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# Plausible propulsion methods given a virtually infinite electric power source

Humanity discovered a new energy source, virtually infinite, that can produce gigawatts of electric power using devices of few kilograms.

Land and sea transport obviously will switch to electric motors without major changes, but what about air and space transport?

What are plausible propulsion methods for high speed aircraft and space launch vehicles with a very light and powerful electric energy source available?

• The turbines on a modern jet produce somewhere between 25-100 Megawatts each (it's a bit tricky to convert thrust to power), and in modern high-bypass jets (i.e. pretty much all non-military craft) 80-90% percent comes from the main intake fan and not the engine exhaust. So, given we have electric motors that are already nudging jet engine power to weight ratios, you can absolutely have an all-electric A380 jumbo and no-one would be the wiser. – Samwise Apr 13 at 1:18
• @Samwise Please make that an answer. – StephenS Apr 13 at 3:41
• Does this answer your question? Infinite energy, helicopters and world-wide attention – nick012000 Apr 14 at 4:03
• @nick012000 I do not think this is a duplicate - the older question is mostly about helicopters, this one is about vehicles able to reach space. – Renan Apr 14 at 13:28

Railguns. Railguns for Everything.

I feel like I should put a caveat here and say 'not if you want the inhabitants to survive', but it's entirely survivable as long as you have the right set of circumstances. In fact, it's something NASA considered. A railgun is basically just a cannon which fire projectiles, except instead of using explosions, it uses electromagnetic force, which is derived from massive amounts of electricity. Germany developed AA railguns in WWII, but seeing as each railgun needed enough power to light up half of 1947 Chicago, the project was understandably scrapped. So long as you modulate the amount of power on the launching end, railguns can be used for bullet trains, launching airplanes, and launching scramjets into space.

There is a slight flaw here, and that is that railguns only provide a boost upon firing the projectile. That is, it's a single massive thrust at the beginning. Not so bad if you want to have a mag-lev bullet train which doesn't experience much friction, but if you want to fire a rocket into orbit, you can fire the railgun strong enough to do so, you just might run into problems. As in 'turn the atmosphere for hundreds of feet surrounding the projectile to plasma' problems, not to mention that such drastic acceleration might damage things inside the projectile, especially if they are the squishy fragile human type of things. (The NASA launch program is for probes, not manned flights. But the theory is sound to get human up that way too. Just maybe with a bit less initial thrust.)

You'll end up using a combination of railgun / conventional rockets to get to orbit - though of course as shock-absorbing technology gets better you'll be able to lean more heavily on the railgun aspect. Once you get into space, it's going to be ion engines all the way. Ion engines are basically the reverse of a railgun, they fire small ions from the ship and use the gradual acceleration that accumulates to the reaction of said firing. Ion engines are too slow to use in atmosphere, but they work just fine in space.

• I believe that catapults on the new Ford class aircraft carriers work on similar principles to railguns as well. – Jan Dorniak Apr 13 at 8:57
• Some variations on this, such as launch loops and some types of skyhook, seem like they would benefit massively from infinite electricity. Anything requiring active support, for starters, is suddenly significantly more practical. Once high altitude stuff like this is more doable, bigger structures can be built, and those also will use electricity as launch / land assists. The real trouble is that it will still be about as expensive to build the first as it is without, won't it? – CAE Jones Apr 14 at 11:04
• I feel like I should put a caveat here and say 'not if you want the inhabitants to survive' well you can't make an omelet without some high G related deaths. – Renan Apr 14 at 13:28

It depends a lot on how you want your power source to work, like if its has draw backs. Air planes could hypothetically use electrically powered turbines to generate thrust, but that would be a lot of power. If you want something a little more creative but realistically less efficient you could have the craft absorb moisture out of the air then super heat it with electricity forcing it out the back it to generate thrust. If you are more inclined to air on the realistic side I'd recommend reading up on modern attempts to make electric air travel, I know they exist, but obviously have not taken off sufficiently.

Space again depends on the power source, I suggest you look at the theory of ion engines or solar sails as both rely on low cost particles (ions and photons) that could be generated from your power source depending on how it functions and are both real world thoughts on how things could be in the future. Again, a more fantastical solution would be having your ships generate magnetic fields using their power to "ride" the fields of the planets and stars. The magnetic fields could also theoretically work for the air craft, but I think if it were employed the travel would be limited to the polar directions, North and South, which could also pose interesting restrictions if you're interested in that.

I hope those ideas help, I'm no expert on them, but I am vaguely familiar with the fact that they have potential

• I'd think that propeller-driven planes would be straight-forward. They may seem old-fashioned, but for shorter hops, they're still used. High-speed trains might also become more interesting. However, note also this and this. – Matthew Apr 13 at 15:34

You can absolutely have an all-electric jumbo jet RIGHT NOW... and here's why:

First off, let's pick an Airbus A380, because if we can do it with one of the largest ever commercial planes then we can do it with any plane.

Now an A380 can carry some 250 tonnes of fuel (which is nearly half the maximum take-off weight), so, provided we can build an electric jet engine that's the same size and weight as a traditional one, then you have a staggering quarter of a million kilograms to play with for the power source, cooling, motor drive electronics, 1/5th of an Olympic swimming pool...

According to this jet engine question on Quora, some 80-odd percent of the thrust of a modern high-bypass engine comes from the intake fan alone (I was a little off in my earlier comment). This means we'll need to run out electric motors 25% harder as we won't have a raging inferno to help with thrust generation as we would in a traditional engine.

Now I couldn't find much info on the weight of modern jet engine cores vs. the whole assembly but this article indicates that the intake fan section can be up to a third of the total weight, which given that the engines on the A380 are nearly 7 tonnes gives over 4 tonnes for the motor and cowling.

Thanks to this KLM blog page they recon the A380's four engines generate 230MW of power, or ~58MW per engine, but considering our need for 25% overdrive gives us ~73MW. The worlds largest electric motors are bigger than that (which in 2018 was 80MW according to GE) so we know that a suitably sized motor can be built (in fact, if you consider generators to be motors - and they kinda are - then you can get them up to at least 700MW... flying aircraft carrier anyone?)

So, we know powerful enough motors exist, we know we have ~4 tonnes to play with, so can we build an electric jumbo? Well, 73MW/4 tonnes is 18kW/kg which is definitely up there in terms of power density for an electric motor (although it's only 11% the power density of the space shuttle's rocket-powered fuel pump!). The EMRAX 268 motor already hits 11.5kW/kg so that's not too bad but not enough. Recently another company claimed to have a motor than suits our needs and if Equipmake's numbers are to be believed then 20kW/kg is doable which is perfect for what we want.

And with that, we've proved you can make an electric jet engine that's just as good as it's gas-guzzling cousins (although, given that we saved 250t on fuel because the power source weighs effectively nothing, the engine could have weight 10 times more and the plane still could have flown), given that half the cost of a flight is fuel, the airlines are going to be your new best friends as you've just come up with a way to reduce maintenance (seeing as an electric jet has only one moving part) while doubling their profits.


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# Tag Info

Just to expand on whuber's comment (and give you an official answer), suppose you take $Z=X+Y$ and then find the mean and variance of this random variable. You have mean: $$\mathbb{E}(Z) = \mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y) = \mu_X + \mu_Y,$$ and variance: $$\mathbb{V}(Z) = \mathbb{V}(X+Y) = \mathbb{V}(X) + 2 \cdot \mathbb{C}(X,Y) + \mathbb{... 1 From Holder's and Markov's inequalities$$E(\lvert X \rvert I(\lvert X \rvert>C))\leq \lVert X\rVert_p (P(\lvert X\rvert>C))^{1-1/p}\leq \lVert X\rVert_p [E(\lvert X \rvert^p/C^p)]^{1-1/p}=E\lvert X \rvert^pC^{1-p} the desired result. The powers confused me a bit...

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# Set that keeps unique categories of objects

I would like to know if this type of special set operator exists, and if yes what is it called and if it has any other special properties.

Lets say I have this set $S$ of items. Like all sets, if the same item is added twice in the set, the set will not add it again. However, when adding an item, using this special operator, it also checks if it is 'better' than another item already in the set, and if yes it replaces it with the new item rather than letting both.

For a more concrete example:

$S_1 = \{apple, orange, pear\}$

$S_2 = \{betterApple, orange, banana\}$

$S = S_1 \cup_\succ S_2 = \{betterApple, orange, pear, banana\}$

So in this case $\cup_\succ$ checked some relationship, lets say $\succ$ where $betterApple \succ apple$, and if such a relationship existed it dropped $apple$ rather than allowing both in the set, keeping the stronger one in the set.

• Any dictionary does what you want, don't they? (I don't know of a corresponding mathematical construct.) – Raphael Apr 30 '14 at 17:39
• What is the structure of the partial order $\prec$? Does it have finite height, finite width, etc.? – D.W. Apr 30 '14 at 19:24
• @D.W. not sure what you mean about finite height and width. The $\succ$ binary relation just says that an item is larger, or better, or stronger than another item of the same type. However if such a relation does not exist this special union operator will add both to the set. – jbx May 1 '14 at 12:32
• @Raphael How is a dictionary applicable in this case? It is just a mapping from keys to values. It does not imply a partial order between some elements. – jbx May 1 '14 at 12:43

There is no official operation that accomplishes what you want, but you can always define an operation $\cup_{\succ}$ which does so. Just define the operation in words.
If for some reason you want a "physical" implementation, then here is one way. Each set is a collection of pairs, i.e. \begin{align*} S_1 &= \{ (apple,apple), (orange,orange), (pear,pear) \}, \\ S_2 &= \{ (apple,betterApple), (orange,orange), (banana,banana) \}. \end{align*} Furthermore, there is some partial order $\prec$ such that $(x,y),(z,w)$ are comparable iff $x = z$. When you take the union $S_1 \cup S_2$, you get $$S_1 \cup S_2 = \{ (apple,apple), (apple,betterApple), (orange,orange), (pear,pear), (banana,banana) \}.$$ Then you can define a function $$\phi(S) = \{ x \in S : \text{ no y \in S satisfies y \succ x } \}.$$ What you want is $\phi(S_1 \cup S_2)$.
• Thanks. Yes I just wanted to check if there exists some term which is widely known that achieves this. Its no problem to define $\succ$ as a partial function that takes two items of the same type and returns the stronger one. Or I could have a function $\phi(x, y) = \{\{x\} if x \succ y; \{y\} if y \succ; \{x,y\} otherwise \}$ – jbx May 1 '14 at 12:38


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Reinforcement Learning

Reinforcement Learning is where the agent is not presented with target outputs, but is given a reward signal, which it aims to maximize.

There are many applications for which it is difﬁcult, inappropriate, or even impossible to provide a “training set”.

• optimal control
• mobile robots, pole balancing, ﬂying a helicopter
• resource allocation
• job shop scheduling, mobile phone channel allocation
• mix of allocation and control
• elevator control, backgammon

# Reinforcement Learning Framework

• An agent interacts with its environment.
• There is a set S of states and a set A of actions.
• At each time step t, the agent is in some state $S_t$. It chooses an action $a_t$ and then goes to the next state $S_{t+1} = \delta(S_t, a_t)$ and receives a reward of $r(S_t, a_t)$.
• reward and delta can be multi-valued
• The aim is to find an optimal policy $\pi : S \rightarrow A$ which will maximise the cumulative reward.

# Models of Optimality

To work out if a fast nickel is worth a slow dime (reward now vs bigger reward later), we can use the following methods:

## Finite Horizon Reward

Computationally simple.

(1)
\begin{align} \sum^h_{i=0}r_{t+i} \end{align}

## Average Reward

Hard to deal with because we can't sensibly choose between a small reward soon and a large one very far in the future.

(2)
\begin{align} lim_{h \rightarrow \infty} \frac{1}{h}\sum_{i=0}^{h-1} r_{t+i} \end{align}

## Infinite Discounted Reward

Easier for proving theorems.

