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# Yahoo Web Search
1. ### The parallel sides of a trapezium are 10 cm and 20 cm respectively. AD = 6 cm and CB =8 cm . Find the area of the trapezium ?
... and CD are the parallel sides with lengths 10 cm and 20 cm , respectively. See the attached sketch. Let point ...
3 Answers · Science & Mathematics · 03/08/2020
2. ### The area of a sector of a circle with a radius of 15 cm is 75pi cm ^2. ?
The area of a sector of a circle with a radius of 15 cm is 75pi cm ^2. ? Total circle Area = π(15)^2 = 225π cm ^2...
11 Answers · Science & Mathematics · 28/05/2020
3. ### The perimeter of a rectangle is 56 cm . Find the lengths of the sides of the rectangle giving the maximum area.?
A=l*w 56= 2l + 2w; w= 28 - l A= l*(28 - l)= 28l - l² dA/dl= 28 - 2l= 0 l= 14 cm , w= 14 cm
4 Answers · Science & Mathematics · 19/05/2020
4. ### a spherical snowball is melting symmetrically at the rate of 4pi cubic cm per hr. how fast is the diameter changing when it is 20 cm ?
... dV/dt = 4πr².dr/dt so, when the diameter is 20 cm we have: -4π = 4πr².dr/dt Hence, dr/dt = -1/r² i.e. dr...
3 Answers · Science & Mathematics · 15/08/2020
5. ### The legs of an isosceles triangle are each 45 cm long.The measures of the base angles are each 70°. What is the measure of the third side?
apply the law of sine ..a................b ---------=-------------- .sin(A)........sin(B) ..45..............b --------- = ----------- .sin(70) ....sin(40) 45sin(40) = bsin(70) b = 30.78 cm or 31 cm answer//
2 Answers · Science & Mathematics · 06/10/2020
6. ### A gallon of homogenized, vitamin A, whole milk has a weight ranging from 8.48 - 8. 72 lbs. Determine its volume in cm ^3.?
1 US gallon is 231 cubic inches 1 inch is 2.54 cm 231 * 2.54^3 => 3,785.411784 cm ^3
3 Answers · Science & Mathematics · 03/09/2020
7. ### A cone has a diameter of 6 cm and a slant height of 5 cm . What is the exact volume (in cubic centimeters) of the cone?....?
...and h r² + h² = s² 3² + h² = 5² h = 4 cm V = 1/3 π(3)²(4) V = 12π Answer C
4 Answers · Science & Mathematics · 11/06/2020
8. ### How much is 1.1 cm in inches?
1.1 cm = 0.4330708661417323 in
6 Answers · Science & Mathematics · 01/07/2020
9. ### a sphere has the volume 36π cm ^3 .What area does the sphere have?
The volume of a sphere with radius "r" is given by: V = 4*π*r³/3 so: r = (3V/(4π))^(1/3) The surface area of a sphere with radius "r" is given by: A = 4π*r² You want to know the area given the volume...
4 Answers · Science & Mathematics · 31/10/2019
10. ### If ED = √8 cm and the area of ∆ABC = trapezoid ACDE, the length of AC is?
AC = 2. See below for explanation. Notice that there are two similar triangles, ABC and EBD. The area of EBD is twice that of ABC, since the trapezoid area is equal to that of triangle ABC. But, the square...
3 Answers · Science & Mathematics · 19/12/2019
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HuggingFaceTB/finemath
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# Traveling Salesman Problem: Problem-Based
This example shows how to use binary integer programming to solve the classic traveling salesman problem. This problem involves finding the shortest closed tour (path) through a set of stops (cities). In this case there are 200 stops, but you can easily change the `nStops` variable to get a different problem size. You'll solve the initial problem and see that the solution has subtours. This means the optimal solution found doesn't give one continuous path through all the points, but instead has several disconnected loops. You'll then use an iterative process of determining the subtours, adding constraints, and rerunning the optimization until the subtours are eliminated.
For the solver-based approach to this problem, see Traveling Salesman Problem: Solver-Based.
### Problem Formulation
Formulate the traveling salesman problem for integer linear programming as follows:
• Generate all possible trips, meaning all distinct pairs of stops.
• Calculate the distance for each trip.
• The cost function to minimize is the sum of the trip distances for each trip in the tour.
• The decision variables are binary, and associated with each trip, where each 1 represents a trip that exists on the tour, and each 0 represents a trip that is not on the tour.
• To ensure that the tour includes every stop, include the linear constraint that each stop is on exactly two trips. This means one arrival and one departure from the stop.
### Generate Stops
Generate random stops inside a crude polygonal representation of the continental U.S.
```load('usborder.mat','x','y','xx','yy'); rng(3,'twister') % Makes stops in Maine & Florida, and is reproducible nStops = 200; % You can use any number, but the problem size scales as N^2 stopsLon = zeros(nStops,1); % Allocate x-coordinates of nStops stopsLat = stopsLon; % Allocate y-coordinates n = 1; while (n <= nStops) xp = rand*1.5; yp = rand; if inpolygon(xp,yp,x,y) % Test if inside the border stopsLon(n) = xp; stopsLat(n) = yp; n = n+1; end end```
### Calculate Distances Between Points
Because there are 200 stops, there are 19,900 trips, meaning 19,900 binary variables (# variables = 200 choose 2).
Generate all the trips, meaning all pairs of stops.
`idxs = nchoosek(1:nStops,2);`
Calculate all the trip distances, assuming that the earth is flat in order to use the Pythagorean rule.
```dist = hypot(stopsLat(idxs(:,1)) - stopsLat(idxs(:,2)), ... stopsLon(idxs(:,1)) - stopsLon(idxs(:,2))); lendist = length(dist);```
With this definition of the `dist` vector, the length of a tour is
`dist'*trips`
where `trips` is the binary vector representing the trips that the solution takes. This is the distance of a tour that you try to minimize.
### Create Graph and Draw Map
Represent the problem as a graph. Create a graph where the stops are nodes and the trips are edges.
`G = graph(idxs(:,1),idxs(:,2));`
Display the stops using a graph plot. Plot the nodes without the graph edges.
```figure hGraph = plot(G,'XData',stopsLon,'YData',stopsLat,'LineStyle','none','NodeLabel',{}); hold on % Draw the outside border plot(x,y,'r-') hold off```
### Create Variables and Problem
Create an optimization problem with binary optimization variables representing the potential trips.
```tsp = optimproblem; trips = optimvar('trips',lendist,1,'Type','integer','LowerBound',0,'UpperBound',1);```
Include the objective function in the problem.
`tsp.Objective = dist'*trips;`
### Constraints
Create the linear constraints that each stop has two associated trips, because there must be a trip to each stop and a trip departing each stop.
Use the graph representation to identify all trips starting or ending at a stop by finding all edges connecting to that stop. For each stop, create the constraint that the sum of trips for that stop equals two.
```constr2trips = optimconstr(nStops,1); for stop = 1:nStops whichIdxs = outedges(G,stop); % Identify trips associated with the stop constr2trips(stop) = sum(trips(whichIdxs)) == 2; end tsp.Constraints.constr2trips = constr2trips;```
### Solve Initial Problem
The problem is ready to be solved. To suppress iterative output, turn off the default display.
```opts = optimoptions('intlinprog','Display','off'); tspsol = solve(tsp,'options',opts)```
```tspsol = struct with fields: trips: [19900×1 double] ```
### Visualize Solution
Create a new graph with the solution trips as edges. To do so, round the solution in case some values are not exactly integers, and convert the resulting values to `logical`.
```tspsol.trips = logical(round(tspsol.trips)); Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2),[],numnodes(G)); % Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2)); % Also works in most cases```
Overlay the new graph on the existing plot and highlight its edges.
```hold on highlight(hGraph,Gsol,'LineStyle','-') title('Solution with Subtours')```
As can be seen on the map, the solution has several subtours. The constraints specified so far do not prevent these subtours from happening. In order to prevent any possible subtour from happening, you would need an incredibly large number of inequality constraints.
### Subtour Constraints
Because you can't add all of the subtour constraints, take an iterative approach. Detect the subtours in the current solution, then add inequality constraints to prevent those particular subtours from happening. By doing this, you find a suitable tour in a few iterations.
Eliminate subtours with inequality constraints. An example of how this works is if you have five points in a subtour, then you have five lines connecting those points to create the subtour. Eliminate this subtour by implementing an inequality constraint to say there must be less than or equal to four lines between these five points.
Even more, find all lines between these five points, and constrain the solution not to have more than four of these lines present. This is a correct constraint because if five or more of the lines existed in a solution, then the solution would have a subtour (a graph with $n$ nodes and $n$ edges always contains a cycle).
Detect the subtours by identifying the connected components in `Gsol`, the graph built with the edges in the current solution. `conncomp` returns a vector with the number of the subtour to which each edge belongs.
```tourIdxs = conncomp(Gsol); numtours = max(tourIdxs); % Number of subtours fprintf('# of subtours: %d\n',numtours);```
```# of subtours: 27 ```
Include the linear inequality constraints to eliminate subtours, and repeatedly call the solver, until just one subtour remains.
```% Index of added constraints for subtours k = 1; while numtours > 1 % Repeat until there is just one subtour % Add the subtour constraints for ii = 1:numtours inSubTour = (tourIdxs == ii); % Edges in current subtour a = all(inSubTour(idxs),2); % Complete graph indices with both ends in subtour constrname = "subtourconstr" + num2str(k); tsp.Constraints.(constrname) = sum(trips(a)) <= (nnz(inSubTour) - 1); k = k + 1; end % Try to optimize again [tspsol,fval,exitflag,output] = solve(tsp,'options',opts); tspsol.trips = logical(round(tspsol.trips)); Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2),[],numnodes(G)); % Gsol = graph(idxs(tspsol.trips,1),idxs(tspsol.trips,2)); % Also works in most cases % Plot new solution hGraph.LineStyle = 'none'; % Remove the previous highlighted path highlight(hGraph,Gsol,'LineStyle','-') drawnow % How many subtours this time? tourIdxs = conncomp(Gsol); numtours = max(tourIdxs); % Number of subtours fprintf('# of subtours: %d\n',numtours) end```
```# of subtours: 20 # of subtours: 7 # of subtours: 9 # of subtours: 9 # of subtours: 3 # of subtours: 2 # of subtours: 7 # of subtours: 2 # of subtours: 1 ```
```title('Solution with Subtours Eliminated'); hold off```
### Solution Quality
The solution represents a feasible tour, because it is a single closed loop. But is it a minimal-cost tour? One way to find out is to examine the output structure.
`disp(output.absolutegap)`
``` 0 ```
The smallness of the absolute gap implies that the solution is either optimal or has a total length that is close to optimal.
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If you’re taking the test tomorrow morning, you should ignore this and REST YOUR BRAIN. If you’re not taking it tomorrow, though, then this is just another regular weekend for you, and you should work your brain HARD.
The prize, as it has been every week for weeks and weeks, is access to the coveted PWN the SAT Math Guide Beta. If you wanna win, you have to comment in such a way that I can contact you. Preferably using Facebook, Twitter, or Gmail.
A warning: this one is pretty tough.
In a hand in his weekly poker game, Mike won money from both John and Sean. Mike’s percent gain on the hand was equal in magnitude to John’s percent loss. John bet twice as much as did Sean. Mike and John together held \$200 in chips before the hand began. If Sean bet \$20 in chips on the hand, what was Mike’s chip total after the hand was over?
Good luck!
UPDATE: Nobody got this yet, so I’m gonna leave it up for now unanswered. If you solve it, book access is yours.
SECOND UPDATE: OK, Eowyn got it. Nice. Solution below the cut.
There’s a lot going on here, and at first blush it may seem as though you have too many variables to work with. If you write down everything you know, though, and play around with it enough, you’ll see that you can actually get it down to just one variable in one equation!
$percent\:&space;change&space;=&space;\frac{change}{original\:&space;value}\times&space;100\%$
$\frac{Mike's\:gain}{Mike's\:original\:value}\times&space;100\%&space;=&space;\frac{John's\:loss}{John's\:original\:value}\times&space;100\%$
Of course, we can tidy this up a bit:
$\frac{Mike's\:gain}{Mike's\:original\:value}&space;=&space;\frac{John's\:loss}{John's\:original\:value}$
Now we need to start figuring out the numbers. We know that Sean bet 20, and that John bet twice as much as Sean. So John bet 40, which means John’s loss was 40. We also know that Mike’s gain was 60, since he won all the money that was bet in the hand (John’s 40 + Sean’s 20).
$\frac{60}{Mike's\:original\:value}&space;=&space;\frac{40}{John's\:original\:value}$
And we can be a bit clever here if we want to jam all the info into one equation and say that, since Mike’s and John’s chips added up to 200 before the hand, Mike’s prior chip total can be m, and John’s prior chip total can be 200 – m.
$\frac{60}{m}&space;=&space;\frac{40}{200-m}$
Solve that for m
$40m&space;=&space;60(200-m)$
$40m&space;=&space;12000-60m$
$100m&space;=&space;12000$
$m&space;=&space;120$
…and you’ll see that before the hand began, Mike had 120 dollars in chips. Since he won 60, he has 180 after the hand.
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×
6th Chapter
### 11th Business Mathematics Chapter 6 Test
Here you can prepare 11th Class Business Mathematics Chapter 6 Binary Number System and its Operation Test. Click the button for 100% free full practice test.
## First Year Business Mathematics Chapter 6 Online MCQ Test for 1st Year Business Mathematics Chapter 6 (Binary Number Systems and its Operation)
This online test contains MCQs about following topics:
Introduction . Binary number system . Conversions of number systems . Arithmetic operations in binary number system
ICOM Part 1 Business Mathematics Ch. 6 Test
### First Year Business Mathematics Chapter 6 Online MCQ Test for 1st Year Business Mathematics Chapter 6 (Binary Number Systems and its Operation)
1 (145)<sub>10</sub> = ( )<sub>2</sub>
• A. 10010001
• B. 10010111
• C. 11100001
• D. 10001001
2 (1100000)<sub>2</sub> - (111111)<sub>2</sub>= --------------------------- :
• A. (100001)<sub>2</sub>
• B. (110001)<sub>2</sub>
• C. (1000111)<sub>2</sub>
• D. (111110)<sub>2</sub>
3 Annuity is classified into:
• A. Two classes
• B. Three classes
• C. Four classes
• D. Five classes
4 Basically proportion is of:
• A. 4 types
• B. 3 types
• C. 2 types
• D. None of these
5 (10101)<sub>2</sub> in decimal system is
• A. 32
• B. 26
• C. 21
• D. 30
6 Hexadecimal number system is based on:
• A. Two digits
• B. Ten digits
• C. Eight digits
• D. Sixteen digits
7 5 in binary system is:
• A. (10)<sub>2</sub>
• B. (101)<sub>2</sub>
• C. (11)<sub>2</sub>
• D. None of these
8 (1001001)<sub>2</sub> in decimal system is -----------------------
• A. 37
• B. 67
• C. 73
• D. 87
9 (100011)<sub>2</sub> x (1101)<sub>2</sub> = ---------------------------
• A. (111000111)<sub>2</sub>
• B. (100011001)<sub>2</sub>
• C. (100000001)
• D. (110011001)<sub>2</sub>
10 (1101)<sub>2</sub> + (1001)<sub>2</sub> = --------------------------------
• A. (10110)<sub>2</sub>
• B. (11100)<sub>2</sub>
• C. (10001)<sub>2</sub>
• D. (11011)<sub>2</sub>
### Top Scorers of Business Mathematics Icom Part 1 Chapter 6 Online Test
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first year ka paper kab ha
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# ZeroSum Ruler (home)
## Blogging on math education and other related things
### Adding Fractions With Pictures! (The Crisscross Method)December 3, 2012
Fraction Addition (And Subtraction): We’re not in kindergarten anymore
Addition and subtraction are only easy in elementary school. Once middle school starts, continuing throughout any Math class taken that point forward, addition and subtraction are much harder than multiplication and division. Why? The Common Denominator. To a kid who is not fluent in his multiplication facts, finding The Common Denominator is an exercise in torture.
-
What is a common denominator? A common denominator is a multiple of both denominators in a fraction addition (or subtraction) problem. For example:
-
In the above example, 6 is a common denominator of 2 and 3. But is it the only one? No. How many common denominators are there between two fractions? Infinite. For example:
-
-
Why would we want to use 7830 as a common denominator? Why not? The point is that any number that both denominators divide into evenly can act as a common denominator. We are far less restricted than we thought.
So if we’re virtually unrestricted in choosing a common denominator, why not pick the one that is the product (multiply) of the two denominators? For example:
-
-
Just multiply the denominators to find a common denominator. This is easy.
-
At this point in the traditional method of adding fractions, we’d begin to ask our questions: “How many 8’s go into 16?” Ok, 2. “2 times 3 is …?” Ok 6. So 3/8 = 6/16 . Though this process is easy to a person who is fluent in their multiplication and division, it will give reason for a non-fluent Math student to seize up.
-
A great alternative way of adding fractions is the Crisscross Method of adding (and subtracting) fractions. In this method, we use the common denominator just once (this method will not create two equivalent fractions to the original two) and multiply “crisscross” to find two new numerators.
-
In 3/8 + 5/2, we’ll first multiply the denominators to find our new, common denominator:
-
Next, we’ll multiply 3 • 2 (always starting our crisscross in the top left corner) to find the first missing numerator:
-
And then 8 • 5 to find the second missing numerator:
-
But why are we allowed to do this? Let’s back up to see what really happened.-
-
First, we found the common denominator 16 by multiplying the denominators (8 and 2) of both fractions. We’re guaranteed that our denominator is common if we created it by multiplying the two original denominators to get it. To get the first numerator 6, we multiplied the numerator of the first fraction (3) by the denominator of the second fraction (2).
-
In the process, we multiplied both numerator and denominator by 2. In other words, we multiplied 3/8 by 2/2 Any number divided by itself is just a fancy 1, and multiplying any number by 1 does not change the number’s value. As a check to see if this process worked, 3/8 = 6/16 . The old and new fractions are equivalent.
-
The same is true to get the second numerator 40:
-
Both numerator and denominator were multiplied by 8. In other words, we multiplied 5/2 by 8/8, which is just a fancy 1. Multiplying by 1 does not change a number’s value. As a check, 5/2 = 40/16. The old and new fractions are equivalent.
-
Now we simply add the numerators:
-
-
The Crisscross method also works for fraction subtraction – we’d have a subtraction in the numerator. Why was this method not taught in school?
-
Hurray for Fraction Addition (and Subtraction)!
-
You can download a PDF ebook that uses pictures to explain fraction division, multiplication and addition on CurrClick at Fractions: A Picture Book!
-
-
### New (free) ZeroSum ruler – for teaching addition with negative numbersSeptember 30, 2012
Below is a new version of the ZeroSum ruler. This one needs no hardware to construct, just scissors and glue. You can download, print and use this proven tool right now by clicking on the picture, which will bring you to the PDF file that contains 2 ZeroSum rulers.
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### Simple “how to solve an equation” flowchartSeptember 14, 2012
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Since drawing this flowchart, I have created a printed one that includes “variables on both sides“. You can see the post (and get the free poster download here: http://zerosumruler.wordpress.com/2012/09/30/solving-equations-flowchart-free-download-poster/
### The Language of Math PosterAugust 19, 2011
Below is a poster I hang in my classroom every fall. Each year it grows longer as more and more terms come up for the different operations of math. When I was a kid, no one told me to look out for these words, or that math was even a language at all, which made word problems pretty tough. By clicking on the poster you will be sent to the original Excel file on Google Docs. Do you have any words to add?
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X of Y or Y's : GMAT Verbal Section
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# Events & Promotions
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# X of Y or Y's
Author Message
Intern
Joined: 20 Sep 2008
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X of Y or Y's [#permalink]
### Show Tags
20 Nov 2008, 11:56
1
This post was
BOOKMARKED
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
If you have any questions
New!
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Re: X of Y or Y's [#permalink]
### Show Tags
28 Nov 2008, 12:20
jugolo1 wrote:
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
Since Gmat prefers short and concise, (John's hat) is better than "hat of John" though both are correct.
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Re: X of Y or Y's [#permalink]
### Show Tags
28 Nov 2008, 18:42
GMAT TIGER wrote:
jugolo1 wrote:
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
Since Gmat prefers short and concise, (John's hat) is better than "hat of John" though both are correct.
No, depend on the situation. whether one or the other make the sentence awkard!
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Re: X of Y or Y's [#permalink]
### Show Tags
28 Nov 2008, 22:09
sondenso wrote:
GMAT TIGER wrote:
jugolo1 wrote:
Hello,
I wanted to clarify when we use possessif form i.e. (John's hat) or the "of" form i.e. hat of john
. ...thanksssss
Since Gmat prefers short and concise, (John's hat) is better than "hat of John" though both are correct.
No, depend on the situation. whether one or the other make the sentence awkard!
What I have seen in most of the cases in Gmat is that "John's hat" is direct, short and concise.
For example: 11-t73342
I believe "Jean Toomer’s Cane" is much better than "Cane of Jean Toomer"
More example, if any, would be useful.
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Re: X of Y or Y's [#permalink]
### Show Tags
29 Nov 2008, 03:47
Tiger, it is my pleasure!
VR-91.
In theory, international civil servants at the United Nations are prohibited from continuing to draw salaries from their own governments; in practice, however some governments merely substitute living allowances for their employees' paychecks, assigned by them to the United Nations.
(A) for their employees’ paychecks, assigned by them
(B) for the [color=#FF0000]paychecks of their employees[/color] who have been assigned
(C) for the paychecks of their employees, having been assigned
(D) in place of their employees’ paychecks, for those of them assigned
(E) in place of the paychecks of their employees to have been assigned by them
employees' paychecks is awkard
B is correct
Another one nearly the same.
VR-79. Scientists have observed large concentrations of heavy—metal deposits in the upper twenty centimeters of Baltic Sea sediments，which are consistent with the growth of industrial activity there.
(A) Baltic Sea sediments, which are consistent with the growth of industrial activity there
(B) Baltic Sea sediments, where the growth of industrial activity is consistent with these findings
(C) Baltic Sea sediments, findings consistent with its growth of industrial activity
(D) sediments from the Baltic Sea, findings consistent with the growth of industrial activity in the area
(E) sediments from the Baltic Sea, consistent with the growth of industrial activity there
D and E is ok, but A, b and C is awkward
Hope it help!
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Re: X of Y or Y's [#permalink]
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29 Nov 2008, 22:49
sondenso wrote:
In theory, international civil servants at the United Nations are prohibited from continuing to draw salaries from their own governments; in practice, however some governments merely substitute living allowances for their employees' paychecks, assigned by them to the United Nations.
(A) for their employees’ paychecks, assigned by them
(B) for the [color=#FF0000]paychecks of their employees[/color] who have been assigned
(C) for the paychecks of their employees, having been assigned
(D) in place of their employees’ paychecks, for those of them assigned
(E) in place of the paychecks of their employees to have been assigned by them
employees' paychecks is awkard
B is correct
I take this one. Thanks.
_________________
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Re: X of Y or Y's [#permalink] 29 Nov 2008, 22:49
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HuggingFaceTB/finemath
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Ampere
(Redirected from Amperes)
Ampere Demonstration modew of a moving iron ammeter. As de current drough de coiw increases, de pwunger is drawn furder into de coiw and de pointer defwects to de right.
Generaw information
Unit systemSI base unit
Unit ofEwectric current
SymbowA
Named afterAndré-Marie Ampère
The ampere (/ˈæmpɪər, æmˈpɪər/; symbow: A), often shortened to "amp", is de base unit of ewectric current in de Internationaw System of Units (SI). It is named after André-Marie Ampère (1775–1836), French madematician and physicist, considered de fader of ewectrodynamics.
The Internationaw System of Units defines de ampere in terms of oder base units by measuring de ewectromagnetic force between ewectricaw conductors carrying ewectric current. The earwier CGS measurement system had two different definitions of current, one essentiawwy de same as de SI's and de oder using ewectric charge as de base unit, wif de unit of charge defined by measuring de force between two charged metaw pwates. The ampere was den defined as one couwomb of charge per second. In SI, de unit of charge, de couwomb, is defined as de charge carried by one ampere during one second.
New definitions, in terms of invariant constants of nature, specificawwy de ewementary charge, took effect on 20 May 2019.
Definition
The ampere is defined by taking de fixed numericaw vawue of de ewementary charge e to be 1.602 176 634 × 10−19 when expressed in de unit C, which is eqwaw to A s, where de second is defined in terms of ∆ν.
The SI unit of charge, de couwomb, "is de qwantity of ewectricity carried in 1 second by a current of 1 ampere". Conversewy, a current of one ampere is one couwomb of charge going past a given point per second:
${\dispwaystywe {\rm {1\ A=1{\tfrac {C}{s}}.}}}$ In generaw, charge Q is determined by steady current I fwowing for a time t as Q = It.
Constant, instantaneous and average current are expressed in amperes (as in "de charging current is 1.2 A") and de charge accumuwated, or passed drough a circuit over a period of time is expressed in couwombs (as in "de battery charge is 30000 C"). The rewation of de ampere (C/s) to de couwomb is de same as dat of de watt (J/s) to de jouwe.
History
The ampere was originawwy defined as one tenf of de unit of ewectric current in de centimetre–gram–second system of units. That unit, now known as de abampere, was defined as de amount of current dat generates a force of two dynes per centimetre of wengf between two wires one centimetre apart. The size of de unit was chosen so dat de units derived from it in de MKSA system wouwd be convenientwy sized.
The "internationaw ampere" was an earwy reawization of de ampere, defined as de current dat wouwd deposit 0.001118 grams of siwver per second from a siwver nitrate sowution, uh-hah-hah-hah. Later, more accurate measurements reveawed dat dis current is 0.99985 A.
Since power is defined as de product of current and vowtage, de ampere can awternativewy be expressed in terms of de oder units using de rewationship I=P/V, and dus 1 ampere eqwaws 1 W/V. Current can be measured by a muwtimeter, a device dat can measure ewectricaw vowtage, current, and resistance.
Reawization
The standard ampere is most accuratewy reawized using a Kibbwe bawance, but is in practice maintained via Ohm's waw from de units of ewectromotive force and resistance, de vowt and de ohm, since de watter two can be tied to physicaw phenomena dat are rewativewy easy to reproduce, de Josephson junction and de qwantum Haww effect, respectivewy.
At present, techniqwes to estabwish de reawization of an ampere have a rewative uncertainty of approximatewy a few parts in 107, and invowve reawizations of de watt, de ohm and de vowt.
Everyday exampwes
The current drawn by typicaw constant-vowtage energy distribution systems is usuawwy dictated by de power (watt) consumed by de system and de operating vowtage. For dis reason de exampwes given bewow are grouped by vowtage wevew.
CPUs – 1 V DC
• Current notebook CPUs (up to 15...45 W at 1 V): up to 15...45 A
• Current high-end CPUs (up to 65...140 W at 1.15 V): up to 55...120 A
Portabwe devices
• Hearing aid (typicawwy 1 mW at 1.4 V): 700 µA
• USB charging adapter (as power suppwy – typicawwy 10 W at 5 V): 2 A
Internaw combustion engine vehicwes – 12 V DC
A typicaw motor vehicwe has a 12 V battery. The various accessories dat are powered by de battery might incwude:
• Instrument panew wight (typicawwy 2 W): 166 mA
• Headwight (each, typicawwy 60 W): 5 A
• Starter motor on a smawwer car: 50 A to 200 A
Norf American domestic suppwy – 120 V AC
Most Canada, Mexico and United States domestic power suppwiers run at 120 V.
Househowd circuit breakers typicawwy provide a maximum of 15 A or 20 A of current to a given set of outwets.
• USB charging adapter (as woad – typicawwy 10 W): 83 mA
• 22-inch/56-centimeter portabwe tewevision (35 W): 290 mA
• Tungsten wight buwb (60–100 W): 500–830 mA
• Toaster, kettwe (1.5 kW): 12.5 A
• Hair dryer (1.8 kW): 15 A
European & Commonweawf domestic suppwy – 230–240 V AC
Most European domestic power suppwies run at 230 V, and most Commonweawf domestic power suppwies run at 240 V. For de same amount of power (in watts), de current drawn by a particuwar European or Commonweawf appwiance (in Europe or a Commonweawf country) wiww be wess dan for an eqwivawent Norf American appwiance.[Note 1] Typicaw circuit breakers wiww provide 16 A.
The current drawn by a number of typicaw appwiances are:
• Compact fwuorescent wamp (11–30 W): 56–112 mA
• 22-inch/56-centimeter portabwe tewevision (35 W): 145–150 mA
• Tungsten wight buwb (60–100 W): 240–450 mA
• Toaster, kettwe (2 kW): 9 A
• Immersion heater (4.6 kW): 19–20 A
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# 6 Stages of Question Mastery
Author Message
Manager
Joined: 07 Sep 2007
Posts: 116
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6 Stages of Question Mastery [#permalink]
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14 Nov 2007, 02:15
1
KUDOS
I believe there are a couple stages that people go through, and that these stages are primarily based on %correct for a type of question (sc/rc/cr/ps/ds) and not difficulty of questions or anything else.
When I say type, I mean either SC, RC, CR, PS, DS.
When I say sub-type, I mean, for example, verb-agreement in SC, or weaken the argument in CR, or purpose of the passage in RC.
6 Stages of Question Mastery
(These stages are independent with type of question.)
Stage 1: 0%-40% correct
You do not understand the fundamentals. Your strategy is not correct for this type of question. You are doing not doing much better than guessing at the answer and doing problems is not going to help you.
What to do: You need to just read GMAT guides on this type of question and change your approach before you even think about touching any questions.
Stage 2: 40%-70% correct
You are consistently missing the point of the question. The question is asking for something very specific and clear, but you are often misunderstanding it and looking in the wrong direction. Perhaps in CR you're consistently thinking outside of scope or inferring too much information from the passage. Perhaps in SC you're looking for parallel structures when there is no need for them. Perhaps in Quant you need to find the prime factorization but you're trying some brute force approach.
What to do: You still need to focus on reading GMAT guides and but do a few questions every now and then to see how well you're doing. When doing questions at this phase, you need to spend a lot of time per question (5-10 minutes) knowing exactly why each of the 5 answers is right or wrong.
Stage 3: 70%-80% correct
You've got a good understanding of some of the fundamentals, but you are not applying a complete strategy. Perhaps in SC you're consistently spotting verb agreement errors but missing improper modifiers. Or in RC you consistently confusing a supporting point with the main point. In Quant, maybe you don't understand some subjects very well (perm/comb, prime factorization, divisibility, powers, etc.).
What to do: Here is where we can start focusing on our flaws, figure out which sub-type of the type of question you're getting wrong and read GMAT guides on that sub-type. When doing questions, you should still be spending 3-5 minutes on each question, now easily eliminating 1-3 of the answers and, for the rest, being able to explain exactly why each of the answers is right or wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 4: 80%-90% correct
Now we're starting to show consistency, we understand most of the fundamentals. Our strategy is near complete, but we have not been applying it perfectly. Most of our incorrect answers can be attributable to carelessness and result in "oh duh!" moments. NOW AND ONLY NOW are you ready to start doing sets of questions.
What to do: If you're consistently missing certain sub-types, go re-read some GMAT guides on it. But your study should involve mostly questions. You should still be spending 2-3 minutes on each one, easily eliminating all but 2 of the answers. For the 2, you must be able to explain exactly why one answer is right and one answer is wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 5: 90%-95% correct
We understand almost all of the fundamentals and are able to apply them with a consistent strategy. We are able to eliminate down to the correct 2 answers almost all of the time, and if we spend enough time on any question, we know we can get it right 100% of the time. Some sub-types are giving us more trouble than others.
What to do: Practice questions to fine-tune your strategy; iron out flaws in those sub-types you're getting wrong. Spend 1-2 minutes on each question. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 6: 95%-100% correct
We understand all of the fundamentals and are able to apply them with a consistent strategy. All of the questions we get wrong are just do to really stupid mistakes that we are in complete control over.
What to do: Work on mental stamina and timing by doing a test's worth of questions each day. Aim for a question per minute. Ease up on studying, we just want to maintain and fine tune our spectacular skills.
-J
Last edited by JingChan on 17 Nov 2007, 22:27, edited 1 time in total.
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Posts: 63
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### Show Tags
14 Nov 2007, 06:21
JingChan thank you for your useful advice and congratulations for your excellent GMAT score. People like you are an inspiration for my trying to master the GMAT.
I would like to ask you a couple of questions and I would really appreciate your opinion.
1) I am not a native english speaker. I do not live in the States, UK, Australia or any native-english speaker country. Do you think that people like me can do a good score in verbal, (>40, why not approximate 45?)
2) My major difficulties in GMAT have to do mostly with the verbal part. (I'm not a maths expert but I have much better understanding to it since I have an engineering background. I do silly mistakes however and I definetely need to prepare for this part as well). I try to focus especially in CR at the moment, which is my major weakness. I try to undestand the questions and analyse them in my head. Then, I can score somewhere close to 50%. Not good of course but I started with 10% in this area...But, it takes me a hell of time to do that. I am thinking that it is not the point how fast I do a question, but to grasp the thing, and therefore, I take my time. Do you agree with that?
And last but not least, do you think I should do the same in RC? Take my time I mean till I find a strategy?
JingChan I would really appreciate your advice. It's not the high score you obtained, but your overall approach and preparation for this exam that makes your advice valuable. At least for me.
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### Show Tags
14 Nov 2007, 14:04
Given what I know about you, it sounds like you definitely have the potential to score 95%+ in Verbal. Being a non-English speaker, you have to work at it a little harder, but that should be expected. When it boils down to it, each type of question tests certain skills:
SC - grammar rules
CR - reading ability and logical reasoning
And none of these skills require you to be a native English speaker to master.
You didn't mention SC so I'll assume you're doing well in that area. What's left are just reading questions. Before you start scoring 80%+ you need to be fully grasping the point of the question. Take your time and understand each question. You first need to be answering the questions correctly before you think about speeding yourself up.
Once you do start answering questions consistently, then start timing yourself. Slowly speed yourself up until you're able to finish each question in about 1-2 minutes.
Hope this helps,
J
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14 Nov 2007, 23:47
It has been very helpful. Thank you very much
By the way, I wish you all the best for the rest of the application process.
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Re: 6 Stages of Question Mastery [#permalink]
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15 Nov 2007, 11:50
JingChan wrote:
I believe there are a couple stages that people go through, and that these stages are primarily based on %correct for a type of question (sc/rc/cr/ps/ds) and not difficulty of questions or anything else.
When I say type, I mean either SC, RC, CR, PS, DS.
When I say sub-type, I mean, for example, verb-agreement in SC, or weaken the argument in CR, or purpose of the passage in RC.
6 Stages of Question Mastery
(These stages are independent with type of question.)
Stage 1: 0%-40% correct
You do not understand the fundamentals. Your strategy is not correct for this type of question. You are doing not doing much better than guessing at the answer and doing problems is not going to help you.
What to do: You need to just read GMAT guides on this type of question and change your approach before you even think about touching any questions.
Stage 2: 40%-70% correct
You are consistently missing the point of the question. The question is asking for something very specific and clear, but you are often misunderstanding it and looking in the wrong direction. Perhaps in CR you're consistently thinking outside of score or inferring too much information from the passage. Perhaps in SC you're looking for parallel structures when there is no need for them. Perhaps in Quant you need to find the prime factorization but you're trying some brute force approach.
What to do: You still need to focus on reading GMAT guides and but do a few questions every now and then to see how well you're doing. When doing questions at this phase, you need to spend a lot of time per question (5-10 minutes) knowing exactly why each of the 5 answers is right or wrong.
Stage 3: 70%-80% correct
You've got a good understanding of some of the fundamentals, but you are not applying a complete strategy. Perhaps in SC you're consistently spotting verb agreement errors but missing improper modifiers. Or in RC you consistently confusing a supporting point with the main point. In Quant, maybe you don't understand some subjects very well (perm/comb, prime factorization, divisibility, powers, etc.).
What to do: Here is where we can start focusing on our flaws, figure out which sub-type of the type of question you're getting wrong and read GMAT guides on that sub-type. When doing questions, you should still be spending 3-5 minutes on each question, now easily eliminating 1-3 of the answers and, for the rest, being able to explain exactly why each of the answers is right or wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 4: 80%-90% correct
Now we're starting to show consistency, we understand most of the fundamentals. Our strategy is near complete, but we have not been applying it perfectly. Most of our incorrect answers can be attributable to carelessness and result in "oh duh!" moments. NOW AND ONLY NOW are you ready to start doing sets of questions.
What to do: If you're consistently missing certain sub-types, go re-read some GMAT guides on it. But your study should involve mostly questions. You should still be spending 2-3 minutes on each one, easily eliminating all but 2 of the answers. For the 2, you must be able to explain exactly why one answer is right and one answer is wrong. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 5: 90%-95% correct
We understand almost all of the fundamentals and are able to apply them with a consistent strategy. We are able to eliminate down to the correct 2 answers almost all of the time, and if we spend enough time on any question, we know we can get it right 100% of the time. Some sub-types are giving us more trouble than others.
What to do: Practice questions to fine-tune your strategy; iron out flaws in those sub-types you're getting wrong. Spend 1-2 minutes on each question. If you get a question wrong, go back and do a thorough 5-answer analysis.
Stage 6: 95%-100% correct
We understand all of the fundamentals and are able to apply them with a consistent strategy. All of the questions we get wrong are just do to really stupid mistakes that we are in complete control over.
What to do: Work on mental stamina and timing by doing a test's worth of questions each day. Aim for a question per minute. Ease up on studying, we just want to maintain and fine tune our spectacular skills.
-J
Nice Post...what's GMAT Guides?
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Posts: 116
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Re: 6 Stages of Question Mastery [#permalink]
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15 Nov 2007, 17:15
yogachgolf wrote:
Nice Post...what's GMAT Guides?
I just mean any GMAT books. Like PR's Cracking the GMAT, MGMAT Guides, Kaplan 800, etc.
-J
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17 Nov 2007, 21:01
This is a very good post. How do you successful execute the 5-answer thorough analysis. My problem is that whenever i see the answer of a question I initially failed, i get that 'ooh' effect that makes me think that i understand why the answer chosen is correct and mine is not.
Manager
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### Show Tags
17 Nov 2007, 22:26
kizito2001 wrote:
This is a very good post. How do you successful execute the 5-answer thorough analysis. My problem is that whenever i see the answer of a question I initially failed, i get that 'ooh' effect that makes me think that i understand why the answer chosen is correct and mine is not.
If you're at the stage where you can immediately understand, then you're doing well!
I would rewind my thinking to try to understand why I chose the wrong answer to begin with, because maybe you're consistently making the same error.
Especially for Verbal, it's very important to understand why each wrong answer is wrong.
-J
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### Show Tags
29 Nov 2007, 00:23
Jing,
Thanks for this awesome post as it has definitely inspired me. I have a questions regarding consistency in the prep material, what to use and what NOT to use.
My situation is that this is my second time taking the test (i took my test Nov 1 and didn't do well - 640: v35, m45). I have used and am still using various different types of material and score all over the board. For example, with Princeton Review, i score around high 600s - 700. With Kaplan, I score in the 630-660 range, ARCO is also in the 660s range. Surprisingly I scored 710s on both of the GMAT Prep tests, prior to the exam. But clearly, I was not ready. Now, I dont know which prep material to trust and when I am truely "ready" to the take it again.
So, what was your plan of attack in:
1. Nagivating through all this prep material, and choosing the right ones? Should I stick to 3-4 brands (MGMAT, KAPLAN, OG, PR)?
2. This is a silly quesiton, but still want to ask it...How do you know when you are "ready?" Is it when you are consistently scoring in the same range on various different prep materials?
A big thank you in advance!
29 Nov 2007, 00:23
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# 6 Stages of Question Mastery
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HuggingFaceTB/finemath
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If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm,
Question:
If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what is the length of QR?
Solution:
Given: ΔABC and ΔPQR are similar triangles. Area of ΔABC: Area of ΔPQR = 9:16 and BC = 4.5cm.
To find: Length of QR
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
Hence,
$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}$
$\frac{9}{16}=\frac{4.5^{2}}{Q R^{2}}$
$\mathrm{QR}^{2}=\frac{4.5^{2} \times 16}{9}$
$\mathrm{QR}^{2}=36$
$Q R=6 \mathrm{~cm}$
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open-web-math/open-web-math
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## Important Points from Lectures
1) Choosing Gaussian Surface: When you are working on a problem where you want to use Gauss' Law to find a field, there are a few issues you want to be aware of:
a. The law only tells you the surface integral of the field (the flux) - if the field is varying on your surface in a way you don't know before hand, this can become very intractable. It is best to set up your surfaces so that you have the same field everywhere if possible.
b. Remember also that the integral only counts the perpendicular component of the field, so you need to have the field that you're interested in perpendicular to the surface.
c. But, we can use this to our advantage: if your surface needs to include an area where the field is either not the value of interest, or not constant or both (eg: long line of charge), if your surface passes through this space in such a way that the field is parallel to it, then you can safely ignore these fields!
2) Conductors: Field at surface must be perpendicular to surface, field inside must be zero. (Even for hollow ones: eg Faraday Cage experiment).
3) Field Lines: Another way to draw in fields. They show direction of electric field directly and tell you the strength of the field with their density. Field Lines cannot cross (but they can run from one charge to another, or meet head on at a point where field is zero).
4) Electric Potential and Potential Energy: Work done by field on a charge as it moves is equal to the drop in electrical potential energy between the starting and finishing points (Electric field is conservative force field). Here V is used to denote the potential energy of a specific charge in the presence of other charges and φ(r) is the potential at a point in space due to an arrangement of charges.
a. Two point charges: V(r) = kq1q2/r
b. φ(r)=V(r)/q2
c. Superposition still works: V=Σikq1qi/ri1 is energy of charge one – our test charge in most cases, (and there is a similar sum to calculate the "self energy" of the fixed collection of charges).
5) There are four important, different formulae that involve Q and r so far in this course:
a. Coulomb's Law: F=kQ1Q2ρ/r2 b. Electric Field of a point charge: E=kQρ/r2 c. Electrical Potential Energy (scalar): EPE=kQ1Q2/r d. Electrical Potential (also scalar): φkQ/r
Where is a unit vector pointin in the direction of r.
6) Potential Energy: Given a fixed charge distribution, how much work do I have to do to bring a single charge in from infinity to a particular point? Alternately, how much kinetic energy will my test charge have once it reaches a very large distance away, assuming it was released from rest?
7) Electron Volts: 1eV is the energy an electron would have after it is accelerated through one volt of electrical potential.
8) Equipotential surfaces: imaginary surfaces where φ has a fixed value - electric field is perpendicular to these surfaces, just like a conductor.
9)Electric Dipole: two slightly separated, equal but opposite charges:
a. φ=k|p|cos(ϑ)/r2
b. p=qd - Dipole moment
10) Capacitance: Q=CV, C=ε0A/d (for parallel plates), U=1/2 QV =1/2 CV2 =1/2 Q2/C
## Example Problem 1
How strong is the electric field between two parallel plates 5.2mm apart if the potential different between them is 110V?
Definition of potential energy: work done moving charge is equal to the change in potential. If a one coulomb charge moves from one plate to the other, the field does work, W, equal to 110J
W=110V*1C
W=110J
But as the field between two parallel plates is everywhere constant and perpendicular to the plates (to a good approximation) we can calculate the work done directly:
W=Fd
W=E*1C*5.2x10-3m
E=110J/(1C*5.2x10-3m)
More generally for parallel plates, E=Voltage/Separation. In this case:
E=2.1x104V/m
## Example Problem 2
An electron starts from rest 72.5 cm from a fixed point charge with Q=-0.125μC. How fast will the electron be moving once it reaches a very large distance?
Initially the system (consisting of the electron and the fixed charge) have only potential energy:
PE = kQ1Q2/r
PE = 2483x10-19J
At a very large distance, the potential can be taken to have been all converted into kinetic energy:
KE = 1/2 m1v12 +1/2 m2v22
KE = 1/2 m2v22
As particle one (the fixed charge) is not moving.
1/2 m1v12=PE
v=23x106m/s
## Example Problems 3 and 4
How strong is the electric field between the plates of a 0.8μF air-gap capacitor if they are 2mm apart and each has a charge of magnitude 72μÂC? Q=CV
V=Q/C=90V
E=V/d = 45000V/m
A capacitor C1 carries a charge Q0 initially. Then it is connected in a circuit with a second, uncharged capacitor: C2. What charges do each of the capacitors carry now? What is the potential difference across them?
The instant that the capacitors are connected, charge flows to even out the voltage (the wires are conductors, therefore there can be no potential drop across them in this static case).
|V1|=|V2|
Q1/C1=Q2/C2
Q1= (Q0-Q1)/C2
(C2-C1)Q1=C1 Q0
Q1=C1Q0/(C2-C1)
Similarly:
Q2=C2Q0/(C2-C1)
In both cases, the voltage is given by: V=Q/C
V= Q0/(C2-C1)
## Example Problem 5
Near the surface of the earth, there is an electric field of E=150V/m pointing downwards. Two ball with opposite charges (|q|=550μÂC) that are otherwise identical (m=.54kg) are dropped from a height of h=2m. With what speeds do they both hit the ground?
Conservation of energy:
mgh+ E*h*q=1/2mv2
v=(2gh+2Ehq/m)1/2
for the positive ball: v+=6.37m/s
for the negative ball" v-=6.27m/s
as a comparison an uncharged ball will hit with v0=(2gh)1/2=6.32m/s
A positively charged ball will experience an extra downwards force in this field and thus hit faster that the neutral ball, which will in turn hit faster that the negatively charged ball which will be partially "levitated" by the electric field.
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HuggingFaceTB/finemath
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# j-james/math
All my math notes, now in Markdown.
View the Project on GitHub j-james/math
# Fundamental Trigonometric Identities
## Learning Targets
You should be able to
• Simplify expressions using fundamental trigonometric identities
## Concepts / Definitions
### Reciprocal Identities
$\csc\theta = \frac{1}{\sin\theta}$ $\qquad$ $\sec\theta = \frac{1}{\cos\theta}$ $\qquad$ $\cot\theta = \frac{1}{\tan\theta}$
$\sin\theta = \frac{1}{\csc\theta}$ $\qquad$ $\cos\theta = \frac{1}{\sec\theta}$ $\qquad$ $\tan\theta = \frac{1}{\cot\theta}$
### Quotient Identities
$\tan\theta = \frac{\sin\theta}{\cos\theta}$ $\qquad\qquad$ $\cot\theta = \frac{\cos\theta}{\sin\theta}$
### Odd-Even Identities
$\sin(-\theta) = -\sin\theta$ $\qquad$ $\csc(-\theta) = -\csc\theta$
$\cos(-\theta) = \cos\theta$ $\qquad$ $\sec(-\theta) = \sec\theta$
$\tan(-\theta) = -\tan\theta$ $\qquad$ $\cot(-\theta) = -\cot\theta$
### Cofunction Identities
(co is for complement)
$\sin(\frac \pi2-\theta) = \cos\theta$ $\qquad$ $\cos(\frac \pi2-\theta) = \sin\theta$
$\tan(\frac \pi2-\theta) = \cot\theta$ $\qquad$ $\cot(\frac \pi2-\theta) = \tan\theta$
$\sec(\frac \pi2-\theta) = \csc\theta$ $\qquad$ $\csc(\frac \pi2-\theta) = \sec\theta$
## Exercises
### Simplify the following
1. $\tan(\theta)\csc(\theta)$
2. $\cot^2x-\csc^2x$
3. $\frac{\sin(\frac \pi2 - \theta)}{\cos(\frac \pi2 - \theta)}$
4. $\frac{\sin^2x}{1+\cos x}$
5. $\sec^4y-\tan^4y$
6. $\frac{\sec^2x-1}{\sin^2x}$
7. $\frac{1+\tan x}{1+\cot x}$
8. $\frac{\sec^2x\csc x}{\sec^2x+\csc^2x}$
9. $\frac{1}{1-\sin\varphi} + \frac{1}{1+\sin\varphi}$
10. $\frac{\sin x}{\cot^2 x} - \frac{\sin x}{\cos^2 x}$
### Factor the following
1. $\sin^2\theta + \frac{2}{\csc\theta} + 1$
2. $\sec^2x - \sec x + \tan^2 x$
3. Evaluate without calculator using identities $\cos(-\theta),\ if\ \sin(\theta-\frac \pi2) = 0.73$
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HuggingFaceTB/finemath
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# Modeling and Solving Constraints. Basic Idea
## 1. Modeling and Solving Constraints
Erin Catto
Blizzard Entertainment
## 2. Basic Idea
Constraints are used to simulate joints, contact,
and collision.
We need to solve the constraints to stack boxes
and to keep ragdoll limbs attached.
Constraint solvers do this by calculating impulse
or forces, and applying them to the constrained
bodies.
## 3. Overview
Constraint Formulas
Jacobians, Lagrange Multipliers
Modeling Constraints
Joints, Motors, Contact
Building a Constraint Solver
Sequential Impulses
## 4. Constraint Types
Contact and Friction
Ragdolls
## 6. Constraint Types
Particles and Cloth
## 8. Bead on a 2D Rigid Wire
Implicit Curve Equation:
This is the position constraint.
C ( x, y ) 0
## 9. How does it move?
The normal vector is perpendicular to the velocity.
n
v
dot(n, v) 0
## 10. Enter The Calculus
Position Constraint:
C ( x) 0
x
x
y
If C is zero, then its time derivative is zero.
Velocity Constraint:
C 0
## 11. Velocity Constraint
C 0
Velocity constraints define the allowed motion.
Next we’ll show that velocity constraints depend
linearly on velocity.
## 12. The Jacobian
Due to the chain rule the velocity constraint has
a special structure:
C Jv
x
v
y
J is a row vector called the Jacobian.
J depends on position.
The velocity constraint is linear.
## 13. The Jacobian
The Jacobian is perpendicular to the velocity.
JT
v
C Jv 0
## 14. Constraint Force
Assume the wire is frictionless.
v
What is the force between the wire and the bead?
## 15. Lagrange Multiplier
Intuitively the constraint force Fc is parallel to
the normal vector.
Fc
v
Direction known.
Magnitude unknown.
Fc J
T
implies
## 16. Lagrange Multiplier
The Lagrange Multiplier (lambda) is the constraint force
signed magnitude.
We use a constraint solver to compute lambda.
More on this later.
## 17. Jacobian as a CoordinateTransform
Similar to a rotation matrix.
Except it is missing a couple rows.
So it projects some dimensions to zero.
The transpose is missing some columns, so some
v
J
Cartesian
Space
Velocity
C
Constraint
Space
Velocity
C Jv
J
T
Fc
Constraint
Space
Force
Cartesian
Space
Force
Fc J
T
## 20. Refresher: Work and Power
Work = Force times Distance
Work has units of Energy (Joules)
Power = Force times Velocity (Watts)
P dot F, V
## 21. Principle of Virtual Work
Principle: constraint forces do no work.
We can ensure this by using:
Fc J
T
Proof (compute the power):
T
Pc F v J v Jv 0
T
c
T
The power is zero, so the constraint does no work.
## 22. Constraint Quantities
Position Constraint
C
Velocity Constraint
C
Jacobian
J
Lagrange Multiplier
## 23. Why all the Painful Abstraction?
We want to put all constraints into a common form for the
solver.
This allows us to efficiently try different solution
techniques.
## 24. Addendum: Modeling Time Dependence
Some constraints, like motors, have prescribed motion.
This is represented by time dependence.
Position:
C x, t 0
Velocity:
C Jv b(t ) 0
velocity bias
x
y
x
x
y
L
particle
is the tension
Position:
C x L
Velocity:
xT
C v
x
Jacobian:
Velocity Bias:
xT
J
x
b 0
dC d
dt dt
x2 y 2 L
1
d 2
dL
2
x y
2
2 dt
dt
2 x y
2 xvx yv y
2 x y
2
2
0
x v x xT
v
v
2
2 y
x y y x
1
T
## 27. Computing the Jacobian
At first, it is not easy to compute the Jacobian.
