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(ANSWER) MATH 225N-Week 6 Assignment Confidence Interval for Mean Population Standard Deviation Known - Essay Prowess
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# (ANSWER) MATH 225N-Week 6 Assignment Confidence Interval for Mean Population Standard Deviation Known

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## Question1

Suppose finishing time for cyclists in a race are normally distributed and have a known population standard deviation of 6 minutes and an unknown population mean. A random sample of 18 cyclists is taken and gives a sample mean of 146 minutes.

Find the confidence interval for the population mean with a 99% confidence level.

z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005
1.282 1.645 1.960 2.326 2.576

You may use a calculator or the common z values above.

• Round the final answer to two decimal places.

## Question 2

The number of square feet per house are normally distributed with a population standard deviation of 154 square feet and an unknown population mean. If a random sample of 16 houses is taken and results in a sample mean of 1550 square feet, find a 80% confidence interval for the population mean.

z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005
1.282 1.645 1.960 2.326 2.576

You may use a calculator or the common z values above.

• Round the final answer to two decimal places.

## Question 3

Suppose the finishing times for cyclists in a race are normally distributed. If the population standard deviation is 16 minutes, what minimum sample size is needed to be 90% confident that the sample mean is within 5 minutes of the true population mean?

z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576

Use the table above for the z-score, and be sure to round up to the nearest integer.

## Question 4

Suppose the finishing times for cyclists in a race are normally distributed. If the population standard deviation is 16 minutes, what minimum sample size is needed to be 90% confident that the sample mean is within 5 minutes of the true population mean?

z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576

Use the table above for the z-score, and be sure to round up to the nearest integer.

## Question 5

The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches.

Identify the parameters needed to calculate a confidence interval at the 99% confidence level. Then find the confidence interval.

z0.10z0.10 z0.05z0.05 z0.025z0.025 z0.01z0.01 z0.005z0.005
1.282 1.645 1.960 2.326 2.576

You may use a calculator or the common z values above.

• Round the final answer to two decimal places.

## Question 6

The lengths, in inches, of adult corn snakes are normally distributed with a population standard deviation of 8 inches and an unknown population mean. A random sample of 25 snakes is taken and results in a sample mean of 58 inches.

What is the correct interpretation of the confidence interval?

We can estimate with 99% confidence that the true population mean length of adult corn snakes is between 53.88 and 62.12 inches.

We can estimate with 99% confidence that the sample mean length of adult corn snakes is between 53.88 and 62.12 inches.

We can estimate that 99% of adult corn snakes will have a length that is between 53.88 and 62.12 inches.

We can estimate with 99% confidence that the true population mean length of adult corn snakes is between 53.88 and 62.12 inches.

Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval.

We can estimate with 99% confidence that the true population mean length of adult corn snakes is between 53.88 and 62.12 inches.

## Question 7

Suppose the germination periods, in days, for grass seed are normally distributed. If the population standard deviation is 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean?

z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576

Use the table above for the z-score, and be sure to round up to the nearest integer.

## Question 8

Suppose the weights, in pounds, of the dogs in a city are normally distributed. If the population standard deviation is 3 pounds, what minimum sample size is needed to be 95% confident that the sample mean is within 1 pound of the true population mean?

z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576

Use the table above for the z-score, and be sure to round up to the nearest integer.


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# 2.6 Measures of the spread of the data  (Page 3/7)

 Page 3 / 7

In these formulas, $f$ represents the frequency with which a value appears. For example, if a value appears once, $f$ is 1. If a value appears three times in the data set or population, $f$ is 3.

## Sampling variability of a statistic

The statistic of a sampling distribution was discussed in Descriptive Statistics: Measuring the Center of the Data . How much the statistic varies from one sample to another is known as the sampling variability of a statistic . You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean in The Central Limit Theorem (not now). The notation for the standard error of the mean is $\frac{\sigma }{\sqrt{n}}$ where $\sigma$ is the standard deviation of the population and $n$ is the size of the sample.

In practice, USE A CALCULATOR OR COMPUTER SOFTWARE TO CALCULATE THE STANDARD DEVIATION. If you are using a TI-83,83+,84+ calculator, you need to select the appropriate standard deviation ${\sigma }_{x}$ or ${s}_{x}$ from the summary statistics. We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean.

In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of $n=\mathrm{20}$ fifth grade students. The ages are rounded to the nearest half year:

• 9
• 9.5
• 9.5
• 10
• 10
• 10
• 10
• 10.5
• 10.5
• 10.5
• 10.5
• 11
• 11
• 11
• 11
• 11
• 11
• 11.5
• 11.5
• 11.5

$\overline{x}=\frac{9+9.5×2+10×4+10.5×4+11×6+11.5×3}{20}=10.525$

The average age is 10.53 years, rounded to 2 places.

The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating $s$ .

Data Freq. Deviations ${\mathrm{Deviations}}^{2}$ (Freq.)( ${\mathrm{Deviations}}^{2}$ )
$x$ $f$ $\left(x-\overline{x}\right)$ ${\left(x-\overline{x}\right)}^{2}$ $\left(f\right){\left(x-\overline{x}\right)}^{2}$
$9$ $1$ $9-10.525=-1.525$ ${\left(-1.525\right)}^{2}=2.325625$ $1×2.325625=2.325625$
$\mathrm{9.5}$ $2$ $9.5-10.525=-1.025$ ${\left(-1.025\right)}^{2}=1.050625$ $2×1.050625=2.101250$
$\mathrm{10}$ $4$ $10-10.525=-0.525$ ${\left(-0.525\right)}^{2}=0.275625$ $4×.275625=1.1025$
$\mathrm{10.5}$ $4$ $10.5-10.525=-0.025$ ${\left(-0.025\right)}^{2}=0.000625$ $4×.000625=.0025$
$\mathrm{11}$ $6$ $11-10.525=0.475$ ${\left(0.475\right)}^{2}=0.225625$ $6×.225625=1.35375$
$\mathrm{11.5}$ $3$ $11.5-10.525=0.975$ ${\left(0.975\right)}^{2}=0.950625$ $3×.950625=2.851875$

The sample variance, ${s}^{2}$ , is equal to the sum of the last column (9.7375) divided by the total number of data values minus one (20 - 1):

${s}^{2}=\frac{9.7375}{20-1}=0.5125$

The sample standard deviation $s$ is equal to the square root of the sample variance:

$s=\sqrt{0.5125}=.0715891$ Rounded to two decimal places, $s=0.72$

Typically, you do the calculation for the standard deviation on your calculator or computer . The intermediate results are not rounded. This is done for accuracy.

Verify the mean and standard deviation calculated above on your calculator or computer.

