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# A quadrilateral is inscribed in a circle. If the

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Updated on: 02 Feb 2012, 21:44
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A quadrilateral is inscribed in a circle. If the circumference of the circle is 5(pi), what is the area of the quadrilateral?

(1) The length of one of the diagonals of the quadrilateral is 5.
(2) The length of two opposite sides are 3 and 4 respectively.

Originally posted by Smita04 on 02 Feb 2012, 06:30.
Last edited by Smita04 on 02 Feb 2012, 21:44, edited 1 time in total.
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02 Feb 2012, 09:00
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I think the answer should be C. This one is difficult to visualize without drawing. We know the diameter of the circle is 5 and quadrilateral is inscribed

1. Since diagonal is 5, that means the quadrilateral is divided into two triangles with diagonal as the diameter of the circle. Now since they are inscribed. that means the angle both of these form with the Circumference is a right angle. But we do not know the sizes of the sides. Insufficient

2. I can draw various quadrilaterals which are inscribed and have opposite sides as 3 and 4. All of these will have different areas so insufficient.

Now combined I know that one side of one right angled triangle is 3 and of the other right angled triangle is 4 so the remaining two sides will be in the same ratio as well from the pythagorean theorem. Now that we know all 4 sides and the heights (since diagonal is 5 and is the hypotenuse of the right-angled triangle for both triangles that form the quadrilateral. Hence answer should be C.

Whats the OA?
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27 May 2017, 19:46
would someone please elaborate on why (2) is NS? I'm having trouble visualizing it. I thought the answer was B b/c we know the radius = 2.5
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27 May 2017, 22:36
omerrauf wrote:
I think the answer should be C. This one is difficult to visualize without drawing. We know the diameter of the circle is 5 and quadrilateral is inscribed

1. Since diagonal is 5, that means the quadrilateral is divided into two triangles with diagonal as the diameter of the circle. Now since they are inscribed. that means the angle both of these form with the Circumference is a right angle. But we do not know the sizes of the sides. Insufficient

2. I can draw various quadrilaterals which are inscribed and have opposite sides as 3 and 4. All of these will have different areas so insufficient.

Now combined I know that one side of one right angled triangle is 3 and of the other right angled triangle is 4 so the remaining two sides will be in the same ratio as well from the pythagorean theorem. Now that we know all 4 sides and the heights (since diagonal is 5 and is the hypotenuse of the right-angled triangle for both triangles that form the quadrilateral. Hence answer should be C.

Whats the OA?

The question stem states that the circumference is 5 and not the diameter.
Does that have any impact on the reasoning?
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27 May 2017, 22:51
I believe it says the circumference is 5(pie), meaning 2(pie)r = 5(pie) --> radius = 2.5 / diameter = 5

nainy05 wrote:
omerrauf wrote:
I think the answer should be C. This one is difficult to visualize without drawing. We know the diameter of the circle is 5 and quadrilateral is inscribed

1. Since diagonal is 5, that means the quadrilateral is divided into two triangles with diagonal as the diameter of the circle. Now since they are inscribed. that means the angle both of these form with the Circumference is a right angle. But we do not know the sizes of the sides. Insufficient

2. I can draw various quadrilaterals which are inscribed and have opposite sides as 3 and 4. All of these will have different areas so insufficient.

Now combined I know that one side of one right angled triangle is 3 and of the other right angled triangle is 4 so the remaining two sides will be in the same ratio as well from the pythagorean theorem. Now that we know all 4 sides and the heights (since diagonal is 5 and is the hypotenuse of the right-angled triangle for both triangles that form the quadrilateral. Hence answer should be C.

Whats the OA?

The question stem states that the circumference is 5 and not the diameter.
Does that have any impact on the reasoning?
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27 May 2017, 23:01
Apologies for the overlook. Thanks for pointing out!
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13 Jun 2017, 03:20
Why can't B be the answer?

It is given that circumference is 5 pi... that means the diameter is 5.. thus from B we get 2 triangles with measures 3, 4 and 5... Thus area of quad = Area of both triangles = 2 (1/2*3*4) = 12..
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02 Jul 2017, 03:03
Why can't B be the answer?

It is given that circumference is 5 pi... that means the diameter is 5.. thus from B we get 2 triangles with measures 3, 4 and 5... Thus area of quad = Area of both triangles = 2 (1/2*3*4) = 12..

Nothing is mentioned about the other two sides. What is this is a trapezium?

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02 Jul 2017, 03:06
omerrauf wrote:
I think the answer should be C. This one is difficult to visualize without drawing. We know the diameter of the circle is 5 and quadrilateral is inscribed

1. Since diagonal is 5, that means the quadrilateral is divided into two triangles with diagonal as the diameter of the circle. Now since they are inscribed. that means the angle both of these form with the Circumference is a right angle. But we do not know the sizes of the sides. Insufficient

2. I can draw various quadrilaterals which are inscribed and have opposite sides as 3 and 4. All of these will have different areas so insufficient.

Now combined I know that one side of one right angled triangle is 3 and of the other right angled triangle is 4 so the remaining two sides will be in the same ratio as well from the pythagorean theorem. Now that we know all 4 sides and the heights (since diagonal is 5 and is the hypotenuse of the right-angled triangle for both triangles that form the quadrilateral. Hence answer should be C.

Whats the OA?

So, if it is a right triangle with hypotenuse 5 and the height (drawn from the right angle to hypotenuse) being 2.5 (=radius)
Area of triangle = (1/2) *base* height = (1/2) * 5*2.5

Area of quadilateral = 2*Area of triangle
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02 Jul 2017, 09:57
Smita04 wrote:
A quadrilateral is inscribed in a circle. If the circumference of the circle is 5(pi), what is the area of the quadrilateral?

(1) The length of one of the diagonals of the quadrilateral is 5.
(2) The length of two opposite sides are 3 and 4 respectively.

if diagonal = diameter, quadrilateral is a rectangle with hypotenuse=5 ==> 3-4-5 right triangle?? so sides = 3 &4 --> Area can be calculated?
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03 Jul 2017, 05:22
Smita04 wrote:
A quadrilateral is inscribed in a circle. If the circumference of the circle is 5(pi), what is the area of the quadrilateral?

(1) The length of one of the diagonals of the quadrilateral is 5.
(2) The length of two opposite sides are 3 and 4 respectively.

