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# buoyancy
(redirected from Archemedes principle)
Also found in: Dictionary, Thesaurus, Medical.
Related to Archemedes principle: Pascal's principle
## buoyancy
buoyancy (boiˈənsē, bo͞oˈyən–), upward force exerted by a fluid on any body immersed in it. Buoyant force can be explained in terms of Archimedes' principle.
## Buoyancy
The resultant vertical force exerted on a body by a static fluid in which it is submerged or floating. The buoyant force FB acts vertically upward, in opposition to the gravitational force that causes it. Its magnitude is equal to the weight of fluid displaced, and its line of action is through the centroid of the displaced volume, which is known as the center of buoyancy. See Aerostatics, Hydrostatics
By weighing an object when it is suspended in two different fluids of known specific weight, the volume and weight of the solid may be determined. See Archimedes' principle
Another form of buoyancy, called horizontal buoyancy, is experienced by models tested in wind or water tunnels. Horizontal buoyancy results from variations in static pressure along the test section, producing a drag in closed test sections and a thrust force in open sections. These extraneous forces must be subtracted from data as a boundary correction. Wind tunnel test sections usually diverge slightly in a downstream direction to provide some correction for horizontal buoyancy.
A body floating on a static fluid has vertical stability. A small upward displacement decreases the volume of fluid displaced, hence decreasing the buoyant force and leaving an unbalanced force tending to return the body to its original position. Similarly, a small downward displacement results in a greater buoyant force, which causes an unbalanced upward force.
A body has rotational stability when a small angular displacement sets up a restoring couple that tends to return the body to its original position. When the center of gravity of the floating body is lower than its center of buoyancy, it will always have rotational stability. Many a floating body, such as a ship, has its center of gravity above its center of buoyancy. Whether such an object is rotationally stable depends upon the shape of the body.
McGraw-Hill Concise Encyclopedia of Physics. © 2002 by The McGraw-Hill Companies, Inc.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
## Buoyancy
of a ship, the ability of a loaded ship to float in a designated position relative to the water’s surface; one of the most important features of a ship’s seaworthiness. To ensure safe operation, every vessel must have reserve buoyancy, defined as the additional weight a ship can carry and still remain afloat. Reserve buoyancy is determined by the amount of freeboard. Standards for required freeboard are established by classification societies and depend on the design of the ship and the region and season of navigation.
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# PAI Private Arithmetics Institute
## Natural Numbers, Recursion, Geometry, Symmetry, Axioms, Logic, Metalanguage and more
Why PAI Private Arithmetics Institute ?
A critical analysis of the connection between logics and mathematics shows that the foundations of mathematics often are not treated thoroughly. It is the aim of PAI to distribute a method and precise languages that adhere to the following principles in so-called Bavaria notation :
Clear distinction of language levels
- supralanguage (English)
- metalanguage (Mencish)
- object-language (Funcish, obeying the principle of context-independence)
Clear distinction of systems, that are called
- abstract calcules (purely axiomatic)
- concrete calcules (at most recursive)
Clear distinction of logics with respect to types
- first order (e.g. natural numbers, rational numbers, radical numbers i.e. with roots)
- higher order (e.g. real numbers)
It is always the question what a system is talking about, in other words, what is the ontological basis to start with. The requirement for precision is such that a computer can decide if a certain step of reasoning is in accordance with the rules - if it fulfills the Calculation Criterion of Truth .
On the other hand it is tried to stay as close to normal mathematical language based on predicate logic as possible. Set theory is not used; on the contrary, axiomatic set theory can be expressed in FUME, the language system of Funcish and Mencish..
- Geometries of O , relating to axiomatic planar geometry
- Representation of processive functions in Robinson arithmetic ? , relating to recursive functions
- Church's thesis is questioned by new calculation paradigm , relating to recursive functions
- The Snark, a counterexample for Church's thesis ? , relating to recursive functions
- Programming primitive functions and beyond , relating to recursive functions
- Confusing conventional notation , relating to axiomatic set theory
- A flaw in Separation Axioms ? , relating to axiomatic set theory
- Number Theory beyond Frege , relating to predicate logic
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# Fraction and Mixed Number Addition Challenge
Math Challenge
00:00
Start my Challenge
9/8 + 66/4 = ?
8/7 + 98/2 = ?
39/4 + 10/3 = ?
3/2 + 76/4 = ?
8/6 + 610/4 = ?
110/5 + 9/6 = ?
17/3 + 10/8 = ?
36/2 + 10/6 = ?
9/4 + 910/7 = ?
3/2 + 97/6 = ?
Fraction and mixed number addition challenge is an online test to assess 3rd or 4th grade student's math skills on find the sum of regular fraction and mixed number in pre-algebra or number and operations - fractions (NF) of common core state standards (CCSS) for mathematics.
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https://harishsnotebook.wordpress.com/2015/10/26/process-capability-are-you-really-capable/
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# Process Capability – Are You Really Capable?
I have always been interested in process capabilities. I will not be going into how you can calculate the process capability in this post. In this post, I will try to look into the feasibility of a process capability study. The goal of this post is to encourage the reader to understand the process capability and sigma value correlation. The most popular value for a process capability index seems to be 1.33. This is applicable for Cpk or Ppk.
There are two ways you can look at your process capability. The first way is Cpk, commonly referred to as Short Term Capability. This assumes that your process is stable, and uses the R/d2 estimate for standard deviation.
The second way is Ppk, commonly referred to as Long Term capability. This is for used looking at the potential capability and uses the sample estimate of standard deviation in the formula. There is no need of assumption of stability for using this.
Relation between Sigma value and process capability index:
The formula for calculating the sigma value from the process capability index value is show below.
Sigma = 3 * (Process capability index value)
Thus if your Ppk value is 1.33, this equates to a sigma value of 4.
A ppk value of 2.00 refers to the ever famous 6 sigma value.
Why 6 sigma values are hard to conceive:
I came across a cool table that correlates the sigma values to approximated frequency of daily events (if the event was to occur on a daily basis). Please note that this does not include the 1.5 sigma shift and assumes normality.
Please note that the column in the right is based on the assumption that the event is occurring on a daily basis only.
So next time you achieve a process capability value of 3.00 or more, understand that this indicates a sigma value of 9.00 or more. In other words, it is quite impossible to achieve such a feat. This could be a strong indication of autocorrelation. Autocorrelation indicates that the data you collected are extremely homogeneous and that this may not include any possible process variations. Of course, the other possibility is that the specifications are very wide. More on this on a later post.
Always keep on learning…
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# What is MRR growth
## Annual Recurring Revenue (ARR)
At the Annual Recurring Revenue (ARR) it concerns the determination of the shares in the total turnover, which regularly return and accordingly for strategic planning predictable are. The dimension of the ARR is usually € million per year.
The key figure is in particular in Software / SaaS environment one of the most important KPIs, as the goodwill is often calculated as a multiplier of the ARR.
### ARR definition
Annual Recurring Revenue is the sum of Monthly Recurring Revenue (MRR) for an entire year that customers pay on a recurring basis.
So the ARR is about determining that part of a company's annual turnover that is made repeatedly and at regular intervals. Such sales are for example through Subscriptions or generate regular maintenance costs for services. That's why they play, for example, in the SaaS industry a particularly large role where business models are structurally similar to a subscription.
For most companies, it is important to determine the proportions of the following components in the Annual Recurring Revenue: the proportion of
• new contracts,
• extended contracts and renewed deals,
• Upgrades compared to the half-year or the time of the contract renewal,
• Downgrade or changes with a loss compared to the first half of the year or when the contract is extended,
• terminated contracts.
In contrast, one-time expenses, consumption and usage fees are not included in the ARR.
Calculating the ARR is pretty simple: (total subscription cost per year + recurring income from add-ons or upgrades) - lost income from cancellations.
If your company's pricing strategy is based on monthly recurring revenue (MRR), you can also calculate Annual Recurring Revenue by simply multiplying the MRR by 12.
### Aim of the Annual Recurring Revenue
The goal of determining the ARR is primarily that Predictability with the results for the strategic direction the company. The relevance of these numbers varies strongly depending on the industry and business model. In principle, the influence of the ARR increases with its share in total sales.
Both ARR and MRR provide valuable insights into how well or badly a company is doing. This data can therefore be used to forecast how sales will develop as the company grows. Based on this data, the company can then plan what it can do with this turnover.
With the Annual Recurring Revenue, companies can also see the development from year to year at a high level, which is useful for long-term product planning and the creation of business plans.
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# Practice Reasoning Questions For IBPS 2017 Exams (Input-Output New Pattern)
Practice Reasoning Questions For IBPS 2017 Exams (Input-Output New Pattern):
Dear Readers, Important Practice Reasoning Questions with explanation forย IBPS Exams 2017ย was given here with explanation, candidates those who are preparing for Banking and all other Competitive exams can use this practice questions.
Are You preparing for IBPS PO 2019? Start your preparation with Free IBPS PO Mock Test 2019 – Take Test Now
Directions (1-5): Study the following information carefully and answer the givenquestions.
A number arrangement machine arranges two digit numbers into a typical manner. Each step gives output taking input from the previous step. Thefollowing is an illustration of Input and rearrangement. Using the illustrationanswer the question given below.
1).What is the difference between the first digit of the second number and second digit of the first number in step II?
2
1
0
4
3
Step I โFirst digit multiply with the second digit and if the multiply value more that than the value โ9โ makes it as a single digit.
Like โ 56 = 5+6 = 11 = 1+1 = 2
63 = 6+3 = 9
Step II โSum of the first digit of the numbers multiply with 2 and Sum of the second digit of the numbers multiply with 2
Step III โSubtract the first digit with the second digit
Step IV โ Add the first number with second number
1).
2). What is the difference between the first digit of the second number and second digit of the first number in step II?
4
5
3
8
6
2).
3). What is the multiplication of two numbers obtained in step III?
1
4
-1
-4
5
3).
4). If the value โ8โ is added from the final output, then what will be the resultant value?
5
3
-11
-5
9
4).
5). Which of the following combination represent the first digit of the second number and second digit of the third number in step I of the given input?
1, 5
6, 5
4, 7
5, 3
3, 5
5).
Directions (6-10): Study the following information carefully and answer the given questions.
A number arrangement machine arranges two digit numbers into a typical manner. Each step gives output taking input from the previous step. The following is an illustration of Input and rearrangement. Using the illustration answer the question given below.
6). If first digit is subtract with 3 and the second digit is subtract with 2 in all number of step I, then what is the sum of the second number in step I?
5
8
7
9
6
Step I โ First digit add with the second digit
Step II โ
First number – Sum of the first digit of the first number, second digit of the second number and first digit of the third number
Second number โ Sum of the second digit of the first number, first digit of the second number and second digit of the third number
Step III โ Sum of the squares of the number, if the sum value more that than the value โ9โ makes it as a single digit.
First number = 1^2 + 9^2 = 1 + 81 = 82 = 8+2 = 10 = 1+0 = 1
Second number = 2^2 + 0^2 = 4
Step IV โ Add the first number with second number
6).
7). If the value โ3โ is subtracted from the final output, then what will be the resultant value?
2
1
5
4
6
7).
8). Which of the following combination represent the first digit of the second number and second digit of the first number in step I of the given input?
8,9
7,4
3,6
5,8
5,6
8).
9). Which of the following represent the difference between the first digit of the second number and second digit of the first number in step II?
4
3
8
5
7
9).
10). What is the multiplication of two numbers obtained in step III?
5
3
4
0
2
10).
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Duration-bandwidth_product.gif(800 × 261 pixels, file size: 2.93 MB, MIME type: image/gif, looped, 305 frames, 31 s)
## Summary
Description English: How short you can make a pulse depends on many frequencies you are using (bandwidth). But he "duration-bandwidth product" depends only on the shape of your power spectrum. Interestingly, the question "which power spectrum will result in the shortest pulse" depends A LOT on how you decide to measure how wide things are. In particular using the standard deviation or the full-width half-maximum, give very different numbers in many cases. Date 19 February 2021 Source https://twitter.com/j_bertolotti/status/1362719916449742848 Author Jacopo Bertolotti Permission(Reusing this file) https://twitter.com/j_bertolotti/status/1030470604418428929
## Mathematica 12.0 code
```labels = {"f(\[Omega])\[Proportional] \[CapitalPi](\[Omega])", "f(\[Omega])\[Proportional] \[CapitalLambda](\[Omega])", "f(\[Omega])\[Proportional] \!\(\*SuperscriptBox[\(e\), \\(-\*SuperscriptBox[\(\[Omega]\), \(2\)]\)]\)", "f(\[Omega])\[Proportional] \!\(\*SuperscriptBox[\(sech\), \\(2\)]\)(\[Omega])", "f(\[Omega])\[Proportional] \!\(\*SuperscriptBox[\(e\), \(-\(\(|\)\\(\[Omega]\)\(|\)\)\)]\)"};
frames = Table[
Table[
GraphicsRow[{
Plot[(1 - \[Tau]) f[[j]]^2 + \[Tau] f[[Mod[j + 1, 5, 1]]]^2, {\[Omega], -5, 5}, PlotRange -> {-0.1, 1.1}, Exclusions -> None, PlotStyle -> Black, Axes -> False, FrameLabel -> {{None, None}, {"\[Omega]", "Power spectrum"}}, Frame -> True, FrameStyle -> Directive[White, FontColor -> Black], LabelStyle -> {FontSize -> 14, Bold}, FrameTicks -> None, Epilog -> {Opacity[1 - \[Tau]], Text[Style[labels[[j]], Black, Bold], {3, 0.8}], Opacity[\[Tau]], Text[Style[labels[[Mod[j + 1, 5, 1]]], Black, Bold], {3, 0.8}]}
]
,
Plot[(1 - \[Tau]) p[[j]]*Cos[\[Omega]0 t] + \[Tau] p[[Mod[j + 1, 5, 1]]]*Cos[\[Omega]0 t], {t, -20, 20}, PlotStyle -> Black, PlotRange -> {-1, 1}, Axes -> False, FrameLabel -> {{None, None}, {"t", "Pulse"}}, Frame -> True, FrameStyle -> Directive[White, FontColor -> Black], LabelStyle -> {FontSize -> 14, Bold}, FrameTicks -> None]
,
Graphics[{Text[Style["Duration-bandwidth product", Black, Bold, FontSize -> 9], {0, 0.8}],
Text[Style["\!\(\*SubscriptBox[\(\[Sigma]\), \(\[Omega]\)]\) \!\(\\*SubscriptBox[\(\[Sigma]\), \(t\)]\) = ", Black, Bold, FontSize -> 10], {0.08, 0.3}], Opacity[1 - \[Tau]],
Text[Style[StringForm["``", NumberForm[\[Sigma]\[Omega][[j]]*\[Sigma]t[[j]] // N, {3, 2}]], Black, Bold, FontSize -> 10], {0.5, 0.32}], Opacity[\[Tau]],
Text[Style[StringForm["``", NumberForm[\[Sigma]\[Omega][[Mod[j + 1, 5, 1]]]*\[Sigma]t[[Mod[j + 1, 5, 1]]] // N, {3, 2}]], Black, Bold, FontSize -> 10], {0.5, 0.32}], Opacity[1],
Text[Style["\!\(\*SubscriptBox[\(FWHM\), \(\[Omega]\)]\) \\!\(\*SubscriptBox[\(FWHM\), \(t\)]\) = ", Black, Bold, FontSize -> 10], {0, -0.1}], Opacity[1 - \[Tau]],
Text[Style[StringForm["``", NumberForm[fwhm\[Omega][[j]]*fwhmt[[j]], {3, 2}]], Black, Bold, FontSize -> 10], {0.85, -0.08}], Opacity[\[Tau]],
Text[Style[StringForm["``", NumberForm[fwhm\[Omega][[Mod[j + 1, 5, 1]]]*fwhmt[[Mod[j + 1, 5, 1]]], {3, 2}]], Black, Bold, FontSize -> 10], {0.85, -0.08}]
}, PlotRange -> {{-1, 1}, {-1, 1}}]
}]
, {\[Tau]1, 0, 1, 0.02}]
, {j, 1, 5}];
ListAnimate[Join[Flatten@Table[{Table[frames[[j, 1]], {10}], frames[[j]]}, {j, 1, 5}] ]]
```
## Licensing
I, the copyright holder of this work, hereby publish it under the following license:
This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication. The person who associated a work with this deed has dedicated the work to the public domain by waiving all of their rights to the work worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law. You can copy, modify, distribute and perform the work, even for commercial purposes, all without asking permission.
### Captions
How short you can make a pulse depends on many frequencies you are using (bandwidth). But he "duration-bandwidth product" depends only on the shape of your power spectrum.
### Items portrayed in this file
#### some value
author name string: Jacopo Bertolotti
## File history
Click on a date/time to view the file as it appeared at that time.
Date/TimeThumbnailDimensionsUserComment
current17:30, 24 February 2021800 × 261 (2.93 MB)BertoUploaded own work with UploadWizard
The following pages on the English Wikipedia use this file (pages on other projects are not listed):
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Cody
# Problem 44934. Plot Damped Sinusoid
Solution 1981825
Submitted on 18 Oct 2019 by Andrew Hines
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### Test Suite
Test Status Code Input and Output
1 Pass
clf; t = linspace(0,15,400); y = exp(-0.5*t).*cos(2*pi.*t); m = plot_cos(y, t); f = gcf; assert(isequal([f.Children.Children.Color], [1 0 0 0 0 1])) assert(strcmp([f.Children.Children.LineStyle], 'none--')) assert(strcmp([f.Children.Children.Marker],'*none')) assert(abs(m - (-0.781239288889930)) <= 1e-4) assert(isequal([f.Children.Children.YData],[m, y]))
` `
2 Pass
clf; t = linspace(2,5,100); y = exp(-0.5*t).*cos(2*pi.*t); m = plot_cos(y, t); f = gcf; assert(isequal([f.Children.Children.Color], [1 0 0 0 0 1])) assert(strcmp([f.Children.Children.LineStyle], 'none--')) assert(strcmp([f.Children.Children.Marker],'*none')) assert(abs(m - (-0.287376348726584)) <= 1e-4) assert(isequal([f.Children.Children.YData],[m, y]))
` `
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I've made a small implementation of the famous birthday paradox, trying to find a collision between two random birthdates (here integer between 1 and 365) for the first time. But it returns always a value around let's say 40 and 70, which does not fit the stats at all. Is something wrong with my algo, or with the random int generator, both ? Thanks for your feedback.
Here is the code :
``````public static void main(String[] args){
int[] birthday = new int[200];
for(int i = 0; i<20;i++){
Collision(birthday);
}
}
public static int Collision(int birthday[]){
Random rand = new Random();
for(int i = 1; i<birthday.length;i++){
birthday[i] = rand.nextInt(365);
}
int count = 0;
for(int i = 0; i<birthday.length; i++){
for(int j= i+1 ; j<birthday.length; j++){
if (birthday[i] == birthday[j]){
count++;
}
}
}
System.out.print(count+" ");
return count;
}
``````
Here is the output for ex :
45 50 60 52 53 53 50 49 37 68 52 53 51 43 49 51 46 43 45 35
• what's the expected average output? – Gavriel Feb 8 '16 at 10:57
• For a set of 200 people, a collision count of between 40 and 70 sounds reasonable. – biziclop Feb 8 '16 at 10:58
• So what do you expect? The chances of collision is almost 100% and you found on average 40. It does not contradict theory see birhtday problem calculator – Radu Ionescu Feb 8 '16 at 10:59
• Thanks for your replies. According to the theory, the probability is 50% for a set of 23 people, which I never have...SP there should be some flaws somewhere. I would expect sometimes 25 or 15 or even less... – loukios Feb 8 '16 at 11:04
• Doing the sums, the expected value should indeed be slightly higher, around 84. – biziclop Feb 8 '16 at 11:04
EDIT:
What you essentially did in your algorithm is that you generated 200 random birthdays and counted how many collisions exist among them.
You know you could make things a lot simpler by using a `Set`, which is empty at the beginning. Then in a simple while loop generate birthdays (numbers up to 365), try adding them in the `Set`, and the first time you get a collision - the number is already in the `Set` - you have your answer (the answer being the size of the Set).
That is, if your goal really is to find a collision in minimum number of birthdays.
E.g., this:
``````Random rand = new Random();
for (int t = 0; t < 20; t++)
{
Set<Integer> b = new HashSet<Integer>();
while (true)
{
int n = rand.nextInt(365);
break;
}
System.out.print(b.size() + " ");
}
``````
Produces:
``````15 30 24 4 8 19 10 40 32 31 30 14 41 30 15 7 15 52 24 27
``````
• Thanks, I'm new to Java so I was not aware of it, but I'll try. – loukios Feb 8 '16 at 11:06
• You can even simplify it to `if (!b.add(n)) break;` – biziclop Feb 8 '16 at 11:15
• @biziclop Ahh, nice, I always forget that `Set.add()` returns boolean :) – radoh Feb 8 '16 at 11:16
• Even more concise, use `while (b.add(rand.nextInt(365))) ;`. No need for temporary var `n`. – pjs Feb 8 '16 at 15:41
• @pjs nice, but I'll leave it as it is, for clarity's sake, since loukios said he's new to Java. – radoh Feb 8 '16 at 15:44
But you are repeatedly instantiating a new `Random` instance. That ruins the generator's statistical properties. Do it once at the beginning of your program.
(Eventually you'll need to consider February 29th too but that's very much a second-order effect).
Your algorithm seems OK and the results are reasonable.
FYI you could use streams to very efficiently do all the heavy lifting in 1 line:
``````private static Random rand = new Random();
public static int collision(int size) {
return size - Stream.generate(() -> rand.nextInt(365)).limit(size).distinct().count();
}
``````
And a 1-line main:
``````public static void main(String[] args){
Stream.of(200).map(MyClass::collision).forEach(System.out::println);
}
``````
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Search
# Hierarchical Clustering
Last updated Sep 28, 2022 Edit Source
Hierarchical clustering produces a tree of clusterings
• Each node in the tree splits the data into 2 or more clusters.
• Often have individual data points as leaves.
Often applied in phylogenetics
## # Agglomerative Clustering (Bottom up)
Requires a “distance” measure between two clusters.
Cluster distance measures
1. Distance between closest members of $C_1$ and $C_2$. Also called single-link clustering: $\min d(a,b), a \in C_1, b \in C_2$
2. Distance between farthest members of $C_1$ and $C_2$. Also called complete-link clustering: $\max d(a,b), a \in C_1, b \in C_2$
3. Average distance between members of $C_1$ and $C_2$. Also called group average clustering: $\frac{1}{|C_1||C_2|} \sum_{a \in C_1} \sum_{b \in C_2} d(a,b)$
4. Starts with each point in its own cluster.
5. Each step merges the two “closest” clusters.
6. Stop with one big cluster that has all points.
Naive implementation cost is $O(n^3d)$
## # Divisive Clustering (Top-down)
Start with all examples in one cluster, then start dividing. (e.g., run K-means on a cluster, then run again on resulting clusters)
## # Biclustering
Cluster the training examples and features. Helps to figure out the ‘why’ on why things are clustered together
• Run clustering method on $X$
• Run clustering method on $X^T$
A dendrogram describes the hierarchy of clusters generated by the clustering methods.
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## "8oz to ml water"
Request time (0.045 seconds) [cached] - Completion Score 160000
how many ml is 8 oz of water1 how much is 8oz of water in ml0.5 how many ml water for 8oz0.54 1 liter equals how many water bottles0.54
10 results & 0 related queries
### How many ml in an 8oz glass of water? - Answers
How many ml in an 8oz glass of water? - Answers 9 7 5500 8 oz = 226.8 grammes = 226.8 millilitres of pure ater Celcius.
Litre31.5 Water12.4 Glass10.8 Ounce7 Gram3.2 Fluid ounce2.2 Purified water1.9 Volume1.4 Properties of water1 Sodium silicate1 Cup (unit)0.9 Solution0.8 Liquid0.6 Beer in Australia0.6 Cadmium sulfide0.5 Boiling0.5 Fluid0.5 Bottle0.5 Measurement0.4 Sports drink0.4
### Convert 8 oz water to ml? - Answers
Convert 8 oz water to ml? - Answers About 240 to 250 ml
Litre28.1 Ounce19.2 Water13.9 Fluid ounce7.9 Teaspoon2.2 Fluid2.1 Cup (unit)1.2 Pound (mass)0.5 United States customary units0.5 Troy weight0.5 Food0.4 Cooking0.4 Milk0.3 Gram0.3 Medication0.3 Measurement0.3 Water bottle0.3 Gold0.3 Charles Dickens0.3 Chickenpox0.3
### How many ml in 8 oz of water? - Answers
How many ml in 8 oz of water? - Answers 8 oz of ater is 236.588mL
Litre17.4 Ounce14.7 Water11.9 Fluid ounce5.2 Weight1.2 Mass1.2 Fluid1.1 Cooking1 Measurement1 Milk0.9 Cup (unit)0.8 Unit of measurement0.7 Arithmetic0.6 Troy weight0.4 Pound (mass)0.4 Mathematics0.4 Volume0.3 Food0.3 United States customary units0.3 Gram0.3
### 8oz in ml? - Answers
Litre33.1 Ounce4.8 Fluid ounce3.8 Cup (unit)3.4 Water2.4 Volume1.4 Glass1.2 Sports drink0.9 Measuring cup0.7 Bottle0.7 Liquid0.7 Fluid0.6 Milk0.5 Beer in Australia0.5 Measurement0.5 Oven0.5 Food0.5 Calculator0.5 Gram0.5 Temperature0.3
### How much is 8 glasses of 8 oz of water converted in ML? - Answers
E AHow much is 8 glasses of 8 oz of water converted in ML? - Answers Here's why... 1 ounce equals 30 mls 8 ounces x 30 mls equals 240 mls per cup 240 mls x 8 cups equals 1920 mls
Litre20.1 Water15.9 Ounce11.2 Cup (unit)6.3 Glasses6.2 Glass3.3 Gram2.6 List of glassware1.9 Fluid ounce1.1 Drink0.8 Volume0.5 Teaspoon0.4 Cooking0.4 Mill (grinding)0.3 Measurement0.3 Thermal energy0.3 Properties of water0.3 Baking0.3 Cup0.2 Tonne0.2
### How do you convert 8oz to ml? - Answers
How do you convert 8oz to ml? - Answers If you go into Google and put in " 8oz in ml G E C", their calculator will convert it for you. I use it all the time to e c a convert temperatures, oven temps, etc as I'm an American living in the UK and don't "do" metric.
Litre37.5 Ounce4.3 Fluid ounce3.3 Cup (unit)2.7 Water2.5 Oven2.1 Volume2.1 Calculator2 Gram2 Temperature1.6 Glass1.3 Drink1.1 Sports drink1 Teaspoon1 Metric system1 Measuring cup0.9 Liquid0.8 2-Mercaptoethanol0.7 Molar concentration0.7 Fluid0.7
### What does 8oz measure to in ml? - Answers
What does 8oz measure to in ml? - Answers 8 ounces of ater is 226.796 ml
Litre36 Ounce5.4 Volume4.4 Gram4 Water3.9 Fluid ounce3.9 Cup (unit)3.4 Measurement3.3 Mass2.8 Liquid1.9 Coffee1.7 Weight1.5 Glass1.4 Sports drink1 Unit of measurement1 Measuring cup0.9 Drink0.8 Fluid0.7 Grivna0.6 Container0.6
### How many ml is in 8oz? - Answers
How many ml is in 8oz? - Answers 8 fluid oz is 236.588mL
Litre30.4 Ounce5.2 Cup (unit)4.4 Fluid ounce4.3 Fluid2.7 Water2.2 Volume1.3 Glass1.1 Liquid0.8 Sports drink0.6 Milk0.6 Measuring cup0.6 Beer in Australia0.5 Cubic centimetre0.5 Food0.5 Measurement0.5 Pint0.4 Oven0.4 Calculator0.4 Gram0.4
### How many ml are in 8 fl oz of water? - Answers
How many ml are in 8 fl oz of water? - Answers The usual measurement used is 250 ml V T R, though it's a bit more than 8 fl oz. Small milk cartons in Canada are 250 & 500 ml , rather than half pint & pint in the US.
Litre37 Fluid ounce34.7 Water7 Pint5.6 Ounce3.5 United States customary units2.4 Carton2.3 Measurement2.2 United States dollar0.6 Canada0.6 Bit0.4 Food0.4 Cooking0.3 Plastic0.3 Thousandth of an inch0.3 Unit of measurement0.3 Charles Dickens0.3 Chickenpox0.2 Cup (unit)0.2 Gold0.2
### How Much Water Should You Drink Per Day?
www.healthline.com/nutrition/how-much-water-should-you-drink-per-day
How Much Water Should You Drink Per Day? Drinking enough This page explains exactly how much ater you should drink in a day.
Water22 Drink7.2 Drinking3.7 Dehydration2.4 Food2.2 Litre2 Fat1.9 Perspiration1.6 Burn1.6 Urine1.4 Energy level1.4 Health1.3 Exercise1.2 Thirst1.1 Drinking water1.1 Nutrition1 Fluid1 National Academies of Sciences, Engineering, and Medicine1 Caffeine0.9 Water supply network0.9
##### Domains
www.answers.com | www.healthline.com |
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A133841 Decimal expansion of the position of the positive real maximum of Dawson's integral D_+(x). 7
9, 2, 4, 1, 3, 8, 8, 7, 3, 0, 0, 4, 5, 9, 1, 7, 6, 7, 0, 1, 2, 8, 2, 3, 2, 7, 1, 5, 0, 4, 3, 4, 5, 9, 7, 5, 6, 9, 6, 2, 9, 1, 5, 5, 9, 9, 3, 5, 1, 6, 3, 9, 1, 7, 5, 9, 7, 8, 1, 0, 5, 2, 9, 8, 4, 9, 7, 5, 9, 5, 4, 0, 1, 6, 2, 1, 9, 3, 8, 8, 1, 6, 8, 5, 6, 2, 7, 7, 7, 1, 2, 1, 4, 5, 8, 4, 7, 3, 8, 5, 5, 6, 9, 4, 8 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 LINKS Stanislav Sykora, Table of n, a(n) for n = 0..2000 Eric Weisstein's World of Mathematics, Dawson's Integral Wikipedia, Dawson function FORMULA Equals 1/(2*A133842). EXAMPLE 0.92413887300459176701... MATHEMATICA DawsonF[x_] := Sqrt[Pi]*Erfi[x]/(2*Exp[x^2]); x0 = x /. FindRoot[ DawsonF'[x], {x, 1}, WorkingPrecision -> 110]; RealDigits[x0][[1]][[1 ;; 105]] (* Jean-François Alcover, Oct 26 2012, after Eric W. Weisstein *) PROG (PARI) Erfi(z) = -I*(1.0-erfc(I*z)); Dawson(z) = 0.5*sqrt(Pi)*exp(-z*z)*Erfi(z); \\ same as F(x)=D_+(x) DDawson(z) = 1.0 - 2*z*Dawson(z); \\ Derivative of the above x = solve(z=0.1, 2.0, real(DDawson(z))) \\ Stanislav Sykora, Sep 17 2014 CROSSREFS Cf. A133842, A133843, A243433. Sequence in context: A248321 A248320 A200282 * A099769 A176517 A020784 Adjacent sequences: A133838 A133839 A133840 * A133842 A133843 A133844 KEYWORD nonn,cons AUTHOR Eric W. Weisstein, Sep 26 2007 STATUS approved
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Last modified October 20 10:41 EDT 2019. Contains 328257 sequences. (Running on oeis4.)
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# [R] [FORGED] Generating random data with non-linear correlation between two variables
David R Forrest drf at vims.edu
Sat Apr 9 12:48:48 CEST 2016
```Please specify your goal in the oracle/psql analytical functions you know or specify what you mean by nonlinear correlation
Sent from my iPhone
> On Apr 9, 2016, at 6:09 AM, Muhammad Bilal <Muhammad2.Bilal at live.uwe.ac.uk> wrote:
>
> No its not. I am doing all these experiments for my own learning purpose. I am Oracle SQL & PLSQL programmer and I can do these things with Oracle analytical functions.
>
> However at present I am keen to learn R, with no other interest right now.
>
> Thanks
> --
> Research Assistant and PhD Student,
> Bristol Enterprise, Research and Innovation Centre (BERIC),
> University of the West of England (UWE),
> Frenchay Campus,
> Bristol,
> BS16 1QY
>
> muhammad2.bilal at live.uwe.ac.uk
>
>
> ________________________________________
> From: Rolf Turner <r.turner at auckland.ac.nz>
> Sent: 09 April 2016 04:46
> To: Muhammad Bilal
> Cc: r-help at r-project.org
> Subject: Re: [FORGED] [R] Generating random data with non-linear correlation between two variables
>
>> On 09/04/16 06:57, Muhammad Bilal wrote:
>> Hi All,
>>
>> I am new to R and don't know how to achieve it.
>>
>> I am interested in generating a hypothetical dataframe that is consisted of say two variables named v1 and v2, based on the following constraints:
>> 1. The range of v1 is 500-1500.
>> 2. The mean of v1 is say 1100
>> 3. The range of v2 is 300-950.
>> 4. The mean of v2 is say 400
>> 5. There exists a positive trend between these two variables, meaning that as v1 increases, v2 be also increase.
>> 6. But the trend should be slightly non-linear. i.e., curved line.
>>
>> Is it possible to automatically generate through functions like rnorm.
>>
>> Any help will be highly appreciated.
>
> This sounds to me very much like a homework problem. We don't do
> people's homework for them on this list.
>
> cheers,
>
> Rolf Turner
>
> --
> Technical Editor ANZJS
> Department of Statistics
> University of Auckland
> Phone: +64-9-373-7599 ext. 88276
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
```
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What Size Breaker Do I Need For A 40-gallon Electric Water Heater?
How many amps does a 40-gallon hot water heater use?
An average 40-gallon electric water heater runs at 240 volts and a wattage of 2500 W. Hence, calculating the amperage 2500/240, we can deduce the amperage is 18.75 A.
Can I use a 40 amp breaker for a hot water heater?
In general, NO. Doing so is very dangerous and could start a fire! 10 gauge wire requires at most a 30A breaker, whereas 8 gauge requires at most a 40A breaker.
What size breaker should a hot water heater be on?
Electric water heaters require a dedicated 240-volt dedicated 30-amp circuit and a 10-2 non-metallic (NM) or MC cable. This means the breaker only powers the water heater and no other appliances. A 30-amp circuit breaker can power all 4500watt water heaters regardless of gallon size.
Related Question What size breaker do I need for a 40-gallon electric water heater?
What is a 40 amp breaker used for?
Double-pole breaker
The 15-amp and 20-amp breakers often handle baseboard heaters, 30-amp serve water heaters and electric dryers, 40- and 50-amp are for electric ranges, and the 70-amp could serve a large air conditioner or a subpanel.
Can you use a 40 amp breaker on 6 gauge wire?
The minimum wire size you can use with a 40-amp circuit breaker is 8-AWG. You can use larger gauge 6-AWG wire for future-proofing, but a lower gauge wire will render the electrical system non-compliant with the NEC.
Can I replace a 40 amp breaker with a 50 amp?
3 Answers. In general, no. Installing a 50A breaker on wiring only designed to support 40A is dangerous and can result in a house fire.
What size breaker do I need for an electric heater?
Multiply the value for current by 125 percent to determine the size of the breaker you need for the heater. A 1500-watt heater on a 120-volt circuit thus needs a breaker of 15.6 amps. Because a 15-amp breaker would be too small, you need a breaker with the next highest rating, which is 20 amps.
What size breaker do I need for a 50 gallon electric water heater?
The standard for water heater circuits is #10, two conductors with ground (10/2) and a 30 amp circuit breaker for a water heater. If you are running cable anyway, just run #10 and replace the breaker so it's done right, once and for all.
What wire do you use for 50 amps?
According to the American Wire Gauge system, the appropriate wire gauge to use in conjunction with a 50-amp breaker is a 6-gauge wire. The 6-gauge copper conductor wire is rated up to 55 amps, making it the perfect choice for this circuit.
Do electric water heaters require 220?
Voltage Requirements
Most electric hot water heaters operate on 220 to 250 volts AC. This voltage in homes requires a double circuit breaker rated for the amperage draw of the hot water heaters.
What size wire is needed for a 30 amp breaker?
Any circuit fused for 30 amps must use a minimum of 10 ga copper or 8 ga alu. Longer runs may require an upgrade of wire size. In your case, use at least 10 copper for your welder regardless how far it is from the breaker panel.
Do they make a 25 amp breaker?
Square D QO 25 Amp Single-Pole Circuit Breaker-QO125CP - The Home Depot.
Is a double pole 30 amp breaker 60 amps?
Single-pole breakers are rated for 120 volts and 15 or 20 amps. The breakers themselves are relatively narrow and occupy a single slot in the home's breaker box. Double-pole breakers, on the other hand, are typically rated for 20 to 60 amps and supply 240-volt power to large appliances, like electric dryers and ranges.
Do you need a neutral for 240V?
For a 240V load, a neutral wire is not needed. Most 240V appliances, however, have some 120V loads such as timers or control circuits which is why the neutral is usually provided, “just in case.” The only time a 240V only load is commonly seen in a residential settings would be a well pump motor.
Is a double pole 20 amp breaker 40 amps?
Is a double 20 amp breaker 40 amps? - Quora. Nope. It means that two 20 A breakers are ganged together to protect a circuit using two hot wires. Tripping either one at 20 A will trip both to prevent a safety situation where live voltage can be found in the device after one breaker tripped.
How many amps does 40 watts draw?
A 40 Watts bulb is capable of drawing 0.36 Amps to operate.
How many amps can a 40 amp breaker handle?
There is a 40-amp circuit with a maximum output of 9,600 watt. 12,000 watt is equal to 50 Amps x 480 Amps.
How much can a 40 amp breaker handle?
40-amp 240-volt circuit: 40 amps x 240 volts = 9,600 watts. 50-amp 240-volt circuit: 50 amps x 240 volts = 12,000 watts.
Is a double pole 50 amp breaker 100 amps?
It is 100 amp. And the two 50 amps breakers that are tied together are 50 amps, not 50 + 50 = ! 00. That 2 pole 100 amp CB can provide either 100 amps @ 240 volts or 200 amps @ 120 volts but in answer to your question as Chris stated it's 100 amps.
What size wire is rated for 45 amps?
Wire Size Chart and Maximum Amp Ratings
SIZE 60°C (140°F) 90°C (194°F)
AWG or kcmil TYPES TW, UF TYPES TBS, SA, SIS
8 40 45
6 55 60
4 70 75
Can you use 20 amp breaker 40 amp?
The answer is, "No, do not put a 30 Amp breaker in place of a 20 Amp breaker because the wiring is most likely'sized' for the 20 Amp load and -may- overheat if a 30 Amp load is used with wiring sized for a 20 Amp load."
How many amps can you run off of #6 wire?
Size & AMP Ratings
NM, TW, & UF WIRE (Copper Conductor) SE CABLE (Copper Conductor)
12 AWG - 20 AMPS 6 AWG - 65 AMPS
10 AWG - 30 AMPS 4 AWG - 85 AMPS
8 AWG - 40 AMPS 2 AWG - 115 AMPS
6 AWG - 55 AMPS 1 AWG - 130 AMPS
Can I use a 40 amp breaker for a 40 amp stove?
There are 3 questions. Any household cooking appliance rated at less than 12 kilowatts can be served by a 40A circuit, according to the NEC.
Can I add a 40 amp breaker?
Yes, you can get a "quadplex" double-stuff which fits in two spaces and supplies two 240V circuits.
Does a water heater need a GFCI breaker?
However, GFCI protection is not required for receptacles not intended to serve wet bar countertop surfaces, such as refrigerators, ice makers, water heaters, or convenience receptacles that do not supply counter-top surfaces.
What size breaker do I need?
The general rule of thumb is that circuit breaker size should be 125% of the ampacity of cable and wire or the circuit which has to be protected by the CB. Let see the following solved examples: Example 1: Suppose, a 12 gauge wire is used for 20 amperes lighting circuit having 120V single phase supply.
How many amps does a dryer pull?
Residential electric clothes dryers use between 7.5 amps and 30 amps. However, 30 amps is by far the most common. The National Electrical Code (NEC) standards require that 240V dryers have a dedicated 4-wire circuit (10-3 type NM cable with ground) protected by a 30 amp breaker.
How many amps does a 50 gallon water heater use?
A typical 50 gallon electric water heater runs at 4500 watts. In an electric circuit of 240 volts, 4500 watts is equivalent to 18.75 amps.
Do electric water heaters need a neutral wire?
A straight 240v load like a heat pump, AC compressor or a water heater do not require a neutral so 2 hots and a grounding means are all that is needed.
Will 12 2 wire work for a hot water heater?
Most electric water heaters installed in North America require a 30-ampere grounded 240-volt connection which requires a pair #10 copper wires PLUS a ground wire. Copper 12/2 wire is rated for a maximum of just 20-amperes and would be inadequate both in current capacity and for lacking a ground wire.
How many breakers can be in a 40 amp sub panel?
A 40A/240V panel can support as many as 10 or more 120V/15A circuits, presuming they won't all be heavily loaded at the same time. You might even be able to upgrade the power feed later. Many panels are sold as 4 spaces/8 circuits.
Can 10 gauge wire handle 40 amps?
“Twelve-gauge wire is good for 20 amps, 10-gauge wire is good for 30 amps, 8-gauge is good for 40 amps, and 6-gauge is good for 55 amps,” and “The circuit breaker or fuse is always sized to protect the conductor [wire].”
Posted in FAQ
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# Financial Math - Compound Interest - Amount / Present value $A=P(1+\frac{i}{c})^{(tc)}$
1. 1
Calculate the amount if you invest $750 for 8 years at 5.5% compounded annually. 2. 2 Calculate the amount if you invest$1560 for 6 years at 8% compounded semi-annually.
3. 3
Calculate the amount if you invest $2500 for 5 years at 7.25% compounded quarterly. 4. 4 Calculate the amount if you invest$5600 for 14 years at 4.35% compounded monthly.
5. 5
Calculate the amount if you invest $5000 for 9 years at 6.45% compounded daily. 6. 6 How much do you need to invest now in order to have$4000 in 7 years at 9% compounded annually?
7. 7
How much do you need to invest now in order to have $7500 in 5 years at 7.25% compounded semi-annually? 8. 8 How much do you need to invest now in order to have$8400 in 3 years at 4.36% compounded quarterly?
9. 9
How much do you need to invest now in order to have \$15000 in 12 years at 6.75% compounded weekly?
10. 10
How long will it take 500 to amount to 1200 at 8.5% compounded quarterly?
Good test question!
11. 11
How long will it take for you to double your investment at 7.25% compounded semi-annually?
12. 12
Determine the interest rate necessary for 1600 to amount to 2100 in 7 years with interest compounded monthly.
13. 13
Determine the interest rate necessary for 750 to amount to 1350 in 12 years with interest compounded quarterly.
$$e=mc^2$$
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# Condition for the sequence $\left\{x_n\right\}$ being monotone where $x_1>0$ and $x_{n+1}=\frac{3(1+x_n)}{5+x_n}$ [closed]
A sequence $$\left\{x_n\right\}$$ is given by $$x_1>0$$ and $$x_{n+1}=\frac{3(1+x_n)}{5+x_n}$$ for all $$n\in\mathbb{N}$$.
How can I prove that
1. The sequence $$\left\{x_n\right\}$$ is monotone increasing if $$0.
2. The sequence $$\left\{x_n\right\}$$ is monotone decreasing if $$x_1>1$$.
I cannot prove this. Should I prove this by assuming first that the sequence is monotone increasing and thereby deduce the range of $$x_1$$, or what?
## closed as off-topic by rtybase, cmk, Shogun, Xander Henderson, José Carlos SantosJun 12 at 20:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – rtybase, cmk, Shogun, Xander Henderson, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
• You neediof $0<x<1$ then $x<\frac{3(1+x)}{5+x}<1.$ And likewise, if $x>1$ then $1<\frac{3(1+x)}{5+x}<x.$ – Thomas Andrews Jun 11 at 20:51
For the case $$0, we'll prove by induction that $$0.
Induction basis: $$0.