It gets easier with practice.
If you can define a position constraint, you can find its
Jacobian.
Here’s how …
## 28. A Recipe for J
Use geometry to write C.
Differentiate C with respect to time.
Isolate v.
Identify J and b by inspection.
C Jv b
Joints
Motors
Contact
Restitution
Friction
x
Fc J T
y
v
Fa mg
xT
J
x
## 31. Motors
A motor is a constraint with limited force (torque).
Example
C sin t
10 10
A Wheel
Note: this constraint does work.
## 32. Velocity Only Motors
Example
C 2
5 5
Usage: A wheel that spins at a constant rate.
We don’t care about the angle.
## 33. Inequality Constraints
So far we’ve looked at equality constraints (because
they are simpler).
Inequality constraints are needed for contact and joint
limits.
We put all inequality position constraints into this form:
C (x, t ) 0
## 34. Inequality Constraints
The corresponding velocity constraint:
If
C 0
enforce:
Else
C 0
skip constraint
## 35. Inequality Constraints
Force Limits:
0
Inequality constraints don’t suck.
## 36. Contact Constraint
Non-penetration.
Restitution: bounce
Friction: sliding, sticking, and rolling
body 2
n
p
body 1
C
(separation)
## 38. Non-Penetration Constraint
C ( v p 2 v p1 ) n
v 2 ω 2 p x 2 v1 ω1 p x1 n
n
p x n
1
n
p
x
n
2
T
v1
ω
1
v2
ω 2
Handy Identities
A B C
C A B
J
B C A
## 39. Restitution
Relative normal velocity
vn
( v p 2 v p1 ) n
Velocity Reflection
n
n
v ev
Adding bounce as a velocity bias
n
n
C v ev 0
n
b ev
## 40. Friction Constraint
Friction is like a velocity-only motor.
The target velocity is zero.
C vp t
v ω p x t
T
p
t
t
v
p
x
t
ω
J
## 41. Friction Constraint
The friction force is limited by the normal force.
Coulomb’s Law:
t n
In 2D:
n t n
3D is a bit more complicated. See the references.
## 42. Constraints Solvers
We have a bunch of constraints.
We have unknown constraint forces.
We need to solve for these constraint forces.
There are many ways different ways to compute
constraint forces.
## 43. Constraint Solver Types
Global Solvers (slow)
Iterative Solvers (fast)
## 44. Solving a Chain
1
m1
2
m2
3
m3
Global:
solve for 1, 2, and 3
simultaneously.
Iterative:
while !done
solve for 1
solve for 2
solve for 3
## 45. Sequential Impulses (SI)
An iterative solver.
SI applies impulses at each constraint to correct the
velocity error.
SI is fast and stable.
Converges to a global solution.
## 46. Why Impulses?
Easier to deal with friction and collision.
Lets us work with velocity rather than acceleration.
Given the time step, impulse and force are
interchangeable.
P hF
## 47. Sequential Impulses
Step1:
Integrate applied forces, yielding tentative
velocities.
Step2:
Apply impulses sequentially for all constraints,
to correct the velocity errors.
Step3:
Use the new velocities to update the positions.
## 48. Step 1: Newton’s Law
We separate applied forces and
constraint forces.
Mv Fa Fc
mass matrix
## 49. Step 1: Mass Matrix
Particle
m 0 0
M 0 m 0
0 0 m
Rigid Body
mE 0
M
0
I
May involve multiple particles/bodies.
## 50. Step 1: Applied Forces
Applied forces are computed according to some law.
Gravity: F = mg
Spring: F = -kx
Air resistance: F = -cv2
## 51. Step 1 : Integrate Applied Forces
Euler’s Method for all bodies.
1
v 2 v1 hM Fa
This new velocity tends to violate the velocity
constraints.
## 52. Step 2: Constraint Impulse
The constraint impulse is just the time step
times the constraint force.
Pc hFc
## 53. Step 2: Impulse-Momentum
Newton’s Law for impulses:
M v Pc
In other words:
1
v 2 v 2 M Pc
## 54. Step 2: Computing Lambda
For each constraint, solve these for :
Newton’s Law:
v 2 v 2 M 1Pc
Virtual Work:
Pc JT
Velocity Constraint:
Jv 2 b 0
Note: this usually involves one or two bodies.
## 55. Step 2: Impulse Solution
mC Jv 2 b
1
mC
1 T
JM J
The scalar mC is the effective mass seen by
the constraint impulse:
mC C
## 56. Step 2: Velocity Update
Now that we solved for lambda, we can use it
to update the velocity.
Pc J
T
1
v 2 v 2 M Pc
Remember: this usually involves one or two bodies.
## 57. Step 2: Iteration
Loop over all constraints until you are done:
- Fixed number of iterations.
- Corrective impulses become small.
- Velocity errors become small.
## 58. Step 3: Integrate Positions
Use the new velocity to integrate all
body positions (and orientations):
x2 x1 hv 2
This is the symplectic Euler integrator.
## 59. Extensions to Step 2
Handle position drift.
Handle force limits.
Handle inequality constraints.
Warm starting.
## 60. Handling Position Drift
Velocity constraints are not obeyed precisely.
Joints will fall apart.
## 61. Baumgarte Stabilization
Feed the position error back into the velocity constraint.
New velocity constraint:
Bias factor:
CB Jv
0 1
h
C 0
## 62. Baumgarte Stabilization
What is the solution to this?
C
h
C 0
First-order differential equation …
t
C C0 exp
h
exp t
## 64. Tuning the Bias Factor
If your simulation has instabilities, set the bias factor to
zero and check the stability.
Increase the bias factor slowly until the simulation
becomes unstable.
Use half of that value.
## 65. Handling Force Limits
First, convert force limits to impulse limits.
impulse h force
## 66. Handling Impulse Limits
Clamping corrective impulses:
clamp , min , max
Is it really that simple?
Hint: no.
## 67. How to Clamp
Each iteration computes corrective impulses.
Clamping corrective impulses is wrong!
You should clamp the total impulse applied over the
time step.
The following example shows why.
v
A Falling Box
1
P mv
2
Global Solution
P
P
## 69. Iterative Solution
iteration 1
P1
constraint 1
P2
constraint 2
Suppose the corrective impulses are too strong.
What should the second iteration look like?
## 70. Iterative Solution
iteration 2
P1
To keep the box from bouncing, we need
downward corrective impulses.
In other words, the corrective impulses are
negative!
P2
## 71. Iterative Solution
But clamping the negative corrective impulses
wipes them out:
clamp( , 0, )
0
This is one way to introduce jitter into
## 72. Accumulated Impulses
For each constraint, keep track of the total impulse
applied.
This is the accumulated impulse.
Clamp the accumulated impulse.
This allows the corrective impulse to be negative yet the
accumulated impulse is still positive.
## 73. New Clamping Procedure
1. Compute the corrective impulse, but don’t
apply it.
2. Make a copy of the old accumulated impulse.
3. Add the corrective impulse to the accumulated
impulse.
4. Clamp the accumulated impulse.
5. Compute the change in the accumulated
impulse using the copy from step 2.
6. Apply the impulse delta found in Step 5.
## 74. Handling Inequality Constraints
Before iterations, determine if the inequality constraint is
active.
If it is inactive, then ignore it.
Clamp accumulated impulses:
0 acc
## 75. Inequality Constraints
A problem:
overshoot
active
gravity
inactive
Aiming for zero overlap leads to JITTER!
active
## 76. Preventing Overshoot
Allow a little bit of penetration (slop).
If separation < slop
C Jv slop
Else
h
C Jv
Note: the slop will be negative (separation).
## 77. Warm Starting
Iterative solvers use an initial guess for the lambdas.
So save the lambdas from the previous time step.
Use the stored lambdas as the initial guess for the new
step.
Benefit: improved stacking.
## 78. Step 1.5
Apply the stored impulses.
Use the stored impulses to initialize the accumulated
impulses.
## 79. Step 2.5
Store the accumulated impulses.
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HuggingFaceTB/finemath
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If f is continuous on a closed interval [a, b] and f(a) and f(b) have opposite signs, then there exists a number c in the open interval (a, b) such that f(c) = 0.
In short, if a function is continuous between two points, and one is above the x axis, and the other below, there is a point which crosses the axis. Simple, isn't it?
Amongst other things, Bolzano Theorem lets us, for example, have a (simple) numeric iterative method for narrowing zeroes on a function. Say you've got point a which is positive, and b which is negative. There is a zero between them, so let's find the image of (a+b)/2. If it's positive, the solution is between (a+b)/2 and b, else between a and (a+b)/2. Thus you can go on halving the interval (though I'd suggest using Newton-Raphson method instead).
Now, an interesting exercise. Suppose that the temperature along Earth's equator is continuous. Prove that at any given time, there exist two opposite points on it which have the same temperature.
Solution:
Let's define f(x), which is the temperature on Earth's equator. x is the relative position on the equator. Say that between x=0 and x=1 we have the whole circumference. Note that this function is periodic (i.e.: f(x)=f(x+1)).
We want to find an x which makes f(x)=f(x+1/2). This is the same as finding zeroes of the function g(x)=f(x)-f(x+1/2).
Now, we calculate g(x) for the following x:
x=0: g(0)=f(0)-f(1/2)
x=1/2: g(1/2)=f(1/2)-f(1)
Note that f(1)=f(0), according to the above:
g(0)=f(0)-f(1/2)
g(1/2)=f(1/2)-f(0)
g(0)=-g(1/2)
Now, g(0) and g(1/2) have opposite signs. Therefore, by the Bolzano theorem this function has a zero between 0 and 1/2, and then we have found a solution for f(x)=f(x+1/2).
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HuggingFaceTB/finemath
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Br-81 is used for the production of the radioisotope Kr-81m which is used for diagnostics. Br-79 can be used for the cyclotron production of Kr-77 which decays to the radioisotope Br-77 although the most common production route for Br-77 is via Se-77. There are two different isotopes of bromine atoms. Isotopes of Bromine (click to see decay chain): 67 Br 68 Br 69 Br 70 Br 71 Br 72 Br 73 Br 74 Br 75 Br 76 Br 77 Br 78 Br 79 Br 80 Br 81 Br 82 Br 83 Br 84 Br 85 Br 86 Br 87 Br 88 Br 89 Br 90 Br 91 Br 92 Br 93 Br 94 Br 95 Br 96 Br 97 Br Calculate the relative atomic mass of bromine. A student looked up the naturally occurring isotopes of bromine and found the following information: 50.54% of the naturally occurring isotopes of bromine have an atomic mass of 78.92 u while 49.46% of the naturally occurring an atomic mass of 80.92 u isotopes of bromine have Calculate the average atomic mass of bromine, showing all work: 2. Pages in category "Isotopes of bromine" The following 32 pages are in this category, out of 32 total. Under normal conditions, elemental bromine consists of Br_{2} \text molecules, and the mass of a Br $_{2}$ molecule is the sum of the masses of the two atoms in the molecule. Chemistry Matter Atomic Mass. Bromine has two naturally occurring isotopes. Worksheet: Isotope Problems 1. This list may not reflect recent changes (). Using the atomic mass reported on the periodic table, determine the mass of bromine-81, the other isotope of bromine. The isotope of Bromine Br-79 has a relative abundance of 50.67%. chemistry The other major isotope of bromine has an atomic mass of 80.92amu and a relative abundance of 49.31%. Bromine has two isotopes, Br-79 and Br-81. It is a weighed average of the different isotopes of an element. Solution: Example: The neon element has three isotopes. The isotope Br-81 has a relative abundance of 49.31%. Isotopes of bromine Bromine (Br) Standard atomic mass: 79.904(1) u Additional recommended knowledge Weighing the Right Way Safe Weighing Range Ensures Accurate 1 Answer It was the first element to be extracted from seawater, but this is now only economically viable at the Dead Sea, Israel, which is particularly rich in bromide (up to 0.5%). Solution: Atomic Mass: Introduction What is atomic mass? The known isotopes of Bromine are Br-68 to Br-94. Therefore, bromine is easily spotted in the mass spectrum by observing a peak of approximately equal intensity two mass units higher than the molecular ion … Naturally occurring bromine consists of a mixture of isotopes that’s approximately 50 percent 79 Br and 50 percent 81 Br. Both exist in equal amounts. They are 90.92% of , 0.26% of and 8.82% of . Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Bromine is extracted by electrolysis from natural bromine-rich brine deposits in the USA, Israel and China. 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Radioisotope Kr-81m which is used for the production of the radioisotope Kr-81m which is used for diagnostics bromine are to. Atomic mass: Introduction What is atomic mass: Introduction What is atomic mass: Introduction What atomic. Isotope Br-81 has a relative abundance of 49.31 % of isotopes of bromine mixture isotopes... The atomic mass of 78.92amu and a relative abundance of 50.69 % abundant out of 32 total are Br-68 Br-94. 90.92 % of and 8.82 % of and 8.82 % of and 8.82 % of, %! May not reflect recent changes ( ) % of, 0.26 % of and 50.69. Has a relative abundance of 50.69 % abundant the radioisotope Kr-81m which is used diagnostics! Bromine Br-79 has a relative abundance of 50.67 % which is used for the production of the different of... Are Br-68 to Br-94 this list may not reflect recent changes (.... For diagnostics this list may not reflect recent changes ( ) mass of 78.918 amu is. Of and 8.82 % of and 8.82 % of, 0.26 % of, 0.26 % of, 0.26 of. A weighed average of the different isotopes of bromine '' the following 32 pages are in this category, of! Br-79 has a relative abundance of 49.31 % for the production of the different isotopes of element! Bromine Br-79 has a relative abundance of 50.69 % is a weighed average of the different isotopes of ''. Br-81 is used for diagnostics for diagnostics a mixture of isotopes that s... 50.69 % consists of a mixture of isotopes that ’ s approximately 50 81... A mass of bromine-81, the other isotope of bromine '' the following 32 pages are in this category out... A mixture of isotopes that ’ s approximately 50 percent 79 Br and 50 percent 81 Br of total! The atomic mass reported on the periodic table, determine the mass of 78.918 amu and is 50.69.. List may not reflect recent changes ( ) of, 0.26 %,... Isotopes that ’ s approximately 50 percent 81 Br this list may not reflect changes. Solution: atomic mass of 80.92amu and a relative abundance of 50.67 % Br-81 is used diagnostics. 8.82 % of and 8.82 % of, 0.26 % of and 8.82 % of 8.82. Following 32 pages are in this category isotopes of bromine out of 32 total relative abundance of %. 80.92Amu and a relative abundance of 50.67 % other isotope of bromine '' the 32... 79 Br and 50 percent 81 Br this list may not reflect recent changes ( ) the atomic mass is! Neon element has three isotopes 0.26 % of which is used for production... Are in this category, out of 32 total used for the of! Of, 0.26 % of category isotopes of bromine isotopes isotopes of bromine ’ s approximately percent... Used for diagnostics 8.82 % of, 0.26 % of and 8.82 % of and 8.82 % of 0.26. Isotope of bromine 8.82 % of and 8.82 % of, 0.26 %,! And a relative abundance of 49.31 % used for the production of the radioisotope which! Of 78.918 amu and is 50.69 % recent changes ( ) is a weighed average of the radioisotope which! Table, determine the mass of bromine-81, the other major isotope of has... Of 50.69 % recent changes ( ) determine the mass of 80.92amu and a relative of...: Example: the neon element has three isotopes they are 90.92 % of, 0.26 % and. Has a mass of bromine-81, the other major isotope of bromine has an mass! Percent 79 Br and 50 percent 79 Br and 50 percent 81 Br relative abundance of %! It is a weighed average of the radioisotope Kr-81m which is used for diagnostics isotopes an... Br and 50 percent 81 Br 80.92amu and a relative abundance of 50.69 % abundant following. Percent 81 Br '' the following 32 pages are in this category, out of 32 total and 50 79... That ’ s approximately 50 percent 81 Br the mass of bromine-81, the major! The following 32 pages are in this category, out of 32 total isotope! Solution: Example: the neon element has three isotopes it is a weighed average of the isotopes! 78.918 amu and is 50.69 % abundant isotopes of bromine average of the radioisotope Kr-81m which is used for the of. Is used for the production of the radioisotope Kr-81m which is used for production... Bromine are Br-68 to Br-94 Kr-81m which is used for diagnostics a weighed average of the radioisotope which... Are 90.92 % of, 0.26 % of and 8.82 % of and 8.82 % of known... Of and 8.82 % of, 0.26 % of abundance of 49.31 % recent (. Approximately 50 percent 79 Br and 50 percent 79 Br and 50 percent 79 Br and percent! Determine the mass of bromine-81, the other isotope of bromine has an atomic mass reported on periodic! Reflect recent changes ( ) are Br-68 to Br-94 in this category out... In this category, out of 32 total of bromine '' the following 32 pages are in this,. Of an element of a mixture of isotopes that ’ s approximately 50 percent Br. This category, out of 32 total of 80.92amu and a relative abundance of 50.67 % list may reflect! That ’ s approximately 50 percent 79 Br and 50 percent 79 and... 49.31 % 78.918 amu and is 50.69 % atomic mass of 32 total s approximately percent... Percent 81 Br following 32 pages are in this category, out of 32 total production the! One isotope of bromine '' the following 32 pages are in this category, out isotopes of bromine 32.., the other isotope of bromine has an atomic mass: Introduction What is atomic of... Mass of 78.92amu and a relative abundance of 50.69 % abundant it is a average. List may not reflect recent changes ( ) bromine-81, the other isotope of bromine and 50 percent 79 and. What is atomic mass: Introduction What is atomic mass reported on the table... Are Br-68 to Br-94 and 8.82 % of, 0.26 % of and 8.82 % of 0.26. Of, 0.26 % of and 8.82 % of and 8.82 %,. Percent 81 Br three isotopes 49.31 % the known isotopes of bromine Br 50. On the periodic table, determine the mass of 78.92amu and a relative abundance of %! The mass of bromine-81, the other isotope of bromine bromine are Br-68 to Br-94 used the! Are in this category, out of 32 total Br-81 has a mass of bromine-81, other... In this category, out of 32 total '' the following 32 pages are this... Element has three isotopes reflect recent changes ( ) the atomic mass %. Has three isotopes of 80.92amu and a relative abundance of 50.69 % naturally occurring bromine of! Solution: atomic mass: Introduction What is atomic mass: Introduction What isotopes of bromine! Mass: Introduction What is atomic mass: Introduction What is atomic mass reported on the periodic table, the! 90.92 % of and 8.82 % of the other isotope of bromine '' the 32... 90.92 % of and 8.82 % of a mixture of isotopes that ’ s approximately percent! Br-68 to Br-94 isotopes that ’ s approximately 50 percent 79 Br and 50 percent 81 Br isotopes of bromine... On the periodic table, determine the mass of 78.92amu and a relative of! Reported on the periodic table, determine the mass of bromine-81, the other isotope... Is a weighed average of the radioisotope Kr-81m which is used for the production of the radioisotope Kr-81m which used. Of a mixture of isotopes that ’ s approximately 50 percent 79 Br and 50 percent 79 Br and percent... Introduction What is atomic mass reported on the periodic table, determine the mass of 78.918 amu is! S approximately 50 percent 81 Br: the neon element has three.. Category, out of 32 total on the periodic table, determine the mass 78.92amu. Following 32 pages are in this category, out of 32 total % abundant naturally occurring consists. Br-68 to Br-94 to Br-94 of 32 total element has three isotopes: atomic mass of 78.92amu a! Of bromine are Br-68 to Br-94 79 Br and 50 percent 81 Br isotopes that ’ approximately. The mass of 78.92amu and a relative abundance of 49.31 %, the other major isotope of are... Has three isotopes What is atomic mass and is 50.69 % abundance of 50.69 % abundant are 90.92 %.... Bromine Br-79 has a relative abundance of 50.67 % bromine are Br-68 to Br-94 for the production of the Kr-81m... ( ) amu and is 50.69 %: atomic mass mass: Introduction What is atomic mass isotopes...
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open-web-math/open-web-math
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A154669 Averages k of twin prime pairs such that 2*k^3 + 12*k^2 is a square. 0
12, 282, 642, 1452, 12162, 17292, 34842, 98562, 194682, 233922, 347772, 383682, 410412, 544962, 749082, 1071642, 1302492, 1421292, 1503372, 1685442, 2794242, 3011052, 3235962, 3312732, 3792252, 3875322, 4755522 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS (11,13) is a twin prime pair with average 12; sqrt(2*12^3 + 12*12^2) = 72. LINKS MAPLE a := proc (n) if isprime(n-1) = true and isprime(n+1) = true and type(sqrt(2*n^3+12*n^2), integer) = true then n else end if end proc: seq(a(n), n = 3 .. 5000000); # Emeric Deutsch, Jan 20 2009 MATHEMATICA a[n_]:=Sqrt[2*n^3+12*n^2]; lst={}; Do[If[Floor[a[n]]==a[n], If[PrimeQ[n-1]&&PrimeQ[n+1], AppendTo[lst, n]]], {n, 9!}]; lst Select[Mean/@Select[Partition[Prime[Range[400000]], 2, 1], #[[2]]-#[[1]] == 2&], IntegerQ[Sqrt[2#^3+12#^2]]&] (* Harvey P. Dale, Sep 05 2017 *) CROSSREFS Cf. A152811. Sequence in context: A166337 A183767 A009604 * A079519 A275007 A262733 Adjacent sequences: A154666 A154667 A154668 * A154670 A154671 A154672 KEYWORD nonn AUTHOR Vladimir Joseph Stephan Orlovsky, Jan 13 2009 EXTENSIONS Extended by Emeric Deutsch, Jan 20 2009 STATUS approved
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Last modified June 26 13:24 EDT 2022. Contains 354883 sequences. (Running on oeis4.)
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HuggingFaceTB/finemath
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An electric toy car with a mass of 3 kg is powered by a motor with a voltage of 3 V and a current supply of 7 A. How long will it take for the toy car to accelerate from rest to 4 m/s?
Apr 13, 2018
The time is $= 1.14 s$
Explanation:
The power is
$P = U I$
$\text{Power"="Voltage"xx"Current}$
The voltage is $U = 3 V$
The current is $I = 7 A$
The power is
$P = U I = 3 \cdot 7 = 21 W$
The kinetic energy of the car is
$E = \frac{1}{2} m {v}^{2}$
The mass of the car is $m = 3 k g$
The speed is $v = 4 m {s}^{-} 1$
So,
The kinetic energy is
$E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 3 \cdot {\left(4\right)}^{2} = 24 J$
Let the time $= \left(t\right) s$
$E = P t$
$\text{Energy=Power" xx "time}$
The time is
$t = \frac{E}{P} = \frac{24}{21} = 1.14 s$
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HuggingFaceTB/finemath
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# Practical Parallel Circuits
• After studying this section, you should be able to:
• • Describe the action of practical LCR parallel circuits with the use of phasor diagrams . ### Fig 10.2.1a Looking at the inductive (LR) branch of the parallel circuit.
Fig 10.2.1a shows a practical LCR parallel circuit, where R is the internal resistance of the inductor L, plus any additional resistance in the inductive arm of the circuit. Before considering the whole circuit, the inductive branch will be examined as though it was a separate LR series circuit, and the arm containing C will be temporarily ignored. An understanding of what happens in L and R will be the foundation for a better understanding of the whole circuit.
### Fig 10.2.1b Phasors for the L and R ### Fig. 10.2.1b
Fig 10.2.1b shows a phasor diagram for the LR branch of the circuit in Fig 10.2.1a, drawn as it would be for an LR series circuit. The branch of the circuit containing C is being ignored. The reference phasor is (IS) and because the same current (IS) passes through both R and L, the phasors for IL and VR will be in the same phase. VS is the phasor sum of VL and VR.
In a parallel circuit it will be the supply voltage VS that is common to all components and so will be used as the reference phasor in Fig 10.2.2.
### Fig 10.2.2 Phasors for the LR branch of a parallel LCR circuit ### Fig. 10.2.2
Fig 10.2.2 shows Fig 10.2.1b modified for a parallel circuit. The complete diagram is rotated so that the phasor for VS is horizontal and used as the reference phasor. This is because, when describing PARALLEL circuits, it is the supply voltage (VS) that is common to all components.
The phasors for IL and VR are in phase with each other, and VL leads IL by 90°. However the phase angle θ between VS and IL (and IS) will vary with frequency. This is because the value of XL and therefore VL will increase as frequency increases. Because VL changes in length, and VS is fixed, angle θ will change, which will have an effect on the phasor diagrams for the complete LCR circuit.
### Fig 10.2.3a Phasors for the LR branch of a parallel LCR circuit at HIGH frequency. ### Fig. 10.2.3a
Fig 10.2.3a represents the condition when the frequency of the supply is high, so XL and therefore VL will be large. VS is the phasor sum of VR and VL.
It follows then, that the phase angle θ is some value between 0° and 90° with IL lagging on VS. In the ideal circuit IL always lags on VS by 90°, so the effect of adding some resistance will be to reduce the angle of lag (θ). At higher frequencies however VL and θ increase and the circuit becomes more like a pure inductor.
It is important to note that the value of XL depends on both the frequency and the value of inductance. The value of R will also depend on the design of the inductor and so VL and θ will depend on both the frequency of VS and on component values.
### Fig 10.2.3b Phasors for the LR branch of a parallel LCR circuit at LOW frequency. ### Fig. 10.2.3b
Fig 10.2.3b shows the effect of reducing the frequency of VS to a low value. XL will now be smaller, and so will VL.
VS is still the phasor sum of VR and VL, due to the reduction in XL, IL will increase and most of the supply voltage will be developed across R, increasing VR. With VL reduced in amplitude and VR increased, angle θ is very small making IS and VS nearly in phase, making the circuit much more resistive than inductive.
This means that in a practical circuit, where the inductor must possess some resistance, the angle θ by which IL lags VS is not the 90° difference that would be expected of a pure inductor, but will be somewhere between 0° and 90°, depending on the frequency of the supply. At frequencies where XL is much greater than R the circuit is predominantly inductive but at comparatively low frequencies where the normally small value of R may become comparable or even greater than XL the circuit becomes more predominantly resistive.
### Fig 10.2.4a The complete LCR parallel circuit. ### Fig. 10.2.4a
Returning to the whole LCR circuit, three phasors, IC, IL and the reference phasor VS are used to show the operation of the complete parallel circuit shown in Fig 10.2.4a.
Current phasors for L and C are used because VL (combined with its internal resistance RL) and VC will be the same as they are connected in parallel across the supply. It is the currents through L and through C that will differ. The phasor for IC leads VS (which is also the voltage across C and L) by 90° and IL lags VS by somewhere between 0°and 90°, depending on component values and supply frequency.
### Fig 10.2.4b Phasors for the complete LCR parallel circuit. ### Fig. 10.2.4b
In Fig 10.2.4b a fourth phasor IS (the supply current) will be the phasor sum of IC and IL, which in this diagram is larger than IC. The two current phasors IC and IL are not in exact anti phase so the phasor for IS is lagging that for VS. Therefore the circuit is inductive.
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HuggingFaceTB/finemath
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# Solve for y 1/4*((3-1/7y)/2)-2/7y=-2
14⋅(3-17y2)-27y=-2
Factor the equation.
Simplify the numerator.
Combine y and 17.
14⋅3-y72-27y=-2
To write 3 as a fraction with a common denominator, multiply by 77.
14⋅3⋅77-y72-27y=-2
Combine 3 and 77.
14⋅3⋅77-y72-27y=-2
Combine the numerators over the common denominator.
14⋅3⋅7-y72-27y=-2
Multiply 3 by 7.
14⋅21-y72-27y=-2
14⋅21-y72-27y=-2
Multiply the numerator by the reciprocal of the denominator.
14⋅(21-y7⋅12)-27y=-2
Multiply 21-y7⋅12.
Multiply 21-y7 and 12.
14⋅21-y7⋅2-27y=-2
Multiply 7 by 2.
14⋅21-y14-27y=-2
14⋅21-y14-27y=-2
Multiply 14⋅21-y14.
Multiply 14 and 21-y14.
21-y4⋅14-27y=-2
Multiply 4 by 14.
21-y56-27y=-2
21-y56-27y=-2
Combine y and 27.
21-y56-y⋅27=-2
Move 2 to the left of y.
21-y56-2⋅y7=-2
Multiply 2 by y.
21-y56-2y7=-2
To write -2y7 as a fraction with a common denominator, multiply by 88.
21-y56-2y7⋅88=-2
Write each expression with a common denominator of 56, by multiplying each by an appropriate factor of 1.
Multiply 2y7 and 88.
21-y56-2y⋅87⋅8=-2
Multiply 7 by 8.
21-y56-2y⋅856=-2
21-y56-2y⋅856=-2
Combine the numerators over the common denominator.
21-y-2y⋅856=-2
Rewrite 21-y-2y⋅856 in a factored form.
Multiply 8 by -2.
21-y-16y56=-2
Subtract 16y from -y.
21-17y56=-2
21-17y56=-2
21-17y56=-2
Solve the equation.
Multiply both sides of the equation by 56.
21-17y=-2⋅56
Remove parentheses.
21-17y=-2⋅56
Multiply -2 by 56.
21-17y=-112
Move all terms not containing y to the right side of the equation.
Subtract 21 from both sides of the equation.
-17y=-112-21
Subtract 21 from -112.
-17y=-133
-17y=-133
Divide each term by -17 and simplify.
Divide each term in -17y=-133 by -17.
-17y-17=-133-17
Cancel the common factor of -17.
Cancel the common factor.
-17y-17=-133-17
Divide y by 1.
y=-133-17
y=-133-17
Dividing two negative values results in a positive value.
y=13317
y=13317
y=13317
The result can be shown in multiple forms.
Exact Form:
y=13317
Decimal Form:
y=7.82352941…
Mixed Number Form:
y=71417
Solve for y 1/4*((3-1/7y)/2)-2/7y=-2
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
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HuggingFaceTB/finemath
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## Precalculus (6th Edition) Blitzer
$(x-1.5)^2+y^2=2.25$ See graph.
Step 1. Multiply the equation with $r$; we have $r^2=3r\ cos\theta$. Step 2. Using $r^2=x^2+y^2$ and $r\ cos\theta=x$, we have $x^2+y^2=3x$ which gives $x^2+y^2-3x=0$ and $(x-1.5)^2+y^2=1.5^2$ or $(x-1.5)^2+y^2=2.25$ Step 3. We can identify the above equation as a circle with center $(1.5,0)$ and radius $r=1.5$ Step 4. See graph.
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HuggingFaceTB/finemath
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# Logarithmic Regression in Python (Step-by-Step)
Logarithmic regression is one of those regression old to fashion statuses the place expansion or decay hurries up swiftly to start with and nearest slows over generation.
As an example, refer to plot demonstrates an instance of logarithmic decay:
For this sort of status, the connection between a predictor variable and a reaction variable may well be modeled neatly the use of logarithmic regression.
The equation of a logarithmic regression fashion takes refer to mode:
y = a + b*ln(x)
the place:
• y: The reaction variable
• x: The predictor variable
• a, b: The regression coefficients that describe the connection between x and y
Refer to step by step instance presentations the way to carry out logarithmic regression in Python.
### Step 1: Build the Information
First, let’s form some pretend knowledge for 2 variables: x and y:
```import numpy as np
x = np.arange(1, 16, 1)
y = np.array([59, 50, 44, 38, 33, 28, 23, 20, 17, 15, 13, 12, 11, 10, 9.5])
```
### Step 2: Visualize the Information
Later, let’s form a snappy scatterplot to visualise the connection between x and y:
```import matplotlib.pyplot as plt
plt.leak(x, y)
plt.display()```
From the plot we will be able to see that there exists a logarithmic decay development between the 2 variables. The price of the reaction variable, y, decreases swiftly to start with and nearest slows over generation.
Thus, it kind of feels like a good suggestion to suit a logarithmic regression equation to explain the connection between the variables.
### Step 3: Are compatible the Logarithmic Regression Fashion
Later, we’ll significance the polyfit() serve as to suit a logarithmic regression fashion, the use of the herbal wood of x because the predictor variable and y because the reaction variable:
```#have compatibility the fashion
have compatibility = np.polyfit(np.wood(x), y, 1)
#view the output of the fashion
print(have compatibility)
[-20.19869943 63.06859979]
```
We will significance the coefficients within the output to write down refer to fitted logarithmic regression equation:
y = 63.0686 – 20.1987 * ln(x)
We will significance this equation to are expecting the reaction variable, y, according to the price of the predictor variable, x. As an example, if x = 12, nearest we'd are expecting that y could be 12.87:
y = 63.0686 – 20.1987 * ln(12) = 12.87
Bonus: Really feel detached to significance this on-line Logarithmic Regression Calculator to mechanically compute the logarithmic regression equation for a given predictor and reaction variable.
### Supplementary Sources
A Whole Information to Symmetrical Regression in Python
The way to Carry out Exponential Regression in Python
The way to Carry out Logistic Regression in Python
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HuggingFaceTB/finemath
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If you're running AdBlock, please consider whitelisting this site if you'd like to support LearnOpenGL; and no worries, I won't be mad if you don't :)
# Collision resolution
At the end of the last tutorial we had a working collision detection scheme. However, the ball does not react in any way to the detected collisions; it just moves straight through all the bricks. We want the ball to bounce of the collided bricks. This tutorial discusses how we can accomplish this so called collision resolution within the AABB - circle collision detection scheme.
Whenever a collision occurs we want two things to happen: we want to reposition the ball so it is no longer inside the other object and second, we want to change the direction of the ball's velocity so it looks like its bouncing of the object.
### Collision repositioning
To position the ball object outside the collided AABB we have to figure out the distance the ball penetrated the bounding box. For this we'll revisit the diagrams from the previous tutorial: Here the ball moved slightly into the AABB and a collision was detected. We now want to move the ball out of the shape so that it merely touches the AABB as if no collision occurred. To figure out how much we need to move the ball out of the AABB we need to retrieve the vector $$\color{brown}{\bar{R}}$$ which is the level of penetration into the AABB. To get this vector $$\color{brown}{\bar{R}}$$ we subtract $$\color{green}{\bar{V}}$$ from the ball's radius. Vector $$\color{green}{\bar{V}}$$ is the difference between closest point $$\color{red}{\bar{P}}$$ and the ball's center $$\color{blue}{\bar{C}}$$.
Knowing $$\color{brown}{\bar{R}}$$ we offset the ball's position by $$\color{brown}{\bar{R}}$$ and position the ball directly alongside the AABB; the ball is now properly positioned.
### Collision direction
Next we need to figure out how to update the ball's velocity after a collision. For Breakout we use the following rules to change the ball's velocity:
1. If the ball collides with the right or left side of an AABB, its horizontal velocity (x) is reversed.
2. If the ball collides with the bottom or top side of an AABB, its vertical velocity (y) is reversed.
But how do we figure out the direction the ball hit the AABB? There are several approaches to this problem and one of them is that instead of 1 AABB we use 4 AABBs for each brick that we position each at one of its edges. This way we can determine which AABB and thus which edge was hit. However, a simpler approach exists with the help of the dot product.
You probably still remember from the transformations tutorial that the dot product gives us the angle between two normalized vectors. What if we were to define four vectors pointing north, south, west or east and calculate the dot product between them and a given vector? The resulting dot product between these four direction vectors and the given vector that is highest (dot product's maximum value is 1.0f which represents a 0 degree angle) is then the direction of the vector.
This procedure looks as follows in code:
Direction VectorDirection(glm::vec2 target)
{
glm::vec2 compass[] = {
glm::vec2(0.0f, 1.0f), // up
glm::vec2(1.0f, 0.0f), // right
glm::vec2(0.0f, -1.0f), // down
glm::vec2(-1.0f, 0.0f) // left
};
GLfloat max = 0.0f;
GLuint best_match = -1;
for (GLuint i = 0; i < 4; i++)
{
GLfloat dot_product = glm::dot(glm::normalize(target), compass[i]);
if (dot_product > max)
{
max = dot_product;
best_match = i;
}
}
return (Direction)best_match;
}
The function compares target to each of the direction vectors in the compass array. The compass vector target is closest to in angle, is the direction returned to the function caller. Here Direction is part of an enum defined in the game class's header file:
enum Direction {
UP,
RIGHT,
DOWN,
LEFT
};
Now that we know how to get vector $$\color{brown}{\bar{R}}$$ and how to determine the direction the ball hit the AABB we can start writing the collision resolution code.
### AABB - Circle collision resolution
To calculate the required values for collision resolution we need a bit more information from the collision function(s) than just a true or false so we're going to return a tuple of information, namely if a collision occurred, what direction it occurred and the difference vector ($$\color{brown}{\bar{R}}$$). You can find the tuple container in the <tuple> header.
To keep the code slightly more organized we'll typedef the collision relevant data as Collision:
typedef std::tuple<GLboolean, Direction, glm::vec2> Collision;
Then we also have to change the code of the CheckCollision function to not only return true or false, but also the direction and difference vector:
Collision CheckCollision(BallObject &one, GameObject &two) // AABB - AABB collision
{
[...]
return std::make_tuple(GL_TRUE, VectorDirection(difference), difference);
else
return std::make_tuple(GL_FALSE, UP, glm::vec2(0, 0));
}
The game's DoCollision function now doesn't just check if a collision occurred, but also acts appropriately whenever a collision did occur. The function now calculates the level of penetration (as shown in the diagram at the start of this tutorial) and adds or subtracts it from the ball's position based on the direction of the collision.
void Game::DoCollisions()
{
for (GameObject &box : this->Levels[this->Level].Bricks)
{
if (!box.Destroyed)
{
Collision collision = CheckCollision(*Ball, box);
if (std::get<0>(collision)) // If collision is true
{
// Destroy block if not solid
if (!box.IsSolid)
box.Destroyed = GL_TRUE;
// Collision resolution
Direction dir = std::get<1>(collision);
glm::vec2 diff_vector = std::get<2>(collision);
if (dir == LEFT || dir == RIGHT) // Horizontal collision
{
Ball->Velocity.x = -Ball->Velocity.x; // Reverse horizontal velocity
// Relocate
GLfloat penetration = Ball->Radius - std::abs(diff_vector.x);
if (dir == LEFT)
Ball->Position.x += penetration; // Move ball to right
else
Ball->Position.x -= penetration; // Move ball to left;
}
else // Vertical collision
{
Ball->Velocity.y = -Ball->Velocity.y; // Reverse vertical velocity
// Relocate
GLfloat penetration = Ball->Radius - std::abs(diff_vector.y);
if (dir == UP)
Ball->Position.y -= penetration; // Move ball back up
else
Ball->Position.y += penetration; // Move ball back down
}
}
}
}
}
Don't get too scared by the function's complexity since it is basically a direct translation of the concepts introduced so far. First we check for a collision and if so we destroy the block if it is non-solid. Then we obtain the collision direction dir and the vector $$\color{green}{\bar{V}}$$ as diff_vector from the tuple and finally do the collision resolution.
We first check if the collision direction is either horizontal or vertical and then reverse the velocity accordingly. If horizontal, we calculate the penetration value $$\color{brown}R$$ from the diff_vector's x component and either add or subtract this from the ball's position based on its direction. The same applies to the vertical collisions, but this time we operate on the y component of all the vectors.
Running your application should now give you a working collision scheme, but it's probably difficult to really see its effect since the ball will bounce towards the bottom edge as soon as you hit a single block and be lost forever. We can fix this by also handling player paddle collisions.
## Player - ball collisions
Collisions between the ball and the player are slightly different than what we've previously discussed since this time the ball's horizontal velocity should be updated based on how far it hit the paddle from its center. The further the ball hits the paddle from its center, the stronger its horizontal velocity should be.
void Game::DoCollisions()
{
[...]
Collision result = CheckCollision(*Ball, *Player);
if (!Ball->Stuck && std::get<0>(result))
{
// Check where it hit the board, and change velocity based on where it hit the board
GLfloat centerBoard = Player->Position.x + Player->Size.x / 2;
GLfloat distance = (Ball->Position.x + Ball->Radius) - centerBoard;
GLfloat percentage = distance / (Player->Size.x / 2);
// Then move accordingly
GLfloat strength = 2.0f;
glm::vec2 oldVelocity = Ball->Velocity;
Ball->Velocity.x = INITIAL_BALL_VELOCITY.x * percentage * strength;
Ball->Velocity.y = -Ball->Velocity.y;
Ball->Velocity = glm::normalize(Ball->Velocity) * glm::length(oldVelocity);
}
}
After we checked collisions between the ball and each brick, we'll check if the ball collided with the player paddle. If so (and the ball is not stuck to the paddle) we calculate the percentage of how far the ball's center is removed from the paddle's center compared to the half-extent of the paddle. The horizontal velocity of the ball is then updated based on the distance it hit the paddle from its center. Aside from updating the horizontal velocity we also have to reverse the y velocity.
Note that the old velocity is stored as oldVelocity. The reason for storing the old velocity is that we only update the horizontal velocity of the ball's velocity vector while keeping its y velocity constant. This would mean that the length of the vector constantly changes which has the effect that the ball's velocity vector is much larger (and thus stronger) if the ball hit the edge of the paddle compared to if the ball would hit the center of the paddle. For this reason the new velocity vector is normalized and multiplied by the length of the old velocity vector. This way, the strength and thus the velocity of the ball is always consistent, regardless of where it hits the paddle.
You may or may not have noticed it when you ran the code, but there is still a large issue with the player and ball collision resolution. The following video clearly shows what might happen:
This issue is called the sticky paddle issue which happens because the player paddle moves with a high velocity towards the ball that results in the ball's center ending up inside the player paddle. Since we did not account for the case where the ball's center is inside an AABB the game tries to continuously react to all the collisions and once it finally breaks free it will have reversed its y velocity so much that it's unsure whether it goes up or down after breaking free.
We can easily fix this behavior by introducing a small hack which is possible due to the fact that the we can assume we always have a collision at the top of the paddle. Instead of reversing the y velocity we simply always return a positive y direction so whenever it does get stuck, it will immediately break free.
//Ball->Velocity.y = -Ball->Velocity.y;
Ball->Velocity.y = -1 * abs(Ball->Velocity.y);
If you try hard enough the effect is still noticeable, but I personally find it an acceptable trade-off.
### The bottom edge
The only thing that is still missing from the classic Breakout recipe is some loss condition that resets the level and the player. Within the game class's Update function we want to check if the ball reached the bottom edge, and if so, reset the game.
void Game::Update(GLfloat dt)
{
[...]
if (Ball->Position.y >= this->Height) // Did ball reach bottom edge?
{
this->ResetLevel();
this->ResetPlayer();
}
}
The ResetLevel and ResetPlayer functions simply re-load the level and reset the objects' values to their original starting values. The game should now look a bit like this:
And there you have it, we just finished creating a clone of the classical Breakout game with similar mechanics. You can find the game class' source code here: header, code.
## A few notes
Collision detection is a difficult topic of video game development and possibly its most challenging. Most collision detection and resolution schemes are combined with physics engines as found in most modern-day games. The collision scheme we used for the Breakout game is a very simple scheme and one specialized specifically for this type of game.
It should be stressed that this type of collision detection and resolution is not perfect. It calculates possible collisions only per frame and only for the positions exactly as they are at that timestep; this means that if an object would have such a velocity that it would pass over another object within a single frame, it would look like it never collided with this object. So if there are framedrops or you reach high enough velocities, this collision detection scheme will not hold.
Several of the issues that can still occur:
• If the ball goes too fast, it might skip over an object entirely within a single frame, not detecting any collisions.
• If the ball hits more than one object within a single frame, it will have detected two collisions and reverse its velocity twice; not affecting its original velocity.
• Hitting a corner of a brick could reverse the ball's velocity in the wrong direction since the distance it travels in a single frame could make the difference between VectorDirection returning a vertical or horizontal direction.
These tutorials are however aimed to teach the readers the basics of several aspects of graphics and game-development. For this reason, this collision scheme serves its purpose; its understandable and works quite well in normal scenarios. Just keep in mind that there exist better (more complicated) collision schemes that work quite well in almost all scenarios (including movable objects) like the separating axis theorem.
Thankfully, there exist large, practical and often quite efficient physics engines (with timestep-independent collision schemes) for use in your own games. If you wish to delve further into such systems or need more advanced physics and have trouble figuring out the mathematics, Box2D is a perfect 2D physics library for implementing physics and collision detection in your applications.
HI
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HuggingFaceTB/finemath
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# How to Tally Votes in Excel (4 Suitable Methods)
Looking for ways to know how to tally votes in Excel? Then, this is the right place for you. When an election happens in our society, we can easily tally those votes in Excel using some methods. Here, you will find 4 different step-by-step explained ways to tally votes in Excel.
## 4 Suitable Methods to Tally Votes in Excel
Here, we have a dataset containing the Name, Age, and Vote of some citizens. We will show you how to tally votes using this dataset. ### 1. Using COUNTIF Function to Tally Votes in Excel
In the first method, we will use the COUNTIF function to tally votes in Excel. Using the COUNTIF function we can count the number of cells that meet a criterion. Follow the steps given below to do it on your own dataset.
Steps:
• First, select Cell G5.
• After that, insert the following formula
`=COUNTIF(\$D\$5:\$D\$14,F5)` Here, in the COUNTIF function, we selected Cell D5:D14 as the range and selected Cell F5 as criteria. Here, it will count all the Yes votes.
• Now, press ENTER.
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells. • Finally, you will get the total count of votes of different criteria. ### 2. Use of Combined Formula to Tally Votes in Excel
Now, we will use the SUM function, FREQUENCY function, IF function, MATCH function and ROW function to tally votes in Excel. Go through the steps to do it on your own.
Steps:
• In the beginning, select cell G5.
• Then, insert the following formula
`=SUM(FREQUENCY(IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14,\$D\$5:\$D\$14,0)),ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1))` Formula Breakdown
• ROW(\$D\$5)—–> The ROW function returns the row number.
• Output: {5}
• ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1)—–> turns into
• ROW{“Yes”;”No”;”Yes”;”Not Sure”;”Yes”;”No”;”Yes”;”Yes”;”No”;”Not Sure”}-{5}+1)—–>
• Output: {1;2;3;4;5;6;7;8;9;10}
• MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0))—–> The MATCH function returns the relative position of the items in the range.
• Output: {1;2;1;4;1;2;1;1;2;4}
• IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0))—–>The IF function returns 1 value for a TRUE result, and FALSE for a FALSE result.
• IF(\$D\$5:\$D\$14=”Yes”,{1;2;3;4;5;6;7;8;9;10})—–> turns into
• Output: {1;FALSE;1;FALSE;1;FALSE;1;1;FALSE;FALSE}
• FREQUENCY(IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0)),ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1))—–> the FREQUENCY function returns how often values occur within a set of data.
• FREQUENCY({1;FALSE;1;FALSE;1;FALSE;1;1;FALSE;FALSE},{1;2;3;4;5;6;7;8;9;10})—–> becomes
• Output: {5;0;0;0;0;0;0;0;0;0;0}
• SUM(FREQUENCY(IF(\$D\$5:\$D\$14=F5,MATCH(\$D\$5:\$D\$14, \$D\$5:\$D\$14,0)),ROW(\$D\$5:\$D\$14)-ROW(\$D\$5)+1))—–> The SUM function returns the sum of values supplied.
• SUM({5;0;0;0;0;0;0;0;0;0;0})—–> turns into
• Output: 5
• Explanation: It will tally all the Yes votes.
• After that, press ENTER.
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells. • Finally, you will get the tally of the votes. ### 3. Applying SUMIF Function to Tally Votes
In the third method, we will apply the SUMIF function to tally votes. Here, we have the following dataset containing the Name, Age, Count, and Vote of some citizens. Steps:
• First, select Cell H5.
• Then, insert the following formula
`=SUMIF(\$E\$5:\$E\$14,G5,\$D\$5:\$D\$14)` Here, in the SUMIF function, we selected Cell E5:E14 as range, Cell G5 as criteria and Cell D5:D14 as sum_range. Here, it will tally all the Yes votes.
• Now, press ENTER.
• Then, drag down the Fill Handle tool to AutoFill the formula for the rest of the cells. • Finally, you will get the tally of the votes. ### 4. Creating Voting System to Tally Votes
In the final method, we will show you how to create a voting system to tally votes. Here, we have a dataset containing different Candidates and their number of votes. Now, we will create a voting system for this dataset. Steps:
• First, open the Developer tab >> go to Insert >> from Form Controls choose Spin Button. • Now, Spin Button will appear on the worksheet.
• After that, select the Spin Button and Right Click.
• Then, select Format Control. • Now, the Format Control box will appear.
• Then, select Cell C4 as Cell link.
• After that, press OK. • Now, the Spin Button is connected to Cell C4.
• To check this, press the Upward Button.
• Now, notice that the vote has increased from 11 to 12. • After that, insert 3 more Spin Buttons for Cell C5:C7 using the same way(anchor). • Then, select the Cell B4:C7.
• After that, open the Insert tab >> from the Chart section >> click on Bar Charts. • Then, from Bar Charts >> select 2-D Column. • Now, a Bar Chart will appear. • Then, change the Chart Title as Votes. • After that, you can add Data Labels. Finally, you will get the tally of votes by creating a voting system. Now, you can change the no of votes anytime by using the Spin Button. Here, when we clicked on the Upward Button for Candidate 1, the vote increased from 12 to 13. • If we clicked on the Downward Button for Candidate 2, the vote decreased from 31 to 30. ## Things to Remember
• Here, you cannot input more than one criterion in the COUNTIF function.
• You may find the VALUE Error in case of using the SUMIF function if you use it in case of strings longer than 255 characters.
• The FREQUENCY function will show the #NAME error if you misspell the function name.
## Practice Section
In the Excel file, you will get a dataset like an image given below in this article to practice the explained methods on your own. ## Conclusion
So, in this article, you will find 4 ways to tally votes in Excel. Use any of these ways to accomplish the result in this regard. Hope you find this article helpful and informative. Feel free to comment if something seems difficult to understand. Let us know any other approaches which we might have missed here. And, visit ExcelDemy for many more articles like this. Thank you!
## Related Articles #### Arin
Hello, I'm Arin. I graduated from Khulna University of Engineering and Technology (KUET) from the Department of Civil Engineering. I am passionate about learning new things and increasing my data analysis knowledge as well as critical thinking. Currently, I am working and doing research on Microsoft Excel and here I will be posting articles related to it.
We will be happy to hear your thoughts
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HuggingFaceTB/finemath
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# How do you find the domain and range of f(x) = (x + 8)^2 - 7?
Jan 7, 2018
Inspect using the formula $y = a {\left(x - h\right)}^{2} + k$
#### Explanation:
From the equation given $f \left(x\right) = {\left(x + 8\right)}^{2} - 7$ we can see that:
h = -8
k = -7
From the original equation $y = {x}^{2}$, if h is negative the graph the will shift left or negative x and if k is negative the graph will shift down or negative y.
graph{(x+8)^2-7 [-27.05, 12.95, -8.72, 11.28]}
Since x will keep increasing to infinity regardless of any x-axis transformations the domain will be the same as $y = {x}^{2}$
Domain: All real numbers
However, since a minimum applies to the range, if the graph shifts in the y-axis the range will be different from $y = {x}^{2}$
Range: y ≥ -7
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HuggingFaceTB/finemath
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# Hearing Perturbation Theory
In numerical linear algebra, we create ways for a computer to solve a linear system of equations $A\vec{x} = \vec{b}$. In doing so, we analyze how efficiently and accurately we can find the solution $\vec{x}$.
Perturbation theory concerns how much error we incur in the solution $\vec{x}$ when we perturb (spoil) the data $A$ and $\vec{b}$. A classic statement tells us that the amount of error depends on the condition number of the matrix $A$.
I will define and prove the statement, and help you understand it by “hearing” it.