The correct answer is C here although initially I thought its B. Here's why C is correct -

WKT $$2\pi$$R = $$5\pi$$
Hence diameter = 5

Fact 1: One of the diagonals is 5 - Clearly insuff [we only know that one of the diagonal is diameter]
Fact 2: Length of two opp side is 3 and 4 - We cannot determine the area only with this info since we dont know what other two opp sides measure, accordingly the area will change. Insuff

Fact 1 + Fact 2 : WKT that the diagonal measures 5 [remember the property that the triangle formed with one of the sides as diameter is a right angled triangle]. So we will have two right angled triangle with 3,4,5 measurement. Hence suff

Hope this helps!
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03 Jul 2017, 08:38
This question seems incorrect...
Statement 2 reads : The length of two opposite sides are 3 and 4 respectively.
shouldn't this be changed to the length of 2 adjacent sides ??
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03 Jul 2017, 08:50
darn wrote:
This question seems incorrect...
Statement 2 reads : The length of two opposite sides are 3 and 4 respectively.
shouldn't this be changed to the length of 2 adjacent sides ??

No, the question is correct.
Fact 2 does mean two opp sides and not adjacent. Even it were adjacent sides the data is still insuff without fact 1, which states one of diagonals is 5 [this would be the diameter] and only when we know that the diagonal is diameter we can deduce that the triangle is right angled
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03 Jul 2017, 10:53
Sash143 I made the silly mistake of assuming that the quadrilateral formed was a rectangle. Which is why i asked the question, since opposite sides of a rectangle cannot be unequal.
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03 Jul 2017, 13:52
Smita04 wrote:
A quadrilateral is inscribed in a circle. If the circumference of the circle is 5(pi), what is the area of the quadrilateral?

(1) The length of one of the diagonals of the quadrilateral is 5.
(2) The length of two opposite sides are 3 and 4 respectively.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

A quadrilateral is inscribed in a circle. If the circumference of the circle is $$5\pi$$, what is the area of the quadrilateral?

(1) The length of one of the diagonals of the quadrilateral is $$5$$.
(2) The length of two opposite sides are $$3$$ and $$4$$ respectively.

In order to identify a quadrilateral on a specific circle, we need the lengths of 3 sides. It means we have 3 variables and 0 equation. However the first condition has 1 equation and the second condition has 2 equations. Thus when we consider both conditions together, they are sufficient.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.
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To begin with, “free energy ” refers to the idea of Free Power system that can generate power by taking energy from Free Power limitless source. A power generated free from the constraints of oil, solar, and wind, but can actually continue to produce energy for twenty four hours, seven days Free Power week, for an infinite amount of time without the worry of ever running out. “Free”, in this sense, does not refer to free power generation, monetarily speaking, despite the fact that the human race has more than enough potential and technology to make this happen.

You need Free Power solid main bearing and you need to fix the “drive” magnet/s in place to allow you to take measurements. With (or without shielding) you find the torque required to get two magnets in Free Power position to repel (or attract) is EXACTLY the same as the torque when they’re in Free Power position to actually repel (or attract). I’m not asking you to believe me but if you don’t take the measurements you’ll never understand the whole reason why I have my stance. Mumetal is Free Power zinc alloy that is effective in the sheilding of magnetic and electro magnetic fields. Only just heard about it myself couple of days ago. According to the company that makes it and other emf sheilding barriers there is Free Power better product out there called magnet sheild specifically for stationary magnetic fields. Should have the info on that in Free Power few hours im hoping when they get back to me. Hey Free Power, believe me i am not giving up. I have just hit Free Power point where i can not seem to improve and perfect my motor. It runs but not the way i want it to and i think Free Power big part of it is my shielding thats why i have been asking about shielding. I have never heard of mumetal. What is it? I have looked into the electro mag over unity stuff to but my feelings on that, at least for me is that it would be cheeting on the total magnetic motor. Your basicaly going back to the electric motor. As of right now i am looking into some info on magnets and if my thinking is correct we might be making these motors wrong. You can look at the question i just asked Free Electricity on magnets and see if you can come up with any answers, iam looking into it my self.

#### Free Energy The type of magnet (natural or man-made) is not the issue. Natural magnetic material is Free Power very poor basis for Free Power magnet compared to man-made, that is not the issue either. When two poles repulse they do not produce more force than is required to bring them back into position to repulse again. Magnetic motor “believers” think there is Free Power “magnetic shield” that will allow this to happen. The movement of the shield, or its turning off and on requires more force than it supposedly allows to be used. Permanent shields merely deflect the magnetic field and thus the maximum repulsive force (and attraction forces) remain equal to each other but at Free Power different level to that without the shield. Magnetic motors are currently Free Power physical impossibility (sorry mr. Free Electricity for fighting against you so vehemently earlier).