## Using the ti-83,83+,84+ calculators

• Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.
• Put the data values (9, 9.5, 10, 10.5, 11, 11.5) into list L1 and the frequencies (1, 2, 4, 4, 6, 3) into list L2. Use the arrow keys to move around.
• Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.
• $\overline{x}$ =10.525
• Use Sx because this is sample data (not a population): $\mathrm{Sx}$ =0.715891

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1. ## relative velocity angle

question 1.

part i distance traveled by man d=s*t .6*20=12 distance traveled by woman =.4*20=8 i cant do part ii. ive tried setting up a triangle with there distances i just worked out and you are given the angle tan^-1 3/4 and i have tried the sine rule to work it out but the right answer is 60 degrees and i am not getting it. previous similar worked examples were solved using the sine rule

2. ## Re: relative velocity angle

The distances you have calculated are correct but they are not the sides of a triangle.

Redraw your diagram, so that the woman ends up due south of the man. Your diagram will consist of 2 triangles. Then try again.

Come back if you get stuck again.

4. ## Re: relative velocity angle

Do you know if they are the exact sizes of the sides triangles. 10/sinx = 8/sin(tan^-1 3/4). Does not work

5. ## Re: relative velocity angle

No it doesn't work because the triangle(s) you need to work with does not include 8 and 10 as sides.

Angle ACB = arctan(3/4)

CA = 10 (distance apart at start)

Using basic trig (or recognising a 3:4:5 = 6:8:10 right angled triangle). AB=6 and BC=8

CP = 12 (distance swum by man) and AR = 8 (distance swum by woman)

Now concentrate on triangles ABQ and RPQ (which are similar).

See if that helps.

6. ## Re: relative velocity angle

man swimmer starts at (0,0) and swims due W at 0.6 m/s ...

$x_1(t)=-0.6t$

$y_1(t)=0$

woman swimmer starts at (-8,6) and swims at an angle $\alpha$ S of W at 0.4 m/s ...

$x_2(t) = -8 -0.4\cos{\alpha} \cdot t$

$y_2(t) = 6 - 0.4\sin{\alpha} \cdot t$

at $t=20$ sec, the woman swimmer is due S of the man swimmer ...

$x_2(20)=x_1(20)$

$-8-8\cos{\alpha}=-12 \implies \cos{\alpha}=\dfrac{1}{2} \implies \alpha = 60^\circ$

for part (iii) ...

$d=\sqrt{[x_2(t)-x_1(t)]^2 + [y_2(t)-y_1(t)]^2}$

expand and simplify the resulting time quadratic under the radical and find the time for minimum distance by either calculus or non-calculus means.

7. ## Re: relative velocity angle

Ok thanks. But is there not any way to work it out with the triangle's

8. ## Re: relative velocity angle

There is no way to tell distance AQ or PQ. All i know is the size of AB


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For the region R below, write \int \int_Z f dA

For the region R below, write $\int {\int }_{Z}fdA$ as an integral in polar coordinates.

Use t for $\theta$ in your expressions.

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estenutC
From the given figure, it is seen that the region R is bounded by two circles, ${x}^{2}+{y}^{2}=9\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{2}+{y}^{2}=16$
Therefore, region R can be defined as
$R=\left\{{\left(r,\theta \right)}^{3}\le r\le 4,0\le \theta \le \pi \right\}$
it is known that,
$\int {\int }_{R}fdA={\int }_{{\theta }_{1}}^{{\theta }_{2}}f\left(r\mathrm{cos}\theta ,r\mathrm{sin}\theta \right)rdrd\theta$
Therefore, in polar co-ordinate,
$\int {\int }_{R}fdA={\int }_{0}^{\pi }{\int }_{3}^{4}f\left(r\mathrm{cos}\theta ,r\mathrm{sin}\theta \right)rdrd\theta$ (i)
Thus, comparing (i) with $\int {\int }_{R}fdA={\int }_{0}^{\pi }{\int }_{3}^{4}f\left(r\mathrm{cos}\theta ,r\mathrm{sin}\theta \right)rdrd\theta$ it is inferred that
a = 0
$b=\pi$
c = 3
d = 4
With $dA=rdrd\theta$


text:
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## ConvertVirtualToRealMover

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Author: Jared Adolf-Bryfogle (jadolfbr@gmail.com) and Sebastian Raemisch (raemisch@scripps.edu) Mover for switching from VIRTUAL residues to REAL residues. A VIRTUAL residue is one that is not scored or output.

<ConvertVirtualToRealMover name="(&string;)" residue_selector="(&string;)" />
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This mover is useful for building specific protocols or for (not)-ignoring parts of your pose. A virtual residue is one that is not scored or output to a PDB.

## Example Use case

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# 随机算法 (Fall 2015)/Problem Set 2

## Problem 1

For any ${\displaystyle \alpha \geq 1}$, a cut ${\displaystyle C}$ in an undirected (multi)graph ${\displaystyle G(V,E)}$is called an ${\displaystyle \alpha }$-min-cut if ${\displaystyle |C|\leq \alpha |C^{*}|}$ where ${\displaystyle C^{*}}$ is a min-cut in ${\displaystyle G}$.

• Give a lower bound to the probability that a single iteration of Karger's Random Contraction algorithm returns an ${\displaystyle \alpha }$-min-cut in a graph ${\displaystyle G(V,E)}$ of ${\displaystyle n}$ vertices.
• Use the above bound to estimate the number of distinct ${\displaystyle \alpha }$-min cuts in ${\displaystyle G}$.

## Problem 2

Suppose that we flip a fair coin ${\displaystyle n}$ times to obtain ${\displaystyle n}$ random bits. Consider all ${\displaystyle m={n \choose 2}}$ pairs of these bits in some order. Let ${\displaystyle Y_{i}}$ be the exclusive-or of the ${\displaystyle i}$th pair of bits, and let ${\displaystyle Y=\sum _{i=1}^{m}Y_{i}}$ be the number of ${\displaystyle Y_{i}}$ that equal 1.

1. Show that the ${\displaystyle Y_{i}}$ are NOT mutually independent.
2. Show that the ${\displaystyle Y_{i}}$ satisfy the property ${\displaystyle \mathbf {E} [Y_{i}Y_{j}]=\mathbf {E} [Y_{i}]\mathbf {E} [Y_{j}]}$.
3. Compute ${\displaystyle \mathbf {Var} [Y]}$.
4. Using Chebyshev's inequality, prove a bound on ${\displaystyle \Pr[|Y-\mathbf {E} [Y]|\geq n]}$.

## Problem 3

Show that the maximum load when ${\displaystyle n}$ balls are hashed into ${\displaystyle n}$ bins using a hash function chosen from a 2-universal family of hash functions is at most ${\displaystyle O({\sqrt {n}})}$ with probability at least 0.99. Generalize this argument to ${\displaystyle k}$-universal hash functions.