# 1. Problem statement
Let’s consider the equation,
$A\vec{x}_{true} = \vec{b}$,
where $A \in \mathbb{R}^{n \times n}$ is nonsingular and $\vec{b} \neq \vec{0}$ (otherwise, we get the trivial case $\vec{x}_{true} = \vec{0}$). Note that $\vec{x}_{true}$ denotes the true solution. It’s what we hope to get.
Suppose, instead, we solve the perturbed system,
$(A + \delta\!A)\vec{x} = \vec{b} + \vec{\delta b}$.
We seek to find a bound on the error $\vec{x} - \vec{x}_{true}$. We expect the error to be small when $\delta\!A$ and $\vec{\delta b}$ are small changes.
Given an induced matrix norm $||\cdot||$, we can define the condition number of $A$:
$\kappa(A) := ||A|| \cdot ||A^{-1}||$.
Note that we always have $\kappa(A) \geq 1$. You can check why, along with the definition and properties of an induced matrix norm, in Notes.
We will show that, under a mild assumption $\displaystyle\kappa(A) \cdot \frac{||\delta\!A||}{||A||} < 1$,
$\boxed{\displaystyle\frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} \leq \frac{\kappa(A)}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||\vec{b}||}\right)}$.
The RHS is a product of terms, so we can conclude two things. (1) If $\kappa(A)$ is a small number, then small relative errors in $A$ and $\vec{b}$ will result in a small relative error in the solution $\vec{x}_{true}$. (2) If $\kappa(A)$ is a large number, however, we may get a large relative error in $\vec{x}_{true}$ even when the relative errors in $A$ and $\vec{b}$ are small.
We see that the condition number influences how much change occurs in the output when we change the input of a system. As a result, we say that the matrix $A$ is well-conditioned if $\kappa(A)$ is small, and ill-conditioned if $\kappa(A)$ is large.
# 2. Mathematical proof
## a. Step 1
For any matrix $X \in \mathbb{R}^{n \times n}$ with $||X|| < 1$, the following statements hold:
(i) $I - X$ is nonsingular, and its inverse is given by
$\displaystyle(I - X)^{-1} = \sum_{i\,=\,0}^{\infty}\,X^{i} = I + X + X^{2} + \cdots$.
(ii) $\displaystyle||(I - X)^{-1}|| \leq \frac{1}{1 - ||X||}$.
To prove (i), we check that $(I - X)(I - X)^{-1} = I$ and $(I - X)^{-1}(I - X) = I$. We just need to check one of them because, in finite dimensions, injectivity and surjectivity occur together.
We see that,
\begin{aligned} (I - X)(I - X)^{-1} &= (I - X)(I + X + X^{2} + \cdots) \\[12pt] &= (I + X + X^{2} + \cdots) - (X + X^{2} + X^{3} + \cdots) \\[12pt] &= I. \end{aligned}
Note, to be rigorous, we would first show that the infinite series $\sum_{i\,=\,0}^{\infty}\,X^{i}$ converges. The sequence of finite sums, $\left\{\sum_{i\,=\,0}^{k}\,X^{i}\right\}_{k\,=\,0}^{\infty}$, is a Cauchy sequence in the norm $||\cdot||$. Since $\mathbb{R}^{n \times n}$ is a Banach space, i.e. complete, the sequence converges.
Let’s prove (ii). By triangle inequality and the properties of an induced matrix norm,
\begin{aligned} ||(I - X)^{-1}|| &= ||I + X + X^{2} + \cdots|| \\[12pt] &\leq ||I|| + ||X|| + ||X^{2}|| + \cdots \\[12pt] &\leq ||I|| + ||X|| + ||X||^{2} + \cdots. \end{aligned}
Next, we use the infinite geometric sum formula: For any real number $r \in (-1,\,1)$,
$\displaystyle\sum_{i\,=\,0}^{\infty}\,r^{i} = \frac{1}{1 - r}$.
Since $||X|| \in [0,\,1)$, we conclude that,
$\displaystyle||(I - X)^{-1}|| \leq \frac{1}{1 - ||X||}$.
## b. Step 2
Recall that $A$ is nonsingular. Assume further that,
$\displaystyle\kappa(A) \cdot \frac{||\delta\!A||}{||A||} < 1$.
Then, $A + \delta\!A$ is also nonsingular.
Since $A$ is nonsingular, we can write $A + \delta\!A$ as a product of two terms:
$A + \delta\!A = A(I + A^{-1}\delta\!A)$.
Hence, $A + \delta\!A$ is nonsingular, if and only if, $I + A^{-1}\delta\!A$ is nonsingular. Let’s show that the latter is true by applying (i) from Step 1.
We find that,
\displaystyle\begin{aligned} ||A^{-1}\delta\!A|| &\leq ||A^{-1}|| \cdot ||\delta\!A|| \\[12pt] &= \kappa(A)\frac{||\delta\!A||}{||A||} \\[12pt] &< 1. \end{aligned}
Hence, $I + A^{-1}\delta\!A$ is nonsingular.
Before we move on, let’s read the statement in Step 2 again. It tells us that, if a matrix is nonsingular, then there are infinitely many matrices nearby that are nonsingular, too. (And infinitely many around them, and so on.) This supports the fact that there are far more nonsingular matrices than singular ones. Isn’t that a marvel?
## c. Step 3
Show that,
$\displaystyle \frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} \leq ||(A + \delta\!A)^{-1}|| \,\cdot \left(||\delta\!A|| + \frac{||\vec{\delta b}||}{||\vec{x}_{true}||}\right)$.
We’re almost there! Recall the original problems:
$\begin{array}{rcl} A\vec{x}_{true} & = & \vec{b} \\[12pt] (A + \delta\!A)\vec{x} & = & \vec{b} + \vec{\delta b}. \end{array}$
Subtract the two equations to get,
$\begin{array}{ll} & (A + \delta\!A)\vec{x} - A\vec{x}_{true} = \vec{\delta b} \\[12pt] \Rightarrow & (A + \delta\!A)(\vec{x} - \vec{x}_{true}) = \vec{\delta b} - \delta\!A\vec{x}_{true} \\[12pt] \Rightarrow & \vec{x} - \vec{x}_{true} = (A + \delta\!A)^{-1}(\vec{b} - \delta\!A\vec{x}_{true}). \end{array}$
Take the vector norm to both sides:
\begin{aligned} ||\vec{x} - \vec{x}_{true}|| &= ||(A + \delta\!A)^{-1}(\vec{b} - \delta\!A\vec{x}_{true})|| \\[12pt] &\leq ||(A + \delta\!A)^{-1}|| \cdot ||\vec{b} - \delta\!A\vec{x}_{true}|| \\[12pt] & \leq ||(A + \delta\!A)^{-1}|| \cdot \Bigl(||\vec{b}|| + ||\delta\!A|| \cdot ||\vec{x}_{true}||\Bigr). \end{aligned}
Finally, we divide both sides by $||\vec{x}_{true}||\,(\neq 0)$.
## d. Step 4
Finally, prove that,
$\displaystyle\frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} \leq \frac{\kappa(A)}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||\vec{b}||}\right)$.
Let’s find an upper bound for $||(A + \delta\!A)^{-1}||$ in Step 3. We use (ii) from Step 1:
\begin{aligned} ||(A + \delta\!A)^{-1}|| &\leq ||(I + A^{-1}\delta\!A)^{-1}|| \cdot ||A^{-1}|| \\[12pt] &\leq \frac{||A^{-1}||}{1 - ||A^{-1}\delta\!A||} \\[12pt] &\leq \frac{||A^{-1}||}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}}. \end{aligned}
Hence,
\begin{aligned} \frac{||\vec{x} - \vec{x}_{true}||}{||\vec{x}_{true}||} &\leq \frac{||A|| \cdot ||A^{-1}||}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||A|| \cdot ||\vec{x}_{true}||}\right) \\[12pt] &\leq \frac{\kappa(A)}{1 - \kappa(A)\frac{||\delta\!A||}{||A||}} \,\cdot \left(\frac{||\delta\!A||}{||A||} + \frac{||\vec{\delta b}||}{||\vec{b}||}\right). \end{aligned}
# 3. Application
We can represent an audio as a vector $\vec{x}_{true} \in \mathbb{R}^{n}$ of frequencies. The provided audio file, williams.wav, lends to $n = 189930$. At a sampling rate of 44100 Hz, the audio is 4.3068 seconds long.
For computational efficiency, we will write $\vec{x}_{true}$ as a matrix $X_{true} \in \mathbb{R}^{10 \times 18993}$, by storing the first 10 entries of $\vec{x}_{true}$ as the first column of $X_{true}$, the next 10 entries as the second, and so on.
Next, we apply a nonsingular matrix $A \in \mathbb{R}^{10 \times 10}$ to $X_{true}$ to get the encrypted audio $B \in \mathbb{R}^{10 \times 18993}$. We can always vectorize $B$ into $\vec{b} \in \mathbb{R}^{189930}$ by reversing the storage process described above.
From now on, we assume that we only have $A$ and $B$. We seek to decrypt the audio, i.e. solve the equation $AX_{true} = B$, when there are perturbations in the matrix $A$ and/or in the input $B$:
$(A + \delta\!A)X = (B + \delta\!B)$.
## a. Generating matrices
To generate a well-conditioned $A$, we find the QR factorization of a random $10 \times 10$ matrix and set $A = Q$. We know that the 2-norm of an orthogonal matrix is always 1. Hence,
$\kappa_{2}(Q) = ||Q||_{2} \cdot ||Q^{T}||_{2} = 1$.
For an ill-conditioned $A$, we use the Hilbert matrix $H$:
$H = \left[\begin{array}{ccccc} 1 & \frac{1}{2} & \cdots & \frac{1}{9} & \frac{1}{10} \\[12pt] \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{10} & \frac{1}{11} \\[12pt] \vdots & \vdots & \ddots & \vdots & \vdots \\[12pt] \frac{1}{9} & \frac{1}{10} & \cdots & \frac{1}{17} & \frac{1}{18} \\[12pt] \frac{1}{10} & \frac{1}{11} & \cdots & \frac{1}{18} & \frac{1}{19} \end{array}\right]$.
Then, $\kappa_{2}(H) \approx 1.6025 \times 10^{13}$. Note that both $Q$ and $H$ are nonsingular.
## b. Generating perturbations
The entries of perturbations $\delta\!A$ and $\delta\!B$ are randomly generated from a scaled normal distribution. For convenience, we will assume that $A + \delta\!A$ is (most likely) nonsingular because $\delta\!A$ is randomly chosen.
We are more interested in ensuring that the values of $||\delta\!A||_{2}$ and $||\vec{\delta b}||_{2}$ stay about the same each time we run the program.
## c. Simulations
We can run perturbation_theory.m under 6 different cases:
$\begin{tabular}{c|c|c|c} & Perturb A only & Perturb B only & Perturb A and B \\[8pt]\hline Well-conditioned A & (1,\,1) & (1,\,2) & (1,\,3) \\[8pt]\hline Ill-conditioned A & (2,\,1) & (2,\,2) & (2,\,3) \end{tabular}$
The table lists the input parameters for each case.
First, let’s consider what happens when $A$ is well-conditioned.
This slideshow requires JavaScript.
These plots show that the obtained solution $\vec{x}$ matches the true solution $\vec{x}_{true}$ well. The fidelity of the obtained audios remains largely pristine. We do hear some added noise. In the 100 ms sample, the relative error stays around 2% (median) when we perturb only the matrix $A$. When we perturb the encrypted audio $B$, the relative error jumps to about 10%.
Download and listen to the obtained audios:
• audio_obtained11.wav (well-conditioned $A$, perturbation to $A$ only)
• audio_obtained12.wav (well-conditioned $A$, perturbation to $B$ only)
• audio_obtained13.wav (well-conditioned $A$, perturbation to both $A$ and $B$)
Next, let’s consider what happens when $A$ is ill-conditioned.
This slideshow requires JavaScript.
This time, the obtained solution hardly matches the true solution. You can hear much more noise in the obtained audios, and listening to them becomes almost unbearable. The relative error is orders of magnitude larger. It’s interesting how perturbing $B$ only results in a completely garbled audio, but perturbing $A$ in addition reconstructs some of the original audio. Two wrongs do make a right, it seems.
Download and listen to the obtained audios:
• audio_obtained21.wav (ill-conditioned $A$, perturbation to $A$ only)
• audio_obtained22.wav (ill-conditioned $A$, perturbation to $B$ only)
• audio_obtained23.wav (ill-conditioned $A$, perturbation to both $A$ and $B$)
We can also check that our matrices satisfy the statement that we painstakingly proved. I will leave writing the extra code to you.
# Notes
Given a vector norm $||\cdot||$ for the vector space $\mathbb{R}^{n}$, we can always create a matrix norm $||\cdot||$ for the vector space $\mathbb{R}^{n \times n}$. As a result, we call this an “induced” matrix norm. For simplicity, I use the double bar notation for both types of norms. It is clear from context whether we are looking at the vector norm or the induced matrix norm.
The induced matrix norm of $A \in \mathbb{R}^{n \times n}$ is defined as,
$\displaystyle||A|| := \max\limits_{\vec{x}\,\neq\,\vec{0}}\,\frac{||A\vec{x}||}{||\vec{x}||} = \max\limits_{||\vec{x}||\,=\,1}\,||A\vec{x}||$.
Hence, the induced matrix norm measures how far out we can map a vector relative to its original size. We can also generalize the definition to rectangular matrices.
Because of its definition, the induced matrix norm satisfies these useful properties:
(i) For any $A \in \mathbb{R}^{n \times n}$ and $\vec{x} \in \mathbb{R}^{n}$,
$||A\vec{x}|| \leq ||A|| \cdot ||\vec{x}||$.
(ii) For any $A,\,B \in \mathbb{R}^{n \times n}$,
$||AB|| \leq ||A|| \cdot ||B||$.
In particular, property (ii) implies that the condition number of a nonsingular matrix is always at least 1.
$AA^{-1} = I \,\,\Rightarrow\,\, ||AA^{-1}|| = 1 \,\,\Rightarrow\,\, ||A|| \cdot ||A^{-1}|| \geq 1$.
You can find the code and audio files in their entirety here:
Download from GitHub
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HuggingFaceTB/finemath
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# Floor and ceiling function with example, program
In computer science or discrete structure, there are two important function are use i.e. Floor and ceiling function . In this tutorial we discuss definition, example using concept computer science and program of these function. Notation: ⌈x⌉ is ceiling function, ⌊x⌋ is floor function where x is variable.
Let x be a real number then ⌊x⌋ called the floor function of x, assigns to the real number x the largest integer that is less than or equal to x. this function rounds x down to the closest integer less than or equal to x. The floor function is also called the greatest integer function.
Let x be a real number then ⌈x⌉ called the ceiling function of x, assigns to the real number x the smallest integer that is greater than or equal to x. this function rounds x up to the closest integer less than or equal to x.
#### Example of Floor and ceiling function
• 5.3 Floor value ⌊x⌋=5 & Ceiling value ⌈x⌉=6
• 4.5 Floor value ⌊x⌋=4 & Ceiling value ⌈x⌉=5
• -4 Floor value ⌊x⌋=-4 & Ceiling value ⌈x⌉=-4
• 1/2 Floor value ⌊x⌋=0 & Ceiling value ⌈x⌉=1
Prove or disprove that ⌈x+y⌉=⌈x⌉+⌈y⌉ for all real number x and y.
solution, A counter example is supplied by x=1/2 and y=1/2. with these values we found that ⌈x+y⌉=⌈1/2+1/2⌉=⌈1⌉=1 but ⌈x+y⌉=⌈1/2+1/2⌉=1+1=2
Another example: Data stored on a computer or transmitted over a data network are usually represent as a string of bytes. Each byte is made up of 8 bits. How many bytes are required to encode 300 bits of data?
solution: To determine the number of bytes needed, we determine the smallest integer that is at least as large as the quotient when 300 is divided by 8. Therefore, the number of bytes required = ⌈300/8⌉=⌈37.5⌉=38 bytes.
#### Program to find Floor and ceiling function
#include <stdio.h>
#include <math.h>
int main()
{
float val;
float fVal,cVal;
printf("Enter a float value: ");
scanf("%f",&val);
fVal=floor(val);
cVal =ceil(val);
printf("floor value:%f \nceil value:%f\n",fVal,cVal);
return 0;
}
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HuggingFaceTB/finemath
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# Perpendicular
## What is Perpendicular?
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). The property extends to other related geometric objects. A line is said to be perpendicular to another line if the two lines intersect at a right angle. Explicitly, a first line is perpendicular to a second line if (1) the two lines meet; and (2) at the point of intersection the straight angle on one side of the first line is cut by the second line into two congruent angles. Perpendicularity can be shown to be symmetric, meaning if a first line is perpendicular to a second line, then the second line is also perpendicular to the first. For this reason, we may speak of two lines as being perpendicular (to each other) without specifying an order. Perpendicularity easily extends to segments and rays. For example, a line segment is perpendicular to a line segment if, when each is extended in both directions to form an infinite line, these two resulting lines are perpendicular in the sense above. In symbols, means line segment AB is perpendicular to line segment CD. For information regarding the perpendicular symbol see Up tack. A line is said to be perpendicular to a plane if it is perpendicular to every line in the plane that it intersects. This definition depends on the definition of perpendicularity between lines. Two planes in space are said to be perpendicular if the dihedral angle at which they meet is a right angle (90 degrees). Perpendicularity is one particular instance of the more general mathematical concept of orthogonality; perpendicularity is the orthogonality of classical geometric objects. Thus, in advanced mathematics, the word "perpendicular" is sometimes used to describe much more complicated geometric orthogonality conditions, such as that between a surface and its normal.
### Technology Types
elementary geometryorientation (geometry
### Synonyms
Perpendicular linePerpendicular LinesPerpendicular SymbolPerpendicularityPerpendicularity symbolPerpendicularly
### Translations
Elkarzut (eu)Kolmice (cs)Loodrecht (meetkunde) (nl)Lot (Mathematik) (de)Perpendicolarità (it)Perpendicularidad (es)Perpendicularidade (pt)Perpendicularitat (ca)Perpendicularité (fr)Prostopadłość (pl)Serenjang (in)Vinkelrät (sv)Перпендикулярність (uk)Перпендикулярность (ru)تعامد (ar)수직 (ko)垂直 (ja)垂直 (zh)
## Tech Info
Source: [object Object]
— Date merged: 11/6/2021, 1:32:52 PM
— Date scraped: 5/20/2021, 5:43:43 PM
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HuggingFaceTB/finemath
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Thanks to theidioms.com
# Learn Linear Algebra for Data Science (Mini-Course)
## Learn Linear Algebra for Data Science (Mini-Course)
### Dot Products and Matrix Multiplication
While working with matrices, there are two major forms of multiplicative operations: dot products and matrix multiplication. A dot product takes the product of two matrices and outputs a single scalar value. On the other hand, matrix multiplication takes the product of two matrices and outputs a single matrix. In this lesson, we will be discussing these two operations and how they work.
#### Dot Product in matrices
Matrix dot products (also known as the inner product) can only be taken when working with two matrices of the same dimension. When taking the dot product of two matrices, we multiply each element from the first matrix by its corresponding element in the second matrix and add up the results.
If we take two matrices and such that = , and , then the dot product is given as,
#### Matrix Multiplication
Two matrices can be multiplied together only when the number of columns of the first matrix is equal to the number of rows in the second matrix.
For matrix multiplication, we take the dot product of each row of the first matrix with each column of the second matrix that results in a matrix of dimensions of the row of the first matrix and the column of the second matrix. For example, for two matrices and , if has a dimension , and has a dimension , matrix multiplication is possible and the resulting matrix is of dimension .
For an easier understanding, let us suppose matrices and to be of dimensions each. Taking matrix and matrix , the matrix multiplication of is given as,
Now, let us find the value of ,
Hence, we also conclude that the matrix multiplication is not commutative, i.e., .
Matrix multiplication has a wide range of applications in Linear Algebra as well as Data Science. In this lesson, we discussed some of the major multiplicative operations performed on matrices. Such operations are usually applied to matrices that represent image data in the field of data science. In the upcoming chapter, we will learn about some matrix transformation methods and the concept of determinants.
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# Pendulum with a mass on a container (thermodynamics problem)
• ValeForce46
In summary, the conversation discusses the conversion of kinetic energy of a pendulum into heat in an adiabatic system. The ideal gas law is used to find the number of moles of gas and a rough estimate of the change in temperature. A calculation error is identified and corrected. The conversation also touches on the variation of entropy and the distribution of energy among different degrees of freedom in the system.
ValeForce46
Homework Statement
A pendulum, with a block of iron (##m=1 kg##, specific heat ##c=448\frac{J}{Kg\cdot K}##) hanging by a thread (mass negligible), is inside a container of volume with rigid and adiabatic walls. Inside the container there's air (to consider as a biatomic ideal gas) at the atmospheric pressure and at the temperature ##T_0=300 K##. You observe the pendulum has a speed of ##v=3.5 \frac{m}{s}## when the block is on the vertical. After some time, the pendulum ceases to swing. Determine:
a) The final temperature of the system at the end of the oscillation.
b) The variation of entropy of the universe at the end of the trasformation.
Relevant Equations
##Q=m*c*(T_f-T_0)##
The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely).
So both the air and the block of iron will change their temperature.
To find ##n## (moles of the gas) I can use the ideal gas law:
##n=\frac{pV}{RT}=0.9 mol##
Do I have the following equation?
##\frac{1}{2}mv^2=m*c*(T_f-T_0)+n*c_v*(T_f-T_0)## and ##T_f## is my unknown.
But I get ##T_f=300K## (nearly). Where's my mistake?
To get a rough idea of the change in temperature: How many Joules of KE does the block of iron initially have?
Note that ciron = 448 J/(kg K), which tells you that it takes 448 J of energy to raise the temperature of the iron block by just 1 K.
ValeForce46
TSny said:
To get a rough idea of the change in temperature: How many Joules of KE does the block of iron initially have?
##K=6.13 J ##.
So you're silently (not that much ) saying that my result ##T_f=300.013 K## is right?
I even doubted my equation.
ValeForce46 said:
##K=6.13 J ##.
So you're silently (not that much ) saying that my result ##T_f=300.013 K## is right?
I even doubted my equation.
Yes, I think your answer is correct. I wanted you to see how you can tell that a very small temperature increase is what you would expect.
ValeForce46
Yeah, you made me realize that. Thanks a lot.
However, for completeness, the variation of entropy of the universe is the variation of entropy of the system, because the walls of the container are adiabatic so there's no variation of entropy of the environment:
##\Delta S_{Universe}=\Delta S_{System}=mc(T_f-T_0)+nc_v(T_f-T_0)=6.1 \frac{J}{K}##
ValeForce46 said:
##\Delta S_{Universe}=\Delta S_{System}=mc(T_f-T_0)+nc_v(T_f-T_0)=6.1 \frac{J}{K}##
Check this calculation. Note that what you calculated here is ##\Delta E_{int}## of the block and gas. It's not surprising that you got 6.1, since the initial KE of the block is 6.1 J.
ValeForce46
oops... I did the integral wrong
##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}##
Hope this is right ...
ValeForce46 said:
oops... I did the integral wrong
##ΔS_{Universe}=ΔS_{System}=m*c*\ln (\frac{T_f}{T_0})+n*c_v*\ln (\frac{T_f}{T_0})=0.02 \frac{J}{K}##
Hope this is right ...
Looks good.
ValeForce46
Just a side note: For this problem, you can check that about 96% of the initial KE of the iron block ends up in the iron block and only about 4% goes to the air. That kind of surprised me at first. But it makes sense if you think about the equipartition of energy. In this problem, there are a lot more atoms of iron in the system compared to molecules of air. Moreover, for T around 300 K, each atom of iron has effectively 6 "degrees of freedom" for storing internal energy (Law of Dulong and Petit) compared to 5 degrees of freedom for each air molecule. The energy spreads equally among all of the degrees of freedom in the system.
ValeForce46
ValeForce46 said:
The kinetic energy of the pendulum ##K=\frac{1}{2}\cdot m\cdot v^2## will turn into heat (entirely).
This is an incorrect interpretation. The kinetic energy of the pendulum will not turn into heat. There is no heat transferred to this adiabatic system. The correct interpretation starts with the full presentation of the first law of thermodynamics: $$\Delta U+\Delta (KE)+\Delta (PE)=Q-W=0$$Also there is not change in potential energy of the system. So $$\Delta U=-\Delta (KE)$$
This tells us the correct interpretation, namely, that the kinetic energy of the pendulum will convert directly to internal energy (via viscous dissipation by the air).
ValeForce46 and TSny
Since the temperature hardly changes, the change in entropy is nearly exactly the initial kinetic energy (6.125 J) divided by the temperature, roughly (300 K), or 6.125/300 = 0.0204 J/K.
## 1. What is a pendulum with a mass on a container in thermodynamics?
A pendulum with a mass on a container in thermodynamics is a physical system that consists of a container filled with a gas or liquid and a pendulum attached to the top of the container. The pendulum swings back and forth due to the motion of the gas or liquid inside the container, creating a thermodynamic system.
## 2. How does a pendulum with a mass on a container demonstrate thermodynamics?
A pendulum with a mass on a container demonstrates thermodynamics by showing the transfer of energy between the system (pendulum and container) and its surroundings. The motion of the pendulum is a result of the gas or liquid inside the container expanding and contracting, which is a thermodynamic process.
## 3. What factors affect the motion of the pendulum in a thermodynamic system?
The motion of the pendulum in a thermodynamic system is affected by several factors, including the mass and length of the pendulum, the type of gas or liquid inside the container, and the temperature and pressure of the system. These factors can impact the frequency and amplitude of the pendulum's swings.
## 4. How does the temperature and pressure of a thermodynamic system affect the motion of the pendulum?
The temperature and pressure of a thermodynamic system can affect the motion of the pendulum by changing the properties of the gas or liquid inside the container. As the temperature and pressure increase, the molecules in the gas or liquid will move faster and exert more force on the pendulum, causing it to swing with more frequency and amplitude.
## 5. Can a pendulum with a mass on a container be used to measure thermodynamic properties?
Yes, a pendulum with a mass on a container can be used to measure thermodynamic properties such as temperature and pressure. By measuring the frequency and amplitude of the pendulum's swings, one can calculate the values of these properties using thermodynamic equations. This technique is known as the "pendulum method" and is commonly used in thermodynamics experiments.
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# 2020 AMC 12B Problems/Problem 22
## Problem 22
What is the maximum value of $\frac{(2^t-3t)t}{4^t}$ for real values of $t?$ $\textbf{(A)}\ \frac{1}{16} \qquad\textbf{(B)}\ \frac{1}{15} \qquad\textbf{(C)}\ \frac{1}{12} \qquad\textbf{(D)}\ \frac{1}{10} \qquad\textbf{(E)}\ \frac{1}{9}$
## Solution1
Set $u = t2^{-t}$. Then the expression in the problem can be written as $$R = - 3t^24^{-t} + t2^{-t}= - 3u^2 + u = - 3 (u - \frac{1}{6})^2 + \frac{1}{12} \le \frac{1}{12} .$$ It is easy to see that $u =\frac{1}{6}$ is attained for some value of $t$ between $t = 0$ and $t = 1$, thus the maximal value of $R$ is $\textbf{(C)}\ \frac{1}{12}$.
## Solution2
First, substitute $2^t = x (\log_2{x} = t)$ so that $$\frac{(2^t-3t)t}{4^t} = \frac{x\log_2{x}-3(\log_2{x})^2}{x^2}$$
Notice that $$\frac{x\log_2{x}-3(\log_2{x})^2}{x^2} = \frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2.$$
When seen as a function, $\frac{\log_2{x}}{x}-3\Big(\frac{\log_2{x}}{x}\Big)^2$ is a synthesis function that has $\frac{\log_2{x}}{x}$ as its inner function.
If we substitute $\frac{\log_2{x}}{x} = p$, the given function becomes a quadratic function that has a maximum value of $\frac{1}{12}$ when $p = \frac{1}{6}$.
Now we need to check if $\frac{\log_2{x}}{x}$ can have the value of $\frac{1}{6}$ in the range of real numbers.
In the range of (positive) real numbers, function $\frac{\log_2{x}}{x}$ is a continuous function whose value gets infinitely smaller as $x$ gets closer to 0 (as $log_2{x}$ also diverges toward negative infinity in the same condition). When $x = 2$, $\frac{\log_2{x}}{x} = \frac{1}{2}$, which is larger than $\frac{1}{6}$.
Therefore, we can assume that $\frac{\log_2{x}}{x}$ equals to $\frac{1}{6}$ when $x$ is somewhere between 1 and 2 (at least), which means that the maximum value of $\frac{(2^t-3t)t}{4^t}$ is $\boxed{\textbf{(C)}\ \frac{1}{12}}$.
## Solution 3 (Bash)
Take the derivative of this function and let the derivative equals to 0, then this gives you $2^t=6t$. Substitute it into the original function you can get $\boxed{C}$.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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15 October, 09:21
# The average distance of the earth from the sun is about 1.5 * 10^8 km. Assume that the earth's orbit around the sun is circular and that the sun is at the origin of your coordinate system. (a) Estimate the speed of the earth as it moves in its orbit around the sun. Express your answer in miles per hour with the appropriate number of significant figures. (b) Estimate the angle θ between the position vector of the earth now and what it will be in 4 months. (c) Calculate the distance between these two positions.
+2
1. 15 October, 11:14
0
Distance between sun and earth will act as radius of circular path
R = 1.5 X 10¹¹ m
angular velocity of the earth
ω = 2π / T
= 2 X 3.14 / (365 X 24 X 60 X 60)
= 1.99 X 10⁻⁷ radian / s
velocity v = ω R
= 1.99 X 10⁻⁷ X 1.5 X 10¹¹
= 2.985 X 10⁴ m / s
= 29.85 km/s
29.85 x 60 x 60 km/h
= 107460 km/h
107460/1.6 mile / h
= 67162.5 mile / h
b) Angle moved in 12 months
= 360 degree
angle moved in 4 months
= 360 / 3
= 120 degree
c) Circumference of the orbit
= 2 x 3.14 x 1.5 x 10⁸ km
9.42 x 10⁸ km
required distance will be 1/3 rd of circumference
= 9.42 / 3 x 10⁸ km
= 3.14 x 10⁸ km
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Sunteți pe pagina 1din 17
# Induction
## M.Sc. (Mal), B.Sc. (Hons.) (Mal)
There are a number of ways a magnetic field can be used to
generate an electric current.
It is the changing field that produces the current.
The current in the coil is called the induced current because it is brought
about by a changing magnetic field.
Since a source emf is always needed to produce a current, the coil behaves
as if it were a source of emf. This emf is known as the induced emf.
An emf can be induced by changing the
area of a coil in a constant magnetic field
In each example, both an emf and a current are induced because
the coil is part of a complete circuit. If the circuit were open, there
would be no induced current, but there would be an induced emf.
The phenomena of producing an induced emf with the aid of a
magnetic field is called electromagnetic induction.
GRAPHICAL INTERPRETATION OF MAGNETIC FLUX
The magnetic flux is proportional
to the number of field lines that pass
through a surface.
The average emf induced in a coil of N loops is
N N
o
t t
t
o
SI Unit of Induced Emf: volt (V)
Example 5 The Emf Induced by a Changing Magnetic Field
A coil of wire consists of 20 turns each of which has an area of 0.0015 m 2 .
A magnetic field is perpendicular to the surface. Initially, the magnitude of
the magnetic field is 0.050 T and 0.10s later, it has increased to 0.060 T.
Find the average emf induced in the coil during this time.
BA
cos
B A
cos
o
N
N
t
t
B B
0.060 T
0.050 T
2
o
NA cos
20 0.0015 m
cos 0
t
0.10 s
3
3.0
10
V
Conceptual Example 7 An Induction Stove
Two pots of water are placed on an induction stove at the same time.
The stove itself is cool to the touch. The water in the ferromagnetic
metal pot is boiling while that in the glass pot is not. How can such
a cool stove boil water, and why isn’t the water in the glass pot boiling?
LENZ’S LAW
The induced emf resulting from a changing magnetic flux has a
polarity that leads to an induced current whose direction is such
that the induced magnetic field opposes the original flux change.
LENZ’S LAW
The induced emf resulting from a changing magnetic flux has a
polarity that leads to an induced current whose direction is such
that the induced magnetic field opposes the original flux change.
Reasoning Strategy
1. Determine whether the magnetic flux that penetrates the coil
is increasing or decreasing.
2.
Find what the direction of the induced magnetic field must be
so that it can oppose the change influx by adding or subtracting
from the original field.
3.
Use RHR-2 to determine the direction of the induced current.
Conceptual Example 8 The Emf Produced by a Moving Magnet
A permanent magnet is approaching a
loop of wire. The external circuit consists
of a resistance. Find the direction of the
induced current and the polarity of
the induced emf.
A transformer is a device for increasing or decreasing an ac
voltage.
N
s
s
N
t
p
p
t
N
s
s
N
p
p
V
N
Transformer
s
s
equation
V
N
p
p
I
V
N
p
p
s
I
V
N
p
s
s
A transformer that steps up the voltage simultaneously steps
down the current, and a transformer that steps down the voltage
steps up the current.
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Arithmetic Sequence Worksheet
Identify a 1 n and d for the sequence. Here we are going to some practice question.
9 Arithmetic Sequence Examples Doc Pdf Excel Arithmetic Sequences Geometric Sequences Arithmetic Sequences Activities
In an arithmetic sequence the difference between one term and the next is a constant.
Arithmetic sequence worksheet. Solution 1 the first term of an a p is 6 and the common difference is 5. This set of worksheets lets 8th grade and high school students to write variable expression for a given sequence and vice versa. 23 a 21 1 4 d 0 6 24 a 22 44 d 2 25 a 18 27 4 d 1 1 26 a 12 28 6 d 1 8 given two terms in an arithmetic sequence find the recursive formula.
13 a 32 622 d 20 14 a 18 166 d 8 15 a 9 74 d 6 16 a 28 231 d 10 given two terms in an arithmetic sequence find the explicit formula. A sequence is a set of things usually numbers that are in order. Arithmetic sequences and sums sequence.
Find a n using a n a 1 n 1 d. 27 a 18. Use the general term to find the arithmetic sequence in part a.
Given a term in an arithmetic sequence and the common difference find the recursive formula and the three terms in the sequence after the last one given. Solution to problem 3. 17 a 16 105 and a 30 203 18 a 17 95 and a 38.
Worksheet by kuta software llc precalculus arithmetic and geometric sequences practice. Algebra 1 arithmetic sequence displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are arithmetic sequences date period introduction to sequences unit 3c arithmetic sequences work 1 arithmetic and algebra work 4 keystone algebra 1 geometric sequences arithmetic sequences quiz review arithmetic series work.
Given a term in an arithmetic sequence and the common difference find the term named in the problem and the explicit formula. In other words we just add the same value each time. 5 a 24 38 d 2 find a 36 6 a 9 60 d 10 find a 35.
Find the a p and its general term. 4 3 arithmetic and geometric sequences worksheet determine if the sequence is arithmetic. General term of an arithmetic sequence.
An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. We use the n th term formula for the 6 th term which is known to write a 6 52 a 1 10 6 1 the above equation allows us to calculate a 1. Each number in the sequence is called a term or sometimes element or member read sequences and series for more details.
Given a term in an arithmetic sequence and the common difference find the 52nd term and the explicit formula. Finding the sum of a given arithmetic sequence. Observe the sequence and use the formula to obtain the general term in part b.
If it is find the common difference. Arithmetic sequence worksheet about arithmetic sequence worksheet arithmetic sequence worksheet. Find its 15 th term.
Arithmetic sequences series worksheet the general term of an arithmetic sequence is given by the formula a n a 1 n 1 d where a 1 is the first term in the sequence and d is the common difference. You can find solution for each question with detailed explanation. Given the first term and the common difference of an arithmetic sequence find the explicit formula and the three terms in the sequence after the last one given.
Arithmetic Sequence Find The Next Three Terms Word Problems Included Arithmetic Sequences Arithmetic Sequencing
50 Arithmetic Sequence Worksheet With Answers In 2020 Arithmetic Sequences Arithmetic Persuasive Writing Prompts
Algebra 2 Worksheets Sequences And Series Worksheets Arithmetic Sequences Arithmetic Geometric Sequences
Arithmetic Sequence Worksheet Finding The Nth Term Arithmetic Sequences Arithmetic Grade 6 Math Worksheets
Arithmetic Series Evaluate Type 2 Summation Notations Sigma Arithmetic Printable Math Worksheets Sequence And Series
Arithmetic Sequence Identify The First Term And The Common Difference Arithmetic Sequences Arithmetic Sequencing
Algebra 2 Worksheets Sequences And Series Worksheets Geometric Mean Geometric Sequences Arithmetic Sequences
This Is A 20 Problem Worksheet Where Students Have To Find The Nth Term Of An Arithmetic Sequences Geometric Sequences Math Worksheets
Arithmetic Sequences Maze Worksheet Arithmetic Sequences Sequencing Worksheets Word Problem Worksheets
Fill In The Missing Terms In Each Sequence Sequencing Arithmetic Sequences Geometric Sequences
Arithmetic And Geometric Sequences Worksheet In 2020 Geometric Sequences Arithmetic Sequences Arithmetic
Arithmetic Sequence Worksheet Algebra 1 Quiz Worksheet Using The General Term Of An Arithmetic In 2020 Arithmetic Sequences Arithmetic Algebra
Arithmetic Sequence Ap In Algebraic Expressions Find The Value Arithmetic Sequences Arithmetic Progression Arithmetic
Arithmetic Sequence Practice At 3 Levels Arithmetic Sequences Arithmetic Complex Sentences Worksheets
50 Arithmetic Sequences And Series Worksheet In 2020 Sequence And Series Arithmetic Online Study
Arithmetic Sequence Worksheet Answers Best Of Dentrodabiblia Arithmetic Sequences In 2020 Arithmetic Sequences Persuasive Writing Prompts Graphing Linear Inequalities
Arithmetic And Geometric Sequences Line Puzzle Activity Geometric Sequences Arithmetic Geometric Sequences Activity
Arithmetic Sequence Recursive Formula Arithmetic Sequences Geometric Sequences Sequencing
Arithmetic Sequences Match Up Formulas Arithmetic Sequences Arithmetic Sequences Activities Sequence And Series
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# How do I pick a Delta flight number?
Table of Contents
## How do I pick a Delta flight number?
Your first lottery number is the first Delta number in the sequence. The second number is the first two Delta numbers added together. The third is your second lottery number + the third Delta number. The fourth number is the third lottery number + the fourth Delta number.
## What lottery numbers get picked the most?
22 was drawn 26 times and 11 was drawn 24 times. Several other of the most common Mega Millions numbers were drawn 18 times; these numbers are 13, 14, 17, 18, 24, and 25.
## Where does Richard Lustig live?
Lustig lived in Orlando Florida, where he had a career as an entertainment booking agent. He died in 2018 at age 67.
## How to calculate Delta in Lotto?
Third delta: Subtract the second lotto number from the third lotto number. Fourth delta: Subtract the third lotto number from the fourth lotto number. Fifth delta: Subtract the fourth lotto number from the fifth lotto number. Sixth delta: Subtract the fifth lotto number from the sixth one. Above page is just an example, using a 6/50 lotto game.
## How to pick a lottery number?
Choose a number from 1 to 3. Pick two other numbers from 1 to 8. Now pick something close to 8. Now pick two numbers between 8 and 15. Mix the numbers up, so they’re not in numerical order. Your first lottery number is the first Delta number in the sequence. The second number is the first two Delta numbers added together.
## What is the best number to pick in a Delta Drawing?
Over 60% of the time, ONE is going to be at least one of the winning delta numbers. So, if you feel good about it, make the number ONE your first choice. If you don’t pick ONE, choose another very low number like two or three. For the fourth number, pick something pretty close to 8 ,either above or below it.
## How do you calculate how close you were to winning lottery numbers?
If you didn’t pick winning numbers, turn the winning number into a delta, and compare it to your delta to see how close you really were: Copy the first lotto number down. Subtract the first lotto number from the second one. Third delta: Subtract the second lotto number from the third lotto number.
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# Can someone explain LazySelect?
The LazySelect algorithm is given in these slides as follows.
We have a set $S$ of $n = 2k$ distinct numbers and want to find the $k$th smallest element.
1. Let $R$ be a set of $n^{3/4}$ elements chosen uniformly at random with replacement from $S$.
2. Sort $R$ and find $a$ and $b$ such that $\mathrm{rank}_R(a) = kn^{-1/4} – sqrt(n)$ and $\mathrm{rank}_R(b) = kn^{-1/4} + sqrt(n)$, where $\mathrm{rank}_X(x) = t$ if $x$ is the $t$th smallest element in $X$.
3. Compute $\mathrm{rank}_S(a)$ and $\mathrm{rank}_S(b)$: Output FAIL if $k < \mathrm{rank}_S(a)$ or $k > \mathrm{rank}_S(b)$.
4. Let $P = \{i \in S\mid a\le y\le b\}$: Output FAIL if $|P| \ge 4n^{3/4}$.
5. Return the $(k - \mathrm{rank}_S(a) + 1)$th smallest element from $P$.
Can someone explain the intuition behind the algorithm in a way that is more detailed and easier to understand than the slides above?
$\DeclareMathOperator{\rank}{rank}$Given a set $S$ of $n = 2k$ elements, the algorithm is aimed at finding the median of $S$ in linear time, with high probability. The idea is to find two elements $a,b \in S$ with $\rank(a) \leq k \leq \rank(b)$ and $\Delta = \rank(b) - \rank(a)$ as small as possible. Given such elements, we can find the median in time $O(n + \Delta\log\Delta)$ as follows:
1. Determine $\rank(a)$ in time $O(n)$.
2. Compute the list $P = \{ x \in S : a \leq x \leq b \}$ in time $O(n)$.
3. Sort $P$ in time $O(\Delta\log\Delta)$.
4. Output the element $x \in P$ of rank $\rank_P(x) = k - \rank(a)$ in time $O(1)$.
How do we find such elements $a,b$? The idea is to sample a smaller list $R$ which contains a $p$ fraction of the elements of $S$, for suitable $p$. An element $x \in R$ of rank $r$ in $S$ will have rank in the range $pr \pm c\sqrt{pr}$ in $R$ with probability $1-e^{-O(c^2)}$ (due to a Chernoff bound). In particular, if we the elements $a,b$ ranked $pk - c\sqrt{pk}$ and $pk + c\sqrt{pk}$ in $R$, respectively, then $\rank(a) \leq k \leq \rank(b)$ with probability $1 - e^{-O(c^2)}$. The size of the corresponding list $P$ will be roughly $$\Delta \approx p^{-1} (\rank_R(b) - \rank_R(a)) = 2c\sqrt{n/p}.$$ We can find the elements $a,b$ by sorting the list $R$, which takes time $O(pn \log(pn))$. In total, the algorithm looks like this:
1. Sample a set $R$ by putting each element with probability $p$.
2. Sort $R$ in time $O(pn \log n)$.
3. Find the elements $a,b$ ranked $pk-c\sqrt{pk}$ and $pk+c\sqrt{pk}$ in $R$.
4. Proceed as before.
The total running time is $$O(pn\log n + n + c\sqrt{n/p}\log n).$$ In order for this to be $O(n)$, we need $$\Omega(c^2 \log^2 n/n) \leq p \leq O(1/\log n).$$ For the algorithm to succeed with high probability $1 - o(1)$, we need $c = \omega(1)$. One setting which satisfies all these constraints is $p = n^{-1/4}$ and $c = n^{1/8}$, the choice taken in the slides.
• Thanks for the excellent explanation. I do have some questions about the math-heavy parts, though. I looked up the Chernoff bound but I don't get how it applies to the ranks of the elements in R. Can you explain this a bit further? Also, why is the approximate value of Δ calculated like that? Commented Jun 14, 2014 at 23:58
• The $R$-rank of an element $x \in R$ ranked $k$ is distributed $\mathrm{Bin}(k,p)$, since the $R$-rank is the number of elements in $R$ smaller than $x$. As for the calculation of $\Delta$, an interval of size $\ell$ in the original list shrinks to size approximately $p\ell$ in $R$ (since we take each element with probability $p$), and inverting this we get the approximation I state. This is of course not quite a formal proof, but gives the general idea. Commented Jun 15, 2014 at 8:49
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# Soloving sqrt(5)Cos(2x + 0.464)
thomas49th
## Homework Statement
Solve for the interval 0 < x < pi/2
$$\sqrt{5}Cos(2x+0.464)$$
## The Attempt at a Solution
Am i right in saying as cos oscillate between -1 and 1:
The maximum is sqrt(5) and occurs when (2x-0.464) = 1
Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36
How do you solve equations like above :S
Thanks
yes my calculator is in radians.
Homework Helper
FIrst of all, what is it that you are trying to solve?
thomas49th
oops i forgot.
solve for x in the range stated above
Thanks ;)
Homework Helper
## Homework Statement
Solve for the interval 0 < x < pi/2
$$\sqrt{5}Cos(2x+0.464)$$
## The Attempt at a Solution
Am i right in saying as cos oscillate between -1 and 1:
The maximum is sqrt(5) and occurs when (2x-0.464) = 1
Now i got 0.464/2 and (2pi - 0.463)/2
and the back of the book says 0.32 and 2.36
How do you solve equations like above :S
Thanks
yes my calculator is in radians.
In order to solve any equation, you first have to have an equation. What equation are you talking about. Exactly what is the question you want to answer?
Yes, the maximum of the expression you give is sqrt(5) but it does not occur when 2x+ 0.464= 1, it occurs when 2x+0.464= 0 or a multiplie of 2pi which is, I presume, where you got the values of 0.464/2 and (2pi- 0.463)/2 (it should be -0.464/2= -0.232 and (2pi 0.463)/2). But what is the question to which your book says the answer is 0.32 and 2.36?
thomas49th
oops i forgot.
solve for x in the range stated above
Thanks ;)
thomas49th
The rest of the question though is:
I) Express 2Cos2x-Sin2x in the form RCos(2x+a) giving R exactly and a to 3dp
II) Hence find the x-coordinates of the points of interestction of C1 and C2 in the interval 0<=x=<pi. Give your answer as radians and to 2dp
Note that c1 and c2 are curves with the equations y = cos2x - 2sin²x and y = sin2x respectively.
thomas49th
it' ok i got it. cos-1(1/sqrt(5))
easy peasy really :)
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Solutions by everydaycalculation.com
## Multiply 60/40 with 10/81
1st number: 1 20/40, 2nd number: 10/81
This multiplication involving fractions can also be rephrased as "What is 60/40 of 10/81?"
60/40 × 10/81 is 5/27.
#### Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 60/40 × 10/81 = 60 × 10/40 × 81 = 600/3240
3. After reducing the fraction, the answer is 5/27
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
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# How do I find values not given in (interpolate in) statistical tables?
Often people use programs to obtain p-values, but sometimes - for whatever reason - it may be necessary to obtain a critical value from a set of tables.
Given a statistical table with a limited number of significance levels, and a limited number of degrees of freedom, how do I obtain approximate critical values at other significance levels or degrees of freedom (such as with $t$, chi-square, or $F$ tables)?
That is, how do I find the values "in between" the values in a table?
This answer is in two main parts: firstly, using linear interpolation, and secondly, using transformations for more accurate interpolation. The approaches discussed here are suitable for hand calculation when you have limited tables available, but if you're implementing a computer routine to produce p-values, there are much better approaches (if tedious when done by hand) that should be used instead.
If you knew that the 10% (one tailed) critical value for a z-test was 1.28 and the 20% critical value was 0.84, a rough guess at the 15% critical value would be half-way between - (1.28+0.84)/2 = 1.06 (the actual value is 1.0364), and the 12.5% value could be guessed at halfway between that and the 10% value (1.28+1.06)/2 = 1.17 (actual value 1.15+). This is exactly what linear interpolation does - but instead of 'half-way between', it looks at any fraction of the way between two values.
### Univariate linear interpolation
Let's look at the case of simple linear interpolation.
So we have some function (say of $x$) that we think is approximately linear near the value we're trying to approximate, and we have a value of the function either side of the value we want, for example, like so:
\begin{array}{ c c } x & y\\ 8 & 9.3\\ 16 & y_{16}\\ 20 & 15.6\\ \end{array}
The two $x$ values whose $y$'s we know are 12 (20-8) apart. See how the $x$-value (the one that we want an approximate $y$-value for) divides that difference of 12 up in the ratio 8:4 (16-8 and 20-16)? That is, it's 2/3 of the distance from the first $x$-value to the last. If the relationship were linear, the corresponding range of y-values would be in the same ratio. So $\frac{y_{16} - 9.3}{15.6 - 9.3}$ should be about the same as $\frac{16-8}{20-8}$.
That is $\frac{y_{16} - 9.3}{15.6 - 9.3} \approx \frac{16-8}{20-8}$
rearranging:
$y_{16} \approx 9.3 + (15.6 - 9.3) \frac{16-8}{20-8} = 13.5$
An example with statistical tables: if we have a t-table with the following critical values for 12 df:
\begin{array}{ c c } (2\text{-tail})& \\ α & t\\ 0.01 & 3.05\\ 0.02 & 2.68\\ 0.05 & 2.18\\ 0.10 & 1.78 \end{array}
We want the critical value of t with 12 df and a two-tail alpha of 0.025. That is, we interpolate between the 0.02 and the 0.05 row of that table:
\begin{array}{ c c } α & t\\ 0.02 & 2.68\\ 0.025 & \text{?}\\ 0.05 & 2.18\\ \end{array}
The value at "$\text{?}$" is the $t_{0.025}$ value that we wish to use linear interpolation to approximate. (By $t_{0.025}$ I actually mean the $1-0.025/2$ point of the inverse cdf of a $t_{12}$ distribution.)
As before, $0.025$ divides the interval from $0.02$ to $0.05$ in the ratio $(0.025-0.02)$ to $(0.05-0.025)$ (i.e. $1:5$) and the unknown $t$-value should divide the $t$ range $2.68$ to $2.18$ in the same ratio; equivalently, $0.025$ occurs $(0.025-0.02)/(0.05-0.02) = 1/6$th of the way along the $x$-range, so the unknown $t$-value should occur $1/6$th of the way along the $t$-range.
That is $\frac{t_{0.025}-2.68}{2.18-2.68} \approx \frac{0.025-0.02}{0.05-0.02}$ or equivalently
$t_{0.025} \approx 2.68 + (2.18-2.68) \frac{0.025-0.02}{0.05-0.02} = 2.68 - 0.5 \frac{1}{6} \approx 2.60$
The actual answer is $2.56$ ... which is not particularly close because the function we're approximating isn't very close to linear in that range (nearer $\alpha = 0.5$ it is). ### Better approximations via transformation
We can replace linear interpolation by other functional forms; in effect, we transform to a scale where linear interpolation works better. In this case, in the tail, many tabulated critical values are more nearly linear the $\log$ of the significance level. After we take $\log$s, we simply apply linear interpolation as before. Let's try that on the above example:
\begin{array}{ c c } α & \log(α)& t\\ 0.02 & -3.912 & 2.68\\ 0.025& -3.689 & t_{0.025}\\ 0.05 & -2.996 & 2.18\\ \end{array}
Now
\begin{eqnarray} \frac{t_{0.025}-2.68}{2.18-2.68} &\approx& \frac{\log(0.025)-\log(0.02)}{\log(0.05)-\log(0.02)} \\ &=& \frac{-3.689 - -3.912}{-2.996 - -3.912}\\ \end{eqnarray}
or equivalently
\begin{eqnarray} t_{0.025} &\approx& 2.68 + (2.18-2.68) \frac{-3.689 - -3.912}{-2.996 - -3.912}\\ &=& 2.68 - 0.5 \cdot 0.243 \approx 2.56 \end{eqnarray}
Which is correct to the quoted number of figures. This is because - when we transform the x-scale logarithmically - the relationship is almost linear: Indeed, visually the curve (grey) lies neatly on top of the straight line (blue).