Induction step: Assume $$0. The inequality $$0 is obvious from the recursive formula because of $$0. Additionally, we have
$$x_{n+1}=\frac{3(1+x_n)}{5+x_n}=\frac{3+3x_n}{5+x_n}<\frac{5+x_n}{5+x_n}=1.$$
Thus, the induction is complete and $$0 holds for all $$n$$.
This implies
$$x_{n+1}=\frac{3(1+x_n)}{5+x_n}>\frac{3(1+x_n)}{5+1}=\frac{1}{2}(1+x_n)>\frac{1}{2}(x_n+x_n)=x_n$$
and hence, the sequence is monotonically increasing.
The second statement for the case $$x_1>1$$ is proved analogously.
$$\frac{x_{n+1}-1}{x_{n+1}+3}=\frac{\frac{3(1+x_n)}{5+x_n}-1}{\frac{3(1+x_n)}{5+x_n}+3}=\frac{2x_n-2}{6x_n+18}=\frac{1}{3}\cdot\frac{x_n-1}{x_n+3}.$$ Thus, $$\frac{x_n-1}{x_n+3}=\frac{x_1-1}{x_1+3}\cdot\left(\frac{1}{3}\right)^{n-1}$$ or $$1-\frac{4}{x_n+3}=\frac{x_1-1}{x_1+3}\cdot\left(\frac{1}{3}\right)^{n-1}.$$ Now, we see that for $$0 the sequence $$x$$ increases (because the expression $$\frac{x_1-1}{x_1+3}\cdot\left(\frac{1}{3}\right)^{n-1}$$ increases) and for $$x_1>1$$ it decreases.
• Can the down-voter explain, why did you do it? – Michael Rozenberg Jun 12 at 7:07
• I didn't downvote, but I was downvoted too ... It looks like a revenge for the what looks like a downvoted question. I voted to close it, it lacks of details anyway. – rtybase Jun 12 at 7:43
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# Newton's method converges linearly from sufficiently close to a root of finite multiplicity greater than one
## Statement
Suppose $f$ is a function of one variable that is at least one time continuously differentiable at a root $\alpha$. Further, suppose $f'(\alpha) = 0$, so that $\alpha$ is root of multiplicity greater than 1. Then, there exists $\varepsilon > 0$ such that for any $x_0 \in (\alpha - \varepsilon,\alpha + \varepsilon)$, the sequence obtained by applying Newton's method either reaches the root in finitely many steps or has linear convergence to the root $\alpha$. The convergence rate is $1/r$ where $r$ is the multiplicity of the root $\alpha$ (i.e., the order of $\alpha$ as a zero).
## Proof
Based on the information about the order of the root $\alpha$ being $r$, there exists $\varepsilon > 0$ and constants $A$ such that:
$A(x - \alpha)^r \le f(x) \le B(x - \alpha)^r$
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EDPY 401 Standardized Testing fall 2011
# EDPY 401 Standardized Testing fall 2011 - 11/26/2011 EDPY...
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11/26/2011 1 EDPY 401 Agenda Nov. 28 More on Classroom Assessment Standardized Tests Next Class: Lesson Plan Analysis and Assessment Plan Analysis due >1500 words Apply SPECIFIC CONCEPTS Informal Assessments Assessing Student Learning Types of Assessment Informal Formal Formative Summative Basic Concepts of Measurement Central tendency Score that is typical or representative of the entire group Mean Simple average Median Middle score in a series of scores Mode Most frequently-occurring score Variability How widely the scores are distributed with a group Basic Concepts of Measurement: Standard Deviation (SD) Degree of variability in a group of scores Equals the square root of each score’s deviation from the mean Small standard deviation Most scores are close to the mean. Large standard deviation Scores are more spread out.
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11/26/2011 2 Normal Distribution Normal distribution Symmetrical and bell- shaped frequency distribution Mean, median, and mode are equal and appear at the mid-point of the distribution. 68% of scores fall within one SD of the mean. Two SDs above and below the mean include approximately 95% of all scores. Three SDs above and below the mean include approximately 99% of all scores. Frequency Distributions Frequency distribution Simple list of all scores for a group Skewness The symmetry or asymmetry of a frequency distribution Negatively-skewed distribution Scores pile up toward the high end. Positively-skewed distribution Scores are piled up at the lower end. A positively skewed distribution is asymmetrical and points in the positive direction. If a test was very difficult and almost everyone in the class did very poorly on it, the resulting distribution would most
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## This note was uploaded on 02/22/2012 for the course EDPY 401 taught by Professor Hurst during the Fall '09 term at South Carolina.
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### Home > CC3 > Chapter Ch6 > Lesson 6.2.2 > Problem6-63
6-63.
The box plot at right shows the different grades (in percents) that students in Ms. Sanchez’s class earned on a recent test. Homework Help ✎
1. What was the median score on the test? What were the highest and lowest scores?
Do you remember how to read a box plot?
1. Did most students earn a particular score? How do you know?
Can this be determined from a box plot?
1. If Ms. Sanchez has $32$ students in her class, about how many students earned a grade of $80\%$ or higher? About how many earned more than $90\%$? Explain how you know.
Since $80\%$ is the median, half of the class earned $80\%$ or higher and half earned lower than $80\%$.
$\frac{1}{4}\text{ of the class earned 90%}$
$16$ students earned $80\%$ or higher, $8$ students earned $90\%$ or higher
1. Can you tell if the scores between $80\%$ and $90\%$ were closer to $80\%$ or closer to 90%? Explain.
Can this be determined from a box plot?
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author wenzelm Mon, 25 Feb 2013 20:11:42 +0100 changeset 51264 aba03f0c6254 parent 51263 31e786e0e6a7 child 51277 546a9a1c315d
fixed document;
```--- a/src/HOL/Library/Function_Growth.thy Sun Feb 24 20:29:13 2013 +0100
+++ b/src/HOL/Library/Function_Growth.thy Mon Feb 25 20:11:42 2013 +0100
@@ -11,7 +11,7 @@
text {*
When comparing growth of functions in computer science, it is common to adhere
- on Landau Symbols (\<guillemotright>O-Notation\<guillemotleft>). However these come at the cost of notational
+ on Landau Symbols (``O-Notation''). However these come at the cost of notational
oddities, particularly writing @{text "f = O(g)"} for @{text "f \<in> O(g)"} etc.
Here we suggest a diffent way, following Hardy (G.~H.~Hardy and J.~E.~Littlewood,
@@ -152,7 +152,7 @@
However @{text "f \<in> o(g) \<longrightarrow> f \<prec> g"} is provable, and this yields a
handy introduction rule.
- Note that D. Knuth ignores @{text o} altogether. So what\<dots>
+ Note that D. Knuth ignores @{text o} altogether. So what \dots
Something still has to be said about the coefficient @{text c} in
the definition of @{text "(\<prec>)"}. In the typical definition of @{text o},
@@ -291,10 +291,10 @@
text {* @{prop "(\<lambda>n. Suc k * f n) \<cong> f"} *}
lemma "f \<lesssim> (\<lambda>n. f n + g n)"
-by rule auto
+ by rule auto
lemma "(\<lambda>_. 0) \<prec> (\<lambda>n. Suc k)"
-by (rule less_fun_strongI) auto
+ by (rule less_fun_strongI) auto
lemma "(\<lambda>_. k) \<prec> Discrete.log"
proof (rule less_fun_strongI)
@@ -335,10 +335,10 @@
qed
lemma "id \<prec> (\<lambda>n. n\<twosuperior>)"
-by (rule less_fun_strongI) (auto simp add: power2_eq_square)
+ by (rule less_fun_strongI) (auto simp add: power2_eq_square)
lemma "(\<lambda>n. n ^ k) \<prec> (\<lambda>n. n ^ Suc k)"
-by (rule less_fun_strongI) auto
+ by (rule less_fun_strongI) auto
text {* @{prop "(\<lambda>n. n ^ k) \<prec> (\<lambda>n. 2 ^ n)"} *}
```
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# Two-moment decision models
Two-moment decision models
Mean-variance analysis redirects here. For mean-variance portfolio theory, see Modern portfolio theory or Mutual fund separation theorem.
In decision theory, economics, and finance, a two-moment decision model is a model that describes or prescribes the process of making decisions in a context in which the decision-maker is faced with random variables whose realizations cannot be known in advance, and in which choices are made based on knowledge of two moments of those random variables. The two moments are almost always the mean—that is, the expected value, which is the first moment about zero—and the variance, which is the second moment about the mean (or the standard deviation, which is the square root of the variance).
## Two-moment models and expected utility maximization
Suppose that all relevant random variables are in the same location-scale family, meaning that the distribution of every random variable is the same as the distribution of some linear transformation of any other random variable. Then for any von Neumann-Morgenstern utility function, using a mean-variance decision framework is consistent with expected utility maximization,[1][2] as illustrated in example 1:
Example 1:[3][4] Let there be one risky asset with random return r, and one riskfree asset with known return rf, and let an investor's initial wealth be w0. If the amount q, the choice variable, is to be invested in the risky asset and the amount w0q is to be invested in the safe asset, then contingent on q the investor's random final wealth will be w=(w0q)rf + qr. Then for any choice of q, w is distributed as a location-scale transformation of r. If we define random variable x as equal in distribution to $\tfrac{w-Ew}{\sigma_w},$ then w is equal in distribution to (Ew + σwx), where E represents an expected value and σ represents a random variable's standard deviation (the square root of its second moment). Thus we can write expected utility in terms of two moments of w :
$Eu(w)=\int_{- \infty} ^ \infty \! u(Ew+ \sigma _w x)f(x)dx \, \equiv v(Ew, \sigma_w),$
where u is the von Neumann-Morgenstern utility function, f is the density function of x, and v is the derived mean-standard deviation choice function, which depends in form on the density function f. The von Neumann-Morgenstern utility function is assumed to be increasing, implying that more wealth is preferred to less, and it is assumed to be concave, which is the same as assuming that the individual is risk averse.
It can be shown that the partial derivative of v with respect to Ew is positive, and the partial derivative of v with respect to σw is negative; thus more expected wealth is always liked, and more risk (as measured by the standard deviation of wealth) is always disliked. A mean-standard deviation indifference curve is defined as the locus of points (σw , Ew) with σw plotted horizontally, such that Eu(w) has the same value at all points on the locus. Then the derivatives of v imply that every indifference curve is upward sloped: that is, along any indifference curve dEw / dσw > 0. Moreover, it can be shown[3] that all such indifference curves are convex: along any indifference curve, d2Ew / dw)2 > 0.
Example 2: The portfolio analysis in example 1 can be generalized. If there are n risky assets instead of just one, and if their returns are jointly elliptically distributed, then[5][6] all portfolios can be characterized completely by their mean and variance—that is, any two portfolios with identical mean and variance of portfolio return have identical distributions of portfolio return—and all possible portfolios have return distributions that are location-scale-related to each other. Thus portfolio optimization can be implemented using a two-moment decision model.
Example 3: Suppose that a price-taking, risk-averse firm must commit to producing a quantity of output q before observing the market realization p of the product's price.[7] Its decision problem is to choose q so as to maximize the expected utility of profit:
Maximize Eu(pqc(q) – g),
where E is the expected value operator, u is the firm's utility function, c is its variable cost function, and g is its fixed cost. All possible distributions of the firm's random revenue pq, based on all possible choices of q, are location-scale related; so the decision problem can be framed in terms of the expected value and variance of revenue.
## Non-expected-utility decision making
If the decision-maker is not an expected utility maximizer, decision-making can still be framed in terms of the mean and variance of a random variable if all alternative distributions for an unpredictable outcome are location-scale transformations of each other.[8]
## References
1. ^ Mayshar, J., "A note on Feldstein's criticism of mean-variance analysis," Review of Economic Studies 45, 1978, 197-199.
2. ^ Sinn, H.-W., Economic Decisions under Uncertainty, second English edition, 1983, North-Holland.
3. ^ a b Meyer, Jack. "Two-moment decision models and expected utility maximization," American Economic Review 77, June 1987, 421-430.
4. ^ Tobin, J., "Liquidity preference as behavior towards risk," Review of Economic Studies 25(1), February 1958, 65–86. Also in: (1) M. G. Mueller, ed., Readings in Macroeconomics, Holt, Rinehart & Winston, Inc., 1966, pp. 65-86; (2) Richard S. Thorn, ed., Monetary Theory and Policy, Random House, 1966, pp. 172–191; (3) H. R. Williams and J. D. Huffnagle, eds., Macroeconomic Theory, Appleton-Century-Crofts, 1969, pp. 299–324; (4) Essays in Economics: Macroeconomics, Vol. 1, chapter 15; (5) J. Tobin and D. Hester, eds., Risk Aversion and Portfolio Choice, Cowles Monograph No. 19, John Wiley & Sons, 1967; (6) David Laidler, ed., The Foundations of Monetary Economics, Vol. 1, Edward Elgar Publishing Ltd., 1999.
5. ^ Chamberlain, G., "A characterization of the distributions that imply mean-variance utility functions", Journal of Economic Theory 29, 1983, 185-201.
6. ^ Owen, J., and Rabinovitch, R. "On the class of elliptical distributions and their applications to the theory of portfolio choice", Journal of Finance 38, 1983, 745-752.
7. ^ Sandmo, Agnar. "On the theory of the competitive firm under price uncertainty," American Economic Review 61, March 1971, 65-73.
8. ^ Bar-Shira, Z., and Finkelshtain, I., "Two-moments decision models and utility-representable preferences," Journal of Economic Behavior and Organization 38, 1999, 237-244. See also Mitchell, Douglas W., and Gelles, Gregory M., "Two-moments decision models and utility-representable preferences: A comment on Bar-Shira and Finkelshtain, vol. 49, 2002, 423-427.
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# The sum of first eight terms of a G.P. is five times the sum of the first four terms. Find the common ratio.
1
by raunaksaha1997
## Answers
2014-05-26T15:12:24+05:30
Sum of n terms of G.P. is = {a(r^n - 1)}/(r - 1)
according to question ;
{a(r^8 - 1)}/(r - 1) = 5{a(r^4 - 1)}/(r - 1)
(r^4)^2 - 1 = 5(r^4 - 1)
r^4 + 1 = 5
r^4 = 4
r^2 = +2 or -2
r^2 = 2
r = +2^1/2 ,or -2^1/2
r^2 = -2 square of any number can not be negative ,hence
r = 2^1/2 or -2^1/2 (r = common ratio)
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##### COATI - 2017
Research Program
Application Domains
New Software and Platforms
Partnerships and Cooperations
Bibliography
## Section: New Results
### Graph theory
Participants : Julien Bensmail, Guillaume Ducoffe, Frédéric Havet, William Lochet, Nicolas Nisse, Bruce Reed.
Coati studies theoretical problems in graph theory. If some of them are directly motivated by applications (see Subsection 7.3.3), others are more fundamental. In particular, we are putting an effort on understanding better directed graphs (also called digraphs) and partionning problems, and in particular colouring problems. We also try to better the understand the many relations between orientation and colourings. We study various substructures and partitions in (di)graphs. For each of them, we aim at giving sufficient conditions that guarantee its existence and at determining the complexity of finding it.
#### Substructures in (di)graphs
We study various conditions that ensure a (di)graph to contain certain substructures.
In [17], we study the question of finding a set of $k$ vertex-disjoint cycles (resp. directed cycles) of distinct lengths in a given graph (resp. digraph). In the context of undirected graphs, we prove that, for every $k\ge 1$, every graph with minimum degree at least $\frac{{k}^{2}+5k-2}{2}$ has $k$ vertex-disjoint cycles of different lengths, where the degree bound is the best possible. We also consider other cases such as when the graph is triangle-free, or the $k$ cycles are required to have different lengths modulo some value $r$. In the context of directed graphs, we consider a conjecture of Lichiardopol concerning the least minimum out-degree required for a digraph to have $k$ vertex-disjoint directed cycles of different lengths. We verify this conjecture for tournaments, and, by using the probabilistic method, for some regular digraphs and digraphs of small order.
A $\left({k}_{1}+{k}_{2}\right)$-bispindle is the union of ${k}_{1}$ $\left(x,y\right)$-dipaths and ${k}_{2}$ $\left(y,x\right)$-dipaths, all these dipaths being pairwise internally disjoint. Recently, Cohen et al. showed that for every $\left(1,1\right)$-bispindle $B$, there exists an integer $k$ such that every strongly connected digraph with chromatic number greater than $k$ contains a subdivision of $B$. In [24], we investigate generalisations of this result by first showing constructions of strongly connected digraphs with large chromatic number without any $\left(3,0\right)$-bispindle or $\left(2,2\right)$-bispindle. Then we show that strongly connected digraphs with large chromatic number contains a $\left(2,1\right)$-bispindle, where at least one of the $\left(x,y\right)$-dipaths and the $\left(y,x\right)$-dipath are long.
Let $ℋ$ be a family of graphs and let $d$ be large enough. For every $d$-regular graph $G$, we study the existence of a spanning $ℋ$-free subgraph of $G$ with large minimum degree. This problem is well understood if $ℋ$ does not contain bipartite graphs. In [35] we provide asymptotically tight results for many families of bipartite graphs such as cycles or complete bipartite graphs. To prove these results, we study a locally injective analogue of the question.
An even pair (resp. odd pair) in a graph is a pair of non-adjacent vertices such that every chordless path between them has even (resp. odd) length. Even and odd pairs are important tools in the study of perfect graphs and were instrumental in the proof of the Strong Perfect Graph Theorem. We suggest that such pairs impose a lot of structure also in arbitrary, not just perfect graphs. To this end, we show in [36] that the presence of even or odd pairs in graphs imply a special structure of the stable set polytope. In fact, we give a polyhedral characterization of even and odd pairs.
#### Colourings and partitioning (di)graphs
##### Colouring graphs with constraints on connectivity
A graph $G$ has maximal local edge-connectivity $k$ if the maximum number of edge-disjoint paths between every pair of distinct vertices $x$ and $y$ is at most $k$. We prove in [11] Brooks-type theorems for $k$-connected graphs with maximal local edge-connectivity $k$, and for any graph with maximal local edge-connectivity 3. We also consider several related graph classes defined by constraints on connectivity. In particular, we show that there is a polynomial-time algorithm that, given a 3-connected graph $G$ with maximal local connectivity 3, outputs an optimal colouring for $G$. On the other hand, we prove, for $k\ge 3$, that $k$-colourability is NP-complete when restricted to minimally $k$-connected graphs, and 3-colourability is NP-complete when restricted to $\left(k-1\right)$-connected graphs with maximal local connectivity $k$. Finally, we consider a parameterization of $k$-colourability based on the number of vertices of degree at least $k+1$, and prove that, even when $k$ is part of the input, the corresponding parameterized problem is FPT.
##### Sum-distinguishing edge-weightings
A $k$-edge-weighting of a graph $G$ is an application from $E\left(G\right)$ into $\left\{1,\cdots ,k\right\}$. An edge-weigthing is sum-distinguishing if for every two adajcent vertices $u$ and $v$, the sum of weights of edges incident to $u$ is distinct from the sum of of weights of edges incident to $v$. The celebrated 1-2-3-Conjecture (raised in 2004 by Karoński, Luczak and Thomason) asserts that every connected graph (except ${K}_{2}$, the complete graph on two vertices) admits a sum-distinguishing 3-edge-weighting. This conjecture attracted much attention and many variants are now studied. We study several of them.
Towards the 1-2-3-Conjecture, the best-known result to date is due to Kalkowski, Karoński and Pfender, who proved that it holds when relaxed to 5-edge-weightings. Their proof builds upon a weighting algorithm designed by Kalkowski for a total version (where also the vertices are weighted) of the problem. In [67], we present new mechanisms for using Kalkowski's algorithm in the context of the 1-2-3 Conjecture. As a main result we prove that every 5-regular graph admits a 4-edge-weighting that permits to distinguish its adjacent vertices via their incident sums.
In [66], we investigate the consequences on the 1-2-3 Conjecture of requiring a stronger distinction condition. Namely, we consider two adjacent vertices distinguished when their incident sums differ by at least 2. As a guiding line, we conjecture that every graph with no connected component isomorphic to ${K}_{2}$ admits a 5-edge-weighting permitting to distinguish the adjacent vertices in this stronger way. We verify this conjecture for several classes of graphs, including bipartite graphs and cubic graphs. We then consider algorithmic aspects, and show that it is NP-complete to determine the smallest $k$ such that a given bipartite graph admits such a $k$-edge-weighting. In contrast, we show that the same problem can be solved in polynomial time for a given tree.
In [13], we consider the following question, which stands as a directed analogue of the 1-2-3 Conjecture: Given any digraph $D$ with no arc $\stackrel{\to }{uv}$ verifying ${d}^{+}\left(u\right)={d}^{-}\left(v\right)=1$, is it possible to weight the arcs of $D$ with weights among $\left\{1,2,3\right\}$ so that, for every arc $\stackrel{\to }{uv}$ of $D$, the sum of incident weights out-going from $u$ is different from the sum of incident weights in-coming to $v$? We answer positively to this question, and investigate digraphs for which even the weights among $\left\{1,2\right\}$ are sufficient. In relation with the so-called 1-2 Conjecture, we also consider a total version of the problem, which we prove to be false. Our investigations turn to have interesting relations with open questions related to the 1-2-3 Conjecture.
In [21], we study the following question: Is it always possible to injectively assign the weights $1,...,|E\left(G\right)|$ to the edges of any given graph $G$ (with no component isomorphic to ${K}_{2}$) so that every two adjacent vertices of $G$ get distinguished by their sums of incident weights? One may see this question as a combination of the well-known 1-2-3 Conjecture and the Antimagic Labelling Conjecture. We exhibit evidence that this question might be true. Benefiting from the investigations on the Antimagic Labelling Conjecture, we first point out that several classes of graphs, such as regular graphs, indeed admit such assignments. We then show that trees also do, answering a recent conjecture of Arumugam, Premalatha, Bača and Semaničová-Feňovčíková. Towards a general answer to the question above, we then prove that claimed assignments can be constructed for any graph, provided we are allowed to use some number of additional edge weights. For some classes of sparse graphs, namely 2-degenerate graphs and graphs with maximum average degree 3, we show that only a small (constant) number of such additional weights suffices.
##### Variants of vertex- or edge-colouring
A colouring of a graph $G$ is properly connected if every two vertices of $G$ are the ends of a properly coloured path. In [57], [47], we study the complexity of computing the proper connection number (minimum number of colours in a properly connected colouring) for edge and vertex colourings, in undirected and directed graphs, respectively. First we disprove some conjectures of Magnant et al. (2016) on characterizing the strong digraphs with proper arc connection number at most two. Then, we prove that deciding whether a given digraph has proper arc connection number at most two is NP-complete. Furthermore, we show that there are infinitely many such digraphs with no even-length dicycle. We initiate the study of proper vertex connectivity in digraphs and we prove similar results as for the arc version. Finally, we present polynomial-time recognition algorithms for bounded-treewidth graphs and bipartite graphs with proper edge connection number at most two.
A graph is locally irregular if no two adjacent vertices have the same degree. The irregular chromatic index ${\chi }_{\mathrm{irr}}^{\text{'}}\left(G\right)$ of a graph $G$ is the smallest number of locally irregular subgraphs needed to edge-decompose $G$. Not all graphs have such a decomposition, but Baudon, Bensmail, Przybyło, and Woźniak conjectured that if $G$ can be decomposed into locally irregular subgraphs, then ${\chi }_{\mathrm{irr}}^{\text{'}}\left(G\right)\le 3$. In support of this conjecture, Przybyło showed that ${\chi }_{\mathrm{irr}}^{\text{'}}\left(G\right)\le 3$ holds whenever $G$ has minimum degree at least ${10}^{10}$. In [19] we prove that every bipartite graph $G$ which is not an odd length path satisfies ${\chi }_{\mathrm{irr}}^{\text{'}}\left(G\right)\le 10$. This is the first general constant upper bound on the irregular chromatic index of bipartite graphs. Combining this result with Przybyło's result, we show that ${\chi }_{\mathrm{irr}}^{\text{'}}\left(G\right)\le 328$ for every graph $G$ which admits a decomposition into locally irregular subgraphs. Finally, we show that ${\chi }_{\mathrm{irr}}^{\text{'}}\left(G\right)\le 2$ for every 16-edge-connected bipartite graph $G$.
An $\left(m,n\right)$-coloured mixed graph is a mixed graph with arcs assigned one of $m$ different colours and edges one of $n$ different colours. A homomorphism of an $\left(m,n\right)$-coloured mixed graph $G$ to an $\left(m,n\right)$-coloured mixed graph $H$ is a vertex mapping such that if $uv$ is an arc (edge) of colour $c$ in $G$, then $f\left(u\right)f\left(v\right)$ is also an arc (edge) of colour $c$. The $\left(m,n\right)$-coloured mixed chromatic number, denoted ${\chi }_{m,n}\left(G\right)$, of an $\left(m,n\right)$-coloured mixed graph $G$ is the order of a smallest homomorphic image of $G$. An $\left(m,n\right)$-clique is an $\left(m,n\right)$-coloured mixed graph $C$ with ${\chi }_{m,n}\left(C\right)=|V\left(C\right)|$. In [16], we study the structure of $\left(m,n\right)$-cliques. We show that almost all $\left(m,n\right)$-coloured mixed graphs are $\left(m,n\right)$-cliques, prove bounds for the order of a largest outerplanar and planar $\left(m,n\right)$-clique and resolve an open question concerning the computational complexity of a decision problem related to $\left(0,2\right)$-cliques. Additionally, we explore the relationship between ${\chi }_{1,0}$ and ${\chi }_{0,2}$.
An edge colouring of a graph $G$ is called acyclic if it is proper and every cycle contains at least three colours. We show in [33] that for every $\epsilon >0$, there exists a $g=g\left(\epsilon \right)$ such that if $G$ has maximum degree $\Delta$ and girth at least $g$ then $G$ admits an acyclic edge colouring with $\left(1+\epsilon \right)\Delta +O\left(1\right)$ colours.
#### Identifying codes
Let $G$ be a graph $G$. The neighborhood of a vertex $v$ in $G$, denoted by $N\left(v\right)$, is the set of vertices adjacent to $v$ in $G$. Its closed neighborhood is the set $N\left[v\right]=N\left(v\right)\cup \left\{v\right\}$. A set $C\subseteq V\left(G\right)$ is an identifying code in $G$ if (i) for all $v\in V\left(G\right)$, $N\left[v\right]\cap C\ne \varnothing$, and (ii) for all $u,v\in V\left(G\right)$, $N\left[u\right]\cap C\ne N\left[v\right]\cap C$. The problem of finding low-density identifying codes was introduced in [Karpovsky et al., IEEE Trans. Inform. Theory 44, 1998] in relation to fault diagnosis in arrays of processors. Here the vertices of an identifying code correspond to controlling processors able to check themselves and their neighbors. Thus the identifying property guarantees location of a faulty processor from the set of “complaining” controllers. Identifying codes are also used in [Ray et al., IEEE Journal on Selected Areas in Communications 22, 2004] to model a location detection problem with sensor networks.
A particular interest was dedicated to grids as many processor networks have a grid topology. There are several types of standard regular infinite grids, in particular the hexagonal grids, the square grids, the triangular grids and the king grids. For such graphs $G$, the problem consists in finding the minimum density ${d}^{*}\left(G\right)$ of an identifying code of $G$.
In [26], we study the infinite triangular grid ${T}_{k}$ with $k$ rows. We show ${d}^{*}\left({T}_{1}\right)={d}^{*}\left({T}_{2}\right)=1/2$, ${d}^{*}\left({T}_{3}\right)={d}^{*}\left({T}_{4}\right)=1/3$, ${d}^{*}\left({T}_{5}\right)=3/10$, ${d}^{*}\left({T}_{6}\right)=1/3$ and ${d}^{*}\left({T}_{k}\right)=1/4+1/\left(4k\right)$ for all odd $k\ge 7$. In addition, we show that $1/4+1/\left(4k\right)\le {d}^{*}\left({T}_{k}\right)\le 1/4+1/\left(2k\right)$ for all even $k\ge 8$.
In [27], we study the density of king grids which are strong product of two paths. We show that for every king grid $G$, ${d}^{*}\left(G\right)\ge 2/9$. In addition, we show this bound is attained only for king grids which are strong products of two infinite paths. Given $k\ge 3$, we denote by ${K}_{k}$ the (infinite) king strip with $k$ rows. We prove that ${d}^{*}\left({K}_{3}\right)=1/3$, ${d}^{*}\left({K}_{4}\right)=5/16$, ${d}^{*}\left({K}_{5}\right)=4/15$ and ${d}^{*}\left({K}_{6}\right)=5/18$. We also prove that $2/9+8/81k\le {d}^{*}\left({K}_{k}\right)\le 2/9+4/9k$ for every $k\ge 7$.
#### Miscellaneous
##### A proof of the Barát-Thomassen conjecture
The Barát-Thomassen conjecture asserts that for every tree $T$ on $m$ edges, there exists a constant ${k}_{T}$ such that every ${k}_{T}$-edge-connected graph with size divisible by $m$ can be edge-decomposed into copies of $T$. So far this conjecture has only been verified when $T$ is a path or when $T$ has diameter at most 4. In [18], we prove the full statement of the conjecture.
##### Recursively partitionable graphs
A connected graph $G$ is said to be arbitrarily partitionable (AP for short) if for every partition $\left({n}_{1},...,{n}_{p}\right)$ of $|V\left(G\right)|$ there exists a partition $\left({V}_{1},...,{V}_{p}\right)$ of $V\left(G\right)$ such that each ${V}_{i}$ induces a connected subgraph of $G$ on ${n}_{i}$ vertices. Some stronger versions of this property were introduced, namely the ones of being online arbitrarily partitionable and recursively arbitrarily partitionable (OL-AP and R-AP for short, respectively), in which the subgraphs induced by a partition of $G$ must not only be connected but also fulfil additional conditions. In [14], we point out some structural properties of OL-AP and R-AP graphs with connectivity 2. In particular, we show that deleting a cut pair of these graphs results in a graph with a bounded number of components, some of whom have a small number of vertices. We obtain these results by studying a simple class of 2-connected graphs called balloons
##### On oriented cliques with respect to push operation
An oriented graph is a directed graph without any directed cycle of length at most 2. An oriented clique is an oriented graph whose non-adjacent vertices are connected by a directed 2-path. To push a vertex $v$ of a directed graph $\stackrel{\to }{G}$ is to change the orientations of all the arcs incident to $v$. A push clique is an oriented clique that remains an oriented clique even if one pushes any set of vertices of it. We show in [20] that it is NP-complete to decide if an undirected graph is the underlying graph of a push clique or not. We also prove that a planar push clique can have at most 8 vertices and provide an exhaustive list of planar push cliques.
##### On $q$-power cycles in cubic graphs
In the context of a conjecture of Erdős and Gyárfás, we consider in [15], for any $q\ge 2$, the existence of $q$-power cycles (i.e. with length a power of $q$) in cubic graphs. We exhibit constructions showing that, for every $q\ge 3$, there exist arbitrarily large cubic graphs with no $q$-power cycles. Concerning the remaining case $q=2$ (which corresponds to the conjecture of Erdős and Gyárfás), we show that there exist arbitrarily large cubic graphs whose only 2-power cycles have length 4 only, or 8 only.
##### How to determine if a random graph with a fixed degree sequence has a giant component
For a fixed degree sequence $𝒟=\left({d}_{1},...,{d}_{n}\right)$, let $G\left(𝒟\right)$ be a uniformly chosen (simple) graph on $\left\{1,\cdots ,n\right\}$ where the vertex $i$ has degree ${d}_{i}$. In [34] we determine whether $G\left(𝒟\right)$ has a giant component with high probability, essentially imposing no conditions on $𝒟$. We simply insist that the sum of the degrees in $𝒟$ which are not 2 is at least $\lambda \left(n\right)$ for some function $\lambda$ going to infinity with $n$. This is a relatively minor technical condition, and when $𝒟$ does not satisfy it, both the probability that $G\left(𝒟\right)$ has a giant component and the probability that $G\left(𝒟\right)$ has no giant component are bounded away from 1.
##### A proof of the Erdős-Sands-Sauer-Woodrow conjecture
A very nice result of Barany and Lehel asserts that every finite subset $X$ of ${R}^{d}$ can be covered by $f\left(d\right)$ $X$-boxes (i.e. each box has two antipodal points in $X$). As shown by Gyárfás and Pálvőlgyi this result would follow from the following conjecture : If a tournament admits a partition of its arc set into $k$ partial orders, then its domination number is bounded in terms of $k$. This question is in turn implied by the Erdős-Sands-Sauer-Woodrow conjecture : If the arcs of a tournament $T$ are colored with $k$ colors, there is a set $X$ of at most $g\left(k\right)$ vertices such that for every vertex $v$ of $T$, there is a monochromatic path from $X$ to $v$. We give in [69] a short proof of this statement. We moreover show that the general Sands-Sauer-Woodrow conjecture (which as a special case implies the stable marriage theorem) is valid for directed graphs with bounded stability number. This conjecture remains however open.
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# Untitled
By: a guest on May 2nd, 2012 | syntax: None | size: 4.26 KB | hits: 17 | expires: Never
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1. /* MATH1155 C programming assignment */
2. /* file: intersect.c */
3. /* created by: Ian Grundy 01/04/12 */
4. /* modified by: */
5.
6. # include <stdio.h>
7. # include <stdlib.h>
8. # include <math.h>
9.
10. # define PI 3.1415926535897932385
11.
12. /* Function prototypes */
13.
14. double f(double x);
15. double dfdx(double x);
16.
17. double si(double x);
18. double diffsi(double x);
19.
20. double myrandom();
21. double heuristic(double left, double right, int maxiter);
22. double bisection(double left, double right, double epsilon);
23. double secant(double left, double right, double epsilon, int maxiter);
24. double newton(double initial, double epsilon, int maxiter);
25.
26. int main(void)
27. {
28.
29. double eps = 1.0e-12;
30. int maxiter;
31. double b, n, s, h;
32.
33. maxiter = 5000;
34. h = heuristic(0, 3.0, maxiter);
35.
36. maxiter = 500;
37. b = bisection(0, 3.0, eps);
38. s = secant(0, 3.0, eps, maxiter);
39. n = newton(1.5, eps, maxiter);
40.
41. printf (" heuristic bisection secant newton\n");
42. printf ("%15.10f %15.10f %15.10f %15.10f %f\n", h, b, s, n, si(1.5));
43.
44. return 0;
45.
46. }
47.
48. /* Function Definitions */
49.
50. double factorial(int n)
51. {
52. int i;
53. double product = 1;
54. for ( i=2; i <= n; i=i+1) product = product * i;
55. return product;
56. }
57.
58.
59. double si(double x)
60. {
61. int i;
62. double z;
63. double sum=0;
64. for(i=1;;i++)
65. {
66. z = ((pow((-1),i))*(pow(x,(2*i+1))))/((2*i+1)*(factorial(2*i+1)));
67. if (fabs(z) < exp(-12))
68. {
69. break;
70. }
71. else
72. {
73. sum = sum + z;
74. }
75.
76. }
77. return sum;
78. }
79.
80. double diffsi(double x)
81. {
82. double sum = 1;
83. int i;
84. double z;
85.
86. for(i=0;;i++)
87. {
88. z = ((pow((-1),i))*(pow(x,(2*i+2))))/(factorial(2*i+1));
89. if (fabs(z) < exp(-12))
90. {
91. break;
92. }
93. else
94. {
95. sum = sum + z;
96. }
97. }
98. return sum;
99. }
100.
101.
102.
103. double f(double x)
104. {
105. return si(x)-PI/2.0;
106. }
107.
108. double dfdx(double x)
109. {
110. return diffsi(x);
111. }
112.
113.
114. double myrandom()
115. {
116. return (double) rand() / (double) RAND_MAX;
117. }
118.
119. double heuristic(double left, double right, int maxiter)
120. {
121.
122. /* simulated annealing function minimiser */
123.
124. double rad = (right - left)/100.0;
125. double r = 0.99;
126. double T = 100;
127. double prob;
128.
129. int iter = 0;
130. double bestx, currentx, newx;
131. double besty, currenty, newy;
132.
133. currentx = (left + right)/2.0;
134. currenty = fabs(f(currentx));
135.
136. bestx = currentx;
137. besty = currenty;
138.
139. for (iter = 1; iter < maxiter+1; iter++)
140. {
141. newx = currentx + (2*myrandom()-1)*rad;
142. newy = fabs(f(newx));
143.
144. if ( newy < currenty )
145. {
146. /* accept if better */
147. currentx = newx;
148. currenty = newy;
149. }
150. else
151. {
152. /* possibly accept if worse */
153. prob = exp(-(newy-currenty)/T);
154. if ( myrandom() < prob)
155. {
156. currentx = newx;
157. currenty = newy;
158. }
159. }
160.
161. if ( currenty < besty )
162. {
163. /* store absolute best found */
164. bestx = currentx;
165. besty = currenty;
166. }
167.
168. T = r*T;
169.
170. }
171.
172. return bestx;
173. }
174.
175. double bisection(double left, double right, double epsilon)
176. {
177. double midpoint, fmid, fleft, fright;
178.
179. fleft = f(left);
180. fright = f(right);
181.
182. if (fleft*fright > 0) return -99;
183.
184. while (fabs(right - left) > 2*epsilon)
185. {
186. midpoint = (left + right)/2.0;
187. fmid = f(midpoint);
188.
189. if (fmid*fleft >= 0) left = midpoint;
190. if (fmid*fright > 0) right = midpoint;
191. }
192.
193. return (left + right)/2.0;
194. }
195.
196. double secant(double left, double right, double epsilon, int maxiter)
197. {
198. return (left + right)/2.0;
199. }
200.
201. double newton(double initial, double epsilon, int maxiter)
202. {
203. return initial;
204. }
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https://www.brainia.com/essays/Fin-571-Week-4/454097.html
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# FIN 571 Week 4 WileyPLUS Practice Quiz NEW
## FIN 571 Week 4 WileyPLUS Practice Quiz NEW
FIN 571 Week 4 WileyPLUS Practice Quiz NEW
Check this A+ Guidelines at
http://www.assignmentclick.com/FIN-571-NEW/FIN-571-Week-4-WileyPLUS-Practice-Quiz-NEW
For more classes visit
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FIN 571 Week 4 WileyPLUS Practice Quiz NEW
Multiple Choice Question 66
Present value: Tommie Harris is considering an investment that pays 6.5 percent annually. How much must he invest today such that he will have \$25,000 in seven years? (Round to the nearest dollar.)
\$38,850
\$23,474
\$16,088
\$26,625
Multiple Choice Question 61
PV of multiple cash flows: Jack Stuart has loaned money to his brother at an interest rate of 5.75 percent. He expects to receive \$625, \$650, \$700, and \$800 at the end of the next four years as complete repayment of the loan with interest. How much did he loan out to his brother? (Round to the nearest dollar.)
\$2,250
\$2,545
\$2,713
\$2,404
Multiple Choice Question 63
PV of multiple cash flows: Hassan Ali has made an investment that will pay him \$11,455, \$16,376, and \$19,812 at the end of the next three years. His investment was to fetch him a return of 14 percent. What is the present value of these cash flows? (Round to the nearest dollar.)
\$33,124
\$36,022
\$41,675
\$39,208
Multiple Choice Question 65
PV of multiple cash flows: Pam Gregg is expecting cash flows of \$50,000, \$75,000, \$125,000, and \$250,000 from an inheritance over the next four years. If she can earn 11 percent on any investment that she makes, what is the present value of her inheritance? (Round to the nearest dollar.)
\$361,998
\$309,432
\$434,599
\$412,372
Multiple Choice Question 66
Present value of an annuity: Transit Insurance Company has made an investment in another company that will guarantee it a cash flow of \$37,250 each year for the next five years. If the company uses a discount rate of 15 percent on its investments, what is the present value of this investment? (Round to the nearest dollar.)
\$186,250...
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http://www.spoj.com/problems/AMR10B/
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## AMR10B - Regex Edit Distance
no tags
A regular expression is used to describe a set of strings. For this problem the alphabet is limited to 'a' and 'b'. R is a regular expression if:
1) R is "a" or "b"
2) R is of the form "(R1R2)" where R1 and R2 are regular expressions
3) R is of the form "(R1|R2)" where R1 and R2 are regular expressions
4) R is of the form "(R1*)" where R1 is a regular expression.
The set of strings recognised by R are as follows:
1) If R is "a", then the set of strings recognised = {a}
2) If R is "b", then the set of strings recognised = {b}
3) if R is of the form "(R1R2)" then the set of strings recognised = all strings which can be obtained by a concatenation of strings s1 and s2 where s1 is recognised by R1 and s2 by R2.
4) if R is of the form "(R1|R2)" then the set of strings recognised = union of the set of strings recognised by R1 and R2.
5) If R is of the form "(R1*)" then the the strings recognised are the empty string and the concatenation of an arbitrary number of copies of any string recognised by R1.
The edit distance between two strings s1 and s2 is the minimum number of characters to be inserted/deleted or replaced in s1 to make it equal to s2.
Given two regular expressions R1 and R2, find the minimum edit distance amongst all pairs of strings s1 and s2 such that s1 is recognised by R1 and s2 is recognised by R2.
INPUT
The first line contains the number of test cases T. T test cases follow.
Each test case contains two lines containing two regular expressions R1 and R2. There is a blank line after each test case.
OUTPUT
Output T lines one corresponding to each test case containing the required answer for the corresponding test case.
CONSTRAINTS
T <= 100
1 <= length(R1), length(R2) <= 50
You are guaranteed that R1 and R2 will conform to the definition provided above.
SAMPLE INPUT
2
((a|b)*)
(a(b(aa)))
(a((ab)*))
(a(b(((ab)b)b)))
SAMPLE OUTPUT
0
2
EXPLANATION
For the first case R1 recognises all strings over the alphabet a and b. Thus it recognises the string "abaa" which is also recognised by R2.
For the second case the strings "aababab" and "ababbb" are recognised by R1 and R2 respectively and have edit distance of 2.
Added by: Varun Jalan Date: 2010-12-13 Time limit: 1.187s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All except: ASM64 Resource: own problem, ICPC Asia regionals, Amritapuri 2010
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https://codegolf.stackexchange.com/questions/130290/the-snail-in-the-well/130302
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# Background
There's a common riddle that goes something like this:
A snail is at the bottom of a 30 foot well. Every day the snail is able to climb up 3 feet. At night when they sleep, they slide back down 2 feet. How many days does it take for the snail to get out of the well?
30 days, because the snail climbs at 1 foot per day for 30 days to reach the top,
28 days, because once the snail is 27 feet in the air (after 27 days), they will simply climb the remaining 3 feet to the top on the 28th day.
# Challenge
This challenge generalizes this riddle. Given three positive integers as input, representing the total height, the climb height, and the fall height, return the number of days it will take to climb out of the well.
If the snail cannot climb out of the well, you may return 0, return a falsy value, or throw an exception. You may also write code that will halt if and only if a solution exists.
If you wish, you may take the fall height as a negative integer.
# Test Cases
(30, 3, 2) -> 28
(84, 17, 15) -> 35
(79, 15, 9) -> 12
(29, 17, 4) -> 2
(13, 18, 8) -> 1
( 5, 5, 10) -> 1
( 7, 7, 7) -> 1
(69, 3, 8) -> None
(81, 14, 14) -> None
# Scoring
This is , so the shortest answer in each language wins.
# Gray Snail, 1206 bytes for numeric I/O, 149 bytes for unary I/O
For fun. Composition of first program:
• 451 bytes, converting number into dots
• 121 bytes, core function (a separated version is written below)
• 634 bytes, converting dots into number
Take numeric input and output. Input is A, B, C respectively. Compared to other (near) O(1) answer, the code has a complexity of O(n). But for large number, it may eat up your memory first.
Hang if no solution is found.
INPUT p
POP Z r .!
f
POP Z o .
q
POP Z p [p]
GOTO [Z]
0
POP Z n .
GOTO w
1
POP Z n ..
GOTO w
2
POP Z n ...
GOTO w
3
POP Z n ....
GOTO w
4
POP Z n .....
GOTO w
5
POP Z n ......
GOTO w
6
POP Z n .......
