arcosh(1 x)arcschx=arsinh(1 x)arcothx=artanh(1 x){\displaystyle {\begin{aligned}\operatorname {arsech} x&=\operatorname {arcosh} \left({\frac {1}{x}}\right)\\operatorname {arcsch} x&=\operatorname {arsinh} \left({\frac {1}{x}}\right)\\operatorname {arcoth} x&=\operatorname {artanh} \left({\frac {1}{x}}\right)\end{aligned}}}
Hyperbolic sine and cosine satisfy:
coshx+sinhx=e x coshx−sinhx=e−x cosh 2x−sinh 2x=1{\displaystyle {\begin{aligned}\cosh x+\sinh x&=e^{x}\\cosh x-\sinh x&=e^{-x}\\cosh ^{2}x-\sinh ^{2}x&=1\end{aligned}}}
the last of which is similar to the Pythagorean trigonometric identity.
One also has
sech 2x=1−tanh 2x csch 2x=coth 2x−1{\displaystyle {\begin{aligned}\operatorname {sech} ^{2}x&=1-\tanh ^{2}x\\operatorname {csch} ^{2}x&=\coth ^{2}x-1\end{aligned}}}
for the other functions.
Comparison with circular functions
[change | change source]
Circle and hyperbola tangent at (1,1) display geometry of circular functions in terms of circular sector area u and hyperbolic functions depending on hyperbolic sector area u.
The hyperbolic functions represent an expansion of trigonometry beyond the functions defined on unit circle. Both types depend on one argument, either circular angle or hyperbolic angle.
Since the area of a circular sector with radius r and angle u (in radians) is r 2 u/2, it will be equal to u when r = √2. In the diagram, such a circle is tangent to the hyperbola xy = 1 at (1,1). The yellow sector depicts an area and angle magnitude. Similarly, the yellow and red sectors together depict an area and hyperbolic angle magnitude.
The legs of the two right triangles with hypotenuse on the ray defining the angles are of length 2{\displaystyle {\sqrt {2}}} times the circular and hyperbolic functions.
Relationship to the exponential function
[change | change source]
The decomposition of the exponential function in its even and odd parts gives the identities
e x=coshx+sinhx,{\displaystyle e^{x}=\cosh x+\sinh x,}
and
e−x=coshx−sinhx.{\displaystyle e^{-x}=\cosh x-\sinh x.}
The first one is analogous to Euler's formula
e i x=cosx+i sinx.{\displaystyle e^{ix}=\cos x+i\sin x.}
Additionally,
e x=1+tanhx 1−tanhx=1+tanhx 2 1−tanhx 2{\displaystyle e^{x}={\sqrt {\frac {1+\tanh x}{1-\tanh x}}}={\frac {1+\tanh {\frac {x}{2}}}{1-\tanh {\frac {x}{2}}}}}
Related pages
[change | change source]
e (mathematical constant)
Trigonometric functions
References
[change | change source]
↑ 1.01.1"Comprehensive List of Algebra Symbols". Math Vault. 2020-03-25. Retrieved 2020-08-29.
↑ 2.02.1Weisstein, Eric W. "Hyperbolic Functions". mathworld.wolfram.com. Retrieved 2020-08-29.
↑(1999) Collins Concise Dictionary, 4th edition, HarperCollins, Glasgow, ISBN0 00 472257 4, p. 1386
↑ 4.04.1Collins Concise Dictionary, p. 328
↑ 5.05.1"Hyperbolic Functions". www.mathsisfun.com. Retrieved 2020-08-29.
↑Collins Concise Dictionary, p. 1520
↑Collins Concise Dictionary, p. 1340
↑Collins Concise Dictionary, p. 329
↑tanh
↑Woodhouse, N. M. J. (2003), Special Relativity, London: Springer, p.71, ISBN978-1-85233-426-0
↑Abramowitz, Milton; Stegun, Irene A., eds. (1972), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, New York: Dover Publications, ISBN978-0-486-61272-0
↑Robert E. Bradley, Lawrence A. D'Antonio, Charles Edward Sandifer. Euler at 300: an appreciation. Mathematical Association of America, 2007. Page 100.
Other websites
[change | change source]
Wikimedia Commons has media related to Hyperbolic functions.
Hyperbolic functions on PlanetMath
GonioLab: Visualization of the unit circle, trigonometric and hyperbolic functions (Java Web Start)
Web-based calculator of hyperbolic functions
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12698 | https://erich-friedman.github.io/mathmagic/0401.html | Math Magic
Problem of the Month (April 2001)
This month we investigate adjacency graphs of sets of congruent polygons. Take a set of congruent polygons. Form a graph using the polygons as vertices, and draw an edge between two polygons if they share part of an edge. For example, the collection of squares below on the left has the adjacency graph below on the right.
Which graphs can be adjacency graphs? Can you find 4 congruent polygons that have K 4 (the complete graph on 4 vertices) as an adjacency graph? How about K 5 minus an edge? What are the simplest polygons (in terms of number of sides, or aspect ratio, or lack of small angles) that can realize a given adjacency graph? Can every tree be an adjacency graph?