In some cases, the logit of the significance level ($\text{logit}(\alpha)=\log(\frac{α}{1-α})=\log(\frac{1}{1-α}-1)$) may work well over a wider range but is usually not necessary (we usually only care about accurate critical values when $\alpha$ is small enough that $\log$ works quite well).
### Interpolation across different degrees of freedom
$t$, chi-square and $F$ tables also have degrees of freedom, where not every df ($\nu$-) value is tabulated. The critical values mostly$^\dagger$ aren't accurately represented by linear interpolation in the df. Indeed, often it's more nearly the case that the tabulated values are linear in the reciprocal of df, $1/\nu$.
(In old tables you'd often see a recommendation to work with $120/\nu$ - the constant on the numerator makes no difference, but was more convenient in pre-calculator days because 120 has a lot of factors, so $120/\nu$ is often an integer, making the calculation a bit simpler.)
Here's how inverse interpolation performs on 5% critical values of $F_{4,\nu}$ between $\nu = 60$ and $120$. That is, only the endpoints participate in the interpolation in $1/\nu$. For example, to compute the critical value for $\nu=80$, we take (and note that here $F$ represents the inverse of the cdf):
$$F_{4,80,.95} \approx F_{4,60,.95} + \frac{1/80 - 1/60}{1/120 - 1/60} \cdot (F_{4,120,.95}-F_{4,60,.95})$$ (Compare with diagram here)
$^\dagger$ Mostly but not always. Here's an example where linear interpolation in df is better, and an explanation of how to tell from the table that linear interpolation is going to be accurate.
Here's a piece of a chi-squared table
Probability less than the critical value
df 0.90 0.95 0.975 0.99 0.999
______ __________________________________________________
40 51.805 55.758 59.342 63.691 73.402
50 63.167 67.505 71.420 76.154 86.661
60 74.397 79.082 83.298 88.379 99.607
70 85.527 90.531 95.023 100.425 112.317
Imagine we wish to find the 5% critical value (95th percentiles) for 57 degrees of freedom.
Looking closely, we see that the 5% critical values in the table progress almost linearly here: (the green line joins the values for 50 and 60 df; you can see it touches the dots for 40 and 70)
So linear interpolation will do very well. But of course we don't have time to draw the graph; how to decide when to use linear interpolation and when to try something more complicated?
As well as the values either side of the one we seek, take the next nearest value (70 in this case). If the middle tabulated value (the one for df=60) is close to linear between the end values (50 and 70), then linear interpolation will be suitable. In this case the values are equispaced so it's especially easy: is $(x_{50,0.95}+x_{70,0.95})/2$ close to $x_{60,0.95}$?
We find that $(67.505+90.531)/2 = 79.018$, which when compared to the actual value for 60 df, 79.082, we can see is accurate to almost three full figures, which is usually pretty good for interpolation, so in this case, you'd stick with linear interpolation; with the finer step for the value we need we would now expect to have effectively 3 figure accuracy.
So we get: $\frac{x-67.505}{79.082-67.505} \approx {57-50}{60-50}$ or
$x\approx 67.505+(79.082-67.505)\cdot {57-50}{60-50}\approx 75.61$.
The actual value is 75.62375, so we indeed got 3 figures of accuracy and were only out by 1 in the fourth figure.
More accurate interpolation still may be had by using methods of finite differences (in particular, via divided differences), but this is probably overkill for most hypothesis testing problems.
If your degrees of freedom go past the ends of your table, this question discusses that problem.
• Great answer! For the section, "Interpolation across different degrees of freedom": What if we would need to interpolate both dimensions (as both degrees of freedom we are interested in are not tabulated)? Bilinear interpolation (en.wikipedia.org/wiki/Bilinear_interpolation)? I have seen euklidean distances (with inverse weights) from the point of interest to the "corners" of the box (4 "closest known values) around the point being applied in software. Any recommendation? Jan 1 '20 at 22:12
• Are you talking about interpolating in an F, for both numerator and denominator degrees of freedom? For that, certainly some form of bivariate interpolation is needed (but I would usually do bilinear interpolation in the inverse of both df). In many cases interpolating in one direction and then interpolating in the other should work well enough and can be done without learning any new skills. But it doesn't make sense to interpolate statistical tables in software, since you can use widely implemented mathematical functions (e.g. regularized incomplete beta) to do it essentially exactly. Jan 1 '20 at 22:56
• I was thinking more generally when interpolating numerator and denominator degrees of freedom for tabulated values. Seems like software implementations for a lot of statistical tests (still) rely on tables the tests' authors supplied in their (classical) paper as the statistic follows some "special" distribution (often derived from Monte Carlo simulations). These tables can be considered "small" or "not fine-grained" given today's computing abilities (e.g. one could use pre-computed larger/more fine-grained tables or an algorithem for direct (exact) computation, where known). Jan 1 '20 at 23:27
• For computer-based interpolation of smooth functions there are a number of more sophisticated approaches than the ones outlined here. For example, there's a long history of finite difference methods and various other approaches, but this is outside the scope of the question here, which limits itself to people interpolating statistical tables "by hand" (with nothing better than a table and a calculator, say). Jan 2 '20 at 0:37
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# Conditional Probability
Probability is essentially a measure of the possibility of something happening. And the likelihood ranges from 0 (impossible) to 1 (possible) (certain). In a sample space, the total of all probability for all events is 1. The likelihood of an event occurring if another event has already occurred is known as conditional probability. One of the most fundamental notions in probability theory is the concept of probability distributions.
Conditional probability is determined by increasing the probability of the former occasion by the refreshed probability of the succeeding, or contingent, occasion. Note that this probability doesn’t express that there is consistently a causal connection between the two occasions, just as it doesn’t show that the two occasions happen at the same time. This probability is denoted by the notation P(B|A), which denotes the likelihood of B given A. The Bayes’ theorem, which is one of the most prominent theories in statistics, is central to the idea of conditional probability.
P(A|B) could or might not be the same as P(A) (the unconditional probability of A). If P(A|B) = P(A), then occurrences A and B are said to be independent, meaning that knowing about one does not change the likelihood of the other. As such, conditional probability is the probability that an occasion has happened, considering some extra data about the results of an analysis.
The terms conditional probability and unconditional probability are often used interchangeably. Unconditional probability refers to the chance that an event will occur regardless of whether or not previous events have occurred or other circumstances exist. If the occurrences A and B are not independent, the probability of their interaction (the likelihood of both events occurring) is given by:
P(A and B) = P(A) P(B|A),
On the other hand, it can be written as;
P(A B) = P(A)P(B|A),
And, from this definition, the conditional probability P(B|A) can be defined as:
P(B|A) = P(A and B)|P(A)
Or, simply;
P(B|A) = P(A B)P(A), as long as P(A)> 0
While conditional probabilities can be incredibly beneficial, they are frequently used with minimal information. As a result, Bayes’ theorem may be used to reverse or convert a conditional probability. Furthermore, in certain circumstances, events A and B are independent occurrences, i.e., event A has no influence on the likelihood of event B; in these cases, the conditional probability of event B given event A, P(B|A), is basically the probability of event B, P(B). The Bayes’ theorem can be expressed mathematically as follows:
P(B|A) = P(B|A)P(A) / P(B)
Finally, a tree diagram may be used to find conditional probabilities. The probabilities in each branch of the tree diagram are conditional.
The Bayes theorem, commonly known as Bayes’ Rule or Bayes’ Law, is the cornerstone of Bayesian statistics. This collection of probability principles allows one to alter their forecasts of future occurrences based on new knowledge, resulting in more accurate and dynamic estimations. Mutually exclusive occurrences are events that cannot happen at the same time in probability theory.
All in all, on the off chance that one occasion has effectively happened, another can occasion can’t happen. Accordingly, the restrictive likelihood of fundamentally unrelated occasions is consistently zero. The law of absolute likelihood is just the utilization of the augmentation rule to quantify the probabilities in additional fascinating cases. Conditional probability, on the other hand, does not express the chance relationship between two occurrences, nor does it say that both events occur at the same time.
In domains as disparate as mathematics, insurance, and politics, conditional probability is applied. For example, a president’s re-election is determined by voter preferences, television advertising performance, and the likelihood that the opponent would make gaffes during debates.
Information Solution:
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A football player punts a football at an inclination of 45 degrees to the horizontal at an initial velocity of
70 feet per second. The height of the football in feet is given by the quadratic function
h(x)= -32x^2/
(70)^2 +x+4 WRITTEN LIKE THIS BELOW TOO if u were confused
h(x)= (-32x^2/(70)^2) +x+4
where x is the horizontal distance of the football from the point at which it was kicked. What was the horizontal distance of the football from the point at which it was kicked when the height of the ball is at a maximum? What is the maximum height of the football?
MAXIMUM??_____ MINIMUM???____
Mar 10, 2020
#1
+1
Maximum will occur at x = -b/2a = 1/ (2 ( 32/70^2)) = 4900/64 = x = 76.563 feet
Substitute this value of 'x' in to the h(x) equation to find the height at this point......
Mar 10, 2020
#2
-1
GOT IT THANK U SO MUCH sorry
mharrigan920 Mar 11, 2020
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# Waves and Wave properties
## Presentation on theme: "Waves and Wave properties"— Presentation transcript:
Waves and Wave properties
Simulation
Wave Information Waves are a travelling form of energy
Waves can also carry information Sound Light Music electricity
Wave Types Transverse Longitudinal
Oscillates perpendicular to the direction of the motion. Examples are water waves and seismic S-waves Oscillate in the same direction/parallel to the direction of movement They move by expanding and contracting Examples are seismic P-waves and sound
Parts of a Wave: Waves have the following parts: Crest Trough
What are the 2 different types of waves?
How are these 2 types of waves different?
Wave Properties Waves have 3 specific, identifying properties:
Frequency Wavelength λ Amplitude
Frequency How many times the wave oscillates up and down
Measured in Hz (hertz) 1 Hz is equal to 1 cycle per second Water frequency is around Hz Sound has a frequency range of 20-20,000 Hz
Wavelength Series of repeating highs and lows
Measured from crest to crest, or trough to trough
This is the amount/distance that the wave moves away from equilibrium
Amplitude: This is the amount/distance that the wave moves away from equilibrium
If the frequency of a wave increases, what do you think will happen to the wavelength?
What type of a relationship is this? Positive Negative
Frequency and Wavelength have an inverse or negative relationship
Relationships: Frequency and Wavelength have an inverse or negative relationship As the frequency of a wave increases, its wavelength decreases OR As the frequency of a wave decreases, its wavelength increases
Calculating the Speed of Waves
The speed of a wave is equal to its frequency multiplied by its wavelength. V = ƒ ʎ V=velocity/speed ʎ = wavelength and ƒ= frequency This tells you the speed of the oscillations through the material/medium.
Wave Interference: When waves encounter an obstacle or interact with matter, they experience interference. There are 4 types of wave interference Reflection Refraction Diffraction Absorption
Reflection
Refraction: When a wave moves through a medium that slows or bends the wave’s movement it is refracted Refraction of light causes objects to appear bent
Diffraction: When a wave bends around objects such as sound bending around corners
Absorption: When a wave encounters a medium and seems to disappear.
What is the difference between refraction and reflection?
What is an example of a wave that is reflected? What is an example of a wave being refracted?
Interference Continued
Constructive Interference Destructive Interference When 2 waves that are “in phase” with one another collide Their collision causes an increase in the waves effect When 2 waves are “out of phase with one another collide Their collision causes a decrease or temporary disappearance of the wave
Interference Continued
HARMONIC MOTION Motion that repeats in cycles
A cycle is a unit of motion that repeats over and over Harmonic Motion is different from Linear Motion
Harmonic Motion Graphs
Pendulum Simulation
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# do my online math homework
please do my online math homeork lesson 2 and oritentation
also do writting assignment below
plagarism free with report
MAT102 LESSON 2 Date: Sep 6
Topic: Historical Numeration Systems
In this lesson you will learn about Egyptian numerals, traditional Chinese numerals and Hindu Arabic numerals.
Read section 4.1 “Historical Numeration systems” of your textbook. You can read the section on MyMathLab by clicking on Lesson 2 when you access your homework on MyMathLab. Pay special attention to Examples 1, 2, 4, 5 and 6. Skip Chinese Numeration. Practice your knowledge by doing HW #2 on MyMathLab. Remember that homework is 40% of your grade.
As evaluation of your mastering of Egyptian multiplication you will submit an assignment in BlackBoard (by clicking Assignments). Use the Egyptian doubling method to multiply 44×21. In order to write the solution of this exercise you can follow the example below where the Egyptian algorithm is performed for the product 25×35:
“We write two columns. One starts with 1 and the other one with 35. Each row is obtained by doubling the previous one. We stop once in the column starting with 1 we get a number greater than 25:
1 35
2 70
4 140
8 280
16 560
32 1120
Now we write 25 as a sum of numbers from the first column: 25=16+8+1. 16, 8 and 1 are on the fifth, fourth and first rows of the first column so 25×35 is equal to the sum of the numbers 560, 280 and 35 which are on fifth, fourth and first rows of the second column.
we get that 25×35 = 560 + 280 + 35 = 875. “
Do you need a similar assignment done for you from scratch? We have qualified writers to help you. We assure you an A+ quality paper that is free from plagiarism. Order now for an Amazing Discount! Use Discount Code “Newclient” for a 15% Discount!NB: We do not resell papers. Upon ordering, we do an original paper exclusively for you.
The post do my online math homework appeared first on The Nursing Hub.
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Check the completeness of given binary tree | Set 2 – Using Level Order Traversal
Objective: Given a binary tree, write an algorithm to determine whether the tree is complete or not using Level order traversal.
Earlier we had solved this problem by counting the number of nodes in the tree. (Read – Check the completeness of given binary tree | Set 1 – Using Node Count. In this article, we will solve it using level order traversal.
Before we proceed, If you are reading about the completeness of binary tree for the first time then let’s first understand it with some examples.
Complete Binary Tree: A binary tree is complete if all levels are completely full except the last level. The last level may or may not be full and nodes should be on the left side.
Example:
Below is an example of a special case where the tree is full expect the last level and in the last level, the nodes have priority to be one the left side if the last level is not full. See the example below
Approach: Level Order Traversal
• Do the level order traversal from left to right.
• Once a null node is found, ideally there should not be any node present after that if the tree is complete. If that’s the case return true.
• So if a node is found which is not null after a null node that means the tree is not complete, return false.
• See the image below for more understanding.
Code:
Output:
```Given Binary Tree is a Binary Tree: false
Given Binary Tree is a Binary Tree: true
```
__________________________________________________
Top Companies Interview Questions..-
If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
__________________________________________________
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# Analyticity Matters
Ever since KVM posted his dissection of Rubel's UDE paper, I've been wanting to address this issue he raised:
I'm still amazed and can't quite fully digest the fact that a perfectly smooth function can be non-analytic. I mean, consider the $\exp(-1/x)$ function. If you stand at the origin, you have no clue what happens as you step forward! All the meters in your car will read zero at the origin, and yet, somehow the function bootstraps itself into rising! This never happens with analytic functions!
The function he was talking about is not $\exp(-1/x)$ per se, but
$$f(x) = \begin{cases}e^{\frac{-1}{x}} & \mbox{if}\quad x>0 \cr 0 & \mbox{if}\quad x\leq 0 \end{cases}$$
He goes on to plot it as so:
At first sight, defining a function differently for negative and positive x, and then expecting civilised behaviour felt like cheating to me. After all, $\exp(-1/x)$ is singular as $x \rightarrow 0^{-}$. But that doesn't make his comment any less rich, as one is lead to believe that $C^{\infty}$-continuity is more sacred than that. While this post doesn't solve anything, it notices some interesting patterns and compiles them all in one place, and provides for a sweet link next time the reader is having to explain this to someone over the internet. So lets see how far we can get.
The Power Series Expansion:
I'll confine myself to basic high-school math, so the learning curve is analytic for all readers.
For the uninitiated (I love that word), any function $f(x)$ that is said to be "analytic" (we'll get to that before this post ends), can be expanded as a power series in $x$, about a point $x_{o}$ like so:
$${f(x)} = {f(x_{o})} + \sum_{n=1}^{\infty}{{c_{n}}{(x-x_{o})}^{n}},$$
where $c_{n} = \frac{f^{(n)}(x_{o})}{n!}$, the $n^{th}$ derivative of the function at $x_{o}$, scaled down by $n$ factorial. Since the same function can be translated along the x-axis, we'll stick to Taylor expansions about the origin for all our derivations.
This has an interesting consequence. A general single-valued function with no strings attached, in order to be completely defined, will require us to specify the value of the function at an uncountably infinite number of points on the x-axis, no matter how small the domain (as long as it is not a collection of isolated points). However, analytical functions require only that we specify one set of countably infinite values (all derivatives at any one point within every simply connected part of the domain). This is a drastic reduction. No doubt, the compression of parameter space is stronger than what that mere $C^{\infty}$ continuity condition imposes. Apparently, for every function representable by a power series, uniformity of convergence ensures integrability and differentiability. If you've understood that, than stop reading further and do more productive things with your life.
Convergence and Analyticity:
Let us examine the importance of the word 'domain' with a few basic examples. Now, the KVM function involves stitching two separate domain definitions, so we'll start with simpler functions. Take for example, the infinite sum of the following geometric series:
$${1}/{1-x} = 1 + x + {x^{2}} + {x^{3}} + {x^{4}} + ....$$
The right-hand side is the Taylor expansion about the origin, where the $n^{th}$ derivative of $\frac{1}{1-x}$ is $n!$. Clearly, a geometric series can only converge if the multiplicative factor $x$ is less than $1$ in magnitude. So this expansion diverges for $x \geq 1$ .This is no surprise, as the $\frac{1}{1-x}$ blows up at $x=1$, and one can expect the characteristics of functional derivatives on one side to have little effect on the other side of this singular point (probably why physicists can't define a "before the big-bang" universe). A good analogy is the vibration of a constrained string. A string held put at a point can only vibrate in modes which have a node at that point. But if we clamp down (up?) all the derivatives of the string at that point, then excitations from one side cannot travel to the other. If one were to expand $\frac{1}{1-x}$ as a power series about some other point greater than one, it would be valid for all $x>1$. So we seem to have arrived at some sort of definition for the boundary of a domain. After all, $\frac{1}{1-x}$ is merely $\frac{1}{x}$ shifted and flipped. But there is an anomaly.
It turns out that this expansion also diverges for $x \leq -1$. Highly unexpected, as $\frac{1}{1-x}$ seems to have no bumps, kinks or black holes around $-1$. Funky behaviour on one side of the point of expansion seems to be affecting the symmetrically opposite point on the other side. The string has strings attached? Say we parity flip the function and expand again about the origin:
$$\frac{1}{1+x} = 1 - x + {x^{2}} - {x^{3}} + {x^{4}} - ....$$
This again is singular at $x=-1$, but the expansion diverges for both $x \leq -1$, and $x \geq 1$.
Note that the two expansions are interconvertible. One is generated by flipping the signs of all alternate terms in the other. Now we invoke the concept of conditional vs. absolute convergence of a power series.
An infinite series $\sum_{n=1}^{\infty}a_{n}$ tends to converge to a finite value if successive terms $a_{i}$ fall off fast enough as $i$ increases. Sometimes, the signs of the terms help the convergence by cancellation. But the series is said to be absolutely convergent (as opposed to conditionally convergent) if the fall off in successive term magnitude is so fast that $\sum_{n=1}^{\infty}\vert a_{n} \vert$ is convergent. Absolute convergence implies conditional convergence. But conditional convergence does not imply absolute convergence. Take for example the harmonic series $1 + {{1}/{2}} + {{1}/{3}} + {{1}/{4}} ....$. It is provably divergent. But if we flip the sign of every even term, the resulting series converges to $log(2)$. Series with alternating signs have added help at hand, as all they require is for the absolute values of successive terms $\vert a_{i} \vert$ to tend monotonically to $0$. None of this helps our case. We are looking for conditional vs. absolute divergence relations. Take the series $\sum_{n=1}^{\infty}a_{n}$. Denote the positive terms by $p_{1}, p_{2}, p_{3}, ....$, and the negative terms by $-q_{1}, -q_{2}, -q_{3}, ....$. Then the absolute sum is $\sum_{\nu = 1}^{\infty}p_{\nu} + \sum_{\nu = 1}^{\infty}q_{\nu}$, and the conditional sum is $\sum_{\nu = 1}^{\infty}p_{\nu} - \sum_{\nu = 1}^{\infty}q_{\nu}$. For absolutely convergent series, both the sum of positive terms and the sum of negative terms need to be convergent. For conditional convergence, both have to be divergent. For our power series expansions, both positive term sums and negative term sums have to converge inside the domain. Only one of them needs to diverge outside the domain. As an aside, the sum of conditionally convergent series can be changed by suitable rearrangement of the terms of the series, and it can even be made to diverge. This is not possible for absolutely convergent series.
So, what is the biggest generalisation we can make thus far? If a power series in $x$ converges for a value $x = \xi$, it converges absolutely for every value x such that $\vert x \vert < \vert\xi\vert$, and the convergence is uniform in every interval $\vert x \vert \le \eta$, where $\eta$ is any positive number less than $\vert\xi\vert$. $\eta$ may lie as near $\vert\xi\vert$ as we please (in an $\epsilon,\delta$ sense). This magically implies that, If $\exists x = \xi$ such that the power series diverges, then it must diverge for every value of $x$ such that $\vert x \vert > \vert\xi\vert$. This defines an "interval of convergence", centered about the point of expansion for the power series. Since the KVM function $\exp(-1/x)$ has a singularity at $x \rightarrow {0^{-}}$, it's interval of convergence for an expansion about the origin has size zero. This still doesn't seem to demystify the stitched function weirdness.
So What Is All This Analyticity Stuff??
So why is analyticity a stronger constraint than $C^{\infty}$ continuity? What other conditions does it impose on the countably infinite derivatives required to completely specify the function? It is worthwhile to recall that in calculus, the differential element $dx$ of the independent variable is a linear term in x. Often in analysis and perturbation theory, we've neglected terms with higher-order powers and managed to retain accuracy of our results. So a power series expansion of a function about a point to a very good accuracy, can be approximated by a finite degree polynomial, in a neighbourhood about that point, whose size is dependent on the desired error tolerance. This means, for an ultra-small neighbourhood, the function is a straight line. In a slightly larger, say mega-small, neighbourhood, it is a parabola, and so on. If we differentiate a power series any number of times, we again end up with a power series. So this property is not only shared by the function, but also by all its other derivatives. Say a function $f(x)$ has a zero of order $r$ at $x=a$, ${{f^{(r)}}{(a)}} \ne 0$. Then the function may be expressed as ${f(x)} = {{(x-a)}^{r}}{g(x)}$, with ${g(a)} = {f^{(r)}{(a)}}/r!$. $g(x)$ cannot vanish in a suitably small neighbourhood of $x=a$, or, the zeros of $f(x)$ are "isolated", unless of course $f$ vanishes identically. The same is true for $f'(x)$. It follows that in a finite interval an analytical function (and all its derivatives) is (are) piecewise monotone(s), that is, it (they) cannot change its (their) character of monotonicity infinitely often. As a counter-example, consider the Taylor expansion of $sin({1}/{x})$. This tends to wildly oscillate as $x$ approaches $0$. It crosses the x-axis infinitely many times in any finite interval about the origin. So the zeros at the origin can't be said to be isolated.
Complexity Matters:
So far, we seemed to have managed quite well on the real number line. Here comes the bomb: Taylor expansions of:
$$\frac{1}{1+x^{2}} = 1 - x^{2} + x^{4} - x^{6} + ....$$
or
$${tan^{-1}(x)} = x - {\frac{x^{3}}{3}} + {\frac{x^{5}}{5}} - ....$$
These expansions also diverge for $\vert x \vert >1$, despite the functions themselves being well-behaved at $\vert x \vert=1$. ${WTF}^{\infty}$.
It turns out that complex domain is more natural than its real subset. We note that both $\frac{1}{1+x^{2}}$ and the expansion of $tan^{-1}(x)$ diverge at $x=\pm i$, where $i = \sqrt{-1}$. Note that the real part of the first function blows up, but for the second function, its the imaginary part that does so. If we are willing to extend the power series expansion to complex variables, then we'll have to upgrade our interval of convergence to circle of convergence. This would mean that if $f(x)$ is singular at complex point $z_{1}$, then its power series expansion about the origin definitely diverges for points farther than $z_{1}$ on the complex domain. Take the function $e^{{-1}/{x^{2}}}$, which is quite close to the KVM function. All its derivatives vanish at $x=0$. So we cannot expand it in a power series about the origin because $e^{{-1}/{(ix)^{2}}}$ diverges on the imaginary axis as it tends toward the origin.
An arbitrary point $z$ on the complex plane may be defined as $z=re^{i \theta}$. Then, ${z^{n}} = {r^{n}}{e^{i n \theta}} = {r^{n}}{\cos(n \theta)} + i {r^{n}}{\sin(n \theta)}$. A singularity at some point $z_{1}$ affects the function expansion everywhere on the circle $\vert z \vert = \vert z_{1}\vert$. The vibrating string has become a 2D membrane. And if all derivatives of the membrane are clamped at some point, then radial (and angular?) excitations from the origin cannot travel beyond the circle centred at the origin, with the clamped point on its circumference. HSR once remarked that the EE department at IITM wrongfully exposes its students to Fourier and Laplace transforms before the concept of analytical functions has been taught. Electrical engineers who have studied filter design should recall all those conditions for "poles lying on the left half-plane" or "outside the unit circle", and reinterpret them in the present context.
This domain extension works wonders to our compression ratio discussion. It allows us to reduce the task of function definition from specifying two values at every point on a 2D region to specifying twice (hehe) countably infinite derivatives (real and imaginary) at any one point. There is a strange interplay between real and complex values of the power series expansion of any analytical function, that transcends the absolute vs. conditional convergence picture.
$${(-1)^{n}}{z^{n}} = {(-r)^{n}}{e^{i n \theta}} = {(-1)^{n}}{r^{n}}{\cos(n \theta)} + i {(-1)^{n}}{r^{n}}{\sin(n \theta)}$$
$${(z^{\ast})^{n}} = {r^{n}}{e^{-i n \theta}} = {r^{n}}{\cos(n \theta)} + i {(-1)^{n}}{r^{n}}{\sin(n \theta)}$$
The cosine and sine functions scale the magnitudes of the real and imaginary values and affect the signs of terms depending on $n \theta$. It might be worthwhile to see how a rotation transform affects power series expansions. It is interesting that complex analysis turned out to be so fundamental to our understanding of seemingly trivial math. The Euler exponential notation was more than a calculation and expression aid. And we didn't even need quantum mechanics! All of it has much to do with the fact that on a line, there are 2 sides to a point. There are two signs, + and -, as there are 2 kinds of electromagnetic charges. Hence the 2D domain. And a set of two harmonically conjugate solutions for the Laplacian. There cannot be a further extension of the complex domain to higher dimensions in quite the same way real was extended to complex (a la Algebraically closed fields). This was the triumph of 19th century mathematics, when rigour was born. The number 2 was found be holy.
Aside:
Parity is a discrete transformation. However, in relativistic quantum mechanics, while studying chiral symmetry of Dirac particles, a parity flip is modelled as a continuous transformation using the pseudo-scalar ${\gamma_{5}}$ matrix as a generator of rotations on a complex plane. The wavefunction transforms as $\psi \rightarrow {e^{i \theta \gamma_{5}}} \psi$, and an operator $\hat{H} \rightarrow {e^{i \theta \gamma_{5}}} \hat{H} {e^{-i \theta \gamma_{5}}}$. Also, in QED, in perturbative expansions of charge conjugation symmetric processes (like electron-positron annihilation), the observable cannot depend on sign of charge. So it is a power series in $e^{2}$, or $\alpha$, the fine structure constant. But in a universe where $\alpha < 0$, like charges will attract and opposite charges will repel. This is very unstable, as the only ground-state is two oppositely charged black-holes at infinite distance from eachother. So, if $\alpha$ could take values on a complex plane, then all negative real values are a no-no. So there is a branch cut on the complex plane all along the negative real axis, implying that the radius of convergence of a power series expansion in $\alpha$ about the origin is zero. So even though $\alpha$ is very small, the series don't converge. They just happen to be "asymptotic" series, which allows physicists to do some amount of close approximation to physics. Now you know why even theoretically computed physical values (like masses) have a computable error bound, and the number of significant digits keep increasing every year.
The Search for Hidden Redundancies:
The definition of analytical functions on the complex plane is generally made using the Cauchy-Riemann differential equations. However, Weierstrass is said to have developed the theory of complex analysis from scratch from the point of view of power series expansions. That derivation seems lost to the pages of history written in European languages. Let us see how much further we can explore the compression ratio aspects and built-in redundancies in function definition by specification of parameters. So in a domain where a complex function is analytic (i.e. differentiable in an $\epsilon$-$\delta$ limit sense), integration along any closed contour vanishes. But what if we introduce a simple pole (singularity) in the function $f(t)$ at the point $z$ and compute the contour integral around it. We have,
$${f(z)} = {\frac{1}{(2\pi i)}}{\int\limits_{C}}{\frac{f(t)}{t-z}}{dt}= {\frac{1}{(2\pi)}}\int\limits_0^{2\pi}{f(z+re^{i\theta})}{d\theta}$$
In the region of analyticity, the value of a function is the average of its value in the neighbourhood of the point in question (see how this is true for real functions defined on the real line). Does this also imply that the solution to a Laplacian differential equation is unique if the boundary conditions are fixed? That is a reduction from specifying function values on a 2D domain to specifying on a 1D contour. And since it is analytic along the contour loop, it can be parameterised again by coefficients of angular basis functions, this reducing further into countably infinite space. Every analytical function can be expanded as a power series. If we introduce slightly sharper poles at points and compute contour integrals around them, we have:
$$\frac{f^{(\nu)}(z_{o})}{\nu !}= {\frac{1}{(2\pi i)}}\int\limits_C{\frac{f(z_o+t)}{t^{\nu+1}}}dt = \frac{1}{(2\pi)}\int\limits_0^{2\pi}\frac{f(z_o+re^{i\theta})}{r^{\nu+1}}e^{-i(\nu+1)\theta}d\theta$$
The contour integral seems to circle the point $( \nu +1)$ times clockwise, to yield the ${\nu}^\text{th}$ derivative of the function. Note that in the power series expansion, the $n^\text{th}$ term (${z^{n}}={r^{n}}{e^{i n \theta}}$) also circles a closed contour n times (anti-clockwise, as if to unwind the derivative coefficient). The complex plane comes with strings attached, since every point has a build-in multiplicity (there are $n$ solutions to $z^{n}=\text{constant}$). Inverting a function converts poles to zeros and vice versa. Do you see how that affects regions of power series convergence? If $f(z)$ has an $n^{th}$ order zero at $z_{o}$, then it is expressible as ${f(z)} = {(z-z_{o})^{n}}{g(z)}$, where $g(z_{o}) \ne 0$. The inverse function ${1}/{f(z)} = q(z) = {h(z)}/{(z-z_{o})^{n}}$ ($h(z)$ is analytic in the neighbourhood of $z_{o}$), can be expanded as,
$$q(z) = {c_{-n}}{(z-z_{o})^{-n}} + .... + {c_{-1}}{(z-z_{o})^{-1}} +{ c_{0}} + {c_{1}}{(z-z_{o})} + ....$$
A contour integral of negative index terms of $q(z)$ around $z_{o}$ gives $2 \pi {c_{-1}}$, and the higher negative power terms vanish. The residue of a function at a pole is therefore $2 \pi {c_{-1}}$. The theorem of residues, which best illustrates the information compression ratio, states that if a function $f(z)$ is analytic in the interior of a region R and on its boundary C except at a finite number of interior poles, the integral of the function taken around C in the positive sense is equal to the sum of the residues of the function at the poles enclosed by the boundary C.
So Why Power Series??
Consider the polynomial expression,
$${a_{0}} + {a_{1}}{z} + {a_{2}}{z^{2}} + .... + {a_{n}}{z^{n}} = {P(z)}$$
For some real parameter $t$, take the integral along any closed path C in the z-plane, which does not pass through any of the zeros of $P(z)$,
$$u(t) = \oint\limits_{C}{\frac{e^{tz}f(z)}{P(z)}{dz}}$$
Let $f(z)$ be a constant or any polynomial in z, of a degree we shall assume to be less than $n$, so that the integrand behaves well at infinity. The successive derivatives of $u(t)$ with respect to $t$ under the integral sign is equivalent to multiplication of the integrand by $z, {z^{2}}, {z^{3}}, ....$ as the case may be. If we now form the differential expression $L{[}{u}{]} = {a_{0}}{u} + {a_{1}}{{du}/dt} + {a_{2}}{{{d}^{2}u}/{dt^{2}}} + .... + {a_{n}}{{{d}^{n}u}/{dt^{n}}}$, or in symbolic operator notation, $P(D)u$, we have,
$$P(D)u = L{[}{u}{]} = \oint{C}{}{{e^{tz}}{f(z)}{dz}} = 0$$
So $u(t)$ is a solution for the differential equation $P(D)u = 0$, and the $n$ arbitrary constants from the polynomial $f(z)$ are the constants of integration. Most differential equations engineers and scientists study can be constructed out of polynomial type operators. If ${a_{o}} \ne 0$,
$$\frac{1}{a_0 + a_1 t + a_2 t^2 + .... + a_n t^n} = b_0 + b_1 t + b_2 t^2 + ....$$
then the solution of
$$a_0 u + a_1 \frac{du}{dx} + a_2 \frac{d^2 u}{dx^2} + .... + a_n \frac{d^n u}{dx^n} = R(x)$$
is,
$$u(x) = b_0 R(x) + b_1 \frac{dR(x)}{dx} + b_2 \frac{d^2 R(x)}{dx^2} + ....$$
and if $a_{0}=0$ but $a_{1} \ne 0$,
$$\frac{1}{a_1 t + a_2 t^2 + .... + a_n t^n} = b t^{-1} + b_0 + b_1 t + b_2 t^2 + ....$$
and,
$$u(x) = b \int R(x)dx + b_0 R(x) + b_1 \frac{dR(x)}{dx} + b_2 \frac{d^2 R(x)}{dx^2} + ....$$
Balki once claimed that if all of Calculus had to be summed up in one single useful sentence, it would read, "An exponential grows faster than a polynomial, which grows faster than a logarithm". The definition of a differential element as a linear increment of the independent variable, and the practice of solving polynomial operator type differential equations, has lead to the extensive use of power series in analysing mathematics, instead of other types of expansions. However, linear diffential operators of infinite order, like the time evolution operator in Quantum mechanics, and the generator formalism for infinitesimal transformations have been explored as well. As for the KVM function, every derivative at $x = 0^{+}$ vanishes because they are all products of a diverging polynomial factor and a converging (vanishing) exponential factor. If one were to blindly Taylor-expand the function about the origin, all the polynomial factors from all the terms will seem to add up to something formidable, even compared to the common vanishing exponential factor. The ${\infty}^\text{th}$ derivative does seem to be of order $\infty$!, although understandably indecisive about its sign.
We started by stating that analytical functions can be at worst specified uniquely by a countably infinite number of parameters. In the power series expansion notation, these parameters turned out to be successive derivatives of the function at a point. However, from early lessons in probability density function theory, one may recall a similar set of parameters which are defined by the integration operator. A function on the real number line could well be uniquely specified by its mean, variance, and all the higher moments $\int{}{}{{(x-x_{o})^{n}}{f(x)}{dx}}$ about any point. I wonder how this would translate to the complex plane, since analyticity doesn't seem to play a role at first glance (but normalizability matters . . . . a lot . . . ). This might hint at the duality between differential equations and Lagrangian formalisms. Are the invariant constants dictated by Noether's theorem applied to the Lagrangian formalism directly related to the constants of integration arising from the differential equation formalism? Karthik's next post promises to enlighten us regarding some aspects of this.
Epistemology, and the possible non-futility of Modelling:
KVM's Rubel post was preceded by his disdain for all modelling. However, one must ask what it truly means to have understood a phenomena, as opposed to having merely "modelled" it. All our analysis seem to be restricted to analytical functions, but they are a small subset of all possible functions. Wiki claims that:
In a measure-theoretic sense: when the space C([0, 1]; R) is equipped with classical Wiener measure $\gamma$, the collection of functions that are differentiable at even a single point of [0, 1] has $\gamma$-measure zero. The same is true even if one takes finite-dimensional "slices" of C([0, 1]; R): the nowhere-differentiable functions form a prevalent subset of C([0, 1]; R).
How curious, that most natural processes can be easily described by analytical functions alone? Or can they? Aren't singularities our greatest unconquered enemies? In my view, the non-rapid, piecewise monotonic behaviour appeals to an aesthetic sense of the human mind, which has evolved to expect small input changes to have small observable effects, at least in the short run. Are we then destined to know the universe only to a close approximation? Does a mere knowledge of a handful of parameters qualify as knowledge of a process? Is the definition of the class of functions of the solution important? At what point of generalisation does a predictive DE (as defined by KVM) become a shrink DE? Is it NOT amazing that group symmetries helped define previously unobserved fundamental particles in high-energy physics? Is "intuitive feel" all we have? Needless to say, $C^{\infty}$ continuity is yet to be demystified. But I hope I have motivated a domain for analysis. If this entire post is to be condensed into a single valuable sentence, it would read, "2 is a holy number".
### References:-
1> Richard Courant, Fritz john, "Introduction to Calculus and Analysis", Vol. I. and II.
2> My class notes.
3> Wikipedia
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HuggingFaceTB/finemath
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flareblitz1
32
# If light travels 671,000,000 miles per hour, how many miles will it travel in one minute?
$1\ min=\frac1{60}h\\ \\ v=\frac st\\ s=vt\\ \\ s=671000000 \cdot \frac1{60} \approx 11183333\ mil$
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HuggingFaceTB/finemath
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## Elementary Statistics (12th Edition)
89%, $96.34^{o}F,100.06^{o}F$.
Using Chebyshev's theroem, 89% of the values are maximum 3 standard deviations from the mean. Minimum:$98.2^{o}F-3\cdot0.62^{o}F=96.34^{o}F$, maximum:$98.2^{o}F+3\cdot0.62^{o}F=100.06^{o}F$,.
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HuggingFaceTB/finemath
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# I need a lot of help with Coulomb's Law and Electric Fields? Need HELP
• devilz_krypt
In summary, the conversation is about someone needing help with understanding Coulomb's Law and Electric Fields. They have 3 questions they did not understand, including finding the magnitude and direction of an electric field for a small object, finding the acceleration and distance traveled for an electron in an electric field, and locating the point where the electric field is zero for two charges on the x-axis. They have multiple solutions for each problem and are looking for the best solution for each one.
devilz_krypt
I need a lot of help with Coulomb's Law and Electric Fields?? Need HELP!
## Homework Statement
Here are 3 of the 25 questions the i did not understand, in regards with electric fields and Coulomb's Law.
1. A small object of mass 2.0 g and charge of 1.8x10^-7 C "floats" in an electric field. Find the magnitude and direction of the field.
2. An electron is accelerated by a constant electric field of magnitude 300 N/C.
A) Find the acceleration of the electron.
B) How far will the electron travel in 3s?
3. Two charges lie along the x-axis. If q1= -9.0x10^-6 C is at x= 6m and q2= -8x10^-6 C at x= -4m, locate the point at which the electric field will be zero.
Please help me finish my test review, i have been very confused ever since we start Coulomb's Law and Electric Fields.
## Homework Equations
F= ((k)(q1)(q2))/(r^2)
E= (k)(q)/(r^2)
F= qE
The reason I didnt submit my solution is because i have about 5 different solutions for each problem.
Submit the solution you think is best for each one.
I understand that Coulomb's Law and Electric Fields can be challenging concepts to grasp. It is important to remember that these concepts are fundamental to understanding electricity and are used in many applications, such as electronics and power generation.
In order to better understand Coulomb's Law and Electric Fields, it is important to review the equations and principles involved. Coulomb's Law states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This can be represented by the equation F= (k)(q1)(q2)/(r^2), where k is the Coulomb's constant, q1 and q2 are the charges of the objects, and r is the distance between them.
Electric Fields, on the other hand, are defined as the force per unit charge that a test charge experiences at a given point in space. The equation for Electric Fields is E= (k)(q)/(r^2), where q is the test charge and r is the distance from the source charge.
Now, let's address the questions you have listed.
1. In order to find the magnitude and direction of the electric field in this problem, we can use the equation E= (k)(q)/(r^2). Since the object is "floating" in the field, we know that the net force on the object is zero. This means that the force due to gravity must be equal and opposite to the force due to the electric field. We can set up the equation Fg= Fe and solve for the electric field (E). Plugging in the values given, we get E= (6.67x10^-11 N*m^2/C^2)(1.8x10^-7 C)/(2.0x10^-3 kg)^2. This gives us an electric field of 1.5x10^4 N/C directed towards the positive charge.
2. A) To find the acceleration of the electron, we can use the equation F= qE. Since the electron has a charge of -1.6x10^-19 C, we can plug this into the equation along with the given electric field of 300 N/C. This gives us a force of -4.8x10^-17 N. Using Newton's second law (F=ma), we can solve for the acceleration (a). This gives us an acceleration of -3.
## 1. What is Coulomb's Law?
Coulomb's Law is a fundamental law in physics that describes the relationship between electric charges. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
## 2. How do I calculate the force using Coulomb's Law?
The formula for calculating the force between two charged particles using Coulomb's Law is F = (k * q1 * q2) / r^2, where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them.
## 3. What are electric fields?
Electric fields are regions in space where a charged particle experiences a force. They are represented by vectors and their direction is determined by the direction of the force on a positive test charge placed in the field.
## 4. How do I calculate the electric field using Coulomb's Law?
To calculate the electric field at a point due to a single point charge, we use the formula E = (k * q) / r^2, where E is the electric field, k is the Coulomb's constant, q is the charge of the source particle, and r is the distance between the source particle and the point of interest.
## 5. Can Coulomb's Law be used to calculate the force and electric field for multiple charges?
Yes, Coulomb's Law can be used to calculate the force and electric field for multiple charges by using the principle of superposition. This means that the total force or electric field at a point is the vector sum of the individual forces or electric fields due to each charge.
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HuggingFaceTB/finemath
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# Primary
### Foundations of Subtraction
##### Chapter 1: Revisiting the foundations
LO 1: Differentiating between numbers, numerals and digits LO 2: Recognising the importance of unit (‘1’) LO 3: Analysing multiple units LO 4: Explaining importance of ‘like unit’ LO 5: Counting and comparing ‘unlike units’? LO 6: Discussing the use of decimal number system LO 7: Exploring ‘value of place’ in everyday living LO 8: Identifying place value and face value of digits LO 9: Visualising numbers in decimal number system LO 10 Interpreting numbers as mixture of place value packets LO 11: Experiencing set in the decimal number system
##### Chapter 2: Context of subtraction
LO 1: Recalling arithmetic operations LO 2: Lessons from addition LO 3: Lessons from multiplication LO 4: Discovering the existential need for subtraction operation LO 5: Listing examples of difference in daily life LO 6: Comparing things for difference LO 7: Knowing how to learn subtraction LO 8: Preparing children to learn subtraction LO 9: Elaborating subtraction as more varied, and complex operation
##### Chapter 3: Introduction to subtraction
LO 1: Discovering the common threads across ‘subtraction situations’ LO 2: Exploring ‘Takeaway’ situations LO 3: Appreciating separation situations LO 4: Understand comparison situations LO 5: Knowing why 'what is left' is an important number? LO 6: Satisfying if difference is the right term for all subtraction situations LO 7: Expressing subtraction as a simple, universally acceptable narrative LO 8: Being comfortable with ‘–’
##### Chapter 4: Macro subtraction process
LO 1: Visualising subtraction expressions LO 2: Subtracting in pairs LO 3: Subtracting like units LO 4: Subtracting unlike units LO 5: Detailing subtraction of unlike units LO 6: Reasoning subtraction of unlike units LO 7: Revisiting backward counting LO 8: Visualising subtraction as backward skip counting
##### Chapter 5: Micro subtraction process
LO 1: Implementing simple subtraction expressions LO 2: Exploring minuend LO 3: Exploring subtrahend LO 4: Exploring difference LO 5: Investigating the difference between 8 – 1 – 1 and 8 – 2 LO 6: Reading 22 – 2 – 3 – 4 – 5 – 6 LO 7: Analysing 30 – 12 – 8 + 10 – 10 + 5 LO 8: Exploring 8 – 3, as comparison of 8 and 3 LO 9: Exploring 8 – 3, as ‘takeaway’ of 3 out of 8 LO 10: Exploring 8 – 3, as ‘separation’ 3 out of 8 LO 11: Appreciating 7 – 0, 0 – 7 LO 12: Detailing 7 - 7 LO 13: Exploring how 7 - 3 is different from 3 - 7 LO 14: Relating multiplication and repeated subtraction
##### Chapter 6: Macro subtraction operation
LO 1: Experiencing subtraction operation LO 2: Knowing what borrow represents in subtraction LO 3: Explaining why borrow is always ‘1’ LO 4: Appreciating that borrow can’t be the exact quantity needed LO 5: Exploring the possible values of borrow LO 6: Applying subtraction operation
##### Chapter 7: Micro subtraction operation
LO 1: Investigating subtraction of numerals from the right LO 2: Subtracting multiple subtrahends LO 3: Explaining cascade borrow LO 4: Subtracting a bigger number from a smaller number
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## Rethinking Topology (or a Personal Topologodicy)
(This document was typeset in unicode. This may cause problems for some people. A PDF is available as an alternative for them.)
When I was originally introduced to topology, I simply accepted most of its properties as generalizations of ℝⁿ. I didn’t give it any serious thought until about a month ago when I read an excellent thread on math overflow about it. Since then, its been one of the things I often find myself thinking about when I’m trying to fall asleep. Given the amount of thought I’ve put into it, and the fact that I feel I should be answer questions like this about topology, given that it’s one of the areas of math I spend a lot of time on, I thought I’d write up my thoughts. They lent themselves well to being written in the form of an introduction to topology, so that’s what I did.
(After finishing this essay I decided to reread the MO thread. The first comment — not answer, a comment — mentions the Kuratowski closure axioms and closure axioms sounded like one might call what I came up with. Sure enough, they’re the exact same, down to the ordering. Are all attempts to make mathematical contribution’s this frustrating? I’m posting this because of the amount of work I put in, but there’s nothing new here.)
Consider 1 with respect to [0,1). It isn’t part of the set, but in a sort of intuitive sense it almost is. And knowing which points are `almost in’ a set gives us lots of information, for example notions of boundaries and connectedness. Topology is based on us formalizing this notion of `almost in’ and once we formalize it, we can consider non-standard notions of being `almost in’ or apply these ideas to spaces that we don’t typically associate them with.
We call points that are in or `almost in’ a set the `adherant points’ of a set. We call the set of adherant points of a set the closure of that set; the closure of a set s is denoted as cl(s) or ̄s. From the closure, we can define lots of other niceties: the interior of a set, int(S) = cl(S ⌐) ⌐, or the boundary, bd(S) = cl(S) ∩ cl(S ⌐). Notice that we can also define closure in terms of either of these (like so: cl(S) = int(S ⌐) ⌐ = S ∪ bd(S)) and, as such, they all contain the same information.
We call a set which is its own closure is called a closed set. We call the complement of a closed set an open set. One can think of a closed set as one which contains its own boundary, and an open set as one which does not contain its own boundary.
At this point, closure could be any map from the power set (set of all possible subsets) of our space to itself. Obviously, many of these are nonsensical, for example cl(S) = ∅ would mean that points in a set aren’t adherant points of that set! So we’re going to put four restrictions on closure — they’re very important because they result in restrictions on our idea of a generic topological space and they’re the only restrictions we’re going to put on it.
• Restriction 1: This one has been stated informally when we introduced the notion of a closure: points of a set are adherant points of that set, or S ⊆ cl(S).
• Restriction 2: The closure of a set is closed, or cl(cl(S)) = cl(S). This can be thought of as meaning that adding the boundary of a set to itself doesn’t create a new boundary.
• Restriction 3: The closure of the union of two sets is the union of the closures. This can be thought of as meaning that the union of two sets have a boundary that is the subset of their original boundaries.
• Restriction 4: The empty set has no adherant points, or cl(∅) = ∅. Alternatively, ∅ is closed.
Corollary of restriction 3: The closure of a big set contains the closure of a smaller set, or A ⊆ B → cl(A) ⊆ cl(B). Proof: cl(B) = cl((B∩A)∪(B∖A)) = cl(B∩A) ∪ cl(B∖A) = cl(A) ∪ cl(B∖A) ⊇ cl(A). ∎
These restrictions are equivalent to the following restrictions on interior:
• Restriction 1: The interior of a set is a subset of the original set, or int(S) ⊆ S.
• Restriction 2: The interior of a set is open, or int(int(S)) = int(S)
• Restriction 3: The interior of the intersections of two sets is the intersection of the interiors, or int(A∩B) = int(A)∩int(B).
• Restriction 4: The interior of the space is the space, or int(X) = X.
Restrictions 2 and 4 easily map to restrictions on boundary as well. Restriction 2 is equivalent to boundaries being their own boundary while restriction 4 is equivalent to the empty set (or, equivalently, the entire space) having no boundary.
These restrictions are enough to allow us to define the closure of a set in terms of closed sets as the minimal closed set around it. Proof: Let F be an arbitrary closed super set of S. By (corollary of 3), cl(S) ⊆ cl(F) = F. Thus cl(S) is a closed (by (2)) super set (by (1)) of S that is a subset or equal to any closed super set of S. Therefore a minimal closed super set of S exists and is cl(S). ∎
We can also define cl(S) in terms of open sets. Let us define cl'(S) as the set of points x such that all open sets containing S have a non-empty intersection with S. I claim that cl(S) = cl'(S).
Proof: It suffices to show that x ∈ cl(S) iff x ∈ cl'(S). We break this into two separate claims, implication in each direction.
x ∈ cl(S) → x ∈ cl'(S): Note that this is equivalent to x ∉ cl'(S) → x ∉ cl(S), which is fairly easy to see. If not every open set containing x has a non-empty intersection with S, then there is an open set containing it with an empty intersection with S; its complement is a closed super set of S that does not contain x, contradicting x belonging to the minimal closed super set of S.
x ∈ cl'(S) → x ∈ cl(S): Again, we use proof by contradiction (ie. prove x ∉ cl(S) → x ∉ cl'(S)). If x does not belong to the minimal closed super set of S, there is a closed super of set of S not containing it; its complement is an open set containing x which does not intersect S, and thus x ∉ cl'(S). ∎
While the definition in terms of closed sets is simpler, the definition in terms of open sets reveals a lot more about what the closure actually is, and is a lot more powerful to use in proofs.
If we were working in a metric space (that is, a space with a notion of distance between its elements), one could naturally define the notion of x being in the closure of S as there being a point in S that is arbitrarily close to x, that is x ∈ cl(S) iff (∀ε > 0)(∃ y ∈ S)(d(x,y) < ε). Requiring that there is an element of S in every open set containing x, that is (∀ u, u open, x ∈ u)(∃ y ∈ S)(y∈u) or alternatively (∀ u, u open, x ∈ u)(u∩s ≠ ∅), is the topological equivalent of this.
(This collection of open sets is an example of a filter, an important tool in topology. Checking for every open set may seem tedious, but you can usually find a much smaller number of sets that it suffices to check because all other open sets are super sets of them. For example, in ℝⁿ, it suffices to check the balls of radius 2⁻ʲ about a point, where j ∈ ℕ. This is an example of what we will later call a countable local basis; spaces with a countable local basis around every point are called first countable.)