Hint: Perhaps the only information we can use about a 2-universal hash function is the number of collisions. What does it become for ${\displaystyle k}$-universal hashing?

• (Optional) Can this ${\displaystyle O({\sqrt {n}})}$ be improved if the only thing we assume about the hash function is the 2-universality? Why?

## Problem 4

The maximum directed cut problem (MAX-DICUT).

We are given as input a directed graph ${\displaystyle G=(V,E)}$, with each directed edge ${\displaystyle (u,v)\in E}$ having a nonnegative weight ${\displaystyle w_{uv}\geq 0}$. The goal is to partition ${\displaystyle V}$ into two sets ${\displaystyle S\,}$ and ${\displaystyle {\bar {S}}=V\setminus S}$ so as to maximize the value of ${\displaystyle \sum _{(u,v)\in E \atop u\in S,v\not \in S}w_{uv}}$, that is, the total weight of the edges going from ${\displaystyle S\,}$ to ${\displaystyle {\bar {S}}}$.

• Give a randomized ${\displaystyle {\frac {1}{4}}}$-approximation algorithm based on random sampling.
• Prove that the following is an integer programming for the problem:
{\displaystyle {\begin{aligned}{\text{maximize}}&&\sum _{(i,j)\in E}w_{ij}z_{ij}\\{\text{subject to}}&&z_{ij}&\leq x_{i},&\forall (i,j)&\in E,\\&&z_{ij}&\leq 1-x_{j},&\forall (i,j)&\in E,\\&&x_{i}&\in \{0,1\},&\forall i&\in V,\\&&0\leq z_{ij}&\leq 1,&\forall (i,j)&\in E.\end{aligned}}}
• Consider a randomized rounding algorithm that solves an LP relaxation of the above integer programming and puts vertex ${\displaystyle i}$ in ${\displaystyle S}$ with probability ${\displaystyle f(x_{i}^{*})}$. We may assume that ${\displaystyle f(x)}$ is a linear function in the form ${\displaystyle f(x)=ax+b}$ with some constant ${\displaystyle a}$ and ${\displaystyle b}$ to be fixed. Try to find good ${\displaystyle a}$ and ${\displaystyle b}$ so that the randomized rounding algorithm has a good approximation ratio.

## Problem 5

The set cover problem is defined as follows:

• Let ${\displaystyle U=\{u_{1},u_{2},\ldots ,u_{n}\}}$ be a set of ${\displaystyle n}$ elements, and let ${\displaystyle {\mathcal {S}}=\{S_{1},S_{2},\ldots ,S_{m}\}}$ be a family of subsets of ${\displaystyle U}$. For each ${\displaystyle u_{i}\in U}$, let ${\displaystyle w_{i}}$ be a nonnegative weight of ${\displaystyle u_{i}}$. The goal is to find a subset ${\displaystyle V\subseteq U}$ with the minimum total weight ${\displaystyle \sum _{i\in V}w_{i}}$, that intersects with all ${\displaystyle S_{i}\in {\mathcal {S}}}$.

This problem is NP-hard.

(Remark: There are two equivalent definitions of the set cover problem. We take the hitting set version.)

Questions:

• Prove that the following is an integer programming for the problem:
{\displaystyle {\begin{aligned}{\text{minimize}}&&\sum _{(i,j)\in E}w_{i}x_{i}\\{\text{subject to}}&&\sum _{i:u_{i}\in S_{j}}x_{i}&\geq 1,&1\leq j\leq m,\\&&x_{i}&\in \{0,1\},&1\leq i\leq n.\end{aligned}}}
• Give a randomized rounding algorithm which returns an ${\displaystyle O(\log m)}$-approximate solution with probability at least ${\displaystyle {\frac {1}{2}}}$. (Hint: you may repeat the randomized rounding process if there remains some uncovered subsets after one time of applying the randomized rounding.)


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## Algebra: A Combined Approach (4th Edition)

$x^{2} + 7x + 12$ To factor this, we need two factors of 12 such that their sum is 7. Thus the factors are 4 and 3. $x^{2} + 7x + 12$ = $x^{2} + 3x + 4x + 12$ =x(x + 3) + 4(x + 3) = (x + 4)(x + 3) .


text:
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## appleduardo 3 years ago how can i derivate the following function?

• This Question is Closed
1. appleduardo

$\frac{ 1 }{ x \sqrt{x} }$

2. JVitato1

(-3)/(2x^5/2)

3. appleduardo

how did u get that?

4. JVitato1

take the sqrt x which equals x^1/2 adding in x you get x^3/2. Then do the derivative. You could even move the denominator to the top making it x^-3/2. that might make it easier.

5. Kuromeru

$\frac{ 1 }{x^{n} } = x^{-n}$ You can also multiply the terms in the denominator since $\sqrt[n]{x} = x^{\frac{ 1 }{ n }}$ and $x^{n}x^{m} = x ^{n+m}$

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# Intuitionistic logic

Intuitionistic logic, sometimes more generally called constructive logic, is a system of symbolic logic that differs from classical logic by replacing the traditional concept of truth with the concept of constructive provability. For example, in classical logic, propositional formulae are always assigned a truth value from the two element set of trivial propositions $\{\top, \bot\}$ ("true" and "false" respectively) regardless of whether we have direct evidence for either case. In contrast, propositional formulae in intuitionistic logic are not assigned any definite truth value at all and instead only considered "true" when we have direct evidence, hence proof. (We can also say, instead of the propositional formula being "true" due to direct evidence, that it is inhabited by a proof in the Curry-Howard sense.) Operations in intuitionistic logic therefore preserve justification, with respect to evidence and provability, rather than truth-valuation.

Unproved statements in Intuitionistic logic are not given an intermediate truth value (as is sometimes mistakenly asserted). Indeed, you can prove that it has no third truth value, a result dating back to Glivenko in 1928.[1] Instead they remain of unknown truth value, until they are either proved or disproved. Statements are disproved by deducing a contradiction from them.

A consequence of this point of view is that intuitionistic logic has no interpretation as a two-valued logic, nor even as a finite-valued logic, in the familiar sense. Although intuitionistic logic retains the trivial propositions $\{\top, \bot\}$ from classical logic, each proof of a propositional formula is considered a valid propositional value, thus by Heyting's notion of propositions-as-sets, propositional formulae are (potentially non-finite) sets of their proofs.

Semantically, intuitionistic logic is a restriction of classical logic in which the law of excluded middle and double negation elimination are not admitted as axioms. Excluded middle and double negation elimination can still be proved for some propositions on a case by case basis, however, but do not hold universally as they do with classical logic.

Several semantics for intuitionistic logic have been studied. One semantic mirrors classical Boolean-valued semantics but uses Heyting algebras in place of Boolean algebras. Another semantic uses Kripke models.