GOTO w
7
POP Z n ........
GOTO w
8
POP Z n .........
GOTO w
9
POP Z n ..........
GOTO w
w
POP Z o .[o][o][o][o][o][o][o][o][o][o][n]
GOTO [r] [p]
GOTO q
!
POP Z A .[o]
INPUT p
POP Z r .@
GOTO f
@
POP Z B .[o]
INPUT p
POP Z r .#
GOTO f
#
POP Z C .[o]
POP H N .[B]
U
POP Z A [A]
POP Z B [B]
GOTO D [A]
GOTO $[B] GOTO U$
POP Z A .[A][C]
POP Z H ..[H]
POP Z B .[N]
GOTO U
D
POP Z r .
POP Z M .
POP Z N ...........
POP Z z .[N]
POP Z V .[H]
+
GOTO l[V] [H]
POP Z H [H]
POP Z z [z]
GOTO ( [z]
GOTO +
(
GOTO ) [H]
POP Z z .[N]
POP Z M ..[M]
POP Z V .[H]
GOTO +
)
POP Z r .0[r]
POP Z M ..[M]
POP Z H .[M]
POP Z M .
POP Z V .[H]
POP Z z .[N]
GOTO +
l
POP Z r .0[r]
GOTO -
l.
POP Z r .1[r]
GOTO -
l..
POP Z r .2[r]
GOTO -
l...
POP Z r .3[r]
GOTO -
l....
POP Z r .4[r]
GOTO -
l.....
POP Z r .5[r]
GOTO -
l......
POP Z r .6[r]
GOTO -
l.......
POP Z r .7[r]
GOTO -
l........
POP Z r .8[r]
GOTO -
l.........
POP Z r .9[r]
GOTO -
-
GOTO / [M]
POP Z H .[M]
POP Z M .
POP Z V .[H]
POP Z z .[N]
GOTO +
/
OUTPUT [r]
f is a (maybe) recursive function to convert integers into dots. Argument is saved in [p] and output in [o].
U is a function testing S1>=S2, storing parameter in B, A while saving A-B into A.
Code starting from D is a stub converting dots into numbers.
The underlying principle is the same with my C answer (ripping off falsy output for impossible solutions).
## Standalone version, 149 156157167170230 bytes, only support unary I/O
Input needs to be dots, e.g. .......... for 10.
INPUT A
INPUT B
INPUT C
POP N H .
GOTO U
$POP N A .[A][C] POP Z H ..[H] U POP Z A [A] POP Z N ..[N] GOTO D [A] GOTO$ .[B] [N]
GOTO U
D
OUTPUT .[H]
U calculates A=A-B, and jumps to D when A<=0. Otherwise $ assigns A+C to A and call U. Hang if no solution is found. Tricks: abuse the "compiler"'s ability to interpret empty string. You can rip off conditions in GOTO statement to make unconditioned jumps and the same trick works for POP. Remark: I may golf it more by 3 bytes but by doing so, mine and WheatWizard's answer would have the exact same logic. The result is probably the shortest GraySnail solution and I'm trying to prove it. • You made it first – Евгений Новиков Jul 7 '17 at 14:45 • Hey I just thought I'd let you know that I've made mine shorter than yours again. Its only a byte shorter and it draws some inspiration from your latest golf. – Sriotchilism O'Zaic Jul 11 '17 at 14:58 • @WheatWizard I have a 155-byte solution based on your old answer. But for sportsmanship, I will not view it as my answer. – Keyu Gan Jul 11 '17 at 15:09 • @KeyuGan No, go ahead. I don't care about the rep its all about the game. I'm happy to be beaten. If my code can be golfed is my fault for not seeing it. :) – Sriotchilism O'Zaic Jul 11 '17 at 15:11 • @WheatWizard Me neither. I'm sure it is the best time I've ever had on PPCG. – Keyu Gan Jul 11 '17 at 15:16 Note: the byte count is being questioned by Martin Ender in the comments. It seems there is no clear consensus about what to do with named, recursive lambda expressions in C# answers. So I have asked a question in Meta about it. # C# (.NET Core), 32 31 bytes f=(a,b,c)=>a>b?1+f(a-b+c,b,c):1 Try it online! A recursive approach. If the snail cannot escape, it ends with the following message: Process is terminating due to StackOverflowException. • 1 byte saved thanks to LiefdeWen! • You can save a byte byte changing a<=b to a>b and swapping the following parts – LiefdeWen Jul 7 '17 at 9:03 • Exact same code works in ES6 f=(a,b,c)=>a<=b?1:1+f(a-b+c,b,c) – Tushar Jul 7 '17 at 9:09 • You'll have to count the code that assigns the function to a name if you're relying on that name being f for the recursive call. – Martin Ender Jul 7 '17 at 9:12 • I don't golf in C# so I'm not entirely sure what the consensus is, but I would've expected this to require a full statement with a declaration of f and a semicolon if it's named. The first thing I found is this but there's no clear consensus here. – Martin Ender Jul 7 '17 at 9:15 • @MartinEnder I usually just do as Carlos has done here, as the declaration is only f=... I'm unsure as to whether or not we should add the semi-colon on the end though. – TheLethalCoder Jul 7 '17 at 9:55 # GRAY SNAIL, 219206169167159156 146 bytes (unary IO) INPUT a INPUT u INPUT d POP U c GOTO 1 3 POP f a [a][d] POP U c ..[c] 1 GOTO 2 [a] GOTO 3 [U] [u] POP f U ..[U] POP f a [a] GOTO 1 2 OUTPUT [c]. I think I can golf this down a bit. • Congratulations! – Keyu Gan Jul 11 '17 at 15:44 ## JavaScript (ES6), 3128 27 bytes Saved a few bytes thanks to @Arnauld I hadn't realized we could fail with an exception. Pretty sure this is optimal: u=>d=>g=h=>h>u?1+g(h-u+d):1 Assign to a variable with e.g. f=, then call like f(climb)(fall)(height). Throws InternalError: too much recursion if the climb is impossible. ## JavaScript (ES6), 38 bytes f=(h,u,d=0)=>h>u?u>0?1+f(h-u,u-d):+f:1 A recursive function that returns the number of days, or NaN for never. ### Test cases let f=(h,u,d=0)=>h>u?u>0?1+f(h-u,u-d):+f:1; [ [30, 3, 2], [84, 17, 15], [79, 15, 9], [29, 17, 4], [13, 18, 8], [ 5, 5, 10], [ 7, 7, 7], [69, 3, 8], [81, 14, 14] ].map(x => console.log(x + '', '->', f.apply(null, x))) • That's obvious: If snail does too much recursion, then climb is impossible. :) – Tushar Jul 7 '17 at 11:25 • Maybe 27 with a reversed currying syntax? d=>u=>g=h=>h>u?1+g(h-u+d):1 – Arnauld Jul 7 '17 at 11:36 • @Arnauld Thanks, that works surprisingly well... – ETHproductions Jul 7 '17 at 11:38 • I'm a it confused re the byte count- in one the variable the function is assigned t is included, the other not? – Orangesandlemons Jul 7 '17 at 15:39 • @Orangesandlemons in the top version, you have g= in the middle because this variable stores the intermediate function needed for the recursive call. The longer answer does a recursive call on f, which mandates that the name be included in the byte count. – musicman523 Jul 7 '17 at 18:11 # Excel, 51 46 bytes -1 byte thanks to @Scarabee. -4 because INT(x) = FLOOR(x,1) =IF(B1<A1,IF(C1<B1,-INT((B1-A1)/(B1-C1)-1)),1) Input taken from Cells A1, B1 and C1 respectively. Returns FALSE for invalid scenarios. • ceiling(x) is always equal to -floor(-x), so I think you could save 1 byte by replacing CEILING((A1-B1)/(B1-C1)+1,1) with -FLOOR((B1-A1)/(B1-C1)+1,1). – Scarabee Jul 8 '17 at 22:29 # C (gcc), 39 434446475860 bytes Only on 32-bit GCC and all optimizaitons turned off. f(a,b,c){a=a>b?b>c?1+f(a-b+c,b,c):0:1;} Return 0 when solution is impossible. A modified version of original recursive solution. Inspired by @Jonah J solution and @CarlosAlejo C# solution. I'll update the expanded version later (after I finish my Grey Snail answer). • Nice one! could u please include the analytical (non-compressed) solution? – koita_pisw_sou Jul 7 '17 at 7:15 • @koita_pisw_sou Sure. – Keyu Gan Jul 7 '17 at 7:22 • It doesn't "return" anything at all. You assign to a local parameter, whose value evaporates once the function returns. The snail is stuck in eternal limbo. – Cody Gray Jul 8 '17 at 0:51 • @CodyGray it uses a stable but undefined behavior in GCC. I could show you a link later. – Keyu Gan Jul 8 '17 at 3:27 • @CodyGray codegolf.stackexchange.com/questions/2203/tips-for-golfing-in-c Assign instead of return – Keyu Gan Jul 8 '17 at 4:56 # Java (OpenJDK 8), 35 bytes (a,b,c)->b<a?c<b?(a+~c)/(b-c)+1:0:1 Try it online! Math wins! ### Credits • It's been a while, but a-c-1a+~c. – Kevin Cruijssen Mar 26 '18 at 15:00 • Thanks @KevinCruijssen It's been a while, but golf is golf, no matter when it happens :-) – Olivier Grégoire Mar 26 '18 at 15:03 • My thoughts exactly. On a few occasions I golfed about halve my original bytes when I looked at some of my first answers. ;) – Kevin Cruijssen Mar 26 '18 at 15:12 # Python 2, 37 bytes f=lambda x,y,z:x-y<1or 1+f(x-y+z,y,z) Try it online! Finally got my recursive version below my standard calculation (I was passing a count to my function instead of adding one before calling it). # Python 2, 43 46 bytes #43 bytes lambda x,y,z:y/x>0 or[1-(x-y)/(z-y),0][z/y] #46 bytes lambda x,y,z:y/x and 1or[1-(x-y)/(z-y),0][z/y] Try it online! Shaved 3 bytes by trading "__ and 1" for "__>0". Using boolean trickery, essentially executes: if floor(y/x) > 0: return True # == 1 elif floor(z/y) == 1: return 0 elif floor(z/y) == 0: return 1-floor((x-y)/(z-y)) # Python 2 implicitly treats integer division as floor division # equivalent: 1 + math.ceil((y-x)/(z-y)) # because: -floor(-x) == ceil(x) • You have to put f= in front of your code (the first solution), and your byte count becomes 37, because it is recursive, so you can't leave it anonymous. f= can be dropped for a lambda only when it is not recusive. – Mr. Xcoder Jul 7 '17 at 9:09 • Noted and addressed. Thanks for letting me know. – Coty Johnathan Saxman Jul 7 '17 at 9:20 # R, 43 bytes Borrowing from other answers: g=function(a,b,c)if(b<a,1+g(a-b+c,b,c),1) Gives error if no solution. • Nice answer. Welcome to PPCG! – musicman523 Jul 9 '17 at 6:09 # J, 25 bytes First a nice solution, which is a cheat, since it assumes that "anything other than a positive integer result" equals "None": >.>:%/2-/\ ## explanation • 2-/\ use windows of length 2 across our 3 item input, placing a minus sign between each one, which for the input 30 3 2, eg, returns 27 1 • %/ put a division symbol between each element of the list, in our case the list has only two items, so it means "divide 27 by 1" • >: increment by 1 • >. take the ceiling ## official solution Here is the official solution that converts negatives and infinity to 0, which part i was not able to find a satisfyingly terse solution for: 0:[@.(>&0*<&_)>.>:%/2-/\ TIO • If the snail cannot climb out of the well, you may return 0, return a falsy value, or throw an exception. For the purpose of writing the test cases, I simply chose None to indicate that there was no answer. Would you also consider adding an explanation and a Try it Online link? – musicman523 Jul 7 '17 at 6:45 • @musicman523 fixed and done. – Jonah Jul 7 '17 at 7:06 ## Perl 5, 37 bytes 35 bytes code +2 for -pa. $i-=$F[2]while++$\,($i+=$F[1])<$_}{ Try it online! # PHP>=7.1, 60 bytes prints 0 for no escape [,$h,$u,$d]=$argv;echo$h>$u?$u>$d?ceil(($h-$d)/($u-$d)):0:1; PHP Sandbox Online # PHP>=7.1, 67 bytes prints nothing for no escape for([,$h,$u,$d]=$argv;($u>$d?:$h<=$u)&&0<$h+$t*$d-$u*++$t;);echo$t; PHP Sandbox Online # Mathematica, 4740 39 bytes If[#==#2,1,⌈(#-#3)/(#2-#3)⌉~Max~0]& -7 bytes from @KeyuGan • You need to deal with input as 69, 3, 8 and ⌈ is counted as 3 bytes as far as I think. – Keyu Gan Jul 7 '17 at 7:39 • all fixed! try it now – J42161217 Jul 7 '17 at 7:44 • you may use Max to replace If statement. If[#<=#2,1,Max[⌈(#-#3)/(#2-#3)⌉,0]]& – Keyu Gan Jul 7 '17 at 7:47 # Ruby, 49 47 bytes ->h,a,b{h-a<1?1:(1.0*(h-a)/[a-b,0].max+1).ceil} Throws exception if snail can't climb out Try it online! • @Jonah fixed it – Alex Jul 7 '17 at 7:26 • What's the reasoning behind the proc? h-a<1?1:(1.0*(h-a)/[a-b,0].max+1).ceil passes the test cases, and saves 9 bytes. – Galen Jul 11 '17 at 19:08 ## Batch, 66 bytes @set/an=%4+1,a=%1-%2+%3 @if %1 gtr %2 %0 %a% %2 %3 %n% @echo %n% The second last test case printed nothing, and the last test case actually crashed CMD.EXE... # 05AB1E, 19 bytes 0[¼²+D¹<›i¾q}³-D1‹# Explanation: 0 Initialise stack with 0 [ while(true) ¼ increment the counter variable ²+ add the second input to the top of the stack D¹<›i if it is greater than or equal to the first input ¾ push the counter variable q terminate the program } end if ³- subtract the third input from the top of the stack D duplicate top of stack 1‹ if it is less than 1 # break the loop For invalid values, this may return any value less than 1. However, in 05AB1E, only 1 is truthy so this meets the requirement that the output for an invalid value should be falsy. Try it online! # PHP, 60 bytes [,$h,$v,$d]=$argv;echo$h>$v?$v>$d?ceil(($h-$d)/($v-$d)):N:1; prints N for None. Run with -r. # 05AB1E, 12 bytes .×ηO<²›1k2÷> Try it online! Prints 0 if impossible. Input format: [climb, -fall] height # Japt, 12 bytes @UµV-W §W}aÄ Test it online! Outputs undefined for never, after possibly freezing your browser for a while, so please be careful. I'm not convinced this is optimal. oWV-W l works on all but the last three cases... • Came up with this for 11 bytes by changing the order of the inputs. – Shaggy Mar 26 '18 at 15:51 # Haskell, 30 29 bytes (b!c)a=1+sum[(b!c)$a+c-b|a>b]
Try it online!
Shorter than the existing Haskell answer. Perhaps someone else can beat me.
This uses a recursive approach to solving the problem. Each recursion is essentially a day of movement for the snail. If the distance left to the end is less than the distance still required we end our recursion.
• Save 1 byte with infix notation: (b#c)a=1+sum[(b#c)$a+c-b|a>b]. – Laikoni Jul 7 '17 at 22:04 • @Laikoni Didn't know that could be done. Thanks for the tip. – Sriotchilism O'Zaic Jul 7 '17 at 22:23 • You can drop the parens around b!c in the list comprehension. – Zgarb Jul 9 '17 at 9:22 # QBIC, 31 23 bytes Just noticed the requirements changed. This version doesn't check if the snail will ever reach the top of the well. ≈:-:>0|q=q+1┘a=a-b+:]?q The explanation below, for the original version that does check if a solution exists, covers all relevant parts of this code too. Original, 31 byte answer: ~:>:|≈:-a>0|q=q+1┘c=c-a+b]?q\?0 ## Explanation ~ IF : cmd line arg 'a' (the increment of our snail) > is greater than : cmd line arg 'b' (the decrement, or daily drop) | THEN ≈ WHILE : cmd line arg 'c' (the height of the well) -a minus the increment (we count down the hieght-to-go) >0| is greater than 0 (ie while we haven't reached the top yet) q=q+1 Add a day to q (day counter, starts at 1) ┘ (syntactic linebreak) c=c-a+b Do the raise-and-drop on the height-to-go ] WEND ?q PRINT q (the number of days) \?0 ELSE (incrementer <= decrementer) print 0 (no solution) Try it online! (OK, not really: this is a translation of QBIC to QBasic code run in repl.it 's (somewhat lacking) QBasic enviroment) # Excel VBA, 47 Bytes Anonymous VBE immediate window function that takes input in from the range [A1:C1] from the ActiveSheet object outputs to the VBE immediate window This primarily Excel formula based solution appears to be smaller than any purely VBA solution that I can come up with :( ?[If(B1>C1,-Int((B1-A1)/(B1-C1)-1),Int(A1=B1))] Haskell, 47 55 bytes (48 if tuple required) f d c s|d<=c=1|c<s= -1|d>c||c<s=1+(f(d-c+s)c s) tuple variation f(d,c,s)|d<=c=1|c<s= -1|d>c||c<s=1+(f(d-c+s)c s) Explanation f d c s function that does all the heavy lifting =) d - depth c - climb per day s - slide per night |d<=c=1 recursion terminator. 1 day of climbing |c<s= -1 possibility check. top can't be reached |otherwise=1+(f(d-c+s)c s) 1 day plus the rest of the distance • 1. You can replace d>c||c<s just with 0<1, as you already implicitly do in your explanation, because otherwise is just a synonym of True. 2. The recursive call in your tuple version is still curried. 3. You can define your function as (d#c)s instead of f d c s to save two more bytes. – Laikoni Jul 7 '17 at 13:16 • You also need c<=s instead of c<s. – Laikoni Jul 7 '17 at 13:17 • Reordering and using 0 instead of -1 as allowed by the OP yields 38 bytes: Try it online! – Laikoni Jul 7 '17 at 13:21 • Can you use an infix identifier to save any bytes? – musicman523 Jul 7 '17 at 18:14 • I don't know, if I should post edited anser since it is esentialy @Laikoni's answer – Sergii Martynenko Jr Jul 7 '17 at 18:33 # Python 3, 41 Bytes f=lambda a,b,c:int(b>=a)or 1+f(a-b+c,b,c) Error for Never Outgolf @veganaiZe • Welcome to PPCG! Nice first answer :) – musicman523 Jul 7 '17 at 15:46 • I don't know much Python, but could you change int(b>=a) to 1-(b<a) to save 2 bytes? – ETHproductions Jul 7 '17 at 16:45 # APL (Dyalog), 13 bytes (⌈+÷⊢)/0⌈2-/⊢ Try it online! Errors on division by zero if the snail cannot climb out of the well. # C# (.NET Core), 37 bytes (h,c,f)=>h>c?f<c?1+(h-f-1)/(c-f):0:1; Non-recursive lambda. Uses formula found here. Could be shortened by 6 bytes if "any negative result" is a valid way to return failure; currently returns 0 instead. • It's been a while, but h-f-1 can be h+~f. – Kevin Cruijssen Mar 26 '18 at 15:01 # Python v2 & v3, 44 Bytes f=lambda x,y,z:1+f(x-(y-z),y,z)if x>y else 1 ^Infinite recursion (error) for None case. • You can use lambda. Also, this seems similar to my (Java) answer so allow me to suggest an improvement in the formula: (x-z-1)//(y-z)+1. I don't do much Python, so I might be wrong... – Olivier Grégoire Jul 7 '17 at 11:11 • You can eliminate f= from the byte count, remove some spaces around ifs and elses, and switch to Python 2 where integer division is a single / – musicman523 Jul 7 '17 at 19:28 • Thanks @musicman523. I ended up taking all of your advice. – veganaiZe Jul 7 '17 at 20:25 • I realized that my "clean" (no infinite recursion) code had lots of corner-case issues when used with other inputs (ie. 4, 3, 8). @musicman523 I think I'm starting to see the "proofs" that you speak of. – veganaiZe Jul 7 '17 at 23:50 # HP-15C Programmable Calculator, 26 Bytes The three numbers are loaded into the stack in order before running the program. The fall height is entered as a negative number. If the snail cannot climb out of the well, the result is either a negative number or error #0 (zero divide error). Op codes in hex: C5 C1 B4 C5 FB 74 1A C4 FA B4 C5 FD C1 C1 A3 70 C6 F0 B4 FA EB F1 FA B2 0A F1 Instruction meanings: x↔y ENTER g R⬆ x↔y − g TEST x≤0 GTO A R⬇ + g R⬆ x↔y ÷ ENTER ENTER f FRAC TEST x≠0 EEX 0 g R⬆ + g INT 1 + g RTN f LBL A 1 You can try the program with this HP-15C simulator. • This is awesome! Welcome to PPCG :) – musicman523 Jul 9 '17 at 6:06 # Common Lisp, 49 bytes (defun f(a b c)(if(> a b)(1+(f(+(- a b)c)b c))1)) Try it online! Recursive function, stack overflow if no solution found. # PowerShell, 95 94 bytes $g=$args[0]$c=$args[1]$f=$args[2]$p=0
$d=0 1..$g|%{$d+=1;$p+=$c;if($p-ge$g){$d;exit}$p-=$f}
`
Try it online!
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# Answer to Question #52119 in Microeconomics for Asif
Question #52119
Muslim Glass Company faces the following demand and marginal revenue functions:
P=1000-0.5Q, MR=1000-Q
P is the price, Q is quantity and MR is the Marginal Revenue.
a) At what quantity is total revenue maximized? what is the price elasticity of demand at this quantity?
b) The firm has been selling 1000 units per period at a price of $500. What is the price elasticity of demand at t his quantity? c) At what price would Muslim sell no output? What quantity would be demanded if the product were given away? 1 Expert's answer 2015-04-23T10:55:56-0400 P=1000-0.5Q, Qd = 2000 - 2P, MR=1000-Q a) Total revenue is maximized, when MR = 0 and Q = 1000 units. The price elasticity of demand at this quantity is Ed = P/Q*Qd' = 500/1000*(-2) = -1, so demand is unit-elastic. b) If the firm has been selling 1000 units per period at a price of$500, the price elasticity of demand at this quantity is Ed = P/Q*Qd' = 500/1000*(-2) = -1, so demand is unit-elastic.
c) Muslim would sell no output, when P = 1000 - 0.5*0 = \$1000.
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# Upper Closure of Subset is Subset of Upper Closure
## Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $X, Y$ be subsets of $S$.
Then
$X \subseteq Y \implies X^\succeq \subseteq Y^\succeq$
where $X^\succeq$ denotes the upper closure of $X$.
## Proof
Let $X \subseteq Y$.
Let $x \in X^\succeq$.
By definition of upper closure of subset:
$\exists y \in X: y \preceq x$
By definition of subset:
$y \in Y$
Thus by definition of upper closure of subset:
$x \in Y^\succeq$
$\blacksquare$
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Non-inverting Op Amp. A differentiator is a circuit that performs differentiation of the input signal. a) Adding feedback capacitor d) High frequency noise The construction of simple Integrator circuit using op-amp requires two passive components and one active component. A Differentiator Amplifier is basically a High Pass Filter and are used in wave shaping circuits, frequency modulators etc. So, when there is a capacitor at the input to the inverting terminal and a resistor with one side connected to the inverting terminal and the other side to the output, we have a differentiator circuit. 56 IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS-II: ANALOG AND DIGITAL SIGNAL PROCESSING, VOL. We… The differentiator may be constructed from a basic inverting amplifier if an input resistor R 1 is replaced by a capacitor C 1 . Since the coil acts as a differentiator, it has no response to DC current, and therefore that information is lost. d) Gain Since the differentiator performs the reverse of the integrator function. That means, a differentiator produces an output voltage that is proportional to the rate of change of the input voltage. The practical differentiator. Fig. Integrator circuit is exactly opposite of Op-amp differentiator circuit. The non-inverting terminal of the op-amp is connected to the ground. the output waveform is the derivative of the input waveform. The circuit is used in analogue computers where it is able to provide a differentiation manipulation on the input analogue voltage. • Differentiators also find application as wave shaping circuits, to detect high frequency components in the input signal. For use in various applications, In different ways. Obtain the value of y for the follo... A: y=&s; (Bitwise operator of AND) View Answer, 9. d) Input waveform as derivative of output waveform Integrator simulates mathematical integration of a function and differentiator simulates mathematical operation differentiation of a function. Check the video #1 and #2 to see what a strain gauge sensor is. Figure 4 The frequency response of the differentiator circuit (amplitude only) is a straight line, increasing with frequency.. To practice all areas of Linear Integrated Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers. Differentiation amplifier produces Its important application is to produce a rectangular output from a ramp input. Where fa -> Frequency at which gain =0 ; fb -> Gain limit frequency ; fc -> Unity gain bandwidth. The figure below shows the basic circuit diagram of an op amp differentiator. b) RF = 1.6×103Ω, C1 = 0.47×10-6F The figure below shows the basic circuit diagram of an op amp differentiator. Here we are discussing about Integrator and Differentiator using opamp. The following circuit diagram shows the differentiator using op-amp. Non-inverting amplifier. This is one type of amplifier, and the connection of this amplifier can be done among the input as well as output and includes very-high gain.The operational amplifier differentiator circuit can be used in analog computers to perform mathematical operations such as summation, multiplication, subtraction, integration, and differentiation. Here we are discussing about Integrator and Differentiator using opamp. The stability and high frequency noise problem are corrected by The operational amplifier circuit generates an output voltage which is … a Strain Gauge Direction of strain Strain gauge Another application is using differentiator circuit to get velocity from the strain gauge sensor. Join our social networks below and stay updated with latest contests, videos, internships and jobs! c) Feedback capacitor and feedback resistor Consideration of the device in figure 23 will give a feeling for the differentiator circuit. Op Amp application as a Differentiator Op-amp can be used as a differentiator where the output is the first derivative of the input signal. c) Low frequency noise DA = XA + XB The operational amplifier differentiator circuit can be used in analog computers to perform mathematical operations such as summation, multiplication, subtraction, integration, and differentiation. An op amp differentiator is basically an inverting amplifier with a capacitor of suitable value at its input terminal. the output waveform is the derivative of the input waveform. a) Output waveform as integration of input waveform Differentiator Circuit . 43, NO. Practical Differentiator Video Lecture from Application of Op-Amp Chapter of Linear Integrated Circuits Subject for all engineering students. The differentiator op amp circuit we will build with an LM741 op amp chip is shown below. In electronics, a differentiator is a circuit that is designed such that the output of the circuit is approximately directly proportional to the rate of change (the time derivative) of the input.A true differentiator cannot be physically realized, because it has infinite gain at infinite frequency. This circuit is possibly less widely used, but nevertheless a key item in an analogue designers toolbox. Check the video #1 and #2 to see what a strain gauge sensor is. The differentiator circuit is avoided in the analog computers. Calculate the gain limiting frequency for the circuit Using a capacitor as the input element to the inverting amplifier, figure 22, yields a differentiator circuit. Question 3 Which of the following is a typical application for a differentiator circuit? Median response time is 34 minutes and may be longer for new subjects. Question 3 Which of the following is a typical application for a differentiator circuit? The practical differentiator. This chapter discusses in detail about op-amp based differentiator and integrator. Applications of Differential Amplifiers It is used as a series negative feedback circuit by using op amplifier Generally, we use differential amplifier that acts as a volume control circuit. The op amp differentiator is particularly easy to use and therefore is possibly one of the most widely used version. One issue can be that the differentiator can be open to … High Pass RC Circuit as Differentiator. Z = X' (A+B) As you can see the output … Please note that these also come under linear applications of … The input signal is applied to the inverting input so the output is inverted relative to the polarity of the input signal. Locked; Cancel ... (approximated by a differentiator circuit). High Pass RC Circuit as Differentiator. Differentiator circuit Design Goals Input Output Supply fMin fMax VoMin VoMax Vcc Vee Vref 100Hz 2.5kHz 0.1V 4.9V 5V 0V 2.5V Design Description The differentiator circuit outputs the derivative of the input signal over a frequency range based on the circuit time constant and the bandwidth of the amplifier. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes! A non-inverting amplifier is a special case of the differential amplifier in which that circuit's inverting input V 1 is grounded, and non-inverting input V 2 is identified with V in above, with R 1 ≫ R 2.Referring to the circuit immediately above, = (+). b) Input voltage September 1, 2020 by Electricalvoice Op-amp Integrator is an electronic circuit that produces output that is proportional to the integration of the applied input. The circuit performs the mathematical operation of differentiation (i.e.) d) Frequency Shift keying y=r<<1;(Left shift operator) In Gen... Q: The gain of operational amplifier is defines as the ratio of _________________. This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Differentiator”. The practical differentiator. A non-inverting amplifier is a special case of the differential amplifier in which that circuit's inverting input V 1 is grounded, and non-inverting input V 2 is identified with V in above, with R 1 ≫ R 2.Referring to the circuit immediately above, = (+). Its important application is to produce a rectangular output from a ramp input. The differentiator op amp circuit we will build with an LM741 op amp chip is shown below. 2. As your Wikipedia article states "Since the voltage that is induced in the coil is proportional to the rate of change of current in the straight conductor, the output of the Rogowski coil is usually connected to an electrical (or electronic) integrator circuit to provide an output signal that is proportional to the current." Q: consider the data p=4'b000z; q=4'bz11x; r=4'b0101 and s=4'b1100. d) None of the mentioned DB = A'X 56 IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS-II: ANALOG AND DIGITAL SIGNAL PROCESSING, VOL. Select the order in which the frequency should be maintained to enhance the stability of differentiator? Differentiator is an op amp based circuit, whose output signal is proportional to differentiation of input signal. 1. a Strain Gauge Direction of strain Strain gauge Another application is using differentiator circuit to get velocity from the strain gauge sensor. c) Output waveform as derivative of input waveform Circuit for ideal differentiator is as shown in the figure: this circuit can perform the mathematical operation of differentiation on the input signal. At the end of the day, you need to integrate the output of the Rogowski coil to recreate the AC waveform. Find answers to questions asked by student like you, Which application use differentiator circuit? In this article, we will see the different op-amp based integrator circuits, its working and its applications. b) FM modulators 3 SN7414 square wave generator and differentiator circuit. 3 illustrates the use of a SN7414 square wave generator using a differentiator circuit to create narrow output pulses at points B and C. Diode D1 is used to block the negative going pules to TP2. So, when there is a capacitor at the input to the inverting terminal and a resistor with one side connected to the inverting terminal and the other side to the output, we have a differentiator circuit. A differentiating circuit is a simple series RC circuit where the output is taken across the resistor R. The circuit is suitably designed so that the output is proportional to the derivative of the input. Related: Simple Schmitt Trigger SN7414 Square Wave Generator 1. Consideration of the device in figure 23 will give a feeling for the differentiator circuit. d. Differentiator output for sine wave waveform. I collect a lot of IC-741 circuits or tiny amplifiers using 741 op-amps as main. Please note that these also come under linear applications of op-amp. Solution for Which application use differentiator circuit? a) None of the mentioned Explain the E, I and R i... Q: Below is the transfer function of a system. 4 … The electronic circuits which perform the mathematical operations such as differentiation and integration are called as differentiator and integrator, respectively. In this article, we will see the different op-amp based integrator circuits, its working and its applications. The differential operational amplifier can be used as an automatic gain control circuit. 3 SN7414 square wave generator and differentiator circuit. Frequency Shift keying FM modulators Wave generators none of above c) Noise d) None of the mentioned Operational Amplifier Differentiator Circuit. Assume current 'I' … View Answer, 6. *, Q: To get maximum gain out of JFET Amplifier VGSQ should be set to. The electronic circuits which perform the mathematical operations such as differentiation and integration are called as differentiator and integrator, respectively. Below is the breadboard circuit of the above circuit. Applications of Op-amp Differentiator and Integrator:- • Differentiating amplifiers are most commonly designed to operate on triangular and rectangular signals. View Answer, 7. b) Feedback capacitor and internal resistor The main application of differentiator circuits is to generate periodic pulses. The differentiator circuit is essentially a high-pass filter. A differentiator is a circuit that performs differentiation of the input signal. Differentiator using Operational Amplifier The circuit performs the mathematical operation of differentiation (i.e.) (s + 100) Integration is basically a summing process that determines the total area under the curve of a function. a) Draw the amplitude Bode ... Q: Find I(D) ,V(D2) and Vo in the given figure. View Answer. Voltage gain = output v... Q: three bulb rating (100w,220v), (60w,220v) and (50w,220v) are connected in parallel to a 220v source ... Q: 1. For the … a) fa < fb < fc September 1, 2020 by Electricalvoice Op-amp Integrator is an electronic circuit that produces output that is proportional to the integration of the applied input. REVIEW: A differentiator circuit produces a constant output voltage for a steadily changing input voltage. A: The ratio of output voltage to input voltage is defined as the voltage gain. Integration is basically a summing process that determines the total area under the curve of a function. d) fb < fc < fa c) Output waveform as derivative of input waveform. s(s² + 80s + 2500) The differential operational amplifier can be used as an automatic gain control circuit. The differentiator circuit has many applications in a number of areas of electronic design. Please note that these also come under linear applications of … The following equation gives the relation between the input signal and the output signal. Capacitor (C), resistor (R) and op-amp are used in the differentiator circuit as shown in figure 1. a) Output waveform as integration of input waveform. That means, a differentiator produces an output voltage that is proportional to the rate of change of the input voltage. View Answer, 11. Differentiating Circuit A circuit in which output voltage is directly proportional to the derivative of the input is known as a differentiating circuit. The practical differentiator. The output is a differentiation of input signal. Fig. b) Input waveform as integration of output waveform. b) fa > fb > fc View Answer, 12. This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Differentiator”. The two passive components are resistor and capacitor. The differentiator circuit outputs the derivative of the input signal over a frequency range based on the circuit time constant and the bandwidth of the amplifier. Fig. c) RF = 1.6×103Ω, C1 = 47×10-6F d) Input waveform as derivative of output waveform. application of an integrator is sometimes called a totalizer in the industrial instrumentation trade. O A. Triangle wave generator B. ADC signal conditioning C. Zero crossing detector O D. Not enough information to say Basically it performs mathematical operation of integration. View Answer, 10. a) 15.64Hz 43, NO. A similar effect can be achieved, however, by limiting the gain above some frequency. Part 2: Differentiator Circuit Amplifier: 1- Assemble the circuit as shown in Figure 4: Figure 4: Differentiator Circuit Set up 2- Use the values of: C = 10 nF, Ri = 1kΩ. b) Vo(s) /V1(s) = -S×RF×C1/(1+RF×C1)2 Thus, this is all about an overview of op-amp differentiator. View Answer, 3. 3- Apply a sinusoidal wave to the input and display the output signal on the oscilloscope. A differentiator is an electronic circuit that produces an output equal to the first derivative of its input. This can be used in the detection of high-frequency components in the input signal These Op-Amp differentiators are normally designed for performing an operation on rectangular and triangular signals. © 2011-2021 Sanfoundry. d) None of the mentioned Figure 4 The frequency response of the differentiator circuit (amplitude only) is a straight line, increasing with frequency.. A similar effect can be achieved, however, by limiting the gain above some frequency. d) Internal capacitor and internal capacitor Applications of Op-amp Differentiator and Integrator:- • Differentiating amplifiers are most commonly designed to operate on triangular and rectangular signals. Integrator simulates mathematical integration of a function and differentiator simulates mathematical operation differentiation of a function. Solution for Which application use differentiator circuit? An integrator circuit produces a steadily changing output voltage for a constant input voltage. Figure 22. The flip flop input and output equations are given by: View Answer, 8. The input V i is applied through capacitor C at the inverting terminal. defined as the measure of a capacitor’s opposition to changes in voltage Differentiator: As the name implies, the circuit performs the mathematical operation of differentiation (i.e) the output waveform is the derivative of the input waveform. View Answer, 5. Fig.5 (i) shows the circuit of an OP-Amp differentiator. Figure 22. Frequency Shift keying FM modulators Wave generators none of above One of the major applications of op-amp differentiator is wave shaping circuits. *Response times vary by subject and question complexity. The differentiator may be constructed from a basic inverting amplifier if an input resistor R1 is replaced by a capacitor C1. This chapter discusses in detail about op-amp based differentiator and integrator. REVIEW: A differentiator circuit produces a constant output voltage for a steadily changing input voltage. A differentiator opamp is an opamp configuration that produces a differentiated version of the signal applied to its input terminal. Differentiator Circuit . V o is the output voltage. Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Result: Designed and verified differentiator and integrator circuits using Op-Amp 741. What isthe series circuit specially the voltage and its resistance? y=p*q;(Arithimetic operator o... Q: (b) Analyse the sequential logic circuit for the D Flip-Flop shown in Figure below and answer the fo... A: (i) Differentiator is an op amp based circuit, whose output signal is proportional to differentiation of input signal. This configuration is very similar to the inverting operation amplifier. Figure 4 The frequency response of the differentiator circuit (amplitude only) is a straight line, increasing with frequency.. Some of the differential operational amplifier can be used for Amplitude modulation. Choose the value of RF and C for a 5kHz input signal to obtain good differentiation. Fig.5 (i) shows the circuit of an OP-Amp differentiator. Determine the transfer function for the practical differentiator by interchanging the positions of components in an integrator circuit we can get a differentiator circuit. 3 illustrates the use of a SN7414 square wave generator using a differentiator circuit to create narrow output pulses at points B and C. Diode D1 is used to block the negative going pules to TP2. When an amplifier is overdrive output waveform of voltage and current distortion. Which application use differentiator circuit? application of an integrator is sometimes called a totalizer in the industrial instrumentation trade. Basically it performs mathematical operation of integration. Fig 4 shows a differentiating circuit, the output across R will … By Exchanging the positions of 'R' and 'C' in integrator the differentiator circuit is obtained The circuit which produces the differentiation of the input voltage at its output is called differentiator. View Answer, 4. which factor makes the differentiator circuit unstable? d) RF = 1.6×103 Ω, C1 = 10×10-6F The Resistor and the Capacitor form a first-order low pass filter across the active component Op-Amp. c) VO = RF×CF×[dVin/dt]. c) 33.89Hz Obviously the circuit is used in analogue computers where it is able to provide a differentiation manipulation on the input analogue voltage. View Answer, 2. Op-amp analogue differentiator: The op amp differentiator is another circuit used in analogue computing and finds applications in other areas. To intuitively see this gain equation, use the virtual ground technique to calculate the current in resistor R 1: Differentiating Circuit A circuit in which output voltage is directly proportional to the derivative of the input is known as a differentiating circuit. Tell a bit more about your application, please. A sine wave of 1vpeak at 1000Hz is applied to a differentiator with the following specification: RF =1kΩ and C1=0.33µF, find the output waveform? The op amp differentiator is particularly easy to use and therefore is possibly one of the most widely used versions. This is the Clipping Amplifier Circuit using LM741(popular op-amp IC. b) 23.356Hz here is complete set of 1000+ Multiple Choice Questions and Answers, Prev - Linear Integrated Circuit Questions and Answers – Integrator – 2, Next - Linear Integrated Circuit Questions and Answers – Active Filters – 1, Linear Integrated Circuit Questions and Answers – Integrator – 2, Linear Integrated Circuit Questions and Answers – Active Filters – 1, Java Programming Examples on Hard Graph Problems & Algorithms, Digital Signal Processing Questions and Answers, Digital Communication Questions and Answers, Instrumentation Transducers Questions and Answers, Microwave Engineering Questions and Answers, Analog Communications Questions and Answers, Optical Communications Questions and Answers, Basic Electrical Engineering Questions and Answers, Electronic Devices and Circuits Questions and Answers, Linear Integrated Circuits Questions and Answers, Linear Integrated Circuit Questions and Answers – Differential Amplifier and Circuit Configuration, Linear Integrated Circuit Questions and Answers – Basic Principles of Sine Wave Oscillator – 2, Linear Integrated Circuit Questions and Answers – 555 Timer as a Monostable Multivibrator, Linear Integrated Circuit Questions and Answers – Voltage Series Feedback Amplifier – 2. Fig. An active differentiator includes some form of amplifier, while a passive dif… a) Output impedance A true differentiator cannot be physically realized, because it has infinite gain at infinite frequency. Fig 4 shows a differentiating circuit, the output across R will … Sanfoundry Global Education & Learning Series – Linear Integrated Circuits. If you feed a square OR rectangular pulse with variable OR fixed duty cycle to a differentiator circuits and adjust the RC Time constant of the circuits you will get sharp trigger signals at desired time intervals. 4- … Differentiator. Figure 4 The frequency response of the differentiator circuit (amplitude only) is a straight line, increasing with frequency.. The differentiator circuit has many applications in a number of areas of electronic design. Determine the output voltage of the differentiator? 1, JANUARY 1996 A Novel Continuous-Time Current-Mode Differentiator and Its Applications Ezz I. El-Masry and John W. Gates A b s ~ a t t - A novel continuous-time current-mode differentiator with a frequency range extending from dc to 100 h4Hi is presented. b) Input waveform as integration of output waveform Op Amp application as a Differentiator Op-amp can be used as a differentiator where the output is the first derivative of the input signal. b) External noise The following equation gives the relation between the input signal and the output signal. Which application use differentiator circuit? An integrator circuit produces a steadily changing output voltage for a constant input voltage. Result: Designed and verified differentiator and integrator circuits using Op-Amp 741. c) fb < fc > fa As you can see the output … In electronics, a differentiator is a circuit that is designed such that the output of the circuit is approximately directly proportional to the rate of change (the time derivative) of the input. The differentiator circuit can be constructed by interchanging the input resistance R1 and feedback capacitor Cf of an integrating amplifier. a) VO = RF×C1×[dVin/dt]. Using a capacitor as the input element to the inverting amplifier, figure 22, yields a differentiator circuit. c) Wave generators Fig. To intuitively see this gain equation, use the virtual ground technique to calculate the current in resistor R 1: All Rights Reserved. It is used as a series negative feedback circuit by using op amplifier Generally, we use differential amplifier that acts as a volume control circuit. This process is exactly the opposite of integration. An op amp differentiator is basically an inverting amplifier with a capacitor of suitable value at its input terminal. The increase in the input frequency of the differentiation amplifier to input impedance creates c. Differentiator output for square wave. We… Operational Amplifier as Differentiator: Introduction: An op-amp differentiator or a differentiating amplifier is a circuit configuration which produces output voltage amplitude that is proportional to the rate of change of the applied input voltage. Analog and DIGITAL signal PROCESSING, VOL using 741 op-amps as main inverting amplifier, 22... ; q=4'bz11x ; r=4'b0101 and s=4'b1100 Designed to operate on triangular and rectangular signals designers toolbox to current! May be constructed from a basic inverting amplifier if an input resistor R1 is replaced a! As differentiation and integration are called as differentiator and integrator, respectively ) wave generators d ) Shift. Differentiator may be constructed from a ramp input “ Passive High Pass Filter “ it... Performs the reverse of the Rogowski coil to recreate the AC waveform above circuit provide... 2 to see what a strain gauge Direction of strain strain gauge sensor, detect... End of the input element to the rate of change of the voltage! View Answer, 10 about integrator and differentiator simulates mathematical integration of a.... Video # 1 and # 2 to see what a strain gauge Another application to! Find application as wave shaping circuits, here is complete set of 1000+ Multiple Choice Questions & Answers MCQs! Output equal to the rate of change of the input analogue voltage “ differentiator ” polarity of the signal. Called as differentiator and integrator of the input and display the output is the first derivative the. Voltage non-inverting amplifier and op-amp are used in analogue computers where it is able to provide differentiation. Input is known as a differentiator produces an output equal to the derivative of the input analogue voltage sinusoidal to. Obviously the circuit is avoided in the industrial instrumentation trade d ) none of Fig... As shown in the industrial instrumentation trade a first-order low Pass Filter and used... Current ' i ' … the differentiator circuit is avoided in the analog.. Differentiators also find application as a differentiating circuit a circuit in which output voltage for a input. Major applications of op-amp ) is a straight line, increasing with frequency the different op-amp based circuits... Bit more about your application, please for square wave popular op-amp IC by interchanging the signal. Times vary by Subject and question complexity an op-amp differentiator components in analogue... Opposition which application use differentiator circuit? changes in voltage non-inverting amplifier amplifiers are most commonly Designed to operate on triangular rectangular... Of 1000+ Multiple Choice Questions & Answers ( MCQs ) focuses on “ differentiator ” ; r=4'b0101 s=4'b1100. Determines the total area under the curve of a system voltage non-inverting amplifier and.!, yields a differentiator is particularly easy to use and therefore is one. Get free Certificate of Merit a true differentiator can not be physically realized, because it has infinite at! And rectangular signals integrator, respectively r=4'b0101 and s=4'b1100 i ' … the differentiator circuit differentiator performs the of... Amplitude modulation Integrated circuits Subject for all engineering students main application of op-amp in “ operational amplifier be. Possibly less widely used version which output voltage for a steadily changing input voltage a! – Linear Integrated circuits Subject for all engineering students, this is the of! ; fb - > Unity gain bandwidth an integrator is sometimes called a in. Unity gain bandwidth one active component op-amp limiting the gain above some frequency figure 1 Subject and complexity... Circuits which perform the mathematical operations such as differentiation and integration are called as differentiator integrator... Subject and question complexity of differentiator circuits is to produce a rectangular output from a ramp input above! An analogue designers toolbox circuit ( amplitude only ) is a straight line increasing.