For a real challenge, notice that the graph above is regular, meaning each square touches exactly three other squares. What are the smallest regular adjacency graphs? And what are the simplest polygons that create them?
ANSWERS
Joseph DeVincentis found chevrons with adjacency graphs K 4 and K 5 minus an edge. They are shown below.
Here are my polyomino solutions to these problems.
The first figure above was also found by John Hoffman who thinks that it cannot be done with a convex shape. Sasha Ravsky agrees. John Hoffman also asks whether every planar graph is realizable as an adjacency graph of a set of congruent polygons? He guesses no, but I guess yes. Can anyone find a counterexample?
Joseph DeVincentis found that a rectangle that can have an adjacency graph that is 4-regular. His arrangement is on the left. My arrangement is on the right.
In 2021, Isky Mathews proved that every tree is an adjacency graph.
If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 1/10/21. |
12699 | https://math.stackexchange.com/questions/849604/whats-the-importance-of-a-formula-for-the-real-and-imaginary-parts-of-a-complex | What's the importance of a formula for the real and imaginary parts of a complex number? - Mathematics Stack Exchange
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What's the importance of a formula for the real and imaginary parts of a complex number?
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I've learned that
\bbox[8px,border:1px solid black]{\operatorname{Re}(z)= \frac{z+\overline{z}}{2} \qquad \qquad \operatorname{Im}(z)=\frac{z-\overline{z}}{2i}}
And that in the number z=a+bi, a is the real part and b is the imaginary part. The formulas I mentioned above are used to get a and b alone. But by looking at z, I could get the real part just by taking a and ignoring the rest. The same is valid for b and in both cases, without using the formulas. So why are these formulas important? I just learned the basics of complex numbers and still don't know why one needs those formulas.
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edited Jun 27, 2014 at 19:33
Hakim
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asked Jun 27, 2014 at 16:38
Red BananaRed Banana
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9 \begingroup"But by looking at z, I could get the real part just by taking a and ignoring the rest. The same is valid for b and in both cases, without using the formulas." What if you're not given a complex number in the form a+ib? What if you're given, say, \sqrt{17}e^{i\pi/13}?\endgroup –Git Gud Commented Jun 27, 2014 at 16:40
7 \begingroup _But by looking at z, I could get the real part just by taking a and ignoring the rest_: this statement is true only if you're given z in algebraic form. What if you're given z=\mathrm{e}^{2i}? (Edit: beaten by @GitGud).\endgroup –gniourf_gniourf Commented Jun 27, 2014 at 16:41
1 \begingroup Well for one, you have an algebraic (as opposed to geometric or intuitive) proof that |Re(z)|=|\frac{z+\bar{z}}{2}|<\frac{1}{2}(|z|+|\bar{z}|) = \frac{2}{2}|z|=|z|\endgroup –Squirtle Commented Jun 27, 2014 at 16:45
\begingroup I see. I still didn't read about complex numbers expressed in that way. I just discovered a few seconds ago that they are called complex numbers in the exponential form.\endgroup –Red Banana Commented Jun 27, 2014 at 16:47
3 \begingroup@GitGud In that case we use \mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta and conclude that a=\sqrt{17}\cos\frac{\pi}{13} and b=\sqrt{17}\sin\frac{\pi}{13}.\endgroup –Fly by Night Commented Jun 27, 2014 at 17:00
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This is a nice question. We often learn formulae without asking why they are useful.
I've re-typed this post half a dozen times. Every time, I thought I had a nice use, but then found out that I didn't need the formulae after all. Having said that, I think that I have found one.
Using the exponential form z=\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta, we see that
\begin{eqnarray} \cos\theta &=& \frac{1}{2}!\left(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}\right) \ \ \sin\theta &=& \frac{1}{2\mathrm{i}}!\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right) \end{eqnarray} We can use these formulae to evaluate sine and cosine over the complex plane: \begin{eqnarray} \cos(\mathrm{i}) &=& \frac{1}{2}!\left(\mathrm{e}^{\mathrm{i}\mathrm{i}} + \mathrm{e}^{-\mathrm{i}\mathrm{i}}\right) \ \ &=& \frac{1}{2}!\left(\mathrm{e}^{-1} + \mathrm{e}^{1}\right) \ \ &=& \frac{1+\mathrm{e}^2}{2\mathrm{e}}\approx 1.543 \end{eqnarray} I have ignored the multi-valued problem, i.e. \mathrm{e}^{\mathrm{i}\theta} = \mathrm{e}^{\mathrm{i}(\theta+2\pi k)} for all k \in \mathbb{Z}.