Intuitionistic logic is practically useful because its restrictions produce proofs that have the existence property, making it also suitable for other forms of mathematical constructivism. Informally, this means that if you have a constructive proof that an object exists, you can turn that constructive proof into an algorithm for generating an example of it.

Formalized intuitionistic logic was originally developed by Arend Heyting to provide a formal basis for Brouwer's programme of intuitionism.

## Syntax

The Rieger–Nishimura lattice. Its nodes are the propositional formulas in one variable up to intuitionistic logical equivalence, ordered by intuitionistic logical implication.

The syntax of formulas of intuitionistic logic is similar to propositional logic or first-order logic. However, intuitionistic connectives are not definable in terms of each other in the same way as in classical logic, hence their choice matters. In intuitionistic propositional logic it is customary to use →, ∧, ∨, ⊥ as the basic connectives, treating ¬A as an abbreviation for (A → ⊥). In intuitionistic first-order logic both quantifiers ∃, ∀ are needed.

Many tautologies of classical logic can no longer be proved within intuitionistic logic. Examples include not only the law of excluded middle p ∨ ¬p, but also Peirce's law ((pq) → p) → p, and even double negation elimination. In classical logic, both p → ¬¬p and also ¬¬pp are theorems. In intuitionistic logic, only the former is a theorem: double negation can be introduced, but it cannot be eliminated. Rejecting p ∨ ¬p may seem strange to those more familiar with classical logic, but proving this propositional formula in intuitionistic logic would require producing a proof for the truth or falsity of all possible propositional formulae, which is impossible for a variety of reasons.

Because many classically valid tautologies are not theorems of intuitionistic logic, but all theorems of intuitionistic logic are valid classically, intuitionistic logic can be viewed as a weakening of classical logic, albeit one with many useful properties.

### Sequent calculus

Main article: Sequent calculus

Gentzen discovered that a simple restriction of his system LK (his sequent calculus for classical logic) results in a system which is sound and complete with respect to intuitionistic logic. He called this system LJ. In LK any number of formulas is allowed to appear on the conclusion side of a sequent; in contrast LJ allows at most one formula in this position.

Other derivatives of LK are limited to intuitionisitic derivations but still allow multiple conclusions in a sequent. LJ'[2] is one example.

### Hilbert-style calculus

Intuitionistic logic can be defined using the following Hilbert-style calculus. This is similar to a way of axiomatizing classical propositional logic.

In propositional logic, the inference rule is modus ponens

• MP: from $\phi$ and $\phi \to \psi$ infer $\psi$

and the axioms are

• THEN-1: $\phi \to (\chi \to \phi )$
• THEN-2: $(\phi \to (\chi \to \psi )) \to ((\phi \to \chi ) \to (\phi \to \psi ))$
• AND-1: $\phi \land \chi \to \phi$
• AND-2: $\phi \land \chi \to \chi$
• AND-3: $\phi \to (\chi \to (\phi \land \chi ))$
• OR-1: $\phi \to \phi \lor \chi$
• OR-2: $\chi \to \phi \lor \chi$
• OR-3: $(\phi \to \psi ) \to ((\chi \to \psi ) \to (\phi \lor \chi \to \psi ))$
• FALSE: $\bot \to \phi$

To make this a system of first-order predicate logic, the generalization rules

• $\forall$-GEN: from $\psi \to \phi$ infer $\psi \to (\forall x \ \phi )$, if $x$ is not free in $\psi$
• $\exists$-GEN: from $\phi \to \psi$ infer $(\exists x \ \phi ) \to \psi$, if $x$ is not free in $\psi$

are added, along with the axioms

• PRED-1: $(\forall x \ \phi (x)) \to \phi (t)$, if the term t is free for substitution for the variable x in $\phi$ (i.e., if no occurrence of any variable in t becomes bound in $\phi (t)$)
• PRED-2: $\phi (t) \to (\exists x \ \phi (x))$, with the same restriction as for PRED-1

#### Optional connectives

##### Negation

If one wishes to include a connective $\lnot$ for negation rather than consider it an abbreviation for $\phi \to \bot$, it is enough to add:

• NOT-1': $(\phi \to \bot ) \to \lnot \phi$
• NOT-2': $\lnot \phi \to (\phi \to \bot )$

There are a number of alternatives available if one wishes to omit the connective $\bot$ (false). For example, one may replace the three axioms FALSE, NOT-1', and NOT-2' with the two axioms

• NOT-1: $(\phi \to \chi ) \to ((\phi \to \lnot \chi ) \to \lnot \phi )$
• NOT-2: $\phi \to (\lnot \phi \to \chi )$

as at Propositional calculus#Axioms. Alternatives to NOT-1 are $(\phi \to \lnot \chi ) \to (\chi \to \lnot \phi )$ or $(\phi \to \lnot \phi ) \to \lnot \phi$.

##### Equivalence

The connective $\leftrightarrow$ for equivalence may be treated as an abbreviation, with $\phi \leftrightarrow \chi$ standing for $(\phi \to \chi ) \land (\chi \to \phi )$. Alternatively, one may add the axioms

• IFF-1: $(\phi \leftrightarrow \chi ) \to (\phi \to \chi )$
• IFF-2: $(\phi \leftrightarrow \chi ) \to (\chi \to \phi )$
• IFF-3: $(\phi \to \chi ) \to ((\chi \to \phi ) \to (\phi \leftrightarrow \chi ))$

IFF-1 and IFF-2 can, if desired, be combined into a single axiom $(\phi \leftrightarrow \chi ) \to ((\phi \to \chi ) \land (\chi \to \phi ))$ using conjunction.

#### Relation to classical logic

The system of classical logic is obtained by adding any one of the following axioms:

• $\phi \lor \lnot \phi$ (Law of the excluded middle. May also be formulated as $(\phi \to \chi ) \to ((\lnot \phi \to \chi ) \to \chi )$.)
• $\lnot \lnot \phi \to \phi$ (Double negation elimination)
• $((\phi \to \chi ) \to \phi ) \to \phi$ (Peirce's law)

In general, one may take as the extra axiom any classical tautology that is not valid in the two-element Kripke frame $\circ{\longrightarrow}\circ$ (in other words, that is not included in Smetanich's logic).

Another relationship is given by the Gödel–Gentzen negative translation, which provides an embedding of classical first-order logic into intuitionistic logic: a first-order formula is provable in classical logic if and only if its Gödel–Gentzen translation is provable intuitionistically. Therefore intuitionistic logic can instead be seen as a means of extending classical logic with constructive semantics.

In 1932, Kurt Gödel defined a system of Gödel logics intermediate between classical and intuitionistic logic; such logics are known as intermediate logics.