(3)
\begin{align} \sum^{\infty}_{i=0} \gamma ^i r_{t+i}, 0 \leq \gamma < 1 \end{align}

# Delayed Reinforcement

If you start at 2 the best policy is to stay there and take the green arrow.
If you start at 1 with a low discount factor then it's better to stay at 1, otherwise go to 2.

# Exploration Exploitation Tradeoff

Most of the time we should choose what we think is the best action. However, in order to ensure convergence to the optimal strategy, we must occasionally choose something different from our preferred action, e.g.

• Choose a random action 5% of the time, or
• Use a Bolzmann distribution to choose the next action:
(4)
\begin{align} P(a) = \frac{e^{\frac{\vec{V}(a)}{T}}}{\sum_{b \in A} e^{\frac{\vec{V}(b)}{T}}} \end{align}

# Temporal Difference Learning (TDL)

In TDL we expect that over time we have better information rather than worse. It's where the (discounted) value of the next state, plus the immediate reward, is used as the target value for the current state.

A more sophisticated version, called TD(λ), uses a weighted average of future states. This can sometimes help, but isn't always needed.

The problem with TD learning is having to learn a value for every state. For problems with very large numbers of possible states it becomes in feasible.

TD(0): use $V_{k+1}$as the training value for $v_k$

$TD(\lambda )$: use $T_k$ as the training value for $V_k$ where:

(5)
\begin{align} T_k = (1 - \lambda ) \sum^m_{t = k + 1} \lambda^{t - 1 - k} V_t + \lambda^{m - k} V_{m+1} \end{align}

# Q-Learning

Q-Learning is a variation on T-Learning, where instead of learning a value function we learn a 'Q' function.

For each state s in the set of states S, let V*(s) be the maximum discounted reward obtainable from s, and let Q(s,a) be the discounted reward available by ﬁrst doing action a and then acting optimally.

A policy has to be prepared to start at any state, and hence we have to learn a value for not only every state, but for every action with every state.

## Math

If $0 \leq \gamma < 1$

(6)
\begin{align} x = reward_0 + \gamma reward_1 + \gamma^2 reward_2 + \gamma^3 reward_3 + ... + \gamma^n reward_n \end{align}

If all rewards are 1:

(7)
\begin{align} \gamma x = \gamma + \gamma^2 + \gamma^3 + ... + \gamma^n \end{align}
(8)
\begin{align} x(1 - \gamma) = 1 x = \frac{1}{1 - \gamma} \end{align}

# Example: Backgammon

Backpropagation:

(9)
\begin{align} w \leftarrow w + \eta (T-V_\frac{\delta V}{\delta w} \end{align}

The different learnings choose T differently:

• We could learn moves from example games
• We could put T as the final outcome of the game
• Could use Temporal Difference Learning

It worked for backgammon because:

• random dice rolls in Backgammon force self-play to explore a much

larger part of the search space than it otherwise would.

• humans are bad at probabilistic reasoning?

* evolutionary algorithm can also produce a surprisingly strong player, but a gradient-based method such as TD-learning is better able to ﬁne-tune the rarely used weights, and exploit the limited non-linear capabilities of the neural network.

# Limits of Theoretical Results

• Delayed reinforcement
• Reward resulting from an action may not be received until several time steps later, which also slows down the learning
• Search space must be finite
• convergence is slow if the search space is large
• Relies on visiting every state inﬁnitely often
• For “real world” problems, we can’t rely on a lookup table
• Need to have some kind of generalisation (e.g. TD-Gammon)
• It only knows about a state if it's been there, and when there are infinite states (chess) then it's impractical
page revision: 9, last edited: 03 May 2012 00:37


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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Equations with Variables on One Side | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts - Honors Go to the latest version.

# 2.1: Equations with Variables on One Side

Created by: CK-12

Erin, Jillian, Stephanie and Jacob went to the movies. The total bill for the tickets and snacks came to $72.00. What is an equation that represents this situation? How much should each teen pay to split the bill evenly? ### Watch This ### Guidance When solving any equation, your job is to find the value for the letter that makes the equation true. Solving equations with variables on one side can be done with the help of models such as a balance or algebra tiles. When solving equations with variables on one side of the equation there is one main rule to follow: whatever you do to one side of the equals sign you must do the same to the other side of the equals sign. For example, if you add a number to the left side of an equals sign, you must add the same number to the right side of the equals sign. #### Example A $5a + 2 = 17$ The problem can be solved if we think about the problem in terms of a balance. We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance. In order to solve the equation, we have to get the variable $a$ all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side. Let’s first subtract 2 from both sides to get rid of the 2 on the left. Since 5 is multiplied by $a$ , we can get a by itself (or isolate it) by dividing by 5. Remember that whatever we do to one side, we have to do to the other. If we simplify this expression, we get: Therefore $a = 3$ . We can check our answer to see if we are correct by substituting our answer back into the original equation. $5a + 2 &= 17\\5({\color{red}3}) + 2 &= 17\\15 + 2 &= 17\\17 &= 17 \ \$ #### Example B $7b - 7 = 42$ Again, we can solve the problem if we think about the problem in terms of a balance (or a seesaw). We know that the two sides are equal so the balance has to stay horizontal. We can place each side of the equation on each side of the balance. In order to solve the equation, we have to get the variable $b$ all by itself. Always remember that we need to keep the balance horizontal. This means that whatever we do to one side of the equation, we have to do to the other side. Let’s first add 7 from both sides to get rid of the 7 on the left. Since 7 is multiplied by $b$ , we can get a by itself (or isolate it) by dividing by 7. Remember that whatever we do to one side, we have to do to the other. If we simplify this expression, we get: Therefore $b = 7$ . We can check our answer to see if we are correct. $7b - 7 &= 42\\7({\color{red}7}) - 7 &= 42\\49 - 7 &= 42\\42 &= 42 \ \$ #### Example C This same method can be extended by using algebra tiles. If we let rectangular tiles represent the variable, square tiles represent one unit, green tiles represent positives numbers, and white tiles represent the negative numbers, we can solve the equations using an alternate method. The green algebra $x-$ tiles represent variables; therefore, there are 3 $c$ blocks for the equation. The other green blocks represent the numbers or constants. There is a 2 on the left side of the equation so there are 2 square green blocks. There is an 11 on the right side of the equation so there are 11 square green blocks on the right side of the equation. To solve, add two negative tiles to the right and left hand sides. The same rule applies to this problem as to all of the previous problems. Whatever we do to one side we have to do to the other. This leaves us with the following: We can reorganize these to look like the following: Organizing the remaining algebra tiles allows us to realize the answer to be $x = 3$ or for our example $c = 3$ . Let’s do our check as with the previous two problems. $3c + 2 &= 11\\3({\color{red}3}) + 2 &= 11\\9 + 2 &= 11\\11 &= 11 \ \ Y$ #### Concept Problem Revisited There are four teens going to the movies (Erin, Jillian, Stephanie, and Jacob). The total bill was$72.00. Therefore our equation is $4x = 72$ . We divide by 4 to find our answer.

$x &= \frac{72}{4}\\x &= 18$

Therefore each teen will have to pay \$18.00 for their movie ticket and snack.

### Vocabulary

Constant
A constant is also a numerical coefficient but does not contain a variable. For example in the equation $4x + 72 = 0$ , the 72 is a constant.
Equation
An equation is a mathematical equation with expressions separated by an equals sign.
Numerical Coefficient
In mathematical equations, the numerical coefficients are the numbers associated with the variable. For example, with the expression $4x$ , 4 is the numerical coefficient and $x$ is the variable.
Variable
A variable is an unknown quantity in a mathematical expression. It is represented by a letter. It is sometimes referred to as the literal coefficient.

### Guided Practice

1. Use a model to solve for the variable in the problem $x-5=12$ .

2. Use a different model than used in question (1) to solve for the variable in the problem $3y+9=12$ .

3. Using one of the models from the concept, solve for $x$ in the equation $3x-2x+16=-3$ .

1. $x-5=12$

Therefore, $x=17$ . Let's do a check to make sure.

$x - 5 &= 12\\({\color{red}17}) - 5 &= 12\\17 - 5 &= 12\\12 &= 12 \ \$

2. $3y+9=12$

First you have to subtract 9 from both sides of the equation in order to start to isolate the variable.

Now, in order to get $y$ all by itself, you have to divide both sides by 3. This will isolate the variable $y$ .

Therefore, $y=1$ . Let's do a check to make sure.

$3y + 9 &= 12\\3({\color{red}1}) +9 &= 12\\3 + 9 &= 12\\12 &= 12 \ \$

3. $3x-2x+16=-3$

You can use any method to solve this equation. Remember to isolate the $x$ variable. You will notice here that there are two $x$ values on the left. First let’s combine these terms.

$3x-2x+16 &= -3\\x+16 &= -3$

Now you can use any method to solve the equation. You now should just have to subtract 16 from both sides to isolate the $x$ variable.

$x+16 {\color{red}-16} &= -3 {\color{red}-16}\\x &= -19$

Let's do a check to make sure.

$3x-2x+16 &=-3 \quad \text{(original problem)}\\x+16 &= -3 \quad \text{(simplified problem)}\\{\color{red}-19}+16 &= -3\\-3 &= -3 \ \$

### Practice

Use the model of the balance to solve for each of the following variables.

1. $a+3=-5$
2. $2b-1=5$
3. $4c-3=9$
4. $2-d=3$
5. $4=3e=-2$

Use algebra tiles to solve for each of the following variables.

1. $x+3=14$
2. $2y-7=5$
3. $3z+6=9$
4. $5+3x=-3$
5. $2x+2=-4$

Use the models that you have learned to solve for the variables in the following problems.

1. $-4x+13=5$
2. $3x-5=22$
3. $11-2x=5$
4. $2x-4=4$
5. $5x+3=28$

For each of the following models, write a problem involving an equation with a variable on one side of the equation expressed by the model and then solve for the variable.

## Date Created:

Dec 19, 2012

Apr 29, 2014
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Graphing on a number line Question

• January 14th 2009, 09:55 AM
Carolyng66
Graphing on a number line Question
Hello;

I'm sorry Im trying to learn to use Latex while simultaneously mastering precalc so far it's not working.

I'm currently going over, graphing numbers on a number line and my book is doing only a fair job in explaining specific rules.

I understand that; [1<4] but why do I have to write the answer as; 1< or equal to x < or equal to 4

$1\1eqx\1eq4$ if I can understand this reasoning, it would make the rest of my homework easier.

Thank You
Carolyn
• January 14th 2009, 10:27 AM
masters
Quote:

Originally Posted by Carolyng66
Hello;

I'm sorry Im trying to learn to use Latex while simultaneously mastering precalc so far it's not working.

I'm currently going over, graphing numbers on a number line and my book is doing only a fair job in explaining specific rules.

I understand that; [1<4] but why do I have to write the answer as; 1< or equal to x < or equal to 4

$1\1eqx\1eq4$ if I can understand this reasoning, it would make the rest of my homework easier.

Thank You
Carolyn

Hello Carolyn,

I think, maybe, you're talking about "interval notation" and "set builder notation". Let's see if I can explain.

If you want to graph all the elements between and including 1 and 4, we can write the solution set in two ways.

Interval Notation: [1, 4]

The brackets mean the endpoints are included in the graph. Had they not been included, we would have used parentheses.

Set Builder Notation: $\{x | 1 \leq x \leq 4\}$

This is read "The set of all x, such that x is greater than or equal to 1 and less than or equal to 4".

Does this help? If not, post specific examples.
• January 14th 2009, 12:42 PM
stapel
Quote:

Originally Posted by Carolyng66
I understand that; [1<4] but why do I have to write the answer as; 1< or equal to x < or equal to 4

If you mean that you understand that the notation "[1, 4]" indicates the interval from 1 to 4 inclusive, but are wondering why you "have to write the answer" also in the form " $1\, \leq\, x\, \leq\, 4$", I think the reason is simply that you be able to understand the various notations.