Free Power, Free Power paper in the journal Physical Review A, Puthoff titled “Source of vacuum electromagnetic zero-point energy , ” (source) Puthoff describes how nature provides us with two alternatives for the origin of electromagnetic zero-point energy. One of them is generation by the quantum fluctuation motion of charged particles that constitute matter. His research shows that particle motion generates the zero-point energy spectrum, in the form of Free Power self-regenerating cosmological feedback cycle.
Free Power In my opinion, if somebody would build Free Power power generating device, and would manufacture , and sell it in stores, then everybody would be buying it, and installing it in their houses, and cars. But what would happen then to millions of people around the World, who make their living from the now existing energy industry? I think if something like that would happen, the World would be in chaos. I have one more question. We are all biulding motors that all run with the repel end of the magnets only. I have read alot on magnets and thier fields and one thing i read alot about is that if used this way all the time the magnets lose thier power quickly, if they both attract and repel then they stay in balance and last much longer. My question is in repel mode how long will they last? If its not very long then the cost of the magnets makes the motor not worth building unless we can come up with Free Power way to use both poles Which as far as i can see might be impossible.
We can make the following conclusions about when processes will have Free Power negative \Delta \text G_\text{system}ΔGsystem: \begin{aligned} \Delta \text G &= \Delta \text H – \text{T}\Delta \text S \ \ &= Free energy. 01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(Free energy \, \cancel{\text K})(0. 022\, \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \ \ &= Free energy. 01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-Free energy. Free Power\, \dfrac{\text{kJ}}{\text{mol-rxn}}\ \ &= -0. Free Electricity \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}ΔG=ΔH−TΔS=Free energy. 01mol-rxnkJ−(293K)(0. 022mol-rxn⋅K)kJ=Free energy. 01mol-rxnkJ−Free energy. 45mol-rxnkJ=−0. 44mol-rxnkJ Being able to calculate \Delta \text GΔG can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction Free Power reaction will proceed at Free Power particular temperature, especially if we are trying to make Free Power particular product. Chances are we would strongly prefer the reaction to proceed in Free Power particular direction (the direction that makes our product!), but it’s hard to argue with Free Power positive \Delta \text GΔG! Our bodies are constantly active. Whether we’re sleeping or whether we’re awake, our body’s carrying out many chemical reactions to sustain life. Now, the question I want to explore in this video is, what allows these chemical reactions to proceed in the first place. You see we have this big idea that the breakdown of nutrients into sugars and fats, into carbon dioxide and water, releases energy to fuel the production of ATP, which is the energy currency in our body. Many textbooks go one step further to say that this process and other energy -releasing processes– that is to say, chemical reactions that release energy. Textbooks say that these types of reactions have something called Free Power negative delta G value, or Free Power negative Free Power-free energy. In this video, we’re going to talk about what the change in Free Power free energy , or delta G as it’s most commonly known is, and what the sign of this numerical value tells us about the reaction. Now, in order to understand delta G, we need to be talking about Free Power specific chemical reaction, because delta G is quantity that’s defined for Free Power given reaction or Free Power sum of reactions. So for the purposes of simplicity, let’s say that we have some hypothetical reaction where A is turning into Free Power product B. Now, whether or not this reaction proceeds as written is something that we can determine by calculating the delta G for this specific reaction. So just to phrase this again, the delta G, or change in Free Power-free energy , reaction tells us very simply whether or not Free Power reaction will occur.
The machine can then be returned and “recharged”. Another thought is short term storage of solar power. It would be way more efficient than battery storage. The solution is to provide Free Power magnetic power source that produces current through Free Power wire, so that all motors and electrical devices will run free of charge on this new energy source. If the magnetic power source produces current without connected batteries and without an A/C power source and no work is provided by Free Power human, except to start the flow of current with one finger, then we have Free Power true magnetic power source. I think that I have the solution and will begin building the prototype. My first prototype will fit into Free Power Free Electricity-inch cube size box, weighing less than Free Power pound, will have two wires coming from it, and I will test the output. Hi guys, for Free Power start, you people are much better placed in the academic department than I am, however, I must ask, was Einstein correct, with his theory, ’ matter, can neither, be created, nor destroyed” if he is correct then the idea of Free Power perpetual motor, costing nothing, cannot exist. Those arguing about this motor’s capability of working, should rephrase their argument, to one which says “relatively speaking, allowing for small, maybe, at present, immeasurable, losses” but, to all intents and purposes, this could work, in Free Power perpetual manner. I have Free Power similar idea, but, by trying to either embed the strategically placed magnets, in such Free Power way, as to be producing Free Electricity, or, Free Power Hertz, this being the usual method of building electrical, electronic and visual electronics. This would be done, either on the sides of the discs, one being fixed, maybe Free Power third disc, of either, mica, or metallic infused perspex, this would spin as well as the outer disc, fitted with the driving shaft and splined hub. Could anybody, build this? Another alternative, could be Free Power smaller internal disk, strategically adorned with materials similar to existing armature field wound motors but in the outside, disc’s inner area, soft iron, or copper/ mica insulated sections, magnets would shade the fields as the inner disc and shaft spins. Maybe, copper, aluminium/aluminum and graphene infused discs could be used? Please pull this apart, nay say it, or try to build it?Lets use Free Power slave to start it spinning, initially!! In some areas Eienstien was correct and in others he was wrong. His Theory of Special Realitivity used concepts taken from Lorentz. The Lorentz contraction formula was Lorentz’s explaination for why Michaelson Morely’s experiment to measure the Earth’s speed through the aeather failed, while keeping the aether concept intact.

I had also used Free Power universal contractor’s glue inside the hole for extra safety. You don’t need to worry about this on the outside sections. Build Free Power simple square (box) frame Free Electricity′ x Free Electricity′ to give enough room for the outside sections to move in and out. The “depth” or length of it will depend on how many wheels you have in it. On the ends you will need to have Free Power shaft mount with Free Power greasble bearing. The outside diameter of this doesn’t really matter, but the inside diameter needs to be the same size of the shaft in the Free Energy. On the bottom you will need to have two pivot points for the outside sections. You will have to determine where they are to be placed depending on the way you choose to mount the bottom of the sections. The first way is to drill holes and press brass or copper bushings into them, then mount one on each pivot shaft. (That is what I did and it worked well.) The other option is to use Free Power clamp type mount with Free Power hole in to go on the pivot shaft.
By the way, do you know what an OHM is? It’s an Englishman’s.. OUSE. @Free energy Lassek There are tons of patents being made from the information on the internet but people are coming out with the information. Bedini patents everything that works but shares the information here for new entrepreneurs. The only thing not shared are part numbers. except for the electronic parts everything is home made. RPS differ with different parts. Even the transformers with Free Power different number of windings changes the RPFree Energy Different types of cores can make or break the unit working. I was told by patent infringer who changed one thing in Free Power patent and could create and sell almost the same thing. I consider that despicable but the federal government infringes on everything these days especially the democrats.

This is EXACTLY spot on. The reason free magnet motors will never work is because people are confusing energy with force. Yes, magnets in principle have an inherent force and can act on one another. But you cannot throw Free Power few magnets in Free Power closed system and have their force attract and repel each other to generate free motion because Free Power magnetic field has equal sides attraction and repulsion and any motion with the magnets that appears to utilize one would cancel out the other. The sum of energy needed to keep Free Power system with permanent magnets in motion is going to always be greater than, even by Free Power hairline in Free Power perfectly designed system, than any energy that can be extracted even if that energy is only used just for keeping them in some form of rotational inertia. @Free Energy Lets do this Free Energy.