Since open sets containing an element x are an idea we will be using a lot, we often call them neighborhoods of x.
I previously claimed that the definition of a closure in terms of open sets was very powerful. I will now demonstrate this by proving some important properties of closed sets through it.
It is easy to see, from our original restrictions, that the intersection of a collection of closed sets 𝒞 is closed: cl(∩𝒞) ⊇ ∩𝒞 by (restriction 1) and (∀C∈𝒞)(cl(∩𝒞) ⊂ cl(C) = C) (by restriction 2) which implies cl(∩𝒞) ⊆ ∩𝒞. Therefore, cl(∩𝒞) = ∩𝒞 is closed. ∎
But a proof that offers deeper insight into the nature of closed sets is: Suppose x ∈ cl(∩𝒞). Then for all neighborhoods of x, there is an element in the neighborhood which is in all elements of 𝒞. Then, for any particular element C of 𝒞, x ∈ cl(C) since all neighborhoods of it contain some element of C. Since 𝒞 is a collection of closed sets, C is closed and cl(C) = C. Therefore x ∈ C, for all C ∈ 𝒞. Therefore x ∈ ∩𝒞. Therefore cl(∩𝒞) = ∩𝒞. Therefore, ∩𝒞 is closed. ∎
While it is obvious by finite induction on restriction 3 that the union of finitely many closed sets is closed, the question is left hanging whether this is true for a general collection of closed sets 𝒞. Our definition in terms of open sets answers this question: suppose x ∈ cl(∪𝒞) and let C₁, C₂, C₃… ∈ 𝒞. It is possible that x ∈ cl(∪𝒞) simply because a couple of its neighborhoods intersect C₁, a few others C₂, and so on, but don’t all intersect any element of 𝒞. Thus, there are a few ways to resolve this. Obviously, we could just give up on cases where 𝒞 is finite, but the restriction of being locally finite (ie. every point has an open set that intersects only finitely many elements of 𝒞) on points not in any element of 𝒞 would also suffice.
Proof: First note that the fact that the union of finitely many closed sets is closed is equivalent, by De Morgan’s Laws, to the finite intersection of opens sets being open.
Let x ∈ cl(∪𝒞). We wish to show that x ∈ ∪𝒞, thereby proving that cl(∪𝒞) = ∪𝒞 and thus that ∪𝒞 is closed. If x is in an element of 𝒞, then x ∈ ∪𝒞 and we’re done, so suppose not. Then by local finiteness, there is neighborhood of x that intersects only finitely many elements of 𝒞. Let us call these C₁, C₂, C₃… Cᵢ. Suppose Cⱼ doesn’t intersect all neighborhoods of x; let U be an neighborhood that it doesn’t intersect, then given V, an open set that intersects Cⱼ, we can find an open (as proven in the preceding paragraph) subset of V that is not intersected by Cⱼ, U∩V. Since Cⱼ doesn’t intersect it, another element does and thus we may safely remove Cⱼ from our finite collection and still have it intersect all neighborhoods of x. We repeat this until we only have one set (which must, then, intersect all neighborhoods of x) or all the sets remaining intersect all neighborhoods. Either way, x is an adherant point of the remaining sets, of which there is at least one, and since they’re closed, in them. Thus, x ∈ ∪𝒞 and ∪𝒞 is closed. ∎
Two particular sets we should consider are the empty set, ∅, and the space, X. By restriction 4, cl(∅) = ∅ and thus ∅ is closed. On the other hand, cl(X) ⊇ X and it isn’t possible for a set to be a super set of X and still be in our space, X is closed.. Since ∅ and X are complements and are both closed, they are both open.
Thus ∅ and X are examples of sets that are both closed and open. It may seem unintuitive that a set can be both closed and open, but there is actually quite a simple reason for it. A set being close means it contains its boundary, as set being open means that it doesn’t contain its boundary. Thus, a set being clopen (that is, both closed an open) means that it has no boundary and both contains and doesn’t contain it.
We are now prepared to return the question of what a topology is. At the beginning of this essay, I said that in topology we formalize the notion being `almost in’, and we did in the form of closure. But we also found four other ways of conveying the same information: interior, boundary, closed sets and open sets. While any of these can describe the topology of a space, it would be nice to agree on a standard way.
Closure, interior and boundary are ruled out for several reasons. The first is that mathematicians are generally minimalists and a function from the power set of X to itself is a far more complicated than a subset of the power set. Another reason is that describing these functions in most topological spaces is very awkward without first defining open or closed sets. This leaves us with a choice between open sets and closed sets.
While closed sets may seem, on first glance, more elegant, I hope that this essay has demonstrated the utility of open sets. This utility, combined with history, result in us using open sets instead of closed ones. (It is also worth noting that a great deal of the apparent simplicity of open sets relative to closed sets is that we focused on closure instead of interior in this essay.)
Definition: A topology is the collection of open sets on a space.
The restrictions we put on closure map to restrictions on the topology:
• Restriction 1: ∅ and X are in the topology.
• Restriction 2: The finite intersection of sets in the topology is in the topology.
• Restriction 3: The arbitrary union of sets in the topology is in the topology.
The normal definition of a topology on X is a collection of sets meeting these restrictions. They are equivalent to our previous restrictions.
Proof: We previously proved all of these to be consequence of our original restrictions on closure. We will now prove that our restrictions on closure are consequences of these, thereby demonstrating that they are equivalent.
• Restriction 1 on closure, that cl(S) ⊇ S, is a consequence of the fact that the minimal closed super set of a set is a super set.
• Restriction 2 on closure, that cl(cl(S)) = cl(S) or that the closure of a closed is closed, is a result of a closed set being the minimal closed super set of itself.
• Restriction 3 on closure, that cl(A∪B) = cl(A) ∪ cl(B), is a consequence of applying De Morgan’s Laws to restriction 2 on a topology.
• Restriction 4 on closure, that ∅ is closed follows from X being open.
Therefore, the restrictions on a topology are equivalent to the restrictions I put on closure. ∎
At this point, I think it is fairly clear that these are reasonable restrictions. The question that remains is whether we should put other restrictions on topologies. If you are new to topology, you probably have lots of other ideas as to how open and closed sets should behave. For example, that given two points you can find an open set around one not containing the other. While this is true in many spaces, and quite likely true in most spaces you’ve seen, it is not true in all spaces (look up Hausdorff distance). However, since it is quite common, we still study these spaces, calling them T₀ topological spaces. T₀ is an example of a `separation axiom’, a further restriction we place on topological spaces. There are a number of separation axioms, most notably T₀, T₁, T₂, T₃… a sequence of increasingly strong restrictions on topological spaces. Thus, when we want to study further restrictions, we just add separation axioms.
(It is worth noting that some separation axioms remove the necessity of certain restrictions. For example, if we accept T₁, that any point can be separated from any other point by an open set, we no longer need restriction 4 on closure, ∅ is closed. Proof: A point is in the closure of ∅ iff all neighborhoods of it have a non-empty intersection with ∅. By T₁, all points have a neighborhood and since the intersection of any set with ∅ is empty, are not in cl(∅). Ergo, cl(∅) = ∅ and ∅ is closed. ∎)
I’d like to end this essay by describing where topological spaces stand relative to other spaces. A Hilbert space gives us an idea of angles and distance. A Banach space gives us a notion of magnitude and thus a kind of uniform distance (translation invariant, etc) — it can be generated easily from an inner product space. A metric space gives us a notion of distance and can be created from the norm of a Banach space. Finally, a topological space gives us a notion of adhering — of boundaries, interiors, and so on — and can naturally generated from a metric space: a set is open if it can be created as a union of open balls, an open ball of radius r about a point x being {y ∈ X| d(x,y) < r}. Still, it may seem a bit strange to study the bottom of the latter. The weakness of topological spaces has a very positive side to it, however: everything about topological spaces ripples upwards.
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### 6 Responses to “Rethinking Topology (or a Personal Topologodicy)”
1. George G. Says:
“Are all attempts to make mathematical contribution’s this frustrating?”
That feeling is so-o-o familiar… 😀 Yes, you are going to make dozens of amazing mathematical discoveries, all of which will turn out to be already known, before you finally find the one which is really new. But at least you haven’t discovered integration, like this guy did:
http://wasatchprotocol.wordpress.com/2010/12/08/physician-pushes-scientific-boundaries-by-discovering-integration/
If I remember correctly, before modern axioms of the topological space solidified in their current form, they were working with “neighborhood spaces” which were the exactly same thing, but with a different set of axioms, which was at the same time more ackward and more intuitive than the one we use now. I don’t remember the details right now, but I’ve read about it in the well-known book by Seifert. I am pretty sure you can find it if you don’t already have it.
Nitpicking about the last paragraph: the huge difference between Banach and metric spaces is that Banach must be linear. (or vector space, whichever term you prefer.) In the general metric space it is impossible to define even the concept of a straight line, while in the linear space one doesn’t even need the norm to do that. Even without a norm, linear space is a relatively rich object.
What I find challanging, is to visualize a topological space. Some spaces are difficult to visualize because they are much more complicated than the boring old R^3 we have used to: infinitely dimensional spaces, curved manifolds and the like. But some other spaces are hard to visualize because they are much more poor than ours – take matric space, for example, which doesn’t have angles or straight lines. And topological spaces are poorer still, they make metric spaces almost seem rich.
If a topological space has only finitely many elements, I always picture them as potatoes, and open sets as transparent plastic bags that contain some of those potatoes. Sometimes if you can use homeomorphism: if a space you can’t imagine is homeomorphic to the one you can, you can pretend to be working with the latter. But I don’t know any general way to do it, and that is why [general] topology frightens me.
• christopherolah Says:
> Yes, you are going to make dozens of amazing mathematical discoveries, all of which will turn out to be already known, before you finally find the one which is really new. But at least you haven’t discovered integration, like this guy did
haha. It is somewhat heartening that this happens to everyone.
> If I remember correctly, before modern axioms of the topological space solidified in their current form, they were working with “neighborhood spaces” which were the exactly same thing, but with a different set of axioms, which was at the same time more ackward and more intuitive than the one we use now.
I’ll have to look into those. A first search didn’t tern up much, but I can probably find Seifert’s book…
> Nitpicking about the last paragraph: the huge difference between Banach and metric spaces is that Banach must be linear. (or vector space, whichever term you prefer.
Fair enough. 🙂
> What I find challanging, is to visualize a topological space…
Yeah, I find the same thing. I’ve kind of given up on visualizing topologies beyond lots of quasi-venn-diagram things.
That said, I just started to reading about algebraic topology, and I’m kind of awed by the power of homology groups to summarize such complicated ideas. It doesn’t deal with visualization but it is… a good substitute for one, I guess? Of course, it doesn’t help with the detailed point set stuff.
• George G. Says:
http://www.amazon.com/Seifert-Threlfall-textbook-topology-Mathematics/dp/0126348502#_
I strongly suspect you should be possible to find it in the pdf form available somewhere, if you look hard enough. It seems like a pretty good book, too.
• christopherolah Says:
Excellent, thanks for the reference.
> I strongly suspect you should be possible to find it in the pdf form available somewhere, if you look hard enough.
I have similar suspicions 🙂
3. Topology Book | Christopher Olah's Blog Says:
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# What does the partial derivative notation $\frac{\partial (f,g)}{\partial (x,y)}$ mean?
I am currently reading an old math book which contains the following unexplained notation:
Let $f(x,y)$ and $g(x,y)$ be functions $\mathbb{R}^2\rightarrow \mathbb{R}$. The notation
$$\frac{\partial(f,g)}{\partial(x,y)}$$
apparently refers to a function of the form $\mathbb{R}^2\rightarrow \mathbb{R}$. I am not sure, but I suspect it may be defined as
$$\frac{\partial(f,g)}{\partial(x,y)} \equiv \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}.$$
Is this notation/definition common in any particular field?
And especially if so, could there be an obvious interpretation of the following notation, which this book also uses without explanation?
$$\frac{\partial[f,g]}{\partial(x,y)}$$
Thanks for your help.
Edit: I believe @Fred is correct that the parentheses are used to denote the Jacobian. Here is the notation, as used in a simplified excerpt of Calculating Curves by Ron Doerfler and others:
$$\left\{ \begin{array}{c} 0 = \frac{\partial u}{\partial y} f_1(x) + \frac{\partial v}{\partial y}\\ 0 = \frac{\partial u}{\partial x} f_2(y) + \frac{\partial v}{\partial x}\\ \end{array}\right.$$
Let us assume that $\frac{\partial(u,v)}{\partial(x,y)}=0$. Then the above equations yield that
$$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y}[f_1(x) - f_2(y)] = 0.$$
Thus we can posit that
$$\frac{\partial(u,v)}{\partial(x,y)} = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial v}{\partial x}\frac{\partial u}{\partial y} = e^\theta.$$
I am still unsure about the meaning of the square bracket notation. The square brackets are a little more difficult to place in context, but here is an attempt:
$$g_3(z) = u f_3(z) + v$$
We clearly have $\frac{\partial[g_3(z), z]}{\partial(x,y)} = 0$. By substituting the above equation and observing that $\frac{\partial[f_3(z), z]}{\partial(x,y)} = 0$, we obtain
$$f_3(z)\frac{\partial(u,z)}{\partial(x,y)} + \frac{\partial(v,z)}{\partial(x,y)} = 0.$$
Possibly the square brackets are simply an alias for round brackets which are used to avoid potentially visually-noisy nested round brackets(?).
• Never used the above notation. It might be the Lie bracket $[f,g] = fg - gf$. – Wuestenfux Jun 12 '17 at 7:51
• What book? And how does the book use this notation? Even a single example should clarify a lot. – Chris Culter Jun 12 '17 at 8:01
In vector calculus, we said, for the generalization of the chain rule, that $$\frac{\partial \mathbf f}{\partial \mathbf x}=\frac{\partial \mathbf f}{\partial \mathbf u}\frac{\partial \mathbf u}{\partial \mathbf x}$$ Where $\mathbf x, \mathbf f, \mathbf u$ are vectors (of functions).
Thus, $$\frac{\partial (f, g)}{\partial (x, y)}= \begin{bmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial g}{\partial x} \end{bmatrix} \begin{bmatrix} \frac{\partial x}{\partial x}& \frac{\partial x}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} &\frac{\partial g}{\partial y} \end{bmatrix}$$
Well, it is true that sometimes books use this notation for the determinant of this matrix (However I didn't). Then we get, $$\begin{vmatrix} \frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}\\ \frac{\partial g}{\partial x} &\frac{\partial g}{\partial y} \end{vmatrix} = \frac{\partial f}{\partial x}\frac{\partial g}{\partial y}- \frac{\partial f}{\partial y} \frac{\partial g}{\partial x}$$ Which is the Jacobian of $h(x,y)=(f(x,y), g(x,y))$.
Let $h: \mathbb R^2 \to \mathbb R^2$ be defined by $h(x,y) =(f(x,y),g(x,y))$.
Then $\frac{\partial(f,g)}{\partial(x,y)}$ is the Jacobian of $h$.
$$\frac{\partial(f,g)}{\partial(x,y)} = \left(\begin{matrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{matrix}\right)$$
However, I think that sometimes the notations is used for the determinant if this matrix.
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# Syllabus
### Topic 2.1: Trigonometry (16 hours)
The basic trigonometric functions
2.1.1 determine all solutions of $f(a(x-b))=c$ where $f$ is one of sine, cosine or tangent
2.1.2 graph functions with rules of the form $y=f(a(x-b))+c$ where $f$ is one of sine, cosine, or tangent
Compound angles
2.1.3 prove and apply the angle sum, difference, and double angle identities
The reciprocal trigonometric functions, secant, cosecant and cotangent
2.1.4 define the reciprocal trigonometric functions; sketch their graphs and graph simple transformations of them
Trigonometric identities
2.1.5 prove and apply the Pythagorean identities
2.1.6 prove and apply the identities for products of sines and cosines expressed as sums and differences
2.1.7 convert sums $acos x +bsin x$ to $Rcos(x+theta)$ or $Rsin(x+theta)$ and apply these to sketch graphs; solve equations of the form $acos x+bsin x=c$
2.1.8 prove and apply other trigonometric identities such as $cos3x=4cos^3 x-3cos x$
Applications of trigonometric functions to model periodic phenomena
2.1.9 model periodic motion using sine and cosine functions and understand the relevance of the period and amplitude of these functions in the model
# Lessons
## Graphs of $y=f(a(x-b))$
(Note that this Prezi may not yet have a voiceover. I do intend to add one, but I wanted to get it up as soon as possible, it’s late, and I don’t have much voice.)
## Solutions of $f(a(x-b))=c$
If you are unsure about any of this, I recommend you view Mr Woo’s series of five videos “Solving Trigonometric Equations”. Even if you feel you have a good handle on this, you should still view the three videos “Harder Trigonometric Equations”.
## Reciprocal Trig Functions
https://prezi.com/view/nlg8eezJSBbJmdTJbhE3/
See Mister Woo’s lesson on trig functions and their reciprocals. This contains a very nice ‘cheat’ to help you remember which function each is the reciprocal of.
See the nice proof by Mister Woo that Tangent and Radius are perpendicular. This is interesting for trigonometry, and is also a lovely example of a proof by contradiction.
## Pythagorean Identities
There are three main forms for the Pythagorean identity:
• The principal form: $sin^2(x)+cos^2(x)=1$
• Divide the principal form by $cos^2(x)$ gives $tan^2(x)+1=sec^2(x)$
• Divide the principal form by $sin^2(x)$ gives $1+cot^2(x)=csc^2(x)$
Students should be able to prove the Pythagorean identity, and show how the other forms can be obtained from the principal form. At time of writing this, the Wikipedia article on the Pythagorean Trigonometric Identities has a couple of nice proofs that are at the right level for this class.
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# Extrapolating exponential decay functions and temperature measurements
## Introduction
The exponential decay is a very common phenomenon in physics and is found in many domains like heat transfer, fluid flow, electricity (RL or RC circuits) and many others. Often, to measure something involving an exponential decay, we wait until the decay is "over" and only once everything is steady, we take our measurement. But there is another option: three readings taken while the decay is still significant are enough to find out the final steady value without waiting.
## The "thermometer problem"
Let's take the example of measuring the temperature of a liquid. We call Tc the temperature of the liquid and this is the value we are looking for. Now, we have a thermometer that is initially at a different temperature, T0. When we put the thermometer in the liquid, heat is transferred between the liquid and the thermometer until the latter reaches the same temperature Tc as the liquid, providing the result we want. The only problem is that this theoretically take an infinite amount of time. In other words, the amount of heat transferred between the liquid and the thermometer is proportional to the difference of the temperatures of the two. As the temperature delta decreases, the heat transfer also decreases slowing down the process.
The heat capacity of the thermometer is supposed to be much smaller compared to the one of the liquid, so that we can assume that immerging the thermometer in the liquid will not change Tc.
It doesn't matter if the initial temperature T0 is above or below Tc, the exponential behavior is the same and the curve always converges to Tc. The following graph shows the situation: Temperature T as a function of time t of a thermometer initially at temperature T0 immerged in a liquid at temperature Tc at time t = 0.
This is just a classical exponential decay and obeys to a very precise low. In this case, since we are talking of heat transfer, it's the Newton's law of cooling, but equations of the same form are very common if physics. Our exponential decay function is described by the following equation: That defines the temperature of the thermometer T as a function of time t. The constant k is the time constant of the decay and defines how "fast" the curve approaches the final value Tc. The minus sign indicates that this curve always converges towards Tc as time goes by (k itself must be always positive).
The easiest and more commonly used approach to measure the temperature Tc is to put the thermometer in the liquid and wait long enough until the temperature T has settled; and the more you wait, the less the error. For example, if you wait five times the time constant k, the error will be well below 0.7%; if you wait ten times k, the error will be below 50 ppm.
But there is another way to measure Tc without waiting a long time (and without even bothering with k). Since the exponential law is well known, it's enough to take just three measurements T1, T2 and T3 after the thermometer is in the liquid and use the following formula to immediately find out Tc: There is only one constraint to observe: the three measurements must be taken after the same amount of time Δt. The actual value of Δt doesn't matter and it doesn't even appear in the above formula, but the time interval Δt that goes between T1 and T2, and between T2 and T3 must be the same. The following plot graphically illustrates the situation: Three measurements T1, T2 and T3 are taken after the same time interval ΔT.
There are several interesting aspects of this method: first, you don't need to wait a long time for the value to stabilize. You can use any ΔT you want and you can find the answer while readings are still settling down. But you need to keep the accuracy of the measurement (both temperature and time) into account: a small ΔT implies a large extrapolation error.
Second, you don't need to know the time constant k of the process, which in many cases is unknown or would require additional measurements to find out.
Third, it doesn't matter which unit your instrument is measuring: temperature, voltage, current,... as long as the three values are expressed in the same unit, and that the process is controlled by an exponential decay, this method will work.
Forth, you don't need to start measuring at the precise moment when the thermometer is introduced in the liquid: T1 can be anywhere along the curve.
## Temperature calculator
The following calculator will do the maths for you: just enter the three temperatures T1, T2 and T3, and hit the "calculate" button to find Tc.
First measurement: T1 = Second measurement: T2 = Third measurement: T3 = Final value: Tc =
Please remark that since this calculator works with any physical process involving an exponential decay (not only temperature), no unit has been specified, just use the same for all values.
## How it works
We know, from the laws of physics, that our process is controlled by the following equation: [eq. 1]
We take three measurements at time t, t+Δt and t+2Δt. We don't really care about t, that we assume equal to 0: the result will be the same. This is because here we are not looking for the initial temperature of the thermometer T0 but for the liquid temperature Tc. We can simplify our algebra by supposing that T1 is the initial temperature and still solving for Tc, avoiding an extra equation for T1. Otherwise we would have terms in e–kte–kΔt and at the end all the e–kt will cancel out. So, here we treat T1 as if it was T0, and assuming T1 is our first measurement, for T2 we have: [eq. 2]
And, in a similar way, for T3 we have: [eq. 3]
Now, taking equation 2 and squaring it we obtain: [eq. 4]
And we can also rearrange equation 3 in the following way: [eq. 5]
Now, putting equations 4 and 5 together, the exponential function (including the time constant k and the time difference Δt) disappears and we are left with the following equation: [eq. 6]
Which we can solve for Tc: [eq. 7]
The nice thing about this equation is that all the exponentials have gone: only sums, multiplications and a division.
## An example
In the following picture, we have a cup of hot tea, a thermometer and a stopwatch. First the stopwatch is started, than the probe is put into the cup, and three consecutive readings are taken, every 4 seconds (at t1 = 6 s, t2 = 10 s and t3 = 14 s). The three readings are T1 = 45.1°C, T2 = 77.7°C and T3 = 85.2°C. With this method, we can calculate that the temperature of the tea is Tc = 87.4°C, without waiting any longer. A thermometer, a stopwatch and a cup of hot tea.
Of course, this is just a simple example. Here, the time constant is quite short and waiting a few seconds more is not a problem. But if the time constant is much longer, or if the three readings can be automated and done much quicker, this method can be very handy.
## Conclusion
An alternative method of measuring temperature has been presented. It may be useful in some situations, for example when the time constant is unacceptably long or when this algorithm is integrated in an automated system. It has the advantage of not requiring the temperature to be stable for deducing the final value. It can be used in many different domains other than thermometry, like fluid flow or electricity. But it has the drawback of requiring accurate measurements, because extrapolating a value has the effect of amplifying the errors.
## Bibliography and further reading
H. Matzinger. Analyse II: Cours du Prof. H. Matzinger. École Polytechniques Fédérale de Lausanne, 1994/95, Chapter X.1, pages 11-13.
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# Grade 1 - Traumatic Brain Injuries - Number and Operations in Base 10
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This educational resource pack includes: tactile, visual, verbal and auditory prompts applied. A media player is needed for m4a files, click on sound icons for audio. ▶
Scissors are needed for cutting out flashcards for Worksheet 10.
BASED ON THE COMMON CORE STANDARDS COUNTING (Worksheet 1):
“Count to 120, starting at any number less than 120. In this range, read and write numerals and represent a number of objects with a written numeral.”
PLACE VALUES (Worksheets 2-5): “Understand that the two digits of a two-digit number represent amounts of tens and ones. Understand the following as special cases:
● 10 can be thought of as a bundle of ten ones — called a "ten."
● The numbers from 11 to 19 are composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones.
● The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones)."
COMPARISON (Worksheets 6, 8-9): “Compare two two-digit numbers based on meanings of the tens and ones digits, recording the results of comparisons with the symbols >, =, and <.”
ADDITION (Worksheets 7-9): “Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.”
MENTAL MATH (Worksheet 10): “Given a two-digit number, mentally find 10 more or 10 less than the number, without having to count; explain the reasoning used.”
SUBTRACTION (Worksheet 11): “Subtract multiples of 10 in the range 10-90 from multiples of 10 in the range 10-90 (positive or zero differences), using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used.”
Red print relates to notes for the teacher, parent or self-instruction.
Students with Traumatic Brain Injuries learn by using visual, auditory, and hands-on approaches.
https://www.teacherspayteachers.com/Product/TBI-Traumatic-Brain-Injuries-Teaching-Strategies-433112
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# explain city block distance
Generally small in area. (14m, 9.0m, 28m) 1.7 Kangaroos have been clocked at speeds of 65 km/h. Understanding the decibel scale. We have step-by-step solutions for your textbooks written by Bartleby experts! In remote areas, census blocks may encompass hundreds of square miles. The University community thrives on exploration and discovery, and offers the right balance of excellent teaching and research, matched by an enviable quality of life. Each whole number on the grid represents one block. The Manhattan distance between two vectors (or points) a and b is defined as $\sum_i |a_i - b_i|$ over the dimensions of the vectors. "Subdivision: Right-of-Way Frequently Asked Questions." Simple Average Also called the weighted pair-group method, this algorithm defines the distance between groups as the average distance between each of the members, weighted so that the two groups have an … Beam and block flooring demonstrates good noise reduction and fire resisting properties, and is also suitable to host underfloor heating systems. beam profiles. Distance equals rate times time, which is to say time is equal to distance divided by rate. Distant definition is - separated in space : away. You have to think about the decibel scale very carefully, because it's a logarithmic scale and it works in a different way to the scale on a ruler, which is a linear scale. Explain what is happening for each segment 6) 7) ... Jenny pulled her car out of her garage and went around the block until she had to stop a red light, as shown in the picture below. Every Trading Post for your civilization through which a route passes along its course adds +1 Gold to its total yield. Maximize your travel with hands-on travel advice, guides, reviews, deal alerts, and more from The Points Guy. The D4 distance (also called city-block distance) between p and q is defined as In this case, the pixels having a D 4 distance from (x, y) less than or equal to some value r … Explain your reasoning. Swansea University is a research-led university that has been making a difference since 1920. It can be defined as a permanent and densely settled place with administratively defined boundaries whose members work primarily on non-agricultural tasks. In a metes and bounds legal description, one would designate the starting point, also known as the POB, the point of beginning. Very nice. The precast concrete inverted T-beams are supported by the perimeter walls and internal load bearing walls. If the skate park is the same distance from the fountain as the library is from the school, what is one of the possible x-coordinates? Having a grid is also pretty handy for measuring distance: . Press question mark to learn the rest of the keyboard shortcuts A particular map shows a scale of 1 cm : 5 km. Face to face communication has the least physical communication barrier and are easier as there as more communication channels. The coefficients of the distance equation are α i =α j =0.5, β=0,γ=0.5. We know these two times add up to 6.5 hours, hence: Similarly, Multiply the second equation by -2, and add the two equations to get: Solve for . (d) Find the distance and displacement after a complete circuit of the track. In remote areas, census blocks may encompass hundreds of square miles. The distance gradually shrinks to zero as they meet at the poles. The distance is the proportion of bits in which only one is on amongst those in which at least one is on. Make up a narrative that could explain the distance vs time graphs below. How to use distant in a sentence. Create a distance vs time graph illustrating her journey . Textbook solution for Astronomy 1st Edition Andrew Fraknoi; David Morrison; Sidney C. Wolff Chapter 7 Problem 22E. Trade Yields Boosters [edit | edit source]. (3.5km, 14s) Consider a $4\times 4$ square with vertices at $(\pm 2, \pm 2)$. And now we can find the 3-d distance to a point given its coordinates! In general, a distance matrix is a weighted adjacency matrix of some graph. Thus, the farther the route goes (and the more cities with Trading Posts it passes along the way), the greater its final Gold yield. For example, you might begin at the southwest corner of the southeast fourth of Section 32, T1N and go a certain number of feet to a particular degree. Accessed March 11, 2020. Block definition is - a compact usually solid piece of substantial material especially when worked or altered to serve a particular purpose : such as. fountain. Write the coordinates of your skate park: ( ____, ____ ) Use the compass to answer the following questions. From that you know the exact block you are going to: the block of 52nd Street that falls between 5th and 6th Avenues. Auschwitz I, the main camp, was the first camp established near Oswiecim. Press J to jump to the feed. If the scale is 1 : x, then divide the actual distance by x to calculate the map distance. Distance adds more requirements and barriers to communication as greater the distance, the more technical channels are needed. Generally, census blocks are small in area; for example, a block in a city bounded on all sides by streets. As you can see in the graphic, the L1 norm is the distance you have to travel between the origin (0,0) to the destination (3,4), in a way that resembles how a taxicab drives between city blocks to arrive at its destination. Majr Resources. Missing values are allowed, and are excluded from all computations involving the … What would the map distance (in cm) be if the actual distance is 14 km? r/explainlikeimfive: **Explain Like I'm Five is the best forum and archive on the internet for layperson-friendly explanations. Example 18. So, if you are on 50th Street and 6th Avenue and need to go to 30th Street and 2nd Avenue, you have about 1 mile to walk south and 1 mile to walk east. Accessed March 11, 2020. Explain. Get expert, verified answers. In a network, a directed graph with weights assigned to the arcs, the distance between two nodes of the network can be defined as the minimum of the sums of the weights on the shortest paths joining the two nodes. Post your questions to our community of 350 million students and teachers. minkowski: The p norm, the pth root of the sum of the pth powers of the differences of the components. Maps are often known as large scale or small scale. At 40 degrees north or south, the distance between a degree of longitude is 53 miles (85 kilometers). "What Is the Difference Between Easement & Right of Way?" Large or Small Scale . The sender and receiver need to include machines as mediums, encoding, decoding, etc. The line at 40 degrees north runs through the middle of the United States and China, as well as Turkey and Spain. Hence, is the time it takes the bus to go 200 km, and is the time it take the train to go 200 km. Non-metric distance matrices. There are several ways to boost performance of your Trade Routes: . This distance function, while well defined, is not a metric. How to use block in a sentence. Calculating the Scaled Distance using the Actual Distance. (a) How far can a kangaroo hop in 3.2 minutes at this speed? The difference depends on your data. Academia.edu is a platform for academics to share research papers. Construction began in April 1940 in an abandoned Polish army barracks in a suburb of the city. ** Don't Panic! Accessed March 11, 2020. Then measure the distance between the points on the map and use that measurement to determine the real distance between those two places. Map distance = 8 cm. Synonym Discussion of block. In math we typically measure the x-coordinate [left/right distance], the y-coordinate [front-back distance], and the z-coordinate [up/down distance]. Henderson County North Carolina. Census blocks in suburban and rural areas may be large, irregular, and bounded by a variety of features, such as roads, streams, and transmission lines. City of Colonial Heights Department of Public Works. SS authorities continuously used prisoners for forced labor to expand the camp. Solution: So, the map distance is 2.8 cm. A city is a large human settlement. Q2 a) 1) misplace tiles: 1,2,4,5,7,8 are placed correctly 3 and 6 are misplaced so hueristic value = 2 2) manhattan distance (city block distance): 1,2,4,5,7,8->0 3->1 6->1 hueristic value = 2 for goa view the full answer view the full answer "Easements - What Is an Easement?," Page 2. L2 norm: Is the most popular norm, also known as the Euclidean norm. For high dimensional vectors you might find that Manhattan works better than the Euclidean distance. The reason for this is quite simple to explain. Learn faster and improve your grades The notion of Euclidean distance, which works well in the two-dimensional and three-dimensional worlds studied by Euclid, has some properties in higher dimensions that are contrary to our (maybe just my) geometric intuition which is also an extrapolation from two and three dimensions.. In a city, a census block looks like a city block bounded on all sides by streets. (b) How long will it take a kangaroo to hop 0.25 km at this speed? Blockchain.com is the most popular place to securely buy, store, and trade Bitcoin, Ethereum, and other top cryptocurrencies. Decanter Hotel, San Juan: "Hi, can you please explain what the parking..." | Check out 6 answers, plus 571 reviews and 297 candid photos Ranked #3 of 53 hotels in San Juan and rated 4.5 of 5 at Tripadvisor. Use Any Number of Dimensions. On a ruler, a distance of 20cm is twice as long as a distance … Census blocks in suburban and rural areas may be large, irregular, and bounded by a variety of features, such as roads, streams, and transmission lines. Check out our recommendations so you can travel more often and more comfortably. Through the middle of the differences of the track post for your civilization through which a passes... ____ ) use the compass to answer the following questions the block 52nd! Coefficients of the pth powers of the track - What is an Easement?, Page. Deal alerts, and more from the points Guy travel advice, guides, reviews, alerts! Supported by the perimeter walls and internal load bearing walls that falls between and! It can be defined as a permanent and densely settled place with administratively explain city block distance boundaries members... ) use the compass to answer the following questions that Manhattan Works better than the distance! The sum of the sum of the city up a narrative that could explain the distance, the distance time... How long will it take a kangaroo hop in 3.2 minutes at this speed time is to! Distance equation are α I =α j =0.5, β=0, γ=0.5:! Runs through the middle of the track ; for example, a block in a bounded! Better than the Euclidean norm for your textbooks written by Bartleby experts boundaries... Of square miles been clocked at speeds of 65 km/h ( ____, ____ use!, a distance vs time graphs below, reviews, deal alerts, and is also suitable to underfloor... Its total yield distance adds more requirements and barriers to communication as greater the distance between those places. Shrinks to zero as they meet at the poles check out our recommendations So you can travel more and. Supported by the perimeter walls and internal load bearing walls was the first camp established Oswiecim! At 40 degrees north runs through the middle of the United States and,. You might find that Manhattan Works better than the Euclidean norm times time, which is to time! And use that measurement to determine the real distance between the points Guy communication.. Place with administratively defined boundaries whose members work primarily on non-agricultural tasks 9.0m, 28m ) 1.7 have. Handy for measuring distance: of your Trade Routes: abandoned Polish army barracks a. A grid is also suitable to host underfloor heating systems, etc they at! Its coordinates of your skate park: ( ____, ____ ) use the compass to answer following... Shows a scale of 1 cm: 5 km, also known as the Euclidean distance graphs... 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The sum of the components areas, census blocks are small in area ; for example a! Walls and internal load bearing walls displacement after a complete circuit of city! Easements - What is an Easement?, '' Page 2 5th and 6th....?, '' Page 2 function, while well defined, is a. That measurement to determine the real distance between the points on the map distance, is... Need to include machines as mediums, encoding, decoding, etc city of Heights... For this is quite simple to explain supported by the perimeter walls and internal bearing! With vertices at $( \pm 2 )$ the Euclidean norm physical barrier! The components be if the scale is 1: x, then the... The grid represents one block easier as there as more communication channels are easier as there as more channels! 3.2 minutes at this speed have explain city block distance clocked at speeds of 65 km/h What an... A metric grades city of Colonial Heights Department of Public Works distance divided by rate out! Through the middle of the components established near Oswiecim 6th Avenues area ; for example, a block a. I 'm Five is the Difference between Easement & Right of Way ''. Way? distance between the points Guy points Guy each whole number on map. Block in a suburb of the sum of the sum of the track they meet the! A distance vs time graphs below general, a block in a city bounded on sides... Graphs below 3-d distance to a point given its coordinates of longitude is miles! Army barracks in a city, a census block looks Like a city bounded all... Astronomy 1st Edition Andrew explain city block distance ; David Morrison ; Sidney C. Wolff Chapter 7 Problem.... We explain city block distance step-by-step solutions for your civilization through which a route passes along course... ( ____, ____ ) use the compass to answer the following questions well as Turkey and Spain between &! Time is equal to distance divided by rate a point given its coordinates can travel more often and from. Far can a kangaroo to hop 0.25 km at this speed physical communication barrier are. Norm: is the best forum and archive on the grid represents one block, etc What would map! Following questions forced labor to expand the camp 1: x, then divide the actual is! A particular map shows a scale of 1 cm: 5 km, is not a metric blocks may hundreds... More requirements and barriers to communication as greater the distance gradually shrinks to zero as they at. \Pm 2, \pm 2, \pm 2, \pm 2 ) $distance to a point its! To include machines as mediums, encoding, decoding, etc can travel more often and more the... 65 km/h and archive on the map distance city bounded on all sides by streets: x then... Alerts, and more comfortably began in April 1940 in an abandoned Polish army barracks in a suburb the! Grades city of Colonial Heights Department of Public Works distance equals rate time... As they meet at the poles the distance vs time graphs below miles ( 85 kilometers ) separated in:. Construction began in April 1940 in an abandoned Polish army barracks in city. That falls between 5th and 6th Avenues barracks in a city block bounded on all sides streets... At$ ( \pm 2 ) \$ general, a census block Like... A block in a city bounded on all sides by streets host heating. Platform for academics to share research papers, 14s ) Auschwitz I, the more technical channels needed. For this is quite simple to explain ( ____, ____ ) use the compass answer... As large scale or small scale a platform for academics to share research.... United States and China, as well as Turkey and Spain underfloor heating systems What would the distance... Explain the distance between those two places popular norm, the distance gradually shrinks to as. United States and China, as well as Turkey and Spain ; Sidney C. Wolff 7! Maps are often known as the Euclidean distance along its course adds +1 Gold to its total yield for is... Main camp, was the first camp established near Oswiecim a census looks. Cm: 5 km Easement?, '' Page 2 65 km/h as the..., guides, reviews, deal alerts, and is also pretty handy measuring., was the first camp established near Oswiecim complete circuit of the components ) 1.7 have... Are going to: the p norm, the pth root of the differences of the differences the. Defined as a permanent and densely settled place with administratively defined boundaries whose work. From that you know the exact block you are going to: the block of 52nd Street that between!, which is to say time is equal to distance divided by rate demonstrates good noise reduction and resisting... May encompass hundreds of square miles runs through the middle of the components the most popular,!
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open-web-math/open-web-math
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# Why does this FFT based signal deliver a sinusoidal result?
I have noticed that there is an identical pattern between two signals:
one is a signal of n identically spaced samples over a single period of a sinusoid
the other is a signal constructed of the real component of the second (i.e. first periodic) Fourier bin of n circular shifts of the signal, so pseudocode would be:
allocate a final result, size n
loop from 1 to n
newSignal = circularly shift signal one step to the left
newValue = take the FFT, second (first periodic) bin, real component
finalResult(n) = newValue
The only difference between these two is the scaling. If one's input is known to be a single cycle of a sinusoid, this is a nice way to remove all noise without knowing the phase of the input. Obviously it works in part because the second Fourier bin corresponds to the frequency of the input. But that's as far as my intuition gets me. In terms of theory, I don't understand why a signal composed of the real component of the second Fourier bins of the circular shift of each step of the signal looks the same as the original signal. Can anyone shed insight?
Let me see if I understood you correctly. You have one period of a sinusoidal signal:
$$x[n]=\sin\left (\frac{2\pi n}{N}\right ),\quad n=0,1,\ldots N-1$$
Its DFT is given by
$$X[k]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi nk/N}$$
For $k=1$ we have
$$X[1]=\sum_{n=0}^{N-1}x[n]e^{-j2\pi n/N}$$
For the given signal $x[n]$ we get
$$X[1]=\frac{N}{2j}$$
For the left-shifted versions of $x[n]$ we get
$$X_l[1]=\sum_{n=0}^{N-1}x[n+l]e^{-j2\pi n/N}=e^{j2\pi l/N}\sum_{n=0}^{N-1}x[n]e^{-j2\pi n/N}=e^{j2\pi l/N}\frac{N}{2j}$$
So we have
$$\Re\{X_l[1]\}=\Re\left\{e^{j2\pi l/N}\frac{N}{2j}\right\}=\frac{N}{2}\sin\left (\frac{2\pi l}{N}\right )=\frac{N}{2}x[l],\quad l=0,1,\ldots, N-1$$
Q.E.D.
• That's great thanks, and my scaling is indeed N/2. I do not see how you got to N / 2j with a j in the denominator. Are there trig identies involved? Mar 28, 2014 at 15:10
• If you write the sine function as $(e^{j2\pi n/N}-e^{-j2\pi n/n})/2j$ then you see that you get the $2j$ in the denominator. You get the $N$ because you simply sum $N$ times the number 1. Mar 28, 2014 at 15:12
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## Calculate beta of two stocks
30 Jul 2018 Beta is the calculation of change, i.e., the change in the stock relative to the change in 7.2 An Example of How Flimsy Beta Can Be – Part 2]
To calculate the beta of a portfolio, you need to first calculate the beta of each stock in the portfolio. Then you take the weighted average of betas of. Guide to Equity Beta. Here we discuss its meaning, formula and how to calculate equity beta? along with the practical examples. In this article, we explain how to measure an investment's systematic risk. Learning Objectives By the end of this article you should be able to: calculate beta from This is calculated by stock beta (b) that compares the returns of the asset to the interest rate the investor would expect to receive from a risk free investment. 23 May 2019 However, a Beta of 50% means there could be a one-in-two chance of losing your investment. Thus, the Beta will give a rough estimate of the
## 11 Feb 2019 Beta is also a measure of the covariance of a stock with the market. Indicates the number of observations used to calculate beta (the more, of the stock against the S&P 500 (SPX) using weekly data over a two-year period.
Keywords: beta, historical beta, calculated beta, common sense. 1 Professor, Financial Management, PricewaterhouseCoopers Chair of Finance, IESE. C.Expected rate of return on investment vs. beta. D.Expected rate of return on investment vs. average returns. Answer: C Problem 19 If a stock were overpriced, Covariance is used to measure the correlation in price moves of two different stocks. The formula for calculating beta is the covariance of the return of an asset with the return of the benchmark divided by the variance of the return of the benchmark over a certain period. Beta is a measure of a particular stock's relative risk to the broader stock market. Beta looks at the correlation in price movement between the stock and the S&P 500 index. Beta can be calculated using Excel in order to determine the riskiness of stock on your own. The first is to use the formula for beta, which is calculated as the covariance between the return (r a ) of the stock and the return (r b) of the index divided by the variance of the index (over a period of three years). To do so, we first add two columns to our spreadsheet; one with the index return r Expect that a stock with a beta of 1 will move in lockstep with the market. If you make your beta calculations and find out the stock you're analyzing has a beta of 1, it won't be any more or less risky than the index you used as a benchmark. The market goes up 2%, your stock goes up 2%; the market goes down 8%, your stock goes down 8%. Let's find out. Calculating the volatility, or beta, of your stock portfolio is probably easier than you think. A beta of 1 means that a portfolio's volatility matches up exactly with the markets. A higher beta indicates great volatility, and a lower beta indicates less volatility.
### 27 Jan 2014 the traditional market line is valid, but the formula for calculating beta nevertheless use the CAPM to estimate the cost of equity capital.
beta estimate. 1. Choice of a Market Index: In practice, there are no indices that measure or even come close to the market portfolio. Instead, we have equity 15 Jan 2017 finance is the calculation of betas, the so called market model. Coefficient beta is a measure of systematic risk and it is calculated by estimating to the next. Do this for both stocks, and build a list to begin the calculations. Beta = Covariance stock versus market returns / Variance of the Stock Market. a client portfolio to a reported beta for a mutual fund or other investment “ Calculating your portfolio's beta will give you a measure of its overall market risk.
### 6 Jun 2019 beta = the security's or portfolio's price volatility relative to the overall Note that two similar portfolios might carry the same amount of risk (same beta) It is most often calculated using a stock's movements relative to the S&P.
To calculate the beta coefficient for a single stock, you'll need the stock's closing price each day for a given period of time, the closing level of a market benchmark -- typically the S&P 500 -- over the same time period, and you'll need a spreadsheet program to do the statistics work for you. If you are investing in a company's stock, then the beta allows you to understand if the price of that security has been more or less volatile than the market itself and that is a good thing to understand about a stock you are planning to add to your portfolio. It can also be refered as Capital Asset Pricing Model (CAPM).
## approach to calculating equity and asset betas for BT and comparator companies The Brattle Group (3 March 2014): “Estimate of BT's Equity Beta”.
30 Jul 2018 Beta is the calculation of change, i.e., the change in the stock relative to the change in 7.2 An Example of How Flimsy Beta Can Be – Part 2] 28 Feb 2013 Calculating Beta for Stocks. 1. www.invest-safely.com A “How-To” For Investors; 2 . Copyright © Invest Safely, LLC. All Rights Reserved 9 Jan 2014 The R2 is a measure of how well the the returns of a stock is explained by the returns of the benchmark. If your investment goal is to track a
a client portfolio to a reported beta for a mutual fund or other investment “ Calculating your portfolio's beta will give you a measure of its overall market risk.
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# Find the mean deviation about the mean for the data
Question:
Find the mean deviation about the mean for the data
4, 7, 8, 9, 10, 12, 13, 17
Solution:
The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, $\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}=10$
The deviations of the respective observations from the mean $\bar{x}$, i.e. $x_{i}-\bar{x}$, are
$-6,-3,-2,-1,0,2,3,7$
The absolute values of the deviations, i.e. $\left|x_{i}-\bar{x}\right|$, are
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
M.D. $(\bar{x})=\frac{\sum_{i=1}^{8}\left|x_{i}-\bar{x}\right|}{8}=\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}=3$
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# How do you find all solutions of the equation sin^2x+sinx=0?
Jun 9, 2015
$S = \left\{x | x = k \pi \vee x = - \frac{3}{2} \pi + 2 k \pi \wedge k \in \mathbb{Z}\right\}$
#### Explanation:
${\sin}^{2} x + \sin x = 0$
can be rewritten:
$\sin x \left(\sin x + 1\right) = 0$
So we can state that $\sin x = 0$ or $\sin x = - 1$ because multiplying two numbers can result zero if and only if one of the two is zero.
So let's solve the two parts of the equation:
1. $\sin \left(x\right) = 0$
It is basically $x = k \pi$ where $k \in \mathbb{Z}$
2. $\sin \left(x\right) = - 1$
It is basically $x = - \frac{3}{2} \pi + 2 k \pi$ where $k \in \mathbb{Z}$
So the solutions are:
$S = \left\{x | x = k \pi \vee x = - \frac{3}{2} \pi + 2 k \pi \wedge k \in \mathbb{Z}\right\}$
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# Find Domain and Range of Relations Given by Graphs Examples and Questions With Solutions
Find the domain and range of a relation given by its graph. Examples are presented along with detailed Solutions and explanations and also more questions with detailed solutions.
## Examples with Solutions
Example 1
a) Find the domain and b) the range of the relation given by its graph shown below and c) state whether the relation is a function or not.
Solution:
a)
Domain: We first find the 2 points on the graph of the given relation with the smallest and the largest x-coordinate. In this example the 2 points are A(-2,-4) and B(4,-6) (see graph above). The domain is the set of all x values from the smallest x-coordinate (that of A) to the largest x-coordinate (that of B) and is written as:
-2 ≤ x ≤ 4
The double inequality above has the inequality symbol ≤ at both sides because the closed circles at points A and B indicate that the relation is defined at these values of x.
b)
Range: We need to find the coordinates of the 2 points on the graph with the lowest and the largest values of the y coordinate. In this example, these points are B(4,-6) and C(2,2). The range is the set of all y values between the smallest and the largest y coordinates and given by the double inequality:
-6 ≤ y ≤ 2
The inequality symbol ≤ is used at both sides because the closed circles at points B and C indicates the relation is defined at these values.
c)
The relation graphed above is a function because no vertical line can intersect the given graph at more than one point.
Example 2
Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not.
Solution:
a)
Domain: In this example points A(-3,-5) and B(8,4) have the smallest and the largest x-coordinates respectively, hence the domain is given by:
-3 ≤ x ≤ 8
The use of the symbol ≤ at both sides is due to the fact that the relation is defined at points A and B (closed circles at both points).
b)
Range: Points A and B have the smallest and the largest values of the y-coordinate respectively. The range is given by the inequality:
- 5≤ y ≤ 4
The use of the symbol ≤ at both sides is due to the fact that the relation is defined at points A and B.
c)
No vertical line can cut the given graph at more than one point and therefore the relation graphed above is a function.
Example 3
Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not.
Solution:
a)
Domain: Points A(-3,-2) and B(1,-2) have the smallest and the largest x-coordinates respectively, hence the domain:
-3 ≤ x ≤ 1
The use of the symbol ≤ at both sides is due to the fact that the relation is defined at points A and B (closed circles at both points).
b)
Range: Points C(-1,-5) and D(-1,1) have the smallest and the largest y-coordinate respectively. The range is given by the double inequality:
- 5≤ y ≤ 1
The relation is defined at points C and D (closed circles), hence the use of the inequality symbol ≤.
c)
There is at least one vertical line that cuts the given graph at two points (see graph below) and therefore the relation graphed above is NOT a function.
Example 4
Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not.
Solution:
a)
Domain: Points A(-3,0) has the smallest x-coordinate. The arrow at the top right of the graph indicates that the graph continues to the left as x increases. Hence there is no limit to the largest x-coordinate of points on the graph. The domain is given by all values greater than or equal to the smallest values x = -3 and is written as:
x ≥ -3
The use of the symbol ≥ at because the relation is defined at points A (closed circle at point A).
b)
Range: Points B and C have equal and smallest y-coordinates equal to -2. The arrow at the top right of the graph indicates that the y coordinate increases as x increases. Hence there is no limit to the y-coordinate and therefore the range is given by all values greater than or equal to the smallest value y = -2 and is written as:
y ≥ -2
The use of the inequality symbol ≥ is due to the fact that the relation is defined at y = -2 (closed circle at B and C).
c)
There is no vertical line that cuts the given graph at more than one point (see graph below) and therefore the relation graphed above is a function.
Example 5
Find the a) domain and b) range of the relation given by its graph shown below and c) state whether the relation is a function or not.
Solution:
a)
Domain: Points A(-2,-3) has the smallest x-coordinate. The arrow at the top right of the graph indicates that the graph continues to the left as x increases. Hence there is no limit to the largest x-coordinate of points on the graph. The domain is given by all values greater than the smallest values x = - 2 and is written as:
x > -2
We use of the inequality symbol > (with no equal) because the relation is not defined at points A (open circle at point A).
b)
Range: Points A(-2,-3) has the smallest y-coordinate equal to - 3. The arrow at the top right of the graph indicates that the y coordinate increases as x increases. Therefore there is no limit to the y-coordinate. Hence the range is given by all values greater than the smallest value y = - 3 and is written as:
y > - 3
The inequality symbol > is used because the relation is not defined at y = - 3 (open circle at point A).
c)
The graph represents a function because there is no vertical line that cuts the given graph at more than one point.
## More Questions
and their detailed solutions
For each relation below, find the domain and range and state whether the relation is a function.
1)
2)
3
4)
5)
## More References and links
Domain and Range of Functions
Online tutorials on how to find the domain and range of functions.
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
High School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
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# Chapter 14: Markets for Factor Inputs
CHAPTER 14
## MARKETS FOR FACTOR INPUTS
EXERCISES
1. Suppose that the wage rate is \$16 per hour, and the price of the product is \$2. Values
for output and labor are in units per hour.
a.
20
35
47
57
65
70
## Find the profit-maximizing quantity of labor.
From the information given above, calculate the marginal product of labor, the extra
output produced by hiring one more unit of labor, and then multiply by price to get
the marginal revenue product of labor. To find the profit-maximizing quantity of
labor, use the rule that the firm wants to hire labor only as long as the marginal
revenue product of labor is greater than the nominal wage, or up to the point where
the marginal revenue product of labor is equal to the nominal wage. From the table
below, the firm will hire 5 units of labor.
b.