Not all books use the same notation, so you need to be able to read each of them. (Wink)


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# American Institute of Mathematical Sciences

December  2008, 1(4): 519-540. doi: 10.3934/dcdss.2008.1.519

## Random walk in the three-body problem and applications

 1 Department of Astrophysical Sciences, Princeton University, Princeton, New Jersey 08544, United States

Received  April 2008 Revised  July 2008 Published  September 2008

The process of random walk is described, in general, and how it can be applied in the three-body problem in a systematic manner. Several applications are considered. The main one which is a focus of this paper is on the evolution of horseshoe orbits and their transition to breakout motion in the restricted three-body problem. This connection is related to their use for an Earth-impactor in a theory on the formation of the Moon. We briefly discuss another application on the instability of asteroid orbits.
Citation: Edward Belbruno. Random walk in the three-body problem and applications. Discrete & Continuous Dynamical Systems - S, 2008, 1 (4) : 519-540. doi: 10.3934/dcdss.2008.1.519
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2019 Impact Factor: 1.233


text:
# A 40 muF capacitor is connected to a 200 V, 50 hzac supply. The rms value of the current in the circuit is, nearly :

more_vert
A 40 µF capacitor is connected to a 200V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly :

(1) 25.1 A

(2) 1.7 A

(3) 2.05 A

(4) 2.5 A


text:
# Combinatorics back-substitution Question

1) an = an-1 + 2n with a0 = 2
The question is Use back-substitution to find a closed formula for the recurrence relations.
I have an-4 + 2(n-3)+2(n-2)+2(n-1)+2n
Then I have a1 + ... +2(n-5)+2(n-4)+2(n-3)+2(n-2)+2(n-1)+2n
First of all, is my way right? If so, could you please help me find the recurrence relations?
If no, could you tell I have done wrong and point me to the right direction?

2) Find and solve a recurrence relation for pn the value of a stock market indicator that obeys the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's
change. Let p0 = 1; p1 = 4.
I have a trouble to translate " the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's" into recurrence relation. Can someone please help?

Homework Helper
Gold Member
1) an = an-1 + 2n with a0 = 2
The question is Use back-substitution to find a closed formula for the recurrence relations.
I have an-4 + 2(n-3)+2(n-2)+2(n-1)+2n
Then I have a1 + ... +2(n-5)+2(n-4)+2(n-3)+2(n-2)+2(n-1)+2n
First of all, is my way right? If so, could you please help me find the recurrence relations?
If no, could you tell I have done wrong and point me to the right direction?

I presume you mean ##a_n = a_{n-1} +2n## with ##a_0=2##. I would write out the first few, taking care not to simplify, which might hide any patterns:

##a_1 = 2 + 2\cdot 1##
##a_2 = 2 + 2\cdot 1 + 2\cdot 2##
##a_3 = 2 + 2\cdot 1 + 2\cdot 2 + 2\cdot 3##

Do you see any patterns or useful expressions you can use to get a nice closed formula for ##a_n##?

Thanks

Last edited:
ArcanaNoir
2) Find and solve a recurrence relation for pn the value of a stock market indicator that obeys the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's
change. Let p0 = 1; p1 = 4.
I have a trouble to translate " the rule that the change this year from the previous year (that is, pn - pn-1) equals twice last year's" into recurrence relation. Can someone please help?

Sounds to me like: $(p_2-p_1)=2(p_1-p_0)$ so, $p_n-p_{n-1}=2(p_{n-1}-p_{n-2})$.

1 person


text:
Drew Haven drew.haven at gmail.com
Wed Jun 23 17:32:10 EDT 2010

I doubt it's an issue of being too stupid.  If you come from more
to think in Haskell.  It's very different.  It appears hard because
there is a lot of work to re-learning things you thought you knew.

As for combining two state monads, is there a reason you cannot
combine them into one large state?

It looks what you are doing is:

importantFunction initialFoo initialBar =
let foo = execState modifiesFoo initialFoo
in (foo, execState (modifiesBar foo) initialBar)

Which would have the type signature you are expecting and wouldn't

Drew Haven
drew.haven at gmail.com

On Wed, Jun 23, 2010 at 2:10 PM, Jordan Cooper <nefigah at gmail.com> wrote:
> I have two pieces of state I'm managing:
> type FooState = State Foo Foo
> type BarState = State Bar Bar
>
> There is a bit of interplay between them, and I want a function that
> does something like:
> importantFunction :: Foo -> Bar -> (Foo, Bar)
>
> The problem is, >>= only works when I'm chaining all of the same type
> of state. So I can't do:
> modifiesFoo >>= \foo ->
> modifiesBar foo >>= \bar ->
> return (foo, bar)
>
> So how do I get the results of both modifications back? I hope that
> makes sense. Programming in Haskell is proving to be very difficult; I
> once again fear I'm too stupid.
> _______________________________________________
> Beginners mailing list


text:
## General Chemistry: Principles and Modern Applications (10th Edition)

(a) $17.1 \%$ (b) $13.9 \%$ (c) $13.7\%$
(a) $$Percent \space by \space volume = \frac{12.8 \space mL}{75.0 \space mL} \times 100\%= 17.1 \%$$ (b) Mass of solute : $12.8 \space mL \times \frac{0.803 \space g}{1 \space mL} = 10.3 \space g$ Mass of solution: $75.0 \space mL \times \frac{0.988 \space g}{1 \space mL} = 74.1 \space g$ $$Percent \space by \space mass = \frac{10.3 \space g}{74.1 \space g} \times 100\% = 13.9 \%$$ (c) $$Percent(mass/volume) = \frac{10.3 \space g}{75.0 \space mL} \times 100\% = 13.7\%$$


text:
# Function families from lattices

On this course, Micciancio talks about function families (functions parametrized by some value) that can be used in cryptography.

On page 2, he presents the following function family parametrized by bases of full-rank lattices:

For any invertible $\mathbf{B} \in \mathbb{R}^{n \times n}$, let $f_\mathbf{B}(\mathbf{x}) = \mathbf{x} \mod \mathbf{B} = \mathbf{B} \left(\mathbf{B}^{-1}\lfloor \mathbf{x} \rfloor \right)$.

Clearly, there is a typo there, because $\mathbf{B} \left(\mathbf{B}^{-1}\lfloor \mathbf{x} \rfloor \right) = \lfloor \mathbf{x} \rfloor$ and this is at the same time independent of the basis and useless for cryptography.

So, maybe the correct definition must be $f_\mathbf{B}(\mathbf{x}) = \mathbf{x} \mod \mathbf{B} = \mathbf{B} \lfloor \mathbf{B}^{-1} \mathbf{x} \rfloor$. But in this case, the codomain $f_\mathbf{B}$ is the lattice $\mathcal{L}(\mathbf{B})$, but the text just below the definition says

In other words, $f_\mathbf{B}(\mathbf{x})$ rounds $\mathbf{x}$ to the corner of the fundamental parallelepiped $f_\mathbf{B}(\mathbf{x}) + \mathcal{P}(\mathbf{B})$ containing $\mathbf{x}$. As defined, $f_\mathbf{B}$ is a function with domain $X \subseteq \mathbb{R}^n$ and codomain $\mathcal{P}(\mathbf{B})$.

So, do you have any idea of what the actual definition should be?

• I think your correction fits the text – Florian Bourse Apr 5 '18 at 15:49
• But P(B) has only one lattice point and the image of the function I've defined has many lattice points, so P(B) cannot be its codomain. – Hilder Vitor Lima Pereira Apr 5 '18 at 16:56
• $$x \bmod B$$ is often defined as $$x-B\lfloor B^{-1} x\rfloor \in P(B)$$, which has the codomain claimed in the text. – Chris Peikert Apr 9 '18 at 1:59

As pointed out by Chris Peikert in his comment, the actual definition must be

$f_\mathbf{B}(\mathbf{x}) = \mathbf{x} \mod \mathbf{B} = \mathbf{x} - \mathbf{B} \lfloor\mathbf{B}^{-1} \mathbf{x} \rfloor$.

we can see that the codomain of such $f_\mathbf{B}$ is the fundamental region $\mathcal{P}(\mathbf{B})$ as follows:

Write any given $\mathbf{x}$ as a linear combination of the columns $\mathbf{b}_1$, ..., $\mathbf{b}_n$ of $\mathbf{B}$, that is:

Let $\mathbf{x} = \mathbf{B}\mathbf{z}$ for some $\mathbf{z} \in \mathbb{R}^n$ (it's possible because $\mathbf{B}$ is fullrank, so its columns span $\mathbb{R}^{n}$).

Let $\mathbf{t} = \mathbf{z} - \lfloor \mathbf{z} \rfloor$, so $\mathbf{t} \in [-1, 1)^n$ and $\mathbf{x} = \mathbf{B}(\mathbf{t} + \lfloor\mathbf{z} \rfloor)$.

Thus,

$f_\mathbf{B}(\mathbf{x}) = \mathbf{x} - \mathbf{B} \lfloor\mathbf{B}^{-1} \mathbf{B}(\mathbf{t} + \lfloor\mathbf{z} \rfloor) \rfloor = \mathbf{x} - \mathbf{B} \lfloor \mathbf{t} + \lfloor \mathbf{z} \rfloor \rfloor = \mathbf{x} - \mathbf{B} \lfloor \mathbf{z} \rfloor = \mathbf{B}\mathbf{t}$

and $\mathbf{B}\mathbf{t} \in \mathcal{P}(\mathbf{B})$ by definition of $\mathcal{P}(\mathbf{B})$.


text:
# What is the minimum N for which there exist N points in the plane that cannot be covered by any number of non-overlapping closed unit discs?

This problem was posed in March 2010 at G4G9 in a talk by the Japanese mathematician Hirokazu "Iwahiro" Iwasawa. He claims there is a simple proof that N > 10, though he did not share it with the audience, since it proving it is apparently an enlightening exercise in its own right.

-
Maybe I am missing something, but the result seems obviously false: take a point surrounded by 9 other points in a small circle, then how are you going to have a disk covering the center point without touching the remaining points? If I interpret the problem as saying not every point has to be in a disk, then it seems rather trivially true, because all I need to make sure is every time I add a new disk, it covers at least one point, which is certainly doable. –  John Jiang Apr 6 '10 at 22:44
Perhaps I am misunderstanding this, but I thought MathOverflow was for asking questions to which you do not already know the answer. –  Charles Staats Apr 6 '10 at 22:58
This doesn't yield a very tight bound, but if you fill, say, a 2x2 area densely enough with evenly-spaced points, then no matter how the discs are placed there will always be some points uncovered in the "inverted trefoil" area where three discs come together. –  Josh Jordan Apr 7 '10 at 5:02
@DouglasS.Stones An upper bound appears in this paper: arxiv.org/abs/1101.3468 –  Yoav Kallus Jan 25 '14 at 21:16
I think this question is also interesting in higher dimensions. How does the minimal number $N(d)$ (defined in the obvious way) grow with the dimension $d$? –  Roland Bacher Apr 15 at 8:39

The trick for N = 10 (which I heard from a friend earlier today) is to check that the density of the triangular packing of unit diameter circles is high enough that some translate of this packing must cover all the points.

-

I was told this puzzle last friday by Peter Winkler (who had mentioned that it was told to him by a japanese fellow who is perhaps the one you are referring to).

The solution in the $n \leq 10$ case is to consider the tiling of the plane by unit height hexagons. Inscribe within each of these hexagons a unit circle. This grid of circles has density > 0.90 on the plane, and so if you randomly place this grid on the plane you accordingly have expected number of points covered > 9 (out of the 10), and this implies exists an arrangement that covers 10. (theres a few details missing from this probabilistic method argument, but you get the basic idea).