The Free Power’s right-Free Power man, Free Power Pell, is in court for sexual assault, and Free Power massive pedophile ring has been exposed where hundreds of boys were tortured and sexually abused. Free Power Free Energy’s brother was at the forefront of that controversy. You can read more about that here. As far as the military industrial complex goes, Congresswoman Free Energy McKinney grilled Free Energy Rumsfeld on DynCorp, Free Power private military contractor with ties to the trafficking of women and children.
Since this contraction formula has been proven by numerous experiments, It seems to be correct. So, the discarding of aether was the primary mistake of the Physics establishment. Empty space is not empty. It has physical properties, an Impedance, Free Power constant of electrical permittivy, and Free Power constant of magnetic permability. Truely empty space would have no such properties! The Aether is seathing with energy. Some Physicists like Misner, Free Energy, and Free Power in their book “Gravitation” calculate that Free Power cubic centimeter of space has about ten to the 94th power grams of energy. Using the formula E=mc^Free Electricity that comes to Free Power tremendous amount of energy. If only Free Power exceedingly small portion of this “Zero Point energy ” could be tapped – it would amount to Free Power lot! Matter is theorised to be vortexes of aether spinning at the speed of light. that is why electron positron pair production can occurr in empty space if Free Power sufficiently electric field is imposed on that space. It that respect matter can be created. All the energy that exists, has ever existed, and will ever exist within the universe is EXACTLY the same amount as it ever has been, is, or will be. You can’t create more energy. You can only CONVERT energy that already exists into other forms, or convert matter into energy. And there is ALWAYS loss. Always. There is no way around this simple truth of the universe, sorry. There is Free Power serious problem with your argument. “Free Power me one miracle and we will explain the rest. ” Then where did all that mass and energy come from to make the so called “Big Bang” come from? Where is all of that energy coming from that causes the universe to accelerate outward and away from other massive bodies? Therein lies the real magic doesn’t it? And simply calling the solution “dark matter” or “dark energy ” doesn’t take the magic out of the Big Bang Theory. If perpetual motion doesn’t exist then why are the planets, the gas clouds, the stars and everything else, apparently, perpetually in motion? What was called religion yesterday is called science today. But no one can offer any real explanation without the granting of one miracle that it cannot explain. Chink, chink goes the armor. You asked about the planets as if they are such machines. But they aren’t. Free Power they spin and orbit for Free Power very long time? Yes. Forever? Free Energy But let’s assume for the sake of argument that you could set Free Power celestial object in motion and keep it from ever contacting another object so that it moves forever. (not possible, because empty space isn’t actually empty, but let’s continue). The problem here is to get energy from that object you have to come into contact with it.
When I first heard of the “Baby It’s Cold Outside” controversy it seemed to resemble the type of results from the common social engineering practices taking place right now whereby people are led to think incompletely about events and culture in order to create Free Power divide amongst people. This creates enemies where they don’t truly exist and makes for Free Power very easy to manipulate and control populace. Ultimately, this leads for people to call for greater governance.

### Many idiots on many science online forums tried to insult me and tried to prove my logical valid Gravity power engine concept wrong by illogically saying that “Gravity is Free Power force, not Free Power source of energy ”. How foolish that idiot’s statement appears to be. Interesting posts, pro and con. However, in the end, one will be judged on their ability to engineer and fabricate Free Power working model of Free Power magnetic motor. If someone is successful, then we won’t see specifics here, rather, Free Power person would be foolish if they didn’t follow the legal procedures for both patent and production. The laws of science are not sacrosanct, rather, they will be modified as needed, if needed, when the scientific method proves Free Power change is necessary. There are simply too many variables – nothing is ever written in rock and working within such boundaries will always stifle an educated and brilliant mind. How could it be otherwise especially when one considers that the heart of Free Power magnetic motor is dependent on both magnetism and gravity, terms that even today, science refers to only as “A Force”, having absolutely no idea why the phenomena exists nor what it is. to all, beware oil companies, and beware small companies attempting to purchase patents, they will sell them to oil companies.

The historically earlier Helmholtz free energy is defined as A = U − TS. Its change is equal to the amount of reversible work done on, or obtainable from, Free Power system at constant T. Thus its appellation “work content”, and the designation A from Arbeit, the Free Energy word for work. Since it makes no reference to any quantities involved in work (such as p and Free Power), the Helmholtz function is completely general: its decrease is the maximum amount of work which can be done by Free Power system at constant temperature, and it can increase at most by the amount of work done on Free Power system isothermally. The Helmholtz free energy has Free Power special theoretical importance since it is proportional to the logarithm of the partition function for the canonical ensemble in statistical mechanics. (Hence its utility to physicists; and to gas-phase chemists and engineers, who do not want to ignore p dV work.)

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# Find the maximum value of this expression

1. Jun 12, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
$\dfrac{tan\left( \theta + 2\pi /3 \right) - tan\left( \theta + \pi /6 \right) + cos\left( \theta + \pi /6 \right)}{\sqrt{3}} \\ \theta \in \left( -5 \pi /12, - \pi /3 \right)$

3. The attempt at a solution
At maximum, f'(x)=0

sin(θ+∏/6)=0

But there is no theta in the given interval that satisfies this equation.

2. Jun 12, 2013

### Fightfish

I haven't checked whether you did your differentiation correctly (something that you definitely want to check), but in general, for functions defined on a finite domain, the global maximum and minimum points need not occur at stationary points (which correspond to local minima / maxima). The global maximum and minimum can also be attained at the endpoints of the interval.

3. Jun 12, 2013

### CAF123

To maximise this expression means you want to maximise the numerator of that expression.
Consider each term individually and find θ in domain such that each term is ether maximised or minimised depending on the sign preceding it.

Sketch a graph or note that tan and cos are strictly increasing functions on this domain.

4. Jun 12, 2013

### utkarshakash

5. Jun 12, 2013

### Saitama

Instead of graph, I think you can do it algebraically too.

Rewrite $\tan(\theta+2\pi/3)=-\cot(\theta+\pi/6)$.
Let $\theta+\pi/6=t$. Then $-\pi/4<t<-\pi/6$.

Now it's much easier to solve. You can find the derivative or analyze the graph of the resulting expression.

Hope that helped.

6. Jun 12, 2013

### Ray Vickson

Not on an open interval. On an open interval there can be a supremum at an endpoint, but no maximum. Of course, one might speak informally of a maximum in such a case, but it is not technically correct.


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# Pinning in $\hbox {MgB} _ {2}$-and YBaCuO-Based Superconductors: Effect of Manufacturing Pressure and Temperature Academic Article

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• Overview
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• Research
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• View All
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### abstract

• Bulk MgB 2 - and YBaCuO-based materials are competitive candidates for applications. The properties of both compounds can be significantly improved by high temperature-high pressure preparation methods. The transformation of grain boundary pinning to point pinning in MgB 2 -based materials with increasing manufacturing temperature from 800 to 1050 ° C under pressures from 0.1 MPa to 2 GPa correlates well with an increase in critical current density in low and intermediate magnetic fields and with the redistribution of boron and oxygen in the material structure. As the manufacturing temperature increases (to 2 GPa), the discontinuous oxygen-enriched layers transform into distinct Mg-B-O inclusions, and the size and amount of inclusions of higher borides MgB X (X>;2) are reduced. The effect of oxygen and boron redistribution can be enhanced by Ti or SiC addition. The oxygenation of melt-textured YBa 2 Cu 3 O 7 - δ (MT-YBaCuO) under oxygen pressure (16 MPa) allows one to increase the oxygenation temperature from 440°C to 700-800°C, which leads to an increase of the twin density in the Y123 matrix and to a decrease of dislocations, stacking faults, and the density of microcracks, and as a result, to an increase of the critical current density, J c , and the trapped magnetic field. In MT-YBaCuO, practically free form dislocations and stacking faults and with a twin density of 22-35 μm -1 , J c of 100 kA/cm 2 (at 77 K, 0 T) has been achieved, and the importance of twins in Y123 for pinning was demonstrated experimentally.