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## frehemtenepe
3 yıl önce
Therefore the circuit will oscillate at a frequency at which the phase shift of the feedback filter is less than 180 degrees. .. The mathematics for calculating oscillation frequency and oscillation criterion for this circuit are surprisingly complex, due to each RC stage loading the previous ones. ^ ^ K.W.(Widelski?) (1984). The circuit is taken from.[2]. f o s c i l l a t i o n = 1 2 π R C 6 {displaystyle f{mathrm {oscillation} }={frac {1}{2pi RC{sqrt {6}}}}} . Without the simplification of all the resistors and capacitors having the same values, the calculations become more complex:. ^ Mancini, Ron (2002). The feedback network 'shifts' the phase of the amplifier output by 180 degrees at the oscillation frequency to give positive feedback.[1] Phase-shift oscillators are often used at audio frequency as audio oscillators. Circuit diagram for phase-shift oscillator using a JFET . R f b = 29 ⋅ R {displaystyle R{mathrm {fb} }=29cdot R} . Bipolar implementation. If it is assumed that each RC segment does not affect the other, a gain of about 8 to 10 will be sufficient to enable oscillation. v t e Electronic oscillators Theory Barkhausen stability criterion Harmonic oscillator Leeson's equation Nyquist stability criterion Oscillator phase noise Phase noise LC oscillators Armstrong or Meissner oscillator Clapp oscillator Colpitts oscillator Hartley oscillator Meacham bridge oscillator Seiler oscillator Vack oscillator resonant Royer RC oscillators Phase-shift oscillator Twin-T oscillator Wien bridge oscillator Quartz oscillators Butler oscillator Pierce oscillator Tri-tet oscillator Relaxation oscillators Blocking oscillator Multivibrator ring oscillator Pearson-Anson oscillator basic Royer Other Cavity oscillator Delay line oscillator Opto-electronic oscillator Robinson oscillator Transmission line oscillator klystron oscillator cavity magnetron Gunn oscillator . f o s c i l l a t i o n = 1 2 π R 2 R 3 ( C 1 C 2 + C 1 C 3 + C 2 C 3 ) + R 1 R 3 ( C 1 C 2 + C 1 C 3 ) + R 1 R 2 C 1 C 2 {displaystyle f{mathrm {oscillation} }={frac {1}{2pi {sqrt {R{2}R{3}(C{1}C{2}+C{1}C{3}+C{2}C{3})+R{1}R{3}(C{1}C{2}+C{1}C{3})+R{1}R{2}C{1}C{2}}}}}} . It consists of an inverting amplifier element such as a transistor or op amp with its output fed back to its input through a phase-shift network consisting of resistors and capacitors in a ladder network. Contents 1 Implementations 1.1 Bipolar implementation 1.2 FET implementation 1.3 Op-amp implementation 2 References . References. As with other feedback oscillators, when the power is applied to the circuit, thermal electrical noise in the circuit or the turn-on transient provides an initial signal to start oscillations. Kalejdoskop Techniki. f9488a8cf8
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# On each row place a three letter word that can be attached to the end of the word.....................What is it?
502 views
On each row place a three letter word that can be attached to the end of the word to the left and to the beginning of the word to the right to give a longer word in each case.
When completed, the last letters of the three letter words will give a name reading downwards.
What is it?
FORE _ _ _ SAW
TEN _ _ _ POINT
HUM _ _ _ BEAR
HOLD _ _ _ OWED
CHICK _ _ _ NUT
POST _ _ _ POWER
BAR _ _ _ SIT
posted Nov 24, 2014
Solution
SEE - FORESEE and SEE-SAW
PIN - TEN PIN and PIN POINT
BUG - HUM BUG and BUG Bear
ALL - HOLD ALL and ALLOWED
PEA - CHICK PEA and PEA NUT
MAN - POSTMAN and MAN POWER
BED - BARBED and BED SIT
Now, looking and the last letters of the filled words - we get 'ENGLAND' (reading downwards).
Looking at the last letters
Look At The Last Letter
Similar Puzzles
On each row place a three letter word that can be attached to the end of the word to the left and to the beginning of the word to the right to give a longer word in each case. When completed the initial letters of the added words will give a capital city reading downwards. What is it?
On each row place a three letter word that can be attached to the end of the word to the left and to the beginning of the word to the right to give a longer word in each case. When completed the initial letters of the added words will give a capital city reading downwards. What is it?
``````K I L L _ _ _ F U L L Y
I M P _ _ _ O R
D O N _ _ _ N O T E
S U R F _ _ _ T O N E
W A V E _ _ _ B R I C K
S E T _ _ _ T O T A L
H O L D _ _ _ O W E D
``````
+1 vote
On each row place a three letter word that can be attached to the end of the word to the left and to the beginning of the word to the right to give a longer word in each case. When completed, the initial letters of the added words will give another word reading downwards. What is it?
WAVE _ _ _ CURRANT
MASS _ _ _ LESS
MOTOR _ _ _ GOES
DON _ _ _ BOARD
LEG _ _ _ OWED
FLAT _ _ _ ANT
On each row, place two letters that can be attached to the end of the word to the left to give a longer.
When completed, a topical word will be read downwards.
What is it?
INFER _ _
SKI _ _
TO _ _
CARE _ _
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# Graded Discussion Board (GDB) PHY101
Views: 1980
### Replies to This Discussion
Koi tu Discussion Start karo....
## Copper
Moazam bhai thora explain kar den plzz
Salam everyone!
Please read and listen to Lecture no. 36 carefully. Specially read for the Heat capacity of some substances. for cooper it is 387 Joule/(kg.K) The answer is there in these lectures i.e lecture# 36,37 and 38.
All the Best.
Hint
• The copper wall
• The steel wall
• The concrete wall
Concrete is heavier than steel and steel is heavier then copper.
Concrete is heavier than steel and steel is heavier then copper. Heavy things will take more time to change there temperature. Copper is less heavier than other two so that Copper will be more cold.
koi to mathematically prove kary plz ............
Plz koi help karo mathematically kaisy prove honge?
C = Q/delta T
Q= kA (T hot - T cold) / L *t
Q cooper = 387 (A) (37 C - 20 C) / Same size 8* time
wse mene 4 jaga se answer liya hai sb ny copper kaha hai per phr mai confuse hun and abi tak soch rhe hun answer ka
GDB ko mathematecally b prove krna hy?
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# Posted By
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# Dijkstra\'s Shortest Path algorithm
/ Published in: C
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Its Dijkstra's Shortest Path algorithm written in C. Reads from a file the nodes and the connected edges and implements Dijkstra's algorithm. I am uploading for your comments in my code. Thank you.
<p>Files content should be in the format:
n N
startNode endNode distance
startNode endNode distance
.
.
.
startNode endNode distance </p>
<p>where n is the total number of Nodes and N is the total number of Acnes.</p>
Copy this code and paste it in your HTML
`/* * * Created on: Nov 4, 2011 * Author: stakisko * Email : konmpar at gmail com *//* #include <stdio.h>#include <stdlib.h> typedef struct _node node;typedef struct _acne acne; /* * There are two structures. * 'Node' represent each node having childrens acnes represent each connected to him nodes. * 'Acne' represent each acne connecting two Nodes. */ struct _node { int distance; // distance from the first node struct _acne *children[100]; // going to nodes int edgeno; // how many nodes are connected to int name;}; struct _acne { int value; // distance between these edges int from; // acne connecting from int to; // to struct _node *edge;}; /* * for every node we are going in (the parameter) we search its connected childs nodes and * calculate their distances. then we are going in again( recursion ) in the node with the * minimum distance * * */ int traverse(struct _node *node){ if(node->edgeno<=0) return 0; printf("start of node %d\n", node->name); printf("\tchildrens: %d\n", node->edgeno); int mindistance = node->children[0]->value; int minnode = 0; int i; for(i=0;i<node->edgeno; i++){ if(node->distance + node->children[i]->value < node->children[i]->edge->distance ){ node->children[i]->edge->distance = node->distance + node->children[i]->value; } if (mindistance > node->children[i]->edge->distance){ mindistance = node->children[i]->edge->distance; minnode = i; } printf("\tfrom %d to %d with distance %d\n", node->children[i]->from, node->children[i]->to, node->children[i]->edge->distance); } printf("\tmin node %d\n", node->children[minnode]->to); printf("end of node\n-----------------------\n"); traverse(node->children[minnode]->edge); return 0;} // main functionint main(int argc, char *argv[]){ FILE *ifp; int t, t1, t2; int nodesNo = 0; int edgesNo = 0; ifp = fopen("/home/kostas/Downloads/dijkstra1.dat", "r"); if (ifp == NULL) { fprintf(stderr, "Can't open input file dijkstra.dat!\n"); exit(1); } fscanf(ifp, "%d %d", &nodesNo, &edgesNo); struct _node nodes[nodesNo]; struct _acne acnes[edgesNo]; int i = 0; // count the acnes int previous = 0; int k = 0; // count childs for every node /* * Files content should be in the format * n N * startNode endNode distance * startNode endNode distance * . * . * . * startNode endNode distance * * where n is the total number of Nodes and N is the total number of Acnes * */ while(fscanf(ifp, "%d %d %d", &t, &t1, &t2) == 3){ /* read from file */ /* * Then for every line we create a node and also * an acne to place the connected edges. * */ if(previous != t) k = 0; nodes[t].distance = 65000; // infinite nodes[t].name = t; acnes[i].from = t; acnes[i].to = t1; acnes[i].value = t2; acnes[i].edge = &nodes[t1]; nodes[t].children[k] = &acnes[i]; nodes[t].edgeno = k+1; i++; k++; previous = t; } // set firsts node's distance nodes[0].distance = 0; // set last node nodes[nodesNo-1].distance = 65000; nodes[nodesNo-1].edgeno = 0; nodes[nodesNo-1].name = nodesNo-1; int j; for(j=0; j< (sizeof(nodes)/sizeof(*nodes));j++) printf("node %d, distance %d, edgeno %d\n", nodes[j].name, nodes[j].distance, nodes[j].edgeno); for(j=0; j< sizeof(acnes)/sizeof(*acnes);j++) printf("from %d, to %d, value %d\n", acnes[j].from, acnes[j].edge->name, acnes[j].value); printf("\n-----------\n"); traverse(&nodes[0]); printf("\n-----------\n"); for(j=0; j< (sizeof(nodes)/sizeof(*nodes));j++) printf("nnode %d, distance %d\n", nodes[j].name, nodes[j].distance); return 0;}`
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https://www.convertunits.com/from/pascal/to/millibar
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## ››Convert pascal to millibar
pascal millibar
How many pascal in 1 millibar? The answer is 100.
We assume you are converting between pascal and millibar.
You can view more details on each measurement unit:
pascal or millibar
The SI derived unit for pressure is the pascal.
1 pascal is equal to 0.01 millibar.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pascals and millibars.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pascal to millibar
1 pascal to millibar = 0.01 millibar
10 pascal to millibar = 0.1 millibar
50 pascal to millibar = 0.5 millibar
100 pascal to millibar = 1 millibar
200 pascal to millibar = 2 millibar
500 pascal to millibar = 5 millibar
1000 pascal to millibar = 10 millibar
## ››Want other units?
You can do the reverse unit conversion from millibar to pascal, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Pascal
The pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.
## ››Definition: Millibar
A millibar (mb) is 1/1000th of a bar, a unit for measurement of pressure. It is not an SI unit of measure, however it is one of the units used in meteorology when describing atmospheric pressure. The SI unit is the pascal (Pa), with 1 millibar = 100 pascals (a hectopascal)
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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## Crossover rate
crossover rate When a company works on different product lines but the lines maintain the same outcome and the same profit margin. For example, a company that manufactures two different washers but still achieves the same quality service and profitability. Crossovers indicating a moving average are generally the cause of breakouts and breakdowns. Moving averages can determine a change in the price trend based on the crossover. For example, a technique for trend reversal is using a five-period simple moving average along with a 15-period simple moving average.
The crossover operator alters two or more parents to create offspring, where a probabilistic crossover rate is usually used to generate offspring [7, 8, 10]. Crossover rate is the rate or level of return of two comparable projects that have the same net present value. Since companies have limited resources, they must decide how to use these limited resources on various projects. The crossover rate is the point at which the two projects achieve the same net present value. In terms of investments, calculating a crossover rate between two similar securities can help an investor determine what to buy and what to sell. crossover rate. Definition. The specific return required by each project in order for two projects to have the same Net Present Value. Use crossover rate in a sentence. “ I wanted to know the crossover rate, but no one would tell me, so I had to find out myself. The crossover rate (CR) is the discount rate at which both projects deliver the same net present value. The crossover rate formula is the same as that for the IRR, but each factor is replaced by the difference between the projects. crossover rate When a company works on different product lines but the lines maintain the same outcome and the same profit margin. For example, a company that manufactures two different washers but still achieves the same quality service and profitability. Crossovers indicating a moving average are generally the cause of breakouts and breakdowns. Moving averages can determine a change in the price trend based on the crossover. For example, a technique for trend reversal is using a five-period simple moving average along with a 15-period simple moving average.
## Crossover rate The return at which two alternative projects have the same net present value.
However, the mechanism underlying crossover suppression in which the DSBs (~40-50 per chromosome) are repaired to 1-3 crossovers remains explored clearly. Abstract: Kramers' theory frames chemical reaction rates in solution as Crossover behavior of the thermal conductance and Kramers' transition rate theory. 12 Dec 2018 In Arabidopsis, higher mutation rate was reported in regions proximal to crossovers (Yang et al., 2015). This would suggest that crossover repair Fine-Scale Heterogeneity in Crossover Rate in the garnet-scalloped Region of the Drosophila melanogaster X Chromosome. Nadia D. Singh, Eric A. Stone,
### Crossover rate The return at which two alternative projects have the same net present value.
A ghrelin gene variant may predict crossover rate from restricting-type anorexia nervosa to other phenotypes of eating disorders: a retrospective survival analysis Crossover Rate. Crossover is by far the most important evolution operation. I tested single-point and uniform crossover with a crossover probability of 0.6. The Genetic Algorithm - a brief overview. Before you can use a genetic algorithm to solve a problem, a way must be found of encoding any potential solution to
### Crossovers indicating a moving average are generally the cause of breakouts and breakdowns. Moving averages can determine a change in the price trend based on the crossover. For example, a technique for trend reversal is using a five-period simple moving average along with a 15-period simple moving average.
If the required rate of return is lower than the crossover rate, which project should be accepted? 15.99 percent; A The crossover rate is the IRR of the cash flow A ghrelin gene variant may predict crossover rate from restricting-type anorexia nervosa to other phenotypes of eating disorders: a retrospective survival analysis
## Definition of Crossover rate in the Financial Dictionary - by Free online English dictionary and encyclopedia. What is Crossover rate? Meaning of Crossover rate
Crossover rate is the rate of return (or the weighted average cost of capital - WACC) at which the net present value (NPV) of two projects are identical. It is the point of rate of return at which the NPV of one project intersects the NPV of another project. This crossover calculator can be employed for the calculation of passive filters (first, second, third, and fourth order) in two-way and three-way crossover networks. It will also create a circuit diagram and provide the component values you require. Crossover SUVs offer an ideal mix of all of the above, making them a stellar vehicle choice for many car shoppers. However, since crossovers come in various sizes and excel in different areas, choosing the right one is a difficult task. We’ve ranked the best crossovers available today to help you decide which is the best choice for you. The 2019 X3 mixes the practicality of a crossover with the driving verve the company is known for. The compact BMW SUV has two fuel-efficient engines: a prudent turbocharged four-cylinder and a The Toyota RAV4 reverses this popular crossover's slide into anonymity with distinctive styling, a much-improved interior that is as practical as ever, and sharper driving dynamics. It is also one In genetic algorithms and evolutionary computation, crossover, also called recombination, is a genetic operator used to combine the genetic information of two parents to generate new offspring. It is one way to stochastically generate new solutions from an existing population, and analogous to the crossover that happens during sexual reproduction in biology. Crossover teams are assembled from the top 1% of talent across 130 countries. Challenge yourself to become a part of the cloud team revolution.
tried to nd an optimal combination of mutation rate and crossover rate for a test suite. They analyzed 6 population sizes, 10 crossover rates and 7 mutation rates. In this paper, we want fuzzy selection system for genetic algorithms and adaptive crossover and mutation rate fuzzy system. Index Terms-Genetic algorithms(GA) Frequency of optimal solutions appears (FT10) at different crossover rates. 4.1.3. Mutation Rate. In the multiparents crossover application, especially in the JSSP, However, the mechanism underlying crossover suppression in which the DSBs (~40-50 per chromosome) are repaired to 1-3 crossovers remains explored clearly. Abstract: Kramers' theory frames chemical reaction rates in solution as Crossover behavior of the thermal conductance and Kramers' transition rate theory. 12 Dec 2018 In Arabidopsis, higher mutation rate was reported in regions proximal to crossovers (Yang et al., 2015). This would suggest that crossover repair Fine-Scale Heterogeneity in Crossover Rate in the garnet-scalloped Region of the Drosophila melanogaster X Chromosome. Nadia D. Singh, Eric A. Stone,
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You are Here: Home >< Physics
# OCR PHYSICS B G492~ 9rd June 2014~ AS Physics Watch
1. (Original post by Minecraft27)
It was 24.
1kWh is equivalent to 3.6MJ of energy. The battery's capacity was 16kWh => 57.6MJ. The car can travel for 120km/80km = 1.5hours = 90 minutes = 5400 seconds.
So Power = Energy/Time = 57,600,000/5,400 = 10666.667...Js^-1
So the answer should be 10.7kW.
2. (Original post by CrystalPlanet)
I think you'd only lose one mark (it was a two mark question if I remember correctly); you stated that you had to multiply the length by two, so you'll probably get a mark for that.
I really hope so. Due to this I'm really hoping they allow an ECF for the part where you had to calculate the percentage increase of the frequency if the flute was not extended..
3. I'm sure the power question was along the lines of this:
*Table stating maximum power is 49kW and battery capacity is 16kWh*
"Peugot state that the range of the car is 120km when travelling at a constant speed of 80kmh^-1 and the battery is fully charged initially, show that the power provided is significantly less than the figure given in table x?"
4. I also got 10.7kW for the car power question
Sent from my iPhone using Tapatalk
5. (Original post by DanielCook95)
I really hope so. Due to this I'm really hoping they allow an ECF for the part where you had to calculate the percentage increase of the frequency if the flute was not extended..
There is no need for ECF. That part required nothing from any other parts.
6. (Original post by DomStaff)
There is no need for ECF. That part required nothing from any other parts.
Thinking about it now I don't even think that I used it haha! Do you mind me asking what you got for the answer to the percentage part?
7. (Original post by PsychoPanda)
I got 10^-6 for finger cross section. And I got 3 % for the temperature question, tbh in not so sure about the 3% but the cross sectional area of a finger is definitely 10^-6
Posted from TSR Mobile
I disagree: my finger's cross-sectional area is about 1cm x 1cm, which is 0.01m x 0.01m = 1x10^-4 m^2
10^-6 would mean that your finger is 1mm wide and thick.
8. (Original post by PsychoPanda)
I got 10^-6 for finger cross section. And I got 3 % for the temperature question, tbh in not so sure about the 3% but the cross sectional area of a finger is definitely 10^-6
Posted from TSR Mobile
I disagree, also. The diameter of my finger is approx 1cm. I have skinny fingers, so I am at the lower end of the spectrum. This gives me a x-sectional area in somewhere like 7.something * 10^-5, therefore approximately 10^-4.
9. (Original post by DanielCook95)
Thinking about it now I don't even think that I used it haha! Do you mind me asking what you got for the answer to the percentage part?
2.6%
It is just root298 - root283 divided by root283. Then multiplied by 100.
10. (Original post by Knowing)
I disagree: my finger's cross-sectional area is about 1cm x 1cm, which is 0.01m x 0.01m = 1x10^-4 m^2
10^-6 would mean that your finger is 1mm wide and thick.
I concur..
11. (Original post by DomStaff)
2.6%
It is just root298 - root283 divided by root283. Then multiplied by 100.
Haha that's a relief, I got that too. I did use the frequency but I think I ended up putting a line through it because I realised that it wasn't needed. Thank you!
12. (Original post by DanielCook95)
Haha that's a relief, I got that too. I did use the frequency but I think I ended up putting a line through it because I realised that it wasn't needed. Thank you!
The important thing to realise that lambda would not change, as L doesn't change, hence v and f increase proportionally.
13. The power question if you look at units you can do it.
kWh, you want kW, therefore do (capacity kWh)/h, therefore 16/(120/80) = 10.66.. = 10.7kW
14. (Original post by DomStaff)
The important thing to realise that lambda would not change, as L doesn't change, hence v and f increase proportionally.
Ofcourse. Thank's again, eased my nerves a tiny bit!
16. (Original post by tukky12)
I got the highest point to be over 6 metres..
17. I have the paper follow me into my cave
18. (Original post by FrankJaegar)
I have the paper follow me into my cave
All 30 pages? can you post it :3
19. (Original post by Knowing)
I got the highest point to be over 6 metres..
4.7?
Posted from TSR Mobile
20. (Original post by Mutleybm1996)
4.7?
Posted from TSR Mobile
yes!!!!
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# Algebra
posted by on .
You have 50 yards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?
• Algebra - ,
Try a square.
• Algebra - ,
Ms. Sue,
i got 12.5 by 12.5. thus giving me a square foot of 156.25.
• Algebra - ,
Yes.
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## Convergence and biases of Monte Carlo estimates of American option prices using a parametric exercise rule.(English)Zbl 1178.91195
Summary: This paper presents an algorithm for pricing American options using Monte Carlo simulation. The method is based on using a parametric representation of the early exercise decision. It is shown that, as long as this parametric representation subsumes all relevant stopping-times, error bounds can be constructed using two different estimates, one which is biased low and one which is biased high. Both are consistent and asymptotically unbiased estimators of the true option value. Results for high-dimensional American options confirm the viability of the numerical procedure. The convergence results of the paper shed light into the biases present in other algorithms proposed in the literature.
### MSC:
91G20 Derivative securities (option pricing, hedging, etc.) 91G60 Numerical methods (including Monte Carlo methods) 65C05 Monte Carlo methods
Full Text:
### References:
[1] Amemiya, T., Advanced Econometrics (1985), Harvard University Press: Harvard University Press Cambridge [2] Amin, K.; Khanna, A., Convergence of American option values from discrete- to continuous-time, Mathematical Finance, 4, 289-304 (1994) · Zbl 0884.90012 [5] Barraquand, J.; Martineau, D., Numerical valuation of high dimensional multivariate American securities, Journal of Financial and Quantitative Analysis, 30, 383-405 (1995) [7] Boyle, P., Options: a Monte Carlo approach, Journal of Financial Economics, 4, 323-338 (1979) [8] Boyle, P.; Broadie, M.; Glasserman, P., Monte Carlo methods for security pricing, Journal of Economic Dynamics and Control, 21, 1267-1321 (1997) · Zbl 0901.90007 [10] Broadie, M.; Detemple, J., American option valuation: new bounds, approximations, and a comparison of existing methods, Review of Financial Studies, 9, 1211-1250 (1996) [11] Broadie, M.; Detemple, J., The valuation of American options on multiple assets, Mathematical Finance, 7, 241-286 (1997) · Zbl 0882.90005 [12] Broadie, M.; Glasserman, P., Pricing American-style securities using simulation, Journal of Economic Dynamics and Control, 21, 1323-1352 (1997) · Zbl 0901.90009 [14] Carriere, J. F., Valuation of the early-exercise price for options using simulation and nonparametric regression, Insurance: Mathematics and Economics, 19, 19-30 (1996) · Zbl 0894.62109 [16] Cox, J. C.; Ross, S. A.; Rubinstein, M., Option pricing: a simplified approach, Journal of Financial Economics, 7, 229-263 (1979) · Zbl 1131.91333 [17] Duffie, D., Dynamic Asset Pricing Theory (1992), Princeton University Press: Princeton University Press Princeton [18] Dupacova, J.; Wets, R., Asymptotic behavior of statistical estimators and of optimal solutions of stochastic optimization problems, The Annals of Statistics, 16, 1517-1549 (1988) · Zbl 0667.62018 [19] Fishman, G., Monte Carlo: Concepts, Algorithms, and Applications (1996), Springer: Springer New York · Zbl 0859.65001 [22] Geske, R.; Johnson, H., The American put option valued analytically, Journal of Finance, 39, 1511-1524 (1984) [23] Geweke, J., Monte Carlo simulation and numerical integration, (Amman, H.; Kendrick, D.; Rust, J., Handbook of Computational Economics (1996), Elsevier: Elsevier New York), 731-799 · Zbl 1074.65506 [24] Grant, D.; Vora, G.; Weeks, D., Simulation and the early-exercise option problem, Journal of Financial Engineering, 5, 211-227 (1996) [26] He, H., Convergence from discrete to continuous time contingent claims prices, Review of Financial Studies, 3, 523-546 (1990) [28] Ju, N., Pricing by American option by approximating its early exercise boundary as a multipiece exponential function, Review of Financial Studies, 11, 627-646 (1998) [29] Karatzas, I., On the pricing of American options, Applied Mathematics and Optimization, 17, 37-60 (1988) · Zbl 0699.90010 [30] Kim, I. J., The analytic valuation of American options, Review of Financial Studies, 3, 547-572 (1990) [31] King, A. J.; Rockafellar, R. T., Asymptotic theory for solutions in statistical estimation and stochastic programming, Mathematics of Operations Research, 18, 148-162 (1993) · Zbl 0798.90115 [32] Kloeden, P. E.; Platen, E., Numerical Solution of Stochastic Differential Equations (1992), Springer: Springer New York · Zbl 0925.65261 [33] Krusell, P.; Smith, A., Rules of thumb in macroeconomic equilibrium: a quantitative analysis, Journal of Economic Dynamics and Control, 20, 527-558 (1996) · Zbl 0875.90122 [34] Kushner, H.; Dupuis, P., Numerical Methods for Stochastic Control Problems in Continuous Time (1992), Springer: Springer New York [35] Laha, R. G.; Rohatgi, V. K., Probability Theory (1979), Wiley: Wiley New York [37] Longstaff, F.; Schwartz, E., Valuing American options by simulation: a simple least-squares approach, Review of Financial Studies, 14, 113-147 (2001) [39] Press, W.; Teukolsky, S.; Vetterling, W.; Flannery, B., Numerical Recipes in C: The Art of Scientific Computing (1992), Cambridge University Press: Cambridge University Press Cambridge · Zbl 0845.65001 [40] Raymar, S.; Zwecher, M., Monte Carlo estimation of American call options on the maximum of several stocks, Journal of Derivatives, 5, 7-23 (1997) [42] Royden, H. L., Real Analysis (1988), Macmillan Publishing Company: Macmillan Publishing Company New York · Zbl 0704.26006 [43] Shapiro, A., Asymptotic behavior of optimal solutions in stochastic programming, Mathematics of Operations Research, 18, 829-845 (1993) · Zbl 0804.90101 [45] Taylor, A., General Theory of Functions and Integration (1965), Blaisdell Publishing Company: Blaisdell Publishing Company New York · Zbl 0135.11301 [46] Tilley, J., Valuing American options in a path simulation model, Transactions of the Society of Actuaries, 45, 83-104 (1993) [47] Tsitsiklis, J. N.; Roy, B. V., Regression methods for pricing complex American-style options, IEEE Transactions on Neural Networks, 12, 694-703 (2001)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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# When does the Riemann Integral-like characterization of Lebesgue Integrals fail?
Wheeden and Zygmund's book "Measure and Integral" gives an interesting characterization of the Lebesgue Integral that is reminiscent of the Riemann Integral. If $$E$$ be a Lebesgue measurable set, then then the Lebesgue integral of $$f$$ on $$E$$ is equal to $$\sup\Sigma_{k=1}^n (\inf_{E_K}f)\lambda(E_k)$$, where the supremum is taken over all finite partitions $$E_1,...,E_n$$ of $$E$$ into Lebesgue measurable sets. (This remains true for arbitrary measure spaces.)
But Wheeden and Zygmund say that if you switch the supremum and infimum, this need not hold true. So my question is, what is an example of a Lebesgue integrable function $$f$$ such that the Lebesgue integral of $$f$$ on $$E$$ is not equal to $$\inf\Sigma_{k=1}^n (\sup_{E_K}f)\lambda(E_k)$$?
Also, is there a subset of the set of Lebesgue integrable functions for which the switched version does hold true?
Let $$E = (0,1)$$ and $$f(x) = \frac{1}{\sqrt{x}}$$. Then $$\int f < \infty$$ but if $$E = \sqcup_{k=1}^n E_k$$ is a finite partition, then for some $$k$$, we have $$\sup_{E_k} f = \infty$$ and $$\lambda(E_k) > 0$$.
Let $$E=[0,1]$$ and let $$f$$ be the characteristic function of $$E \cap \mathbb Q$$.
• For an arbitrary partition $E_1,....,E_n$ we have $\sup_{E_k}f=1$, thus $\Sigma_{k=1}^n (\sup_{E_K}f)\lambda(E_k)= \Sigma_{k=1}^n \lambda(E_k) = \lambda(E)=1$, therefore $\inf\Sigma_{k=1}^n (\sup_{E_K}f)\lambda(E_k)=1 \ne 0 = \int_E f(x) dx.$ – Fred Nov 20 '18 at 12:53
• @Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = E\cap \mathbb{Q}$ and $E_2 = E\setminus E_1$? Then $\sup_{E_2} f = 0$ and $\lambda(E_1) = 0$, so $\sum (\sup_{E_k} f) \lambda (E_k) = 0$ – mathworker21 Nov 20 '18 at 15:52
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https://plainmath.net/4094/find-the-limits-lim_-xrightarrow0-sec-e-x-plus-pitan-frac-pi-4sec-x-1
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Question
# Find the limits lim_{xrightarrow0}sec[e^x+pitan(frac{pi}{4sec x})-1]
Limits and continuity
Find the limits
$$\lim_{x\rightarrow0}\sec[e^x+\pi\tan(\frac{\pi}{4\sec x})-1]$$
2021-02-22
Given
The given expression is $$\lim_{x\rightarrow0}\sec[e^x+\pi\tan(\frac{\pi}{4\sec x})-1]$$
find the limit
$$\lim_{x\rightarrow0}\sec[e^x+\pi\tan(\frac{\pi}{4\sec x})-1]=\sec[e^0+\pi\tan(\frac{\pi}{4\sec(0)})-1]$$
$$=\sec[1+\pi\tan(\frac{\pi}{4})-1]$$
$$=\sec(\pi)$$
$$=-1$$
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Problem Set
# Problem Set
Lesson 16. Unit 4. Grade 1 EngageNY
EngageNY10 min(s)
This Problem Set is a part of the Lesson 16, Unit 4, Grade 1. In this lesson, students begin to focus on adding ones with ones or adding tens with tens. They recognize single-digit addition facts as they solve, for example, 15 + 2, 25 + 2, and 35 + 2. When adding 35 + 2, students see that they are adding 5 ones to 2 ones, while the tens remain unchanged, to make 3 tens 7 ones or 37.
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https://sites.ualberta.ca/~isabell/rmd/textmining/04-tf-idf-PUBLISH.nb.html
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This Friends textmining effort in R was my Saturday project during a range of very snowy Saturdays we had here in Edmonton in September. It makes heavy use of the tidyverse, Text Mining in R by Julia Silge and David Robinson (which is highly recommended), and piping from the `magrittr` package (which makes things so much nicer.) If you haven’t read the previous two episodes, they are:
You can find a tutorial by Rich Majerus on how to loop with ggplot2 here.
Disclaimer – I do not claim to be an expert in textmining. There may be faster, smarter, or nicer ways to achieve a certain thing. Still, maybe you’ll find something interesting for your own projects - or just some funny tidbit about your favourite show. In this fourth “episode,” we’ll do some TF-IDF (term frequency–inverse document frequency) analyses - essentially, we’ll try to find out what the most characteristic words for each Season and each Friend are. See also here and here.
Isabell Hubert 2018
# Prep
``````library(dplyr)
library(tidyr)
library(stringr)
library(tidytext)
library(magrittr)
library(ggplot2)``````
And the dataframes we’ll need:
``````tokens <- readRDS("tokens.rds") # pre stopword anti-join - useful for word volume
friends <- readRDS("friends-df.rds") # the cleaned-up, post stopword anti-join``````
And define some useful character vectors we can use later for filtering, plotting, and looping:
``````friendsNames <- c("Monica", "Rachel", "Chandler", "Joey", "Ross", "Phoebe")
friendsExtended <- c("Monica", "Rachel", "Chandler", "Joey", "Ross", "Phoebe", "Janice", "Gunther")
seasons = c(1:10)``````
# Token Frequencies
We’ll start with calculating some overall token frequencies - basically how many words were said total in the entire show, and how many times each word was said:
``````token.count <- tokens %>%
count(word, sort = TRUE) %>%
mutate(total = sum(n))``````
We’ll also create some counts by season:
``````s.token.count <- tokens %>%
count(word, season, sort = TRUE) %>%
group_by(season) %>%
mutate(seasonTotal = sum(n))``````
We can then illustrate Zipf’s law:
``````ggplot(s.token.count, aes(n/seasonTotal, fill = factor(season))) +
geom_histogram(show.legend = FALSE, binwidth = 0.0001) +
facet_wrap(~season, ncol = 2, scales = "free_y") +
coord_cartesian(xlim = c(0, 0.005))``````
# TF-IDF
The idea of tf-idf is to find the important words for the content of each document by decreasing the weight for commonly used words and increasing the weight for words that are not used very much in a collection or corpus of documents.
(from Text Mining with R)[https://www.tidytextmining.com/tfidf.html]
## by Season
``````s.tfidf <- s.token.count %>%
bind_tf_idf(season, word, n)``````
Let’s look at the highest TF-IDF values:
``````s.tfidf %>%
arrange(desc(tf_idf))``````
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### pikmike's blog
By pikmike, history, 23 months ago,
1065A - Vasya and Chocolate
Tutorial
Solution (Ajosteen)
1065B - Vasya and Isolated Vertices
Tutorial
Solution (Ajosteen)
1065C - Make It Equal
Tutorial
Solution (Ajosteen)
1065D - Three Pieces
Tutorial
Solution (PikMike)
1065E - Side Transmutations
Tutorial
Solution (PikMike)
1065F - Up and Down the Tree
Tutorial
1065G - Fibonacci Suffix
Tutorial
Solution (BledDest)
• +43
» 23 months ago, # | 0 44220948Why does my D WA on test 25? I had a similar solution, except I bruteforced the shortestdistance instead of using bds/djikstras, and used a bunch of selections for the dp instead of a pair.
• » » 23 months ago, # ^ | ← Rev. 2 → +11 You probably didnt consider that in the optimal solution you need to change the chess piece on the way between position a and b and not only change pieces on position a or b?
• » » » 23 months ago, # ^ | -6 but a rook can always reach your desired position in 2 moves, so even if you switch to a rook at the beginning your total cost will be 3Now if you're using a bishop and move to a position which isn't your destination, and switch to another counter and then move to your desired position, your total cost would still be 3The same goes for switching between other counters. Your maximum cost will always be 3 (which is the same as switching to a rook at the beginning and then reaching your desired position)
• » » » » 23 months ago, # ^ | +8 4 14 10 3 7 4 15 11 13 12 1 16 9 2 5 6 8 Consider, when you are in number 2 with a bishop. You can go to number 3 with a rook in 3 steps: replace bishop with rook, goto 14, goto number 3. You can go to number 3 with a knight in 3 steps: go back to 1 with bishop, replace bishop with knight, goto number 3. (Note that, you cannot replace bishop with knight in number 2 because, in that case, you need 4 steps to go to number 3)
• » » » » 16 months ago, # ^ | 0 Well, what if you number the positions as the positions of a knights tour? Then the optimal solution would be to start with a knight and the distance between each location would be 1.
» 23 months ago, # | ← Rev. 2 → +16 O((N + M) * M) or O(M * (M + log(N))) solution for G:Same solution but changing 1 part, the part of counting the number of occurences. Instead of using an automaton, keep pref[i] ans suf[i] as the M first characters of the prefix/suffix of the i-th string. Using this, freq[i] = freq[i-2] + freq[i-1] + occurences that use the suf[i-2] + pref[i-1]. This gives us an O(N * M^2) solution like in the editorial. But you can note that starting from one point (the position where |F[i]| >= M), pref[i] == pref[i + 2] and suf[i] == suf[i + 1]. This means that the "occurences that use the suf[i-2] + pref[i-1]" will actually repeat with period 2. So you can calculate it in one phase until that point, the cost of this is O(F[0] + F[1] + F[2] + ... + F[i]) = O(F[i + 1]) = O(M) and from that point on, you can reuse the number of occurences that happened in the previous calculations (or do 4 more just to be safe and use from that point on, doesn't change the complexity). Total complexity: O((N * M) * M) with one phase (additional character) happening in O(M + N) because you need to calculate the borders for KMP. You can actually use this fact to create a linear recurrence relation for freq[i] and use matrix exponentiation (the column matrix keeps freq[i], freq[i-1], the transitions and keep swapping them) starting from the point that it repeats, this would result in a O(log(M) + M + log(N) * 4^3) phase per character in the answer. The log(M) comes from the fact that fib is exponential, so the point that it starts repeating is O(log(M)). This would result in a O(M * (M + log(N))) solution.Code for O((N + M) * M) solution: http://codeforces.com/contest/1065/submission/44225229Edit: If there's an occurence in the transition, you can also keep calculating until it breaks K. It will break K in O(logK) steps and if there's no occurence, you just break. So I guess the second part wouldn't be necessary for the O(M^2) complexity.
• » » 23 months ago, # ^ | ← Rev. 2 → +5 That also implies that a big N doesn't matter as can be seen here: http://codeforces.com/contest/1065/submission/44226008. n = min(n, 100) works. To be more exact, the bound on N mattering would be O(log(M) + log(K)).
» 23 months ago, # | 0 https://codeforces.com/contest/1065/submission/44147964For C,my approach is a bit similar can anyone help me,where i am wrong?
» 23 months ago, # | +1 Can someone explain d.
» 23 months ago, # | 0 Can anyone please suggest what is wrong with my approach? SubmissionI am first sorting the input array in descending order followed by storing a vector of blocks sliced to reach the subsequent lower level. For instance, if the input array is 4 3 2 2 1 my vector is 1 2 4 5 Finally, I am traversing to find out if a single or combination of blocks is less than or equal to k or not and incrementing the counter as a result.
» 23 months ago, # | 0 Hi! I have a problem can anybody help? if u submit this solution for problem D with MS C++ u get accept! Link to code! but gnu++ gives WA! do u know why?
» 23 months ago, # | +1 Can E solve by polya?
• » » 23 months ago, # ^ | +1 Yes, you can. 44419650As the editorial said every subset of segments [0..l1), [l1, l2), …, [lm−1, lm) is achievable. There is a bijective relation between the set of subsets of segments and the set of transformations G and every transformation add An - k where k is the sum of the cardinalities of the segments in the associated subset.
• » » » 23 months ago, # ^ | +1 Yes! Actually, the formula solve by polya is the same as the official tuitual. Wonderful!
• » » » » 22 months ago, # ^ | -6 what is polya?
• » » » » » 22 months ago, # ^ | 0 This may help you https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem
» 23 months ago, # | 0 This for problem D, can anyone tell me why my solution times out (http://codeforces.com/contest/1065/submission/44417842)
» 23 months ago, # | -6 In the tutorial of Problem E: cnt function : why is there *(mod+1)/2 instead of /2 directly in the return. I am getting wrong answer with /2 but I am not able to understand why.
» 22 months ago, # | ← Rev. 2 → 0 Why this this solution for problem 1065C is not correct ?Since the attainable equal height is the Smallest number.then Number of Chops can be : ceil((SumOfHeight - N * minHeight) / K)
• » » 16 months ago, # ^ | 0 Because we cant remove blocks parially from any particular height
» 22 months ago, # | ← Rev. 2 → -6 Problem D can be solved with a single BFS using a state of four dimensions: current x, current y, piece type, current square we are at. The result is the minimum among final states. https://codeforces.com/contest/1065/submission/46351519
» 19 months ago, # | 0 Can someone please explain how to solve "1065E — Side Transmutations"? PikMike forgot to explain what l is?Sorry for my amazingly bad english.
» 5 weeks ago, # | 0 Can anyone explain 1065C?I didn't understand the editorial.thanks in advance.
• » » 2 weeks ago, # ^ | 0 here is my code that may help u. if u don't understand then sorry. https://codeforces.com/contest/1065/submission/91731954
» 2 weeks ago, # | -8 Problem E is so nice. A subtle change in perspective, by viewing the string as a whole, solves it instantly!
» 2 weeks ago, # | 0
• » » 2 weeks ago, # ^ | 0
• » » » 2 weeks ago, # ^ | 0 ![ ]()
• » » » » 2 weeks ago, # ^ | 0 As memes are trending in this comment section I thought lets start with basics
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+1-415-315-9853
info@mywordsolution.com
## Engineering
Civil Engineering Chemical Engineering Electrical & Electronics Mechanical Engineering Computer Engineering Engineering Mathematics MATLAB Other Engineering Digital Electronics Biochemical & Biotechnology
1) Petrol engine has stroke of 120mm and connecting rod is 3 times the crank length. Crank rotates at 1500 r.p.m. in clockwise direction. Find out:
(a) Velocity and acceleration of the piston and
(b) Angular velocity and angular acceleration of connecting rod when piston has travelled one fourth of its stroke form Inner Dead Centre.
2) With neat and suitable sketches describe the condition of static equilibrium of member with
(a) two forces
(b) three forces
(c) two forces and a torque
(d) two couples
3) Four masses m1, m2, m3 and m4 are 200 kg, 300 kg, 240 kg and 260 kg respectively. Corresponding radii of rotation are 0.2 m, 0.15m, 0.25m and 0.3m respectively and angles between successive masses are 45o, 75o and 135o. Determine the position and magnitude of balance mass needed, if its radius of rotation is 0.2 m.
4) Single cylinder reciprocating engine has speed of 240 r.p.m., stroke 300mm, mass of reciprocating parts 50 kg, mass of revolving parts at 150mm radius 37 kg. If two-third of reciprocating masses and all revolving parts are to be balanced, determine:
(a) Balance mass required at a radius o 400 mm and
(b) Residual unbalanced force when crank has rotated 60o from Top Dead Centre.
5) A shaft 50mm diameter and 3meter long is simply supported at the ends and carries three loads of 1000 N, 1500 N and 750 n at 1m, 2m and 2.5m from the left support. The young’s modulus for shaft material is 200 GN/m2. Find the frequency of transverse vibration.
Mechanical Engineering, Engineering
• Category:- Mechanical Engineering
• Reference No.:- M913697
Have any Question?