I believe for the $n>10$ case we have some way of computing an upper bound on the density of a sphere packing on the plane that rules it out in general. (or something to that extent)

-

In the answer to Open problems in Euclidean geometry? , Alexey Ustinov brings into attention to a 2012 article.

Greg Aloupis, Robert A. Hearn, Hirokazu Iwasawa, Ryuhei Uehara, Covering Points with Disjoint Unit Disks, 24th Canadian Conference on Computational Geometry (2012)

The abstract of that article confirms that it's concerned of the same problem, and gives improved bounds.

We consider the following problem. How many points must be placed in the plane so that no collection of disjoint unit disks can cover them? The answer, k, is already known to satisfy 11 ≤ k ≤ 53. Here, we improve the lower bound to 13 and the upper bound to 50. We also provide a set of 45 points that apparently cannot be covered, although this has been determined via computer search.

The article also claims that the lower bound of 11 was published in 2008

-

To bring this problem back to the attention of MO, I'll make a guess. Consider the following set of 13 points: 12 equally spaced on a circle of radius $1+\epsilon$, the 13th at the center of that circle. Can you cover all 13 points with non-overlapping unit disks?

-
–  Josh Jordan Apr 12 '10 at 5:33


text:
Chapter 7 - Exponents and Exponential Functions - 7-8 Geometric Sequences - Lesson Check - Page 470: 5

Work Step by Step

The starting value $a_1$ is $4$. The common ratio $r$ is $\frac{-8}{4}=-2$. The recursive formula is: $a_1=4; a_n=a_{n-1}.-2$ The explicit formula is: $a_n=4.(-2)^{n-1}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
1.18.0 User Documentation for Apache MADlib
HITS
Contents

Given a graph, the HITS (Hyperlink-Induced Topic Search) algorithm outputs the authority score and hub score of every vertex, where authority estimates the value of the content of the page and hub estimates the value of its links to other pages. This algorithm was originally developed to rate web pages [1].

HITS
hits( vertex_table,
vertex_id,
edge_table,
edge_args,
out_table,
max_iter,
threshold,
grouping_cols
)

Arguments

vertex_table

TEXT. Name of the table containing the vertex data for the graph. Must contain the column specified in the 'vertex_id' parameter below.

vertex_id

TEXT, default = 'id'. Name of the column in 'vertex_table' containing vertex ids. The vertex ids can be of type INTEGER or BIGINT with no duplicates. They do not need to be contiguous.

edge_table

TEXT. Name of the table containing the edge data. The edge table must contain columns for source vertex and destination vertex.

edge_args

TEXT. A comma-delimited string containing multiple named arguments of the form "name=value". The following parameters are supported for this string argument:

• src (INTEGER or BIGINT): Name of the column containing the source vertex ids in the edge table. Default column name is 'src'.
• dest (INTEGER or BIGINT): Name of the column containing the destination vertex ids in the edge table. Default column name is 'dest'.

out_table

TEXT. Name of the table to store the result of HITS. It will contain a row for every vertex from 'vertex_table' with the following columns:

• vertex_id : The id of a vertex. Will use the input parameter 'vertex_id' for column naming.
• authority : The vertex authority score.
• hub : The vertex hub score.
• grouping_cols : Grouping column values (if any) associated with the vertex_id.

A summary table is also created that contains information regarding the number of iterations required for convergence. It is named by adding the suffix '_summary' to the 'out_table' parameter.

max_iter (optional)

INTEGER, default: 100. The maximum number of iterations allowed. Each iteration consists of both authority and hub phases.

threshold (optional)

FLOAT8, default: (1/number of vertices * 1000). Threshold must be set to a value between 0 and 1, inclusive of end points. If the difference between two consecutive iterations of authority AND two consecutive iterations of hub is smaller than 'threshold', then the computation stops. That is, both authority and hub value differences must be below the specified threshold for the algorithm to stop. If you set the threshold to 0, then you will force the algorithm to run for the full number of iterations specified in 'max_iter'.

grouping_cols (optional)
TEXT, default: NULL. A single column or a list of comma-separated columns that divides the input data into discrete groups, resulting in one distribution per group. When this value is NULL, no grouping is used and a single model is generated for all data.
Note
Expressions are not currently supported for 'grouping_cols'.

Examples
1. Create vertex and edge tables to represent the graph:
DROP TABLE IF EXISTS vertex, edge;
CREATE TABLE vertex(
id INTEGER
);
CREATE TABLE edge(
src INTEGER,
dest INTEGER,
user_id INTEGER
);
INSERT INTO vertex VALUES
(0),
(1),
(2),
(3),
(4),
(5),
(6);
INSERT INTO edge VALUES
(0, 1, 1),
(0, 2, 1),
(0, 4, 1),
(1, 2, 1),
(1, 3, 1),
(2, 3, 1),
(2, 5, 1),
(2, 6, 1),
(3, 0, 1),
(4, 0, 1),
(5, 6, 1),
(6, 3, 1);

2. Running HITS with default values for optional parameters:
DROP TABLE IF EXISTS hits_out, hits_out_summary;
'vertex',             -- Vertex table
'id',                 -- Vertex id column
'edge',               -- Edge table
'src=src, dest=dest', -- Comma delimited string of edge arguments
'hits_out');          -- Output table of HITS
SELECT * FROM hits_out ORDER BY id;

 id |      authority       |         hub
----+----------------------+----------------------
0 |    8.43871829093e-07 |    0.338306115082665
1 |    0.158459587238244 |    0.527865350448059
2 |    0.405627969689677 |    0.675800764727558
3 |    0.721775835521825 |    3.95111934817e-07
4 |    0.158459587238244 |    3.95111934817e-07
5 |    0.316385413093048 |    0.189719957843216
6 |    0.405199928761102 |    0.337944978189241
(7 rows)

SELECT * FROM hits_out_summary;

 __iterations__
-----------------+
17
(1 row)

3. Running HITS with max_iter of 3 results in different authority and hub scores:
DROP TABLE IF EXISTS hits_out, hits_out_summary;
'vertex',             -- Vertex table
'id',                 -- Vertex id column
'edge',               -- Edge table
'src=src, dest=dest', -- Comma delimited string of edge arguments
'hits_out',           -- Output table
3);                  -- Max iteration
SELECT * FROM hits_out ORDER BY id;

 id |     authority     |        hub
----+-------------------+--------------------
0 |   0.08653327387778 | 0.375721659592363
1 |   0.18388320699029 | 0.533118571043218
2 |   0.43266636938891 | 0.654974244424525
3 |   0.70308285025699 | 0.040618557793769
4 |   0.18388320699029 | 0.040618557793769
5 |   0.30286645857224 | 0.182783510071961
6 |   0.38939973245002 | 0.330025782074373
(7 rows)

SELECT * FROM hits_out_summary;

 __iterations__
-----------------+
3
(1 row)

4. Running HITS with a low threshold of 0.00001 results in more iterations for convergence:
DROP TABLE IF EXISTS hits_out, hits_out_summary;
'vertex',             -- Vertex table
'id',                 -- Vertex id column
'edge',               -- Edge table
'src=src, dest=dest', -- Comma delimited string of edge arguments
'hits_out',           -- Output table
NULL,                 -- Default max_iter
0.00001);             -- Threshold
SELECT * FROM hits_out ORDER BY id;

 id |      authority       |         hub
----+----------------------+---------------------
0 |    1.15243075426e-09 |     0.33800946769422
1 |    0.158264459912827 |    0.527792117750177
2 |    0.405384672299625 |    0.675965453766535
3 |     0.72186275724613 |    5.39583282614e-10
4 |    0.158264459912827 |    5.39583282614e-10
5 |    0.316493740997913 |    0.189793242747412
6 |    0.405356461070609 |    0.337985666133163
(7 rows)

SELECT * FROM hits_out_summary;

 __iterations__
-----------------+
25
(1 row)

5. Running HITS with both max_iter and threshold:
DROP TABLE IF EXISTS hits_out, hits_out_summary;
'vertex',             -- Vertex table
'id',                 -- Vertex id column
'edge',               -- Edge table
'src=src, dest=dest', -- Comma delimited string of edge arguments
'hits_out',           -- Output table
20,                   -- Default max_iter
0.00001);             -- Threshold
SELECT * FROM hits_out ORDER BY id;

 id |      authority       |         hub
----+----------------------+---------------------
0 |    7.11260011825e-08 |    0.33810307986005
1 |    0.158326035587958 |   0.527815233930963
2 |    0.405461453180491 |   0.675913495026452
3 |    0.721835343230399 |   3.33021322089e-08
4 |    0.158326035587958 |   3.33021322089e-08
5 |    0.316459563893809 |   0.189770119973925
6 |    0.405307074424261 |   0.337972831786458
(7 rows)

SELECT * FROM hits_out_summary;

 __iterations__
-----------------+
20
(1 row)

The algorithm stopped at 20 iterations even though the convergence for threshold of 0.00001 is at 25 iterations. This is because max_iter was set to 20.
6. Running HITS with grouping column and default values for max_iter and threshold. Add more rows to the edge table to create different graphs based on the user_id column.
INSERT INTO edge VALUES
(0, 1, 2),
(0, 2, 2),
(0, 4, 2),
(1, 2, 2),
(1, 3, 2),
(2, 3, 2),
(3, 0, 2),
(4, 0, 2),
(5, 6, 2),
(6, 3, 2);
DROP TABLE IF EXISTS hits_out, hits_out_summary;
'vertex',             -- Vertex table
'id',                 -- Vertex id column
'edge',               -- Edge table
'src=src, dest=dest', -- Comma delimited string of edge arguments
'hits_out',           -- Output table
NULL,                 -- Default max_iter
NULL,                 -- Threshold
'user_id');           -- Grouping column
SELECT * FROM hits_out ORDER BY user_id, id;

 user_id | id |      authority       |         hub
---------+----+----------------------+----------------------
1 |  0 |    8.43871829093e-07 |    0.338306115082665
1 |  1 |    0.158459587238244 |    0.527865350448059
1 |  2 |    0.405627969689677 |    0.675800764727558
1 |  3 |    0.721775835521825 |    3.95111934817e-07
1 |  4 |    0.158459587238244 |    3.95111934817e-07
1 |  5 |    0.316385413093048 |    0.189719957843216
1 |  6 |    0.405199928761102 |    0.337944978189241
2 |  0 |    1.60841750444e-05 |    0.632262085114062
2 |  1 |    0.316079985713431 |    0.632529390899584
2 |  2 |    0.632364174872359 |    0.316347297480213
2 |  3 |    0.632694582987791 |    8.04208767442e-06
2 |  4 |    0.316079985713431 |    8.04208767442e-06
2 |  5 |                    0 |    1.22712519446e-10
2 |  6 |    2.45425034248e-10 |    0.316347297480213
(14 rows)

SELECT * FROM hits_out_summary order by user_id;

 user_id | __iterations__
---------+----------------
1 |             17
2 |             16
(2 rows)

Notes
1. On a Greenplum cluster, the edge table should be distributed by the source vertex id column for better performance.
2. This implementation of the HITS algorithm supports multigraph and each duplicated edge is considered for counting when calculating authority and hub scores.

Literature

[1] Kleinerg, Jon M., "Authoritative Sources in a Hyperlinked Environment", Journal of the ACM, Sept. 1999. https://www.cs.cornell.edu/home/kleinber/auth.pdf


text:
# Can this definite integral involving series be solved without a calculator?

I got this question today but I can't see if there is any good way to solve it by hand.

Evaluate the definite integral

$$\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}\,\mathrm{d}x$$

where the series in the numerator and denominator continue infinitely.