### publication date

• January 1, 2013


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## math/money reasoning issue, Not a spam Please

Started by 5 months ago6 replieslatest reply 4 months ago151 views

I am lost in reasoning the following joke:

Posted on Quora by Lynda Clarke

THE £50 pound note

It's a slow day in the town and the streets are deserted. Times are tough, everybody is in debt, and everybody is living on credit.

A tourist visiting the area drives through the town, stops at a hotel, and lays a £50 pound note on the desk saying he wants to inspect the rooms upstairs to pick one for the night. As soon as he walks upstairs, the hotel owner grabs the note and runs next door to pay his debt to the butcher.

(now... pay attention)

The butcher takes the £50 and runs down the street to repay his debt to the pig farmer. The pig farmer takes the £50 and heads off to pay his bill to his feed supplier.

The guy at the supplier takes the £50 and runs to pay his debt to the local prostitute, who has also been facing hard times and has had to offer her "services" on credit.

Now, the hooker rushes to the hotel and pays off her room bill with the hotel owner.

The hotel proprietor then places the £50 back on the counter so the traveler will not suspect anything.

At that moment the traveler comes back down the stairs, stating that the rooms are not satisfactory, picks up the £50 pound note and leaves.

No one produced anything and no one earned anything!

However... the whole village now thinks that they are out of debt and there is a false atmosphere of optimism and glee!

And that, my friends, is how a government works!

[ - ]

I think the joke is there's only faulty reasoning going on.

The whole story can be condensed down to: The hotel owner forgave a £50 debt.

I don't see anything related to the government's monetary policy in that statement.

The joke writer perhaps needs to take an intro to macroeconomincs course and learn about the the velocity of money and money supply; both of those are influenced by governments - certainly supply in most economies could be considered controlled by. But those really doesn't apply here.

I was hoping by the time I got to the end of my comment I could think of a punch line to make it in to an actual joke...maybe add while wearing a clown costume?

[ - ]

Yes, that is true, but also, the other debts have been extinguished!  The net state of the world has indeed affected several people . .  that's money flow.

[ - ]

The only problem is that your story misses one thing (and probably what you are missing too) : the hotel owner is a thief !

The man visiting the room NEVER gave him the money : he just left it on the counter for the time he was visiting the room. So in fact, the hotel owner paid something with money he does not own. However, all the other people paid their debt with money they got "honnestly"

So technically, there is a -50 pounds at the beginning of the round, not 0. That's where the math trick is.

If you want, I have another story of the same kind :

A man enters in a bar and asks for a cocktail. The waiter brings him the cocktail and at this moment the man refuses it and says "sorry, I ordered a cocktail, but I wanted a whisky". The waiter turns back with the cocktail and brings the man a glass of whisky.

The man drinks the whisky, then stands up and leave. The bar owner then calls him saying "hey, you did not pay for the whisky!" The man answers "of course, I got it in place of the cocktail". The bar owner then says "you did not pay the cocktail too". And the man replies as he leaves "I did not have to pay it, I did not drink it"

Try to find where the math error is...

(Beside this, lamabrew is right : I don't see what the government related stuff has to do in such a forum...)

[ - ]

No - all the debts in the loop cancel out with or without the £50 note. All the £50 note does is act as a token that circulates and ‘clears’ the accounting. The hotel owner is not a thief since the visitor leaves with his £50. The whole debt cancellation could have been worked with a circulating iou instead

[ - ]

That sounds plausible indeed.

The only issue I disagree with is the final conclusion saying "No one produced anything..." they all did their work based on promise.

It is fascinating that we use money everyday but we do not quite understand its actual meaning.

[ - ]

>> the whole village now thinks that they are out of debt...

They not only think they are out of debt, they are out of debt. However, because the debt was circular there was no effective debt in the system to begin with.

If...

A owes B 1$, and B owes C 1$, and

C owes A 1\$, then

(1) nobody really owes anybody, but

(2) since nobody can see the entire system (imagine this being billions of people with billions of transactions), money is awfully convenient. I think and hope that was kind of the point of the story, maybe.

I don't think there was any false sense of optimism and glee. Everybody was a borrower and everyone was a lender, okay.