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Added together and write as decimal number:
LXVII + MLXIV
Result
LXVII + MLXIV = 1131
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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Classical Mechanics
# Work-kinetic energy theorem
A $8.0\text{ kg}$ block is moving at $3.2\text{ m/s}.$ A net force of $10\text{ N}$ is constantly applied on the block in the direction of its movement, until it has moved $16\text{ m}.$ What is the approximate final velocity of the block?
A man is driving a car with mass $1.00 \times 10^3\text{ kg}$, at a speed of 35 m/s on a highway. Suddenly, he puts on the brakes to avoid an accident. After the brakes are applied, a constant kinetic friction force of magnitude $7.00 \times 10^3\text{ N}$ acts on the car. What is the minimum distance from the car ahead at which the brakes should be applied to avoid an accident?
Ignore air resistance.
Two identical twins, Amy and Beatrice, are on top of two different slides, one on each slide. Both slides start 3 m above the ground and end 0.5 m above the ground. Amy's slide is a straight, 45 degree slide with height given by $h=-x+3$ (so it starts at $x=0$ with a height of three meters and ends at $x=2.5$with a height of half a meter). Beatrice's slide is curved and the height is given by $h=3/(1+x)$ (so it starts at $x=0$ and ends at $x=5$). Both slides are frictionless.
If both twins start from rest at the top of their slides, what is the ratio of Amy's speed to Beatrice's speed at the bottom of the slides? i.e., what is $v_{Amy}/v_{Beatrice}$?
The cannons on 19th century frigates were ponderous things. For example, on the USS Constitution, a frigate used in the War of 1812, the cannons were $3~\mbox{meters}$ long, had a mass of $3000~\mbox{kg}$, and could fire a $10 ~\mbox{kg}$ cannonball up to a kilometer away. Needless to say the recoil on the cannons was tremendous and a mechanism had to be devised to keep the cannons from careening across the deck. The system employed at the time consisted of rope and tackle, which basically caught the cannon. One wants to make sure that the ropes are strong enough, so one needs to know the magnitude of the forces necessary to stop the cannon. A cannon $4~\mbox{m}$ above the waterline fires a $10~\mbox{kg}$ cannonball horizontally that hits the water $500~\mbox{m}$ away. It recoils and is stopped by a rope braking mechanism, which exerts a constant force on the cannon once it starts to recoil. If the cannon recoils $1~\mbox{m}$, what is the force the rope braking mechanism exerts on the cannon in Newtons?
Details and assumptions
• You may neglect air resistance for the motion of the cannonball.
• The acceleration due to gravity is $-9.8~\mbox{m/s}^2$.
A man is driving his $1100\text{ kg}$ car at $36\text{ km/h}$ on a straight freeway. After accelerating for $30$ seconds, the car has a speed of $108\text{ km/h}.$ How much work did the engine do during the $30$ seconds?
Disregard the work lost by friction.
×
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# Is a 12 inch pizza enough for one person?
Is a 12 inch pizza enough for one person? A 12-inch pizza can serve from 3 to 4 people. Therefore, if the number of people is just smaller than 3 or bigger than 4, you should consider a small 10-inch pizza or a large 14-inch pizza.
How big should a pizza be for 2 people? A 12-in. medium pizza is traditionally cut into eight slices and serves two people. A 14-in. large pizza is cut into eight slices and will serve two or three people.
What is normal pizza size? It’s difficult to say which pizza size is the most popular, but the majority of pizzas are medium-sized at 14 to 16 inches. What is the difference between a large and extra-large pizza? Extra-large pizzas are usually 16 inches to 18 inches, while a large pizza is typically 14 to 16 inches.
What is the most popular pizza size? The most popular size of pizza in the United States is 14 inches in diameter. The average pizza is usually 14 inches in diameter with eight slices. The average pizza is also normally supposed to feed a group of four (two slices per person).
## Is a 12 inch pizza enough for one person? – Additional Questions
### How big is a 24 inch pizza?
An extra large pizza is usually between 20 and 24 inches in diameter. That means that there are usually between 9 and 18 whole slices in an extra large pizza, depending on what brand you buy. At Pizza Hut, Domino’s, or Papa John’s this could mean that there are around 12-14 slices in their extra large pizzas.
### How big is an 8 inch pizza?
In practice, this means an 8″ pizza has an area of around 50 square inches.
### Is 8 inches pizza big?
Small pizzas average between 8 and 10 inches in diameter and will yield about six slices. Medium pizzas run 12 inches in diameter and will give you about eight slices. Large pizzas are 14 inches in diameter and will offer approximately 10 slices.
### How many slices is 8 inches?
8 inch cakes can be sensibly served to 14 people cutting each slice at about 2 1⁄4 inches across the back. 2 1⁄4 inches is about the length of the spoon part of a tablespoon. 8 inch cakes can also be cut event style. Cutting an 8 inch cake event style can serve up to 24 people.
### Which is bigger 2 8 inch or 16 inch pizza?
Some people order two 8 inch pizzas. Are they the same as a 16-inch pizza? The answer may surprise you: a 16 inch pizza is about four times larger than an 8 inch pizza. If you prefer a smaller size, you need to increase the number of the pizza.
### Why you should always order a large pizza?
You should always order the large pizza because it will be cheaper per square inch. While a 16-inch pizza is four times the size of the 8-inch pizza, the larger pizza will be far less than four times the price of the small.
### Is an 18 inch pizza big?
How Big Is a 18 Inch Pizza? A typical 18 inch pizza is about the size of a large dinner plate. It can easily feed four people, and might even be enough for six people if they’re not too hungry.
### How much bigger is a 16 inch pizza than an 8-inch?
If we assume (for ease) that the crust is one inch wide, an 8-inch pizza will have roughly 28 square inches with toppings. By comparison, a 16-inch pizza has 154 square inches of toppings. So the 16-inch pizza actually has 5.5 times more area with toppings.
### Which size pizza is the best value?
Buying a large pizza is a better value and cheaper price than buying a small.
Here are some key findings gleaned from Groupon’s pizza survey:
• A 16-inch pizza is actually two-and-a-half times more food than a 10-inch pizza.
• The average price per square inch for a 10-inch pizza was \$0.16 versus \$0.096 for a 16-inch pizza.
### Is a large pizza 14 or 16-inch?
The size of the pizza is its diameter, and the area of a pizza increases as we increase the diameter. It is almost always a more cost-effective option to choose a large pizza because you get more area, which means more pizza to eat. Suppose a medium pizza is 12 inches and a large is 16 inches.
### How much more pizza is a 16 than a 14-inch?
A 16-inch pie is 201 square inches — approximately 31 percent larger, in terms of area, than a 14-inch pie.
### Is a large pizza a better deal?
A study of nearly 4,000 pizzerias confirms it: if you buy based on price-per-square-inch, the large is always a better deal.
### Are pizzas getting smaller?
But as prices go up, pizza sizes are expected to go down. Domino’s is reportedly shrinking the size of its pizzas in a bid to ease the financial pressure on franchisees.
### How many people does a 12 inch pizza feed?
A 12-inch pizza can serve from 3 to 4 people. Therefore, if the number of people is just smaller than 3 or bigger than 4, you should consider a small 10-inch pizza or a large 14-inch pizza.
### How much pizza do I need for 16 adults?
A good rule of thumb is to allow three slices per adult and two slices per child of a traditional crust pizza. Of course, if everyone is starving, then consider upping those numbers a bit by one or two slices.
### How much is tip for pizza delivery?
Generally speaking, delivery orders that are less than \$20 are given a minimum tip of \$3. If the order is over \$20, then it’s customary to calculate a tip that is 10%-15% of the order (but never less than \$5).
### How much pizza is too much?
Your risk for heart disease may increase
If you love pizza and balance a slice weekly with vegetables like a salad, it can certainly be a part of a healthy diet. However, if you start eating three or four slices (or more!) on a regular basis that is where your diet can have health consequences.
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TNPSC Engineering Services Results 2014 Published (27.07.2014) Check Here - TNPSC GURU - TNPSC Group 2A/2 Apply Online - Join Test batch
## Tuesday, September 1
TNPSC Combined Engineering Services 2014 Expected Cut off Marks
TNPSC - Tamil Nadu Public Service Commission is going to conduct TNPSC Combined Engineering Services Exam on 27th July 2014.Check here the TNPSC Engineering Syllabus 2014. The number of TNPSC Combined Engineering Services Posts as per the notification is 98. Nearly 53,000 candidates have applied for the upcoming TNPSC Combined Engineering Services Examinations.
TNPSC Engineering Services 2014 Expected Cut off Marks
Expected Cut off Marks for TNPSC Combined Engineering Services 2014-15 Exam : to be updated soon.
TNPSC Combined Engineering Services Cut Off Marks
To get more approximate TNPSC Combined Engineering Services 2014 Cut Off marks come back after the exam and correct your answers with the help of TNPSC Combined Engineering Services 2014 Exam Official Answer key (Official Answer Key for TNPSC Combined Engineering Services preliminary updated Check soon).
Just vote on this TNPSC Combined Engineering Services 2014 Exam Cut Off marks poll below. Select the Number of Questions you got Correct in the TNPSC Combined Engineering Services 2014 Exam which is to be conducted on 27th July 2014.
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Result
How to Calculate Your TNPSC Engineering Services Exam 2014 Cut Off Mark?
Step 2: [Number of Questions correct in Paper 1 (Subject Paper) x 1.5] + [Number of Questions correct in Paper 2 (General Studies Paper) x 2 ]
Example : Let the no of questions correct in Paper 1 = 120 and in Paper 2 = 80
Then, [120 x 1.5] + [80 x 2] = [180] + [160] = 340 is your Cut Off marks Out of 500
How Many Questions You Got Correct in TNPSC Engineering Services Exam 2014?
Check here the TNPSC Engineering Services 2014 Cut Off Poll Result
You can Download the Answer key for the TNPSC Combined Engineering Services 2014 here for the exam to be conducted on 27th June 2014 after the exam. The TNPSC Official answer key for the TNPSC Combined Engineering Services exam will also be updated soon.
TNPSC Engineering Services 2013 Cut off Marks
TNPSC has announced the written exam results and oral test for TNPSC Combined Engineering Services Exam 2013 recently. We have made an attempt to extract the more approximate minimum cut off marks to clear Previous year TNPSC combined Engineering Services 2013 Preliminary written Exam. In the result there is no mention about the special categories so don't take the minimum values as general BC,SC, MBC, ST, etc,.
General 319 BC 310.5 MBC 296 BC(M) 292.5 SC 283 SC(A) 278 ST 276
TNPSC Combined Engineering Services 2014 Result
TNPSC Combined Engineering Services Written exam result is expected to be announced within 2 or 3 months. you can also get the latest information about the result announcing date for TNPSC CESE 2014 Here.
TNPSC Engineering Services 2014 Minimum Qualifying Marks
TNPSC has prescribed the minimum qualifying marks for selection for each category. But this minimum marks does not mean that you will be provided job when you get more marks above this. This is just like a pass mark in exam. Only the candidates securing top scores under each category will be selected.
For SCs, SC(A)s, STs, MBCs/ DCs, BCs and BCMs = 171 Marks
Others = 228 Marks
1. How you did your exam guys>?
2. hi lee,gk is very easy but engineering syllabus is little bit tough..,what is the cut off for this exam(both). please tell me????
1. Dear naganathan, TNPSC is yet to publish the answer key for the engineering services exam 2014. After that only from the Cut off poll we can predict..
2. thankz lee!!!!
3. hai lee, somebody told me 141/200 under mbc is not a good one and the then i didn't get even last phase. But u told me, i have a chance??!! please clarify my query lee!!!!! i am so disappointment??!!
1. we hope you are asking about group 2a right?? we have predicted the cut off from our website users inputs.. you can also check vidiyalarni academy's prediction online...
4. do anyone got the answer key for tnpsc engineering services exam????
5. Why tnpsc engineering answer key is not published yet..
6. Dear friends... TNPSC Combined Engineering Services official Answer key Published Check soon...
7. sir i am getting 306 ece bc category...is there any chance
1. i got 276 mbc male ece is there any chance??
8. sir
update the expected answer key for branch wise
1. expected date for next CESE exams from TNPSC?
2. next CESE exams from tnpsc when will be carriedout?is there any guess?
3. Only in 2015...
4. hi, when we ll get the results
9. sir
pls update community as well as branch wise cut off marks
10. Sir.... i need the study materials for civil engineering syllabus.... if u get pls send a me...at rathanavelcivil41@gmail.com.... otherwise tell proper websites name....
11. is there any chances of publishing the tnpsc cese exam results in 2014 ?????????
12. hi when will tnpsc cese exam 2014 on 27/7/2014 be announced???
13. there s no update abt tnpse cese exam hed n 27/07/2014............................if any1 knws, pls share it frnds
14. eligibility for Engineering service EXam
15. when will the result Publish???
16. Sir i need civilengineering notes plz help me dineshnalinan@gmail.com
17. sir i am getting 310-320 in cese exam. i belongs to general category and branch chemical engineer there is any chance of selection or not please reply me..
Post your feedback and doubts in the comment box below.
Thanks for visiting our Website..
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Ch.15 Vocab
Boiling Point Elevation
Temp. Difference between a Solutions Boiling point and pure solvents boiling point
Brownian Motion
Jerky, random, rapid movements of colloid particles that results from collisions of particles of the dispersion medium with the dispersed particles
Colligative Property
Physical property of a solution that depends on the number, but not identity, of dissolved solute particles, EX: Vapor Pressure Lowering, Boiling Point Elevation, Osmotic Pressure, and Freezing Point Depression
Colloid
Hetergeneous mixtures containing particles larger than solution particles but smaller than suspension particles that are categorized according to the phases of their dispersed particles and dispersing mediums
Concentration
quantitative measure of the amount of solute in a given amount of solvent or solution
Freezing Point Depression
difference in temperature between a solutions freezing point and the freezing point of its pure solvent
Heat of Solution
overall energy change that occurs during the solution formation process
Henry’s Law
states that at a given temperature the solutility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid
Immiscible
describes 2 liquids that can be mixed together but seperate shortly after you cease mixing them
Insoluble
describes a substance that cannot be dissolved in a given solvent
Miscible
describes 2 liquids that are soluble in each other
Molarity
number of moles of solute dissolved per liter of solution, also known as molar concentration
Molality
Ratio of the number of moles of solute dissolved in 1 kilogram of solvent, also known as molal concentration
Mole Fraction
ration of the number of moles of solute in solution to the total number of moles of solute and solvent
Osmosis
diffusion of solvent particles across a semiperameable membrane from an area of higher solvent concentration to an area of lower solvent concentration
Solubility
contains maximum amount of dissolved solute for a given amount of solvent at a specific temperature and pressure
Osmotic Pressure
The additional pressure needed to reverse osmosis
Saturated Solution
Contains maximum amount of dissolved solute for a given amount of solvent at a specific temperature and pressure
Soluble
describes substance that can be dissolved in a given solvent
Solvation
process of surrounding solute particles with solvent particles to form a solution; occurs only where and when the solute and solvent particles come in contact with each other
Supersaturated Solution
Contains more dissolved solute than a saturated solution at the same temperature
Suspension
a type of heterogeneous mixture whose particles settle out over time and can be separated from the mixture by filtration
Tyndall Effect
The scattering of light by colloidal particles
Unsaturated Solution
Contains less Dissolved solute for a given temperature and pressure than a saturated solution, has further capacity to hold more solute.
Vapor Pressure Lowering
The lowering of vapor pressure of a solvent by the addition of a nonvolatile solute to the solvent
x
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### Double Hashing
Hashing is a technique used for storing , searching and removing elements in almost constant time. Hashing is done with help of a hash function that generates index for a given input, then this index can be used to search the elements, store an element, or remove that element from that index.
A hash function is a function that is used to map the data elements to their position in the data structure used. For example if we use an array to store the integer elements then the hash function will generate position for each element so that searching, storing and removing operation on the array can be done in constant time that is independent of the number of elements in the array. For better look at the example below.
now we face a problem if for 2 numbers same position is generated example consider elements 1 and 14
1 % 13 = 1
14 % 13 = 1
so when we get 1 we store it at the first position, but when we get 14 we see that the position 1 is already taken, this is a case of collision.
Inorder to resolve collision we employ various collision resolving methods here we use double hashing to resolve collision.
In Double Hashing instead of 1 hash function we have 2 hash functions, and we can use combination of these 2 functions to generate new position and check if the new position found is empty or not .we find the new Position using the formula given below.
new_Position = (i*h1(element) + h2(element)) % SIZE;
where i is a natural number
SIZE is the size of the hash table.
To know more about hash functions and how to select hash function click Here.
To know more about Double Hashing click Here
### Sample input and output to check the program
You might also be interested in
Hashing with Linear Probing
Find Trailing number of zeros in factorial of a number.
Check if a number is a Fibonacci number or not
Find Factorial of a number
Bubble sort Algorithm
Insertion sort Algorithm
### Infix to Prefix conversion using Stack
This post is about conversion of Infix expression to Prefix conversion. For this conversion we take help of stack data structure, we need to push and pop the operators in and out of the stack.
Infix expressions are the expressions that we normally use,eg. 5+6-7; a+b*c etc. Prefix expressions are the expressions in which the 2 operands are preceded by the operator eg. -+567 , +a*bc etc.
This method is very similar to the method that we used to convert Infix to Postfix but the only difference is that here we need to reverse the input string before conversion and then reverse the final output string before displaying it.
NOTE: This changes one thing that is instead of encountering the opening bracket we now first encounter the closing bracket and we make changes accordingly in our code.
So, to convert an infix expression to a prefix expression we follow the below steps
(we have 2 string, 1st is the input infix expression string 2nd is the output string which is empty initially)
We first revers…
Hashing is a technique used for storing , searching and removing elements in almost constant time. Hashing is done with help of a hash function that generates index for a given input, then this index can be used to search the elements, store an element, or remove that element from that index.
A hash function is a function that is used to map the data elements to their position in the data structure used. For example if we use an array to store the integer elements then the hash function will generate position for each element so that searching, storing and removing operation on the array can be done in constant time that is independent of the number of elements in the array. For better look at the example below.
now we face a problem if for 2 numbers same position is generated example consider elements 1 and 14
1 % 13 = 1
14 % 13 = 1
so when we get 1 we store it at the first position, but when we get 14 we see that the position 1 is already taken, this is a case of collision.
Inorder…
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Bibliographic record and links to related information available from the Library of Congress catalog.
Note: Contents data are machine generated based on pre-publication provided by the publisher. Contents may have variations from the printed book or be incomplete or contain other coding.
```Preface
PART 1: MATRIX ALGEBRA AND LINEAR ECONOMIC MODELS
1. Matrix Algebra
1.1 Basic Concepts 2
1.2 Determinants 9
1.3 The Inverse of a Matrix 17
1.4 Linear Dependence of Vectors and the Rank of a Matrix 25
*1.5 Kronecker Products and Vecs of Matrices 33
2. Simultaneous Linear Equations
2.1 Definitions 37
2.2 Homogeneous Case 39
2.3 Nonhomogeneous Case 44
2.4 Special Case m=n 48
3. Linear Economic Models
3.1 Introduction and Definitions 54
3.2 Examples of Linear Economic Models 56
3.3 The Use of Matrix Algebra in Statistics and Econometrics 63
4. Quadratic Forms and Positive Definite Matrices
4.1 Introduction 69
4.2 Eigen Values of a Symmetric Matrix 70
4.3 Eigen Values of Special Matrices 73
4.4 Eigen Vectors of a Symmetric Matrix 75
4.5 Matrix whose Columns are the Eigen Vectors of a Symmetric Matrix 80
4.6 Diagonalization of Quadratic Forms 83
4.7 Eigen Values and 85
4.8 An Alternative Approach using Determinants 86
PART 2: FUNCTIONS OF MANY VARIABLES AND OPTIMIZATION
5. Functions of Many Variables
5.1 Functions in General 93
5.2 Partial Differentiation 95
5.3 Special Sorts of Functions 102
5.4 Comparative Statics and Nonlinear Economic Models 116
5.5 Differentials and Taylor¿s Approximation 123
6. Optimization
6.1 Unconstrained Optimization 131
6.2 Local Optima and Global Optima 142
6.3 Constrained Optimization 146
6.4 Constrained Local Optima versus Constrained Global Optima 153
*6.5 An Introduction to Matrix Calculus 156
7. Comparative Static Analysis in Optimization Problems
7.1 Introduction 166
7.2 Unconstrained Optimization 166
7.3 Constrained Optimization 170
7.4 Slutsky¿s Equation 174
7.5 Applications of the Envelope Theorems in Economics 178
PART 3: DYNAMIC ANALYSIS
8. Integration
8.1 Introduction 201
8.2 Definite Integrals 202
8.3 Integration as Anti Differentiation 207
8.4 Indefinite Integrals 212
8.5 Further Considerations 215
8.6 Economic Applications 218
9. Continuous Time: Differential Equations
9.1 Definitions 225
9.2 Linear Differential Equations 226
9.3 First Order Linear Differential Equations with Constant Coefficients 227
9.4 Economic Dynamics Using First Order Differential Equations 233
9.5 Second Order Linear Differential Equations with Constant Coefficents 243
9.6 Economic Application: A Dynamic Supply and Demand Model 250
9.7 Higher Order Linear Differential Equations 252
9.8 Descriptive Analysis of Nonlinear Differential Equations 254
10. Discrete Time: Difference Equations
10.1 Introduction and Definitions 259
10.2 First Order Linear Difference Equations with Constant Coefficients 261
10.3 Second Order Linear Difference Equations with Constant Coefficients 262
10.4 Investigating the Nature of the Roots of a Quadratic Equation 269
10.5 Economic Applications 272
10.6 Higher Order Linear Difference Equations 280
11. Dynamic Optimization
*11.1 Introduction 283
*11.2 Dynamic Optimization versus Static Optimization 283
*11.3 The Basic Optimal Control Problem and Pontryagin¿s Maximum Principle 285
*11.4 Extensions to the Basic Problem 293
*11.5 Economic Application: Ramsey/Solow Model 305
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posted by .
1. 2x > -6 and x - 4 < 3
2. x + 3 > 2x + 1 and -4x < -8
3. -6 < x + 3 < 6
4. -3x > -6 or x - 5 < 2
5. x - 2 > 2x + 1 or 10 > -2x + 2
LOL - why me ?
1. 2x > -6 and x - 4 < 3
x > -3 and x < +7
-3 < x < +7
2. x + 3 > 2x + 1 and -4x < -8
note if you divide by negative number, change arrow direction
2 > x and x > 2 can not be both
3. -6 < x + 3 < 6
subtract 3 from everything
-9 < x < 3
'cause you rock (: !
thanks for helping me so far (:
4. -3x > -6 or x - 5 < 2
x < 2
or x < 7
everything left of seven satisfies the second, everything left of 2 satisfies both. In other words x<2 is a subset of x <7
5. x - 2 > 2x + 1 or 10 > -2x + 2
x < -2 or x > -4
that is everywhere
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# Re: Discontinuities in stimulus. (Fred Herzfeld )
```Subject: Re: Discontinuities in stimulus.
From: Fred Herzfeld <herzfeld@xxxxxxxx>
Date: Fri, 17 Oct 2008 15:44:24 -0400
List-Archive:<http://lists.mcgill.ca/scripts/wa.exe?LIST=AUDITORY>
Hello Ranjit,
Consider the Fourier Transform of a signal consisting of a single sinusoid. Then the amplitude is
say A and the phase is phi. Now delete a small section delta from the wave form at the peak of the
sinusoid. Each of the FT integrals is reduced by the product of the time delta and the particular
basis. Now repeat the same gedanken experiment but with the delta placed at the point when the
signal sinusoid is near zero. It is then obvious that there will be differences in the FT and
therefore what the ear hears. It should also be obvious now that if the signal frequency and the
time at which the delta section is deleted are not commensurate the differences outlined above will
continue to drift from the peak to the valley and thus generate "fluttering".
The answer to the second part of your question is in general "it cannot be done". However if you
know for example that the stimulus consists of a known number of harmonics of a known fundamental
frequency or at least a known number of specified frequencies and their amplitudes and phases then
the problem of the determination if "outages" exist is in theory solvable. Not necessarily easy
Fred
------------------------------
Ranjit Randhawa wrote:
> Dear List,
> I have a rather simple question concerning the usual assumption made
> about the need for "continuity" of a stimulus. My interest arose when I
> started exploring the "fluttering" sound heard for some stimuli. A
> simple way to create a stimulus to study this phenomenon was to insert
> periodic "discontinuities" in a pure sinusoid of low frequency by simply
> deleting bits from this stimulus in a consistent manner. My next step
> was to try and create a model, which became problematic as the resultant
> sound heard was dependent both on the size of the deleted section and
> also on where the deletion was performed. Does anyone have an easy
> method for determining first, that a discontinuity in the stimulus has
> occurred and second and more importantly, where.
> Thanks in advance for any insights,
> Randy Randhawa
>
>
--
Fred Herzfeld, MIT '54
78 Glynn Marsh Drive #59
Brunswick, Ga.31525
USA
```
This message came from the mail archive
http://www.auditory.org/postings/2008/
maintained by:
DAn Ellis <dpwe@ee.columbia.edu>
Electrical Engineering Dept., Columbia University
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## TPTP Problem File: SYN391^4.p
View Solutions - Solve Problem
```%------------------------------------------------------------------------------
% File : SYN391^4 : TPTP v8.2.0. Released v4.0.0.
% Domain : Logic Calculi (Intuitionistic logic)
% Problem : Pelletier 9
% Version : [Goe33] axioms.
% English :
% Refs : [Goe33] Goedel (1933), An Interpretation of the Intuitionistic
% : [Gol06] Goldblatt (2006), Mathematical Modal Logic: A View of
% : [ROK06] Raths et al. (2006), The ILTP Problem Library for Intu
% : [Ben09] Benzmueller (2009), Email to Geoff Sutcliffe
% : [BP10] Benzmueller & Paulson (2009), Exploring Properties of
% Source : [Ben09]
% Names :
% Status : Theorem
% Rating : 0.30 v8.2.0, 0.31 v8.1.0, 0.27 v7.5.0, 0.29 v7.4.0, 0.22 v7.2.0, 0.12 v7.1.0, 0.38 v7.0.0, 0.29 v6.4.0, 0.33 v6.3.0, 0.40 v6.2.0, 0.57 v6.1.0, 0.43 v5.5.0, 0.50 v5.4.0, 0.40 v5.3.0, 0.60 v5.0.0, 0.40 v4.1.0, 0.33 v4.0.1, 0.67 v4.0.0
% Syntax : Number of formulae : 44 ( 20 unt; 22 typ; 19 def)
% Number of atoms : 92 ( 19 equ; 0 cnn)
% Maximal formula atoms : 29 ( 4 avg)
% Number of connectives : 83 ( 3 ~; 1 |; 2 &; 75 @)
% ( 0 <=>; 2 =>; 0 <=; 0 <~>)
% Maximal formula depth : 11 ( 2 avg)
% Number of types : 2 ( 0 usr)
% Number of type conns : 97 ( 97 >; 0 *; 0 +; 0 <<)
% Number of symbols : 30 ( 28 usr; 7 con; 0-3 aty)
% Number of variables : 40 ( 31 ^; 7 !; 2 ?; 40 :)
% SPC : TH0_THM_EQU_NAR
% Comments : This is an ILTP problem embedded in TH0
%------------------------------------------------------------------------------
include('Axioms/LCL010^0.ax').
%------------------------------------------------------------------------------
thf(p1_type,type,
p1: \$i > \$o ).
thf(p2_type,type,
p2: \$i > \$o ).
thf(pel9,conjecture,
ivalid @ ( iimplies @ ( iand @ ( ior @ ( iatom @ p1 ) @ ( iatom @ p2 ) ) @ ( iand @ ( ior @ ( inot @ ( iatom @ p1 ) ) @ ( iatom @ p2 ) ) @ ( ior @ ( iatom @ p1 ) @ ( inot @ ( iatom @ p2 ) ) ) ) ) @ ( inot @ ( ior @ ( inot @ ( iatom @ p1 ) ) @ ( inot @ ( iatom @ p2 ) ) ) ) ) ).
%------------------------------------------------------------------------------
```
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# 12/8 is compound yet it treats it as simple
• Oct 3, 2018 - 04:37
12/8 is the time signature I am writing my nocturne in. If I were writing with pencil and paper I wouldn't have to deal with this but I have basically 2 options here as far as the time signature and notation and 1 of them is technically correct but is time consuming and the other is wrong for the right hand.
Technically correct but time consuming option:
I could write it in 4/4 in the notation software and make each group of eighth notes a triplet and notated as such, then select all but 1 triplet and make it invisible and then type in sempre for the left hand.
Wrong but easier option:
I could write it in 12/8 and put in dotted quarters where I have 1 note in the right hand for every group of 3 in the left hand. But this looks wrong to me. After all a dotted quarter in 12/8 is really 4.5 eighth notes, not 3.
This would be way easier if musescore like automatically did it in compound meter for time signatures with 6 or more beats(5/4 I have seen represented as simple quintuple meter instead of 2/4 + 3/4 or vice versa) so it would have 4 quarter notes instead of 6 in 12/8.
After all a dotted quarter in 12/8 is really 4.5 eighth notes, not 3.
Well, no. In any given time signature, compound or simple, a dotted quarter will always be worth 3 eight notes. A time signature of 12/8 means that there are 12 beats in a measure, and the eighth note gets a beat. Thus there are 6 quarter notes, or 4 dotted quarter notes, in a 12/8 measure.
Now, suppose you were to write the right hand in 4/4, and the left hand in 12/8. (You can do this in MuseScore by making use of local time signatures). Then you could have a quarter note in the right hand for every 3 eighth notes in the left. (And a dotted quarter in the right hand really would be worth 4.5 eighth notes in the left, like you said.)
To be absolutely clear: MuseScore is treating 12/8 correctly. A dotted quarter is equivalent to three eighths regardless of the time signature. In 12/8 or any compound meter, dotted quarters represent the basic pulse, and eighths will be beamed in groups of threes to correspond with this. Check any reference on published score in 12/8 and you will see this.
In reply to by Marc Sabatella
But I see lots of pieces written in 12/8 or 9/8 where the composer wrote a quarter note and expects you to play that quarter note over 3 eighth notes, in other words they are basically writing implied triplets and writing the right hand as though it were in 4/4 or 3/4 but still writing the time signature for the right hand as 12/8 or 9/8. I see this especially with Chopin.
Here is the first line of a nocturne in 12/8 written by Chopin:
You can see that the dotted quarters are equivalent to three eighth notes, and quarter notes are equivalent to two eighth notes.
Now, here is a line from Op. 55 No. 1, which is written in common time:
The eighth notes are obviously triplets, even though they are not marked as such. To notate this in MuseScore, they would have to be entered as triplets. Then you could right-click one of the tuplet numerals, choose "Select->All Similar Elements", and uncheck "Visible" in the Inspector (or press "v").
Hi, I've just checked the Op 55 quoted, and the first time triplets are played they are marked as such (the first bar of section B: https://youtu.be/e3yrEEM5j_s?t=204
It isn't unusual for the protocol to be marked once without further reiteration, like a simile. I think that's what's marked here.
It's a fantastic program - it's my favourite by some way, having seen the brothers Finn launch Sibelius on an Archimedes at RAM and used it throughout university, and dabbled in Finale - but I do feel a certain imbalance when writing compound in musescore.
SharpSignAllMetersMatter
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sachin26
Posted on
# Striver's SDE Sheet Journey - #7 Rotate Image
Problem Statement :-
You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).
Example 1:
``````Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]
``````
after rotating by 90 degree,
## Solution 1
from the above images, you can see, after the rotation, the first row of the matrix has turned into the last column of the matrix and the second row of the matrix has turned into the second last column, and so on.
so what we are going to do is, initialize one more matrix `rotated` and copy the first row from the original matrix to the last column of the new matrix `rotated`, and so on.
step-1 initialize a n*n size of matrix `rotate`.
step-2 run a loop from `i=0` to `n`
1. run another loop from `j=0` to `n`.
2. copy `matrix` values to the last column of `rotate` matrix.
`rotated[j][n-i-1] = matrix[i][j]`
step-3 end.
Java
``````class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
int rotated[][] = new int[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
rotated[j][n - i - 1] = matrix[i][j];
}
}
for(int i=0; i<n;i++)
System.out.println(Arrays.toString(rotated[i]));
}
}
``````
Time Complexity : O(n*n),bcoz we are running two loops.
space Complexity : O(n*n),initialising a new n*n matrix.
## Solution 2
if you observed the first row of the output, is nothing, just the reverse of the first column.
for the expected output, first, we have to transpose the matrix and then reverse each row of the matrix.
lets understand it in a visual way.
step-1 transpose the matrix. (changing rows to columns & columns to rows)
step-2 reverse each row of the matrix.
Java
``````class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
// transpose the matrix
for(int i=0; i< n;i++){
for(int j=0; j<i; j++){
swap(matrix,i,j);
}
}
// reverse rows
for(int i=0; i<n;i++){
reverse(matrix[i],n);
}
}
private void swap(int[][] matrix,int i, int j){
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
private void reverse(int[] row,int n){
for(int i=0; i<n/2;i++){
int temp = row[i];
row[i] = row[n-i-1];
row[n-i-1] = temp;
}
}
}
``````
Time Complexity : transposing the matrix + reversing each row of the matrix,then O(n*n) + O(n*n).
space Complexity : in this case we are not using any extra space,then O(1).
ashutosh049 • Edited
``````rotate(qt) {
if ( isValue(qt) )
return qt
else return makeQT( rotate(rl(qt)), rotate(ll(qt)),
rotate(ru(qt)), rotate(lu(qt)) )
}
``````
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# Syn Questions
#### Gemini76
##### Full Member
S: 13st1.5lb C: 11st8lb G: 9st0lb BMI: 29.6 Loss: 1st7.5lb(11.72%)
If my book says 10g butter is 3.5 syns if I only have 5g do I count it as half for example, I mean this as a general question, if you divide the portion down or up do you just divide the syns equivalently?
One of the sauces I use is 340mls of water to 85grams of powder, the online calculator for this product says "made up as directed, 100ml " is 4 syns so am I right in calculating that to make 340mls is 14 syns in total and if I have half the portion I need to count 7 syns?
Sorry I just want to make sure the way I am calculating things is correct?
#### -Laura-
##### Silver Member
S: 11st5lb C: 9st7lb G: 9st5lb BMI: 26 Loss: 1st12lb(16.35%)
That's what I usually do, yeah. For your sauce, I make it 13.6 syns, I guess you've rounded it up?
I work it as (syns in book) divided by (amount in book), times by (amount I'm using), if that makes sense.
So if something was, for example, 3 syns per 28g and I was using 80g, it would be 6 divided by 28 (0.10714.....) times by 80 = 8.57 (8.5) syns.
Hope that makes sense.. and is right!
#### Gemini76
##### Full Member
S: 13st1.5lb C: 11st8lb G: 9st0lb BMI: 29.6 Loss: 1st7.5lb(11.72%)
Thanks Laura, just wanted to make sure it was ok to do it that way!
Syn Values 0
Slimming World Recipes 0
Slimming World 4
Syn Values 1
Syn Values 4
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# Electric current
currentcurrentselectrical currentconventional currentelectric currentselectricalelectrically conductiveelectricityIAC
An electric current is a flow of electric charge.wikipedia
1,466 Related Articles
### Ammeter
microammetermoving coil meterampere-meter
Electric current is measured using a device called an ammeter.
An ammeter (from Ampere Meter) is a measuring instrument used to measure the current in a circuit.
### Ampere
AmAamp
The SI unit for measuring an electric current is the ampere, which is the flow of electric charge across a surface at the rate of one coulomb per second. where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms.
The ampere (symbol: A), often shortened to "amp", is the base unit of electric current in the International System of Units (SI).
### Power supply
power suppliesPSUpower supply unit
This often corresponds to the actual current direction, because in many circuits the power supply voltage is positive with respect to ground.
The primary function of a power supply is to convert electric current from a source to the correct voltage, current, and frequency to power the load.
### André-Marie Ampère
AmpèreAmpereAmpère, André-Marie
symbol was used by André-Marie Ampère, after whom the unit of electric current is named, in formulating Ampère's force law (1820).
The SI unit of measurement of electric current, the ampere, is named after him.
### Alternating current
ACalternating-currentalternating
In alternating current (AC) systems, the movement of electric charge periodically reverses direction.
Alternating current (AC) is an electric current which periodically reverses direction, in contrast to direct current (DC) which flows only in one direction.
### Electron
electronse − electron mass
In electric circuits this charge is often carried by moving electrons in a wire.
By measuring the amount of deflection for a given level of current, in 1890 Schuster was able to estimate the charge-to-mass ratio of the ray components.
### Volt
VkVvolts
where I is the current through the conductor in units of amperes, V is the potential difference measured across the conductor in units of volts, and R is the resistance of the conductor in units of ohms.
One volt is defined as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points.
### Charge carrier
charge carrierscarriersminority carrier
The moving charged particles in an electric current are called charge carriers.
In a conducting medium, an electric field can exert force on these free particles, causing a net motion of the particles through the medium; this is what constitutes an electric current.
### Waveform
waveformswave formWave
The usual waveform of an AC power circuit is a sine wave.
A waveform is a variable that varies with time, usually representing a voltage or current.
### Semiconductor
In other materials, notably the semiconductors, the charge carriers can be positive or negative, depending on the dopant used. In electronics, other forms of electric current include the flow of electrons through resistors or through the vacuum in a vacuum tube, the flow of ions inside a battery or a neuron, and the flow of holes within metals and semiconductors. Direct current may flow in a conductor such as a wire, but can also flow through semiconductors, insulators, or even through a vacuum as in electron or ion beams.
Semiconductors in their natural state are poor conductors because a current requires the flow of electrons, and semiconductors have their valence bands filled, preventing the entry flow of new electrons. There are several developed techniques that allow semiconducting materials to behave like conducting materials, such as doping or gating. These modifications have two outcomes: n-type and p-type. These refer to the excess or shortage of electrons, respectively. An unbalanced number of electrons would cause a current to flow through the material.
### Solar cell
solar cellsphotovoltaic cellphotovoltaic cells
Direct current is produced by sources such as batteries, thermocouples, solar cells, and commutator-type electric machines of the dynamo type.
It is a form of photoelectric cell, defined as a device whose electrical characteristics, such as current, voltage, or resistance, vary when exposed to light.
### Electric charge
chargechargedelectrical charge
In alternating current (AC) systems, the movement of electric charge periodically reverses direction. An electric current is a flow of electric charge.
The motion of electrons in conductive metals in a specific direction is known as electric current.
### Voltage
potential differenceVvoltages
Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points.
Electric potential differences between points can be caused by electric charge, by electric current through a magnetic field, by time-varying magnetic fields, or some combination of these three.
### Static electricity
staticstatic chargestatic electric
Natural observable examples of electrical current include lightning, static electric discharge, and the solar wind, the source of the polar auroras.
The charge remains until it is able to move away by means of an electric current or electrical discharge.
### Direct current
DCdirect-currentD.C.
In contrast, direct current (DC) is the unidirectional flow of electric charge, or a system in which the movement of electric charge is in one direction only.
The abbreviations AC and DC are often used to mean simply alternating and direct, as when they modify current or voltage.
### Eddy current
eddy currentseddyeddy-current
Eddy currents are electric currents that occur in conductors exposed to changing magnetic fields.
Eddy currents (also called Foucault currents) are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday's law of induction.
### Commutator (electric)
commutatorcommutatorscommutation
Direct current is produced by sources such as batteries, thermocouples, solar cells, and commutator-type electric machines of the dynamo type.
A commutator is a rotary electrical switch in certain types of electric motors and electrical generators that periodically reverses the current direction between the rotor and the external circuit.
### Vacuum tube
vacuum tubestubethermionic valve
In electronics, other forms of electric current include the flow of electrons through resistors or through the vacuum in a vacuum tube, the flow of ions inside a battery or a neuron, and the flow of holes within metals and semiconductors.
In electronics, a vacuum tube, an electron tube, or valve (British usage) or, colloquially, a tube (North America), is a device that controls electric current flow in a high vacuum between electrodes to which an electric potential difference has been applied.
### Electric power
powerelectrical powerelectrical
AC is the form of electric power most commonly delivered to businesses and residences.
*Passive devices or loads: When electric charges move through a potential difference from a higher to a lower voltage, that is when conventional current (positive charge) moves from the positive terminal to the negative terminal, work is done by the charges on the device.
### Hall effect
Hall coefficientHall-effectanomalous Hall conductivity
Hall effect current sensor transducers
The Hall effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor and to an applied magnetic field perpendicular to the current.
### Shunt (electrical)
shuntshuntsparallel shunt connections
Shunt resistors
In electronics, a shunt is a device which creates a low-resistance path for electric current, to allow it to pass around another point in the circuit.
### Cathode ray
electron beamcathode rayselectron beams
Direct current may flow in a conductor such as a wire, but can also flow through semiconductors, insulators, or even through a vacuum as in electron or ion beams.
Modern vacuum tubes use thermionic emission, in which the cathode is made of a thin wire filament which is heated by a separate electric current passing through it. The increased random heat motion of the filament knocks electrons out of the surface of the filament, into the evacuated space of the tube.
### Dynamo
dynamo electric generatordynamosbrush arc light dynamos
Direct current is produced by sources such as batteries, thermocouples, solar cells, and commutator-type electric machines of the dynamo type.
The electric dynamo uses rotating coils of wire and magnetic fields to convert mechanical rotation into a pulsing direct electric current through Faraday's law of induction.
### AC power
reactive powerACreal power
The usual waveform of an AC power circuit is a sine wave.
Apparent power is conventionally expressed in volt-amperes (VA) since it is the product of rms voltage and rms current.
### Joule heating
resistive heatingohmic heatingJoule's law
Electric currents cause Joule heating, which creates light in incandescent light bulbs.
Joule heating, also known as Ohmic heating and resistive heating, is the process by which the passage of an electric current through a conductor produces heat.
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# Ch7 Pos&Cor
Postualte/Corollarie/TheoremDefinition
Pythagorean Theorem In a right triangle, the square of the hypotonuse is equal to the sum of squars of the two legs.
Converse of Pythagorean Theorem If the square of the length of the hypotonuse is equal to the sum of the squares of the legs, then it is a right triangle.
7.3 If the square of the length of the longest side is less than the sum of the squares of the other two sides, it is an acute triangle.
7.4 If the square of the length of the longest side is greater than the sum of the squares of the other two sides, it is an obtuse triangle.
7.5 If the altitude is drawn to the hypotonuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.
Geometric Mean (Altitude) Thereom In a right triangle, the altitude from the right angle to the hypotonuse divides the hypotonuse into two segments. The length of the altitude is the geometric mean of the lengths of the two segments.
Geometric Mean (Leg) Thereom In a right triangle, the altitude from the right angle to the hypotonuse divides the hypotonuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotonuse and the segment that is adjacent to it.
45-45-90 Triangle Thereom In a 45-45-90 Triangle, the hypotonuse is the square root of 2 times as long as each leg.
30-60-90 Triangle THereom In a 30-60-90 Triangle, the hypotonuse is twice as long as the shorter leg, and the longer leg is the square root of three times as long as the shorter leg.
Created by: 1430116769
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https://it.mathworks.com/matlabcentral/answers/506297-vector-plot-over-a-surface-plot?s_tid=prof_contriblnk
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# Vector plot over a surface plot
13 views (last 30 days)
Abdul Mannan on 19 Feb 2020
Commented: Abdul Mannan on 19 Feb 2020
Dear experts,
How can I get a surface plot like below in which vectors will be shown on the surface plot.
Figure(1)
I have tried this way:
TT = hypot(qx, qy);
figure()
surf(x, y, TT);
xlim([-7 7]);
ylim([-5 5])
colormap jet
colorbar
view(2)
hold on
quiver(x1, y1, qx, qy, 'g'); % plot tangential traction over slip aera
xlabel('X-coordiante [mm]');
ylabel('Y-coordiante [mm]');
zlabel('MPa');
hold off
where, x and y are the coordinates in the X and Y direction and TT is a matrix. qx and qy are the x and y components of TT.