If you let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$, solving for $y$ we get $y=\frac{1\pm\sqrt{1+4x}}{2}$. And similarly for the denominator we have $z=\sqrt{xz}$. So $z=x$. So the integral simplifies to

$$\int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$

Now my problems are

1. I don't know what to with the $\pm$.

2. I tried to solve the integral by separating it as a sum of two fractions. But I can't solve $$\int_2^{12}\frac{\sqrt{1+4x}}{2x}\,\mathrm{d}x\,.$$

• – lab bhattacharjee Nov 28 '17 at 10:19
• "1.I don't know what to with the $\pm$" Use that you know $x\geq 2$ to help you choose one of the solutions. – Arthur Nov 28 '17 at 10:21
• "without a calculator..." First of all, a calculated value is not a mathematical solution. Second of all, maybe the calculator has an algebraic solver (like Macsyma). Is your real question "does the anti-derivative exist in closed form?" ? – Carl Witthoft Nov 28 '17 at 16:09
• Note that your technique for simplifying the nested radicals only works if the expressions converge. When the expressions don't converge, it will still give you a value, but that value will be meaningless. Now, in this case, you can show that both sequences are increasing and bounded above, and so must converge. But in general, you should be very cautious with the "$y = f(x, y)$, so $y = f(x, f(x, f(x, ...)))$" concept. – Paul Sinclair Nov 28 '17 at 17:32
• @CarlWitthoft I agree with you but I would have liked that better if it was an indefinite integral.;-) – tatan Nov 29 '17 at 3:25

As you did, let $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y}$. Clearly $y\ge0$ and $y$ satisfies $$y^2-y-x=0$$ from which you have $$y=\frac{1\pm\sqrt{4x+1}}{2}.$$ Since $x\in[2,12]$ and $y\ge0$, you must choose "$+$". Since if you choose "$-$", then $$y=\frac{1-\sqrt{4x+1}}{2}<0.$$ Now, under $t=\sqrt{1+4x}$ \begin{eqnarray} &&\int_2^{12}\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}dx\\ &=&\int_2^{12}\frac{1+\sqrt{1+4x}}{2x}dx\\ &=&\int_2^{12}\frac{1}{2x}dx+\int_2^{12}\frac{\sqrt{1+4x}}{2x}dx\\ &=&\frac12\ln x\bigg|_2^{12}+\int_3^7\frac{t^2}{t^2-1}dt\\ &=&\frac12\ln6+\bigg(t+\frac12\ln\frac{t-1}{t+1}\bigg)\bigg|_3^7\\ &=&\frac12\ln6+4+\frac12\ln\frac32\\ &=&\ln3+4. \end{eqnarray}

• Can you explain your substitution of t in the second integral? – tatan Nov 29 '17 at 3:18
• @tatan - if $t = \sqrt{1+4x}$, then $x = \frac{t^2 + 1}4$, so $dx = \frac t2 dt$. – Paul Sinclair Nov 29 '17 at 3:37
• @PaulSinclair Thanks – tatan Nov 29 '17 at 3:43
• correction: $x = \frac{t^2 -1}4$. – Paul Sinclair Nov 29 '17 at 3:44

Hint:

 1.

For real $y,$ $\sqrt{y^2}=|y|\ge0$

and $\sqrt{1+4x}\ge3$ for $2\le x\le12\implies1-\sqrt{1+4x}<0$

2.

Set $\sqrt{1+4x}=u\implies4x=u^2-1$

$\int_2^{12}\frac{1\pm\sqrt{1+4x}}{2x}dx$

Now, notice that the limits are defined for positive values of x. Also,

$y=\frac{1\pm\sqrt{1+4x}}{2}<0$ if you consider the negative sign for $x>0$.

But, $y=\sqrt{x+\sqrt{x+\sqrt{x+...}}}>0$ since the square root of a number is always positive for it to be a function (otherwise, for only one value of $x$, there'll be two values of y i.e., multiple mapping into range for only one value in the domain).

So, $y>0$.

So, $I= \int_2^{12}\frac{1+\sqrt{1+4x}}{2x}dx$

Now, just substitute $4x+1=u^2$ (as mentioned in one of the above answers).

$\implies I=\int \frac{u}{u-1}du$ with appropriate limits

For 2. notice that

$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx=\int_{2}^{12}\frac{1+4x}{2x\sqrt{1+4x}}=\frac{1}{2}\left(\int_{2}^{12}\frac{1}{x\sqrt{1+4x}}dx+4\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}\right).$$

Now the second integral is pretty easy

$$2\int_{2}^{12}\frac{dx}{\sqrt{1+4x}}=\left[\sqrt{1+4x}\right]_{2}^{12}=4,$$

$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{2}^{12}\frac{dx}{x\sqrt{1+\left(2\sqrt{x}\right)^2}},$$

let $2\sqrt{x}=\sinh y$, so $x=\frac{\sinh^2 y}{4}$ and $dx=\frac{\cosh y \sinh y}{2}\ dy$. Hence

$$\int_{2}^{12}\frac{dx}{2x\sqrt{1+4x}}=\frac{1}{2}\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{\frac{\cosh y \sinh y}{2}}{\frac{\sinh^2}{4} y\sqrt{1+\sinh^2 y}}dy=\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{dy}{\sinh y}=$$

$$=2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy.$$

Now let $e^y=z$, so $e^y\ dy=dz$

$$2\int_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}\frac{e^y}{e^{2y}-1}dy=2\int_{\eta}^{\psi}\frac{dz}{z^2-1},$$

where $\eta=e^{\sinh^{-1}{2\sqrt2}}$ and $\psi={e^{\sinh^{-1}{4\sqrt3}}}.$

$$2\int_{\eta}^{\psi}\frac{dz}{z^2-1}=\ln\left|\frac{z-1}{z+1}\right|_{\eta}^{\psi}=\ln\left|\frac{e^y-1}{e^y+1}\right|_{\sinh^{-1}{2\sqrt2}}^{\sinh^{-1}{4\sqrt3}}=$$

$$=\ln\left|\frac{e^{\sinh^{-1}{2\sqrt{x}}}-1}{e^{\sinh^{-1}{2\sqrt{x}}}+1}\right|_{2}^{12},$$

so

$$\int_{2}^{12}\frac{\sqrt{1+4x}}{2x}dx==\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{12}}}-1}{e^{\sinh^{-1}{2\sqrt{12}}}+1}\right|-\ln \left|\frac{e^{\sinh^{-1}{2\sqrt{2}}}-1}{e^{\sinh^{-1}{2\sqrt{2}}}+1}\right|+4.$$

For any:

$$\int_a^{b}\frac{1\pm\sqrt{1+4x}}{2x}dx$$

where $\frac{1\pm\sqrt{1+4x}}{2x}$ is defined for domain {a, b}

$$\frac{1}{2}\int_a^{b}\frac{1}{x}dx \pm \int_a^{b}\frac{\sqrt{1+4x}}{2x}dx$$ $$\frac{1}{2}(\ln{|b|} - \ln{|a|}) + \int_a^{b}\frac{\sqrt{1+4x}}{2x}dx - \int_a^{b}\frac{\sqrt{1+4x}}{2x}dx$$

$$\frac{1}{2}(\ln{|b|} - \ln{|a|})$$

Check if {2, 12} is a valid. I'm tired, so I won't picture what $\frac{1}{2x}\pm\frac{\sqrt{1+4x}}{2x}$ looks like. But you only have to check on the functions' decompositions.

Defined domains for each decompositions: $$\frac{1}{2x} \quad defined \quad \overline{\{0\}}$$

--

$$\frac{\sqrt{1+4x}}{2x} \Rightarrow \frac{1}{2x}\sqrt{1+4x}$$ $$1+4x<0 \Rightarrow x < -\frac{1}{4}$$ $$\frac{1}{2x}\sqrt{1+4x} \quad defined \quad \overline{\{0\} \; \cup \; x < -\frac{1}{4}}$$

therefore, $$\frac{1\pm\sqrt{1+4x}}{2x} \quad defined \quad \overline{\{0\} \; \cup \; x < -\frac{1}{4}}$$

so domain {2, 12} is safe and $\frac{1}{2}(\ln{|b|} - \ln{|a|})$ applies

• I think you should add $=$ signs in between – tatan Nov 28 '17 at 10:35
• I did not understand how you got from the $\pm$ to separating the $+$ and $-$ – tatan Nov 28 '17 at 10:36
• Huh? $a\pm b$ does not mean $a+b-b$. – Jack M Nov 28 '17 at 11:13
• Yes it does. Coz it encompasses 2 functions, exactly opposite of each other $\pm b$, both $b$ and $-b$. Makes sense coz we have a square root function. Gives us 2 possibilities as value. So if we integrate, $b$ and $-b$ integrals cancel out. – Dehbop Nov 28 '17 at 11:18
• OK, I MIGHT be wrong if $\frac{1\pm\sqrt{1+4x}}{2x}$ is the wrong identity. Honestly, I'm not that sure about that identity. Check with the asker. Though $\frac{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}{\sqrt{x\sqrt{x\sqrt{x}...}}}$, without sign, does seem to imply $\pm$ so it should carry out that there is $\pm$ in $\frac{1\pm\sqrt{1+4x}}{2x}$ – Dehbop Nov 28 '17 at 11:23


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# Do classical waves have reflectionless potentials?

In quantum mechanics, there exist reflectionless potentials, i.e. potentials whose transmission coefficient is one regardless of the incoming energy.

For a classical wave, such as a wave on a string, one could imagine creating a 'potential' by adding a term to the wave equation, $$\frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial^2 x} + k(x)y.$$ Physically, this would correspond to attaching the string to vertical springs with strength proportional to $k(x)$. Do there exist reflectionless potentials for this problem?

• The closest I can think of is noise cancellation but I don't think that's the same. – Bill Alsept May 24 '17 at 21:30


text:
by Zondrina
Tags: automorphisms
 P: 1,338 So I've been reading a bit about automorphisms today and I was wondering about something. I'm particularly talking about groups in this case, say a group G. So an automorphism is a bijective homomorphism ( endomorphism if you prefer ) of a group G. We use Aut(G) to denote the set of all automorphisms of G. There is also an inner automorphism induced by a in G. The map f := G → G such that f(x) = axa-1 for all x in G is called the inner automorphism of G induced by a. We use Inn(G) to denote the set of all inner automorphisms of G. My question now : If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite? It makes sense to me that they would be.
P: 1,623
 Quote by Zondrina If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite?
No. It is easy to check that $\mathrm{Aut}(\mathbb{Z}) \cong \mathbb{Z}/(2)$ and that $\mathrm{Inn}(\mathbb{Z}) = 0$.
P: 304
 Quote by Zondrina If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite? It makes sense to me that they would be.
No, not always. The integers, Z, is a group under addition. |Z| is infinite, but there are only two automorphisms on Z, given by f(x)=x (identity map) and g(x)=-x, respectively.

P: 1,338

 Quote by Erland No, not always. The integers, Z, is a group under addition. |Z| is infinite, but there are only two automorphisms on Z, given by f(x)=x (identity map) and g(x)=-x, respectively.
Ahh yes, that makes sense now because every other map doesn't generate all of Z.

So f(x) = 2x wouldn't be an automorphism for example. While it is a homomorphism, it is not injective, only surjective I believe.

Short proof to make sure : So f : Z → Z is already well defined for us. So we assume f(x) = f(y) → 2x = 2y → 2(x-y) = 0. Since 2|0, clearly x = y and f is injective. Now notice that the n elements of Z get mapped to fewer elements, so f is not surjective. Easy to show it's a homomorphism.

So f(x) = x and g(x) = -x are the only things that could be in Aut(Z) since they are the only maps which turn out to be automorphisms.

As for Inn(Z) = 0. That confused me a bit.

EDIT : Wouldn't it be Inn(Z) = { 0 } ?
P: 1,623
 Quote by Zondrina As for Inn(Z) = 0. That confused me a bit. EDIT : Wouldn't it be Inn(Z) = { 0 } ?
They mean the same thing. Both of them are shorthand for the statement that $\mathrm{Inn}(\mathbb{Z})$ is the trivial group.
Mentor
P: 16,701
 Quote by Zondrina So f(x) = 2x wouldn't be an automorphism for example. While it is a homomorphism, it is not injective, only surjective I believe.
You switched the two around. It is injective but not surjective, which is what you proved next.
P: 1,338
 Quote by micromass You switched the two around. It is injective but not surjective, which is what you proved next.
Oh whoops, mistype. Thanks for noticing.