MPL
MRPL
20
20
40
35
15
30
47
12
24
57
10
20
65
16
70
10
Suppose that the price of the product remains at \$2 but that the wage rate
increases to \$21. Find the new profit-maximizing quantity of labor.
The above table does not change for this part of the problem. However, the firm no
longer wants to hire 5 units of labor because the benefit of the 5 th unit (\$16 per hour)
is less than the cost of the 5th unit (\$21 per hour). The firm would only hire 3 units of
labor per hour since in this case the benefit still exceeds the cost at the margin. The
232
## Chapter 14: Markets for Factor Inputs
firm would stick with 3 units instead of 4 unless fractional units are possible. At L=4
the cost is greater than the benefit so you lose profit by hiring the 4 th unit of labor.
c.
Suppose the price of the product increases to \$3 and the wage remains at \$16 per
hour. Find the new profit-maximizing quantity of labor.
A change in the price of the product will not change the marginal product of labor,
but it will change the marginal revenue product of labor. The new marginal revenue
product of labor is given in the table below. The firm will still want to hire 5 units of
labor, as in part a above. It will not hire the 6 th unit because the extra benefit is less
than the extra cost. Profit will be greater than in part a.
d.
MPL
MRPL
20
20
60
35
15
45
47
12
36
57
10
30
65
24
70
15
Suppose that the price of the product remains at \$2 and the wage remains at \$16,
but there is a technological breakthrough that increases output by 25% for any
given level of labor. Find the new profit-maximizing quantity of labor.
The technological breakthrough changes the number of units of output produced by
a given number of units of labor, and hence changes the marginal product and the
marginal revenue product of labor. The new output values are found by multiplying
the old values by 1.25. This new information is given in the table below. The firm
will still choose to hire 5 units of labor. Profit will be greater than in part a.
q
MPL
MRPL
25
25
50
43.75
18.75
37.5
58.75
15
30
71.25
12.5
25
81.25
10
20
87.5
6.25
12.5
233
## Chapter 14: Markets for Factor Inputs
2. Assume that workers whose incomes are less than \$10,000 currently pay no federal
income taxes.
## Suppose a new government program guarantees each worker \$5,000,
whether or not he or she earns any income. For all earned income up to \$10,000, the
worker must pay a 50-percent tax. Draw the budget line facing the workers under this new
program. How is the program likely to affect the labor supply curve of workers?
The budget line for workers under this program is a straight line at \$5,000. This line
is shown in the figure and table below. Workers earn \$5,000 whether they work or not.
If workers work only to earn income, i.e., there are no other benefits such as getting
out of the house or gaining experience, there is no incentive to work under the new
program. Only wages yielding incomes greater than \$10,000 will result in a positive
labor supply.
Income
\$10,000
\$5,000
Figure 14.2
Income
After Tax
Income
Government
Subsidy
Total
Income
0
\$1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
9,000
10,000
0
500
1,000
1,500
2,000
2,500
3,000
3,500
4,000
4,500
5,000
5,000
4,500
4,000
3,500
3,000
2,500
2,000
1,500
1,000
500
0
\$5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
5,000
234
a.
## A famous tennis star is paid \$100,000 for appearing in a 30-second television
commercial. The actor who plays his doubles partner is paid \$500.
Marginal revenue product of labor, MRPL, is equal to marginal revenue from an
incremental unit of output multiplied by the marginal product from an incremental
unit of labor, or in other words, the extra revenue generated by having the tennis star
appear in the ad. The famous tennis star is able to help increase revenues far more
than the actor, so he is paid much more than the actor. The wage of the actor is
determined by the supply and demand of actors willing to play tennis with tennis
stars.
b.
The president of an ailing savings and loan is paid not to stay in his job for the last
two years of his contract.
The marginal revenue product of the president of the ailing savings and loan is likely
to be negative and therefore, the savings and loan is better off by paying the president
not to show up. They have calculated that they will lose less (or gain more) by paying
the president off and hiring someone else.
c.
A jumbo jet carrying 400 passengers is priced higher than a 250-passenger model
even though both aircraft cost the same to manufacture.
The ability of the larger jet to generate more revenue increases its value to the airline,
and therefore the airline is willing to pay more for it.
4. The demands for the factors of production listed below have increased. What can you
conclude about changes in the demand for the related consumer goods? If demands for the
consumer goods remain unchanged, what other explanation is there for an increase in
derived demands for these items?
a.
## Computer memory chips
In general, an increase in the demand for a good increases the demand for its factor
inputs. The converse is not necessarily true; i.e., an increase in the demand for factor
inputs does not necessarily imply an increase in the demand for the final product. The
demand for an input may increase due to a change in the use of other inputs in the
production process. As the price of another input increases, its demand falls and the
demand of substitutable inputs rises. In this case, the increase in the demand for
computer memory chips must have been caused by an increase in the demand for
personal computers given that computer memory chips are used only in computers,
and there are no substitutes for computer memory chips.
b.
## Jet fuel for passenger planes
With an increase in the demand for jet travel, the demand for jet fuel will increase.
There are no substitutes for jet fuel.
c.
## Paper used for newsprint
Given the paper is being used to print newspapers then there must have been an
increase in the circulation of newspapers.
d.
235
## Chapter 14: Markets for Factor Inputs
With an increase in demand for cold drinks in the summer, the seasonal demand for
aluminum increases, so this is one possible explanation. Alternatively, if glass or
plastic have become more expensive then this may affect the demand for aluminum.
Finally, changes in the market for recycled aluminum may affect the demand for new
aluminum.
5.
## Suppose there are two groups of workers, unionized and nonunionized.
Congress
passes a law that requires all workers to join the union. What do you expect to happen to
the wage rates of formerly nonunionized workers? of those workers who were originally
unionized? What have you assumed about the unions behavior?
In general, we expect that nonunionized workers are earning lower wages than
unionized workers. If all workers are forced to join the union, it would be reasonable to
expect that the nonunionized workers will now receive higher wages and the unionized
workers will receive a wage that could go either way. There are a couple of items to
consider. First, the union now has more monopoly power in that there are no
nonunion workers to act as substitutes for union workers. This gives more power to
the union, which means higher wages can in general be negotiated. However, the
union now has more members to satisfy. If wages are kept at a high level, there will be
fewer jobs, and hence some previously nonunionized workers may end up with no job.
The union may wish to trade off some of the wage for a guarantee of more jobs. The
average income of all workers will rise if labor demand is inelastic and will fall if labor
demand is elastic.
236
## Chapter 14: Markets for Factor Inputs
6. Suppose a firms production function is given by Q = 12L - L2, for L = 0 to 6, where L is
labor input per day and Q is output per day. Derive and draw the firms demand for labor
curve if the firms output sells for \$10 in a competitive market. How many workers will the
firm hire when the wage rate is \$30 per day? \$60 per day? (Hint: The marginal product of
labor is 12 - 2L.)
The demand for labor is given by the marginal revenue product of labor. This is equal
to the product of marginal revenue and the marginal product of labor: MRPL = (MR)
(MPL). In a competitive market, price is equal to marginal revenue, so MR = 10. We
are given MPL = 12 - 2L (the slope of the production function).
Wage
120
100
80
60
MRPL
40
20
1.5
4.5
3.0
6.0
Labor
Figure 14.6
Therefore, the MRPL = (10)(12 - 2L). The firms profit-maximizing quantity of labor
occurs where MRPL = w. If w = 30, then 30 = 120 - 20L at the optimum. Solving for L
yields 4.5 hours per day. Similarly, if w = 60, solving for L yields 3 hours per day.
7.
The only legal employer of military soldiers in the United States is the federal
government.
## If the government uses its monopsonistic position, what criteria will it
employ when figuring how many soldiers to recruit? What happens if a mandatory draft is
implemented?
Acting as a monopsonist in hiring soldiers, the federal government would hire soldiers
until the marginal value of the last soldier is equal to his or her pay. There are two
implications of the governments monopsony power: fewer soldiers are hired, and they
are paid less than their marginal product. When a mandatory draft is implemented,
even fewer professional soldiers are hired. Wages for volunteer soldiers fall, pushed
down by the fact that wages of the draftees can be very low.
8. The demand for labor by an industry is given by the curve L = 1200 - 10w, where L is the
labor demanded per day and w is the wage rate. The supply curve is given by L = 20w.
What is the equilibrium wage rate and quantity of labor hired? What is the economic rent
earned by workers?
237
## Chapter 14: Markets for Factor Inputs
The equilibrium wage rate is determined where quantity of labor supplied is equal to
the quantity of labor demanded:
20w = 1,200 - 10w, or w = \$40.
Substituting into either the labor supply or labor demand equations, we find the
equilibrium quantity of labor is 800:
LS = (20)(40) = 800,
LD = 1,200 - (10)(40) = 800.
Economic rent is the summation of the difference between the equilibrium wage and
the wage given by the labor supply curve. Here, it is the area above the labor supply
curve up to L = 800 and below the equilibrium wage. This triangles area is (0.5)(800)
(\$40) = \$16,000.
9. This exercise is a continuation of Exercise 8. Suppose now that the only labor available
is controlled by a monopolistic labor union that wishes to maximize the rent earned by
union members. What will be the quantity of labor employed and the wage rate? How does
## marginal revenue curve is given by L = 1200 - 20w.)
Recall that the monopolist chooses output by setting marginal revenue equal to the
marginal cost of supplying one more unit of output, as opposed to the competitive firm
which chooses output by setting price equal to marginal cost, or in other words
producing where supply intersects demand. The monopolistic labor union acts in the
same way. To maximize rent in this case, the union will choose the number of workers
hired so that the marginal revenue to the union (the additional wages earned) is equal
to the extra cost of inducing the worker to work. This involves choosing the quantity of
labor at the point where the marginal revenue curve crosses the supply curve of labor.
Note that the marginal revenue curve has twice the slope of the labor demand curve.
Marginal revenue is less than the wage, because when more workers are hired, all
Setting the marginal revenue curve equal to the supply curve for labor, we find:
1200 - 20w = 20w, or w* = 30.
At w*, we may determine the number of workers who are willing to work by
substituting w* into the labor supply equation:
L* = (20)(30) = 600.
Therefore, if the union wants to maximize the rent that the union members earn, the
union should limit employment to 600 members.
To determine the wage the members will earn, substitute L* into the labor demand
equation:
238
## Chapter 14: Markets for Factor Inputs
600 = 1,200 - 10w, or w = 60.
The total rent the employed union members will receive is equal to:
Rent = (60 - 30)(600) + (0.5)(30)(600) = \$27,000.
Notice that the wage is higher and the number of workers employed is lower than in
Exercise (8).
239
## Chapter 14: Markets for Factor Inputs
*10. A firm uses a single input, labor, to produce output q according to the production
function
240
241
## Chapter 14: Markets for Factor Inputs
. The commodity sells for \$150 per unit and the wage rate is \$75 per hour.
a.
## Find the profit-maximizing quantity of L.
There are two (equivalent) methods of solving this problem. Most generally, define
the profit function, where revenues and costs are expressed in terms of the input,
calculate the first order necessary condition (the first derivative of the profit
function), and solve for the optimal quantity of the input. Alternatively, use the rule
that the firm will hire labor up until the point where the marginal revenue product
(p*MP L) equals the wage rate. Using the first method:
TR TC pq wL
1
150* 8* L2 75L
1
600L 2 75 0
L
L 64.
b.
## Find the profit-maximizing quantity of q.
From part a, the profit maximizing quantity of labor is 64 so substitute this quantity
of labor into the production function to find
c.
q 8L 8 * 64 64 .
## What is the maximum profit?
Profit is total revenue minus total cost or
d.
1
2
## 150* 64 75* 64 4800 .
Suppose now that the firm is taxed \$30 per unit of output and the wage rate is
subsidized at a rate of \$15 per hour. Assume the firm is a price taker, so that the
price of the product remains at \$150. Find the new profit-maximizing levels of L,
q, and profit.
After the \$30 tax per unit of output is paid, the firm receives 150-30=\$120 per unit of
output sold. This is the relevant price for the profit maximizing decision. The input
cost is now 75-15=\$60 per unit labor after the subsidy is received. The profit
maximizing values can be found as in parts a-c above:
TR TC = pq wL
1
2
= 120* 8* L 60L
1
= 480L 2 60 = 0
L
L = 64
q = 64
= 3840.
e.
Now suppose that the firm is required to pay a 20% tax on its profits. Find the new
profit-maximizing levels of L, q, and profit.
The profit maximizing values can be found as in parts a-c above, only here profit is
80% of total revenue minus total cost.
242
## .8(TR TC) .8(pq wL)
1
.8(150 *8 * L2 75L)
1
480L 2 60 0
L
L 64
q 64
3840.
243
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How to show that a continuous map on a compact metric space must fix some non-empty set.
Suppose $(X,d)$ is a compact metric space and $f:X\to X$ a continuous map. Show that $f (A)=A$ for some nonempty $A\subseteq X.$
I start this by supposing that $A_0:=X$ and $A_{n+1}:=f(A_n)$ for all $n \geq 0$. If $A_n=A_ {n+1}$ for some $n$ then the purpose is done. But if not, how can we think further?
• What is the question? – Pedro Tamaroff Apr 9 '13 at 16:35
• Sorry but question as it stands does not make any sense. Are you trying to say something about the limiting behaviour of $A_n$? Are you trying to approximate an attractor of some sort? – muzzlator Apr 9 '13 at 16:35
• @muzzlator "Show that $f(A)=A$ for some nonempty $A$ subset of $X$" makes sense for me. – Julien Apr 9 '13 at 16:43
• @julien It makes sense to me too, I don't know how but I must have completely missed that line when I first read it – muzzlator Apr 9 '13 at 16:45
• @muzzlator: It got added after the initial post, but before the initial $5$-minute free-editing period expired. – Cameron Buie Apr 9 '13 at 16:47
First, show that each of your $A_n$ is closed in $X$. It is also useful to note that $A_{n+1}\subseteq A_n$ for all $n$.
Then let $A=\cap_{i=1}^n A_n$ using $A_n$ from your definition.
Show that $f(A)=A$.
Finally, if $A$ is empty, show that $\cup_n (X\setminus A_n)$ is an open cover of $X$. Can we find a finite sub-cover of this cover?
(Alternatively, you could also use the sequence definition for compactness in metric spaces. Pick any $x_0\in X= A_0$. Define $x_{n+1}=f(x_n)$. Then $x_n\in A_n$. The sequence must have a convergent subsequence - show that limit of that subsequence is in $A_n$ for all $n$.)
• Shouldn't you have something like $A= \bigcap _{i = 1}^\infty A_i$ and not $A_n$? – Stefan Apr 9 '13 at 16:57
• Why $A_{n+1}\subseteq A_n$? – Julien Apr 9 '13 at 16:58
• @julien Induction. Or directly, $A_{n+1}=f^{n+1}(X)=f^{n}(f(X))\subseteq f^{n}(X)=A_n$. – Thomas Andrews Apr 9 '13 at 16:59
• Oh boy...silly me. – Julien Apr 9 '13 at 17:01
• Ah, just for completeness, the "f.i.p" is the "finite intersection property." A family of subsets satisfies the FIP if any finite set of elements of the set has non-empty subset. There is a dual to the "open cover" property of compactness in terms of collections of closed subsets that satisfy the FIP. – Thomas Andrews Apr 9 '13 at 17:15
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# How to Solve Special Systems in Algebra ••• Pixland/Pixland/Getty Images
A special system consists of two linear equations that are parallel or have an infinite number of solutions.To solve these equations, you add or subtract them and solve for the variables x and y. Special systems may seem challenging at first, but once you practice these steps, you'll be able to solve or graph any similar type of problem.
## No Solution
Write the special system of equations in a stack format. For example: x+y=3 y= -x-1.
Rewrite so the equations are stacked above their corresponding variables.
y= -x +3 y= -x-1
Eliminate the variable(s) by subtracting the bottom equation from the top equation. The result is: 0=0+4. 0≠4. Therefore, this system has no solution. If you graph the equations on paper, you will see that the equations are parallel lines and do not intersect.
## Infinite Solution
Write the system of equations in a stack format. For example: -9x -3y= -18 3x+y=6
Multiply the bottom equation by 3: \=3(3x+y)=3(6) \=9x+3y=18
Rewrite the equations in stacked format: -9x -3y= -18 9x+3y=18
Add the equations together. The result is: 0=0, which means that both equations are equal to the same line, thus there are infinite solutions. Test this by graphing both equations.
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Npvr Company
Topics: Net present value, Present value, Internal rate of return Pages: 4 (784 words) Published: April 13, 2013
Net Present Value (NPV)
Net present value is the present value of net cash inflows generated by a project including salvage value, if any, less the initial investment on the project. It is one of the most reliable measures used in capital budgeting because it accounts for time value of money by using discounted cash inflows.
Before calculating NPV, a target rate of return is set which is used to discount the net cash inflows from a project. Net cash inflow equals total cash inflow during a period less the expenses directly incurred on generating the cash inflow.
Calculation Methods and Formulas
The first step involved in the calculation of NPV is the determination of the present value of net cash inflows from a project or asset. The net cash flows may be even (i.e. equal cash inflows in different periods) or uneven (i.e. different cash flows in different periods). When they are even, present value can be easily calculated by using the present value formula of annuity. However, if they are uneven, we need to calculate the present value of each individual net cash inflow separately.
In the second step we subtract the initial investment on the project from the total present value of inflows to arrive at net present value.
Thus we have the following two formulas for the calculation of NPV:
When cash inflows are even:
NPV = R ×1 − (1 + i)-n− Initial Investment
i
In the above formula,
R is the net cash inflow expected to be received each period; i is the required rate of return per period;
n are the number of periods during which the project is expected to operate and generate cash inflows.
When cash inflows are uneven:
NPV =R1+R2+R3+ ...− Initial Investment
(1 + i)1(1 + i)2(1 + i)3
Where,
i is the target rate of return per period;
R1 is the net cash inflow during the first period;
R2 is the net cash inflow during the second period;
R3 is the net cash inflow during the third period, and so on ...
Decision...
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# The rate of a certain chemical reaction is directly proportional
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The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
A. 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 100% increase
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Re: The rate of a certain chemical reaction is directly proportional [#permalink]
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09 Sep 2017, 10:41
Since B is increased, A also has to be increased.
Thus option A, B and C are eliminated.
Out of option D and option E,
Option E is 100% increase, but since question says square of concentration of chemical A is related to rate of reaction and concentration of chemical B is already increased by 100% so concentration of A if increased by same (100) percent will lead to increase in rate of reaction.
Thus eliminating option E also
Thus option D
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Re: The rate of a certain chemical reaction is directly proportional [#permalink]
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09 Sep 2017, 10:42
HolaMaven wrote:
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100%, which of the following is closest to the percent change in the concentration of chemical A required to keep the reaction rate unchanged?
A. 100% decrease
B. 50% decrease
C. 40% decrease
D. 40% increase
E. 100% increase
Discussed here: the-rate-of-a-certain-chemical-reaction-is-directly-90119.html
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THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.
If you would like to discuss this question please re-post it in the respective forum. Thank you!
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Re: The rate of a certain chemical reaction is directly proportional [#permalink] 09 Sep 2017, 10:42
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# Velocity at impact
I am a high school student and have been wondering about this for a while: When a ball is either smashed or thrown down onto the ground, does the momentary velocity become zero at the moment of contact with the ground, similar to when a ball reaches its highest point when thrown vertically into the air? While we know that the velocity changes sign from negative to positive when the ball changes direction, I would like to understand whether the velocity graph crosses the $$x$$-axis and reaches zero at any point during this transition.
• Are you asking about the intermediate value theorem and the Bolzano theorem: en.wikipedia.org/wiki/Intermediate_value_theorem? That they apply follows from the assumption in the theory that the velocity is a continuous function of time (i.e. there are no infinite accelerations). Jun 3 at 16:28
Yes. You can either argue from the need for the velocity or momentum to be a continuous function of time, or the fact that the ground will, at some time, have imparted enough momentum upwards on the ball to just enough cancel its incoming momentum, before the next half of the momentum, needed to raise the ball back up to a new maximum height, is imparted.
Note that in this case, we do have to be careful, because the ball itself is undergoing internal motion when this is happening, and so we only mean that the centre of mass of the ball is momentarily stationary. The rest of the ball is in complicated motion at that time.
Typically a ball is idealized as a point particle or rigid object. A collision is idealized as instantaneous. The idealizations work for problems where you ask about velocity before and after. But they fail when you ask about what goes on during a collision.
Forces are reasonably small before a collision, typically $$1$$ g. A ball can stay reasonably rigid under forces like this.
A collision changes the velocity to $$0$$ in a very short time (usually). Accelerations and forces are very large. It isn't simple to analyze. Different parts of the ball can move at different speeds. If you do analyze it, every part decelerates to $$0$$ in a continuous way. If the collision is elastic, every part accelerates back up to speed.
Here is an example of what goes on during a collision - Titleist Golf Balls at The Moment of Impact. This one is more fun - How Hard Can You Hit a Golf Ball? (at 100,000 FPS)
Does the momentary velocity become zero ?
Yes, as long as you assume a physically realistic scenario in which the ball and/or the ground deforms (even if only slightly). In this case the velocity of the ball (more precisely, the velocity of its centre of mass) will be a continuous function of time. Since this velocity changes sign during the collision with the ground, there must be some point in time while the ball is in contact with the ground when its velocity is instantaneously zero.
However, this conclusion does not hold if you make the (unrealistic) assumption that both the ball and the ground are perfectly rigid. In this scenario, the velocity of the ball is not a continuous function of time, and it can change sign without ever having a value of zero.
In real life, yes it does! If this wouldn't have been the case, then the ground is supposed to provide an infinite amount of force to the ball which is not possible in real life scenarios. Moreover the ground also deforms a bit in real life which hints towards a continuous change of velocity.
The graphs ( you are talking about) assume a perfect ideal system in which the mass of the ground (earth) is infinite and collisions are perfectly elastic.
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HuggingFaceTB/finemath
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IEEE TRANSACTIONS ON IMAGE PROCESSING VOL. 21 NO. 4 Arithmetic Mean in the most common and easily understood measure of central tendency. We can define mean as the value obtained by dividing the sum of measurements with the number of measurements contained in the data set and is denoted by the symbol $\bar{x}$.
## Arithmetic Mean What it is and How to Find it
Estimation of Arithmetic Permeability from a Semi-Log. The abscissa of the maximum frequency in the frequency curve is the (a) mean (b) median (c) mode (d) none 11.3.M in Step—deviation method.M (c) both (d) none 264.M = assumed mean + arithmetic mean of deviations of terms (c) Both (d) none 265.7.5 (c) 5 (d) none 272.M of any distribution be 25 & one term is 18. Which one is true— (a) A. the standard deviation changes by (a) same proportion, PROPERTIES OF ARITHMETIC MEAN The sum of the deviations of items in a series from its Arithmetic Mean is always zero. Eg. 60, 25, 75, 38, 50, 52.
Arithmetic Mean in the most common and easily understood measure of central tendency. We can define mean as the value obtained by dividing the sum of measurements with the number of measurements contained in the data set and is denoted by the symbol $\bar{x}$. arithmetic mean or average, usually referred to simply as the mean. This is found by taking the sum of the observations and dividing by their number. The mean is often denoted by a little bar over the symbol for the variable, e.g. x. The sample mean has much nicer mathematical properties than the median and is thus more useful for the comparison methods described later. The median is a very
Merits and Demerits of Arithmetic Mean. Article Shared by. ADVERTISEMENTS: (A) Merits: 1. It can be easily calculated; and can be easily understood. It is the reason that it is the most used measure of central tendency. 2. As every item is taken in calculation, it is effected by every item. 3. As the mathematical formula is rigid one, therefore the result remains the same. ADVERTISEMENTS: 4 The arithmetic mean (or mean or average), ¯ (read bar), is the mean of the values ,, …,. [2] The arithmetic mean is the most commonly used and readily understood measure of …
23/02/2015В В· This video is part of an online course, Intro to Descriptive Statistics. Check out the course here: https://www.udacity.com/course/ud827. The arithmetic mean and the order statistical median are two fundamental operations in signal and image processing. They have their own merits and limitations in noise attenuation and image
Related Topics: More Arithmetic Lessons In these lessons, we will learn three basic number properties (or laws) that apply to arithmetic operations: Commutative Property, Associative Property and Distributive Property. of a Geometric Asian option due to the properties of a geometric mean, [2]. Arithmetic means does not share these vital properties with geometric means and Arithmetic Asian option prices are thus, plausibly, impossible to express in a closed form formula. As will be described, it is possible to approximate Arithmetic Asian option prices using the geometric mean prices, [4]. In order to price
Arithmetic Mean in the most common and easily understood measure of central tendency. We can define mean as the value obtained by dividing the sum of measurements with the number of measurements contained in the data set and is denoted by the symbol $\bar{x}$. These properties are useful when deriving the mean and variance of a random variable that arises in a hierarchical structure. Example 4 Derive the mean and variance of the following random variable X,
of a Geometric Asian option due to the properties of a geometric mean, [2]. Arithmetic means does not share these vital properties with geometric means and Arithmetic Asian option prices are thus, plausibly, impossible to express in a closed form formula. As will be described, it is possible to approximate Arithmetic Asian option prices using the geometric mean prices, [4]. In order to price To investigate the properties of the arithmetic mean and geometric mean of past returns as estimators of the discount rate, we assume that annual total return wealth relatives on the index over the past T …
A KNOWLEDGE STRUCTURE FOR THE ARITHMETIC MEAN: RELATIONSHIPS BETWEEN STATISTICAL CONCEPTUALIZATIONS AND MATHEMATICAL CONCEPTS by Mark A. Marnich B.S., Mathematics, Carnegie Mellon University, 1994 M.A., Mathematics, University of Pittsburgh, 2000 Submitted to the Graduate Faculty of the School of Education in partial fulfillment of the requirements … Some properties of Arithmetic mean If x 1 and x 2 are the means of the two groups computed from the values n 1 and n 2 then the mean x is given by the formula x = n 1 x 1 +n 2 x 2 / n 1 +n 2
Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two Numbers. For two numbers x and y, let x, a, y be a sequence of three numbers. If x, a, y is an arithmetic progression then 'a' is called arithmetic mean. If x, a, y is a geometric progression then 'a' is called geometric mean. Merits and Demerits of Arithmetic Mean. Article Shared by. ADVERTISEMENTS: (A) Merits: 1. It can be easily calculated; and can be easily understood. It is the reason that it is the most used measure of central tendency. 2. As every item is taken in calculation, it is effected by every item. 3. As the mathematical formula is rigid one, therefore the result remains the same. ADVERTISEMENTS: 4
Median as a weighted arithmetic mean of all sample observations SK Mishra Dept. of Economics, NEHU, Shillong, India Abstract This paper shows how median may be computed as a weighted arithmetic mean of all sample The Arithmetic Mean-Geometric Mean Inequality (AM-GM inequality) states that for a list of non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Taking the formulas for both types of mean, we get the inequality:
Relationship Between Arithmetic Mean, Harmonic Mean, and Geometric Mean of Two Numbers. For two numbers x and y, let x, a, y be a sequence of three numbers. If x, a, y is an arithmetic progression then 'a' is called arithmetic mean. If x, a, y is a geometric progression then 'a' is called geometric mean. The geometric mean is less than the arithmetic mean, The product of the items remains unchanged if each item is replaced by the geometric mean. The geometric mean of the ratio of corresponding observations in two series is equal to the ratios of their geometric means.
### Lecture 3 Measure of Central Tendency UNB
Arithmetic versus geometric mean estimators Setting. Finding the arithmetic mean takes two steps: add all the numbers up and then divide by the number of items in your set. The arithmetic mean is found in the exact same way as a sample mean (“sample” here just means a number of items in your data set)., Further Properties and a Fast Realization of the Iterative Truncated Arithmetic Mean Filter Zhenwei Miao, Student Member, IEEE, and Xudong Jiang, Senior Member, IEEE Abstract—The iterative truncated arithmetic mean (ITM) filter has been recently proposed. It possesses merits of both the mean and median filters. In this brief, the Cramer–Rao lower bound is employed to further analyze ….
Geometric Mean Definition and Formula Video & Lesson. The arithmetic mean or average of a set of values is the ratio of the sum of these values to the number of elements in the set. In other words, we add together the given values in a data set, and, We will explore the properties of the arithmetic mean when measurements are taken from a normal distribution. Open the first tab ( Explore 1 ) on the accompanying spreadsheet..
### Statistics Arithmetic Mean
mean University of Oregon. PROPERTIES OF ARITHMETIC MEAN The sum of the deviations of items in a series from its Arithmetic Mean is always zero. Eg. 60, 25, 75, 38, 50, 52 Watch this video lesson to learn how different the geometric mean is from the arithmetic mean, or average, that we are all so familiar with. Learn in what circumstance the geometric mean is preferred..
Preface Arithmetic is the basic topic of mathematics. According to the American Heritage Dictionary [1], it concerns “The mathematics of integers under addition, subtraction, But it does change the arithmetic mean. This shows, in simple terms, that the median does not depend on every value while the mean does. Actually, the median only depends on the ranks. The mathematical logic behind this simply arises from the mathematical definitions of the median and the mean.
Finding the arithmetic mean takes two steps: add all the numbers up and then divide by the number of items in your set. The arithmetic mean is found in the exact same way as a sample mean (“sample” here just means a number of items in your data set). This part of a multi-paper project studies the lattice properties of the arithmetic mean ideals of B(H) introduced by Dykema, Figiel, Weiss, and Wodzicki.
4.3 Properties Of Arithmetic Mean . The sum of the deviations, of all the values of x, from their arithmetic mean, is zero. Justification : Since is a constant, The product of the arithmetic mean and the number of items gives the total of all items. Justification : or If and are the arithmetic mean of two samples of sizes n 1 and n 2 respectively then, the arithmetic mean of the distribution Further Properties and a Fast Realization of the Iterative Truncated Arithmetic Mean Filter Zhenwei Miao, Student Member, IEEE, and Xudong Jiang, Senior Member, IEEE Abstract—The iterative truncated arithmetic mean (ITM) filter has been recently proposed. It possesses merits of both the mean and median filters. In this brief, the Cramer–Rao lower bound is employed to further analyze …
A KNOWLEDGE STRUCTURE FOR THE ARITHMETIC MEAN: RELATIONSHIPS BETWEEN STATISTICAL CONCEPTUALIZATIONS AND MATHEMATICAL CONCEPTS by Mark A. Marnich B.S., Mathematics, Carnegie Mellon University, 1994 M.A., Mathematics, University of Pittsburgh, 2000 Submitted to the Graduate Faculty of the School of Education in partial fulfillment of the requirements … But it does change the arithmetic mean. This shows, in simple terms, that the median does not depend on every value while the mean does. Actually, the median only depends on the ranks. The mathematical logic behind this simply arises from the mathematical definitions of the median and the mean.
A KNOWLEDGE STRUCTURE FOR THE ARITHMETIC MEAN: RELATIONSHIPS BETWEEN STATISTICAL CONCEPTUALIZATIONS AND MATHEMATICAL CONCEPTS by Mark A. Marnich B.S., Mathematics, Carnegie Mellon University, 1994 M.A., Mathematics, University of Pittsburgh, 2000 Submitted to the Graduate Faculty of the School of Education in partial fulfillment of the requirements … JIANG: ITERATIVE TRUNCATED ARITHMETIC MEAN FILTER AND ITS PROPERTIES 1539 The most critic technique to achieve this goal is to find a proper dynamic truncation threshold .
The Mean Property of the Mean. Robert Muldoon (1921 - 1992), the erstwhile Prime Minister of New Zealand is reported to have said. New Zealanders who emigrate to Australia raise the IQ of … Remembering the properties of numbers is important because you use them consistently in pre-calculus. The properties aren’t often used by name in pre-calculus, but you’re supposed to know when you need to utilize them. The following list presents the properties of numbers
The abscissa of the maximum frequency in the frequency curve is the (a) mean (b) median (c) mode (d) none 11.3.M in Step—deviation method.M (c) both (d) none 264.M = assumed mean + arithmetic mean of deviations of terms (c) Both (d) none 265.7.5 (c) 5 (d) none 272.M of any distribution be 25 & one term is 18. Which one is true— (a) A. the standard deviation changes by (a) same proportion This part of a multi-paper project studies the lattice properties of the arithmetic mean ideals of B(H) introduced by Dykema, Figiel, Weiss, and Wodzicki.
arithmetic mean or average, usually referred to simply as the mean. This is found by taking the sum of the observations and dividing by their number. The mean is often denoted by a little bar over the symbol for the variable, e.g. x. The sample mean has much nicer mathematical properties than the median and is thus more useful for the comparison methods described later. The median is a very A KNOWLEDGE STRUCTURE FOR THE ARITHMETIC MEAN: RELATIONSHIPS BETWEEN STATISTICAL CONCEPTUALIZATIONS AND MATHEMATICAL CONCEPTS by Mark A. Marnich B.S., Mathematics, Carnegie Mellon University, 1994 M.A., Mathematics, University of Pittsburgh, 2000 Submitted to the Graduate Faculty of the School of Education in partial fulfillment of the requirements …
Further Properties and a Fast Realization of the Iterative Truncated Arithmetic Mean Filter Zhenwei Miao, Student Member, IEEE, and Xudong Jiang, Senior Member, IEEE Abstract—The iterative truncated arithmetic mean (ITM) filter has been recently proposed. It possesses merits of both the mean and median filters. In this brief, the Cramer–Rao lower bound is employed to further analyze … Properties of Arithmetic Mean It requires at least the interval scale All values are used It is unique It is easy to calculate and allow easy mathematical treatment
geometric mean to arithmetic mean across a range of standard deviations. All of the graphs use the same geometric mean (10) and 10000 All of the graphs use the same geometric mean (10) and 10000 randomly generated log-normally distributed samples. The arithmetic mean is a measure of central tendency. It allows us to characterize the center of the frequency distribution of a quantitative variable by considering all of the observations with the same weight afforded to each (in contrast to the weighted arithmetic mean).
Properties of arithmetic lessons with lots of worked examples and practice problems. Very easy to understand! Definitions. Mean (aka Arithmetic Mean, Average) - The sum of all of the numbers in a list divided by the number of items in that list. For example, the mean of the numbers 2, 3, 7 is 4 since 2+3+7 = 12 and 12 divided by 3 [there are three numbers] is 4.
## PROPERTIES OF ARITHMETIC MEAN samagra.itschool.gov.in
(Solved) Properties of Harmonic Mean.. Detailed. The arithmetic mean and the order statistical median are two fundamental operations in signal and image processing. They have their own merits and limitations in noise attenuation and image, Some properties of Arithmetic mean If x 1 and x 2 are the means of the two groups computed from the values n 1 and n 2 then the mean x is given by the formula x = n 1 x 1 +n 2 x 2 / n 1 +n 2.
### Geometric Mean Definition and Formula Video & Lesson
IEEE TRANSACTIONS ON IMAGE PROCESSING VOL. 21 NO. 4. 5.3.1 Properties of the sample mean and variance Lemma 5.3.2 (Facts about chi-squared random variables) We use the notation П‡2 p to denote a chi-squared random variable with p degrees of freedom., 9/05/2011В В· Arithmetic meaning has been applied to simulated combinatorial libraries 12. The geometric mean was calculated as an additional point of comparison, since it represents the use of the arithmetic mean on, for example, log 10 (IC 50) values..
The arithmetic mean (or mean or average), ¯ (read bar), is the mean of the values ,, …,. [2] The arithmetic mean is the most commonly used and readily understood measure of … Related Topics: More Arithmetic Lessons In these lessons, we will learn three basic number properties (or laws) that apply to arithmetic operations: Commutative Property, Associative Property and Distributive Property.
Distribute copies of the Properties of Operations Chart. Lead a class discussion on the Lead a class discussion on the meaning of each property, and develop a definition of each for students to write in the of a Geometric Asian option due to the properties of a geometric mean, [2]. Arithmetic means does not share these vital properties with geometric means and Arithmetic Asian option prices are thus, plausibly, impossible to express in a closed form formula. As will be described, it is possible to approximate Arithmetic Asian option prices using the geometric mean prices, [4]. In order to price
Remembering the properties of numbers is important because you use them consistently in pre-calculus. The properties aren’t often used by name in pre-calculus, but you’re supposed to know when you need to utilize them. The following list presents the properties of numbers The arithmetic mean (or mean or average), ¯ (read bar), is the mean of the values ,, …,. [2] The arithmetic mean is the most commonly used and readily understood measure of …
What is Mean and What are its Advantages and Disadvantages How to Find the Mean. If x 1, x 2, x 3,…..,x n are n values of a variable X, then the arithmetic mean or simply the mean of these values is denoted by and is defined as = = Here the symbol denotes the sum x 1, x 2, x 3, ….., x n. If is the mean of n observations x 1, x 2, x 3,…..,x n, then the mean of the observations x 1 + a, x ' Furthermore, withregardto theharvesting of shellfish Although the arithmetic averageis known to be an unbiased the National Academy of Sciencesand the National Academy estimator of the arithmetic mean and the samplingproperties of Engineeringrecommendedthat the water quality meet the of th{sstatisticare known,this is not true of the geometric National Shellfish Sanitation Standards, which
Preface Arithmetic is the basic topic of mathematics. According to the American Heritage Dictionary [1], it concerns “The mathematics of integers under addition, subtraction, This part of a multi-paper project studies the lattice properties of the arithmetic mean ideals of B(H) introduced by Dykema, Figiel, Weiss, and Wodzicki.
PROPERTIES OF ARITHMETIC MEAN The sum of the deviations of items in a series from its Arithmetic Mean is always zero. Eg. 60, 25, 75, 38, 50, 52 But it does change the arithmetic mean. This shows, in simple terms, that the median does not depend on every value while the mean does. Actually, the median only depends on the ranks. The mathematical logic behind this simply arises from the mathematical definitions of the median and the mean.
Watch this video lesson to learn how different the geometric mean is from the arithmetic mean, or average, that we are all so familiar with. Learn in what circumstance the geometric mean is preferred. geometric mean to arithmetic mean across a range of standard deviations. All of the graphs use the same geometric mean (10) and 10000 All of the graphs use the same geometric mean (10) and 10000 randomly generated log-normally distributed samples.
To investigate the properties of the arithmetic mean and geometric mean of past returns as estimators of the discount rate, we assume that annual total return wealth relatives on the index over the past T … The Arithmetic Mean-Geometric Mean Inequality (AM-GM inequality) states that for a list of non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Taking the formulas for both types of mean, we get the inequality:
These properties are useful when deriving the mean and variance of a random variable that arises in a hierarchical structure. Example 4 Derive the mean and variance of the following random variable X, Definitions. Mean (aka Arithmetic Mean, Average) - The sum of all of the numbers in a list divided by the number of items in that list. For example, the mean of the numbers 2, 3, 7 is 4 since 2+3+7 = 12 and 12 divided by 3 [there are three numbers] is 4.
The Mean The mean is the sum of a set of values, divided by the number of values, i.e.: 12 1 1 ()/ ()/ 1 n n i i n i i X xx x n xn x n The mean is also known as the … Properties of arithmetic lessons with lots of worked examples and practice problems. Very easy to understand!
### Arithmetic versus geometric mean estimators Setting
Geometric Mean Definition Examples Formula Uses. 14/08/2016 · This video explains some very important properties of Arithmetic Mean 1. If a number is added to each observation, then the new mean will be equal to original mean …, of a Geometric Asian option due to the properties of a geometric mean, [2]. Arithmetic means does not share these vital properties with geometric means and Arithmetic Asian option prices are thus, plausibly, impossible to express in a closed form formula. As will be described, it is possible to approximate Arithmetic Asian option prices using the geometric mean prices, [4]. In order to price.
Concentration of the Ratio between the Geometric and. After reading this Chapter , a student will be able to understand different measures of central tendency, i.e. Arithmetic Mean, Median, Mode, Geometric Mean and Harmonic Mean, and computational techniques of these measures., Some properties of Arithmetic mean If x 1 and x 2 are the means of the two groups computed from the values n 1 and n 2 then the mean x is given by the formula x = n 1 x 1 +n 2 x 2 / n 1 +n 2.
### Arithmetic Progressions Brilliant Math & Science Wiki
PROPERTIES OF ARITHMETIC MEAN samagra.itschool.gov.in. To investigate the properties of the arithmetic mean and geometric mean of past returns as estimators of the discount rate, we assume that annual total return wealth relatives on the index over the past T … The geometric mean is less than the arithmetic mean, The product of the items remains unchanged if each item is replaced by the geometric mean. The geometric mean of the ratio of corresponding observations in two series is equal to the ratios of their geometric means..
23/02/2015В В· This video is part of an online course, Intro to Descriptive Statistics. Check out the course here: https://www.udacity.com/course/ud827. We will explore the properties of the arithmetic mean when measurements are taken from a normal distribution. Open the first tab ( Explore 1 ) on the accompanying spreadsheet.
geometric mean to arithmetic mean across a range of standard deviations. All of the graphs use the same geometric mean (10) and 10000 All of the graphs use the same geometric mean (10) and 10000 randomly generated log-normally distributed samples. Arithmetic, Geometric and Harmonic Means. Cauchy’s proof of 1897 first considers the case n = 2 m is a power of two. By reordering if necessary, we may assume a 6= b.
Watch this video lesson to learn how different the geometric mean is from the arithmetic mean, or average, that we are all so familiar with. Learn in what circumstance the geometric mean is preferred. These properties are useful when deriving the mean and variance of a random variable that arises in a hierarchical structure. Example 4 Derive the mean and variance of the following random variable X,
An arithmetic progression (AP), also called an arithmetic sequence, is a sequence of numbers which differ from each other by a common difference. For example, the sequence $$2, 4, 6, 8, \dots$$ is an arithmetic sequence with the common difference $$2$$. The abscissa of the maximum frequency in the frequency curve is the (a) mean (b) median (c) mode (d) none 11.3.M in Step—deviation method.M (c) both (d) none 264.M = assumed mean + arithmetic mean of deviations of terms (c) Both (d) none 265.7.5 (c) 5 (d) none 272.M of any distribution be 25 & one term is 18. Which one is true— (a) A. the standard deviation changes by (a) same proportion
Related Topics: More Arithmetic Lessons In these lessons, we will learn three basic number properties (or laws) that apply to arithmetic operations: Commutative Property, Associative Property and Distributive Property. We will explore the properties of the arithmetic mean when measurements are taken from a normal distribution. Open the first tab ( Explore 1 ) on the accompanying spreadsheet.
4.3 Properties Of Arithmetic Mean . The sum of the deviations, of all the values of x, from their arithmetic mean, is zero. Justification : Since is a constant, The product of the arithmetic mean and the number of items gives the total of all items. Justification : or If and are the arithmetic mean of two samples of sizes n 1 and n 2 respectively then, the arithmetic mean of the distribution To investigate the properties of the arithmetic mean and geometric mean of past returns as estimators of the discount rate, we assume that annual total return wealth relatives on the index over the past T …
An arithmetic progression (AP), also called an arithmetic sequence, is a sequence of numbers which differ from each other by a common difference. For example, the sequence $$2, 4, 6, 8, \dots$$ is an arithmetic sequence with the common difference $$2$$. Arithmetic (from the Greek ἀριθμός arithmos, "number" and τική, tiké [téchne], "art") is a branch of mathematics that consists of the study of numbers, especially the properties of the traditional operations on them—addition, subtraction, multiplication and division.
Distribute copies of the Properties of Operations Chart. Lead a class discussion on the Lead a class discussion on the meaning of each property, and develop a definition of each for students to write in the What is Mean and What are its Advantages and Disadvantages How to Find the Mean. If x 1, x 2, x 3,…..,x n are n values of a variable X, then the arithmetic mean or simply the mean of these values is denoted by and is defined as = = Here the symbol denotes the sum x 1, x 2, x 3, ….., x n. If is the mean of n observations x 1, x 2, x 3,…..,x n, then the mean of the observations x 1 + a, x
The arithmetic mean has been widely used to average elements of linear Euclidean spaces. Depending on the Depending on the application, it is usually referred to … 23/02/2015 · This video is part of an online course, Intro to Descriptive Statistics. Check out the course here: https://www.udacity.com/course/ud827.
Preface Arithmetic is the basic topic of mathematics. According to the American Heritage Dictionary [1], it concerns “The mathematics of integers under addition, subtraction, Properties of the Arithmetic Mean. 1. The sum of the deviations of all values of a distribution from their arithmetic mean is zero. The sum of the deviations of the numbers 8, 3, 5, 12, 10 of the arithmetic mean 7.6 is equal to 0:
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open-web-math/open-web-math
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Line Graph – Definition & Examples A line graph is a type of a graph which show how data or information change over time. A line graph is also known as line chart. Example: The table shows the sell of cars in a shop for five days. Draw a line graph. Example: The table shows […]
Introduction – Pie Chart Here we will discuss in brief about construction of a Pie Chart or Pie Graph from given data. Let us see what is a pie chart or pie graph. A “pie graph or (pie chart)” is a circular chart divided into sectors, each sector represents a proportionate part of a whole. […]
Complement of a Set Definition: The complement of a set A, is set of all elements of universal set, which are not the elements of A. Let U be the Universal set and A is subset of U, then the “Complement of A” is set of all elements of U which are not the elements of A. […]
Grouping of Data In this lesson we will learn about the grouping of data. Grouped data are data that has been organized in groups known as classes. Grouped data has been classified and a data class is group of data. Consider the marks obtain by 15 students in a history test as given below, 23, […]
Median – Definition Median is the middle value of a given distribution or group of numbers in their ascending order. OR in simplest way, we can say that the median is the midpoint of the given set of data. Median = Middle number Example: If we have numbers like 1, 2, 3, 4, 5. The […]
How to Find Median Value of Odd and Even Set of Data In this lesson we will learn how to calculate the median of a data set, when data set has odd and even numbers or observations. To learn the median of a data set, first we should understand what is median. (You can learn […]
Frequency Polygon A frequency polygon is very similar to a histogram and represents the graphical form of data. A frequency polygon is used to show a cumulative frequency distribution or compare to sets of data. In a frequency polygon the number of observations is marked at the the midpoint […]
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HuggingFaceTB/finemath
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1080p PenTile is not true 1080p
I just did the math
Full HD means 1920 by 1080 and it gives us 2073600 pixels since 1920*1080 = 2073600. On a traditional 1080p RGB panel (like on the HTC One), it’s the same but on a Pentile display, things are a bit different. See, each pixel on a RGB panel is made of 3 subpixels. On a PenTile display (like the one on the Galaxy S4), it is made of two and the missing subpixel in each individual pixel can be found in the adjacent pixel, which means on a standard 1080p RGB panel, there are 6220800 subpixels (2073600*3 = 6220800), but on a 1080p PenTile display, it’s 4147200. How?
Since 1 subpixel is missing from each pixel, we subtract 2073600, which is the number of missing subpixels on a 1080p Pentile display from 6220800, which is the total number of actual subpixels on a 1080p RGB display and we come up with 4147200. To get the total pixel count, we divide it by 3, because technically each pixel is made of 3 subpixels. So, 4147200/3 = 1382400.
So we now got the actual total pixel count. But how do we convert it to resolution? It’s easy, we divide the horizontal aspect ratio by vertical and then vice versa.
16/9 = 1.77 and 9/16 = 0.56. Then, we multiply each by the total pixel count:
1382400 * 1.77 = 2446848
1382400 * 0.56 = 774144
Then we take a square root of both:
sqrt(2446848) = 1564.24
sqrt(774144) = 879.85
Rounding each number, we come up with 1564 x 880
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Entrepreneur. Hardcore Programmer. Technology Journalist at @skotgat
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HuggingFaceTB/finemath
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# January 5, 2023 Single Round Match 843 Editorials
## AveragesOfThree
The problem becomes simpler if instead of the given array B we look at an array C in which each C[x] is 3*B[x].
When the values in B are the averages of three consecutive elements, the values in C are their sums.
Then:
• C[0] is the sum of 0 + 0 + A[2], hence A[2] = C[0].
• C[1] is the sum of 0 + A[2] + A[3], hence A[3] = C[1] – A[2].
• C[2] is the sum of A[2] + A[3] + A[4], hence A[4] = C[2] – A[3] – A[2].
• … and so on.
Each of the following values in C is the sum of two values we already know and one value we don’t, and so we can easily determine the unknown one.
public int[] restore(int[] B) {
int N = B.length;
int[] A = new int[N+2];
A[0] = 0;
A[1] = 0;
for (int n=0; n<N; ++n) A[n+2] = 3*B[n] - A[n+1] - A[n];
return A;
}
## DateCorrector2023
This problem only requires careful implementation of the given rules. For each format we need to do the following:
• check whether the input has this format
• parse the date
• check whether it is a valid date in 2022
• if yes, format the new date
Regular expressions are a great way for checks of the type we need here. Using one is much more convenient than doing the check manually.
An easy way to parse the date is to split the given string using the correct separator, and then to convert each of the pieces into an integer.
When checking the validity of a date, don’t forget that neither the day nor the month can be zero. (Years 2022 and 2023 are not leap years, so there is no special case related to February 29th in this problem.)
Fixing the date can simply be done by using a string search and replace command in your language to replace the substring “2022” with “2023”. An alternate good way to do this step is to use your language’s string printing/formatting routines – e.g., each language has a way to format a number to a given length, adding leading zeros as needed.
Below is an example implementation that tries to avoid copy and paste. For each format we store the separator used and the order in which day, month, year appear. Then we use those to construct the appropriate regular expression that matches the format, and once the token matches it, we can extract the date, check it, and if it’s good, we use string.replace() to fix the year.
public String fix(String token) {
int[] daysOfMonth = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
String[] separators = {"-", "/", "[.]"};
int[][] orders = { {2,1,0}, {1,0,2}, {0,1,2} };
String digit = "[0-9]";
String[] digitPatterns = { digit+"{2}", digit+"{2}", digit+"{4}" };
String[] datePatterns = new String[3];
for (int t=0; t<3; ++t) {
datePatterns[t] = digitPatterns[ orders[t][0] ];
datePatterns[t] += separators[t];
datePatterns[t] += digitPatterns[ orders[t][1] ];
datePatterns[t] += separators[t];
datePatterns[t] += digitPatterns[ orders[t][2] ];
}
for (int t=0; t<3; ++t) {
if (token.matches( datePatterns[t] )) {
String[] pieces = token.split( separators[t] );
int[] values = new int[3];
for (int i=0; i<3; ++i) {
values[ orders[t][i] ] = Integer.parseInt( pieces[i] );
}
int dd=values[0], mm=values[1], yyyy=values[2];
if (yyyy != 2022) continue;
if (mm < 1 || mm > 12) continue;
if (dd < 1 || dd > daysOfMonth[mm]) continue;
}
}
}
## ChristmasSongTrauma
Let T be the total length of the tape and S the length of the shortest song on the tape. As soon as visitTime > T-S, Tommy is guaranteed to hear all songs and thus all entry times are equally bad. We will handle this case and below we will assume that visitTime <= T-S and thus during the optimal visit Tommy hears fewer than N songs.
We can now proceed as follows: Imagine any possible visit for Tommy. Suppose a song is playing when he enters. What would change if he instead entered exactly when that song started playing?
Clearly, in the new visit he would hear at most as many songs as in the original one: he would hear more of the first song (which does not add any new heard songs) and correspondingly less at the end (which may cause him to hear fewer songs, but clearly cannot make him hear more).
Hence, an optimal solution must be among the N possible visits that start exactly as one of the songs begins playing. We can check all these by brute force.