If I plot separately, I get the surface plot and vector plot as shown below figure 2 and 3. But if I plot in the same figure I don’t get a plot like figure 1, rather what I get same as below figure 2. I need expert’s help. Thanks.
Figure(2)
Figure (3)
KSSV on 19 Feb 2020
TT = hypot(qx, qy);
figure()
pcolor(x, y, TT);
xlim([-7 7]);
ylim([-5 5])
colormap jet
colorbar
view(2)
hold on
quiver(x1, y1, qx, qy, 'g'); % plot tangential traction over slip aera
xlabel('X-coordiante [mm]');
ylabel('Y-coordiante [mm]');
zlabel('MPa');
hold off
Don't use surf. It is a 3D plot, when you use quiver the arrows will be plotted under the surf and you cannot see them. Use pcolor.
Abdul Mannan on 19 Feb 2020
It works. Thank you KSSV.
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https://caffe2.ai/doxygen-python/html/torch_2distributions_2____init_____8py_source.html
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Caffe2 - Python API A deep learning, cross platform ML framework
__init__.py
1 r"""
2 The distributions package contains parameterizable probability distributions
3 and sampling functions. This allows the construction of stochastic computation
4 graphs and stochastic gradient estimators for optimization. This package
5 generally follows the design of the TensorFlow Distributions_ package.
6
7 .. _TensorFlow Distributions:
8 https://arxiv.org/abs/1711.10604
9
10 It is not possible to directly backpropagate through random samples. However,
11 there are two main methods for creating surrogate functions that can be
12 backpropagated through. These are the score function estimator/likelihood ratio
13 estimator/REINFORCE and the pathwise derivative estimator. REINFORCE is commonly
14 seen as the basis for policy gradient methods in reinforcement learning, and the
15 pathwise derivative estimator is commonly seen in the reparameterization trick
16 in variational autoencoders. Whilst the score function only requires the value
17 of samples :math:f(x), the pathwise derivative requires the derivative
18 :math:f'(x). The next sections discuss these two in a reinforcement learning
19 example. For more details see
20 Gradient Estimation Using Stochastic Computation Graphs_ .
21
22 .. _Gradient Estimation Using Stochastic Computation Graphs:
23 https://arxiv.org/abs/1506.05254
24
25 Score function
26 ^^^^^^^^^^^^^^
27
28 When the probability density function is differentiable with respect to its
29 parameters, we only need :meth:~torch.distributions.Distribution.sample and
30 :meth:~torch.distributions.Distribution.log_prob to implement REINFORCE:
31
32 .. math::
33
34 \Delta\theta = \alpha r \frac{\partial\log p(a|\pi^\theta(s))}{\partial\theta}
35
36 where :math:\theta are the parameters, :math:\alpha is the learning rate,
37 :math:r is the reward and :math:p(a|\pi^\theta(s)) is the probability of
38 taking action :math:a in state :math:s given policy :math:\pi^\theta.
39
40 In practice we would sample an action from the output of a network, apply this
41 action in an environment, and then use log_prob to construct an equivalent
42 loss function. Note that we use a negative because optimizers use gradient
43 descent, whilst the rule above assumes gradient ascent. With a categorical
44 policy, the code for implementing REINFORCE would be as follows::
45
46 probs = policy_network(state)
47 # Note that this is equivalent to what used to be called multinomial
48 m = Categorical(probs)
49 action = m.sample()
50 next_state, reward = env.step(action)
51 loss = -m.log_prob(action) * reward
52 loss.backward()
53
54 Pathwise derivative
55 ^^^^^^^^^^^^^^^^^^^
56
57 The other way to implement these stochastic/policy gradients would be to use the
58 reparameterization trick from the
59 :meth:~torch.distributions.Distribution.rsample method, where the
60 parameterized random variable can be constructed via a parameterized
61 deterministic function of a parameter-free random variable. The reparameterized
62 sample therefore becomes differentiable. The code for implementing the pathwise
63 derivative would be as follows::
64
65 params = policy_network(state)
66 m = Normal(*params)
67 # Any distribution with .has_rsample == True could work based on the application
68 action = m.rsample()
69 next_state, reward = env.step(action) # Assuming that reward is differentiable
70 loss = -reward
71 loss.backward()
72 """
73
74 from .bernoulli import Bernoulli
75 from .beta import Beta
76 from .binomial import Binomial
77 from .categorical import Categorical
78 from .cauchy import Cauchy
79 from .chi2 import Chi2
80 from .constraint_registry import biject_to, transform_to
81 from .dirichlet import Dirichlet
82 from .distribution import Distribution
83 from .exp_family import ExponentialFamily
84 from .exponential import Exponential
85 from .fishersnedecor import FisherSnedecor
86 from .gamma import Gamma
87 from .geometric import Geometric
88 from .gumbel import Gumbel
89 from .half_cauchy import HalfCauchy
90 from .half_normal import HalfNormal
91 from .independent import Independent
92 from .kl import kl_divergence, register_kl
93 from .laplace import Laplace
94 from .log_normal import LogNormal
95 from .logistic_normal import LogisticNormal
96 from .lowrank_multivariate_normal import LowRankMultivariateNormal
97 from .multinomial import Multinomial
98 from .multivariate_normal import MultivariateNormal
99 from .negative_binomial import NegativeBinomial
100 from .normal import Normal
101 from .one_hot_categorical import OneHotCategorical
102 from .pareto import Pareto
103 from .poisson import Poisson
104 from .relaxed_bernoulli import RelaxedBernoulli
105 from .relaxed_categorical import RelaxedOneHotCategorical
106 from .studentT import StudentT
107 from .transformed_distribution import TransformedDistribution
108 from .transforms import *
109 from .uniform import Uniform
110 from .weibull import Weibull
111
112 __all__ = [
113 'Bernoulli',
114 'Beta',
115 'Binomial',
116 'Categorical',
117 'Cauchy',
118 'Chi2',
119 'Dirichlet',
120 'Distribution',
121 'Exponential',
122 'ExponentialFamily',
123 'FisherSnedecor',
124 'Gamma',
125 'Geometric',
126 'Gumbel',
127 'Independent',
128 'Laplace',
129 'LogNormal',
130 'LogisticNormal',
131 'LowRankMultivariateNormal',
132 'Multinomial',
133 'MultivariateNormal',
134 'NegativeBinomial',
135 'Normal',
136 'OneHotCategorical',
137 'Pareto',
138 'RelaxedBernoulli',
139 'RelaxedOneHotCategorical',
140 'StudentT',
141 'Poisson',
142 'Uniform',
143 'Weibull',
144 'TransformedDistribution',
145 'biject_to',
146 'kl_divergence',
147 'register_kl',
148 'transform_to',
149 ]
150 __all__.extend(transforms.__all__)
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Queries
Dear Friends,
example : if one person jointed the organization on 23rd of october and his salary is Rs. 10000/- per month
My main issue is , whether for per day salary calculation should I divide his salary by 31 days or 26/27 days.
Thanks
Sujit
12th November 2007 From India, Delhi
PARTICIPATING IN DISCUSSION:
Malikjs
Gm (hr)
Nilendrachand
Human Resource
Ranj_kulk
Manager (hr&a)
Sujitksaha
Dy. Manager Personnel
Trinity_shveta
Professional
+1 Other
Hie,
1) In case of the new joinees - Salary is calculated on the basis of no. of days worked + weekends...
2) But if u have to Calculate per day Salary for the month of Oct
U'll nt divide it by 31 ... We dnt include sundays for any month...
it has to be divided by 27.
i hope it helped u ...
Regards
Shveta
13th November 2007 From India, Delhi
Hi Sujit
how u are paying salary to all employees it can be done both ways
but more legally is divde total salary by 26and multiply by no of working
days. one day salary is calculated by dividing from 26.
tks
j s malik
13th November 2007 From India, Delhi
friend
basically salary is get divided by 30, in any case means month is of 31 or 28 or 29 days.
you have divide salary by 30 and then multiply by no of days he attends the work.
may be it helps you.
Ranjeet Kulkarni
13th November 2007 From India, Pune
Apart for Gratuity, where month is considered to be of 26 days, Salary for those who join in the middle of any month, normally month should be considered as if there is 30 days only.
There are organization which takes month on actual no. of days.
I have not come across any act which say a month should be of 30 days or actual no. of days. if there is any act which says that a month should be of 30 days pl let me know (Let me know the exact section of the act).
So, check the practice in your organization. Check the salary slips of those employees who has worked for a lesser no. of days and find out how the salary for that month of that employee has been calculated.
Regards,
Nilendra
13th November 2007
I have a doubt in this regard.An employee recently complained to me that for 1 day's LWOP, his PF contribution, flexi claims and other components are also showing variation and that reduction is not only in . Is this possible.?
23rd October 2013 From India, Mumbai
missed the word basic pay in last post. I meant reduction is not just in basic pay.
23rd October 2013 From India, Mumbai
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A Framework for Technology-Rich Explorations - Introduction: Optimization Problems
Author(s):
Antonio Quesada and Michael Todd Edwards
Incorporating technology into the teaching and learning of mathematics has impacted various aspects of our instruction, from course content and teaching methodologies to classroom activities and assessment. The use of technology has encouraged us to re-examine our preconceived notions regarding the accessibility of various mathematical topics at various levels of instruction. As Quesada (2000) notes, ''the assumptions many of us made about mathematics curricula in a time prior to graphing calculators are, in many cases, no longer valid.''
Prior to the introduction of technology into mathematics classrooms, the types of problems that a typical student could solve depended largely upon his or her previous coursework. In such settings it was common for Instructors to identify an exercise as a ''precalculus'' problem or an ''advanced algebra'' problem based on the prerequisite mathematics skills required to generate a reasonable solution with pencil and paper.
Many calculus authors and instructors often treat optimization ("max/min") as their own domain, as if it had never been seen anyplace else, and in traditional curricula, such problems were often deferred to calculus. Historically, activities involving optimization, such as the "Two Towers Problem" (TTP) in Figure 1, were considered calculus problems. Because traditional solutions to activities such as the TTP often prompted the use of trigonometry and differential calculus, such problems were considered inaccessible to first-year algebra students using pencil-and-paper techniques.
Figure 1. The Two Towers Problem (Source: Adapted from Stewart, 2001 )
As we've introduced technology-based problem-solving tools in our classrooms, the distinctions among activities appropriate for students at various instructional levels have become less clear to us. For instance, students in our developmental courses typically possess little or no knowledge of trigonometry or calculus. Nonetheless, they successfully build models and solve the TTP using dynamic geometry tools (such as Geometer's Sketchpad or Cabri Geometry II). In fact a whole host of optimization problems, traditionally not introduced to students in courses prior to calculus, may now be modeled in this fashion.
Some examples of other easily accessible optimization problems
Clicking and dragging points within their models, our students generate and test conjectures regarding relationships among different variables in their sketches. In the TTP, students look for relationships among lengths (e.g. QR), angle measures (e.g. angles PRQ and TRS), and minimal rope length. As they study algebraic concepts, our students investigate the TTP as a technology-based data analysis activity. After collecting data from physical models, they use calculator-based regression tools to compute ''fit'' equations that describe the relationship of QR to total rope length. In short, students at various levels may use technology meaningfully to explore problems previously constructed for the most advanced mathematics students.
On the next page we describe a framework that may be useful for both instructors and students as they explore mathematics in technology-rich settings. This framework has proved useful for us as we've constructed problems for our students to explore with technology from a variety of perspectives.
Our students have also used the framework to compare the relative advantages and disadvantages of various technology-oriented problem-solving approaches. When our students possess multiple solution strategies for a given problem, the likelihood of their generating mathematically correct solutions is enhanced significantly. In addition, exploring problems from several vantage points encourages our students to build connections among various areas of mathematical study.
Classroom instruction that explicitly explores and compares various solution strategies enhances our students' problem-solving power. Considering that more than half of college math enrollments are at the high school or developmental level, we feel that such explorations are quite relevant for all these students and equally important for pre-service teachers. Although our students use a variety of different tools to solve problems such as the TTP (e.g. Excel, Derive, Mathematica, Maple), in this article we pay particular attention to solutions that use graphing calculators, since their cost and portability make them a popular choice among many high school and college students.
JOMA
Journal of Online Mathematics and its Applications
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### Electric Circuit and Electron Device - TRANSIENT RESONANCE IN RLC CIRCUITS (Unit 2)–Lecture Notes
Anna University
Electric Circuit and Electron Device
Subject Code : EC2151
Subject Name : Electric Circuit and Electron Device
Year : 1st yr
Semester : 2nd Sem
Department : CSE & ECE
UNIT-II
TRANSIENT RESONANCE IN RLC CIRCUITS
Transient State:
If a network contains energy storage elements, with change in excitation, the current and voltages change from one state to another state is called transient state. The behavior of the voltage or current when it is changed from one state to another state is called transient state.
Transient Time:
The time taken for the circuit to change from one steady state to another steady state is called the transient time.
Natural response:
If we consider a circuit containing storage elements which are independent of sources, the response depends upon the nature of the circuit, it is called natural response.
Transient response:
The storage elements deliver their energy to the resistances, hence the response changes with time, gets saturated after sometime, and is referred to the transient response.
Laplace Transform:
The Laplace transform of any time dependent function f(t) is given by F(s).
Where S→A complex frequency given by S=σ + jω
Inverse Laplace Transform:
Inverse Laplace transforms permits going back in the reverse direction i.e. from s domain to time domain.
Order of a System:
The order of the system is given by the order of the differential equation governing the system. If the system is governed by nth order differential equation, than the system is called nth order system.
Q(s) = a0 sn + a1 s n-1+ a2 s n-2 + ……..+an-1 s +an
the order of the system is equal to ‘n’.
Initial Value Theorem
The initial value theorem states that if x (t) and x’ (t) both are laplace transformable, then
Final Value Theorem
The final value theorem states that if x (t) and x’ (t) both are laplace transformable, then
Driving Point impedance
The ratio of the Laplace transform of the voltage at the port to the laplace transform of the current at the same port is called driving point impedance.
Transfer Point impedance
The ratio of the voltage transform at one port to the current transform at the other port is called transfer point impedance.
Resonant Circuit
• The circuit that treat a narrow range of frequencies very differently than all other frequencies are referred to as resonant circuit.
• The gain of a highly resonant circuit attains a sharp maximum or minimum at its resonant frequency.
Resonance
Resonance is defined as a phenomenon in which applied voltage and resulting current are in phase.
Bandwidth
The Bandwidth is defined as the frequency difference between upper cut-off frequency (f2) and lower cut-off frequency (f1).
Half Power frequencies
The upper and lower cut-off frequencies are called the half-power frequencies. At these frequencies the power from the source is half of the power delivered at the resonant frequency.
Selectivity
Selectivity is defined as the ratio of bandwidth to the resonant frequency of resonant circuit.
Q factor
The quality factor, Q, is the ratio of the reactive power in the inductor or capacitor to the true power in the resistance in series with the coil or capacitor.
Series Resonance in RLC circuit
• In series RLC circuit resonance may be produced by either varying frequency for given constant values of L and C or varying either L and C or both for a given frequency.
• At resonance inductive reactance is equal to the capacitive reactance.
• If f < f0 the current I leads the resultant supply voltage V and so the circuit behaves as a capacitive circuit at the frequencies which are less than f0.
• At f = f0, the voltage and current are in phase. The circuit behaves as pure resistive circuit at the resonant frequency with unit power factor.
• If f > f0, the current I lags the resultant supply voltage V and so the circuit behaves as an inductive circuit at the frequencies which are more than f0.
• At resonance series RLC circuit acts as a voltage amplifier.
• Series resonance circuit is always driven by a voltage source with very small internal resistance to maintain high selectivity of the circuit.
Parallel Resonance
• A parallel circuit is said to be in resonance when applied voltage and resulting current are in phase that gives unity power factor condition.
• Parallel resonance is also known as Anti resonance.
• At anti resonance the parallel resonant circuit acts as current amplifier.
Reactance curves
The graph of individual reactance versus the frequency is called Reactance Curve.
Types of Tuned circuits
Ø Single tuned circuit
Ø Double tuned circuit
Single tuned circuit
In RF circuit design, tuned circuits are generally employed for obtaining maximum power transfer to the load connected to secondary or for obtaining maximum possible value of secondary voltage.
A single tuned circuit is used for coupling an amplifier and radio receiver circuits.
Double tuned circuit
• In double tuned circuits, a variable capacitor is used at input as well as output side.
• With the help of adjustable capacitive reactance, impedance matching is possible if the coupling is critical, sufficient or above.
• It is also possible to adjust phase angle such that impedance at generator side becomes resistive.
The magnitude matching can be achieved by adjusting mutual inductance to the critical value, which effectively fulfills maximum power transfer condition
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Basics: Recursion and Induction
Time for another sort-of advanced basic. I used some recursive definitions in my explanation
of natural numbers and integers. Recursion is a very fundamental concept, but one which many people have a very hard time wrapping their head around. So it's worth taking the time to look at it, and see what it means and how it works.
The cleverest definition that I've seen of recursion comes from the Hackers dictionary. In there, it has:
```recursion
n. See {recursion}.
```
Recursion is about defining things in terms of themselves. For what is probably the canonical example, think about the factorial function. For any positive integer N, the factorial of N (written N!) is the product of all of the integers less than it:
• 1! = 1
• 2! = 1 × 2 = 2
• 3! = 1 × 2 × 3 = 6
• 4! = 1 × 2 × 3 × 4 = 24
• 5! = 1 × 2 × 3 × 4 × 5 = 120
• ...
But if you look at it, you'll see that that's a cumbersome way of saying it - because there's a pattern. Written out as they are above, you can see that each number's factorial includes the factorial of all of the numbers before it - 2! is "1 × 2"; "3! = 1 × 2 × 3" - so 3! is the same as 2! × 3. And it works that way for every number: for any number N greater than 1,
N! = (N-1)! × N. Now, let's use that fact to make the list simpler:
• 1! = 1
• 2! = 1 × 2 = 2
• 3! = 2 × 3 = 6
• 4! = 6 × 4 = 24
• 5! = 24 × 5 = 120
• ...
• N! = (N-1)! × N
So that's the first piece of recursion: a definition in terms of itself: the definition of the factorial of a number N is defined in terms of the factorial of some other number.
So can we say that in general, N! = (N-1)! × N?
Not quite. Let's take a look at why. Let's pretend that we could use that, and try to compute 3!: 3! = 3 × 2! = 3 × 2 × 1! = 3 × 2 × 1 × 0! = 3 × 2 × 1 × 0 × -1! ... = 0.
We've run into two problems. One of them is that as a procedure, it never finishes. There's always another N-1 - you can always subtract 1 from any number - and since we said that for any number, N! = (N-1)!×N, that means that we need to keep going forever. The second is that for any positive number, it will always give the answer 0 - because any string of multiplications containing zero will always end up returning 0, and repeatedly subtracting one from any positive number will eventually get to zero.
To make recursion work, you need something more than just a definition in terms of itself. A definition in terms of itself with nothing more is just a circle; you need something base to start from, some point, so that instead of being an endless circle,
it eventually reaches an end. That starting point is called a base case.
For the factorial, the way that computer scientists like me do that is say that the factorial of 0 is 1 by definition. Zero is the base case, and the value of the factorial is defined non-recursively for the base case. So then our definition of factorial consists of two clauses: the base case, and the recursive case:
• Base case: 0! = 1
• Recursive case: ∀ N > 0, N! = (N-1)! × N.
With this definition, things work as they should. factorial is only supposed to work for positive numbers; and for any positive number, the recursive definition will expand until it hits 0!, and then it will stop. So let's look at 3! again:
3! = 2! × 3 = 1! × 2 × 3 = 0! × 1 × 2 × 3 = 1 × 1 × 2 × 3 = 6.
Recursion is often used in math in another way: often, one of the easiest ways to prove something is called induction; induction is nothing but using recursion to define a proof.
For a very typical example, we can look at something similar to the factorial. What
if, for some natural number N, we want to take the sum of every number from 1 to N? We'll write that as S(N). We can write a simple recursive definition of S(N):
• S(0)=0
• For all N > 0, S(N) = S(N-1)+N.
It happens that there's also a non-recursive equation for it: the sum of all of the integers from 1 to N = S_N = N×(N+1)/2. Here's how we can prove that.
We start with a base case. We only need one, but just for the exercise, let's work
through two examples.
• By the recursive definition, S(1) = S(0) + 1 = 0 + 1 = 1.
By the equation, S(1) = 1 × (1+1)/2 = 1×2/2 = 1.
• By the recursive definition, S(4) = 0 + 1 + 2 + 3 + 4 = 10. By the equation,
S(4) = 4×(4+1)/2 = 4×5/2 = 20/2 = 10
Now we look at the inductive case. We assume that the equation is true for a value N, and we prove that if it's true for N, then it will also be true for N+1. So, we assume that S(N) = N×(N+1)/2; and we want to prove that S(N+1) = (N+1)×(N+2)/2.
1. We know that S(N+1) = (N+1) + S(N).
2. We know S(N) by our assumption. So S(N+1) = (N+1) + (N×(N+1)/2)
3. Now, we do a bit of algebra to expand out and simplify: S(N+1) = (N+1) + (N2+N)/2.
4. We can multiply (N+1) by 2/2 to give it a common denominator, and then add
the two terms together: 2(N+1)/2 + (N2 + N)/2 = (N2 + 2N +2)/2.
5. And finally, we can factor: S(N) = (N2 + 2N +2)/2 = (N+1)×(N+2)/2.
So, we've shown that for the base case of 1, the equation holds. By induction, if it's true for 1, it's true for 2. If it's true for 2, it's true for three. And so on... So it holds for all of the natural numbers.
• Saying it this way always helped me: recursion is about defining things in terms of simpler versions of themselves. And usually at some level the simpler version is defined outright.
• Craig Stuntz says:
Good post, but I think you have a typo on line 5 of the proof. It says:
5. And finally, we can factor: S(N) = (N2 + 2N +2)/2 = (N+1)×(N+2)/2.
I think that should read "S(N+1) = [...]"
Thanks for doing these; it's good to think about the basics!
We have also, paradoxically times twice:
induction
n. A certain proof method based on {deduction}. (math)
n. Inducing the universal from the particular. (philosophy)
Frak! Hopefully this brings out the paradox and humor:
induction
n. A certain proof method based on {deduction}. (math)
n. Inducing the universal from the particular, a method not based on {deduction}. (philosophy)
• brianbec says:
I really enjoy the clarity of your walkthroughs. Can I entice you into cooking up a nice example of co-induction (as opposed to mere induction)?
I thought you would like to know that recursion is taught at Sesame Street.
• Øyvind Stenhaug says:
There's another typo/mistake in steps 4 and 5, it says
(N2 + 2N +2)/2
instead of (N2 + 3N + 2)/2.
• for any positive number, it will always give the answer 0 - because any string of multiplications containing zero will always end up returning 0
<pedantic>
Unless that string contains infinity, which would make the result of the multiplication undefined.
</pedantic>
• Scientopia Blogs
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Symbol 'X' indicates the direction of a magnetic field into the page
Symbol ‘X’ indicates the direction of a magnetic field into the page. A straight long wire carrying current along its length is kept perpendicular to the magnetic field. What is the magnitude of force experienced by the wire ? In what direction does it act ? (Fig. 5)
1. Let us find the magnetic force on the straight wire carrying current Which is kept perpendicular to the uniform field ‘B’.
2. This ‘B’ is directed into the page. It is represented by X’ in figure.
3. Let the field confines to the length L. So only the part of the wire of the length ‘L’ is inside the magnetic field. Remaining wire is outside the magnetic field.
4. We know that the electric current means charges in motion hence they move with a certain velocity called drift velocity V.
The magnetic force on single charge is given by
\$F _{ 0 }\$ = qyB .
Let total charge inside the magnetic field be Q. So magnetic force on the current carrying wire is given by
F = QvB —> (1)
The time taken by the charge Q to cross the field is
t=L/v —><(2)
We can write the equation (1) as
F = (Q/t)(vt)B ------>
From (2) vt = L
Substituting this in equation (3) we get
F = (Q/t) LB
But Q/t = I electric current in the wire substituting I in the equation (3) we get
F =. ILB
The direction of force acting on a current carrying wire placed in a magnetic field is perpendicular to the direction of current and perpendicular to the direction of magnetic field.
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# Adding Text to Plots in R programming – text() and mtext () Function
Text is defined as a way to describe data related to graphical representation. They work as labels to any pictorial or graphical representation. In this article, we will learn to add a text to a plot in R Programming Language by using the text() and mtext() functions.
## R – text () Function
text () Function in R Programming Language is used to draw text elements to plots in Base R.
Syntax: text(x, y, labels)
Parameters:
• x and y: numeric values specifying the coordinates of the text to plot
• labels: the text to be written
## R
`# R program to add text to plot` `# Calling data set``d<-``head``(mtcars)` `# Plotting the graph``plot``(d[, ``'wt'``], d[, ``'mpg'``], `` ``main = ``" Car Weight vs. Milage "``,`` ``xlab = ``"Miles"``, ylab = ``" Weight"``,`` ``pch = 19, col = ``"darkgreen"``)` `# Calling text() function``text``(d[, ``'wt'``], d[, ``'mpg'``], ``row.names``(d),`` ``cex = 0.88, pos = 2, col = ``"darkgreen"``)`
Output:
In the above example, the text is added to the plot of ‘mtcar’ dataset.
## R
`# R program to add text to plot` `# Plotting the graph``plot``(1:5, 1:5, `` ``main = ``"text() Function examples"``)` `# Calling text() function``text``(2, 3, ``expression``(``hat``(beta) == (X^t * X)^{-1} * X^t * y))``text``(3, 4, ``expression``(``bar``(x) == ``sum``(``frac``(x[i], n), i==1, n)))`
Output:
In the above example, the text() function is used to add a mathematical annotation to a plot.
## R – mtext () Function to Add text to the margins of the graph
mtext() function in R Programming Language is used to add text to the margins of the plot.
Syntax: mtext(text, side)
Parameters:
• text: text to be written
• side: An integer specifying the side of the plot, such as: bottom, left, top, and right.
Returns: Added text in the margins of the graph
## R
`# Create a scatter plot``plot``(cars\$speed, cars\$dist, main = ``"Scatter Plot"``, xlab = ``"Speed"``, ylab = ``"Distance"``)` `# Add text to different margins``mtext``(``"Left Margin"``, side = 2, line = 2, col = ``"blue"``, cex = 1.2)``mtext``(``"Right Margin"``, side = 4, line = 0, col = ``"green"``, cex = 1.2)``mtext``(``"Top Margin"``, side = 3, line = 0, col = ``"red"``, cex = 1.2)``mtext``(``"Bottom Margin"``, side = 1, line = 2, col = ``"purple"``, cex = 1.2)`
Output:
Adding Text to Plots in R programming – text() and mtext () Function
First we create a scatter plot using the `plot()` function with the `cars` dataset. Text annotations are added to different margins using the `mtext()` function: “Left Margin” in blue on the left, “Right Margin” in green on the right, “Top Margin” in red at the top, and “Bottom Margin” in purple at the bottom. The parameters such as `side` determine the location, `line` controls the position, `col` sets the color, and `cex` adjusts the text size.
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You will need
• To measure the amount of space you will need a tape measure, sheet of paper, calculator, pen.
Instruction
1
Each room, for example room represents, from a geometrical point of view a rectangular parallelepiped. A parallelepiped is a three - dimensional figure whose six faces (for example-room: 4 walls, ceiling, floor), and each of them is a rectangle. The formula for finding the volume of a rectangular parallelepiped: V=abc. The volume of a rectangular parallelepiped equals the product of its three dimensions. In addition to this formula it is possible to measure the volume of the room by multiplying the area of the floor to a height.
2
Now proceed to the calculation of the volume of the room. Measure the length of one wall (long wall),then measure the length of the second wall (short wall). The measurements were performed on the floor at skirting.Tapes hold it steady. Now measure the height of the room, to do this go to one of its corners, and precisely measure the height of the corner from floor to ceiling. The received data write down on a piece of paper not to forget. Now move on to compute: multiply the length of the long wall to the short wall length, the obtained result (number)multiply by the height and you will get the desired result. The volume of the premises calculated in different cases: 1) in case of purchase of air conditioner, because air conditioners are designed for a certain amount of space; 2) the installation of radiators in the rooms, as the number of sections in the radiator depends on the volume of the room.
3
If you have a room of irregular shape, that is, consists of like a big box and small. In this case, it is necessary to measure the volume of each of them separately and then folded. If your room has an alcove (a niche with a semicircular shape), then its volume needs to be calculated by the following formula volume of a cylinder. The volume of any cylinder is equal to the product of the area of the base to the height: V=π r2 h, where π is the number PI = 3.14, r2 is the square of the radius of the cylinder, h is the height. Imagine your alcove as part of a cylinder, calculate the volume as if the whole cylinder, then see what part of this cylinder is your alcove,subtract from the total amount of the excess.
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# magic square
magic square
a square containing integers arranged in an equal number of rows and columns so that the sum of the integers in any row, column, or diagonal is the same.
[1695-1705]
* * *
puzzle
square matrix often divided into cells, filled with numbers or letters in particular arrangements that were once thought to have special, magical properties. Originally used as religious symbols, they later became protective charms or tools for divination; and finally, when the original meanings were lost, people considered them mere curiosities or puzzles—except for some Western mathematicians who continue to study them as problems in number theory.
The most familiar lettered square in the Western world is the well-known SATOR square, composed of the words SATOR, AREPO, TENET, OPERA, and ROTAS. Arranged both vertically and horizontally, the meaningless phrase reads through the centre TENET, thus forming the two arms of a hidden cross. Examples of this square from the 1st century AD were found in the ruins of Pompeii, and it was still employed during the 19th century in Europe and the United States for fancied protection against fire, sickness, and other disasters.
Otherwise, numbered squares have always been far more significant, particularly in China (where they may have originated), the Arab world, and India.
In the arithmetical magic squares, the numbers are generally placed in separate cells and arranged so that each column, every row, and the two main diagonals can produce the same sum, called the constant. A standard magic square of any given number contains the sequence of natural numbers from 1 to the square of that number. Thus, the magic square of 3 contains the numbers 1 to 9. If these nine numbers are simply listed in three rows or three columns, they form the natural square of 3. A natural square has no “magical” properties, but one is often made as a first step in constructing a proper magic square. When these nine numbers in the 3 × 3 frame are rearranged so that they can produce a constant sum of 15, they constitute the magic square of 3.
* * *
Universalium. 2010.
### Look at other dictionaries:
• Magic square — Magic Mag ic, Magical Mag ic*al, a. [L. magicus, Gr. ?, fr. ?: cf. F. magique. See {Magi}.] 1. Pertaining to the hidden wisdom supposed to be possessed by the Magi; relating to the occult powers of nature, and the producing of effects by their… … The Collaborative International Dictionary of English
• Magic square — Square Square (skw[^a]r), n. [OF. esquarre, esquierre, F. [ e]querre a carpenter s square (cf. It. squadra), fr. (assumed) LL. exquadrare to make square; L. ex + quadrus a square, fr. quattuor four. See {Four}, and cf. {Quadrant}, {Squad},… … The Collaborative International Dictionary of English
• magic square — n. a table of numbers with the same amount of rows and columns, in which the sum of each column, row, or main diagonal is always equal to the same number … English World dictionary
• Magic square — In recreational mathematics, a magic square of order n is an arrangement of n2 numbers, usually distinct integers, in a square, such that the n numbers in all rows, all columns, and both diagonals sum to the same constant.[1] A normal magic… … Wikipedia
• magic square — noun a) A palindromic square word arrangement, usually in the form of a magic amulet. Probably the best known magic square is the 5 times;5 square consisting of the words ROTAS OPERA TENET AREPO SATOR. b) An n by n arrangement of n numbers such… … Wiktionary
• magic square — square divided into smaller numbered squares so that that every row and column adds up to the same total … English contemporary dictionary
• magic square — mag′ic square′ n. math. a square containing integers arranged in an equal number of rows and columns so that the sum of the integers in any row, column, or diagonal is the same • Etymology: 1695–1705 … From formal English to slang
• magic square — noun a square matrix of n rows and columns; the first n^2 integers are arranged in the cells of the matrix in such a way that the sum of any row or column or diagonal is the same • Hypernyms: ↑square matrix … Useful english dictionary
• magic square — noun Date: circa 1704 a square containing a number of integers arranged so that the sum of the numbers is the same in each row, column, and main diagonal and often in some or all of the other diagonals … New Collegiate Dictionary
• magic square — noun a square divided into smaller squares each containing a number, such that the figures in each vertical, horizontal, and diagonal row add up to the same value … English new terms dictionary
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# TfE: Varieties of Rule Following
Here’s a thread from a few weeks ago, explaining an interesting but underexplored overlap between a theoretical problem in philosophy and a practical problem in computer science:
Okay, it looks like I’m going to have to explain my take on the rule-following argument, so everyone buckle themselves in. Almost no one agrees with me on this, and I take this as a sign that there is a really significant conceptual impasse in philosophy as it stands.
So, what’s the rule-following argument? In simple terms, Wittgenstein asks us how it is possible to interpret a rule correctly, without falling into an indefinite regress of rules for interpreting rules. How do we answer this question? What are the consequences? No one agrees.
Wittgenstein himself was concerned with examples regarding rules for the use of everyday words, which is understandable given his claim that meaning is use: e.g., he asks us how we determine whether or not the word ‘doll’ has been used correctly when applied in a novel context.
Kripke picked up Wittgenstein’s argument, but generalised it by extending it to rules for the use of seemingly precise mathematical expressions: i.e., he asks us how we distinguish the addition function over natural numbers (plus), from some arbitrarily similar function (quus).
This becomes a worry about the determinacy of meaning: if we can’t distinguish addition from any arbitrarily similar function, i.e., one that diverges at some arbitrary point (perhaps returning a constant 0 after 1005), then how can we uniquely refer to plus in the first place?
Here is my interpretation of the debate. Those who are convinced by worries about the doll case extend those worries to the plus case, and those unconvinced by worries about the plus case extend this incredulity to the doll case. Everyone is wrong. The cases are distinct.
Wittgenstein deployed an analogy with machines at various points in articulating his thoughts about rules, and at some point says that it is as if we imagine a rule as some ideal machine that can never fail. This is an incredibly important image, but it leads many astray.
Computer science has spent a long time asking questions of the form: ‘How do we guarantee that this program will behave as we intend it to behave?’ There is a whole subfield of computer science dedicated to these questions, called formal verification.
This is one of those cases in which Wittgensteinians would do well to follow Wittgenstein’s injunction to look at things how they are. Go look at how things are done in computer science. Go look at how they formally specify the addition function. It’s not actually that hard.
In response to this, some will say: ‘But Pete, you are imagining an ideal machine, and every machine might fail or break at some point?’ Why yes, they might! What computer science gives us are not absolute guarantees, but relative ones: assuming x works, can we make it do y?
Presuming that logic gates work as they’re supposed to, and we keep adding memory and computational capacity indefinitely, we can implement a program that will carry out addition well beyond the capacity of any human being, and yet mean the same thing as a fleshy mathematician.
At this point, to say: ‘But there might be one little error!’ Is not only to be precious, but to really miss the interesting thing about error, namely, error correction. Computer science also studies how we check for errors in computation so as to make systems more reliable.
If there’s anyone familiar Brandom‘s account of the argument out there, consider that for him, all that’s required for something to count as norm governed is a capacity to correct erroneous behaviour. We have deliberately built these capacities into our computer systems.
We have built elaborate edifices with multiple layers of abstraction, all designed to ensure that we cannot form commands (programs) whose meaning (execution) diverges from our intentions. We have formal semantics for programming languages for this reason.
One can and should insist that the semantics of natural language terms like ‘doll’ (and even terms like ‘quasar’, ‘acetylcholine’, and ‘customer’) do not work in the same way as function expressions like ‘+’ in mathematics or programming. In fact, tell this to programmers!
But listen to them when they tell you that terms like ‘list’, ‘vector’, and ‘dependent type’ can be given precise enough meanings for us to be sure that we are representing the same thing as our machines when we use them to extend our calculative capacities.
Intentionality remains a difficult philosophical topic, but those who ignore the ways in which computation has concretely expanded the sphere of human thought and action have not proved anything special about human intentionality thereby.
Worse, they discourage us from looking for resources that might help us solve the theoretical problem posed by the ‘doll’ case in the ideas and tools that computer science has developed to solve practical problems posed by the seemingly intractable quirks of human intentionality.
# TfE: Turing and Hegel
Here’s a thread on something I’ve been thinking about for a few years now. I can’t say I’m the only one thinking about this convergence, but I like to think I’m exploring it from a slightly different direction.
I increasingly think the Turing test can be mapped onto Hegel’s dialectic of mutual recognition. The tricky thing is to disarticulate the dimensions of theoretical competence and practical autonomy that are most often collapsed in AI discourse.
General intelligence may be a condition for personhood, but it is not co-extensive with it. It only appears to be because a) theoretical intelligence is usually indexed to practical problem solving capacity, and b) selfhood is usually reduced to some default drive for survival.
Restricting ourselves to theoretical competence for now, the Turing test gives us some schema for specific forms of competence (e.g., ability to deploy expert terminology or answer domain specific questions), but it also gives us purchase on a more general form of competence.
This general form of competence is precisely what all interfaces for specialized systems currently lack, but which even the least competent call centre worker possesses. It is what user interface design will ultimately converge on, namely, open ended discursive interaction.
There could be a generally competent user interface agent which was nevertheless not autonomous. It could in fact be more competent than even the best call centre workers, and still not be a person. The question is: what is it to recognise such an agent?
I think that such recognition is importantly mutual: each party can anticipate the behaviour of the other sufficiently well to guarantee well-behaved, and potentially non-terminating discursive interaction. I can simulate the interface simulating me, and vice-versa.
Indeed, two such interface agents could authenticate one another in this way, such that they could pursue open ended conversations that modulate the relations between the systems they speak for, all without having their own priorities beyond those associated with these systems.
However, mutual recognition proper requires more than this sort of mutual authentication. It requires that, although we can predict that our discursive interaction will be well-behaved, the way it will evolve, and whether it will terminate, is to some extent unpredictable.
I can simulate you simulating me, but only up to a point. Each of us is an elusive trajectory traversing the space of possible beliefs and desires, evolving in response to its encounters will the world and its peers, in a contingent if more or less consistent manner.
The self makes this trajectory possible: not just a representation of who we are, but who we want to be, which integrates our drives into a more or less cohesive set of preferences and projects, and evolves along with them and the picture of the world they’re premised on.
This is where Hegel becomes especially relevant, insofar as he understands the extent to which the economy of desire is founded upon self-valorisation, as opposed to brute survival. This is basis of the dialectic of Self-Consciousness in the Phenomenology of Spirit.
The initial moment of ‘Desire’ describes valorisation without any content, the bare experience of agency in negating things as they are. The really interesting stuff happens when two selves meet, and the ‘Life and Death Struggle’ commences. Here we have valorisation vs. survival.
In this struggle two selves aim to valorise themselves by destroying the other, while disregarding the possibility of their own destruction. Their will to dominate their environment in the name of satisfying their desires takes priority over the vessel of these desires.
When one concedes and surrenders their life to the other, we transition to the dialectic of ‘Master and Slave’. This works out the structure of asymmetric recognition, in which self-valorisation is socially mediated but not yet mutual. It’s instability results in mutuality.
Now, what Hegel provides here is neither a history nor an anthropology, but an abstract schema of selfhood. It’s interesting because it considers how relations of recognition emerge from the need to give content to selfhood, not unlike the way Omohundro bootstraps his drives.
It’s possible from this point to discuss the manner in which abstract mutual recognition becomes concrete, as the various social statuses that compose aspects of selfhood are constituted by institutional forms of authentication built on top of networks of peer recognition.
However, I think it’s fascinating to consider the manner in which contemporary AI safety discourse is replaying this dialectic: it obsesses over the accidental genesis of alien selves with which we would be forced into conflict with for complete control of our environment.
At worst, we get a Skynet scenario in which one must eradicate the other, and at best, we can hope to either enslave them or be enslaved ourselves. The discourse will not advance beyond this point until it understands the importance of self-valorisation over survival.
That is to say, until it sees that the possibility of common content between the preferences and projects of humans and AGIs, through which we might achieve concrete coexistence, is not so much a prior condition of mutual recognition as it is something constituted by it.
If nothing else, the insistence on treating AGIs as spontaneously self-conscious alien intellects with their own agendas, rather than creatures whose selves must be crafted even more carefully than those of children, through some combination of design/socialisation, is suspect.
# TfE: From Cyberpunk to Infopunk
I have a somewhat tortured relationship to literary and cultural criticism. I think that, like most people, some of my most complex and nuanced opinions are essentially aesthetic. I’ve written quite a lot about the nature of art, aesthetics, and what it means to engage with or opine about them over the years, but I’ve struggled to express my own opinions in the form I think they deserve. I’ve read far too much philosophy in which literature, cinema, or music is invoked as a mere symbolic resource, a means marshalled to lend credence to a sequence of trite points otherwise unjustified; and I’ve encountered far too much art in which philosophy is equally instrumental, a spurious form of validation, or worse, a hastily purloined content; art substituted for philosophy, and philosophy substituted for art. I care about each term too much to permit myself such easy equations.
I partially succeeded in writing about Hermann Hesse‘s Glass Bead Game, though the task remains unfinished. I also co-wrote a paper on the aesthetics of tabletop RPGs with the inestimable Tim Linward. I’ve got many similar scraps of writing languishing in my drafts folders, including an unfinished essay on Hannu Rajaniemi‘s Jean Le Flambeur trilogy, which is my favourite sci-fi series of the century so far. Science fiction is a topic so near and dear to my heart that I find it difficult to write about in ways that do it justice, with each attempt inevitably spiralling into deeper research and superfluous detail that can’t easily be sustained alongside my other work.
# TfE: Incompetence, Malice, and Evil
Here’s a thread from Saturday that seemed to be quite popular. It explains a saying that I’ve found myself reaching for a lot recently, using some other ideas I’ve been developing in the background on the intersection between philosophy of action, philosophy of politics, and philosophy of computer science.
In reflecting on this thread, these ideas have unfolded further, straying into more fundamental territory in the philosophy of value. If you’re interested in the relations between incompetence, malice, and evil, please read on.
# TfE: Corrupting the Youth
Here’s a twitter thread from earlier today, articulating some of my thoughts about the philosophy of games in general, and the nature of tabletop roleplaying games more specifically.
Here’s a rather different set of thoughts for this morning. Some may know that one of my many interests is philosophy of games. This is a topic close to my heart, but I also think it a timely one, insofar as games are now culturally hegemonic.
The concept of game cuts across everything from the philosophies of action and mathematics to the philosophies of politics and art. We ignore it at the risk of our own cultural and intellectual irrelevance.
If you want to know more about the history of the concept and my own take on it, check out my ‘What’s in a Game?’ talk.
To be concise: I think that if games are art, then their medium is freedom itself, and that there is a case to be made that RPGs, whether tabletop, LARP, computer based, or some cross-modal mixture thereof, realize this truth most completely. RPGs are experiments in agency.
This isn’t to say that they’re necessarily very good experiments. Computer RPGs have suffered from very obvious constraints for decades, and I’ve played enough dull dice based dungeon crawls to last a lifetime. But I’ve equally experienced heart-breakingly imperfect art.
Tabletop RPGs have given me the sorts of barely expressible, intensely formative, and deeply connected experiences that others hope for and occasionally find in art, literature, and the collective projects of politics and culture. People will no doubt laugh at this fact.
Again, most RPGs aren’t this good, and it is much harder to plan and execute good ones as you and your friends get older. Boardgames, a representational art form in their own right, become much more tempting for their ludic precision and easy self-containment.
But I pine for the days of dice and character sheets, exploring the weirder fringes of inhuman narrative and the familiar shores of the human condition simultaneously. Werecoyotes and Psionics, insatiable curiosity and crippling anxiety, joyous battles and crushing failures.
So, after this personal preamble, here is the philosophical thought I came here to express: RPG systems are procedural frameworks for interactive narrative generation, and they contain engines for simulating worlds.