Thanks for clarifying jgens I think I understand this now.

 Related Discussions Calculus & Beyond Homework 1 Linear & Abstract Algebra 2 Linear & Abstract Algebra 3 Linear & Abstract Algebra 2 Calculus & Beyond Homework 6


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## Derivatives of Logarithmic and Exponential Functions

(This topic is also in Section 4.5 in Applied Calculus 5e and Section or Section 11.5 of Finite Mathematics and Applied Calculus 5e)

### Derivatives of Logarithmic Functions

The derivatives of the logarithmic functions are given as follows:

Derivative of logb and ln

\frac{d}{dx} \log_b(x) = \frac{1}{x\.\ln b}

An important special case is this:

\frac{d}{dx} \ln x = \frac{1}{x\.\ln e} = \frac{1}{x}     Because \ln e = 1

Quick Examples

\frac{d}{dx} \log_3(x) = \frac{1}{x\.\ln 3}
\frac{d}{dx}[4\.\ln x] = 4\frac{d}{dx}[\ln x] = 4\frac{1}{x} = \frac{4}{x}
\frac{d}{dx}[4\.\log_3(x)] = 4\frac{d}{dx}[\log_3(x)] = 4\frac{1}{x\.\ln 3} = \frac{4}{x\.\ln 3}
\frac{d}{dx}[x^2\ln x] = 2x\.\ln x + x^2\frac{1}{x} = 2x\.\ln x + x     By the product rule
Some for you

Note Use graphing calculator format to input your answers (spaces are ignored). Here are some examples of expressions involving logarithms and exponentials:

Mathematical Expression Input Formula
 ln x
 ln(x)
 (4x+2) ln x
 (4x+2)*ln(x)
 ln(ex + 1)
 ln(e^x + 1)
 log3x
 ln(x)/ln(3)
 log3(x+1)2x2
 ln(x+1) / ( ln(3) * 2*x^2 )

 \frac{d}{dx}[] = \frac{d}{dx}[] = \frac{d}{dx}[] = \frac{d}{dx}\Bigleft(\Bigright) =

Q Where do these formulas come from?
A For derivations of these formulas, consult Section 4.5 in Applied Calculus, or Section 11.5 in Finite Mathematics and Applied Calculus.

Q We know how to differentiate expressions that contain the logarithm of x. What about the logarithm of a more complicated quantity, for instance ln(x2-3x+2)?
A To differentiate something like that, we need to use the chain rule. Here is a list of chain rule items that include some involving logarithms.

Original Rule
Generalized Rule
(Chain Rule)
 ddx f(x) = g(x)
 ddx f(u) = g(u) dudx
General form of the Chain Rule
 ddx xn = nx n-1
 ddx un = nun-1 dudx
Generalized Power Rule
 ddx 4x-1/2 = -2x-3/2
 ddx 4u-1/2 = -2u-3/2 dudx
An example of the above rule
 ddx sin x = cos x
 ddx sin u = cos u dudx
Take me to text on trig functions!
 ddx ln x = 1x
 ddx ln (u) = 1u dudx
The derivative of the natural logarithm of a quantity is the
reciprocal of that quantity, times the derivative of that quantity.
 ddx logb(x) = 1x ln(b)
 ddx logb(u) = 1u ln(b) dudx

Quick Examples

 \frac{d}{dx}\ln(3x^2-x) = \frac{1}{3x^2-x}\frac{d}{dx}[3x^2-x] = \frac{1}{3x^2-x}(6x-1) = \frac{6x-1}{3x^2-x} \frac{d}{dx} = \frac{d}{dx} = \frac{d}{dx} =

\frac{d}{dx} = ?

\frac{d}{dx} = ?

### Derivative of Log of the Absolute Value

Something curious happens if we take the derivative of the logarithm of the absolute value of x:
 \frac{d}{dx} \ln |x| = \frac{1}{|x|} \frac{d}{dx} |x| By the chain rule above = \frac{1}{|x|} \frac{|x|}{x} The derivative of |x| is |x|/x = \frac{1}{x} Exactly the same as the derivative of ln x!
In other words, replacing x by the absolute value of x has absolutely no effect on the derivative of the natural logarithm. Similarly it has no effect on the derivative of the logarithm of x to any base, or on the logarithm of any quantity. Example:
\frac{d}{dx}\ln|3x^2-x| = \frac{1}{3x^2-x}(6x-1) = \frac{6x-1}{3x^2-x}
as though the absolute value wasn't there! Thus the answers to all the questions above with "ln (--) replaced by "ln |--|" are exactly the same. See the textbook Applied Calculus for more details.

### Derivatives of Exponential Functions

The derivatives of the exponential functions are given as follows.

Derivative of bx and ex

\frac{d}{dx} b^x = b^x \ln b

An important special case is this:

\frac{d}{dx} e^x = e^x
Quick Examples

\frac{d}{dx}[3^x] = 3^x \ln 3
\frac{d}{dx}[2e^x] = 2\frac{d}{dx}[e^x] = 2e^x
\frac{d}{dx}[x^2e^x] = 2x\.e^x + x^2e^x = e^x(2x + x^2)     By the product rule
Some for you

 \frac{d}{dx}[] = \frac{d}{dx}[] =

Q Where do these formulas come from?
A Consult Section 4.5 of Applied Calculus, Section 11.5 of Finite Mathematics and Applied Calculus.

These formulas allow us to further expand our table of derivatives involving the chain rule:

Original Rule
Generalized Rule
(Chain Rule)
 ddx f(x) = g(x)
 ddx f(u) = g(u) dudx
General form of
Chain Rule
 ddx xn = nx n-1
 ddx un = nun-1 dudx
Generalized Power Rule
 ddx 4x-1/2 = -2x-3/2
 ddx 4u-1/2 = -2u-3/2 dudx
An example of the above rule
 ddx sin x = cos x
 ddx sin u = cos u dudx
Take me to text on trig functions!
 ddx ln x = 1x
 ddx ln (u) = 1u dudx
The derivative of the natural logarithm of a quantity is the
reciprocal of that quantity, times the derivative of that quantity.
 ddx logb(x) = 1x ln(b)
 ddx logb(u) = 1u ln(b) dudx
 ddx ex = ex
 ddx eu = eu dudx
The derivative of e raised to a quantity is e raised to
that quantity, times the derivative of that quantity.
 ddx bx = bx ln(b)
 ddx bu = bu ln(b) dudx

Quick Examples

 \frac{d}{dx}[e^{-x}] = e^{-x}\frac{d}{dx}[-x] = -e^{-x} \frac{d}{dx}[e^{3x2-x}] = e^{3x2-x}\frac{d}{dx}[3x2-x] = (6x-1)e^{3x2-x}

Some for you

 \frac{d}{dx}[] = \frac{d}{dx}[] = \frac{d}{dx}[] =

In this question, several of the choices are based on actual answers students gave in a test. Only one is correct!

\frac{d}{dx} = ?

Top of Page

Last Updated: October 2009


text:
Related Searches
Definitions

# Predicate functor logic

In mathematical logic, predicate functor logic (PFL) is one of several ways to express first-order logic (formerly known as predicate logic) by purely algebraic means, i.e., without quantified variables. PFL employs a small number of algebraic devices called predicate functors (or predicate modifiers) that operate on terms to yield terms. PFL is mostly the invention of the logician and philosopher Willard Quine.

## Motivation

This source for this section, as well as for much of this entry, is Quine (1976). Quine proposed PFL as a way of algebraizing first-order logic in a manner analogous to how Boolean algebra algebraizes sentential logic. He designed PFL to have exactly the expressive power of first-order logic with identity. Hence the metamathematics of PFL are exactly those of first-order logic with no interpreted predicate letters: both logics are sound, complete, and undecidable. Most work Quine published on logic and mathematics in the last 30 year of his life touched on PFL in some way.

Quine took "functor" from the writings of his friend Rudolf Carnap, the first to employ it in philosophy and mathematical logic, and defined it as follows:

"The word functor, grammatical in import but logical in habitat... is a sign that attaches to one or more expressions of given grammatical kind(s) to produce an expression of a given grammatical kind." (Quine 1982: 129)

Ways other than PFL to algebraize first-order logic include:

• Cylindric algebra by Alfred Tarski and his American students. The simplified cylindric algebra proposed in Bernays (1959) led Quine to write the paper containing the first use of the phrase "predicate functor";
• The polyadic algebra of Paul Halmos. By virtue of its economical primitives and axioms, this algebra most resembles PFL;
• Relation algebra algebraizes the fragment of first-order logic consisting of formulas having no atomic formula lying in the scope of more than three quantifiers. That fragment suffices, however, for Peano arithmetic and the axiomatic set theory ZFC; hence relation algebra, unlike PFL, is incompletable. Most work on relation algebra since about 1920 has been by Tarski and his American students. The power of relation algebra did not become manifest until the monograph Tarski and Givant (1987), published after the three important papers bearing on PFL, namely Bacon (1985), Kuhn (1983), and Quine (1976);
• Combinatory logic builds on combinators, higher order functions whose domain is another combinator or function, and whose range is yet another combinator. Hence combinatory logic goes beyond first-order logic by having the expressive power of set theory, which makes combinatory logic vulnerable to paradoxes. A predicate functor, on the other hand, simply maps predicates (also called terms) into predicates.

PFL is arguably the simplest of these formalisms, yet also the one about which the least has been written.

Quine had a lifelong fascination with combinatory logic, attested to by his (1976) and his introduction to the translation in Van Heijenoort (1967) of the paper by the Russian logician Moses Schonfinkel founding combinatory logic. When Quine began working on PFL in earnest, in 1959, combinatory logic was commonly deemed a failure for the following reasons:

• Until Dana Scott began writing on the model theory of combinatory logic in the late 1960s, nearly all work on that logic had been by Haskell Curry, his students, or by Robert Feys in Belgium;
• Satisfactory axiomatic formulations of combinatory logic were slow in coming. In the 1930s, some formulations of combinatory logic were found to be inconsistent. Curry also discovered the Curry paradox, peculiar to combinatory logic;
• The lambda calculus, with the same expressive power as combinatory logic, was seen as a superior formalism.

## Kuhn's formalization

The PFL syntax, primitives, and axioms described in this section are largely Kuhn's (1983). The semantics of the functors are Quine's (1982). The rest of this entry incorporates some terminology from Bacon (1985).

### Syntax

An atomic term is an upper case Latin letter, I and S excepted, followed by a numerical superscript called its degree, or by concatenated lower case variables, collectively known as an argument list. The degree of a term conveys the same information as the number of variables following a predicate letter. An atomic term of degree 0 denotes a Boolean variable or a truth value. The degree of I is invariably 2 and so is not indicated.

The "combinatory" (the word is Quine's) predicate functors, all monadic and peculiar to PFL, are Inv, inv, , +, and p. A term is either an atomic term, or constructed by the following recursive rule. If τ is a term, then Invτ, invτ, τ, +τ, and pτ are terms. A functor with a superscript n, n a natural number > 1, denotes n consecutive applications (iterations) of that functor.

A formula is either a term or defined by the recursive rule: if α and β are formulas, then αβ and ~(α) are likewise formulas. Hence "~" is another monadic functor, and concatenation is the sole dyadic predicate functor. Quine called these functors "alethic." The natural interpretation of "~" is negation; that of concatenation is any connective that, when combined with negation, forms a functionally complete set of connectives. Quine's preferred functionally complete set was conjunction and negation. Thus concatenated terms are taken as conjoined. The notation + is Bacon's (1985); all other notation is Quine's (1976; 1982). The alethic part of PFL is identical to the Boolean term schemata of Quine (1982).