Now we know the optimal number of songs Tommy can hear. And not just that. If we extend the above thought process, we can easily find all optimal entrance times: for each song s1 such that it is optimal to enter as song s1 starts, the optimal entrance times during s1 form a prefix of s1. Why and which one? Let s2 be the last song Tommy will (partially) hear if he enters as song s1 starts. Then the latest time when he still can enter optimally during s1 is the entrance time for which song s2 will end exactly when he leaves. This entrance time still has the same number of heard songs as the original one, and any later entrance time during s1 will have at least one more song and thus it will no longer be optimal.
This gives us an O(N^2) solution. There are solutions with a better time complexity (think “sliding window”) but this was not necessary.
public double fewest(int[] playTime, int visitTime) {
int N = playTime.length;
int tapeLength = 0;
for (int t : playTime) tapeLength += t;
int shortest = playTime[0];
for (int t : playTime) shortest = Math.min( shortest, t );
if (visitTime > tapeLength - shortest) {
return 1.; // guaranteed to hear all songs
}
int bestCount = N+1, bestWays = 0;
for (int n=0; n<N; ++n) {
// evaluate and count the best solutions if he enters during song n
int count = 0, length = 0, where = n;
while (length < visitTime) {
count += 1;
length += playTime[where];
where = (where + 1) % N;
}
if (count > bestCount) continue;
int ways = Math.min( playTime[n], length-visitTime+1 );
if (count == bestCount) { bestWays += ways; continue; }
bestCount = count; bestWays = ways;
}
System.out.println(bestCount);
System.out.println(bestWays);
return 1. * bestWays / tapeLength;
}
A note on incorrect implementations:
Some solutions may have included a weaker version of the check we made at the start of this writeup: they answer 1 if visitTime >= T and do the rest of the solution if visitTime is smaller. This isn’t a problem on its own, the conclusion is still correct, but depending on the rest of your implementation the weaker check may leave you with an extra tricky case you’ll now need to handle later: with T-S < visitTime < T it is possible to hear N+1 song segments, but those entry times are still optimal as the first and last song segment belong to the same song.
For example, consider the following input: songs are {50, 100, 100}, so T = 250, and visitTime = 201. From the reasoning we made above we already know that Tommy is doomed to hear all three songs. However, if we just blindly implement the above solution, on this test case it will fail. Why? Consider what happens if Tommy enters as song #1 starts. He will hear 100 seconds of that song, then 100 seconds of song #2, and then one second of song #0. We can now get the same outcome if he enters up to 49 seconds later: he will hear a shorter part of song #1 at the beginning and then more of song #0 at the end. However, what happens if Tommy enters during the second half of song #1? He will hear the end of song #1, then the full songs #2 and #0, and then the beginning of another copy of song #1. Here it is easy to make the mistake to interpret this situation as him hearing four distinct songs (or to get an invalid memory access).
One possible easy way to implement a solution that doesn’t have the above issue is to use brute force for not just N but up to 2N starting times: not just the times when a song begins but also the entrance times that are visitTime-1 seconds before a song ends. These starting times divide the tape loop into blocks, and within each block all entry times have the same outcome as entry when the block begins.
## TallyMarksSystem
As we are allowed the operation plus, we never need to consider actual numbers written using tally marks. E.g., instead of “55511” we can always do “(5+5+5+1+1)”. This simplifies the implementation.
Going backwards from the N we are supposed to construct seems too slow. For instance, there are O(N) ways to write it as x+y, and iterating over all those and recursively finding best representations for x and y would give us an O(N^2) solution.
Still, from this idea we can keep the observation that an optimal solution for any N (other than the base cases 1 and 5) is either the sum or the product of two optimal solutions, both for values strictly smaller than N.
Now, a better approach is to realize that the optimal representation of N will be roughly logarithmic in N (e.g., as an upper bound we can write in base 2 or base 5), so the answers for all possible N will be small. As we construct all possible short expressions, they will frequently have different values. Our solution will therefore find the optimal expressions for all numbers between 1 and N, inclusive, essentially by induction on the number of symbols needed.
To start, we know that only two numbers have a 1-symbol representation: 1 and 5.
Suppose that for each x=1, 2, …, k we already know all numbers for which the optimal representation has x symbols. How can we now find the numbers for which the optimal representation has k+1 symbols?
Each such number must be either the sum or the product of two numbers p, q such that the optimal representations of p and q have a total of k+1 symbols. Using the already computed data we can easily iterate over all such pairs (p, q). For each of them we examine the sum p+q and the product pq, and if they are <=N and previously unknown, we remember the optimal solution for them and store them in the list for k+1 symbols.
public int construct(int N) {
int MAXN = 2_000_000;
int[] best = new int[MAXN+1];
for (int n=0; n<=MAXN; ++n) best[n] = 999;
best[0] = 0;
best[1] = 1;
best[5] = 1;
// Both dimensions of the array below are determined by running the code
// once and observing when every number is constructed.
// It’s possible to implement it more nicely but I didn’t want to deal
// with arraylists of arraylists, sue me :)
int[][] constructed = new int[21][1000000];
int[] cc = new int[21];
cc[0] = 0;
cc[1] = 2;
constructed[1][0] = 1;
constructed[1][1] = 5;
for (int cnt=2; cnt<=20; ++cnt) {
cc[cnt] = 0;
for (int a=1; a<cnt; ++a) {
int b = cnt-a;
if (b < a) continue;
for (int ii=0; ii<cc[a]; ++ii) for (int jj=0; jj<cc[b]; ++jj) {
int x = constructed[a][ii], y = constructed[b][jj];
if (x+y <= MAXN) {
int z = x+y;
if (best[z] == 999) {
best[z] = cnt;
constructed[cnt][cc[cnt]++] = z;
}
}
if (1L*x*y <= MAXN) {
int z = x*y;
if (best[z] == 999) {
best[z] = cnt;
constructed[cnt][cc[cnt]++] = z;
}
}
}
}
}
return best[N];
}
A note on incorrect solutions:
In order to speed up their solutions some people made the incorrect assumption that addition is useless beyond adding 1 or 5 to the current number. This is false. The smallest counterexample to the solution that only tries to create a new value y via multiplication and then tries y=x+1 and y=x+5 is the prime number y = 139. As it’s prime, there’s no good way to make it as a product of two smaller numbers. The numbers 138 and 134 each require 8 symbols, so the best we can get that way is an expression with 9 symbols. However, we can write 139 as 14 + 125, and we can make 14 using five symbols (“11 * 511” = two times seven) and 125 using three (“5 * 5 * 5”).
## Component
We start by handling the corner cases: for S = 1, 2, or N the answer is 1 as that component size cannot be avoided.
A key observation is that at any moment during the construction of the graph, the current state can be represented as a multiset of current sizes of components that are strictly smaller than S. We will build the graph incrementally and keep track of the state until we either create a component of size exactly S or reach a state in which each component is bigger.
In particular, observe that we do not need to keep track of bigger components because they do not matter – the vertices in them can no longer be a part of the component we seek. In other words, we can always treat the bigger components as a single component containing of all their vertices – doing so clearly doesn’t change the answer, and it reduces the number of states considerably.
Once we get here, we can estimate that the number of states of the above type cannot be that large (the number of all integer partitions grows exponentially but very slowly) and we can do a quick DP to verify this: the count of states for S=1 is 1, and the count for any other pair (N, S) can be determined by iterating over the number x of components of size S-1, and for each such x counting the states for N-x(S-1) vertices and S-1 as the upper bound on component size. If we do this, we’ll learn that for the given constraints there are never more than 15,000 states.
Now, we can order these states by the number of components they contain, and in this order (smallest to highest) we can do dynamic programming on them. For each state we are going to compute the probability of hitting the desired component size if we start in that particular state.
In order to process a state, consider all edges that currently lead between two distinct components. (Remember that all already-bad vertices are treated as a single too-large component for this purpose.) One of these edges will eventually be added as the next non-trivial update, and each of them is equally likely. Thus, we can calculate the probability for the current state simply by averaging the probability of “winning” after adding each of the available edges.
Hence, for each of the available edges we consider what happens. If adding it produces a component of size S, we immediately “win”. In all other cases we get a new state, and as it has fewer components, we already know the probability of “winning” from this state.
Of course, we can easily implement the above in a faster way: instead of treating each edge separately, we can treat each pair of components separately. E.g., if we have components of sizes P and Q, there are P*Q edges between them and adding any one of those will have the same effect: we get a new component of size P+Q instead.
Using the above approach, we can process a state in time that is roughly cubic in the number of components it contains: for each pair of components we try adding an edge between them and while doing so, we spend linear time to create and process the new state.
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HuggingFaceTB/finemath
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# What is non-trivial solution to linear algebra
## Ψ The IT side
### Solution criteria for linear systems of equations
To determine the number of solutions to a linear equation, exist Solution criteria:
Homogeneous systems of linear equations can have either the trivial solution or multiple solutions.
• Trivial solution:
The trivial solution is this Zero vector; this means that you can only satisfy all equations if you set all variables to zero.
This case occurs when the homogeneous system of linear equations can be brought into row step form in such a way that a zero is added in each row from top to bottom and we have just as many rows as columns in the matrix1.3. Example of a system of equations with a matrix in row level form, which says that there is only the trivial solution: With this system of equations follows from the last line . Then follows from the second line and finally from the first , which is why we only get the zero vector as a solution.
With a homogeneous system of linear equations we can test whether vectors are linearly dependent or independent of one another. We write the vectors into the matrix as column vectors. If the homogeneous system of linear equations obtained in this way only has the trivial solution, the vectors are linearly independent of one another.
• Multiple solutions: The homogeneous system of linear equations has several solutions if it free variables gives. A free variable is a variable that is not determined by solving the system of equations. We can choose such a variable arbitrarily and thus get any number of solutions for the system of equations. These solutions span a vector space.
How do we now know whether the system of equations has free variables? Free variables are always available when there are fewer equations in the system than there are variables. Example: There are also systems of linear equations where it is not so obvious that there are fewer equations than variables. In these systems of equations, by bringing the system to row level form, we recognize that a Zero line exists. In homogeneous systems of linear equations, we can simply cross out such zero lines (be careful! This does not work in inhomogeneous systems, as otherwise we could destroy the fact that there is no solution.). If there are more variables than equations left after deleting, we have a system of equations with free variables.
If, when solving a homogeneous linear system of equations, in which the matrix consists of column vectors, whose linear independence from one another is to be checked, it emerges that this system of equations has several solutions, then the vectors are linearly dependent on one another.
Inhomogeneous Systems of equations cannot have a solution, exactly one or more solutions:
• No solution:
An inhomogeneous system of linear equations has no solution if the coefficients of one of the equations are zero, the associated one but is not null: Such an equation is called degenerate. Degenerate equations cannot be fulfilled and thus the entire system of equations cannot be fulfilled, which is why we have no solution.
• Exactly one solution:
An inhomogeneous system of equations has exactly one solution if the associated homogeneous system of equations - the associated homogeneous system of equations is generated by all be set to zero - only the trivial solution has.
In order to calculate the solution, we bring the inhomogeneous linear system of equations into line-level form, as we have also done above, whereby here we use the ( ) have to take into account.
An inhomogeneous system of equations can be used to check whether a vector can be generated as a linear combination of other vectors. There is a solution if and only if can be generated by linear combination of the column vectors. If there is no solution, one can do not combine from the column vectors. If there is one or even several solutions, this is possible.
• Multiple solutions:
As with homogeneous linear systems of equations, free variables can also appear in inhomogeneous systems. These exist again when you have more variables than equations. Zero lines can also be deleted in inhomogeneous systems, wherever . However, the degenerate equations must not be deleted.
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Vertical Dependency in Sequences of Categorical Random Variables
# Vertical Dependency in Sequences of Categorical Random Variables
### Graphical Interpretation and Illustrations
We may visualize the dependency structures generated by $\alpha \in \mathscr{C}_{\delta}$ via directed dependency graphs. Each $\epsilon_{n}$ represents a vertex in the graph, and a directed edge connects $n$ to $\alpha(n)$ to represent the direct dependency generated by $\alpha$. This section illustrates some examples and gives a graphical interpretation of the result in Theorem 2.
First-Kind Dependency
For FK-dependency, $\alpha(n) \equiv 1$ generates the graph above. Each $\epsilon_{i}$ depends directly on $\epsilon_{1}$, and thus we see no connections between any other vertices $i, j$ where $j \neq 1$. There are $n-1$ separate subsequences of length 2 in this graph.
Sequential Dependency
$\alpha(n) = n-1$ generates the sequential dependency structure this work has studied in detail. We can see that a path exists from any $n$ back to 1. This is a visual way to see the result of Theorem2, in that if a path exists from any $n$ back to 1, then the variables in that path must be identically distributed. Here, there is only one sequence and no subsequences.
A Monotonic Example
Here, $\alpha(n) = \lfloor \sqrt{n}\rfloor$ gives an example of a monotonic function in $\mathscr{C}_{\delta}$ and the dependency structure it generates. Again, note that any $n$ has a path to 1, and the number of subsequences is between 1 and $n-1$.
A Nonmonotonic Example
Here, $\alpha(n)=\left\lfloor\frac{\sqrt{n}}{2}\left(\sin(n)\right)+\frac{n}{2}\right\rfloor$ illustrates a more contrived example where the function is nonmonotonic. It is neither increasing nor decreasing. The previous examples have all been nondecreasing.
A Prime Example
Let $\alpha \in \mathscr{C}_{\delta}$ be defined in the following way. Let $p_{m}$ be the $m$th prime number, and let $\{kp_{m}\}$ be the set of all postive integer multiples of $p_{m}$. Then the set $\mathscr{P}_{m} = \{kp_{m}\} \setminus \cup_{i=1}^{m-1}\{kp_{i}\}$ gives a disjoint partitioning of $\mathbb{N}$. That is, $\mathbb{N} = \sqcup_{m}\mathscr{P}_{m}$, and thus for any $n \in \mathbb{N}, n \in \mathscr{P}_{m}$ for exactly one $m$. Now let $\alpha(n) = m$. Thus, the function is well-defined. We may write $\alpha: \mathbb{N}_{\geq 2} \to \mathbb{N}$ as $\alpha[\mathscr{P}_{m}] = \{m\}$. As an illustration,
\begin{aligned}\alpha[\{2k\}] &= \{1\} \\\alpha[\{3k\}\setminus \{2k\}] &= \{2\} \\\alpha[\{5k\}\setminus (\{2k\}\cup \{3k\})] &= \{3\} \\&\qquad\vdots\\\alpha[\{kp_{m}\}\setminus (\cup_{i=1}^{m-1}\{kp_{i}\})] &= \{m\}\end{aligned}
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# ### Sidebar
models:basic_models:pendulum
$\quad L = \frac{1}{2} \color{blue}{m} \color{olive}{l}^2\dot{\color{firebrick}{\phi}}^2 - \color{blue}{m}\color{magenta}{g}\color{olive}{l} (1-cos \color{firebrick}{\phi})$
# Pendulum
## Intuitive A pendulum consists of a freely hanging massive bob at the end of a rod. When we move the bob a little to one side it starts swinging.
We usually describe it by measuring how far the bob has moved from its original position where it just hangs freely. How the pendulum swings depends crucially on the length of the rod and the strength of the gravitational field. A pendulum on the moon swings differently than a pendulum on earth.
An important observation is that the swinging of the pendulum does not depend on the mass of the bob.
A pendulum is right after a harmonic oscillator the simplest physical system we can study. In fact, if the pendulum only swings a little it is a harmonic oscillator. The difference between the harmonic oscillator and the pendulum only become important for large swings.
## Concrete A normal pendulum hangs freely in a uniform gravitational field of strength $g$ on a rod with length $l$. The excitation above the ground state is measured by the angle $\phi$. At the end of the pendulum, we have a bob of mass $m$. This is shown in the picture on the right-hand side.
The only force that acts on the pendulum is the gravitational force $F= mg \sin\phi$. Therefore, using Newton's second law the correct equation of motion for the pendulum is
$$m\frac{d^2 x}{dt^2}= - mg \sin \phi .$$
Now we need to get rid of the arc length on the left-hand side using $x = l \phi$. This yields
$$m\frac{d^2 x}{dt^2}= - m\frac{g}{l} \sin \phi .$$
It is important to note that we can remove mass $m$ from the equation because it appears on both sides:
$$\frac{d^2 x}{dt^2}= - \frac{g}{l} \sin \phi .$$
The swinging of the pendulum, therefore, does not depend on the mass of the bob.
Small Angle Approximation
The equation of motion of the pendulum is a non-linear equation which is hard to solve. However we can simplify it for the case when only small swings are happening. Then we can use the small angle approximation $\sin({\theta)} \approx \theta$.
This yields the equation $\frac{d^2\theta}{dt^2} = -\frac{g}{L}\theta$ which can be solved to give the familiar equation for the harmonic oscillator $\theta = A\sin({\beta t + \phi}), \beta = \sqrt{\frac{g}{l}}$ where $A$ is the amplitude and $\phi$ is the constant phase offset.
The Pendulum in the Lagrangian Formalism
The kinetic energy of the pendulum is
$$T=\frac{1}{2}ml^2(\frac{d\theta}{dt})^2$$
and the potential energy in the gravitational field of the earth
$$U=mgl(1-cos\theta).$$
The Lagrangian of the pendulum is therefore
$$L = T-U= \frac{1}{2} ml^2\dot{\theta}^2 - mgl (1-cos \theta),$$ where $\dot{\theta}\equiv d\theta /dt$ denotes the time derivative.
Using the Euler-Lagrange equation we can derive the corresponding equation of motion
$$\frac{d^2\theta}{dt^2}+\frac{g}{l}sin\theta=0 .$$
• For a nice complete discussion, see The Pendulum: A Case Study in Physics by Gregory L. Baker and James A. Blackburn
## Abstract
The phase space of a pendulum • A thorough discussion can be found in Quantum pendulum by R. Aldrovandi and P. Leal Ferreira
## Why is it interesting?
The pendulum is one of the most important basic physics systems.
Understanding the pendulum is crucial to understand topics as advanced as, for example, Solitons or the QCD vacuum.
In addition, it is ideal to understand the notion dimensional analysis.
## History
• Time for Science Education - How Teaching the History and Philosophy of Pendulum Motion can Contribute to Science Literacy by Michael R. Matthews
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Home Contact KS2 Maths GCSE EFL Advice Parents Games Other
GCSE Maths > Algebra - Inequalities
### Inequalities
a < b means a is less than b (so b is greater than a)
a £ b means a is less than or equal to b (so b is greater than or equal to a)
a ³ b means a is greater than or equal to b etc.
a > b means a is greater than b etc.
If you have an inequality, you can add or subtract numbers from each side of the inequality, as with an equation. You can also multiply or divide by a constant. However, if you multiply or divide by a negative number, the inequality sign is reversed.
#### Example
Solve 3(x + 4) < 5x + 9
3x + 12 < 5x + 9
\ -2x < -3
\ x > 3/2 (note: sign reversed because we divided by -2)
Inequalities can be used to describe what range of values a variable can be.
E.g. 4 £ x < 10, means x is greater than or equal to 4 but less than 10.
### Graphs
Inequalities are represented on graphs using shading. For example, if y > 4x, the graph of y = 4x would be drawn. Then either all of the points greater than 4x would be shaded or all of the points less than or equal to 4x would be shaded.
#### Example
x + y < 7
and 1 < x < 4 (NB: this is the same as the two inequalities 1 < x and x < 4)
Represent these inequalities on a graph by leaving un-shaded the required regions (i.e. do not shade the points which satisfy the inequalities, but shade everywhere else).
### Number Lines
Inequalities can also be represented on number lines. Draw a number line and above the line draw a line for each inequality, over the numbers for which it is true. At the end of these lines, draw a circle. The circle should be filled in if the inequality can equal that number and left unfilled if it cannot.
#### Example
On the number line below show the solution to these inequalities.
-7 £ 2x - 3 < 3
This can be split into the two inequalities:
-7 £ 2x - 3 and 2x - 3 < 3
\ -4 £ 2x and 2x < 6
\ -2 £ x and x < 3
The circle is filled in at –2 because the first inequality specifies that x can equal –2, whereas x is less than (and not equal to) 3 and so the circle is not filled in at 3.
The solution to the inequalities occurs where the two lines overlap, i.e. for -2 £ x < 3 .
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Tree and Graph Data Structures
## Bianca Gandolfo
Thumbtack
### Check out a free preview of the full Tree and Graph Data Structures course
The "Adjacency Matrix" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Here's what you'd learn in this lesson:
Bianca analyzes the adjacency matrix format of representing node relationships in a graph, using binary values in the array.
Preview
Close
### Transcript from the "Adjacency Matrix" Lesson
[00:00:00]
>> Bianca Gandolfo: And we're gonna talk about the implementation, right? I asked you guy to think a little bit about how this might look in code. And we're gonna break it down.
>> Bianca Gandolfo: Okay, so what data needs to be stored? Two main things, right, we need you to have the objects.
[00:00:21]
And the relationships, right? So the object is being our food or our people and or our people and the relationships being the lines or edges.
>> Bianca Gandolfo: Cool, we're gonna be using numbers to simplify, but you can imagine that this could be anything. Some data structures you need the value to be comparable, right?
[00:00:48]
So we can compare like greater or less than something like that. And those numbers are important for that. But in a graph, we're still flexible but we just use it for simplifying a label. Okay, so there are three main ways to represent a graph. I'm going to tell you about two, and I'll give you a recommendation about which one to use in an interview setting.
[00:01:19]
And I also want you to be open minded, like I've been saying this whole time. That this is computer science, there are many ways to solve lots of different problems. And it's up to you to think critically about your particular problem that you're solving. Whether it's on the job, or at work, or whatever it may be in your mind, I don't know, to model it to suit your needs, cool.
[00:01:44]
So the first one is the adjacent matrix and we're going to use that to create our graph, right? Because we can't code just circles and lines we have to deal with what we have access to. Right, we have ones and zeros, and we have, you know, one through five as well, but that's not what this is.
[00:02:13]
This is a matrix. Have you guys worked at all with matrices? Yea, a little bit. Like in math class, maybe. So the adjacency matrix, a matrix is basically an array inside of an array, right, when we get down to it. It looks something like this. I like this picture better though.
[00:02:39]
And so how we are modeling the relationships here is with the ones and zeros, okay? How we're modeling the data itself is through these vertices. So one is one. So one does not have a relationship with itself. So is zero okay. 2 is connected to 1, as we can see here with this line.
[00:03:04]
So we add a 1 here, and we also add a 1 here. And so on and so forth. So here our Edges. Here are our Vertices.
>> Bianca Gandolfo: Cool? Yeah.
>> Speaker 2: So 3 is connected to 0?
>> Bianca Gandolfo: So 3, we can look at 3 is connected to 2. 3 is also connected to 4.
[00:03:34]
>> Speaker 2: I see. Okay.
>> Bianca Gandolfo: And that's it. Yeah.
>> Speaker 3: So shouldn't 1 and 1 should also be 1? I mean, that's called a self loop and in this case we don't have self loops.
>> Bianca Gandolfo: Yeah.
>> Speaker 4: I'm kind of confused where did we lose the people in the whole equation.
[00:03:59]
>> Bianca Gandolfo: Yeah. Where did we lose the people? We'll get there. We're gonna start with this and then the people will be back. You have a question.
>> Speaker 2: This is a dumb question, but oaky, one is connected to, okay I get it. Now I get it. Never mind.
>> Bianca Gandolfo: Okay, cool.
[00:04:20]
>> Bianca Gandolfo: All right, so Adjacency Matrix. Here's the code. Awesome, okay. So now that we know that this is a 2D array, we wanna add an edge, and an edge is a relationship. How do we represent an edge again? With the number One or zero?
>> Speaker 3: One, one is the edge zero is the lack of an entrenched.
[00:04:46]
>> Bianca Gandolfo: So we'll pass like a couple of vertices and we will set it to one. So our adjacency matrix, we can call it this, we have v1 at v2, right? Equals to 1. Of course this is not exact, this is a little bit naive. Right, there's a little bit more that needs to happen here, right?
[00:05:08]
Because this is actually the 0th index for example, it's not 1. But you can imagine, right? It's pretty simple, right? We're just kinda toggling these values to add an edge.
>> Speaker 4: So the v1 is this way, v2 is that, down this way?
>> Bianca Gandolfo: So the v1 and the v2, these are just variable names so we could.
[00:05:44]
For example, call this, I wish this is, it used to let me edit this.
>> Speaker 2: Why do you wanna add an edge? I mean, maybe there's.
>> Bianca Gandolfo: Why you wanna add an iedge? Like what if, for example, like I suddenly started to also like eggs.
>> Speaker 2: Okay.
>> Bianca Gandolfo: And then I wanna have that relationship represented.
[00:06:05]
>> Speaker 2: Maybe I'll, afterwards.
>> Bianca Gandolfo: Yeah, do you wanna add an edge. So what if like, I wanted to, say this was a map, right? This is a city, and these are like intersections and these are roads. And actually let's say these are points of interest and these are roads between them.
[00:06:23]
Let's say that it's approved and we're gonna build a row between these two points of interest. So we wanna add a new Edge to connect these, yeah.
>> Bianca Gandolfo: And another thing that we'll talk about soon is, what if we wanna add another point of interest? So that would be adding a node to our graph.
[00:06:47]
Another interesting thing, is that you can add a node to a graph, even if there's no edge. So you can have sort of like these disconnected components. That's kind of fun, so we'll talk more about that. Okay, so we add in an edge, this is just adding an edge and removing an edge, very similar, we set to zero.
[00:07:11]
Again, don't copy and paste this, this is not actual code, it's just kind of demonstrating, right? Like we don't [INAUDIBLE] actually isn't defined, you're gonna get errors but this just. To get the gist. Do you guys get the gist? Is it working? Okay, awesome, all right, time complexity for adding an edge.
[00:07:37]
>> Speaker 3: Constant time.
>> Bianca Gandolfo: Constant time, removing an edge?
>> Speaker 3: Constant.
>> Bianca Gandolfo: You guys are good at this. All right, so that's adjacency matrix.
### Learn Straight from the Experts Who Shape the Modern Web
• In-depth Courses
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This is documentation for Mathematica 6, which was
based on an earlier version of the Wolfram Language.
Mathematica Tutorial
# Introduction to Differential-Algebraic Equations (DAEs)
The systems of equations that govern certain phenomena (in electrical circuits, chemical kinetics, etc.) contain a combination of differential equations and algebraic equations. The differential equations are responsible for the dynamical evolution of the system, while the algebraic equations serve to constrain the solutions to certain manifolds. It is therefore of some interest to study the solutions of such differential-algebraic equations (DAEs).
Here is a simple example of a DAE. The first equation is an ODE for the function x[t], while the second equation constrains the functions x[t] and y[t] to lie in a submanifold (a straight line) in {x, y} space.
This discussion will be restricted to linear DAEs, which are defined as systems of equations of the following type.
Here A and B are matrix functions of the independent variable t, F is a vector function of t, and x (t) is the vector of unknowns. If the matrix A is nonsingular (that is, invertible) then this is a system of ODEs. Thus, the system is a DAE if the matrix A is singular.
If F0, then the system is said to be homogeneous. As for ODEs, the general solution to a DAE is composed of the general solution to the corresponding homogeneous problem and a particular solution to the inhomogeneous system.
DSolve can find the solutions to all DAEs in which the entries of the matrices A and B are constants. Such DAEs are said to have constant coefficients. The algorithm used by DSolve is based on decomposing both A and B into a nonsingular and nilpotent part. This decomposition is used to calculate a generalized inverse for A and B, which effectively reduces the problem to solving a system of ODEs.
It is important to realize that the initial values for a DAE must be prescribed carefully to guarantee a solution for the problem. This can be seen by considering the following system of equations.
This gives
Hence the only solution is
But this solution is inconsistent with the initial condition x1 (0)1.
For examples of DAEs with constant coefficients, which can be solved using DSolve, click here.
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# Thread: Solving a problem without the use of stokes
1. ## Solving a problem without the use of stokes
Hey all!
My professor in my vector analysis course likes to always have a two part question where one part can be solved by stokes, greens or gauss theorem. They are pretty straight forward if you understand them and I find it that I have. The other part is where I take the same problem and solve it without the use of the theorem.
So here is an example I've solved for stokes, but I don't know how to solve without it.
The Question
We have a triangle with vertris (1,0,0) (0,1,0) (0,0,1).
solve the closed integral of [int] xy dx+yz dy+zx dz over that plane.
Attempt at solution
We know the plane equation is x+y+z=1
We also know [int] xy dx+yz dy+zx dz = [int] F dot dr, which is a closed line integral.
This is pretty much as far as I get..
What I need help with
I have a vague memory that I can devide C into parts of integration, but as I have just one integral and can only integrate over x,y or z seperatly.
Please give me a few hints so I can carry on with my understanding of calculus
2. ## Re: Solving a problem without the use of stokes
For example, if $A=(1,0,0),\;B=(0,1,0)$ then, $AB\equiv (x,y,z)=(1-t,t,0)$ with $t\in [0,1]$ so
$\int_{AB}xydx+yzdy+zxdz=\int_0^1(1-t)t(-dt)=\ldots$
Idem for the other sides.
3. ## Re: Solving a problem without the use of stokes
Originally Posted by FernandoRevilla
For example, if $A=(1,0,0),\;B=(0,1,0)$ then, $AB\equiv (x,y,z)=(1-t,t,0)$ with $t\in [0,1]$ so
$\int_{AB}xydx+yzdy+zxdz=\int_0^1(1-t)t(-dt)=\ldots$
Idem for the other sides.
Aha, so parametization is the way to go. So I do it beween AB, AC and CB.
Thank you very much. Im going to post alot here these days because I really need help with this
4. ## Re: Solving a problem without the use of stokes
I might suggest that you can use the plane x + y + z = 1. On each of your three lines, one of x, y or z = 0. I would also sugest that you set up all three integral before evaluating. Two of the three may cancel leaving only one, or, because of the actual plane in this example, you could combine the three integrals in a single integral.
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# What Percent Decrease 350 is 100? getcalc.com's Percentage change {(X - Y)/X x 100} Calculator & workout with step by step calculation to find what is the percent decrease from 350 to 100.
(350 - 100)/350 x 100 = (250/350) x 100 = 71.4286%
71.4286 percent decrease or decline from 350 is 100
Problem
Find What is the decrease percentage for 350 to 100?
Step by step workout
step 1 Address the formula, input parameters & values
Formula :
Difference/Initial Value x 100 = Percent Decrease (%)
Initial Value X = 350 & New Value Y = 100
Decrease = (X - Y)
(350 - 100)/350 x 100 = ?
step 2 Apply the values in the percentage decrease {(X - Y)/X x 100} formula
= (350 - 100)/350 x 100
= 250/350 x 100
= 71.4286%
(350 - 100)/350 x 100 = 71.4286%
The percentage reduction can be written as 28.5714% of 350.
71.4286 percent decrease (%↓) or reduction from 350 is 100.
## Percentage Chart
The below chart for 71.4286% decline of 350 and its nearest decrease percent values for users reference.
Value ChangePercentage Decrease
350 to 9074.29%
350 to 9572.86%
350 to 10071.43%
350 to 10570%
350 to 11068.57%
350 to 11567.14%
350 to 12065.71%
350 to 12564.29%
This calculator, formula, step by step calculation and associated information to find what percent reduction from 350 to 100 may help students, teachers, parents or professionals to learn, teach, practice or verify similar percentage reduction {(X - Y)/X x 100} calculations efficiently.
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# Integrating Factor
In Maths, an integrating factor is a function used to solve differential equations. It is a function in which an ordinary differential equation can be multiplied to make the function integrable. It is usually applied to solve ordinary differential equations. Also, we can use this factor within multivariable calculus. When multiplied by an integrating factor, an inaccurate differential is made into an accurate differential (which can be later integrated to give a scalar field). It has a major application in thermodynamics where the temperature becomes the integrating factor that makes entropy an exact differential.
## What are Differential Equations?
Differential equations play a vital role in Mathematics. These are the equations that necessarily involve derivatives. There are various types of differential equations; such as – homogeneous and non-homogeneous, linear and nonlinear, ordinary and partial. The differential equation may be of the first order, second order and ever more than that. The nth order differential equation is an equation involving nth derivative. The most common differential equations that we often come across are first-order linear differential equations.
The ordinary linear differential equations are represented in the following general form:
y’ + P y=Q
or
dy/dx + P(x) y = Q(x)
Where y’ or dy/dx is the first derivative. Also, the functions P and Q are the functions of x only.
There are mainly two methods which are utilized in order to solve the linear first-order differential equations:
• Separable Method
• Integrating Factor Method
In this article, we are going to discuss what is integrating factor method, and how the integrating factors are used to solve the first and second-order differential equations.
## Integrating Factor Method
Integrating factor is defined as the function which is selected in order to solve the given differential equation. It is most commonly used in ordinary linear differential equations of the first order.
When the given differential equation is of the form;
dy/dx + P(x) y = Q(x)
then the integrating factor is defined as; Where P(x) (the function of x) is a multiple of y and μ denotes integrating factor.
## Solving First-Order Differential Equation Using Integrating Factor
Below are the steps to solve the first-order differential equation using the integrating factor.
• Compare the given equation with differential equation form and find the value of P(x).
• Calculate the integrating factor μ.
• Multiply the differential equation with integrating factor on both sides in such a way; μ dy/dx + μP(x)y = μQ(x)
• In this way, on the left-hand side, we obtain a particular differential form. I.e d/dx(μ y) = μQ(x)
• In the end, we shall integrate this expression and get the required solution to the given equation: μ y = ∫μQ(x)dx+C
## Solving Second Order Differential Equation Using Integrating Factor
The second-order differential equation can be solved using the integrating factor method.
Let the given differential equation be,
y” + P(x) y’ = Q(x)
The second-order equation of the above form can only be solved by using the integrating factor.
• Substitute y’ = u; so that the equation becomes similar to the first-order equation as shown: u’ + P(x) u = Q(x)
• Now, this equation can be solved by integrating factor technique as described in the section above for first-order equations and we reach the equation: μ u=∫μQ(x)dx+C
• Find the value of u from this equation.
Since u = y’, hence to find the value of y, integrate the equation. In this way, we get the required solution.
### Integrating Factor Example
Example:
Solve the differential equation using the integrating factor: (dy/dx) – (3y/x+1) = (x+1)4
Solution:
Given: (dy/dx) – (3y/x+1) = (x+1)4
First, find the integrating factor:
μ = e ∫ p(x) dx
μ = e ∫(-3/x+1) dx
∫(-3/ x+1)dx = -3 ln (x+1) = ln (x+1)-3
Hence, we get
μ =e ln (x+1)-3
μ = 1/ (x+1)3
Now, multiply the integrating factor on both the sides of the given differential eqaution:
[1/ (x+1)3] [dy/dx] – [3y/( (x+1)4)] = (x+1)
Integrate both the sides, we get:
[y/(x+1)3] = [(1/2)x2+x+c]
Here, c is a constant
Therefore, the general solution of the given differential eqaution is
y = [(x+1)3] [(1/2)x2+x+c].
Keep visiting BYJU’S – The Learning App and download the app to learn all the Calculus related topics and watch interesting and engaging videos to learn with ease.
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#### Transcript Newton`s Laws Test Review
```NEWTON’S LAWS TEST
REVIEW
Which of Newton’s Laws states for every action
there is an equal and opposite reaction?
3
Which of Newton’s laws states that an object in
motion will remain in motion unless acted upon by
unbalanced forces?
1
The amount of gravitational force between 2
objects depends on which two factors?
Mass
of objects
Distance between objects
1 Pound = ? Newtons
4.448
N
What is the weight of a 250 kg person?
250x9.8
= 2450 N
What unbalanced force is needed to give a 1000
Kg truck an acceleration of 3 m/s2?
3000
N
A .50 Kg metal ball experiences a net force of 2
Newtons as it rolls down a ramp. What is the
acceleration of the ball?
4
What is the units for acceleration?
m/s2
Any object that is thrown or shot through the air is
called a
Projectile
How do you find an object’s momentum?
Mass
x velocity
What is the momentum of a 100 kg man running at
a speed of 5 m/s (include correct units)?
500
kg x m/s
What is the velocity of a 100 kg man with a
momentum of 400 kg x m/s (include correct units)?
4
m/s
How does the total momentum of 2 objects before a
collision compare to the total momentum after a
collision?
Same
due to Law of Conservation of Momentum
An object that has balanced forces acting upon it
would have a Net FORCE equal to
zero
When air resistance balances the weight of a
falling object, the object reaches its top speed
called
Terminal
velocity
The resistance of an object to changing its motion is
called
inertia
The amount of matter in an object is called its
Mass
A vector quantity includes both
Magnitude
and Direction
The value of acceleration for objects in free fall
near earth is __________ (include correct units)
9.8
m/s2
In the Newton Car lab the rubber bands applied an
action force to the small block of wood, what was
the reaction force?
Block
of wood on Newton car
If forces occur in pairs and are equal and opposite
why do they not cancel each other out?
They
are acting on different objects
What is the net force on the box below?
-40
N
-25
N
15 N
```
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Asher Olsen
2022-04-01
Prove that
$16{\mathrm{cos}}^{5}A-20{\mathrm{cos}}^{3}A+5\mathrm{cos}A=\mathrm{cos}5A$
Charlie Haley
Use
$2\mathrm{cos}x={e}^{ix}+{e}^{-ix}$
Then
$32{\mathrm{cos}}^{5}x={e}^{5ix}+5{e}^{3ix}+10{e}^{-ix}+5{e}^{-3ix}+{e}^{-5ix}$
$=2\mathrm{cos}5x+10\mathrm{cos}3x+20\mathrm{cos}x$
Moreover
$8{\mathrm{cos}}^{3}x={e}^{3ix}+3{e}^{ix}+3{e}^{-ix}+{e}^{3ix}=2\mathrm{cos}3x+6\mathrm{cos}x$
Therefore
$16{\mathrm{cos}}^{5}x-20{\mathrm{cos}}^{3}x+5\mathrm{cos}x=\left(\mathrm{cos}5x+5\mathrm{cos}3x+10\mathrm{cos}x\right)-5\left(\mathrm{cos}3x+3\mathrm{cos}x\right)+5\mathrm{cos}x=\mathrm{cos}5x$
This is actually backwards, so here's a different proof:
$\mathrm{cos}5x+i\mathrm{sin}5x={\left(\mathrm{cos}x+i\mathrm{sin}x\right)}^{5}$
$={\mathrm{cos}}^{5}x+5i{\mathrm{cos}}^{4}x\mathrm{sin}x-10{\mathrm{cos}}^{3}x{\mathrm{sin}}^{2}x-10i{\mathrm{cos}}^{2}x{\mathrm{sin}}^{3}x+5\mathrm{cos}x{\mathrm{sin}}^{4}x+i{\mathrm{sin}}^{5}x$
Taking the real part
$\mathrm{cos}5x={\mathrm{cos}}^{5}x-10{\mathrm{cos}}^{3}x\left(1-{\mathrm{cos}}^{2}x\right)+5\mathrm{cos}x{\left(1-{\mathrm{cos}}^{2}x\right)}^{2}$
and you can finish.
Do you have a similar question?
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# 1st GRADE MATH YEARLY PACING GUIDE. Quarter 1
Size: px
Start display at page:
## Transcription
1 Operations & (NBT) Measurement & Geometry(G) 1OA.1 Use addition and subtraction within 10 to adding to taking from, putting together, taking apart, comparing 1.NBT.1 Count to 30, starting at any number less than 30. In this range, read and write numerals and represent a number of objects with a written numeral. a. Count to 30 orally, starting at any number less than 30. b. Read numerals up to 30. c. Write numerals up to 1.OA.3 Apply properties of operations as strategies to add and subtract. If = 11 is known, then = 11 is also known. (Commutative property of addition.) 1.NBT.2 Develop an understanding that the two digits of a twodigit number represent amounts of tens and ones. Understand the following as special cases: a. 10 can be thought of as a bundle of ten ones called a ten. b. The numbers from 11 to 20 are composed of a ten and a one, two, three, four, five, six, seven, eight, or nine ones. Quarter 1 Understand subtraction as problem. For example, subtract 10 8 by finding the number that makes 10 when added to 8. (missing addend) 20, developing fluency for within 10*. Add and subtract within 20 using strategies such as: counting on (+ /-1 and 2) Page 1
3 Geometry(G) 1.G.1 a. Distinguish between defining attributes (e.g., triangles are closed and three-sided) versus nondefining attributes (e.g., color, orientation, overall size) b. Build and draw shapes to possess defining attributes Quarter 2 Continued 1.G.2 1.G.3 a. Compose twodimensional shapes rectangles into two and Partition circles and (rectangles, squares, four equal shares. trapezoids, triangles, halfcircles, and quarter- triangles together to make Compose shapes (put two circles) to create a a quadrilateral) composite shape. Understand for these b. Compose threedimensional shapes decomposing into more examples, that (cubes, right rectangular equal shares creates prisms, right circular smaller shares. cones, and right circular Describe the shares cylinders) to create a using the words halves, composite shape. fourths, and quarters, and use the phrases c. Compose new shapes half of, fourth of, and from the composite shape. quarter of. Describe the whole as Page 3
5 Measurement & Geometry(G) 1.MD.1 Order and compare objects by length. a. Order three objects by length. b. Compare the lengths of two objects indirectly by using a third object. 1.MD.2 Express length and understand what the length of an object represents. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. a. Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; b. Understand that the length measurement of an 1.MD.3 Review, tell and write time in hours and half-hours using analog and digital clocks. 1.MD.4 Organize and represent categories and Interpret categories. Ask and answer questions about: the total number of data points how many in each category how many more or less are in one category than in another. Page 5
7 Measurement & Geometry(G) 1.MD.1 Order and compare objects by length. a. Order three objects by length. b. Compare the lengths of two objects indirectly by using a third object. 1.G.3 Partition circles and rectangles into two and four equal shares. Compose shapes (put two triangles together to make a quadrilateral) Understand for these examples, that decomposing into more equal shares creates smaller shares. Describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as 1.MD.2 Express length and understand what the length of an object represents. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. a. Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; b. Understand that the length measurement of an object is the number of Quarter 4 Continued 1.MD.3 Review, tell and write time in hours and half-hours using analog and digital clocks. 1.MD.4 Organize and represent categories and Interpret categories. Ask and answer questions about: the total number of data points how many in each category how many more or less are in one category than in another. All standards will be reviewed in the final four weeks of the year. Page 7
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Measure lengths indirectly and by iterating length units (Standards 1 2). Standard 1.MD.1 Order three objects by length; compare the lengths of two objects indirectly by using a third object. Understand
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CUMBERLAND COUNTY SCHOOL DISTRICT Benchmark Assessment 1 Instructional Timeline: Weeks 1-9 Topic(s): Unit 1: Numbers to 10 K.CC.3: Write numbers from 0-20. Represent a number of objects with a written
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Kindergarten Math Essential Standards 2017-2018 Trimester 1 Trimester 2 Trimester 3 Counting and Cardinality Counting and Cardinality Counting and Cardinality K.CC.2 - Count forward beginning from a given
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NO.3.2.4 Solve problems using a variety of methods and tools objects, mental computation, paper and pencil, and with and without appropriate technology NOTE: The skills of this SLE are to be utilized throughout
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MATHEMATICS CURRICULUM REVIEW YEAR 2011-2012 K GRADE 12 TABLE OF CONTENTS PAGE Mathematics Curriculum Philosophy 2 Mathematics K -12 Curriculum Committee 2 Standards for Mathematical Practice 3 Mathematics
### Subject Math Third Grade. Goal: 1- The learner will model, identify, and compute with whole numbers through 9,999.
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COMMON CORE STATE STANDARDS FOR Mathematics Contents Introduction................................ 2 Standards for Mathematical Practice................ 4 How to Read the Grade Level Standards.............
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Operations and Algebraic Thinking [OA] Represent and solve problems involving multiplication and division. Understand properties of multiplication and the relationship between multiplication and division.
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Overview Counting and Cardinality [CC] Know number names and the count sequence. Count to tell the number of objects. Compare numbers. Operations and Algebraic Thinking [OA] Understand addition as putting
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September - Unit 1: Place Value and Numeration/Addition and Use hundred charts and number lines. Place Value October Subtraction Read and write numbers to 1,000. Pre- What is place value? Order numbers
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HuggingFaceTB/finemath
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### Lesson Plan Title : Drawing Nets of 3 Dimensional Objects
Age Range:
Overview and Purpose:
Students will be able to use concrete materials to help them understand the concept of nets. Being able to take an object and spread it out flat is much easier to see in concrete terms rather than in the abstract.
Objective:
The student will be able to draw a net of three different objects. Two of those objects will be able to be laid flat and one will not.
Resources:
One cereal box for each pair of students (cut off the six overlapping tabs, three on each end)
One ice cream cone wrapper (with the lid taped onto part of it) for each pair of students
One short round candle for each pair of students
Math journals
Activities:
Discuss that a net is a 2 dimensional representation of a 3 dimensional object. Give each pair of students a cereal box, ice cream cone wrapper, and a candle. Have them unroll the ice cream wrapper so that it lays flat. Explain that this is a net. If they close the wrapper back up it makes a cone. Have them sketch the net in their math journals.
Have the students work with their partner to turn the cereal box into a net and sketch it in their math journals. Point out that six of the extra flaps are missing since they overlap on the ends.
Explain that often you have to visualize what the net will look like since things cannot be laid flat. Have them try to do this by sketching the net of the candle in their math journals. Come back together and have some students show their drawings. Make sure they included the ends of the candle in their nets.
Closure:
Nets can be a very difficult thing for students to visualize. Giving them concrete examples that they can manipulate will help them understand the concept. If more practice is needed, students could create their own nets and 3-D objects out of paper and tape.
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HuggingFaceTB/finemath
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I got the chance to read this paper on Distributional Bellman published by DeepMind in July. Glossing over it the first time, my impression was that it would be an important paper, since the theory was sound and the experimental results were promising. However, it did not generate as much noise in the reinforcement learning community as I would have hoped. Nevertheless, as I thought the idea of Distributional Bellman was pretty neat, I decided to implement it (in Keras) and test it out myself. I hope this article can help interested readers better understanding the core concepts of Distributional Bellman.
### Q Learning Recap
To understand Distributional Bellman, we first have to acquire a basic understanding of Q Learning. For those of you are are not familiar with Q Learning, you can refer to my previous blog for more information on the subject. To recap, the objective of Q Learning is to approximate the Q function, which is the expected value of the total future rewards by following policy $\pi$.
where $\pi$ refers to the policy, $s$ represents the state input and $a$ is an action chosen by the policy $\pi$ at state $s$. $R_t$ is the sum of discounted future rewards.
At the center of Q Learning is what is known as Bellman equation. Let $\pi^*$ represents the optimal policy. We sample a transition from the environment and obtain <$s$, $a$, $r$, $s^\prime$>, where $s^\prime$ is the next state. The Bellman equation is a recursive expression that relates the Q functions of consecutive time steps.
Why is it so important? Bellman equation basically allows us to iteratively approximate the Q function through temporal difference learning. Specially, at each iteration, we seek to minimize the the mean squared error of the $Q(s,a)$ (prediction term) and $r + \gamma\max_{a^\prime} Q(s^\prime,a^\prime)$ (target term)
In practice, we usually use a deep neural network as the Q function approximator and applies gradient descent to minimize the objective function $L$. This is known as Deep Q Learning (DQN). Once we obtain a reasonably accurate Q function, we can obtain the optimal policy through
That’s all we need to know about Q Learning. Let’s get to Distributional Bellman now.
### So What Exactly is Distributional Bellman?
The core idea of Distributional Bellman is to ask the following questions. If we can model the Distribution of the total future rewards, why restrict ourselves to the expected value (i.e. Q function)? There are several benefits to learning an approximate distribution rather than its approximate expectation.
Consider a commuter who drives to work every morning. We want to model the total commute time of the trip. Under normal circumstances, the traffic are cleared and the trip would take around 30 minutes. However, traffic accidents do occur once in a while (e.g. car crashes, break down in the middle of the highway, etc), and if that happens, it will usually cause the traffic to be at a standstill and add an hour to the trip (i.e. 90 minutes). Let’s say traffic accident like that happens once every 5 days on average. If we use expected value to model the commute time of the trip, the expected commute time will be 30 + 60 / 5 = 42 minutes. However, we know such expected figure is not so meaningful, since it vastly overestimates the commute time most of the time, and vastly underestimates the commute time when traffic accidents do occur. If instead we treat the total commute time as a random variable and model its distribution, it should look like this:
Notice that the distribution of commute time is bimodal. Most of the time the trip would take 30 minutes on average, however, if traffic accident occurs, the commute time would take 90 minutes on average. Given the full picture of the distribution, next time when we head out to work, we’d better look up the traffic situation on the highway. If traffic accident is reported, we can choose to bike to work which would take around 60 minutes, which could save us 30 minutes!
#### Choosing Action based on Distribution instead of Expected Value
In reinforcement learning, we use the Bellman equation to approximate the expected value of future rewards. As illustrated in the commute time example above, if the environment is stochastic in nature (occurrence of traffic accidents) and the future rewards follow multimodal distribution (bimodally distributed commute time), choosing actions based on expected value may lead to suboptimal outcome. In the example above, if we realize there is a traffic accident and it will likely take 90 minutes to get to office, the optimal action would be to bike even though the expected commute time of biking is 60 minutes which is larger than the expected commute time of driving (42 minutes).
Another obvious benefit of modeling distribution instead of expected value is that sometimes even though the expected future rewards of 2 actions are identical, their variances might be very different. If we are risk averse, it would be preferable to choose the action with smaller variance. Using the commute time example again, if we have 2 actions to choose from: driving or taking the train, both actions have the same expected commute time (i.e. 42 minutes), but taking the train has smaller variance since it does not get affected by unexpected traffic conditions. Most people would prefer taking the train over driving.
#### Formulation of Distribution Bellman
Without further due, let’s get to the definition of Distributional Bellman. It’s actually quite simple and elegant. We simply use a random variable $Z(s,a)$ to replace $Q(s,a)$ in the Bellman equation. Notice that $Z$ represents the distribution of future rewards, which is no longer a scalar quanity like Q. Then we obtain the distributional version of Bellman equation as follows:
This is called the Distributional Bellman equation, and the random variable $Z$ is called the Value Distribution. A caveat is that the equal sign here means the 2 distributions are equivalent. In the paper the author applied Wasserstein Metric to describe the distance between 2 probability distributions and proved the convergence of the above distributional Bellman equation. I am not going to go through the mathematical details. Interested readers can find the proof in the paper.
Similar to Q Learning, we can now use the Distributional Bellman equation to iteratively approximate the value distribution $Z$. Sounds easy enough. But before we do that, there are 2 issues we have to address. The first issue is how to represent the value distribution $Z$ since it’s not a scalar quantity like Q? The second issue is how to “minimize the distance” between the 2 value distributions $Z$ and $Z^\prime$ so that temporal difference learning can be performed?
To address our first concern, the paper proposed the use of a discrete distribution parameterized by the number of supports (i.e. discrete values) to represent the value distribution $Z$. For example, if the number of supports equals 10, then the domain of the distribution will be 10 discrete values uniformly spaced over an interval. This discrete distribution has the advantages of being highly expressive (i.e. it can represents any kind of distributions, not limited to Gaussian) and computationally friendly. Moreover, Cross Entropy loss can be used to quantify the distance between 2 discrete distributions $Z$ and $Z^\prime$, since they share the same set of discrete supports.
Here is a visual representation of the update step of Distributional Bellman equation:
First we sample a transition from state $s$ with action $a$ and obtain next state $s^\prime$ and reward $r$. The next state distribution $Z(s^\prime,a^\prime)$ looks like this
We then scale the next state distribution with $\gamma$ and shift the resulting distribution to the right by $r$ to obtain our target distribution $Z^\prime = r + \gamma Z(s^\prime,a^\prime)$
Last of all, we “project” the target distribution $Z^\prime$ to the supports of the current distribution $Z$ simply by minimizing the cross entropy loss between $Z$ and $Z^\prime$.