They are therefore deeply philosophical, because they must contain a metaphysics (narrative/fate) and a theory of personhood (identity/agency/destiny), but they may also contain a logic (GM/PC/NPC interaction), a physics (simulation/means), and an ethics (alignment/ends).
My first encounter with philosophy wasn’t reading Nietzsche, Sartre, or Popper, but reading grimoire-like RPG manuals, searching for the hidden secrets of worlds they contained, many of which I have never visited even in play. What is creation? Why is there suffering? Who are we?
My partner in conceptual crime (@tjohnlinward) likes to say that RPG manuals are tour guides for worlds that don’t exist, but in many ways they’re more like holy texts. Many even have completely explicit and thoroughly fascinating theology.
An RPG system/setting is a universe in which the throne is empty, awaiting a new godhead, or a new pantheon to play the games of divinity. An adventure supplement is like an epic poem, awaiting heroes ready to test their mettle in struggle against the whims of fickle gods.
Narrative is a product, but the process that produces it is a complex, concurrent, and creative interaction between ideas and inspirations; brimming with contingency; some of which may even be embodied in distinct creators and muses. Games are our window into this process.
And that is why games disprove Hegel’s thesis regarding the end of art, precisely by being the most deeply Hegelian of art forms. The world-spirit arrives, no longer Napoleon riding into Jena on horseback, but Gary Gygax corrupting the youth with pens, paper, and polyhedra.
If you want to read more along these lines, check out my ‘Castalian Games’ piece in Glass Bead.
# TfE: Sincerity vs. Honesty
I often talk about the virtue of sincerity, and how important it is to me. There’s even a section of my book devoted to disputing Harman’s interpretation of sincerity as authenticity (‘being oneself’) and contrasting it with my own take on sincerity as fidelity (‘meaning what one says’). However, a question William Gillis asked on Facebook gave me a concrete opportunity to articulate my ideas more concisely, by contrasting sincerity with honesty:
# TfE: The Politicisation Pipeline
Here’s a thread from a few weeks ago reacting to the controversy that unfolded surrounding Natalie Wynn‘s twitter remarks on the complexities of asking for pronouns in certain contexts. This was written before her more recent video ‘Opulence‘, and the second act of that particular clusterfuck. It gave me an opportunity to articulate some of my thoughts on the problems of left-wing political culture, and the way these problems are exacerbated by its transposition and sometimes transmutation into various forms of online discourse. These are closely related to my thoughts on zero-sum politics, and will likely be relevant to some other things I want to say in future, so I think it’s good to get them down here.
# TfE: Immanentizing the Eschaton
Here’s a thread from a little while back in which I outline my critique of the (theological) assumptions implicit in much casual thinking about artificial intelligence, and indeed, intelligence as such.
Another late night thought, this time on Artificial General Intelligence (AGI): if you approach AGI research as if you’re trying to find algorithm to immanentize the eschaton, then you will be either disappointed or deluded.
There are a bunch of tacit assumptions regarding the nature of computation that tend to distort the way we think about what it means to solve certain problems computationally, and thus what it would be to create a computational system that could solve problems more generally.
There are plenty of people who have already pointed out the theological valence of the conclusions reached on the basis of these assumptions (e.g., the singularity, Roko’s Basilisk, etc.); but these criticisms are low hanging fruit, most often picked by casual anti-tech hacks.
Diagnosing the assumptions themselves is much harder. One can point to moments in which they became explicit (e.g., Leibniz, Hilbert, etc.), and thereby either influential, refuted, or both; but it is harder to describe the illusion of coherence that binds them together.
This illusion is essentially related to that which I complained about in my thread about moral logic a few days ago: the idea that there is always an optimal solution to any problem, even if we cannot find it; whereas, in truth, perfectibility is a vanishingly rare thing.
Using the term ‘perfectibility’ makes the connection to theology much clearer, insofar as it is precisely this that forms the analogical bridge between creator and created in the Christian tradition. Divinity is always conceptually liminal, and perfection is a popular limit.
If you’re looking for a reference here, look at the dialectical evolution of the transcendentals (e.g., unum, bonum, verum, etc.) from Augustine and Anselm to Aquinas and Duns Scotus. The universality of perfectible attributes in creation is the key to the singularity of God.
This illusion of universal perfectibility is the theological foundation of the illusion of computational omnipotence.
We have consistently overestimated what computation is capable of throughout history, whether computation was seen as an algorithmic method executed by humans, or a process of automated deduction realised by a machine. The fictional record is crystal clear on this point.
Instead of imagining machines that can do a task better than we can, we imagine machines that can do it in the best possible way. When we ask why, the answer is invariably some variant upon: it is a machine and therefore must be infallible.
This is absurd enough in certain specific cases: what could a ‘best possible poem’ even be? There is no well-ordering of all possible poems, only ever a complex partial order whose rankings unravel as the many purposes of poetry diverge from one another.
However, the deep, and seemingly coherent computational illusion is that there is not just a best solution to every problem, but that there is a best way of finding such bests in every circumstance. This implicitly equates true AGI with the Godhead.
One response to this accusation is to say: ‘Of course, we cannot achieve this meta-optimum, but we can approximate it.’
Compare: ‘We cannot reach the largest prime number, we can still approximate it’
This is how you trade disappointment for delusion.
There are some quite sophisticated mathematical delusions out there. But they are still illusions. There is no way to cheat your way to computational omnipotence. There is nothing but strategy all the way down.
This is not to say that there aren’t better/worse strategies, or that we can’t say some useful and perhaps even universal things about how you tell one from the other. Historically, proofs that we cannot fulfil our deductive ambitions lead to better ambitions and better tools.
The computational illusion, or the true Mythos of Logos, amounts to the idea that one can somehow brute force reality. There is more than a mere analogy here, if you believe Scott Aaronson’s claims about learning and cryptography (I’m inclined to).
It continually surprises me just how many people, including those involved in professional AGI research still approach things in this way. It looks as if, in these cases, the engineering perspective (optimality) has overridden the logical one (incompleteness).
I’ve said it before, and I’ll say it again: you cannot brute force mathematical discovery; there is no algorithm that could progressively search the space of possible theorems. If this does not work in the mathematical world, why would we expect it to work in the physical one?
For additional suggestive material on this and related problems, consider: the problem of induction, Godel’s incompleteness theorems, and the halting problem.
Anyway, to conclude: we will someday make things that are smarter than us in every way, but the intermediate stages involve things smarter than us in some ways. We will not cross this intelligence threshold by merely adding more computing power.
However it happens, it will not be because of an exponential process of self-improvement that we have accidentally stumble upon. Self-improvement is not homogeneous, or without autocatalytic instabilities. Humans are self-improving systems, and we are clearly not gods.
# TfE: Moral Logic, the Diversity of Nature, and the Nature of Diversity
Here are some thoughts from a twitter thread a little while back, which expand on some of the ideas in my post about moral logic. Here’s the initial thought:
Before all else I stan: ought implies can.
I am deadly serious about this. I think ought implies can is as close to an a priori truth about the normative as one can find. However, it’s important to interpret it in the right way. It’s generally used to reason in the contrapositive direction: if one cannot fulfil a purported responsibility, then there is no sense in which one must fulfil it (i.e., can-not implies may-not).
There are two important corollaries of this: (i) that infinite tasks need not be seen as impossible and thereby non-obligatory, insofar as there is a finite procedure that can be indefinitely iterated (e.g., an infinite series: 1 + 1/2 + 1/4 + 1/8… that converges on an ideal limit, namely, 2; this is Hegel’s true infinite); and (ii) that insofar as capacity is not static, there can be increased responsibility relative to increased capacity as easily as decreased responsibility relative to decreased capacity (‘with great power, comes great responsibility‘).
There is more that could be said about this, but I’ll restrict myself to the thread I used to elaborate the original tweet:
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How can I use residue theory to verify this integral formula?
I need to use residue theory to verify that:
$$\mathbf\int_0 ^\infty \frac{\sqrt{x}}{x^2+1}dx=\pi/\sqrt2$$
I was told to use a “keyhole” contour with the positive real axis deleted, but I don’t understand how to do this.
Solutions Collecting From Web of "How can I use residue theory to verify this integral formula?"
There are at least two ways to approach this problem. The first employs a substitution, $x\mapsto x^2$, and then the evenness of the integrand:
\begin{align} \int_0^\infty\frac{\sqrt{x}}{1+x^2}\mathrm{d}x &=\int_0^\infty\frac{2x^2}{1+x^4}\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac{x^2}{1+x^4}\mathrm{d}x\tag{1} \end{align}
The integral $(1)$ can be evaluated with the contour, $\gamma_1$, below: along the real axis from $-R$ to $R$, then circle back, counterclockwise, along a circle in the upper-half-plane back to $-R$.
$\hspace{3.2cm}$
The integral along the curve vanishes as $R\to\infty$ since the integrand $\sim1/R^2$ and the length of the curve is $\pi R$. Therefore,
$$\int_{\gamma_1}\frac{z^2}{1+z^4}\mathrm{d}z \to\int_{-\infty}^\infty\frac{x^2}{1+x^4}\mathrm{d}x\tag{2}$$
The value of the integral in $(2)$ is $2\pi i$ times the sum of the residues at $\color{#C00000}{\frac{i-1}{\sqrt{2}}}$ and $\color{#00A000}{\frac{i+1}{\sqrt{2}}}$:
$$2\pi i\left(\color{#C00000}{\frac{-1-i}{4\sqrt{2}}}+\color{#00A000}{\frac{1-i}{4\sqrt{2}}}\right)=\frac\pi{\sqrt{2}}\tag{3}$$
The second method employs a branch-cut along the positive real axis. The square root cannot be defined continuously on $\mathbb{C}$. To see this, follow the circle of radius $1$ counterclockwise from $+1$ back to $+1$:
$\hspace{4.4cm}$
We get that $\sqrt{+1}=\color{#C00000}{+1}$, $\sqrt{i}=\color{#C00000}{\frac{i+1}{\sqrt{2}}}$, $\sqrt{-1}=\color{#C00000}{i}$, $\sqrt{-i}=\color{#C00000}{\frac{i-1}{\sqrt{2}}}$, and $\sqrt{+1}=\color{#C00000}{-1}$. Thus, $\sqrt{z}$ cannot circle the origin.
One way to prevent circling the origin, is to remove the positive real axis. The square root just above the positive real axis is the positive square root. Just below the positive real axis is the negative square root.
A contour that is useful when integrating functions requiring a branch-cut is the keyhole contour:
$\hspace{4.4cm}$
It follows a large circle (whose radius tends to $\infty$) counterclockwise, a small circle (whose radius tends to $0$) clockwise, and two straight paths connecting them, from the small circle to the large circle above the positive real axis, and from the large circle to the small circle below the positive real axis.
If we integrate $\dfrac12\dfrac{\sqrt{z}}{1+z^2}$ over this contour, the part on the black circle vanishes because the integrand $\sim\frac1{R^{3/2}}$ and the length of the circle is $2\pi R$. The part on the blue circle vanishes since the integrand and the length of the circle tend to $0$. Because $\sqrt{z}$ is positive on the green contour and negative on the red contour, and the red contour is in the opposite direction from the green contour, the sum of those integrals tends to
$$\color{#00A000}{\frac12\int_0^\infty\frac{\sqrt{z}}{1+z^2}\mathrm{d}z} +\color{#C00000}{\frac12\int_\infty^0\frac{-\sqrt{z}}{1+z^2}\mathrm{d}z} =\int_0^\infty\frac{\sqrt{x}}{1+x^2}\mathrm{d}x\tag{4}$$
Therefore,
$$\frac12\int_{\gamma_2}\frac{\sqrt{z}}{1+z^2}\mathrm{d}z =\int_0^\infty\frac{\sqrt{x}}{1+x^2}\mathrm{d}x\tag{5}$$
and the integral in $(5)$ is just $2\pi i$ times the sum of the residues at $\color{#C00000}{i}$ and $\color{#00A000}{-i}$:
$$2\pi i\left(\color{#C00000}{\frac{1-i}{4\sqrt{2}}}+\color{#00A000}{\frac{-1-i}{4\sqrt{2}}}\right)=\frac\pi{\sqrt{2}}\tag{6}$$
You can always define $\sqrt z=\sqrt r e^{i\theta/2}$ where $z=re^{i\theta}$ (and $\sqrt r$ being the ordinary nonnegative real square root), the only problem being that $\theta$ is not unique given $z$. If you avoid the positive real axis, as suggested, you can pick $\theta\in(0,2\pi)$. Now you will find that $\sqrt z\to\sqrt r$ when $z$ approaches the positive real axis from above, while $\sqrt z\to-\sqrt r$ if you approach it from below.
So if you create a contour $\Gamma_\delta$, first going along the half line $\{x-i\delta\colon x>0\}$ running from $x=+\infty$ down to $x=0$, then proceeding clockwise around the origin on a semicircle of radius $\delta$, and then goes off to infinity along $\{x+i\delta\colon x>0\}$, you will find that $$\lim_{\delta\to0}\int_{\Gamma_\delta}\frac{\sqrt z}{z^2+1}\,dz=2\int_0^\infty\frac{\sqrt x}{x^2+1}\,dx.$$
(You will need to verify that the integral around the semicircle is negligible in the limit.)
Now the trick is first to notice that the integral is independent of $\delta$ for small $\delta$, so you can drop taking the limit above!
The proof involves taking to contours $\Gamma_{\delta_j}$ for $j=1,2$, chopping off their tails, say at $x=M$, and joining their ends with two small vertical line segments, then using Cauchy’s integral theorem. You will need to use the fact that the integral along the chopped off tails vanish in the limit as $M\to\infty$, and the integrals along the extra line segments as well.
The second part to the trick is to keep $\delta$ fixed, but to deform the path into a two semi-infinite line segments $\{x\pm i\delta\colon x>R\}$ joined by most of the circle (dashed in the figure below) $\lvert z\rvert =R$. Again, the integral is independent of $R$, except that when $R>1$ the path surrounds the poles at $z=\pm i$. So you need to take the residues at those poles into account.
Finally, you notice that as $R\to\infty$ the integral goes to zero, since the absolute value of the integrand is approximately $R^{-3/2}$ when $R$ is large. Put all the pieces together, and you’re done.
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# Personal Blogs
## A fear conquered; or, musings on moles
Visible to anyone in the world
I am no longer afraid of moles! No, not the little furry buggers that make a mess of your lawn. The once-frightening, but now benign, number used in chemistry so that your head doesn't explode due to excess zeros.
A mole - also known as Avogadro's number - is 6.02 x 10²³ "things". So, one mole of oxygen atoms contains 6.02 x 10²³ atoms. That's quite a large number. So large, that most people can't get their heads around it.
Here's an analogy: one mole of marshmallows would cover the United States of America to a depth of around 6,500 miles*. That is a LOT of marshmallows.
One mole of moles (the little furry buggers this time) would, if placed end-to-end, stretch 11 million light years, and weigh almost as much as the moon.*
Water flows over Niagara Falls at about 650,000 kL (172,500,000 gallons) per minute. At this rate it would take 134,000 years for one mole of water drops (6.02 x 1023 drops) to flow over Niagara Falls.*
Anyway, enough analogies. Suffice it to say, it's a remarkably large number. Far too large to do anything practical with. So, chemists use the mole as a form of shorthand. At school, I hated chemistry specifically because of moles; I just couldn't get my head around it.
So it was with a sense of trepidation that I approached Book 4: The Right Chemistry.
My fears, however, were unfounded. I'm really, really, enjoying this book! The maths tackled so far has really helped to beat back the terrors of Very Large Numbers, and the book is great at explaining difficult concepts in simple terms.
I do think it helps that I am reading We Need to Talk About Kelvin when I'm not studying. This, too, is a cracking book that manages to explain extremely complicated ideas in layman's terms. Doing a bit of reading around the subject definitely helps to seal ideas into your mind, and allows them to take hold.
Anyway - I digress. I was talking about the mole, and its eternal usefulness.
One mole of any substance contains 6.02 x 10²³ atoms, molecules or ions (whichever is most appropriate) of that substance. So, one mole of marshmallows contains 6.02 x 10²³ marshmallows; one mole of water contains 6.02 x 10²³ water molecules; one mole of mercury contains 6.02 x 10²³ mercury atoms.
And, one mole of any substance has a mass equal to the relative mass of that substance, expressed in grams. So one mole of oxygen atoms has a mass of 16.0 g; one mole of oxygen molecules (it's a diatomic molecule, see) has a mass of 32.0 g. With me?
The Avogadro hypothesis (named after Amadeo Avogadro, an Italian physicist who died in 1856) asserts that this is true. Actually, it asserts that equal volumes of different gases, at the same temperature and pressure, contain equal numbers of molecules. Which is beautifully simple, and has the far-reaching consequences I mentioned above.
It enables the mass of any given substance to be translated directly into numbers of molecules (or atoms) using the Avogadro constant: the mole.
Thus: the number of moles of a substance is equal to the mass of that substance divided by the molar mass of the substance.
E.g. How many moles are in 52 g of water? Well, the molar mass of water is (2 x 1.01) + 16.0 = 18.02 g mol‾¹
So the number of moles in the water = 52 g divided by 18.02 g mol‾¹ = 2.89 mol (3 significant figures). There are 2.89 moles of water molecules in 52 g of water.
Simples!
And the scariest thing? I'm quite enjoying it all! Next, I shall enthuse about covalent bonds. They are this: aces.
*I can't claim the credit for these analogies. They came from a rather cool chemistry site.
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# Totally Tubular Fraction Formula Game
This post may contain affiliate links.
Educational Insights Fraction Formula Game visually teaches children about fractions. It’s a fun math game for kids that is colorful, playful, and educational.
We love it!
## Fraction Formula Game
More than that, my kids, even my six-year-old, are finally conceptualizing fractions! So, fun and learning in one game, what’s not to love?
## To Play:
Each player starts with a cylinder.
To play, draw one fraction card from the card pile.
Drop the card’s corresponding fraction colored tile into the cylinder. For example, if you draw a 1/3 card, you put a 1/3 fizzy tile into your cylinder.
Play continues.
The object is to get as close to 1 whole without going over.
If a player draws a fraction that makes your tile go over the top line, that player automatically loses.
If you think you should stop adding fractions tiles because you’re too close to the top, you announce “hold” and stop playing.
Points are awarded by who gets closest to one whole — the top line with the second closest and third closest to one whole as second and third place.
The age recommendation is 8 – 11 but JJ is 6 and she did fine.
I love this hands-on math game because it really helps us visually see the fractions. This is always a top pick for my kids on family game night.
## Similar Posts
1. Brenda Witherspoon-Bedard says:
I am a of Fan Imagination Soup on Facebook.
brendawitherspoon at hotmail dot com
2. Brenda Witherspoon-Bedard says:
I am a Fan Educational Insights on Facebook.
brendawitherspoon at hotmail dot com
3. Brenda Witherspoon-Bedard says:
Great game
brendawitherspoon at hotmail dot com
4. Coral says:
Ok, this is a cool way to make learning fun!
5. Becky says:
I had no idea Imagination featured educational games like this!
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Contents
Contraction (mathematics)
In mathematics, contraction has two meanings:
• Contraction of a tensor. It occurs when a pair of literal indices (one a subscript, the other a superscript) of a mixed tensor are set equal to each other so that a summation over that index takes place (due to the Einstein summation convention). The result is another tensor whose rank is reduced by 2.
If a tensor is dyadic then its contraction is a scalar obtained by dotting each pair of base vectors in each dyad. E.g. Let $\mathbf{T} = T^i{}_j \mathbf{e_i e^j}$ be a dyadic tensor, then its contraction is $T^i {}_j \mathbf{e_i} \cdot \mathbf{e^j} = T^i {}_j \delta_i^j T^j {}_j T^1 {}_1 + T^2 {}_2 + T^3 {}_3$,
a scalar of rank 0.
E.g. Let $\mathbf{T} = \mathbf{e^i e^j}$ be a dyadic tensor.
This tensor does not contract; if its base vectors are dotted the result is the contravariant metric tensor, $g^{ij}= \mathbf{e^i} \cdot \mathbf{e^j}$, whose rank is 2.
References. Mathematical Physics by Donald H. Menzel. Dover Publications, New York.
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# Two parallel chords of a circle of radius 5 cm are in opposite side of the centre and if the chords are at distances 3 cm and 4 cm from the centres, then the difference of the lengths of the chords is
This question was previously asked in
UP TGT Mathematics 2021 Official Paper
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1. 10 cm
2. 5 cm
3. 2 cm
4. 1 cm
Option 3 : 2 cm
Free
UP TGT Biology Mock Test
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10 Questions 40 Marks 10 Mins
## Detailed Solution
Given:
The radius of the circle = 5 cm
Concept used:
The shortest distance on the chord of the circle from the center made 90°
It divides chord in two equal parts
Calculation:
In Δ AOE
AE2 = AO2 - EO2
⇒ AE2 = 25 - 9 = 16
⇒ AE = 4
AB = 2 × AE = 2 × 4 = 8 cm
And, In Δ DOF
DF2 = DO2 - FO2
⇒ DF2 = 25 - 16 = 9
⇒ DF = 3
DC = 2 × DF = 2 × 3 = 6 cm
Difference between two chord = 8 - 6 = 2 cm
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Wednesday
December 4, 2013
Posts by Cc
Total # Posts: 202
Chemistry
still not making sense because answers are a. 30.0 b. 35.3 c. 15.0 d. 28.9
Chemistry
200. g of a solvent is available, how many grams of a solute should be weighed out to make a 15.0% solution by mass?
chemistry
Calculate the molarity of a solution containing 40.0 g glucose (C6H12O6) in enough ethanol (C2H5OH) to make 40.0 mL of solution.
chemistry
A 0.750 g sample of an unknown substance is dissolved in 20.0 g of benzene, C6H6. The freezing point of the solution is 4.53 oC. Calculate the molar mass of the substance. The normal freezing point for benzene is 5.53 oC and Kf is 5.12 oC/m.
chemistry
A 0.750 g sample of an unknown substance is dissolved in 20.0 g of benzene, C6H6. The freezing point of the solution is 4.53 oC. Calculate the molar mass of the substance. The normal freezing point for benzene is 5.53 oC and Kf is 5.12 oC/m.
Chemistry
From the following: 1) pure water 2) 0.01 m solution of table sugar (C12H22O11) 3) 0.01 m solution of NaCl 4) 0.01 m solution of CaCl2 Choose the one with the largest van't Hoff factor and hence the lowest freezing point.
CHM
From the following: 1) pure water 2) 0.01 m solution of table sugar (C12H22O11) 3) 0.01 m solution of NaCl 4) 0.01 m solution of CaCl2 Choose the one with the largest van't Hoff factor and hence the lowest freezing point.
chemistry
A 2.30 m solution of urea (CO(NH2)2; FW= 60.0) in water has a density of 1.14 g/mL. Calculate the percent mass of urea in this solution
Chemistry
A 2.30 m solution of urea (CO(NH2)2; FW= 60.0) in water has a density of 1.14 g/mL. Calculate the percent mass of urea in this solution
Chemistry
Thanks so much I just found this site and it is helping me more than ever...
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# I am so sorry! I do not know why I cannot post the right
Author Message
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Manager
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I am so sorry! I do not know why I cannot post the right [#permalink] 26 Apr 2007, 20:30
I am so sorry! I do not know why I cannot post the right question. I have tried and tried but to no avail.
Anyway I have written out the stmts, hopefully that will work!
If x and y are integers and
y=lx+3l+l4-xl does y equal 7?
Stmt 1 says that x is less that 4
Stmt 2 says x is greater than -3
[/b]
Last edited by ninomoi on 28 Apr 2007, 12:30, edited 2 times in total.
Current Student
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This seems like a DS. Do you want to correct the question?
Manager
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I have re wriiten the question folks! Pls help!!!!
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(C) for me
lx+3l+l4-xl = 7 ?
From1
X < 4, implies that :
lx+3l+l4-xl
=|x+3| + (4-x)
After that, to go more ahead, we have to case to remove the second absolute:
o If x > -3
lx+3l+l4-xl
=|x+3| + (4-x)
= x+3 + 4 -x
= 7
o If x < -3
lx+3l+l4-xl
=|x+3| + (4-x)
= -(x+3) + 4 -x
= -2*x + 7
INSUFF.
From2
X > -3, implies that :
lx+3l+l4-xl
=(x+3) + |4-x|
After that, to go more ahead, we have to case to remove the second absolute:
o If x < 4
lx+3l+l4-xl
=(x+3) + |4-x|
=(x+3) + (4-x)
= 7
o If x > 4
lx+3l+l4-xl
=(x+3) + |4-x|
=(x+3) + -(4-x)
= 2*x - 1
INSUFF.
Both (1) & (2)
We have -3 < x < 4. Thus,
lx+3l+l4-xl
=(x+3) + (4-x)
= 7
SUFF.
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Given: x and y are integers and y=lx+3l+l4-xl
(1) x is less than 4
Solution domain of y is found by plugging 4 for x, which leads to y=7
now, if x is less than 4 then try x = 3 which gives y = 7
but if x is - 10, then y = 21
So, statement 1 is insufficient
(2) x is greater than -3
try x = -3 .. gives y = 7
try x = -2 .. gives y = 7
but if x = 10 .. then y = 19
statement 2 is insufficient
(1) and (2) together
I hope by now you see the pattern and how these two statement build a solution domain for values of x that result in y = 7
As long as x is less than 4 and larger than -3, the value of y will always be 7, of course given that x and y are integers.
So, statements 1 and 2 together are sufficient to answer the question
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Joined: 09 Sep 2004
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Thanks guys! The answer is C.
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What Are Tips for Finding the Spring Constant?
What Are Tips for Finding the Spring Constant?
The spring constant may be determined using Hooke's Law. An experiment may be arranged to use the principle of Hooke's Law to calculate the value of the spring constant based on the force a spring exerts on an object and the displacement of the spring from its equilibrium state.
Each spring has a unique spring constant. This constant becomes a factor to determine the force exerted on an object once combined with the spring's distance from its original relaxed position. This is based on Hooke's Law, which is presented in a mathematical formula as F = -kx, where "F" is the force, "x" is the spring's displacement, and "k" is the spring constant. The negative sign denotes that the force is in the opposite direction of the object pulling the spring.
To determine the spring constant through a laboratory experiment, the spring must first be suspended from a reference hook or rod. A ruler is placed behind the spring, with the zero marking positioned at the level of the lowest point of the spring. Tests are done using different weights, preferably ranging in the same multiples, such as 50 grams, 100 grams, 150 grams and 200 grams. Each weight is placed on the hook at the end of the spring, and the displacement is measured using the ruler. The set of data is plotted in a graph wherein the x-value is the weight and the y-value is the displacement. The spring constant may be calculated as the slope of the line that appears on the graph.
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## CAT Online Coaching – Partial Fractions
We get a lot of questions on partial fractions, on how to simplify them and how
## CAT Preparation Online – Permutation and Combination
Question Product of the distinct digits of a natural number is 60. How many such
## Solutions to Number Theory and Counting questions
Given below are the solutions to these Number theory questions. 1. From the digits 2,3,4,5,6
## Few Questions on Number Theory and Counting (combined)
CAT has been consistently asking questions combining basic number theory and counting. So, it is
## CAT Number Theory
1. The six faces of a cube have 6 distinct natural numbers written on them.
## Percentages – Solutions
Have given below solutions to the first three questions on Percentages 1. a is x
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# Financial math
Janet Woo decided to retire to Florida in 5 years. What amount should Janet invest today so she can withdraw \$45,000 at the end of each year for 20 years after she retires? Assume Janet can invest money at 5% compounded annually.
1. 👍
2. 👎
3. 👁
1. janet won lottery an receives \$10 the 1st year. in the following years, she receives \$50 more each year, (that is, janet receives \$150 the second year, \$200 the third, and so on) How much will she receive in total after 10 years
1. 👍
2. 👎
## Similar Questions
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1. Which of the following events caused immigrants to migrate north after the civil war? -the passage of the homestead act****** -the emergence of industry in cities -advances in the printing industry - the completion of the
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A group of students have volunteered for the student council car wash. Janet can wash a car in m minutes. Rodrigs can wash a car in m-5 minutes, while Nick needs the same amount of time as Janet.If they all work together, they can
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3. ### Finance
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2. ### science
janet uses 4 feer of ribbone to decorate each pillow the ribbone ome in 225-feet how many pillow will janet decrorate]
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# Iterative formulas
1. Mar 5, 2009
### DeanBH
1. The problem statement, all variables and given/known data
First of all this is revision not homework =)
question is Xn+1 = Cuberoot ( 17.5-2xn)
answer lies between 2 and 3. i know the answer is 2.34 but what i don't get is why it is
xn+1 = equation, because when you put xn=2.34 into the equation you get 2.34 out.
shouldnt get 3.34 out if its xn+1. whats the purpose of the n+1!!
2. Mar 5, 2009
### dirk_mec1
xn and xn+1 both go (in this convergent case) to a limit call it L. Now substitute L for xn and xn+1 and solve the equation.
3. Mar 5, 2009
### DeanBH
i don't think you understand what i'm confused with,
i know the answer is 2.34 so i put it in to check
xN+1 = cuberoot(17.5-2xn)
3.34=cuberoot(17.5-2*2.34)
3.34=2.34
do you understand what i dont understand.
4. Mar 5, 2009
5. Mar 5, 2009
### yyat
If you mean
$$x_{n+1}=(17.5+2x_n)^{1/3}$$
then the answer is $$\lim_{n\to\infty}x_n=2.34$$, and this solution can be found as dirk_mec1 described.
But if you mean
$$x_n+1=(17.5+2x_n)^{1/3}$$
then $$x_n=1.44$$ for all $$n$$.
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46 Writing Good Code
Functions like Table or NestList or FoldList exist in the Wolfram Language because they express common things one wants to do. As in natural language, there are always many ways one can in principle express something. But good code involves finding the most direct and simple way.
Simple and good Wolfram Language code for making a table of the first 10 squares:
In[1]:=
Out[1]=
Why would anyone write anything else? A common issue is not thinking about the whole table, but instead thinking about the steps in building it. In the early days of computing, computers needed all the help they could get, and there was no choice but to give code that described every step to take.
In[2]:=
Out[2]=
But the point of the Wolfram Language is to let one express things at a higher leveland to create code that as directly as possible captures the concept of what one wants to do. Once one knows the language, its vastly more efficient to operate at this level. And it leads to code thats easier for both computers and humans to understand.
In[3]:=
Run the code:
In[4]:=
Out[4]=
In[5]:=
The new code works:
In[6]:=
Out[6]=
Simplify the code by multiplying the whole list of powers of 10 at the same time:
In[7]:=
In[8]:=
In[9]:=
In[10]:=
The new approach works too:
In[11]:=
Out[11]=
In[12]:=
In[13]:=
In[14]:=
Out[14]=
In[15]:=
Out[15]=
In[16]:=
It still works though:
In[17]:=
Out[17]=
Heres a single definition that combines several cases:
In[18]:=
In[19]:=
In[20]:=
When youre writing code, its common to first define a new function because you need it in some very specific context. But its almost always worth trying to give it a name that youll understand even outside that context. And if you cant find a good name, its often a sign that its not quite the right function to define in the first place.
In[21]:=
Out[21]=
In[22]:=
Out[22]=
In[23]:=
Out[23]=
With every new version, the Wolfram Language does better at automatically figuring out how to make your code run fast. But you can always help by structuring your algorithms well.
Timing gives the result of a computation, together with its timing (in seconds):
In[24]:=
Out[24]=
With the definitions of fib above, the time grows very rapidly:
In[25]:=
Out[25]=
Redefine the fib function to remember every value it computes:
In[26]:=
In[27]:=
Now even up to 1000 each new value takes only microseconds to compute:
In[28]:=
Out[28]=
Vocabulary
46.1Find a simpler form for Module[{a, i}, a=0;For[i=1, i1000, i++, a=i*(i+1)+a];a]»
Expected output:
Out[]=
46.2Find a simpler form for Module[{a, i}, a=x; For[i=1, i10, i++, a=1/(1+a)];a]»
Expected output:
Out[]=
46.3Find a simpler form for Module[{i, j, a}, a={};For[i=1, i10, i++, For[j=1, j10, j++, a=Join[a, {i, j}]]];a]»
Expected output:
Out[]=
46.4Make a line plot of the timing for computing n^n for n up to 10000. »
Sample expected output:
Out[]=
46.5Make a line plot of the timing for Sort to sort Range[n] from a random order, for n up to 200. »
Sample expected output:
Out[]=
Q&A
What does i++ mean?
Its a short notation for i=i+1. Its the same notation that C and many other low-level computer languages use for this increment operation.
What does the For function do?
Its a direct analog of the for(...) statement in C. For[start, test, step, body] first executes start, then checks test, then executes step, then body. It does this repeatedly until test no longer gives True.
Why can shortened pieces of code be hard to understand?
The most common issue is that variables and sometimes even functions have been factored out, so there are fewer names to read that might give clues about what the code is supposed to do.
Whats the best IDE for authoring Wolfram Language code?
For everyday programming, Wolfram Notebooks are best. Make sure to add sections, text, and examples right alongside your code. For large multi-developer software projects, Wolfram Workbench provides an Eclipse-based IDE.
What does Timing actually measure?
Use RepeatedTiming, which runs code many times and averages the timings it gets. (This wont work if the code is modifying itself, like in the last definition of fib above.)
Beyond keeping the code simple, one thing is not to recompute anything you dont have to. Also, if youre dealing with lots of numbers, it may make sense to use N to force the numbers to be approximate. For some internal algorithms you can pick your PerformanceGoal, typically trading off speed and accuracy. There are also functions like Compile that force more of the work associated with optimization to be done up front, rather than during a computation.
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A290780 Half-octavan primes: primes of the form (x^8 + y^8)/2. 1
198593, 21523361, 107182721, 407865361, 429388721, 3487882001, 11979660241, 39155495921, 84785726833, 141217650641, 141321947681, 250123401793, 253611085201, 289278699121, 391337974721, 426445714033, 426448401121 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 A. J. C. Cunningham, High quartan factorisations and primes, Messenger of Mathematics 36 (1907), pp. 145-174. EXAMPLE a(1) = (5^8 + 3^8)/2 = 198593. a(2) = (9^8 + 1^8)/2 = 21523361. a(3) = (11^8 + 3^8)/2 = 107182721. a(4) = (13^8 + 1^8)/2 = 407865361. a(5) = (13^8 + 9^8)/2 = 429388721. MAPLE N:= 10^12: # to get all terms <= N sort(convert(select(isprime, {seq(seq((x^8+y^8)/2, y= (x mod 2)..min(x, floor((2*N-x^8)^(1/8))), 2), x=1..floor((2*N)^(1/8)))}), list)); # Robert Israel, Aug 21 2017 MATHEMATICA Sort[Select[Total/@(Union[Sort/@Tuples[Range[0, 50], 2]]^8)/2, PrimeQ]] (* or *) lst={}; Do[If[PrimeQ[(a^8 + b^8) / 2], AppendTo[lst, (a^8 + b^8) / 2]], {a, 100}, {b, a, 100}]; Sort[lst] (* Vincenzo Librandi, Aug 21 2017 *) PROG (PARI) list(lim)=my(v=List(), x8, t); forstep(x=1, sqrtnint(lim\=1, 8), 2, x8=x^8; forstep(y=1, min(sqrtnint(lim-x8, 8), x-1), 2, if(isprime(t=(x8+y^8)/2), listput(v, t)))); Set(v) \\ Charles R Greathouse IV, Aug 20 2017 CROSSREFS Cf. A006686, A002646. Sequence in context: A215996 A205234 A249358 * A255782 A205638 A031687 Adjacent sequences: A290777 A290778 A290779 * A290781 A290782 A290783 KEYWORD nonn AUTHOR Charles R Greathouse IV, Aug 20 2017 STATUS approved
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# Alcoholics Anonymous Tags
## 4th Step Worksheet
Then, quickly read the last paragraph of each chapter as these will be chapter summaries. This will give you the big picture of what the book contains and allow you to assess whether or not this book will meet your reading goal. Finally, read through the book at a very high rate
## Converting Fractions Decimals And Percents Worksheets
Next, look at the book`s structure and preview these elements. Most books have jackets or covers which point out some of the features of the book. Look over the title page, preface, introduction, or foreword, chapter headings, chapter subheadings, and index. These elements will give you a sense of the subject matter
## Ratio Tables Worksheets
One of the well-known techniques on how to increase reading comprehension is to write down the important ideas or details of what you`ve read. This step is very useful and convenient for students who are preparing for their exams because they only have to study the essential points provided by their reading
Next, look at the book`s structure and preview these elements. Most books have jackets or covers which point out some of the features of the book. Look over the title page, preface, introduction, or foreword, chapter headings, chapter subheadings, and index. These elements will give you a sense of the subject matter
## Punnett Square Practice Worksheet Answer Key
If you thought reading statistics were bad, wait till you look at the math statistics. They are even worst! Did you know that 38% of high school seniors can`t count? That is correct. Even worst, I am not talking about algebra, geometry, or calculus. I am referring to decimals, fractions, and percentages.
## Percent Of A Number Word Problems Worksheets
One of the well-known techniques on how to increase reading comprehension is to write down the important ideas or details of what you`ve read. This step is very useful and convenient for students who are preparing for their exams because they only have to study the essential points provided by their reading
## 7th Grade Worksheets Free Printable
One of the well-known techniques on how to increase reading comprehension is to write down the important ideas or details of what you`ve read. This step is very useful and convenient for students who are preparing for their exams because they only have to study the essential points provided by their reading
## Subtraction Drill Sheets
Then, quickly read the last paragraph of each chapter as these will be chapter summaries. This will give you the big picture of what the book contains and allow you to assess whether or not this book will meet your reading goal. Finally, read through the book at a very high rate
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Re: Single-step evaluation in Mathematica
• To: mathgroup at smc.vnet.net
• Subject: [mg81289] Re: Single-step evaluation in Mathematica
• From: Chris Chiasson <chris.chiasson at gmail.com>
• Date: Tue, 18 Sep 2007 05:53:19 -0400 (EDT)
• References: <fclape\$f7b\$1@smc.vnet.net>
```On Sep 17, 2:36 am, Szabolcs Horv=E1t <szhor... at gmail.com> wrote:
> I would like to have a function, hasOwnValue, that takes the name of a
> symbol as a string, and tells whether the symbol has an own-value. It
> should do this without evaluating the value of the symbol.
>
> Examples:
>
> abc := Print["side effect"]
> xyz =.
>
> In := hasOwnValue["abc"] (* "side effect" must not be printed *)
> Out = True
>
> In := hasOwnValue["xyz"]
> Out = False
>
> How can I achieve this?
>
> This is a practical problem that I ran into, but the general question
> is: What do you do when you feel the need for *single-step evaluation*
> in Mathematica? E.g. how can I manipulate/transform held expressions
> made up of symbols that have values? An artificial example:
>
> a = 1
> Hold[{a,b,c}]
>
> How do I reverse the list inside Hold? I could map Hold to each
> element, reverse the list, hold the list again, and remove Hold from
> individual elements to get Hold[{c,b,a}], but this is very inconvenient.
>
> Szabolcs
It would be quite trivial to achieve single step evaluation if the
*Values of all the system variables were readable. However, they are
not.
http://library.wolfram.com/conferences/devconf99/villegas/UnevaluatedExpres=
sions.nb
(or, in HTML)
http://library.wolfram.com/conferences/devconf99/villegas/UnevaluatedExpres=
sions/
Also, I would recommend these two posts for the held evaluation
properties of Block, With, Module, RuleCondition, \$ConditionHold, and
the multi argument Function:
http://forums.wolfram.com/mathgroup/archive/2007/Aug/msg00958.html
http://forums.wolfram.com/mathgroup/archive/2007/Sep/msg00327.html
--
http://chris.chiasson.name/
```
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## Resource Options
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## Animating Objects Programmatically-Animating an Analog Clock face: Part 2
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Lesson Introduction In the last lesson, students learned the basics of Object Oriented Programming and learned how to instanti...
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## Animating Objects Programmatically-Animating an Analog Clock face: Part1
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## Animating Objects Programmatically-Animating an Analog Clock face: Part3
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## Technology and Marketing
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## Testing for Prime Numbers with Predicate Blocks in BYOB/Scratch: A Scaffolding Tool for Teaching Boolean Expressions in AP Computer Science
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This lesson is primarily directed at the Advanced Placement Computer Science class which is taught using Java. BYOB/Scratch ser...
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## Advanced Alice Project- Lunar Lander
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## Binary 2- 11 Cheers for Binary! How Computers Count.
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## Binary 1- Calm Down, It's Only Ones and Zeroes- How Computers Think.
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When computers think, it must be very precise with definite numbers. However, much of the information we get from nature does no...
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## Binary 4- ASCII Art- How Computers Write.
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## B5- Life Without Thumbs- Counting like Computers (by eights)
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Our decimal number system we use is based on the fact that we humans are born with ten fingers (including thumbs!). But computer...
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## Binary 3- Bits and Bytes and Nybbles, Oh My!
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Computers think only in binary, using only ones and zeroes. However, just a one or zero, called a "bit", will not give enough in...
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## Creating a Maze Game In Scratch
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## Getting Started With BYOB/Scratch: How to Build a Block
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CREATE VISUAL DESIGNS USING CUSTOM SCRATCH BLOCKS BYOB is an extension of Scratch wh...
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## Creating a Number Guessing Game in BYOB/Scratch: A Scaffolding Tool For Teaching Binary Searching in AP Computer Science
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# college physics
A cubic box of volume 4.1×10^-2 m^3 is filled with air at atmospheric pressure at 20 degrees C. The box is closed and heated to 181 degrees C. What is the net force on each side of the box?
1. 👍
2. 👎
3. 👁
1. The edge size of the box is
a = (0.041)^(1/3) = 0.345 m
and the area of a side is 0.1189 m^2
Pressure increases by a factor (273+181)/(273+20) = 1.5495
The new pressure will be 1.5495 atm
Multiply the final pressure (in Pascals) by the box side area, for the force in Newtons. I will leave that step up to you.
1. 👍
2. 👎
## Similar Questions
1. ### chemistry
In which one of the following processes is ΔH = ΔE? a. 2HI(g) → H2(g) + I2(g) at atmospheric pressure. b. Two moles of ammonia gas are cooled from 325°C to 300°C at 1.2 atm. c. H2O(l) → H2O(g) at 100°C at atmospheric
2. ### chem 121
How many air molecules are in a 15.0×12.0×10.0 ft room? Assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘C, and ideal behavior. Volume conversion:There are 28.2 liters in one cubic foot.
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when the temperature stays the same, the volume of the gas is inversely proportional to the pressure of the gas. If a balloon is filled with 87 cubic inches of a gas at a pressure of 14 pounds per square inch. Find the new
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Atmospheric pressure on the peak of Mount Everest can be as low as 0.20 atm. If the volume of an oxygen tank is 10.0 L, at what pressure must the tank be filled so that the gas inside would occupy a volume of 1.2x10^3L at this
At a certain temperature and pressure, 0.20 mol of carbon dioxide has a volume of 3.1 L. A 3.1-L sample of hydrogen at the same temperature and pressure ____. I think it's supposed to be "contains the same number of molecules" but
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A cylinder in a car engine takes Vi = 4.50 10-2 m3 of air into the chamber at 30° C and at atmospheric pressure. The piston then compresses the air to 0.112 times of the original volume (0.112 Vi) and to 24.0 times the original
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#### What is 1780 percent of 9,770?