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answered Jun 27, 2014 at 17:38
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\begingroup Using \cos\theta = 1/2(e^{i\theta} + e^{-i\theta}) is also handy for evaluating certain integrals of the form \int{e^{a\theta}\cos{(b\theta)}d\theta}.\endgroup –Hao Ye Commented Jun 27, 2014 at 21:55
\begingroup These presentations of the trigonometric functions also explain (in some sense) the strong parallels between trig functions and hyperbolic functions.\endgroup –user21467 Commented Jun 28, 2014 at 3:59
\begingroup@StevenTaschuk Absolutely. I see them as a single function. If z \in \mathbb{R} then we get the good, old fashioned, cosine function. If z \in \mathrm{i}\mathbb{R} we get the hyperbolic cosine function. I remember first learning this. I got such a buzz.\endgroup –Fly by Night Commented Jun 29, 2014 at 13:19
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For example, you know that e^z=e^{x+iy}=e^x(\cos y + i \sin y) and \overline{e^z}=e^{\overline{z}}=e^x(\cos y -i \sin y), so you have two pretty equations \sin y=\frac{e^{iy}-e^{-iy}}{2i} and \cos y=\frac{e^{iy}+e^{-iy}}{2}.
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edited Jun 27, 2014 at 18:02
answered Jun 27, 2014 at 16:45
aghaagha
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\begingroup Could you justify your claim that \overline{\mathrm{e}^z}=\mathrm{e}^{\overline{z}}? In general \overline{\mathrm{f}(z)} \neq \mathrm{f}(\overline{z}).\endgroup –Fly by Night Commented Jun 27, 2014 at 17:41
\begingroup@Ian The comments section of an answer is to allow people to ask for extra detail or to suggest improvements from the _person answering the question_. My comment was directed at agha and was a request for him to justify his steps. It was not a general appeal for an answer. Had that been the case then I would have posted it as a question.\endgroup –Fly by Night Commented Jun 27, 2014 at 17:49
\begingroup@Fly by night Yes, you'er right, it's property of exponential function, we have e^{x+iy}=e^x(\cos y + i \sin y) by definition, so, \overline{e^z}=e^x(\cos y - i \sin y)=e^x(\cos -y - i \sin -y)=e^{x-iy}=e^{\overline{z}}\endgroup –agha Commented Jun 27, 2014 at 18:01
2 \begingroup Euler's formula for complex exponentials is not a definition, it is a formula. We _do_ have \overline{f(z)}=f(\overline{z}) in any region symmetric about the real axis where f is defined by a real-coefficient power series, which is all of \Bbb C for \exp.\endgroup –anon Commented Jun 27, 2014 at 18:05
\begingroup@blue Let's be careful here. What do you mean by "real-coefficient power series"? The power series of \mathrm{e}^{\mathrm{i}\theta} has real coefficients for all even powered terms and imaginary coefficients for all odd powered terms. 1+\mathrm{i}\theta-\frac{1}{2}\theta^2-\frac{\mathrm{i}}{3!}\theta^3+\frac{1}{4!}\theta^4-\frac{\mathrm{i}}{5!}\theta^5+\cdots\endgroup –Fly by Night Commented Jun 27, 2014 at 18:35
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You have z + \bar{z} =4
and you are asked to find real part of the complex number you get it by this formula. This is most basic use of this formula, later you advance in the course you will see its importance .
One suggestion whenever learning a new topic and its foundation do not think what is its importance unless you are genius it just that let question come think how formula is applied and then rereading the topic you can yourself get to the point. Though it is good that you asked here as it opens the grounds for you to see it's far reaching application.
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edited Jun 27, 2014 at 16:51
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Allow me to explain a couple of _uses_ for the formulas for \sin and \cos in terms of e^{iz}.
We can combine them with the geometric sum formula to compute various trigonometric sums and products without hard-to-remember trig formulas. E.g. \sin(x)+\sin(2x)+\cdots+\sin(nx).
We can interchange between Fourier expansions in terms of real sines and cosines or complex exponentials, or compute integrals involving sines and cosines (e.g. the orthogonality relations).
Additionally, while those formulas for the real and imaginary parts of a complex number may not be entirely ubiquitous, the _idea_ behind them is very important in higher math: decomposing something into its "eigenparts." This appears in linear algebra, where a diagonalizable operator can be decomposed as a direct sum of scalar operators on eigensubspaces. One can symmetrize or antisymmetrize tensors (or perform even more exotic skew-symmetrizations). One can decompose a function into an integral mixture of dual functions (characters in the generalized setting of locally compact abelian groups and abstract harmonic analysis). One can decompose into roots and weights in the setting of Lie algebras and representation theory. And so on.
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edited Jun 27, 2014 at 18:41
answered Jun 27, 2014 at 18:36
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One example of a domain in which these can be useful is differential equations.
Given, say, the equation \frac{d^2y}{dx} + y = 0 there are techniques that allow you to determine that the solution is y = a\cdot e^{ti} + b\cdot e^{-ti} where a and b are arbitrary complex numbers. Knowing that e^{ti} = \overline{e^{-ti}} , and that Re(z) and Im(z) can be written as linear combinations of z and \bar{z} , we can reformulate the solution as y = c \cdot Re(e^{ti}) + d \cdot Im(e^{ti}) or y = c \cdot cos(t) + d \cdot sin(t) (with c and d again arbitrary constants).
Basically, knowing those identities helped us find a solution to the equation using only real numbers.
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answered Jun 28, 2014 at 3:04
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