### Non-interdefinability of operators

In classical propositional logic, it is possible to take one of conjunction, disjunction, or implication as primitive, and define the other two in terms of it together with negation, such as in Łukasiewicz's three axioms of propositional logic. It is even possible to define all four in terms of a sole sufficient operator such as the Peirce arrow (NOR) or Sheffer stroke (NAND). Similarly, in classical first-order logic, one of the quantifiers can be defined in terms of the other and negation.

These are fundamentally consequences of the law of bivalence, which makes all such connectives merely Boolean functions. The law of bivalence does not hold in intuitionistic logic, only the law of non-contradiction. As a result none of the basic connectives can be dispensed with, and the above axioms are all necessary. Most of the classical identities are only theorems of intuitionistic logic in one direction, although some are theorems in both directions. They are as follows:

Conjunction versus disjunction:

• $(\phi \wedge \psi) \to \neg (\neg \phi \vee \neg \psi)$
• $(\phi \vee \psi) \to \neg (\neg \phi \wedge \neg \psi)$
• $(\neg \phi \vee \neg \psi) \to \neg (\phi \wedge \psi)$
• $(\neg \phi \wedge \neg \psi) \leftrightarrow \neg (\phi \vee \psi)$

Conjunction versus implication:

• $(\phi \wedge \psi) \to \neg (\phi \to \neg \psi)$
• $(\phi \to \psi) \to \neg (\phi \wedge \neg \psi)$
• $(\phi \wedge \neg \psi) \to \neg (\phi \to \psi)$
• $(\phi \to \neg \psi) \leftrightarrow \neg (\phi \wedge \psi)$

Disjunction versus implication:

• $(\phi \vee \psi) \to (\neg \phi \to \psi)$
• $(\neg \phi \vee \psi) \to (\phi \to \psi)$

Universal versus existential quantification:

• $(\forall x \ \phi(x)) \to \neg (\exists x \ \neg \phi(x))$
• $(\exists x \ \phi(x)) \to \neg (\forall x \ \neg \phi(x))$
• $(\exists x \ \neg \phi(x)) \to \neg (\forall x \ \phi(x))$
• $(\forall x \ \neg \phi(x)) \leftrightarrow \neg (\exists x \ \phi(x))$

So, for example, "a or b" is a stronger propositional formula than "if not a, then b", whereas these are classically interchangeable. On the other hand, "not (a or b)" is equivalent to "not a, and also not b".

If we include equivalence in the list of connectives, some of the connectives become definable from others:

• $(\phi\leftrightarrow \psi) \leftrightarrow ((\phi \to \psi)\land(\psi\to\phi))$
• $(\phi\to\psi) \leftrightarrow ((\phi\lor\psi) \leftrightarrow \psi)$
• $(\phi\to\psi) \leftrightarrow ((\phi\land\psi) \leftrightarrow \phi)$
• $(\phi\land\psi) \leftrightarrow ((\phi\to\psi)\leftrightarrow\phi)$
• $(\phi\land\psi) \leftrightarrow (((\phi\lor\psi)\leftrightarrow\psi)\leftrightarrow\phi)$

In particular, {∨, ↔, ⊥} and {∨, ↔, ¬} are complete bases of intuitionistic connectives.

As shown by Alexander Kuznetsov, either of the following connectives – the first one ternary, the second one quinary – is by itself functionally complete: either one can serve the role of a sole sufficient operator for intuitionistic propositional logic, thus forming an analog of the Sheffer stroke from classical propositional logic:[3]

• $((p\lor q)\land\neg r)\lor(\neg p\land(q\leftrightarrow r)),$
• $p\to(q\land\neg r\land(s\lor t)).$

## Semantics

The semantics are rather more complicated than for the classical case. A model theory can be given by Heyting algebras or, equivalently, by Kripke semantics. Recently, a Tarski-like model theory was proved complete by Bob Constable, but with a different notion of completeness than classically.

### Heyting algebra semantics

In classical logic, we often discuss the truth values that a formula can take. The values are usually chosen as the members of a Boolean algebra. The meet and join operations in the Boolean algebra are identified with the ∧ and ∨ logical connectives, so that the value of a formula of the form AB is the meet of the value of A and the value of B in the Boolean algebra. Then we have the useful theorem that a formula is a valid proposition of classical logic if and only if its value is 1 for every valuation—that is, for any assignment of values to its variables.

A corresponding theorem is true for intuitionistic logic, but instead of assigning each formula a value from a Boolean algebra, one uses values from a Heyting algebra, of which Boolean algebras are a special case. A formula is valid in intuitionistic logic if and only if it receives the value of the top element for any valuation on any Heyting algebra.

It can be shown that to recognize valid formulas, it is sufficient to consider a single Heyting algebra whose elements are the open subsets of the real line R.[4] In this algebra, the ∧ and ∨ operations correspond to set intersection and union, and the value assigned to a formula AB is int(ACB), the interior of the union of the value of B and the complement of the value of A. The bottom element is the empty set ∅, and the top element is the entire line R. The negation ¬A of a formula A is (as usual) defined to be A → ∅. The value of ¬A then reduces to int(AC), the interior of the complement of the value of A, also known as the exterior of A. With these assignments, intuitionistically valid formulas are precisely those that are assigned the value of the entire line.[4]

For example, the formula ¬(A ∧ ¬A) is valid, because no matter what set X is chosen as the value of the formula A, the value of ¬(A ∧ ¬A) can be shown to be the entire line:

Value(¬(A ∧ ¬A)) =
int((Value(A ∧ ¬A))C) =
int((Value(A) ∩ Value(¬A))C) =
int((X ∩ int((Value(A))C))C) =
int((X ∩ int(XC))C)

A theorem of topology tells us that int(XC) is a subset of XC, so the intersection is empty, leaving:

int(∅C) = int(R) = R

So the valuation of this formula is true, and indeed the formula is valid.

But the law of the excluded middle, A ∨ ¬A, can be shown to be invalid by letting the value of A be {y : y > 0 }. Then the value of ¬A is the interior of {y : y ≤ 0 }, which is {y : y < 0 }, and the value of the formula is the union of {y : y > 0 } and {y : y < 0 }, which is {y : y ≠ 0 }, not the entire line.

The interpretation of any intuitionistically valid formula in the infinite Heyting algebra described above results in the top element, representing true, as the valuation of the formula, regardless of what values from the algebra are assigned to the variables of the formula.[4] Conversely, for every invalid formula, there is an assignment of values to the variables that yields a valuation that differs from the top element.[5][6] No finite Heyting algebra has both these properties.[4]

### Kripke semantics

Main article: Kripke semantics

Building upon his work on semantics of modal logic, Saul Kripke created another semantics for intuitionistic logic, known as Kripke semantics or relational semantics.[7]

### Tarski-like semantics

It was discovered that Tarski-like semantics for intuitionistic logic were not possible to prove complete. However, Robert Constable has shown that a weaker notion of completeness still holds for intuitionistic logic under a Tarski-like model. In this notion of completeness we are concerned not with all of the statements that are true of every model, but with the statements that are true in the same way in every model. That is, a single proof that the model judges a formula to be true must be valid for every model. In this case, there is not only a proof of completeness, but one that is valid according to intuitionistic logic.[8]

### Relation to other logics

Intuitionistic logic is related by duality to a paraconsistent logic known as Brazilian, anti-intuitionistic or dual-intuitionistic logic.[9]

The subsystem of intuitionistic logic with the FALSE axiom removed is known as minimal logic.