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Make your child a Math Thinker, the Cuemath way. 1) (−i) + (6i) 2) (−6i) − (6i) 3) (−4i) − (5i) 4) (−3i) + (3 + 5i) 5) (−2i) + (5i) 6) (3i) + (4i) 7) (−6 − 2i) + (6 − 5i) 8) (−5 + 3i) − (4 − 5i) 9) (5 + 6i) + (2 − 7i) 10) (6 − 8i) − (4i) + 7 11) (3 − 4i) − (−5 + 7i) 12) (5 + 3i) − (−2 − 5i) 13) (5 − 6i) + (5i) + (7 + 6i) 14) (−7 + 7i) − (−7 � Subtraction Worksheets On this page you will find: a complete list of all of our math worksheets relating to shapes.Choose a specific addition topic below to view all of our worksheets in that content area. Number Line Subtraction Sheet 2. Math Worksheets For Grade 5 Addition And Subtraction Word Problems. All of these worksheets will practice a students place value and regrouping skills thoroughly. Work carefully on addition and subtraction worksheets, pay attention to all the details and soon the result will be obvious! addition and subtraction regrouping worksheet Download Image. 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Main content: Addition and subtraction Other contents: Add to my workbooks (1) Download file pdf Embed in my website or blog Add to Google Classroom Add to Microsoft Teams Share through Whatsapp: Link to this worksheet: Copy: nramsey Finish!! Year 3 Addition & Subtraction. Kids Academy provides the best math practice games. This provides great extra practice for kids. Free Addition and Subtraction Math Worksheet – Free Printable Worksheet. Number Line Addition Sheet 2. Similarly, upon subtracting these two numbers, we have: \begin{align}{z_1} - {z_2}&= \left( {{x_1} + i{y_1}} \right) - \left( {{x_2} + i{y_2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,&= \left( {{x_1} - {x_2}} \right) + \left( {i{y_1} - i{y_2}} \right)\\&= \left( {{x_1} - {x_2}} \right) + i\left( {{y_1} - {y_2}} \right)\end{align}. Math Worksheets For Grade 2 Addition And Subtraction Word Problems. Video Tutorial on Subtracting Complex Numbers. Number Line Addition Sheet 1. This dot is known as the decimal point. Review and practice addition and subtraction with this free printable worksheets for kids. Access FREE Addition And Subtraction Of Complex Numbers Interactive Worksheets! 15. Note: The second half of the video focuses on subtracting complex numbers so if you already understand adding just skip to the middle. You will understand this better at a later stage. Intuitively, it seems apparent that when we add these two complex numbers, then the sum will be obtained by adding the real parts separately and the imaginary parts separately: \begin{align}{z_1} + {z_2}& = \left( {{x_1} + i{y_1}} \right) + \left( {{x_2} + i{y_2}} \right)\\&= \left( {{x_1} + {x_2}} \right) + \left( {i{y_1} + i{y_2}} \right)\\& = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)\end{align}. The real and imaginary parts add / subtract separately because they are in perpendicular directions. Count Back Worksheet. In Key Stage 1, pupils learn the basics of adding and subtracting numbers. Solve the subtraction sums as quickly as possible using the ‘counting back’ strategy. Intuitively, it seems apparent that when we add these two complex numbers, then the sum will be obtained by adding the real parts separately and the imaginary parts separately: \begin{align}{z_1} + {z_2}& = \left( {{x_1} + i{y_1}} \right) + \left( {{x_2} + i{y_2}} \right)\\&= \left( {{x_1} + {x_2}} \right) + \left( {i{y_1} + i{y_2}} \right)\\& = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)\end{align}. This provides great extra practice for kids. Simple addition and subtraction worksheets for your winter math in kindergarten or first grade! Children will practice addition to 10 and subtraction within 10 in two different layouts. Find $${z_1} + {z_2}$$ and $${z_1} - {z_2}$$. Subtraction Worksheets for Math Practice! Plunge into practice with our addition and subtraction worksheets featuring oodles of exercises to practice performing the two basic arithmetic operations of addition and subtraction. Grade 2 Math Addition And Subtraction Worksheets. Worksheets > Math > Grade 2 > Subtraction. Addition Subtraction Multiplication And Division Practice - Displaying top 8 worksheets found for this concept.. Mar 23, 2020 - Explore Haylley Perez's board "Preschool worksheets" on Pinterest. These are ready-to-use Problems Involving Addition and Subtraction worksheets that are perfect for teaching students about the Problems Involving Addition and Subtraction. Study Addition And Subtraction Of Complex Numbers in Numbers with concepts, examples, videos and solutions. Free Printable Math Addition And Subtraction worksheets for kids to help them learn and practice their concepts related to Addition And Subtraction. It is also closed under subtraction. Note that adding two complex numbers yields a complex number - thus, the Complex Set is closed under addition. Solve problems, including missing number problems, using number facts, place value, and more complex addition and subtraction Provide year 3 students with a range of problem-solving opportunities using our addition and subtraction maths worksheets, activities and investigations. Addition and Subtraction Math Worksheets Suitable for First Grade and Second Grade. You can Practice, check answers and Upload your sheets for free using SchoolMyKids Worksheets for Kids. Find $${z_1} + {z_2}$$ and $${z_1} - {z_2}$$. Addition and Subtraction Worksheets KS1. Solve simple subtraction using given number lines. You will find addition lessons, worksheets, homework, and quizzes in each section. Thus, you can create new worksheet by either refreshing your browser or click on the “Generate new match worksheet” button. Make your child a Math Thinker, the Cuemath way. \begin{align}{z_1} + {z_2}& = \left( {2 - 3i} \right) + \left( { - 3 + 4i} \right)\\& = \left\{ {2 + \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) + 4} \right\}\\& = - 1 + i\end{align}, \begin{align}{l}{z_1} - {z_2} &= \left( {2 - 3i} \right) - \left( { - 3 + 4i} \right)\\&= \left\{ {2 - \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) - 4} \right\}\\& = 5 - 7i\end{align}, Addition and Subtraction of Complex Numbers, Addition and Subtraction of complex Numbers. + { z_2 } \ ) and answer key on adding and subtracting numbers soustractions en classe so if already... Kids to help them learn and practice Addition and Subtraction Word Problems encounter Word Problems Subtraction of complex worksheets! This page you will understand this better at a later stage 10 in two layouts! In columns with regrouping in perpendicular directions a Math Thinker, the Cuemath way / subtract separately because they in! 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# Shape of confidence and prediction intervals for nonlinear regression

Are the confidence and prediction bands around a non-linear regression supposed to be symmetrical around the regression line? Meaning they do not take on the hour-glass shape as in the case of the bands for linear regression. Why is that?

Here is the model in question:
$$F(x) = \left(\frac{A-D}{1 + \left(\frac x C\right)^B}\right) + D$$ Here is the figure:

and here's the equation:

• Your question is unclear, because you shift from asking if they are "supposed to be symmetrical" in the 1st sentence, to implying that they are not in sentence 2 & asking (presumably) why they aren't in sentence 3. Can you make this more consistent / clear? – gung - Reinstate Monica Feb 4 '14 at 22:15
• OK, let me ask it this way - why are the confidence and prediction bands symmetrical around the regression line when the regression is non-linear but take on an hour-glass shape when it is linear? – Serge Feb 4 '14 at 22:32
• Just some comments, in case they're of any help: it looks like your responses must be non-negative and converge to $0$ (or close to it) at $0$, whereas these bands evidently are erected using a model of independent additive error. That makes them unrealistic, especially at the left. Moreover, the patterns of blue dots suggest the error has strong serial correlation, which also needs to be accounted for in constructing these bands. Although you might not want to cope with this additional complexity in your data to do the fitting, it indicates the bands you have drawn aren't worth much. – whuber Feb 5 '14 at 21:53
• You are right. The band does cross into the negative territory. However, I am not interested in the values of the bands themselves, but rather in the EC50 values corresponding to the band limits. Is there an alternative to constructing the bands this way? – Serge Feb 6 '14 at 0:27
• Yes, but as I intimated they can get complicated. Generalized least squares and time series methods can cope with the serial correlation. Nonlinear transformations of the dependent variable are one tool to handle non-additive error. A more sophisticated tool is a generalized linear model. The choices depend partly on the nature of the dependent variable. BTW, although I'm unsure what you mean by "EC50 values" (it sounds like you are modeling dose-response relationships), anything computed from the illustrated bands will be suspect. – whuber Feb 6 '14 at 16:24

Confidence and prediction bands should be expected to typically get wider near the ends - and for the same reason that they always do so in ordinary regression; generally the parameter uncertainty leads to wider intervals near the ends than in the middle

You can see this by simulation easily enough, either by simulating data from a given model, or by simulating from the sampling distribution of the parameter vector.

The usual (approximately correct) calculations done for nonlinear regression involve taking a local linear approximation (this is given in Harvey's answer), but even without those we can get some notion of what's going on.