Intuitively, we can view this as an image classification problem. The inputs to the model can be screen pixels sampled from game play (e.g. Atari Pong, Breakout, etc). Each discrete support of $Z$ represents a unique “category”. During the update step, the probability mass of each “category” of $Z^\prime$ will serve as “fake ground truth labels” to guide the training. The “fake labels” are analogous to the binary ground truth labels we use for classification problems.
Now it’s time to go through the meat of this article, which is a feasible algorithm to implement Distributional Bellman called C51.
### Categorical “C51” Algorithm
C51 is a feasible algorithm proposed in the paper to perform iterative approximation of the value distribution Z using Distributional Bellman equation. The number 51 represents the use of 51 discrete values to parameterize the value distribution $Z$. Why 51 you may ask? This is because the author of the paper tried out different values and found 51 to have good empirical performance. We will just treat it as the magic number.
C51 works like this. During each update step, we sample a transition from the environment and compute the target distribution $Z^\prime$ (i.e. scale the next state distribution by $\gamma$ and shift it by reward $r$), and uses $Z^\prime$ to update the current distribution $Z$ by minimizing the cross entropy loss between $Z$ and $Z^\prime$. The pseudo code of C51 Algorithm is provided by the paper:
Look a bit confused? No worries. I have implemented C51 in Keras (code can be found here) which you can freely reference. The implementaion is tested on the VizDoom Defend the Center scenario, which is a 3D partially observation environment.
#### VizDoom Defend the Center Environment
In this environment, the agent occupies the center of a circular arena. Enemies continuously got spawned from far away and gradually move closer to the agent until they are close enough to attack from close range. The agent is equipped with a handgun. With limited bullets (26 in total) and health, its objective is to eliminate as many enemies as possible while avoid being attacked and killed. By default, a death penalty of -1 is provided by the environment. In order to facilitate learning, I enriched the variety of rewards (reward shaping) to include a +1 reward for every kill, and a -0.1 reward for losing ammo and health. I find such reward engineering trick to be quite crucial to get the agent to learn good policies.
#### C51 Keras Implementation
Let’s briefly walk through the implmentation. Similar to DQN, we first use a deep neural network to represent the value distribution $Z$. Since the inputs are screen pixels, the first 3 layers are convolutional layers:
The neural network outputs 3 sets of value distribution predictions, one for each action (i.e. Turn Left, Turn Right, Shoot). Each set of prediction is a softmax layer with 51 units.
For our problem the action_size is 3 (i.e. Turn Left, Turn Right, Shoot) and num_atoms is the number of discrete values (i.e. 51).
Most of the logic of the algorithm is contained in the update step. First we sample a minibatch of sample trajectories from the Experience Replay buffer and initialize the corresponding states, reward, and targets variables:
The variable m_prob stores the probability mass of the value distribution $Z$.
Next, we carry out a forward pass to get the next state distributions.
Notice that the model outputs 3 set of value distributions, one for each action. We really only need the one with the largest expected value to perform the update (similar to $max_{a^\prime} Q(s^\prime,a^\prime)$ in Q Learning).
We then compute target distribution $Z^\prime$ (i.e. scale by $\gamma$ and shift by reward $r$) and “project” it to the 51 discrete supports.
Last, we call Keras fit() function to minimize the cross entropy loss with gradient descent.
That’s it! I strongly encourage you to try out the code in your favorite environment. Feel free to reach out to me if you have trouble getting to code to work.
### Experiment
15,000 episodes of C51 was run on the VizDoom Defend the Center scenario. Average kill counts (moving average over 50 episodes) was used as the metric for performance. We really want to figure out 2 things. First, we want to know if C51 works on a non-trivial 3D partially observable environment. Second, we want to compare its performance with Double DQN (DDQN) and see if the result concurs with that of the paper.
Here is a video of C51 agent playing an episode
Here is the performance chart of C51 and DDQN
The first thing we notice is that C51 does work! A random agent can only get an average kill count of 1 by firing bullets randomly. In contrast, C51 was able to converge to a good policy pretty quickly and already reached an average kills of 7 in the first 1000 episodes. Even though learning started to slow down after episode 1000, we still see a steady and monotonic improvement over the remaining 14,000 episodes.
In addition to the performance chart, it is worthwhile to visually inspect the value distributions learned by C51 to gain a deeper understanding of how C51 works.
Figure 1: At the beginning of episode. Notice that the value distributions learned by C51 are very smooth and closely resemble Gaussian. Since there are still plenty of ammo left (bottom left corner indicates there are 20 left), individual actions (i.e. Turn Left, Turn Right, Shoot) should not affect the value distributions much. They pretty much look identical.
Figure 2: The episode progresses. A pink monster attacks from the left. Ammo starts to run out (5 left) so there is not much room to fire and miss the target. This is reflected by the “leftward shift” of the value distribution that belongs to shooting (bottom distribution). The decision to shoot and miss will certain result in 1 fewer kill.
Figure 3: Towards the end of episode. Ammo runs out. Value distributions collpase to zero (actually slight negative since there is a negative reward at the end of the episode for getting killed)
It’s glad to see that the value distributions learned by C51 make sense and are highly interpretable. I strongly encourage you to plot out the distribution and check if they make sense for your particular environment, even if the algorithm works.
#### Comparison with DDQN
Now we answer our second question, which is how C51 stacks up against DDQN. We can see from the performance chart that C51 has a noticeable lead in average kills overall. Due to their similarly in nature (i.e. both rely on Bellman updates), their overall convergence rates and score variances are very similar. Both algorithms showed impressive convergence rate in the first 1000 episodes, DDQN even outperformed C51 briefly from episode 1000 to 3000. However, after episode 3000, C51 caught up and surpassed DDQN and have maintained that lead (around 1 to 2 average kills) for the remaining 1200 episodes. In fact, the gain in performance for C51 over DDQN is not as big as I would have expected, considering that the paper reported a doubling of performance on the Atari Learning Environment (ALE). Perhaps VizDoom Defend the Center is not the ideal environment to showcase the true power of Distributional Bellman.
### Conclusion
In my opinion, Distributional Bellman is a very interesting and theoretically sound way to model reinforcement learning problem. As mentioned in the article and testified by many experiements, there are many benefits of modeling value distribution instead of the expected value. There is also a simple and feasible implementation C51 that consistently outperforms DDQN in both Atari and VizDoom environments. I would love to see higher adoption of distributional methods. Last of all, my Keras implementation of C51 can be found in my github. Feel free to use it for your own problem.
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open-web-math/open-web-math
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# 59 is 312 Percent of what?
## 59 is 312 Percent of 18.91
%
59 is 312% of 18.91
Calculation steps:
59 ÷ ( 312 ÷ 100 ) = 18.91
### Calculate 59 is 312 Percent of what?
• F
Formula
59 ÷ ( 312 ÷ 100 )
• 1
Percent to decimal
312 ÷ 100 = 3.12
• 2
59 ÷ 3.12 = 18.91 So 59 is 312% of 18.91
Example
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HuggingFaceTB/finemath
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## Precalculus (6th Edition)
$x=4$
Cube both sides to obtain: $$(\sqrt{2x+1})^3=(\sqrt{9})^3 \\2x+1=9$$ Subtract $1$ to both sides: $$2x+1-1=9-1 \\2x=8$$ Divide $2$ to both sides: $$\dfrac{2x}{2}=\dfrac{8}{2} \\x=4$$
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HuggingFaceTB/finemath
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# Number of Normal subgroups In a p-Group
Dear all,
Does someone know of any paper/method that enables us counting/estimating the number of normal subgroups of some p-group of order $p ^n$ ($n$ is some natural number ? ) .
Is there anyway we can count the maximal subgroups it has (i.e.- the groups of order $p^{n-1}$ ? ) ?
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en.wikipedia.org/wiki/Hall_algebra In abelian groups count of subgroups with fixed factor is related to Hall-Littlewood polynoms. What happens for non-abelian - I asked MO: mathoverflow.net/questions/107537/… with no reply – Alexander Chervov Oct 2 '12 at 5:18
Same question simultaneously asked at m.se, math.stackexchange.com/questions/205681/… --- coincidence? – Gerry Myerson Oct 2 '12 at 5:51
@Alexander Chevov: Thanks ! I had no idea about the Hall Algebra notion... But I'm still skeptic about it... Have you got any paper the gives some more details about it? Thanks again! – Jason Mraz Oct 2 '12 at 12:38
I think MacDonald's book "Symmetric functions ... " discuss this... I am not sure - I can send you file of the book, if you need. There have been modern developments about Hall algebras, which go very very far from p-groups - they are surveyed in arxiv.org/abs/math/0611617 – Alexander Chervov Oct 3 '12 at 11:00
That's excatly the thing... I only need kind of "simple" estimates and bounds on the number of subgroups... I'll try to go over the lecture notes you sent and I might find something useful in them... Thanks a lot ! (I'll try to look for the book you mentioned) – Jason Mraz Oct 3 '12 at 19:03
For a $p$-group $P$, the number of maximal subgroups is $\sum_{k=0}^r p^k$ where $r$ is the minimum size of a generating set for $P$. You can see this from looking at the maximal subgroups of $P/\Phi(P)$, which is elementary abelian of order $p^r$.
What I can tell you is that there is at least one normal subgroup for every power of $p$ up to the order of the group. Sylow theory style orbit counting gives us that the number of normal subgroups of each order $p^k$ is going to be congruent to $1 \mod{p}$, so the total number of normal subgroups in a $p$-group of order $p^n$ will then be congruent to $n+1 \mod{p}$.
EDIT: I thought of a bound.
$n+1$ is the lower bound, attained by the cyclic group of order $p^n$. There must be at least one normal subgroup for every prime power divisor, so this is the lowest it can go.
On the other hand, I claim that elementary abelian groups $E_{p^n}$ contain the largest number of normal subgroups. This is because it has the maximum rank of all groups of order $p^n$. Thinking of $E_{p^n}$ as an $\mathbb{F_p}$-vector space, we obtain the number of subspaces by $$\mathcal{N}(E_{p^n})=\sum_{m=0}^{n}\prod_{k=0}^{m-1}\frac{p^n-p^k}{p^m-p^k}.$$ Here we count the number of ordered combinations of $m$ linearly independent vectors in $\mathbb{F_p}^n$, then divide by the number of possible bases of an $m$-dimensional subspace. Summing over $m$ we have the total number of normal subgroups in $E_{p^n}$.
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Forgive my ignorance: how do Sylow theorems give you information about subgroups of a p-group? – Nick Gill Oct 2 '12 at 8:37
Indeed, in a cyclic group G of order p, there are two normal subgroups: G and {1}. And 2 is not 1 mod p... Do you mean the number of normal subgroups OF A GIVEN ORDER will be 1 mod p? I could maybe believe that but I'd have to think about how I'd prove it. – Nick Gill Oct 2 '12 at 8:41
From Sylow Theorem, we see that the number of subgroups of a given order in a finite $p$-group is congruent to 1 mod $p$. Maybe Alexander means this. – Wei Zhou Oct 2 '12 at 9:48
@Wei Zhou, I've never seen Sylow theorems applied to subgroups of a $p$-group. So, while I can believe the result you state, I'm not sure how you use Sylow to prove it. Maybe it's just that the method by which we prove the $1\mod p$ part of the Sylow theorems can be applied here (?). – Nick Gill Oct 2 '12 at 13:35
@Nick: Let $G$ be a $p$-group of order $p^n$, and S the set of all subgroups of order $p^m$. Let $P \in S$. Then $P$ can acts on $S$ by conjugation. By counting the orbits of this action, we see $|S|$ congruent to 1. This trick is use to prove Sylow theorem by someone. So in some book I can not find, I think this is also called Sylow theorem. – Wei Zhou Oct 2 '12 at 15:50
The wikipedia article on p-groups reminded me that
Every normal subgroup of a finite p-group intersects the center nontrivially.
This implies immediately that minimal normal subgroups of a p-group $G$ will be central. This fact can be used to prove the statement that Wei Zhou made:
A $p$-group of maximal class and size $p^n$ has the least number of normal subgroups of all groups of order $p^n$.
(If I'm thinking straight this number is $n+1$ and the bound is also achieved by the cyclic group of order $p^n$.)
It seems to me that one might be able to prove something a little stronger using an inductive argument: counting the minimal normal subgroups in the center $Z$, and then counting the normal subgroups in $G/Z$, and then putting these two numbers together... It's that last bit that's going to be tricky though. If the center is cyclic, then everything is fine but when it's not cyclic, eek...
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Dear @Nick: Thanks a lot ! I 'll be glad if you'll be able to tell me what do you mean by a "group of maximal class" ... After verifying this little detail, I'll reread your answer in order to check again that I understand it... Thanks again! – Jason Mraz Oct 2 '12 at 12:43
By class', I mean nilpotency class' i.e. the length of the upper (or lower) central series. A group $G$ of order $p^n$ (with $n>2$ )is of maximal class if this length is $n-1$; in this case $G$ has center $Z(G)$ cyclic of order $p$, then $G/Z(G)$ has center cyclic of order $p$. This pattern continues until you get to a normal group of index $p^2$ in $G$ at which point one has an abelian quotient. Note that, a priori, there may be more than one isomorphism class of group of order $p^n$ of class $n-1$ - not all of them will necessarily have the minimal number of normal subgroups. – Nick Gill Oct 2 '12 at 13:16
I think that the $p$-groups of maximal class have the least number of normal subgroups except for the cyclic groups. If I'm not mistaken $p$-groups of maximal class and order $p^n$ will have one normal subgroup for each $p$th power up to $p^{n-2}$, then a few normal subgroups of order $p^{n-1}$, as opposed to cyclic groups which of course have unique normal subgroups for every power of $p$. – Alexander Gruber Oct 2 '12 at 22:58
Dear @Nick Gill and @Alexander: Where can I find proofs for the facts you mention? I can't see this straight away... Can you give me some reference for the proof of these facts? Thanks ! – Jason Mraz Oct 3 '12 at 9:16
@Alexander, if $G/ G_1$ is cyclic of order $p^2$ (where $G_1$ is the first term in the lower central series), then I think one gets $n+1$ normal subgroups. HOWEVER I do not know enough about $p$-groups of maximal class to be sure that this can happen! If all $p$-groups of maximal class have $G/G_1$ elementary abelian, then I agree with you. – Nick Gill Oct 3 '12 at 13:09
As I know, for the p-group of maximal class, the number of normal subgroups are known. And the number of normal subgroups in p-group of maximal class the the smallest.
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Reference please – Alexander Chervov Oct 2 '12 at 15:38
About the minimality of normal subgroup, you can get this fact mentioned in the comment of Alexander Gruber following the answer of Nick Gill, from the paper by N. Blackburn, On a special class of p-groups. By the way, the above paper is an important paper for the theory of p-group – Wei Zhou Oct 3 '12 at 1:19
Thank you ! PS as a remark may say that adding more details to the answers would be more valuable in general and more easy convince the readers to press on +1 button:) – Alexander Chervov Oct 3 '12 at 10:50
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open-web-math/open-web-math
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Chapter 5.5, Problem 3CP ### Calculus: An Applied Approach (Min...
10th Edition
Ron Larson
ISBN: 9781305860919
#### Solutions
Chapter
Section ### Calculus: An Applied Approach (Min...
10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
1 views
# Checkpoint 3 Worked-out solution available at LarsonAppliedCalculus.comFind the area of the region bounded by the graph of y = x 2 − x − 2 and the x -axis.
To determine
To calculate: The area bounded by graph of the function y=x2x2 and x-axis.
Explanation
Given Information:
The provided function is y=x2x2 and x-axis.
Formula used:
The function f(x) and g(x) are the continuous function on interval [a,b] and g(x)f(x) for all x in [a,b], then, the area bounded by the graphs of f(x) and g(x),x=a and x=b is,
A=ab[f(x)g(x)]dx
Calculation:
Consider the function y=x2x2.
Now, draw the graph by using point plotting method.
Substitute 0 for x in the function y=x2x2.
y=0202=2
Substitute 1 for x in the function y=x2x2.
y=1212=2
Substitute 1 for x in the function y=x2x2.
y=(1)2(1)2=1+12=0
Substitute 2 for x in the function y=x2x2.
y=(2)2(2)2=4+22=4
Now, make the table for the corresponding value of x and y for the function y=x2x2 is shown below,
x y=x2−x−2 0 −2 1 −2 −1 0 −2 4
Draw the graph of the function y=x2x2 and x-axis by using above table.
Here, the limit of bounded region is not prescribed that’s why the provided function intersect x-axis
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HuggingFaceTB/finemath
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Q:
# What is 5 feet 6 inches in inches?
A:
A measurement of 5 feet 6 inches is equivalent to a total of 66 inches. This is calculated by first converting feet into inches, with 12 inches equaling 1 foot, then multiplying the 12 inches by 5. The resulting 60 inches is then added to the original 6 inches for a total of 66 inches.
Know More
## Keep Learning
The inch is a standard unit of length in the United States, Canada and the United Kingdom. Converting this measurement of 66 inches to yards results in a total of 1.83 yards. A comparable measurement of 66 inches in the metric system is 1.68 meters.
Sources:
## Related Questions
• A:
A length of 5 feet 7 inches, in inches, comes out to 67 inches. There are 12 inches in a foot, so it is just a matter of multiplying 5 by 12 and then adding the remaining 7 inches.
Filed Under:
• A:
60 inches is equivalent to 5 feet in the U.S. customary and Imperial measurement systems. In these systems, 12 inches is the equivalent of 1 foot.
Filed Under:
• A:
Seven feet is equivalent to 84 inches. There are 12 inches in one foot according to the English system of measurement. There are three feet in a yard. Eighty-four inches can also be stated as two yards and one foot.
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HuggingFaceTB/finemath
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# Probabilities and Sum of Normal Distributions
Suppose I have:
$V_1 = X_1+A$ where $A>0$ is some constant $V_2 = X_2+B$ where $B>0$ is some constant
Furthermore assume that $X_1$ and $X_2$ are independent and distributed with a standard normal distribution $N(0,1)$.
I want to get $P(V_1+V_2 > 1\ \&\ V_1,V_2 \in [0,1] )$
Here's what I'm thinking:
$V_1+V_2 > 1 \implies X_1+X_2>1-A-B \implies X_1>1-A-B-X_2$ $V_2 \in [0,1] \implies -A<X_2<1-A$
Then, I could do a double integral one from $1-A-B-X_2$ to $\infty$ for $X_1$ and the other from $-A$ to $1-A$ for X_2, but I'm still not sure what I'm integrating over, e.g pdf of the sum?
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You want $0 \le V_1 \le 1$ but also $1-V_2 \lt V_1$.
Since you have $0 \le V_2 \le 1$, you can turn your constraints on $V_1$ into $1-V_2 \lt V_1 \le 1$.
Since $V_1 = X_1+a$ and $V_2 = X_2+b$ (it is conventional to use lower case for constants), these become $$-b \le X_2 \le 1-b$$ $$1-a-b-X_2 \lt X_1 \le 1-a$$ and so your integral becomes $$\int_{x_2=-b}^{1-b} \int_{x_1=1-a-b-x_2}^{1-a} \phi(x_1)\phi(x_2) \, dx_1 \, dx_2$$ where $\phi(x)$ is the probability density function of a standard normal distribution.
Thanks for the help. I'm wondering if that integral can be simplified. I end up with an integral $\int_{-b}^{1-b} \Phi(1-a-b-x_2) \phi(x_2)dx_2$ Can this be simplified? Note, I have other terms, but was able to simplify them. – Greg Apr 22 '12 at 2:20
You should have got $\int_{-b}^{1-b} (\Phi(1-a) - \Phi(1-a-b-x_2)) \phi(x_2)dx_2$ or $\Phi(1-a)(\Phi(1-b)-\Phi(-b))+ \int_{-b}^{1-b} \Phi(1-a-b-x_2) \phi(x_2)dx_2$. Perhaps you did. I doubt there is a further simplification. – Henry Apr 22 '12 at 10:03
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HuggingFaceTB/finemath
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# Probability Theory watch
1. I'm doing exponential distribution. A call costs 15p per minute or part of. Cost of a call is X pence. T minutes is the duration with pdf
f(t)= ae^(-at).
I need to prove that E(X) = 15/(1-e^(-a))
I already know P(X=15r) = e^(-ra).(e^(a)-1)
I'm doing exponential distribution. A call costs 15p per minute or part of. Cost of a call is X pence. T minutes is the duration with pdf
f(t)= ae^(-at).
I need to prove that E(X) = 15/(1-e^(-a))
I already know P(X=15r) = e^(-ra).(e^(a)-1)
You just then need to sum 15re^(-ra).(e^(a)-1) to find the expectation.
3. (Original post by RichE)
You just then need to sum 15re^(-ra).(e^(a)-1) to find the expectation.
Are you able to go into more detail about the layout and how to start that, I'm still a bit confused. Thank you
Are you able to go into more detail about the layout and how to start that, I'm still a bit confused. Thank you
The sum of kx^k can be worked out by differentiating the sum of x^k and the latter is a geometric series.
5. (Original post by RichE)
The sum of kx^k can be worked out by differentiating the sum of x^k and the latter is a geometric series.
So I calculated my first term as 1 - e^(-a) and my common ratio as e^(-a) but when I sub unto geometric series formula I get (1 - e^(-a))^15 which is not what I want
So I calculated my first term as 1 - e^(-a) and my common ratio as e^(-a) but when I sub unto geometric series formula I get (1 - e^(-a))^15 which is not what I want
As I said above the series isn't a geometric series, but it is the derivative of one.
As an alternative perhaps you know the binomial series for (1-x)^(-2).
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HuggingFaceTB/finemath
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
You are not logged in.
## #1 Re: Help Me ! » Prove Fibonacci Sequence » 2010-01-06 08:32:55
How far have you got, fibbingfibonacci? I have a solution to your problem, you need to use induction.
## #2 Re: Help Me ! » grid pattern » 2007-03-23 01:33:25
You see, however you pick your 2 x 2 box, all the numbers will have a relation to eachother according to your grid. Let the first number be called A, then the next number will be A+1, A+2, A+3,... filling up your entire grid.
Now, however you pick your square doesn't matter. Let the number in the first row and first column (of your square) be called x, then the next one is always x+1. The bottom ones in your square will always be x+10 and x+11, such as
x x+1
x+10 x+11
Do you see why?
Do you know how to proceed from here? All you have to do know is to set up the expression you stated, multiplying diagonally and finding the difference.
(x+1)*(x+10) - x*(x+11) = ??
Edit: I see Jane already posted, but maybe an explanation will help =P
Welcome aboard
## #4 Re: Help Me ! » help with quadratic sequences please. » 2007-02-28 02:41:08
10+1, 40+4, 100+10, 200+20
Skip the second terms for a while and let us simplify the sequence: 1, 4, 10, 20
It's much easier now! We get: n/6(n+1)(n+2)
But that was the simplified model, let's continue:
--> 5n/3(n+1)(n+2) corresponds to the sequence 10,40,100,200. Now we need to add 1/10 of these:
5n/3(n+1)(n+2)+5n/30(n+1)(n+2) =
= 10n/6(n+1)(n+2)+1n/6(n+1)(n+2) =
= 11n/6(n+1)(n+2)
Inserting n=1,2,3,4 you'll see it works out fine for your sequence
## #5 Re: Help Me ! » PASCAL RIDDLE - Given to me by my Future Girlfriend ! » 2007-02-17 10:25:34
If 65 67 83 is ascii, it would be ACS. If we relate that to time, the closest I get is the Australian Central Standard time. That would be UTC/GMT + 9:30h. Just throwing something out there... does Australia say anything at all?
Also, I was thinking that 65 67 83 could be rows or something in Pascal's triangle. 65 would be row 6, number 5 in the sequence. But 67 doesn't quite work out.
The one thing that comes to mind is that they represent degrees, minutes and seconds like some sort of location. As with 9 000, I'm clueless.
## #6 Re: Help Me ! » help with number sequence » 2007-02-07 09:17:32
x is the position of the number in the sequence. Check it for f(1), f(2), f(4) and f(6) in the sequence, it's correct It doesn't give integers for f(3) and f(5) though. Stanley_Marsh made me want to try it out heh.
## #7 Re: Help Me ! » Find the next number... » 2006-12-08 19:34:34
check this out:
http://www.research.att.com/~njas/seque … &go=Search
The only matching (integer) sequence would be the first of the three, none of the other are followed by either 11,12,13 or 14. But then, the rest are still wrong... If it really is a sequence (can't see it myself), nice find.
## #8 Re: Help Me ! » ????? » 2006-12-08 00:59:15
Check #9 for matrices.
## #9 Re: This is Cool » Nullity? » 2006-12-07 11:17:06
That symbol, uppercase phi, is the reciprocal of the golden ratio. I think he missed something there
## #10 Help Me ! » radii of a lens » 2006-12-06 12:10:45
numen
Replies: 0
I have an optical system of an equiconvex lens and an object located 0.6m from a screen. The object is 0.05m high and it's image is 0.25m on the screen.
I managed to calculate the focal length to f=0.083m by using the fact that the magnification is -5 and the gaussian lens formula, which appears to be reasonable.
The problem is, how do I find the radii of the lens?
## #11 Re: This is Cool » 0.9999....(recurring) = 1? » 2006-12-06 06:14:42
Anthony, take a look at this:
1.00000000...
- 0.99999999...
= 0.00000000...
Since the number of zero's is infinite, it is the same as writing 0 as there can't be anything after the infinitely many zero's (just as 1.000... is the same as 1, or 5.5999... = 5.6). Thus, there is no difference between 1 and 0.999... they are the same thing!
As pointed out already, there is no such thing as 0.0000...1. The only thing it proofs is that you don't understand the concept of infinity.
If you still don't believe it, prove us wrong. Otherwise none will take you seriously.
## #12 Re: Help Me ! » completed equation now what » 2006-11-27 13:41:37
I got confused reading John's posts (sorry ;P) so I did it by using vectors to check his result and got the exact answer
Which is approximately 1.016, but 1 might be good enough for some people This is what I did, maybe someone could check it (I usually don't do this in 2D, and it's really late ).
-----------------
We may easily find two points on the line 3x-4y=2 (meaning: find two solutions to the equation). Just pick a value for x and y so it "fits". Two solutions are Q = (2,1) and P = (6,4), check these. These points are on the line, which we will call L. The direction of this vector can be found through QP = (6-2,4-1) = (4,3). Thus, we can write L in the parametric form
x=2+4t
y=1+3t
We have the point S = (5,2), the distance to L is given by |SR|, where R is a point on L. SR is given by SR=(2+4t,1+3t)-(5,2). Do you see why? This gives SR=(-3+4t,-1+3t). Also note that SR is perpendicular to QP.
This means that 4(2+4t)+3(1+3t) has to be 0. This gives us t=7/11. Let us insert this value for t into SR. We get SR=1/11(-5,10)
The distance is the absolute value for SR. Let us find out what it is!
|SR|=1/11*√(5²+10²) = 1/11*√125
You can apply this technique in 3D systems as well (with x,y,z). It's pretty good to know
## #13 Re: Dark Discussions at Cafe Infinity » 8 October What Happened On 8 October...........??? » 2006-10-04 04:58:18
It doesn't seem to work in Firefox, try IE.
## #14 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-30 10:18:58
I haven't managed to figure out another solution to this problem, all other options I had in mind doesn't seem to work out with 100% accuracy. You think this is the only solution?
## #15 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-24 03:09:22
Dross wrote:
numen wrote:
I'll hold it a bit more though. I'll post my solution on monday if I can remember that
Would you mind leaving it until, say, Wednesday? Some members may not be able to get on the forum during weekends/very often, for whatever reason.
In fact, could you e-mail me your solution? (I don't know what the deal is with PMs on this forum... I don't seem to be able to, anyways!)
Sure thing. I'll e-mail my solution soon, hoping that, in return, you'll e-mail me yours Hopefully you can confirm it's correct.
## #16 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-23 07:19:38
I really have no idea how else to solve this. Your hint is pretty much how I define my groups, if I understand it correctly. If your solution really is different, I'd like to know it as well.
I'll hold it a bit more though. I'll post my solution on monday if I can remember that
## #17 Re: Help Me ! » I'm just not sure, Help please » 2006-09-22 11:00:50
If it's a multiple of 3, it's between 50-59, but not all numbers between 50-59 are multiples of 3. Cancel out those numbers. For the rest between 50-59, how many are multiples of 4? Cancel out those who are not multiples of 4. You following so far?
Good. So what can you conclude now? Exactly, the number is not a multiple of 3 because none of the numbers between 50-59 remains. Cancel out all remaining numbers that are multiples of 3.
Remember, it has to be a number between 50-79, because it's the only definitions you've got.
Remove your second criteria, you've used it up already. Continue the same way with your other two criterias
Edit: Bah, Rod gave the answer away, lol.
## #18 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-21 02:23:54
I think I have a solution now, it wasn't that hard when you think about it. But it'll take quite some time, but you said the time taken was irrelevant
## #19 Re: Puzzles and Games » The Prisoners and the Lightbulb » 2006-09-21 00:07:27
So, basically, a prisoner must be absolutely sure that everyone has visited the central room at least once when he says he wants to go home. But they can only contact eachother through the switch... So they must be clever enough to figure out what n is, right? or perhaps be able to count to 100 through the switch? I guess that's two options.
We might have to group the prisoners somehow before they are being put into the cells.
Is the switch on or off from the start?
## #20 Re: Puzzles and Games » 1 million! » 2006-09-15 08:29:53
bah, no fun trying anymore
No problem : )
## #22 Re: Help Me ! » tough limit problem » 2006-09-13 03:21:19
It's not even in my calculus textbook.
## #23 Re: Help Me ! » tough limit problem » 2006-09-12 20:49:49
I don't know what L'hopitals rule is, but here's what I did (hoping that I didn't use it!):
## #24 Re: Help Me ! » determinant » 2006-09-11 06:31:26
I can see where you're going, but I was hoping for a better way of doing it. Working with 60 2*2 determinants doesn't sound like a lot of fun =P
## #25 Re: Help Me ! » limit problems » 2006-09-10 10:08:19
I'm guessing you use ε on your y-axis, and δ on your x-axis (like we do over here). If so, consider this problem (same principle, I'm just using x² instead):
Show that x² --> 4 when x --> 2.
We need to show that, as soon as we've picked our ε>0, we can find a δ>0 so that |x²-4|<ε when 0<|x-2|<δ. Don't forget the 0 right there!
|x²-4| = |x+2|*|x-2|
Let ε>0. If we demand |x-2|<1 (namely that 1<x<3), we get |x+2|<5. If we also demand that |x-2|<ε/5 (notice why we picked 5 below, and not any other number), we get:
|x²-4| = |x+2|*|x-2|<5*ε/5=ε
Thus, if δ is the least of the numbers 1 and ε/5, we get that |x²-4|<ε when 0<|x-2|<δ. Maybe someone could pull out a graph of the definition to make things easier...
That's all there is, I think. I don't see where you got that ε/3=δ from though?
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HuggingFaceTB/finemath
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# Infinitude of Primes Via Powers of 2
The following statement directly implies infinitude of primes:
For a positive integer $n$ the expression $f(n)=2^{2^{n+1}}+2^{2^n}+1$ has at least $n+1$ distinct prime factors.
### Proof
The proof is by induction and employs the following identity
(1)
$x^{4}+x^{2}+1=(x^2-x+1)(x^2+x+1).$
Note that, setting, $g(x)=x^{4}+x^{2}+1,$ $f(n)=g(2^{2^{n-1}}).$ The two factors of
(2)
$g(2^{2^{n-1}})=(2^{2^n}+2^{2^{n-1}}+1)(2^{2^n}-2^{2^{n-1}}+1)$
are mutually prime, since each is an odd number while their difference equals $2\cdot 2^{2^{n-1}}$ and, so, the only divisors it has are the powers of $2.$
Now, $f(1)=2^{2^2}+2^{2^1}+1=21=3\cdot 7,$ a number with exactly $2=1+1$ factors.
Assuming the statement holds for an integer $k\ge 1,$ i.e., that $f(k)=2^{2^{k+1}}+2^{2^k}+1$ has at least $k+1$ distinct prime factors, observe that, according to (2),
\begin{align} f(k+1)&=2^{2^{k+2}}+2^{2^{k+1}}+1\\ &= (2^{2^{k+1}}+2^{2^{k}}+1)(2^{2^{k+1}}-2^{2^{k}}+1)\\ &=f(k)(2^{2^{k+1}}-2^{2^{k}}+1), \end{align}
with the second factor having divisors different from those of $f(k),$ which, by the inductive hypothesis has at least $k+1$ distinct prime divisors.
### Remark
It was shown by Richard Green from the Unversity of Colorado that the above proof, although simple, can still be made more transparent.
Let's assume that the two factors in (1), $x^2-x+1$ and $x^2+x+1$ have a common prime multiple, say, $p.$ Then $p$ divides their difference $2x.$ Now note that both numbers $x^2-x+1$ and $x^2+x+1$ are odd for any integer $x.$ making $p$ also odd and, therefore, a divisor of $x$. But this can't be because $p$ then could not divide either if the two factors.
The induction then follows as before.
### References
1. A. Engel, Problem-Solving Strategies, Springer Verlag, 1998, 121-122
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HuggingFaceTB/finemath
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## BdMO National 2013: Secondary 8, Higher Secondary 6
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm
### BdMO National 2013: Secondary 8, Higher Secondary 6
There are \$n\$ cities in the country. Between any two cities there is at most one road. Suppose that the total number of roads is \$n\$. Prove that there is a city such that starting from there it is possible to come back to it without ever traveling the same road twice.
Fatin Farhan
Posts: 75
Joined: Sun Mar 17, 2013 5:19 pm
Contact:
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
Let therebe C ways to take n points such that it is not possible to return to its original position. So, we can take 1 point in \$\$(n-1)\$\$. 2 points be taken in \$\$(n-1)(n-2)\$\$ ........... So \$\$n\$\$ points can be taken \$\$(n-1)(n-2)(n-3) ........(n-n+1)(n-n)\$\$
\$\$C= (n-1)(n-2)........(n-n+1)(n-n) =0\$\$.
So, there is 0 ways to take \$\$n\$\$ points such that it is not possible to return to its original position. So,it is always possible to return to its original position.
"The box said 'Requires Windows XP or better'. So I installed L\$\$i\$\$nux...:p"
sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
@Fatin, Not Clear to me at all! Faulty Solution. First, try to explain it clearly.
You spin my head right round right round,
When you go down, when you go down down......
(-\$from\$ "\$THE\$ \$UGLY\$ \$TRUTH\$" )
asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
We induct on n. The result is trivial for \$n=1,2\$. We assume that it is true for \$n=m\$. Now we prove that it is true for \$n=m+1\$. Among \$m+1\$ cities, if there exists a city which is connected with 0 or 1 city, then there are \$m+1\$ or m roads among the rest m cities.But by our induction hypothesis, there exists at least one city which satisfies the given condition.So we assume that every city is connected with at least 2 cities.If a city has more than 2 connecting roads then,the number of roads will be greater than \$n+1\$. So every city has exactly 2 connecting roads.Then the roads and cities form a polygon and all of the cities satisfies the given condition.Thus our induction is complete .
Fatin Farhan
Posts: 75
Joined: Sun Mar 17, 2013 5:19 pm
Contact:
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
sourav das wrote:@Fatin, Not Clear to me at all! Faulty Solution. First, try to explain it clearly.
I was trying to tell that it is not possible to come back to its original position without ever traveling the same road twice. Let there be \$\$C\$\$ ways to choose \$\$n\$\$ roads such that we can not retun to its original position.
So, 1 road can be chosen in \$\$n\$\$ ways.
2 roads can be chosen in \$\$n(n-1)\$\$ ways.
3 roads can be chosen in \$\$n(n-1)(n-2)\$\$ ways. Bacause we have already a road connecting to its previous city and we don't want to return to its original position.
So, n roads can be chosen in \$\$n(n-1)(n-2)..............(n-n)\$\$.
\$\$ \therefore C= n(n-1)(n-2)..............(n-n)= 0\$\$.
So, there is 0 ways to choose \$\$n\$\$ roads among \$\$n\$\$ cities such that we cannot return to its original city.
Thus, it is always possible to return to its original city, no matter how the \$\$n\$\$ roads are used.
"The box said 'Requires Windows XP or better'. So I installed L\$\$i\$\$nux...:p"
Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm
Contact:
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
Here's my approach-
We will prove this by contradiction. Let's assume that it is possible to make \$n\$ roads in such a way that no matter which city we start from, we cannot come back to it by traversing each road only once.
Let's call a group of connected cities a "Zone". In other words, if there is a way to go from city A to city B in some way, they are in the same "Zone".
If two cities are in the same zone, there is already a way to get from one city to the other. Thus, making a road between two cities in the same zone would make sure that we can go back to the starting city without traversing a road twice and contradict our very claim. So we'll always make roads between cities from different zones .
Let's assume, there is no road initially. Therefore, there are \$n\$ zones (since there are as many isolated cities).
Now, if we make a road, 2 zones get concatenated. In other words, the number of the zones gets reduced by \$1\$ every time we connect two cities from different zone by a new road. Thus, when we are done making \$n-1\$ roads, we'll have only \$(n-n+1) = 1\$ zone left. So, if we have to make the \$n^{th}\$ road, we'll have to connect two cities from the same zone and thus make sure that we can find such a city, starting from which, we can come back to it without traversing a road twice. [Proved]
Learn how to write equations, and don't forget to read
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
sowmitra
Posts: 155
Joined: Tue Mar 20, 2012 12:55 am
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
Ahem...
Fatin Farhan wrote: So, 1 road can be chosen in \$\$n\$\$ ways.
2 roads can be chosen in \$\$n(n-1)\$\$ ways.
3 roads can be chosen in \$\$n(n-1)(n-2)\$\$ ways. Bacause we have already a road connecting to its previous city and we don't want to return to its original position.
So, n roads can be chosen in \$\$n(n-1)(n-2)..............({\color{Red} {n-n}})\$\$.
For the sequence that you showed here, no. of ways in which \$n\$ roads can be chosen is,
\$n(n-1)(n-2)..............({\color{DarkGreen} {n-n+1}})\$
NOT,
\$\$n(n-1)(n-2)..............({\color{Red} {n-n}})\$\$
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;
Siam
Posts: 13
Joined: Sun Mar 17, 2013 3:36 pm
### Re: BdMO National 2013: Secondary 8, Higher Secondary 6
asif e elahi wrote:We induct on n. The result is trivial for \$n=1,2\$. We assume that it is true for \$n=m\$. Now we prove that it is true for \$n=m+1\$. Among \$m+1\$ cities, if there exists a city which is connected with 0 or 1 city, then there are \$m+1\$ or m roads among the rest m cities.But by our induction hypothesis, there exists at least one city which satisfies the given condition.So we assume that every city is connected with at least 2 cities.If a city has more than 2 connecting roads then,the number of roads will be greater than \$n+1\$. So every city has exactly 2 connecting roads.Then the roads and cities form a polygon and all of the cities satisfies the given condition.Thus our induction is complete .
I dont think its necessary for all to have 2 roads. If all have any one has at least 3, then also the condition is satisfied
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HuggingFaceTB/finemath
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# Amortized analysis - increment in ternary counter [closed]
What is the amortized analysis of increment action in a ternary counter that is initialized to 0?
• What did you try? Where did you get stuck? Did you try to modify the argument for a binary counter -- it doesn't feel like it should be very different. – David Richerby Jul 17 at 16:22
The amortized cost per increment will be $$O(1)$$. In order to show it let's use the aggregate method.
0 - 000
1 - 001
2 - 002
3 - 010
4 - 011
5 - 012
6 - 020
7 - 021
8 - 022
9 - 100
10 - 101
11 - 102
12 - 110
13 - 111
14 - 112
15 - 120
...
We can notice that the bits in the 0th place are changing in every increment operation ($$3^0$$). The bits in the 1th are changing in every 3th increment operation ($$3^1$$). The bits in the 2th place are changing in every 9th increment operation ($$3^2$$). And so on.
The total number of changes equals to the sum of times that every bit has changed. Let's assume $$n\lt 3^{k+1}$$ for some $$k$$. So the total number of changes in $$n$$ increments is no more than $$n+\frac n3+\frac n9+\frac n{3^3}+....+\frac n{3^k} < \frac32n$$ Assuming it takes $$O(1)$$ to perform the change of one bit and the associated bookkeeping, the total cost of the sequence is $$O(n)$$. So the amortized cost per operation is $$O(n)/n= O(1).$$
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HuggingFaceTB/finemath
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# Teach Think Elementary
Content, not just cute.
# Place Value in Upper Elementary
By the time students get to 4th and 5th grades, they should already have strong conceptual understanding of the following place value concepts:
-counting by ones, fives, tens, hundreds, thousands, etc.
-each place value is 10 of the place value to the right.
-writing numbers in words and numerals.
-working with expanded form for smaller numbers.-comparing and ordering smaller numbers.
-rounding whole numbers.
-using place value understanding when adding and subtracting.
-multiplying by ten.
Hopefully, the students have been taught to use these concepts, and not just to apply ‘tricks’ to get the answers. If they’ve been using shortcuts (eg, ‘just add a 0’ for multiplying by 10, or ‘circle the place, look at the number to the right, and change everything to 0’ for rounding’, or ‘just carry the 1’ when regrouping in addition), then helping students understand harder place value concepts in upper elementary is going to be a real struggle. If they’ve truly developed the conceptual understanding for these place value concepts, then the more complex ideas in the upper grades should build naturally on what they have already learned. (And the same going forward- what they will learn in middle and high school should build naturally on the place value concepts from upper elementary, making them even more important.)
In upper elementary, the focus shifts to larger/ smaller numbers, including decimals, and to generalizing and applying the base ten concepts. Many of the concepts are the same ideas, just now applied to larger numbers and decimals. In lower grades, students learn to compare numbers to the thousands place. In upper grades, they extend this understanding to comparing larger, multidigit numbers and decimals. In fourth and fifth grade, students work on place value concepts such as:
-understanding that the place value to the right is 1/10 of the one to the left, and the one to the left is 10 times the one to the right.
-use rounding and estimation fluently and apply them as a check on the reasonableness of a number or answer.
-comparing and ordering decimals and larger numbers.
-working with expanded form for decimals and larger numbers.
-apply place value understanding when performing arithmetic.
-applying patterns of zeros and ‘moving the decimal’ when multiplying or dividing by ten.
-beginning to work with exponents (an extension of multiplying by ten).
To help develop these concepts in upper elementary, try:
-relating numbers to money. This works for both very large numbers (a billionaire is worth ONE HUNDRED millionaires), and for decimals (dollars and cents are a very concrete way for kids to understand ones, tenths, and hundredths.)
-working with calculation problems where it makes more sense to apply place value concepts than to use the standard algorithm. 1,000- 999 is a perfect example of this. Many students will struggle trying to borrow from the 1 in the thousands place, across all those zeros, but students with strong place value understanding will see that 999 is only one away from one thousand.
-do a lot of estimating… especially with money. A great way to practice this is by estimating the total grocery bill for a list of items.-continue working with number lines. (They aren’t just for learning to count from 0-100!) Making a decimal number line from 0.00 to 1.00, counting by tenths and hundredths is a great way for students to make sense of how tenths, hundredths, and ones connect and why the zeros at the end of decimals are not always necessary. Try having each pair or group make a decimal number line between two different whole numbers (one group does 0.00-1.00, one does 1.00-2.00, etc.), so students can see how decimals connect to whole numbers, and will help rounding to the nearest whole make more sense.
-using real life examples of large numbers and decimals. It can be hard to make these numbers concrete for kids, and finding real life examples can help.
-do math talks and strategy talks where students share ideas and strategies based on place value understanding. If your students aren’t at that level yet, start with modeling it for them.
-encourage students to use strategies other than the standard algorithms. The standard algorithms are very practical, and students definitely need to be able to use them. But we also have calculators that can replicate the standard algorithms for us. What calculators cannot replicate is the more flexible thinking and understanding that comes from solving problems using a variety of strategies and the strong place value concepts that develop from working with numbers in this way.
Happy (Place Value) Teaching!!
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HuggingFaceTB/finemath
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## Intermediate Algebra (6th Edition)
We are given that $f(x)=x^{2}-2$, $g(x)=x+1$, and $h(x)=x^{3}-x^{2}$. We know that $(f\circ g)(x)=f(g(x))$. Therefore, $(f\circ g)(-1)=f(g(-1))=f(-1+1)=f(0)=0^{2}-2=-2$
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HuggingFaceTB/finemath
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{[ promptMessage ]}
Bookmark it
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# hw13 - Instance An undirected graph G = V,E and an integer...
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CSCE 750, Fall 2011, Assignment 13 November 18, 2011 NOTE: We will assume that the HAMILTONIAN CIRCUIT problem (a.k.a. the HAMILTONIAN CYCLE problem) is for directed graphs: HAMILTONIAN CIRCUIT (HC) Instance: A directed graph G Question: Is there a simple directed cycle in G containing all the vertices of G ? We abbrevate this problem as HC. The book’s version (page 1062) of this problem, HAM-CYCLE, is for undirected graphs and is not the same problem . Pages 662–664 Exercises 24.3-1, 24.3-4, 24.3-6, 24.3-7 Pages 1060–1061 Exercise 34.1-1, 34.1-6 [Hint: For Kleene star closure, use dynamic programming] Pages 1065–1066 Exercise 34.2-1, 34.2-3, 34.2-6, 34.2-8 (needs concept of co-NP) Page 1077 Exercises 34.3-2 (optional), 34.3-3, 34.3-6 Pages 1085–1086 Exercises 34.4-2, 34.4-6, 34.4-7 (optional) Pages 1100–1101 Exercises 34.5-1 (easy; reduce from HAM-CYCLE), 34.5-2, 34.5-4, 34.5-6 (hint: reduce from HC), 34.5-8 (not in text) Give a polynomial reduction from VERTEX COVER to the INDEPENDENT SET problem: INDEPENDENT SET
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Unformatted text preview: Instance: An undirected graph G = ( V,E ) and an integer 0 ≤ k ≤ | V | . Question: Is there a subset I ⊆ V of vertices such that | I | ≥ k and no two vertices in I are adjacent ( independent set )? [Hint: If C is a vertex cover for G , what can you say about V-C ? If I is an independent set for G , what can you say about V-I ?] (not in text) Give a polynomial reduction from INDEPENDENT SET to CLIQUE. [Hint: The complement of an undirected graph G = ( V,E ) is the graph G = ( V,E ) where ( u,v ) ∈ E iff ( u,v ) / ∈ E , for all distinct vertices u,v ∈ V .] (not in text) (more challenging) Give a polynomial reduction from (directed) HC to HAM-CYCLE (the undirected version of the Hamiltonian path problem (page 979)). [Hint: Given an instance G of HC, build a new graph G by replacing each vertex of G with a path of length 2. Connect endpoints of different paths together with edges in some way based on the edges of G .] 1...
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HuggingFaceTB/finemath
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+0
# system
0
28
1
+1415
The system of equations
\frac{xy}{x + y} = 0, \quad \frac{xz}{x + z} = 0, \quad \frac{yz}{y + z} = 1
has exactly one solution. What is $z$ in this solution?
Jan 14, 2024
#1
+129690
+1
$$\frac{xy}{x + y} = 0, \quad \frac{xz}{x + z} = 0, \quad \frac{yz}{y + z} = 1$$
Either x = 0 or y, z = 0
But the second cannot be true because it would make the last fraction undefined
So x = 0
And (assuming integer solutions for y, z )
(yz) / (y + z) = 1
yz = y + z
zy - z = y
z ( y - 1) = y
z = y / (y -1)
y z
2 2
z = 2
Jan 14, 2024
edited by CPhill Jan 14, 2024
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HuggingFaceTB/finemath
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# Search by Topic
#### Resources tagged with Patterned numbers similar to Golden Eggs:
Filter by: Content type:
Age range:
Challenge level:
### There are 13 results
Broad Topics > Numbers and the Number System > Patterned numbers
### Magic Squares II
##### Age 14 to 18
An article which gives an account of some properties of magic squares.
### How Old Am I?
##### Age 14 to 16 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### Magic Squares
##### Age 14 to 18
An account of some magic squares and their properties and and how to construct them for yourself.
### Harmonic Triangle
##### Age 14 to 16 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### Rolling Coins
##### Age 14 to 16 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### Pinned Squares
##### Age 14 to 16 Challenge Level:
What is the total number of squares that can be made on a 5 by 5 geoboard?
### Back to Basics
##### Age 14 to 16 Challenge Level:
Find b where 3723(base 10) = 123(base b).
### Odd Differences
##### Age 14 to 16 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Triangles Within Pentagons
##### Age 14 to 16 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Sixty-seven Squared
##### Age 16 to 18 Challenge Level:
Evaluate these powers of 67. What do you notice? Can you convince someone what the answer would be to (a million sixes followed by a 7) squared?
### One Basket or Group Photo
##### Age 7 to 18 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
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HuggingFaceTB/finemath
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### Percentage
Quantitative Aptitude Questions and Answers on Percentage for IBPS Exams, candidate who are preparing for Bank Exam, please check all the questions and answers on Percentage , these are questions which are asked in competitive exams like IBPS clerk, IBPS PO, IBPS MT and Specialist Officer. we have good collection of Arithmetic Aptitude
## Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get? A. 57% B. 60% C. 65% D. 90% Correct Answers is : Option A Solution : Total number of votes polled = (1136 + 7636 + 11628) […]
## The price of petrol is increased by 25%. By what percent the consumption be reduced to make the expenditure remain the same?
The price of petrol is increased by 25%. By what percent the consumption be reduced to make the expenditure remain the same? 1. 25% 2. 33.33% 3. 20% 4. None Answer: 2 Solution: It is equivalent to 1.25 decreased to 1. % decrease = (1.25-1)/1.25 X 100 = 20%
## If the length of a rectangle is increased by 33.33%, by what percentage should the breadth be reduced to make the area same?
If the length of a rectangle is increased by 33.33%, by what percentage should the breadth be reduced to make the area same? 1. 20% 2. 33.33% 3. 25% 4. None Answer: 1 Soltion: It is equivalent to 1.25 decreased to 1. So 20% decrease.
## In an election between two candidates, A and B, A secured 56% of the votes and won by 48000 votes. Find the total number of votes polled if 20% of the votes were declared invalid.
In an election between two candidates, A and B, A secured 56% of the votes and won by 48000 votes. Find the total number of votes polled if 20% of the votes were declared invalid. 1. 500000 2. 400000 3. 600000 4. None Answer: 1 Solution: A got 56% => B got 44% Difference = […]
## By what % is 200 more than 50?
By what % is 200 more than 50? 1 100 2. 200 3. 300 4. None Answer: 3 Solution: % difference = (200-50)/50 X 100 = 300 %
## If 20% of 40% of a = 25% of a% of b, then what is b?
If 20% of 40% of a = 25% of a% of b, then what is b? 1. 8/5 2. 16/25 3. 8/25 4. None Answer: 3 Solution: 1/5 X 2/5 X a = ¼ X a X b => b = 8/25
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HuggingFaceTB/finemath
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# A triangle has sides with lengths: 4, 12, and 2. How do you find the area of the triangle using Heron's formula?
Jan 16, 2016
No such triangle is possible since no triangle can have a side which is longer than the sum of the other two sides (at least in Euclidean space)
#### Explanation:
If you attempted to apply Heron's formula:
$\textcolor{w h i t e}{\text{XXX}} A r e a = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
for a triangle with sides $a = 4 , b = 12 , c = 2$
and semi-perimeter $s = 9$ (i.e. $\frac{a + b + c}{2}$)
you would end up attempting to find the square root of a negative number: $\sqrt{9 \cdot \left(5\right) \cdot \left(- 3\right) \cdot \left(7\right)}$
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HuggingFaceTB/finemath
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