#### Relation to many-valued logic

Kurt Gödel in 1932 showed that intuitionistic logic is not a finitely-many valued logic. (See the section titled Heyting algebra semantics above for a sort of "infinitely-many valued logic" interpretation of intuitionistic logic.)

#### Relation to intermediate logics

Any finite Heyting algebra which is not equivalent to a Boolean algebra defines (semantically) an intermediate logic. On the other hand, validity of formulae in pure intuitionistic logic is not tied to any individual Heyting algebra but relates to any and all Heyting algebras at the same time.

#### Relation to modal logic

Any formula of the intuitionistic propositional logic may be translated into the modal logic S4 as follows:

 $\bot^*$ = $\ \bot$ $A^*$ = $\ \Box A \qquad \hbox{if } A \hbox{ is prime (a positive literal)}$ $(A \wedge B)^*$ = $\ A^* \wedge B^*$ $(A \vee B)^*$ = $\ A^* \vee B^*$ $(A \rightarrow B)^*$ = $\ \Box(A^* \rightarrow B^*)$ $(\neg A)^*$ = $\ \Box(\neg (A^*)) \qquad \hbox{since } \neg A := A \rightarrow \bot$

and it has been demonstrated[10] that the translated formula is valid in the propositional modal logic S4 if and only if the pre-translated formula is valid in IPC. The above set of formulae are called the Gödel–McKinsey–Tarski translation.

There is also an intuitionistic version of modal logic S4 called Constructive Modal Logic CS4.[11]

### Lambda calculus

There is an extended Curry–Howard isomorphism between IPC and simply-typed lambda calculus.[11]

## Notes

1. ^ Proof that intuitionistic logic has no third truth value, Glivenko 1928
2. ^ Proof Theory by G. Takeuti, ISBN 0-444-10492-5
3. ^ Alexander Chagrov, Michael Zakharyaschev, Modal Logic, vol. 35 of Oxford Logic Guides, Oxford University Press, 1997, pp. 58–59. ISBN 0-19-853779-4.
4. ^ a b c d Sørensen, Morten Heine B; Paweł Urzyczyn (2006). Lectures on the Curry-Howard Isomorphism. Studies in Logic and the Foundations of Mathematics. Elsevier. p. 42. ISBN 0-444-52077-5.
5. ^ Alfred Tarski, Der Aussagenkalkül und die Topologie, Fundamenta Mathematicae 31 (1938), 103–134. [1]
6. ^ Rasiowa, Helena; Roman Sikorski (1963). The Mathematics of Metamathematics. Monografie matematyczne. Warsaw: Państwowe Wydawn. Naukowe. pp. 385–386.
7. ^ Intuitionistic Logic. Written by Joan Moschovakis. Published in Stanford Encyclopedia of Philosophy.
8. ^ R. Constable, M. Bickford, Intuitionistic completeness of first-order logic, Annals of Pure and Applied Logic, to appear, doi:10.1016/j.apal.2013.07.009. Preprint on ArXiv.
9. ^ Aoyama, Hiroshi (2004). "LK, LJ, Dual Intuitionistic Logic, and Quantum Logic". Notre Dame Journal of Formal Logic 45 (4): 193–213. doi:10.1305/ndjfl/1099238445.
10. ^ Lévy, Michel (2011). Logique modale propositionnelle S4 et logique intuitioniste propositionnelle, pp. 4–5.
11. ^ a b Natasha Alechina, Michael Mendler, Valeria de Paiva, and Eike Ritter. Categorical and Kripke Semantics for Constructive S4 Modal Logic

## References

• van Dalen, Dirk, 2001, "Intuitionistic Logic", in Goble, Lou, ed., The Blackwell Guide to Philosophical Logic. Blackwell.
• Morten H. Sørensen, Paweł Urzyczyn, 2006, Lectures on the Curry-Howard Isomorphism (chapter 2: "Intuitionistic Logic"). Studies in Logic and the Foundations of Mathematics vol. 149, Elsevier.
• W. A. Carnielli (with A. B.M. Brunner)."Anti-intuitionism and paraconsistency". Journal of Applied Logic Volume 3, Issue 1, March 2005, pages 161-184.


text:
Area Calculator Feet Area Calculator Feet Square Meters To Meters Convert Sq Meter To Sq Ft Tip Calculate A Area Calculator Feet Cylinder Volume Surface Area Calculator Formula To Find Volume The Base Total Lateral Surface Area Of A Cylinder

area calculator feet area calculator feet square meters to meters convert sq meter to sq ft tip calculate a area calculator feet cylinder volume surface area calculator formula to find volume the base total lateral surface area of a cylinder.


text:
Example 59.56.2. The geometric point $\mathop{\mathrm{Spec}}(\mathbf{C}) \to \mathop{\mathrm{Spec}}(\mathbf{Q})$ is not algebraic.

There are also:

• 2 comment(s) on Section 59.56: Galois action on stalks

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Thread: Trigonometric Equation help need!

1. ## Trigonometric Equation help need!

Solve the equation for solutions over the interval [0, 360).

sec(squared)xtanx = 2tanx

2. Ok I think I got it:

sec(squared)xtanx = 2tanx

sec(squared)xtanx - 2tanx = 0

sec(squared)xtanx - tanx - tanx = 0

tanx(sec(squared)x - 2) = 0

tanx = 0 or sec(squared)x - 2 = 0
sec(squared)x = 2
secx = (squareroot)2

therefore:
a(squared) + b(squared) = c(squared)
1(squared) + b(squared) = ((squareroot)2)(squared)
1 + b(squared) = 2
1 + 1 = 2 (if I'm not mistaken)

therefore tanx = + or - 1

therefore solution set:
{0. 45, 135, 180, 225, 315)

correct me if me wrong please

NEE!!

3. Originally Posted by nee
Ok I think I got it:

sec(squared)xtanx = 2tanx

sec(squared)xtanx - 2tanx = 0

sec(squared)xtanx - tanx - tanx = 0

tanx(sec(squared)x - 2) = 0

tanx = 0 or sec(squared)x - 2 = 0
sec(squared)x = 2
secx = (squareroot)2

therefore:
a(squared) + b(squared) = c(squared)
1(squared) + b(squared) = ((squareroot)2)(squared)
1 + b(squared) = 2
1 + 1 = 2 (if I'm not mistaken)

therefore tanx = + or - 1

therefore solution set:
{0. 45, 135, 180, 225, 315)

correct me if me wrong please

NEE!!
The $\displaystyle 1+b^2=\sqrt{2}$, i dunno where you got that from. It's not needed or necessary to solve this equation.