As is well known, the two alethic functors could be replaced by a single dyadic functor with the following syntax and semantics: if α and β are formulas, then (αβ) is a formula whose semantics are "not (α and/or β)" (see NAND and NOR).

### Axioms and semantics

Quine set out neither axiomatization nor proof procedure for PFL. The following axiomatization of PFL, one of two proposed in Kuhn (1983), is concise and easy to describe, but makes extensive use of free variables and so does not do full justice to the spirit of PFL. Kuhn gives another axiomatization dispensing with free variables, but that is harder to describe and that makes extensive use of defined functors. Kuhn proved both of his PFL axiomatizations sound and complete.

This section is built around the primitive predicate functors and a few defined ones. The alethic functors can be axiomatized by any set of axioms for sentential logic whose primitives are negation and one of ∧ or ∨. Equivalently, all tautologies of sentential logic can be taken as axioms.

For each combinatory functor, we give its semantics, as per Quine (1982), stated in terms of abstraction (set builder notation), followed by either the relevant axiom from Kuhn (1983), or a definition from Quine (1976). The notation $\left\{x_1...x_n : Fx_1...x_n\right\}$ denotes the set of n-tuples satisfying the atomic formula $Fx_1...x_n.$

• Identity, I, is defined as:

$IFx_1x_2...x_n leftrightarrow Fx_1x_1...x_n leftrightarrow Fx_2x_2...x_n.$
Identity is reflexive (Ixx), symmetric (IxyIyx), transitive ((IxyIyz) → Ixz), and obeys the substitution property:
$\left(Fx_1...x_n and Ix_1y\right) rightarrow Fyx_2...x_n.$

• Padding, +, adds a variable to the left of any argument list.

$+F^n =_\left\{def\right\} \left\{x_0x_1...x_n : F^n x_1...x_n\right\}.$
$+Fx_1...x_n leftrightarrow Fx_2...x_n.$

• Cropping, , erases the leftmost variable in any argument list.

$exist F^n =_\left\{def\right\} \left\{x_2...x_n : exist x_1 F^n x_1...x_n\right\}.$
$Fx_1...x_n rightarrow exist Fx_2...x_n.$

Cropping enables two useful defined functors:

• Reflection, S:

$SF^n =_\left\{def\right\} \left\{x_2...x_n : F^n x_2x_2...x_n\right\}.$
$SF^n leftrightarrow exist IF^n.$
S generalizes the notion of reflexivity to all terms of any finite degree greater than 2. N.B: S should not be confused with the primitive combinator S of combinatory logic.

$F^m times G^n leftrightarrow F^m exist^m G^n.$
Here only, Quine adopted an infix notation, because this infix notation for Cartesian product is very well established in mathematics. Cartesian product allows restating conjunction as follows:
$F^mx_1...x_mG^nx_1...x_n leftrightarrow \left(F^m times G^n\right)x_1...x_mx_1...x_n.$
Reorder the concatenated argument list so as to shift a pair of duplicate variables to the far left, then invoke S to eliminate the duplication. Do this as many times as required, resulting is an argument list of length max{m,n).

The next three functors enable reordering argument lists at will.

• Major inversion, Inv, rotates the variables in an argument list to the right, so that the last variable becomes the first.

$InvF^n =_\left\{def\right\} \left\{x_1...x_n : F^nx_nx_1...x_\left\{n-1\right\}\right\}.$
$Inv Fx_1...x_n leftrightarrow Fx_nx_1...x_\left\{n-1\right\}.$

• Minor inversion, inv, swaps the first two variables in an argument list.

$invF^n =_\left\{def\right\} \left\{x_1...x_n : F^n x_2x_1...x_n\right\}.$
$invFx_1...x_n leftrightarrow Fx_2x_1...x_n.$

• Permutation, p, rotates the second through last variables in an argument list to the left, so that the second variable becomes the last.

$pF^n =_\left\{def\right\} \left\{x_1...x_n : F^n x_1x_3...x_nx_2\right\}.$
$pFx_1...x_n leftrightarrow Inv inv Fx_1x_3...x_nx_2.$
Given an argument list consisting of n variables, p implicitly treats the last n-1 variables like a bicycle chain, with each variable constituting a link in the chain. One application of p advances the chain by one link. k consecutive applications of p to Fn moves the k+1 variable to the second argument position in F.

When n=2, Inv and inv merely interchange x1 and x2. When n=1, they have no effect. Hence p has no effect when n<3.

Kuhn (1983) takes Major inversion and Minor inversion as primitive. The notation p in Kuhn corresponds to inv; he has no analog to Permutation and hence has no axioms for it. If, following Quine (1976), p is taken as primitive, Inv and inv can be defined as nontrivial combinations of +, , and iterated p.

The following table summarizes how the functors affect the degrees of their arguments.

Expression Degree
$p; Inv; inv; lnot; I$ no change
$+F^\left\{n-1\right\};$ $exist F^\left\{n+1\right\};$ $SF^\left\{n+1\right\}$ $n$
$alpha^mbeta^n;$ $F^m times G^n$ $max\left(m,n\right)$

### Rules

All instances of a predicate letter may be replaced by another predicate letter of the same degree, without affecting validity. The rules are:

• Modus ponens;
• Let α and β be PFL formulas in which $x_1$ does not appear. Then if $\left(alpha and Fx_1...x_n\right) rightarrow beta$ is a PFL theorem, then $\left(alpha and exist Fx_2...x_n\right) rightarrow beta$ is likewise a PFL theorem.

### Some useful results

Instead axiomatizing PFL, Quine (1976) proposed the following conjectures as candidate axioms.

• $exist I$

n-1 consecutive iterations of p restores the status quo ante:

• $F^n leftrightarrow p^\left\{n-1\right\}F^n$

+ and annihilate each other:

• $F^n rightarrow +exist F^n$

• $F^n leftrightarrow exist +F^n$

Negation distributes over +, , and p:

• $+lnot F^n leftrightarrow lnot +F^n$

• $lnotexist F^n rightarrow exist lnot F^n$

• $plnot F^n leftrightarrow lnot pF^n$

+ and p distributes over conjunction:

• $+\left(F^nG^m\right) leftrightarrow \left(+F^n+G^m\right)$

• $p\left(F^nG^m\right) leftrightarrow \left(pF^npG^m\right)$

Identity has the interesting implication:

• $IF^n rightarrow p^\left\{n-2\right\} exist p+F^n$

Quine also conjectured the rule: If $alpha$ is a PFL theorem, then so are $palpha,$ $+alpha,$ and $lnot exist lnot alpha$.

## Bacon's work

Bacon (1985) takes the conditional, negation, Identity, Padding, and Major and Minor inversion as primitive, and Cropping as defined. Employing terminology and notation differing somewhat from the above, Bacon (1985) sets out two formulations of PFL:

• A natural deduction formulation in the style of Frederick Fitch. Bacon proves this formulation sound and complete in full detail.
• An axiomatic formulation which Bacon asserts, but does not prove, equivalent to the preceding one. Some of these axioms are simply Quine conjectures restated in Bacon's notation.

Bacon also:

## From first-order logic to PFL

The following algorithm is adapted from Quine (1976: 300-2). Given a closed formula of first-order logic, first do the following:

Now apply the following algorithm to the preceding result:

1. Translate the matrices of the most deeply nested quantifiers into disjunctive normal form, consisting of disjuncts of conjuncts of terms, negating atomic terms as required. The resulting subformula contains only negation, conjunction, disjunction, and existential quantification.

2. Distribute the existential quantifiers over the disjuncts in the matrix using the rule of passage (Quine 1982: 119):

$exist x\left[alpha \left(x\right) or gamma \left(x\right)\right] leftrightarrow \left(exist x alpha \left(x\right) or exist x gamma \left(x\right)\right).$

3. Replace conjunction by Cartesian product, by invoking the fact:

4. Concatenate the argument lists of all atomic terms, and move the concatenated list to the far right of the subformula.

5. Use Inv and inv to move all instances of the quantified variable (call it y) to the left of the argument list.

6. Invoke S as many times as required to eliminate all but the last instance of y. Eliminate y by prefixing the subformula with one instance of .

7. Repeat (1)-(6) until all quantified variables have been eliminated. Eliminate any disjunctions falling within the scope of a quantifier by invoking the equivalence:

$\left(alpha or beta or ...\right) leftrightarrow lnot\left(lnotalpha and lnotbeta and ...\right).$

The reverse translation, from PFL to first-order logic, is discussed in Quine (1976: 302-4).

The canonical foundation of mathematics is axiomatic set theory, with a background logic consisting of first-order logic with identity, with a universe of discourse consisting entirely of sets. There is a single predicate letter of degree 2, interpreted as set membership. The PFL translation of the canonical axiomatic set theory ZFC is not difficult, as no ZFC axiom requires more than 6 quantified variables.

## References

• Bacon, John, 1985, "The completeness of a predicate-functor logic," Journal of Symbolic Logic 50: 903-26.
• Paul Bernays, 1959, "Uber eine naturliche Erweiterung des Relationenkalkuls" in Heyting, A., ed., Constructivity in Mathematics. North Holland: 1-14.
• Kuhn, Stephen T., 1983, " An Axiomatization of Predicate Functor Logic," Notre Dame Journal of Formal Logic 24: 233-41.
• Willard Quine, 1976, "Algebraic Logic and Predicate Functors" in Ways of Paradox and Other Essays, enlarged ed. Harvard Univ. Press: 283-307.
• --------, 1982. Methods of Logic, 4th ed. Harvard Univ. Press. Chpt. 45.
• Sommers, Fred, 1982. The Logic of Natural Language. Oxford Univ. Press.
• Alfred Tarski and Givant, Steven, 1987. A Formalization of Set Theory Without Variables. AMS.
• Jean Van Heijenoort, 1967. From Frege to Godel: A Source Book on Mathematical Logic. Harvard Univ. Press.

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# Inequalities – Square inequality – Exercise 1700

Exercise

Solve the inequality:

$$8(1-2x)<{(2x-3)}^2$$

$$x\neq -\frac{1}{2}$$

Solution

$$8(1-2x)<{(2x-3)}^2$$

Open brackets:

$$8-16x< 4x^2-12x+9$$

Move everything to one side:

$$0<-8+16x+ 4x^2-12x+9$$

$$0<4x^2+4x+1$$

It is a square inequality. Its coefficients are

$$a=4, b=4, c=1$$

The coefficient of the square expression (a) is positive, so the parabola (quadratic equation graph) “smiles” (= bowl-shaped). The sign of the inequality means we are looking for the sections the parabola is above the x-axis. We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{-4\pm \sqrt{4^2-4\cdot 4\cdot 1}}{2\cdot 4}=$$

$$=\frac{-4\pm \sqrt{0}}{8}=$$

$$=\frac{-4}{8}=-\frac{1}{2}$$

Hence, the factorizing of the equation is

$$4x^2+4x+1={(x+\frac{1}{2})}^2$$

Hence, we receive that our inequality equals

$${(x+\frac{1}{2})}^2>0$$

The left side is a squared phrase and therefore is always greater than zero except at a point where it is zero. Let’s see when it’s equal to zero:

$$x+\frac{1}{2}=0$$

$$x=-\frac{1}{2}$$

Therefore, the inequality is true for all x except this point.

In other words, inequality solution is

$$x\neq-\frac{1}{2}$$

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!