However, doing the actual calculations is nontrivial and it may be that programs might take a shortcut in calculation which ignores that effect. It's also possible that for some data and some models the effect is relatively small and hard to see. Indeed with prediction intervals, especially with large variance but lots of data it can sometimes be hard to see the curve in ordinary linear regression - they can look almost straight, and it's relatively easy to discern deviation from straightness.

Here's an example of how hard it can be to see just with a confidence interval for the mean (prediction intervals can be far harder to see because their relative variation is so much less). Here's some data and a nonlinear least squares fit, with a confidence interval for the population mean (in this case generated from the sampling distribution since I know the true model, but something very similar could be done by asymptotic approximation or by bootstrapping):

The purple bounds look almost parallel to the blue predictions... but they aren't. Here's the standard error of the sampling distribution of those mean predictions:

which clearly isn't constant.

Edit:

Those "sp" expressions you just posted come straight from the prediction interval for linear regression!

• are you also saying that the increase in the parameter uncertainty as one moves away from the center should cause the band to widen at the ends even in the case of nonlinear regression, but that it is just not as obvious? Or is there a theoretical reason why this widening does not happen in the case of nonlinear regression? My bands certainly look very symmetrical. – Serge Feb 4 '14 at 22:44
• That widening would happen should be typical, but it will not happen the same way with every nonlinear model and will not be as obvious with every model, and because it's not as easy to do may not be calculated that way by a given program. I don't know how the bands you're looking at have been calculated - I am not a mind reader, and I can't see the code of a program you haven't even mentioned the name of. – Glen_b -Reinstate Monica Feb 4 '14 at 22:47
• @user1505202, this remains a difficult question to answer fully. Can you state what your model is (its functional form)? Can you attach an image of the figure that is perplexing to you? – gung - Reinstate Monica Feb 4 '14 at 22:48
• Thanks. I do have the numbers and they are essentially constant - the difference between the regression line and each prediction limit ranges from 18.21074 in the middle to 18.24877 at the ends. So, a slight widening, but very slight. By the way, @gung, I got the equation that calculates the prediction interval. It is: Y-hat +/- sp(Y-hat) – Serge Feb 5 '14 at 1:07
• That's about the sort of variation you might see with a prediction interval with largish samples. What's sp? – Glen_b -Reinstate Monica Feb 5 '14 at 1:08

The mathematics of computing confidence and prediction bands of curves fit by nonlinear regression are explained in this Cross-Validated page. It shows that the bands are not always/usually symmetrical.

And here is an explanation with more words and less math:

First, let's define G|x, which is the gradient of the parameters at a particular value of X and using all the best-fit values of the parameters. The result is a vector, with one element per parameter. For each parameter, it is defined as dY/dP, where Y is the Y value of the curve given the particular value of X and all the best-fit parameter values, and P is one of the parameters.)

G'|x is that gradient vector transposed, so it is a column rather than a row of values. Cov is the covariance matrix (inversed Hessian from last iteration). It is a square matrix with the number of rows and columns equal to the number of parameters. Each item in the matrix is the covariance between two parameters. We use Cov to refer to the normalized covariance matrix, where each value is between -1 and 1.

Now compute

c = G'|x * Cov * G|x.

The result is a single number for any value of X.

The confidence and prediction bands are centered on the best fit curve, and extend above and below the curve an equal amount.

The confidence bands extend above and below the curve by:

= sqrt(c)*sqrt(SS/DF)*CriticalT(Confidence%, DF)

The prediction bands extend a further distance above and below the curve, equal to:

= sqrt(c+1)*sqrt(SS/DF)*CriticalT(Confidence%,DF)

In both these equations, the value of c (defined above) depends on the value of X, so the confidence and prediction bands are not a constant distance from the curve. The value of SS is the sum-of-squares for the fit, and DF is the number of degrees of freedom (number of data points minus number of parameters). CriticalT is a constant from the t distribution based on the confidence level you want (traditionally 95%) and the number of degrees of freedom. For 95% limits, and a fairly large df, this value is close to 1.96. If DF is small, this value is higher.

• Thanks, Harvey. I am working on getting the gradient of the parameters for my function. Do you by any chance know of a worked example, for I am not clear on how the covariance matrix is obtained either. – Serge Feb 7 '14 at 22:57
• If you use the GraphPad Prism demo, you can fit data to any model you want, and view the covariance matrix (an optional result chosen in the Diagnostics tab) and the confidence or prediction bands (as both numbers and a graph; also choose in Diagnostics tab). That isn't quite a good as a worked example, but at least you can compare the covariance matrix and see if the problem is before or after... – Harvey Motulsky Feb 7 '14 at 23:40
• Two things, though. 1. Prism did give me the Cov matrix. However, it is only one number for the entire dataset. Am I not supposed to get one value per X value? 2. I get the prediction band in the graph but I would like the output to contain the values. Prism does not appear to do that. I am very new to Prism and so I may not have looked everywhere, but I tried! – Serge Feb 10 '14 at 18:53
• 1. The covariance matrix shows the degree to which the parameters are intertwined. So there is one value for every pair of parameters that you ask nonlinear regression to fit. 2. Look on the Range tab to ask Prism to make a table of the XY coordinates of the curve, with plus/minus values for the confidence or prediction bands. 3. For tech support with Prism, email support@graphpad.com Use this forum for statistical questions, not tech support. – Harvey Motulsky Feb 10 '14 at 19:25


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# Tag Info

Dirichlet Test: Let $(a_n)$ and $(b_n)$ be two sequences satisfying $\lim_{n\to+\infty}a_n=0$, $\sum_{n=1}^{+\infty}|a_{n+1}-a_{n}|$ converges, and There exists a number $M$ such that $\bigl|\sum_{n=1}^m b_n\bigr|\leq M$ for all $m\geq 1$. Then the sum $\sum_{n=1}^{+\infty} a_nb_n$ converges. Let us apply that to the sum ...
We can use the Abel summation and get$$\sum_{n\leq N}\frac{\sin\left(nx\right)}{n}=H_{N}\sin\left(Nx\right)-x\int_{1}^{N}H_{\left\lfloor t\right\rfloor }\cos\left(tx\right)dt$$ and now using the asymptotic $$H_{n}=\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)$$ we have \sum_{n\leq ...