You got it right up to that point. Which angles does tan x = 0 and where does sec x = $\displaystyle \frac{1}{\sqrt{2}}$, which is actually cos x = $\displaystyle \frac{\sqrt{2}}{2}$, fall on? Find the angles, and you'll have your solution.

4. Gotcha, that does make more sense. Thanks, NEE!!!

5. $\displaystyle \sec^2x \tan x=2 \tan x$

$\displaystyle \sec^2x=2$

$\displaystyle \frac{1}{\cos^2x}=2$

$\displaystyle 2\cos^2x=1$

$\displaystyle \cos^2x=\frac{1}{2}$

$\displaystyle \cos x=\frac{\sqrt{2}}{2}$

$\displaystyle x=\frac{\pi}{4} \ \ or \ \ \frac{7\pi}{4}$

6. Hi,
Originally Posted by masters
$\displaystyle \sec^2x \tan x=2 \tan x$

$\displaystyle \sec^2x=2$

$\displaystyle \frac{1}{\cos^2x}=2$

$\displaystyle 2\cos^2x=1$

$\displaystyle \cos^2x=\frac{1}{2}$

$\displaystyle \cos x=\frac{\sqrt{2}}{2}$

$\displaystyle x=\frac{\pi}{4} \ \ or \ \ \frac{7\pi}{4}$
Consider the solution $\displaystyle \tan(x)=0$

7. Originally Posted by Moo
Hi,

Consider the solution $\displaystyle \tan(x)=0$
Ok, then we have $\displaystyle {0, \frac{\pi}{4}, \pi, \frac{7\pi}{4}}$


text:
## Ivano Ciardelli, Jeroen Groenendijk, and Floris Roelofsen

Print publication date: 2018

Print ISBN-13: 9780198814788

Published to Oxford Scholarship Online: December 2018

DOI: 10.1093/oso/9780198814788.001.0001

Show Summary Details
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PRINTED FROM OXFORD SCHOLARSHIP ONLINE (oxford.universitypressscholarship.com). (c) Copyright Oxford University Press, 2021. All Rights Reserved. An individual user may print out a PDF of a single chapter of a monograph in OSO for personal use. date: 21 June 2021

# Basic operations on propositions

Chapter:
(p.43) 3 Basic operations on propositions
Source:
Inquisitive Semantics
Publisher:
Oxford University Press
DOI:10.1093/oso/9780198814788.003.0003

# Abstract and Keywords

Chapter 3 considers what the basic operations are that can be performed on propositions in inquisitive semantics. In the classical setting, where propositions are simple sets of worlds, one can form the intersection or the union of two propositions, or the complement of a single proposition. These operations play a central role in logic and in semantic analyses of natural languages: conjunction and disjunction are standardly taken to express intersection and union, respectively, while negation is standardly taken to express complementation. It is shown that these operations have natural counterparts in the inquisitive setting, even though propositions are no longer simple sets of worlds.

Now that we have introduced a new notion of propositions, it is natural to consider what the basic operations are that could be performed on such propositions. In the classical setting, where propositions are simple sets of worlds, we can form the intersection or the union of two propositions, or the complement of a single proposition. These operations play a central role in logic and in semantic analyses of natural languages: conjunction and disjunction are standardly taken to express intersection and union, respectively, while negation is standardly taken to express complementation. Do these operations have natural counterparts in the inquisitive setting, where propositions are no longer simple sets of worlds?

We will address this question in Section 3.1, adopting an algebraic perspective. We will find that the basic algebraic operations on classical propositions can indeed be applied to inquisitive propositions as well. This result facilitates a very natural way of dealing with connectives and quantifiers. In particular, in Chapter 4 it will allow us to define an inquisitive semantics for the language of first-order logic which is, from an algebraic perspective, the exact counterpart of classical first-order logic in the inquisitive setting. In Chapter 5 we will suggest that the algebraic operations are also crucial for the semantic analysis of connectives in natural languages. In particular, they will yield a uniform account of conjunction, disjunction, and conditionalization of statements and questions, as illustrated by examples (1)–(3) below, repeated from Section 1.1.3.

(1)

(2)

(p.44)

(3)

Returning now to the roadmap for the present chapter, after having discussed the basic algebraic operations on inquisitive propositions in Section 3.1, we will consider two other natural operations in Section 3.2, namely ones that trivialize the informative or the inquisitive content, respectively, of any given proposition. For reasons that will become clear below, we refer to such operators as projection operators. One projection operator turns any proposition into a corresponding non-inquisitive proposition, while the other turns any proposition into a corresponding non-informative proposition. Clearly, these operations do not have a counterpart in the classical setting, where propositions capture only informative content to begin with; but in the inquisitive setting they naturally arise, and we will suggest that they also have an important role to play in the semantic analysis of natural languages. More specifically, in Chapter 6 we will use these projection operators to capture the semantic contribution of declarative and interrogative complementizers.

# 3.1 Algebraic operations

In this section we will identify the basic algebraic operations that can be applied to inquisitive propositions. To illustrate our approach, we will first briefly review the algebraic perspective on classical logic.

## 3.1.1 The algebraic perspective on classical logic

In the classical setting a proposition P is simply a set of possible worlds. Let us denote the set of all classical propositions as $Pcl$. The proposition expressed by a sentence can be thought of as carving out a certain region in the logical space—the set of all possible worlds—and in asserting a sentence, a speaker is taken to provide the information that the actual world is located within this region. One proposition P entails another proposition Q, PQ, just in case $P⊆Q$, which means that P carves out a smaller region in the logical space than Q does, thereby encoding more information as to what the actual world is like. Entailment forms a partial order on the set of all classical propositions, i.e., it is a reflexive, transitive, and anti-symmetric relation. In Figure 3.1 we have depicted the set of all classical propositions, assuming that the logical space (p.45) consists of two worlds (the diagram on the left) or of three worlds (the diagram on the right). In each case, the arrows indicate entailment.

Figure 3.1 The set of all propositions in classical semantics if the logical space consists of two possible worlds (on the left) or three possible worlds (on the right). Arrows indicate entailment.

Now, given a partially ordered set, we can ask what natural operations can be defined on it—that is, what algebraic structure it has. The set of classical propositions ordered by entailment, $〈Pcl,⊧〉$, forms a so-called Heyting algebra, which comes with four basic operations: meet, join, relative pseudo-complementation and absolute pseudo-complementation.1

To illustrate these operations, we will make use of the diagrams in Figure 3.2. The grid in each diagram represents a space of propositions ordered by entailment: each node in the grid is a proposition, and one proposition entails another just in case there is a path from the first to the second along the grid such that each step of the path goes up. For instance, in Figure 3.2(a) P entails Q, because there is a path going up from P to Q, but P does not entail R because any path from P to R must contain at least one step that goes down. (p.46)

Figure 3.2 Visualization of basic algebraic operations.