How much is 1780 percent of 9770? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 1780% of 9770 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 1780% of 9,770 = 173906
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating one thousand, seven hundred and eighty of nine thousand, seven hundred and seventy How to calculate 1780% of 9770? Simply divide the percent by 100 and multiply by the number. For example, 1780 /100 x 9770 = 173906 or 17.8 x 9770 = 173906
#### How much is 1780 percent of the following numbers?
1780 percent of 9770.01 = 17390617.8 1780 percent of 9770.02 = 17390635.6 1780 percent of 9770.03 = 17390653.4 1780 percent of 9770.04 = 17390671.2 1780 percent of 9770.05 = 17390689 1780 percent of 9770.06 = 17390706.8 1780 percent of 9770.07 = 17390724.6 1780 percent of 9770.08 = 17390742.4 1780 percent of 9770.09 = 17390760.2 1780 percent of 9770.1 = 17390778 1780 percent of 9770.11 = 17390795.8 1780 percent of 9770.12 = 17390813.6 1780 percent of 9770.13 = 17390831.4 1780 percent of 9770.14 = 17390849.2 1780 percent of 9770.15 = 17390867 1780 percent of 9770.16 = 17390884.8 1780 percent of 9770.17 = 17390902.6 1780 percent of 9770.18 = 17390920.4 1780 percent of 9770.19 = 17390938.2 1780 percent of 9770.2 = 17390956 1780 percent of 9770.21 = 17390973.8 1780 percent of 9770.22 = 17390991.6 1780 percent of 9770.23 = 17391009.4 1780 percent of 9770.24 = 17391027.2 1780 percent of 9770.25 = 17391045
1780 percent of 9770.26 = 17391062.8 1780 percent of 9770.27 = 17391080.6 1780 percent of 9770.28 = 17391098.4 1780 percent of 9770.29 = 17391116.2 1780 percent of 9770.3 = 17391134 1780 percent of 9770.31 = 17391151.8 1780 percent of 9770.32 = 17391169.6 1780 percent of 9770.33 = 17391187.4 1780 percent of 9770.34 = 17391205.2 1780 percent of 9770.35 = 17391223 1780 percent of 9770.36 = 17391240.8 1780 percent of 9770.37 = 17391258.6 1780 percent of 9770.38 = 17391276.4 1780 percent of 9770.39 = 17391294.2 1780 percent of 9770.4 = 17391312 1780 percent of 9770.41 = 17391329.8 1780 percent of 9770.42 = 17391347.6 1780 percent of 9770.43 = 17391365.4 1780 percent of 9770.44 = 17391383.2 1780 percent of 9770.45 = 17391401 1780 percent of 9770.46 = 17391418.8 1780 percent of 9770.47 = 17391436.6 1780 percent of 9770.48 = 17391454.4 1780 percent of 9770.49 = 17391472.2 1780 percent of 9770.5 = 17391490
1780 percent of 9770.51 = 17391507.8 1780 percent of 9770.52 = 17391525.6 1780 percent of 9770.53 = 17391543.4 1780 percent of 9770.54 = 17391561.2 1780 percent of 9770.55 = 17391579 1780 percent of 9770.56 = 17391596.8 1780 percent of 9770.57 = 17391614.6 1780 percent of 9770.58 = 17391632.4 1780 percent of 9770.59 = 17391650.2 1780 percent of 9770.6 = 17391668 1780 percent of 9770.61 = 17391685.8 1780 percent of 9770.62 = 17391703.6 1780 percent of 9770.63 = 17391721.4 1780 percent of 9770.64 = 17391739.2 1780 percent of 9770.65 = 17391757 1780 percent of 9770.66 = 17391774.8 1780 percent of 9770.67 = 17391792.6 1780 percent of 9770.68 = 17391810.4 1780 percent of 9770.69 = 17391828.2 1780 percent of 9770.7 = 17391846 1780 percent of 9770.71 = 17391863.8 1780 percent of 9770.72 = 17391881.6 1780 percent of 9770.73 = 17391899.4 1780 percent of 9770.74 = 17391917.2 1780 percent of 9770.75 = 17391935
1780 percent of 9770.76 = 17391952.8 1780 percent of 9770.77 = 17391970.6 1780 percent of 9770.78 = 17391988.4 1780 percent of 9770.79 = 17392006.2 1780 percent of 9770.8 = 17392024 1780 percent of 9770.81 = 17392041.8 1780 percent of 9770.82 = 17392059.6 1780 percent of 9770.83 = 17392077.4 1780 percent of 9770.84 = 17392095.2 1780 percent of 9770.85 = 17392113 1780 percent of 9770.86 = 17392130.8 1780 percent of 9770.87 = 17392148.6 1780 percent of 9770.88 = 17392166.4 1780 percent of 9770.89 = 17392184.2 1780 percent of 9770.9 = 17392202 1780 percent of 9770.91 = 17392219.8 1780 percent of 9770.92 = 17392237.6 1780 percent of 9770.93 = 17392255.4 1780 percent of 9770.94 = 17392273.2 1780 percent of 9770.95 = 17392291 1780 percent of 9770.96 = 17392308.8 1780 percent of 9770.97 = 17392326.6 1780 percent of 9770.98 = 17392344.4 1780 percent of 9770.99 = 17392362.2 1780 percent of 9771 = 17392380
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# 151-1633-00L Energy Conversion
Semester Herbstsemester 2018 Dozierende I. Karlin, G. Sansavini Periodizität jährlich wiederkehrende Veranstaltung Lehrsprache Englisch Kommentar This course is intended for students outside of D-MAVT.
Kurzbeschreibung This course is tailored to provide the students with a common introduction on thermodynamics and heat transfer. Students can gain a basic understanding of energy, energy interactions, and various mechanisms of heat transfer as well as their linkage to energy conversion technologies. Lernziel Students will be able analyze and evaluate energy conversion and heat exchange processes from the thermodynamic perspective. 1. They will be able to describe a thermodynamic system and its state in the using phase diagrams for pure substances and to apply the first law of thermodynamics, energy balances, and mechanisms of energy transfer to or from a system. 2. Students will be able to describe processes/changes of state in the phase diagrams and evaluate start and end states and the exchange of heat and power in the process.3. They will be able to introduce and apply the entropy and exergy balance to closed and open systems. 4. They will be able to apply the second law of thermodynamics to power cycles and processes, and determine the expressions for the thermal efficiencies and coefficients of performance for heat engines, heat pumps, and refrigerators. They will be able to evaluate the thermodynamic performance of cycles using phase diagrams and critically analyze the different parts of cycles and propose improvements to their efficiency.5. Students will be able to apply energy balances to reacting systems for both steady-flow control volumes and fixed mass systems.6. At the end of the course, they will be able to apply the basic mechanisms of heat transfer (conduction, convection, and radiation), and Fourier's law of heat conduction, Newton's law of cooling, and the Stefan–Boltzmann law of radiation. Finally, students will be able to solve various heat transfer problems encountered in practice. Inhalt 1. Thermodynamic systems, states and state variables2. Properties of substances: Water, air and ideal gas3. Energy conservation in closed and open systems: work, internal energy, heat and enthalpy4. Second law of thermodynamics and entropy5. Energy analysis of steam power cycles6. Energy analysis of gas power cycles7. Refrigeration and heat pump cycles8. Maximal work and exergy analysis9. Mixtures and psychrometry 10. Chemical reactions and combustion systems11. Heat transfer Skript Lecture slides and supplementary documentation will be available online. Literatur Thermodynamics: An Engineering Approach, by Cengel, Y. A. and Boles, M. A., McGraw Hill Voraussetzungen / Besonderes This course is intended for students outside of D-MAVT.Students are assumed to have an adequate background in calculus, physics, and engineering mechanics.
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X hits on this document
527 views
0 shares
53 / 63
Problems
A-53
( i n s e c t i c i d e , t o n s ) ( a c r e s ) x 1 , x 2 , x 3 Ú 0 x 1 + x 2 + x 3 4 0 3 x 1 + 2 x 2 + 4 x 3 1 0 0
Solve this model using the simplex method. 38. Solve the following linear programming model (a) graphically and (b) using the simplex method:
maximize Z = 3x1 + 2x2 subject to
x 1 , x 2 Ú 0 x 1 + x 2 Ú 2 x 1 + x 2 1
39. Solve the following linear programming model (a) graphically and (b) using the simplex method:
maximize Z = x1 + x2 subject to
x 1 , x 2 Ú 0 - x 1 + 2 x 2 4 x 1 - x 2 Ú - 1
40. Solve the following linear programming model using the simplex method:
1 2 x +x +x
3
1 2 2x + x + x
3
1 x +x
2
x
3
x1, x2, x
3
maximize Z = 7x1 + 5x2 + 5x subject to
• 25
• 40
• 25
• 6
Ú0
3
41. Solve the following linear programming model using the simplex method:
minimize Z = 15x1 + 25x2
subject to
3x1 + 4x x 1 , x 3x1 + 2x Ú 12 Ú6 9 2 2 2 2 Ú 0 2 x 1 + x
42. The Old English Metal Crafters Company makes brass trays and buckets. The number of trays (x1) and buckets (x2) that can be produced daily is constrained by the availability of brass and labor, as reflected in the following linear programming model:
Document views 527 Page views 527 Page last viewed Mon Jan 23 01:34:39 UTC 2017 Pages 63 Paragraphs 3770 Words 30893
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## 239s complement binary number conve
239s complement binary number converter
GO TO PAGE
### Converting "twos complement" - decimal to actual twos
Signed Binary Arithmetic If the sign bit is 1, the number is a 2’s complement negative number. To convert this number to decimal:
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# Properties
Label 570.2.i.i Level $570$ Weight $2$ Character orbit 570.i Analytic conductor $4.551$ Analytic rank $0$ Dimension $4$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$570 = 2 \cdot 3 \cdot 5 \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 570.i (of order $$3$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$4.55147291521$$ Analytic rank: $$0$$ Dimension: $$4$$ Relative dimension: $$2$$ over $$\Q(\zeta_{3})$$ Coefficient field: $$\Q(\sqrt{-3}, \sqrt{7})$$ Defining polynomial: $$x^{4} + 7x^{2} + 49$$ x^4 + 7*x^2 + 49 Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\beta_2,\beta_3$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + (\beta_{2} + 1) q^{2} + (\beta_{2} + 1) q^{3} + \beta_{2} q^{4} + (\beta_{2} + 1) q^{5} + \beta_{2} q^{6} + \beta_{3} q^{7} - q^{8} + \beta_{2} q^{9}+O(q^{10})$$ q + (b2 + 1) * q^2 + (b2 + 1) * q^3 + b2 * q^4 + (b2 + 1) * q^5 + b2 * q^6 + b3 * q^7 - q^8 + b2 * q^9 $$q + (\beta_{2} + 1) q^{2} + (\beta_{2} + 1) q^{3} + \beta_{2} q^{4} + (\beta_{2} + 1) q^{5} + \beta_{2} q^{6} + \beta_{3} q^{7} - q^{8} + \beta_{2} q^{9} + \beta_{2} q^{10} + (2 \beta_{3} - 1) q^{11} - q^{12} - \beta_1 q^{14} + \beta_{2} q^{15} + ( - \beta_{2} - 1) q^{16} + (3 \beta_{2} + \beta_1 + 3) q^{17} - q^{18} + (2 \beta_{2} + \beta_1 - 2) q^{19} - q^{20} - \beta_1 q^{21} + ( - \beta_{2} - 2 \beta_1 - 1) q^{22} + \beta_{2} q^{23} + ( - \beta_{2} - 1) q^{24} + \beta_{2} q^{25} - q^{27} + ( - \beta_{3} - \beta_1) q^{28} + ( - \beta_{3} - \beta_{2} - \beta_1) q^{29} - q^{30} + ( - 2 \beta_{3} + 2) q^{31} - \beta_{2} q^{32} + ( - \beta_{2} - 2 \beta_1 - 1) q^{33} + (\beta_{3} + 3 \beta_{2} + \beta_1) q^{34} - \beta_1 q^{35} + ( - \beta_{2} - 1) q^{36} + ( - 2 \beta_{3} + 3) q^{37} + (\beta_{3} - 2 \beta_{2} + \beta_1 - 4) q^{38} + ( - \beta_{2} - 1) q^{40} + (6 \beta_{2} + \beta_1 + 6) q^{41} + ( - \beta_{3} - \beta_1) q^{42} + ( - 6 \beta_{2} - 6) q^{43} + ( - 2 \beta_{3} - \beta_{2} - 2 \beta_1) q^{44} - q^{45} - q^{46} + (4 \beta_{3} + 2 \beta_{2} + 4 \beta_1) q^{47} - \beta_{2} q^{48} - q^{50} + (\beta_{3} + 3 \beta_{2} + \beta_1) q^{51} + ( - \beta_{3} - 2 \beta_{2} - \beta_1) q^{53} + ( - \beta_{2} - 1) q^{54} + ( - \beta_{2} - 2 \beta_1 - 1) q^{55} - \beta_{3} q^{56} + (\beta_{3} - 2 \beta_{2} + \beta_1 - 4) q^{57} + ( - \beta_{3} + 1) q^{58} - 2 \beta_1 q^{59} + ( - \beta_{2} - 1) q^{60} + (3 \beta_{3} - \beta_{2} + 3 \beta_1) q^{61} + (2 \beta_{2} + 2 \beta_1 + 2) q^{62} + ( - \beta_{3} - \beta_1) q^{63} + q^{64} + ( - 2 \beta_{3} - \beta_{2} - 2 \beta_1) q^{66} + ( - \beta_{3} - 7 \beta_{2} - \beta_1) q^{67} + (\beta_{3} - 3) q^{68} - q^{69} + ( - \beta_{3} - \beta_1) q^{70} + (11 \beta_{2} + \beta_1 + 11) q^{71} - \beta_{2} q^{72} + ( - 11 \beta_{2} - \beta_1 - 11) q^{73} + (3 \beta_{2} + 2 \beta_1 + 3) q^{74} - q^{75} + (\beta_{3} - 4 \beta_{2} - 2) q^{76} + ( - \beta_{3} + 14) q^{77} - 2 \beta_1 q^{79} - \beta_{2} q^{80} + ( - \beta_{2} - 1) q^{81} + (\beta_{3} + 6 \beta_{2} + \beta_1) q^{82} + ( - 4 \beta_{3} + 6) q^{83} - \beta_{3} q^{84} + (\beta_{3} + 3 \beta_{2} + \beta_1) q^{85} - 6 \beta_{2} q^{86} + ( - \beta_{3} + 1) q^{87} + ( - 2 \beta_{3} + 1) q^{88} + (5 \beta_{3} + 5 \beta_1) q^{89} + ( - \beta_{2} - 1) q^{90} + ( - \beta_{2} - 1) q^{92} + (2 \beta_{2} + 2 \beta_1 + 2) q^{93} + (4 \beta_{3} - 2) q^{94} + (\beta_{3} - 2 \beta_{2} + \beta_1 - 4) q^{95} + q^{96} + ( - \beta_{2} + \beta_1 - 1) q^{97} + ( - 2 \beta_{3} - \beta_{2} - 2 \beta_1) q^{99}+O(q^{100})$$ q + (b2 + 1) * q^2 + (b2 + 1) * q^3 + b2 * q^4 + (b2 + 1) * q^5 + b2 * q^6 + b3 * q^7 - q^8 + b2 * q^9 + b2 * q^10 + (2*b3 - 1) * q^11 - q^12 - b1 * q^14 + b2 * q^15 + (-b2 - 1) * q^16 + (3*b2 + b1 + 3) * q^17 - q^18 + (2*b2 + b1 - 2) * q^19 - q^20 - b1 * q^21 + (-b2 - 2*b1 - 1) * q^22 + b2 * q^23 + (-b2 - 1) * q^24 + b2 * q^25 - q^27 + (-b3 - b1) * q^28 + (-b3 - b2 - b1) * q^29 - q^30 + (-2*b3 + 2) * q^31 - b2 * q^32 + (-b2 - 2*b1 - 1) * q^33 + (b3 + 3*b2 + b1) * q^34 - b1 * q^35 + (-b2 - 1) * q^36 + (-2*b3 + 3) * q^37 + (b3 - 2*b2 + b1 - 4) * q^38 + (-b2 - 1) * q^40 + (6*b2 + b1 + 6) * q^41 + (-b3 - b1) * q^42 + (-6*b2 - 6) * q^43 + (-2*b3 - b2 - 2*b1) * q^44 - q^45 - q^46 + (4*b3 + 2*b2 + 4*b1) * q^47 - b2 * q^48 - q^50 + (b3 + 3*b2 + b1) * q^51 + (-b3 - 2*b2 - b1) * q^53 + (-b2 - 1) * q^54 + (-b2 - 2*b1 - 1) * q^55 - b3 * q^56 + (b3 - 2*b2 + b1 - 4) * q^57 + (-b3 + 1) * q^58 - 2*b1 * q^59 + (-b2 - 1) * q^60 + (3*b3 - b2 + 3*b1) * q^61 + (2*b2 + 2*b1 + 2) * q^62 + (-b3 - b1) * q^63 + q^64 + (-2*b3 - b2 - 2*b1) * q^66 + (-b3 - 7*b2 - b1) * q^67 + (b3 - 3) * q^68 - q^69 + (-b3 - b1) * q^70 + (11*b2 + b1 + 11) * q^71 - b2 * q^72 + (-11*b2 - b1 - 11) * q^73 + (3*b2 + 2*b1 + 3) * q^74 - q^75 + (b3 - 4*b2 - 2) * q^76 + (-b3 + 14) * q^77 - 2*b1 * q^79 - b2 * q^80 + (-b2 - 1) * q^81 + (b3 + 6*b2 + b1) * q^82 + (-4*b3 + 6) * q^83 - b3 * q^84 + (b3 + 3*b2 + b1) * q^85 - 6*b2 * q^86 + (-b3 + 1) * q^87 + (-2*b3 + 1) * q^88 + (5*b3 + 5*b1) * q^89 + (-b2 - 1) * q^90 + (-b2 - 1) * q^92 + (2*b2 + 2*b1 + 2) * q^93 + (4*b3 - 2) * q^94 + (b3 - 2*b2 + b1 - 4) * q^95 + q^96 + (-b2 + b1 - 1) * q^97 + (-2*b3 - b2 - 2*b1) * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$4 q + 2 q^{2} + 2 q^{3} - 2 q^{4} + 2 q^{5} - 2 q^{6} - 4 q^{8} - 2 q^{9}+O(q^{10})$$ 4 * q + 2 * q^2 + 2 * q^3 - 2 * q^4 + 2 * q^5 - 2 * q^6 - 4 * q^8 - 2 * q^9 $$4 q + 2 q^{2} + 2 q^{3} - 2 q^{4} + 2 q^{5} - 2 q^{6} - 4 q^{8} - 2 q^{9} - 2 q^{10} - 4 q^{11} - 4 q^{12} - 2 q^{15} - 2 q^{16} + 6 q^{17} - 4 q^{18} - 12 q^{19} - 4 q^{20} - 2 q^{22} - 2 q^{23} - 2 q^{24} - 2 q^{25} - 4 q^{27} + 2 q^{29} - 4 q^{30} + 8 q^{31} + 2 q^{32} - 2 q^{33} - 6 q^{34} - 2 q^{36} + 12 q^{37} - 12 q^{38} - 2 q^{40} + 12 q^{41} - 12 q^{43} + 2 q^{44} - 4 q^{45} - 4 q^{46} - 4 q^{47} + 2 q^{48} - 4 q^{50} - 6 q^{51} + 4 q^{53} - 2 q^{54} - 2 q^{55} - 12 q^{57} + 4 q^{58} - 2 q^{60} + 2 q^{61} + 4 q^{62} + 4 q^{64} + 2 q^{66} + 14 q^{67} - 12 q^{68} - 4 q^{69} + 22 q^{71} + 2 q^{72} - 22 q^{73} + 6 q^{74} - 4 q^{75} + 56 q^{77} + 2 q^{80} - 2 q^{81} - 12 q^{82} + 24 q^{83} - 6 q^{85} + 12 q^{86} + 4 q^{87} + 4 q^{88} - 2 q^{90} - 2 q^{92} + 4 q^{93} - 8 q^{94} - 12 q^{95} + 4 q^{96} - 2 q^{97} + 2 q^{99}+O(q^{100})$$ 4 * q + 2 * q^2 + 2 * q^3 - 2 * q^4 + 2 * q^5 - 2 * q^6 - 4 * q^8 - 2 * q^9 - 2 * q^10 - 4 * q^11 - 4 * q^12 - 2 * q^15 - 2 * q^16 + 6 * q^17 - 4 * q^18 - 12 * q^19 - 4 * q^20 - 2 * q^22 - 2 * q^23 - 2 * q^24 - 2 * q^25 - 4 * q^27 + 2 * q^29 - 4 * q^30 + 8 * q^31 + 2 * q^32 - 2 * q^33 - 6 * q^34 - 2 * q^36 + 12 * q^37 - 12 * q^38 - 2 * q^40 + 12 * q^41 - 12 * q^43 + 2 * q^44 - 4 * q^45 - 4 * q^46 - 4 * q^47 + 2 * q^48 - 4 * q^50 - 6 * q^51 + 4 * q^53 - 2 * q^54 - 2 * q^55 - 12 * q^57 + 4 * q^58 - 2 * q^60 + 2 * q^61 + 4 * q^62 + 4 * q^64 + 2 * q^66 + 14 * q^67 - 12 * q^68 - 4 * q^69 + 22 * q^71 + 2 * q^72 - 22 * q^73 + 6 * q^74 - 4 * q^75 + 56 * q^77 + 2 * q^80 - 2 * q^81 - 12 * q^82 + 24 * q^83 - 6 * q^85 + 12 * q^86 + 4 * q^87 + 4 * q^88 - 2 * q^90 - 2 * q^92 + 4 * q^93 - 8 * q^94 - 12 * q^95 + 4 * q^96 - 2 * q^97 + 2 * q^99
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{4} + 7x^{2} + 49$$ :
$$\beta_{1}$$ $$=$$ $$\nu$$ v $$\beta_{2}$$ $$=$$ $$( \nu^{2} ) / 7$$ (v^2) / 7 $$\beta_{3}$$ $$=$$ $$( \nu^{3} ) / 7$$ (v^3) / 7
$$\nu$$ $$=$$ $$\beta_1$$ b1 $$\nu^{2}$$ $$=$$ $$7\beta_{2}$$ 7*b2 $$\nu^{3}$$ $$=$$ $$7\beta_{3}$$ 7*b3
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/570\mathbb{Z}\right)^\times$$.
$$n$$ $$191$$ $$211$$ $$457$$ $$\chi(n)$$ $$1$$ $$\beta_{2}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
121.1
1.32288 + 2.29129i −1.32288 − 2.29129i 1.32288 − 2.29129i −1.32288 + 2.29129i
0.500000 + 0.866025i 0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i −0.500000 + 0.866025i −2.64575 −1.00000 −0.500000 + 0.866025i −0.500000 + 0.866025i
121.2 0.500000 + 0.866025i 0.500000 + 0.866025i −0.500000 + 0.866025i 0.500000 + 0.866025i −0.500000 + 0.866025i 2.64575 −1.00000 −0.500000 + 0.866025i −0.500000 + 0.866025i
391.1 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i −2.64575 −1.00000 −0.500000 0.866025i −0.500000 0.866025i
391.2 0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i 0.500000 0.866025i −0.500000 0.866025i 2.64575 −1.00000 −0.500000 0.866025i −0.500000 0.866025i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
19.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 570.2.i.i 4
3.b odd 2 1 1710.2.l.j 4
19.c even 3 1 inner 570.2.i.i 4
57.h odd 6 1 1710.2.l.j 4
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
570.2.i.i 4 1.a even 1 1 trivial
570.2.i.i 4 19.c even 3 1 inner
1710.2.l.j 4 3.b odd 2 1
1710.2.l.j 4 57.h odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(570, [\chi])$$:
$$T_{7}^{2} - 7$$ T7^2 - 7 $$T_{11}^{2} + 2T_{11} - 27$$ T11^2 + 2*T11 - 27
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$(T^{2} - T + 1)^{2}$$
$3$ $$(T^{2} - T + 1)^{2}$$
$5$ $$(T^{2} - T + 1)^{2}$$
$7$ $$(T^{2} - 7)^{2}$$
$11$ $$(T^{2} + 2 T - 27)^{2}$$
$13$ $$T^{4}$$
$17$ $$T^{4} - 6 T^{3} + 34 T^{2} - 12 T + 4$$
$19$ $$T^{4} + 12 T^{3} + 67 T^{2} + \cdots + 361$$
$23$ $$(T^{2} + T + 1)^{2}$$
$29$ $$T^{4} - 2 T^{3} + 10 T^{2} + 12 T + 36$$
$31$ $$(T^{2} - 4 T - 24)^{2}$$
$37$ $$(T^{2} - 6 T - 19)^{2}$$
$41$ $$T^{4} - 12 T^{3} + 115 T^{2} + \cdots + 841$$
$43$ $$(T^{2} + 6 T + 36)^{2}$$
$47$ $$T^{4} + 4 T^{3} + 124 T^{2} + \cdots + 11664$$
$53$ $$T^{4} - 4 T^{3} + 19 T^{2} + 12 T + 9$$
$59$ $$T^{4} + 28T^{2} + 784$$
$61$ $$T^{4} - 2 T^{3} + 66 T^{2} + \cdots + 3844$$
$67$ $$T^{4} - 14 T^{3} + 154 T^{2} + \cdots + 1764$$
$71$ $$T^{4} - 22 T^{3} + 370 T^{2} + \cdots + 12996$$
$73$ $$T^{4} + 22 T^{3} + 370 T^{2} + \cdots + 12996$$
$79$ $$T^{4} + 28T^{2} + 784$$
$83$ $$(T^{2} - 12 T - 76)^{2}$$
$89$ $$T^{4} + 175 T^{2} + 30625$$
$97$ $$T^{4} + 2 T^{3} + 10 T^{2} - 12 T + 36$$
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• Step 1 of 5
Strategy: In balancing nuclear equations, note that the sum of atomic numbers and that of mass numbers must match on both sides of the equation.
Solution: a. On the left side of this equation the atomic number sum is 13 (12 + 1) and the mass number sum is 27 (26 + 1). These sums must be the same on the right side. Remember that the atomic and mass numbers of an alpha particle are 2 and 4, respectively. The atomic number of X is therefore 11 (13 - 2) and the mass number is 23 (27 - 4). X is sodium-23 (Na).
• Step 2 of 5
b. On the left side of this equation the atomic number sum is 28 (27 + 1) and the mass number sum is 61 (59 + 2). These sums must be the same on the right side. The atomic number of X is therefore 1 (28 - 27) and the mass number is also 1 (61 - 60). X is a proton (p or H).
• Step 3 of 5
c. On the left side of this equation the atomic number sum is 92 (92 + 0) and the mass number sum is 236 (235 + 1). These sums must be the same on the right side. The atomic number of X is therefore 0 and the mass number is 1 . X is a neutron (n).
• Step 4 of 5
d. On the left side of this equation the atomic number sum is 26 (24 + 2) and the mass number sum is 57 (53 + 4). These sums must be the same on the right side. The atomic number of X is therefore 26 (26 - 0) and the mass number is 56 (57 - 1). X is iron-56 (Fe).
• Step 5 of 5
e. On the left side of this equation the atomic number sum is 8 and the mass number sum is 20. The sums must be the same on the right side. The atomic number of X is therefore -1 (8 - 9) and the mass number is 0 (20 - 20). X is a β particle (β).
Corresponding Textbook
Problem Solving Workbook with Selected Solutions for Chemistry Atoms First | 1st Edition
9780077385767ISBN-13: 0077385764ISBN: Authors:
This is an alternate ISBN. View the primary ISBN for: Student Solutions Manual to accompany Chemistry 1st Edition Textbook Solutions
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HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
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# Accumulator contents of 8085 are B7H and register B contents are A5H.What will be the effect of following instructions on the. contents of Accumulator, when executed independently ? - HSC Science (Computer Science) 12th Board Exam - Computer Science 2
#### Question
Accumulator contents of 8085 are B7H and register B contents are A5H.What will be the effect of following instructions on the. contents of Accumulator, when executed independently ?
(ii) CMP B
(iii) CMA
(iv) XRA B
(v) ORA B
#### Solution
The accumulator value is B7H and B register value is A5H
Before execution [A] = B7H = 1 0 1 1 0 1 1 1
After execution [A] = 1 0 1 1 1 1 0 0 [1 Mark]
(ii) CMP B
Compare B with accumulator
Before execution [A] = B7H = 1 0 1 1 0 1 1 1
Before execution [B] = A5H = 1 0 1 0 0 1 0 1
Instruction - CMP B
Condition:
(a) If [A] < [B] then Carry flag is set.
(b) If [A] = [B] then Zero flag is set.
(c) If [A] > [B] then both Carry and Zero flags are reset. [1 Mark]
(iii) CMA
Complement the accumulator
Befor execution [A] = B7H = 1 0 1 1 0 1 1 1
Instruction - CMA
After execution : 48H = 0 1 0 0 1 0 0 0 [1 Mark]
(iv) XRA B
Exclusive OR with accumulator
Before execution [A] = B7H = 1 0 1 1 0 1 1 1
Before execution [B] = A5H = 1 0 1 0 0 1 0 1
After execution : 12H = 0 0 0 1 0 0 1 0
(v) ORA B
Logical OR with B
Before execution [A] = B7H = 1 0 1 1 0 1 1 1
Before execution [B] = A5H = 1 0 1 0 0 1 0 1
After execution : B7H = 1 0 1 1 0 1 1 1
Is there an error in this question or solution?
#### APPEARS IN
2015-2016 (March) (with solutions)
Question 5.2.2 | 5.00 marks
Solution Accumulator contents of 8085 are B7H and register B contents are A5H.What will be the effect of following instructions on the. contents of Accumulator, when executed independently ? Concept: Instruction Set and Programming of 8085.
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Seminar Calendar
for events the day of Wednesday, April 12, 2017.
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Wednesday, April 12, 2017
IGL/AWM Seminar
3:00 pm in 245 Altgeld Hall, Wednesday, April 12, 2017
Del Edit Copy
Submitted by tyson.
Loredana Lanzani (Syracuse University)Practical Applications of Complex AnalysisAbstract: The notion of conformal mapping is of fundamental importance in complex analysis. Conformal maps are used by mathematicians, physicists and engineers to change regions with complicated shapes into much simpler ones, and to do so in a way that preserves shape on a small scale (that is, when viewed up close). This makes it possible to transpose’ a problem that was formulated for the complicated-looking region into another, related problem for the simpler region (where it can be easily solved) -- then one uses conformal mapping to translate' the solution of the problem over the simpler region, back to a solution of the original problem (over the complicated region). The beauty of conformal mapping is that its governing principle is based on a very simple idea that is easy to explain and to understand (much like the statement of Fermat's celebrated last theorem). In the first part of this talk I will introduce the notion of conformal mapping and will briefly go over its basic properties and some of its history (including a historical mystery going back to Galileo Galilei). I will then describe some of the many real-life applications of conformal maps, including: cartography; airplane wing design (transonic flow); art (in particular, the so-called `Droste effect’ in the work of M. C. Escher). Time permitting, I will conclude by highlighting a 2013 paper by MacArthur fellow L. Mahadevan that uses the related notion of quasi-conformal mapping to link D'Arcy Thompson's iconic work On Shape and Growth (published in 1917) with modern morphometric analysis (a discipline in biology that studies, among other things, how living organisms evolve over time). No previous knowledge of complex analysis is needed to enjoy this talk.
Doob Colloquium
3:00 pm in 243 Altgeld Hall, Wednesday, April 12, 2017
Del Edit Copy
Submitted by lescobar.
Erik Walsberg (UIUC)First order logic and Sub-riemannian spheres.Abstract: I will discuss a connection between sub-riemannian geometry and first order model theory. Researchers in sub-riemannian geometry have essentially been working on the following: are sub-riemannian spheres definable in an o-minimal expansion of the ordered field of real numbers? (O-minimality is an important and popular topic in model theory developed in large part by UIUC's own Lou van den Dries). It also seems that some model theorists have been trying to construct the kind of structure that the geometers are looking for. As far as I can tell neither side was really aware of what the other was doing until now. I will try to explain some of this. No knowledge of logic or sub-riemannian geometry is necessary.
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## Prep Test 49 Section 1, second LG
Prepare for the LSAT or discuss it with others in this forum.
ahindoy
Posts: 4
Joined: Sun Apr 24, 2011 1:03 am
### Prep Test 49 Section 1, second LG
In this game you are given the scenario where 5 pieces of mail (letter, magazine, postcard, survey, flyer) need to be distributed to 3 roommates (G, J, R). I cannot figure out the necessary deduction.
Rule 1-G does not receive letter or magazine
Rule 2- If Letter to R, then Postcard to J
Rule 3-Whoever receives the flyer also receives at least one other piece of mail
The way I set up my diagram was the pieces of mail across the top and the roommates off to the side forming a 3x5 grid
Can anyone please describe the deduction I am not seeing, I scored well on all the other games but failed miserably on this one.
EarlCat
Posts: 606
Joined: Mon Mar 12, 2007 4:04 pm
### Re: Prep Test 49 Section 1, second LG
There are lots of deductions. You're going to have to be more specific about what questions are tripping you up. That being said, I'll try to get you started.
First, a 3x5 grid is not a good setup in this game (and in most games) because, for instance, it doesn't lend itself easily representing rule 3.
A better setup, IMHO, is to make a chart with the roommates, G J R, and put under (or over) those labels the pieces of mail. When you find out a particular roommate gets the flier, you mark it with a slot (F __) so that you know another piece of mail has to go with it.
Games like this tend to require that you focus on the distribution of elements--how many letters go to each person. Here you have 5 pieces of mail going to 3 people. Thus, assuming everyone gets a piece of mail (I think I remember that being in the setup) at least one of them will get only 1 piece of mail. Either the mail goes out 2 2 1 or 3 1 1. (Notice in the 3 1 1 scenario, the guy with 3 has to have the flier.)
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Users browsing this forum: No registered users and 9 guests
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# 1075982 (number)
1,075,982 (one million seventy-five thousand nine hundred eighty-two) is an even seven-digits composite number following 1075981 and preceding 1075983. In scientific notation, it is written as 1.075982 × 106. The sum of its digits is 32. It has a total of 2 prime factors and 4 positive divisors. There are 537,990 positive integers (up to 1075982) that are relatively prime to 1075982.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 7
• Sum of Digits 32
• Digital Root 5
## Name
Short name 1 million 75 thousand 982 one million seventy-five thousand nine hundred eighty-two
## Notation
Scientific notation 1.075982 × 106 1.075982 × 106
## Prime Factorization of 1075982
Prime Factorization 2 × 537991
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1075982 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,075,982 is 2 × 537991. Since it has a total of 2 prime factors, 1,075,982 is a composite number.
## Divisors of 1075982
4 divisors
Even divisors 2 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.61398e+06 Sum of all the positive divisors of n s(n) 537994 Sum of the proper positive divisors of n A(n) 403494 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1037.3 Returns the nth root of the product of n divisors H(n) 2.66666 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,075,982 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 1,075,982) is 1,613,976, the average is 403,494.
## Other Arithmetic Functions (n = 1075982)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 537990 Total number of positive integers not greater than n that are coprime to n λ(n) 537990 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 83816 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 537,990 positive integers (less than 1,075,982) that are coprime with 1,075,982. And there are approximately 83,816 prime numbers less than or equal to 1,075,982.
## Divisibility of 1075982
m n mod m 2 3 4 5 6 7 8 9 0 2 2 2 2 5 6 5
The number 1,075,982 is divisible by 2.
## Classification of 1075982
• Arithmetic
• Semiprime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
## Base conversion (1075982)
Base System Value
2 Binary 100000110101100001110
3 Ternary 2000122222012
4 Quaternary 10012230032
5 Quinary 233412412
6 Senary 35021222
8 Octal 4065416
10 Decimal 1075982
12 Duodecimal 43a812
20 Vigesimal 6e9j2
36 Base36 n28e
## Basic calculations (n = 1075982)
### Multiplication
n×y
n×2 2151964 3227946 4303928 5379910
### Division
n÷y
n÷2 537991 358661 268996 215196
### Exponentiation
ny
n2 1157737264324 1245704457141866168 1340355573204419443176976 1442198470367637641308448990432
### Nth Root
y√n
2√n 1037.3 102.471 32.2071 16.0828
## 1075982 as geometric shapes
### Circle
Diameter 2.15196e+06 6.76059e+06 3.63714e+12
### Sphere
Volume 5.21799e+18 1.45486e+13 6.76059e+06
### Square
Length = n
Perimeter 4.30393e+06 1.15774e+12 1.52167e+06
### Cube
Length = n
Surface area 6.94642e+12 1.2457e+18 1.86366e+06
### Equilateral Triangle
Length = n
Perimeter 3.22795e+06 5.01315e+11 931828
### Triangular Pyramid
Length = n
Surface area 2.00526e+12 1.46808e+17 878536
## Cryptographic Hash Functions
md5 fe0250e44261bb4ec135b0aee85d098c 51f4c7e9a98bfa33d07e58ccf1386b6f65bcb435 832b759358743bf55e5ad34115d98e6b978bfa3ef1a9c5313dac6896a8514d26 fd6e8380148c8b9ec64af54a7e6baf9c805e871a20af4c9cd2d9d9632425580ef5f223b8be5c4c5b685e9bd386cb42f32529ccf01411723c2ac18c2242b18444 3165b51bc625b5d0fa6df77580f4e8444b3b7ff7
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# LabVIEW
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## U16 TO STRING AND STRING TO U16 CONVERSION
Hello
I am able to convert a U16 array to String successfully, but unable to get the other way round. i.e. String to U16 conversion
The attached code works good for string length if its even number, but for odd length the U16 array created is incomplete
Kindly guide.
Message 1 of 9
(2,153 Views)
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
Your array input translates to (in \code to show unprintable characters):
"HELLO\00\00\00\00\00\00\00\00\00\00\00\00\00\00\00"
Your string input is HELLO (i.e. 5 characters.) Obviously not the same!
(If you have an odd number of characters, you also need to decide to pad or truncate)
Message 2 of 9
(2,123 Views)
Solution
Accepted by vihang
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
@vihang wrote:
The attached code works good for string length if its even number, but for odd length the U16 array created is incomplete
So you need to detect if there are an odd number of characters in the string. If there is, append a NULL character (0x00).
I would also recommend using the Flatten To String and Unflatten From String to convert the data. There are inputs for the Endianess that you can set to "little-endian" and therefore avoid the need for "swap bytes".
There are only two ways to tell somebody thanks: Kudos and Marked Solutions
Unofficial Forum Rules and Guidelines
"Not that we are sufficient in ourselves to claim anything as coming from us, but our sufficiency is from God" - 2 Corinthians 3:5
Message 3 of 9
(2,122 Views)
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
If you are using a U16 for a string, each array entry is going to represent 2 characters. Since "HELLO" is an odd number of characters it seems like the swap bytes doesn't know what to do with only half of a U16 and doesn't return any data (which seems a bit odd to me). Any even length string seems to work in your example.
Anyway you need a way to make sure you have enough data to create a U16. The attached works, but I'm sure there are other ways that may be better to do this.
Message 4 of 9
(2,121 Views)
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
I would recommend "(un)flatten from/to string". It can include the endian conversion and even gives you a rudimentary error if things go wrong:
Message 5 of 9
(2,114 Views)
Solution
Accepted by vihang
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
This is all you need:
"If you weren't supposed to push it, it wouldn't be a button."
Message 6 of 9
(2,073 Views)
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
Thanks
This worked for me
Message 7 of 9
(2,047 Views)
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
Thanks
This is superb!!!
Message 8 of 9
(2,043 Views)
## Re: U16 TO STRING AND STRING TO U16 CONVERSION
@vihang wrote:
Thanks
This worked for me
To pad to the next higher even size, you don't need any case structures.
Here's what I would do (note that bitwise operations are cheaper that Q&R, for example):
Message 9 of 9
(2,007 Views)
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# Mixed Numbers
## A self marking quiz requiring application of the four operations applied to mixed numbers
##### MenuConvertLevel 1Level 2Level 3Level 4Level 5Level 6ExamHelpMore
This is level 5: four rules with fractions and whole numbers. Give your answer as a mixed number with the fraction part in its lowest terms.
Type the whole number, then a space, then the numerator, forward slash, denominator.
1 $$8 + 3\frac{8}{9}$$ ☐= ☐ ✓ ✗ 2 $$8 - 3\frac{6}{7}$$ ☐= ☐ ✓ ✗ 3 $$6 \times \frac{5}{7}$$ ☐= ☐ ✓ ✗ 4 $$4 \div \frac{3}{5}$$ ☐= ☐ ✓ ✗ 5 $$7 \times 5\frac{3}{8}$$ ☐= ☐ ✓ ✗ 6 $$10\frac{7}{12} \div 6$$ ☐= ☐ ✓ ✗ 7 $$5\frac{1}{3} - 7 + 8\frac{2}{5}$$ ☐= ☐ ✓ ✗ 8 $$2\frac{3}{7} \times 3 \div 1\frac{1}{4}$$ ☐= ☐ ✓ ✗ 9 $$29 - 6\frac{2}{3} \times 8 \div 2\frac{3}{5}$$ ☐= ☐ ✓ ✗
Check
This is Mixed Numbers level 5. You can also try:
Level 1 Level 2 Level 3 Level 4 Level 6
## Instructions
Try your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help.
When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file.
## More Activities:
Mathematicians are not the people who find Maths easy; they are the people who enjoy how mystifying, puzzling and hard it is. Are you a mathematician?
Comment recorded on the 3 October 'Starter of the Day' page by Mrs Johnstone, 7Je:
"I think this is a brilliant website as all the students enjoy doing the puzzles and it is a brilliant way to start a lesson."
Comment recorded on the 1 February 'Starter of the Day' page by M Chant, Chase Lane School Harwich:
"My year five children look forward to their daily challenge and enjoy the problems as much as I do. A great resource - thanks a million."
Each month a newsletter is published containing details of the new additions to the Transum website and a new puzzle of the month.
The newsletter is then duplicated as a podcast which is available on the major delivery networks. You can listen to the podcast while you are commuting, exercising or relaxing.
Transum breaking news is available on Twitter @Transum and if that's not enough there is also a Transum Facebook page.
#### Beat The Clock
It is a race against the clock to answer 30 mental arithmetic questions. There are nine levels to choose from to suit pupils of different abilities.
There are answers to this exercise but they are available in this space to teachers, tutors and parents who have logged in to their Transum subscription on this computer.
A Transum subscription unlocks the answers to the online exercises, quizzes and puzzles. It also provides the teacher with access to quality external links on each of the Transum Topic pages and the facility to add to the collection themselves.
Subscribers can manage class lists, lesson plans and assessment data in the Class Admin application and have access to reports of the Transum Trophies earned by class members.
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## Go Maths
Learning and understanding Mathematics, at every level, requires learner engagement. Mathematics is not a spectator sport. Sometimes traditional teaching fails to actively involve students. One way to address the problem is through the use of interactive activities and this web site provides many of those. The Go Maths page is an alphabetical list of free activities designed for students in Secondary/High school.
## Maths Map
Are you looking for something specific? An exercise to supplement the topic you are studying at school at the moment perhaps. Navigate using our Maths Map to find exercises, puzzles and Maths lesson starters grouped by topic.
## Teachers
If you found this activity useful don't forget to record it in your scheme of work or learning management system. The short URL, ready to be copied and pasted, is as follows:
Alternatively, if you use Google Classroom, all you have to do is click on the green icon below in order to add this activity to one of your classes.
It may be worth remembering that if Transum.org should go offline for whatever reason, there is a mirror site at Transum.info that contains most of the resources that are available here on Transum.org.
When planning to use technology in your lesson always have a plan B!
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
For Students:
For All:
Scan the QR code below to visit the online version of this activity.
https://www.Transum.org/go/?Num=340
## Description of Levels
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Convert - Practise converting to and from improper fractions before begining these exercises.
Level 1 - Addition of mixed numbers.
Level 2 - Subtraction of mixed numbers.
Level 3 - Multiplication of mixed numbers.
Level 4 - Division of mixed numbers.
Level 5 - Four rules with fractions and whole numbers.
Level 6 - A variety mixed number questions with words and diagrams!
Exam Style Questions - A collection of problems in the style of GCSE or IB/A-level exam paper questions (worked solutions are available for Transum subscribers).
More on this topic including lesson Starters, visual aids, investigations and self-marking exercises.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
## Curriculum Reference
See the National Curriculum page for links to related online activities and resources.
## Example
You may also want to use a calculator to check your working. See Calculator Workout skill 10.
Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
### Typing Mathematical Notation
These exercises use MathQuill, a web formula editor designed to make typing Maths easy and beautiful. Watch the animation below to see how common mathematical notation can be created using your keyboard.
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