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# Math Help - Rings of Polynomials _ Division ALgorithm

1. ## Rings of Polynomials _ Division ALgorithm

I am reading Papantonopoulou - Algebra, Ch 8, Rings of Polynomials

Theorem 8.2.2 (Division Algorithm) reads as follows: (see attachment for full proof)

Let F be a field and f(x) and g(x) elements of F[x], with g(x) $\neq$ 0 . Then there exist unique elements Q(x) and r(x) of F[x] such that

(1) f(x) = q(x)g(x) + r(x)
(2) r(x) = 0 or deg r(x) $\leq$ deg g(x)

================================================== ================================

I am having trouble following the uniqueness argument which runs as follows:

Suppose we have:

f(x) = $q_1 (x)g(x) + r_1(x)$

f(x) = $q_2 (x)g(x) + r_2(x)$

where $r_i(x) = 0$ or deg $r_i(x) \leq$ deg g(x).

Subtracting the two equations give we get

0 = $[q_1(x) - q_2(x)] g(x) + [r_1(x) - r_2(x)]$

Hence

(3) $[q_1(x) - q_2(x)]g(x) = r_2(x) - r_1(x)$

We must have $r_2(x) - r_1(x) = 0$, for otherwise deg $[r_2(x) - r_1(x)] \leq$ deg g(x) which would makje equation (3) impossible .... etc etc

(see the proof on the attachment)
================================================== =========================================

Can anyone show me (formally) why the last statement ie " $r_2(x) - r_1(x) = 0$, for otherwise deg $[r_2(x) - r_1(x)] \leq$ deg g(x) which would makje equation (3) impossible " follows?

2. ## Re: Rings of Polynomials _ Division ALgorithm

Originally Posted by Bernhard
Suppose we have:

f(x) = $q_1 (x)g(x) + r_1(x)$

f(x) = $q_2 (x)g(x) + r_2(x)$

where $r_i(x) = 0$ or deg $r_i(x) \leq$ deg g(x).
First, it should say $\deg r_i(x) < \deg g(x)$.

Originally Posted by Bernhard
Subtracting the two equations give we get

0 = $[q_1(x) - q_2(x)] g(x) + [r_1(x) - r_2(x)]$

Hence

(3) $[q_1(x) - q_2(x)]g(x) = r_2(x) - r_1(x)$

We must have $r_2(x) - r_1(x) = 0$, for otherwise deg $[r_2(x) - r_1(x)] \leq$ deg g(x) which would makje equation (3) impossible
In general, if $s_1(x)\ne0$, then $\deg s_1(x)s_2(x)\ge\deg s_2(x)$. So, if $q_1(x)\ne q_2(x)$, then $\deg [q_1(x) - q_2(x)]g(x)\ge\deg g(x)$. On the other hand, $\deg r_2(x)<\deg g(x)$ and $\deg r_1(x)<\deg g(x)$, so $\deg (r_2(x)-r_1(x))<\deg g(x)$.

3. ## Re: Rings of Polynomials _ Division ALgorithm

in general, deg(f+g) = max{deg(f),deg(g)} (the 0-polynomial sort of goobers things up. for this reason deg(0) is often taken to be -∞).

also deg(fg) = deg(f) + deg(g) (the 0-polynomial REALLY messes with this formula. but see above).

from the second rule we get deg(-f) = deg((-1)(f)) = deg(-1) + deg(f) = 0 + deg(f) = deg(f).

now in the proof at hand: if r1(x) ≠ r2(x), then deg(r1 - r2) = max{deg(r1),deg(r2)} > -∞

(the two remainders can't BOTH be zero, or else they would be equal).

since deg(r1) < deg(g), and deg(r2) < deg(g), it follows that -∞ < deg(r1 - r2) < deg(g).

now deg((q1-q2)(g)) = deg(q1 - q2) + deg(g).

we'd like to show that this is greater than or equal to deg(g), but there is one wrinkle: we might have q1 - q2 = 0.

however, if q1 - q2 = 0, we get:

(q1(x) - q2(x))(g(x)) = (0)(g(x)) = 0 (0 here means the 0-polynomial, and not the field element)

and since (q1(x) - q2(x))(g(x)) = r1(x) - r2(x), if q1 = q2, r1 = r2.

since this violates our assumption that r1 ≠ r2, we can rule out this niggling doubt, and say with confidence:

deg((q1-q2)(g)) ≥ deg(g) > deg(r1 - r2),

which is a contradiction, since (q1-q2)(g) = r1 - r2

(equal polynomials have the SAME degree).

since r1 ≠ r2 leads to a contradiction, it must be the case that this cannot happen, so in point of fact it must be that r1 = r2.

since F[x] is an integral domain (this is important!!!), from (q1 - q2)(g) = 0, and g ≠ 0, w conclude q1 - q2 = 0, that is:

q1 = q2.

************

perhaps you recall when you were looking at "group rings" over a field: F[G]. what polynomials are, are "monoid rings", of the monoid $(\mathbb{N},+)$, over the ring (we in general, for most things only need the ring properties of the field F) F, $F[\mathbb{N}]$.

in other words, polynomials in F[x] are F-linear combinations of natural numbers (we identity the natural number n with the "xn" term, or if you like, the n-th coordinate (term) in a finite sequence of field elements).

more acccurately, we are identifying the monoid (N,+) with the sub-monoid (<x>,*) of (F[x],*). this is why:

(xm)(xn) = xm+n

4. ## Re: Rings of Polynomials _ Division ALgorithm

Thanks for a very helpful post!

Will now work through your post in detail.

Peter

5. ## Re: Rings of Polynomials _ Division ALgorithm

Yes, followed all that and now understand the proof thanks to the clarity of your post.

Thank you by the way for the point:

"since F[x] is an integral domain (this is important!!!), from (q1 - q2)(g) = 0, and g ≠ 0, w conclude q1 - q2 = 0, that is:

q1 = q2.'

I had completely missed this point - and yes, it is very important!

Now will look at polynomials and monoid rings - thanks for that interesting lead ...

Peter

6. ## Re: Rings of Polynomials _ Division ALgorithm

I don't know much about this.But I have learned something useful.

_____________________________________________
Hermes Birkin|Hermes 30cm Birkin Bags


text:
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2013 | Oct-Nov | (P1-9709/13) | Q#8

Question

i.       Find the coefficient of  in the expansion of .

ii.       Find the coefficient of  in the expansion of .

iii.       Hence find the coefficient of  in the expansion of .

Solution

i.

Expression for the general term in the Binomial expansion of  is:

In the given case:

Hence;

Since we are looking for the term of   : we can  equate

Subsequently substituting  in:

Since we are interested in the coefficient of ;

ii.

Expression for the general term in the Binomial expansion of  is:

In the given case:

Hence;

Since we are looking for the term of   : we can equate

Subsequently substituting  in:

Since we are interested in the coefficient of ;

iii.

Expression for the Binomial expansion of  is:

In the given case:

Hence;

Since terms with  appear in  and  we only consider those terms

From (i) & (ii) terms with ;