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Reducibity of n-ary semigroups: from quasitriviality towards idempotency - Archive ouverte HAL Access content directly
Preprints, Working Papers, ... Year :

## Reducibity of n-ary semigroups: from quasitriviality towards idempotency

(1) , (2) , (2) , (3)
1
2
3
Miguel Couceiro
Jean-Luc Marichal
• Function : Author
• PersonId : 908349
Pierre Mathonet
• Function : Author

#### Abstract

Let $X$ be a nonempty set. Denote by $\mathcal{F}^n_k$ the class of associative operations $F\colon X^n\to X$ satisfying the condition $F(x_1,\ldots,x_n)\in\{x_1,\ldots,x_n\}$ whenever at least $k$ of the elements $x_1,\ldots,x_n$ are equal to each other. The elements of $\mathcal{F}^n_1$ are said to be quasitrivial and those of $\mathcal{F}^n_n$ are said to be idempotent. We show that $\mathcal{F}^n_1=\cdots =\mathcal{F}^n_{n-2}\varsubsetneq\mathcal{F}^n_{n-1}\varsubsetneq\mathcal{F}^n_n$. The class $\mathcal{F}^n_1$ was recently characterized by Couceiro and Devillet, who showed that its elements are reducible to binary associative operations. However, some elements of $\mathcal{F}^n_n$ are not reducible. In this paper, we characterize the class $\mathcal{F}^n_{n-1}\setminus\mathcal{F}^n_1$ and show that its elements are reducible. In particular, we show that each of these elements is an extension of an $n$-ary Abelian group operation whose exponent divides $n-1$.

#### Domains

Mathematics [math]

### Dates and versions

hal-02293908 , version 1 (22-09-2019)

### Identifiers

• HAL Id : hal-02293908 , version 1

### Cite

Miguel Couceiro, Jimmy Devillet, Jean-Luc Marichal, Pierre Mathonet. Reducibity of n-ary semigroups: from quasitriviality towards idempotency. 2019. ⟨hal-02293908⟩

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# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

##### Operators
 a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
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Real linear Kronecker product operations. (English) Zbl 1104.15021
The Kronecker product operations in the real linear matrix analytic setting are developed. More versatile operations are proposed that still possess benefits of the usual Kronecker product. Particular attention is paid to the square matrix case motivated by applications to preconditioning and solving systems of linear equations. A number of generically valid factorizations for the inverse are derived. In this connection, approximating an ${\mathbb R}$-linear operator with the arising Kronecker product structures is quite natural and four associated matrix nearness problems are presented. A respective orthogonal decomposition for real matrices with the arising Kronecker singular value decompositions is introduced. Examples of matrix equations leading to ${\mathbb R}$-linear operators with Kronecker product parts are given. Among many ways of solving these equations, a factorization of QZ-type is derived.

##### MSC:
 15A69 Multilinear algebra, tensor products 65F10 Iterative methods for linear systems 65F35 Matrix norms, conditioning, scaling (numerical linear algebra) 15A23 Factorization of matrices
JDQR; JDQZ
Full Text:
##### References:
 [1] Agler, J.; Helton, W.; Stankus, M.: Classification of hereditary matrices. Linear algebra appl. 274, 125-160 (1998) · Zbl 0939.47014 [2] Bai, Z.; Demmel, J.; Dongarra, J.; Ruhe, A.; Van Der Vorst, H.: Templates for the solution of algebraic eigenvalue problems: A practical guide. (2000) · Zbl 0965.65058 [3] Bevis, J. H.; Hall, F.; Hartwig, R.: The matrix equation AX‾-XB=C and its special cases. SIAM J. Matrix anal. Appl. 9, 348-359 (1988) · Zbl 0655.15013 [4] Bhatia, R.; Rosenthal, P.: How and why to solve the operator equation AX-XB=Y. Bull. London math. Soc. 29, 1-21 (1997) · Zbl 0909.47011 [5] Eirola, T.; Huhtanen, M.; Von Pfaler, J.: Solution methods for R-linear problems in cn. SIAM J. Matrix anal. Appl. 25, 804-828 (2004) · Zbl 1112.65025 [6] Ford, J.; Tyrtyshnikov, E.: Combining Kronecker product approximation with discrete wavelet transforms to solve dense, function-related linear systems. SIAM J. Sci. comput. 25, 961-981 (2003) · Zbl 1046.65020 [7] Gardiner, J.; Laub, A.; Amato, J.; Moler, C.: Solution of the Sylvester matrix equation AXBT+CXDT=E. ACM trans. Math. software 18, 223-231 (1992) · Zbl 0893.65026 [8] George, A.; Ikramov, Kh.; Matushkina, E.; Tang, W. -P.: On a QR-like algorithm for some structured eigenvalue problems. SIAM J. Matrix anal. Appl. 16, 1107-1126 (1995) · Zbl 0837.65035 [9] Golub, G. H.; Van Loan, C. F.: Matrix computations. (1996) · Zbl 0865.65009 [10] Harvey, F. R.: Spinors and calibrations. Perspectives in mathematics 9 (1990) · Zbl 0694.53002 [11] Hong, Y. P.; Horn, R. A.: A canonical form for matrices under consimilarity. Linear algebra appl. 102, 143-168 (1988) · Zbl 0657.15008 [12] Horn, R. A.; Johnson, C. R.: Topics in matrix analysis. (1991) · Zbl 0729.15001 [13] M. Huhtanen, O. Nevanlinna, The real linear resolvent and cosolvent operators, J. Operator Theory, in press. · Zbl 1164.47006 [14] M. Huhtanen, O. Nevanlinna, Approximating real linear operators, submitted for publication. · Zbl 1123.15005 [15] M. Huhtanen, O. Nevanlinna, Real linear matrix analysis, Banach Center Publications, in press. · Zbl 1117.15005 [16] Huhtanen, M.; Von Pfaler, J.: The real linear eigenvalue problem in cn. Linear algebra appl. 394, 169-199 (2005) · Zbl 1070.15006 [17] Mccullough, S. A.; Rodman, L.: Hereditary classes of operators and matrices. Amer. math. Monthly 104, 415-430 (1997) · Zbl 0909.47016 [18] Moler, C.; Stewart, G. W.: An algorithm for generalized matrix eigenvalue problems. SIAM J. Numer. anal. 10, 241-256 (1973) · Zbl 0253.65019 [19] Regalia, P.; Mitra, S.: Kronecker products, unitary matrices and signal processing applications. SIAM rev. 31, 586-613 (1989) · Zbl 0687.15010 [20] Seberry, J.; Zhang, X. -M.: Some orthogonal matrices constructed by strong Kronecker product multiplication. Australas. J. Combin. 7, 213-224 (1993) · Zbl 0779.05013 [21] Van Loan, C. F.: The ubiquitous Kronecker product. J. comput. Appl. math. 123, 85-100 (2000) · Zbl 0966.65039 [22] Van Loan, C. F.; Pitsianis, N.: Approximation with Kronecker products, linear algebra for large scale and real-time applications (Leuven, 1992). NATO adv. Sci. inst. Ser. E appl. Sci. 232 (1993) · Zbl 0813.65078 [23] Youla, D. C.: A normal form for a matrix under the unitary congruence group. Canad. J. Math. 13, 694-704 (1961) · Zbl 0103.25201


text:
Why is continuity required for sheaf-theoretic definitions of a structure on a space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:31:16Z http://mathoverflow.net/feeds/question/25229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25229/why-is-continuity-required-for-sheaf-theoretic-definitions-of-a-structure-on-a-sp Why is continuity required for sheaf-theoretic definitions of a structure on a space Akela 2010-05-19T13:32:42Z 2010-05-20T12:11:36Z
For example, I take differentiability, analyticity, and algebraicity(of a function). All(more or less) imply continuity. So when we define a differentiable function on $\mathbb R^n$ or an analytic function on $\mathbb C^n$, or a regular map on an affine space, we do not explicitly require that the functions are continuous. It follows automatically from the stronger condition.

But, when I look at the definitions in books of a global structure using sheaf theory, for a global definition of a morphism, ie on a differentiable manifold or an analytic space, or an abstract algebraic variety, the definition of a morphism requires a priori that the map be continuous, and then one requires that there is additionally a morphism of sheaves of algebras(of the suitable type of structure sheaves, depending on the local model used).

Why is this so? Is it something done for fancy, or is there a real need for the extra continuity assumption? I mean could things go wrong if this assumption is dropped?
 http://mathoverflow.net/questions/25229/why-is-continuity-required-for-sheaf-theoretic-definitions-of-a-structure-on-a-sp/25230#25230 Answer by Martin Brandenburg for Why is continuity required for sheaf-theoretic definitions of a structure on a space Martin Brandenburg 2010-05-19T14:15:05Z 2010-05-19T14:15:05Z
Let $M,N$ two manifolds and $f : M \to N$ a (set-theoretic) map. Then there are (at least) two definitions for $f$ to be smooth:

(1) For every ball $B \subseteq N$ the preimage $f^{-1}(B)$ can be covered with balls $C \subseteq M$ such that the induced maps $C \to B$ are smooth.

(2) $f$ is continuous and for every ball $B \subseteq N$ and every ball $C \subseteq f^{-1}(B)$ the induced map $C \to B$ is smooth.

Remark that in (1) it follows automatically that $f$ is continuous. However, the second statement in (2) does not imply continuity because it is possible that $f^{-1}(B)$ contains no ball at all, or just not enough.

The same is true for other subsheaves of continuous functions mentioned in the question.
 http://mathoverflow.net/questions/25229/why-is-continuity-required-for-sheaf-theoretic-definitions-of-a-structure-on-a-sp/25260#25260 Answer by Andrew Stacey for Why is continuity required for sheaf-theoretic definitions of a structure on a space Andrew Stacey 2010-05-19T18:36:31Z 2010-05-19T18:36:31Z
As Andrea hints, if you start with sheaves then you need continuity to even begin talking about morphisms of sheaves.

However, if you're interesting in just defining, say, a smooth map between manifolds then you can simply write "$f \colon M \to N$ is smooth if, whenever $c \colon \mathbb{R} \to M$ is a smooth curve then $f \circ c \colon \mathbb{R} \to N$ is smooth". No assumption about continuity is needed there.

Indeed, once one gets to more exotic spaces, continuity becomes a hassle and is best left to one side. For example, the evaluation map $E \times E^* \to \mathbb{R}$ is smooth for any locally convex topological vector space, $E$, but is only continuous for $E$ a normed vector space.
 http://mathoverflow.net/questions/25229/why-is-continuity-required-for-sheaf-theoretic-definitions-of-a-structure-on-a-sp/25352#25352 Answer by David Carchedi for Why is continuity required for sheaf-theoretic definitions of a structure on a space David Carchedi 2010-05-20T12:11:36Z 2010-05-20T12:11:36Z
The simple answer is because you are describing your "spaces" (manifolds, variety etc.) as locally ringed spaces, in particular objects in the category of topological spaces + more data. The arrows in the category of topological spaces are precisely continuous maps, so this is where continuity comes in. To be more specific, let $C$ be any category, and let $S:C \to Cat$ be any pseudo-functor (weak 2-functor). "Pretend" that $S$ is the assignment of each object $c \in C_0$ its category of "sheaves of algebraic objects", e.g. if $C$ is topological spaces you could let $S$ be $S:X \mapsto Sh_{rings}(X)$ which associates to a space $X$ the category of sheaves of local rings over $X$. For any such $S$, one can take its Grothendieck construction, which yields a category fibred over $C$, $\int_C{S} \to C$. The objects of $\int_C{S}$ are pairs $(c,s)$ with $c \in C_0$ and $s \in S(C)_0$ and the maps $(c,s) \to (d,t)$ are pairs $(f,g)$ such that $f:c \to d$ in $C$ and $g:f^*(t) \to s$ in $S(C)$, and the functor $\int_C{S} \to C$ sends $(c,s)$ to $c$ and $(f,g)$ to $f$. If $S$ and $C$ are taken to be $Sh_{rings}$ and $Top$, then you get exactly the category of locally ringed spaces, for example. By construction, the "underlying morphism" of a morphism $(c,s) \to (d,t)$ is a morphism $f:c \to d$. If $C$ is $Top$, then of course, this means it is a morphism in $Top$, hence continuous.