The meet of P and Q is the greatest lower bound of P and Q with respect to entailment, i.e., the weakest proposition that entails both P and Q. As indicated in Figure 3.2(b), this greatest lower bound amounts to the intersection of the two propositions: $P∩Q$. More generally, the meet of a (possibly infinite) set of propositions Σ amounts to the intersection of all the propositions in that set:

$Display mathematics$

If Σ is empty, then $⋂Σ$ is the proposition consisting of all possible worlds, $W$. This is the weakest of all propositions, since it is entailed by all other propositions. It is denoted as ⊤. On the other hand, if Σ is the set of all propositions, then $⋂Σ$ is the empty proposition, . This is the strongest of all propositions, since it entails all other propositions. It is denoted as ⊥.

The join of two propositions P and Q is the least upper bound of P and Q with respect to entailment, i.e., the strongest proposition that is entailed by both P and Q. As indicated in Figure 3.2(b), this least upper bound amounts to the union of the two propositions: $P∪Q$. More generally, the join of a (possibly infinite) set of propositions Σ amounts to the union of all the propositions in that set:

$Display mathematics$

If Σ is empty, then $⋃Σ$ is the empty proposition, ⊥. On the other hand, if if Σ is the set of all propositions, then $⋃Σ$ is the proposition consisting of all possible worlds, ⊤.

The existence of meets and joins for arbitrary sets of propositions implies that $〈Pcl,⊧〉$ forms a complete lattice, bounded by ⊥ and ⊤ as its strongest and weakest elements, respectively.

Now let us turn to relative and absolute pseudo-complementation. The pseudo-complement of a proposition P relative to another (p.47) proposition Q, which we will denote as $P⇒Q$, can be thought of intuitively as the difference between P and Q: it is the weakest proposition R such that P and R together contain at least as much information as Q. More formally, it is the weakest proposition R such that $P∩R⊧Q$. To illustrate this notion, consider the propositions P and Q in Figure 3.2(c). How do we find the pseudo-complement of P relative to Q? The first step is to determine the set of all propositions R which are such that $P∩R⊧Q$. This is the shaded area in Figure 3.2(c). Indeed, if we take the meet of P with any proposition in this area, we obtain a proposition that entails Q. The second step is to select the weakest element of this set, i.e., that proposition which is entailed by all other propositions in the set. In our diagrams, one proposition entails another if there is a path going up from the first to the second. This means that the weakest element of a set of propositions is the topmost one. Thus, the topmost element of the shaded area in Figure 3.2(c) is the pseudo-complement of P relative to Q, which is denoted as $P⇒Q$.

It can be shown that $P⇒Q$ always consists of all possible worlds which, if contained in P, are also contained in Q:

$Display mathematics$

Absolute pseudo-complementation is a limit case of its relative counterpart. The absolute pseudo-complement of a proposition P, which we will denote as P*, is the weakest proposition R such that P and R are incompatible, in the sense that P and R together yield a contradiction. More formally, P* is the weakest proposition R such that $P∩R=⊥$. This means that P* amounts to $P⇒⊥$, the pseudo complement of P relative to ⊥. Within the space of classical propositions, the pseudo-complement of P is simply the set-theoretic complement of P:

$Display mathematics$

In a Heyting algebra it always holds, by definition of P*, that $P∩P*=⊥$. In the specific case of $〈Pcl,⊧〉$, we also always have that $P∪P*=⊤$. This means that in this particular setting, P* is in fact the Boolean complement of P, and that $〈Pcl,⊧〉$ forms a Boolean algebra, a special kind of Heyting algebra.

Thus, classical propositions are amenable to certain basic algebraic operations. Classical first-order logic is obtained by associating these operations with the connectives and the quantifiers. Indeed, the usual definition of truth can be reformulated as a recursive definition of the (p.48) set |φ| of worlds in which φ is true (given a domain D of individuals). The inductive clauses then run as follows:

• φ| = |φ|*

• $|φ∧ψ|=|φ|∩|ψ|$

• $|φ∨ψ|=|φ|∪|ψ|$

• $|φ→ψ|=|φ|⇒|ψ|$

• $|∀x.φ(x)|=⋂d∈D|φ(d)|$

• $|∃x.φ(x)|=⋃d∈D|φ(d)|$

Negation expresses absolute pseudo-complementation; conjunction and disjunction express binary meet and join, respectively; implication expresses relative pseudo-complementation; and quantified formulas, ∀x.φ and ∃x.φ, express the infinitary meet and join, respectively, of {|φ(d)|∣dD}.

Notice that everything started with a notion of propositions and a natural entailment order on these propositions. The entailment order induces certain basic operations on propositions, and classical first-order logic is obtained by associating these basic semantic operations with the connectives and quantifiers.

## 3.1.2 Algebraic operations on inquisitive propositions

Recall that in inquisitive semantics propositions are not sets of worlds, but rather sets of information states, non-empty and downward closed. In this setting, one proposition P entails another proposition Q just in case P is at least as informative and at least as inquisitive as Q. We have seen that this condition is satisfied just in case $P⊆Q$. So technically entailment still amounts to inclusion, just like in classical logic, though now it encompasses both informative and inquisitive strength. In Figure 3.3 we have depicted the set of all propositions in a logical space consisting of two possible worlds, and in Figure 3.4 we have done the same for a logical space consisting of three possible worlds. As before, arrows indicate entailment.

Figure 3.3 The set of all inquisitive semantics propositions if the logical space consists of two possible worlds. Arrows indicate entailment.

Figure 3.4 The set of all inquisitive semantics propositions if the logical space consists of three possible worlds. Arrows indicate entailment.

Now let us consider the algebraic structure of the space of all inquisitive propositions ordered by entailment, $〈P,⊧〉$, in order to determine which operations could be associated with the connectives and the quantifiers in an inquisitive semantics for the language of first-order logic. What kind of algebraic operations can be performed on inquisitive propositions? Does every set of propositions still have a unique greatest (p.49) lower bound (meet) and a unique least upper bound ( join) with regard to entailment? Does every proposition still have a pseudo-complement relative to any other proposition?

It turns out that these questions can be answered in the positive: $〈P,⊧〉$ forms a complete Heyting algebra, just like $〈Pcl,⊧〉$. First, any set of propositions $Σ⊆P$ still has a meet and a join, which can moreover still be characterized in terms of intersection and union.

Fact 3.1 (Meet)

Any set of propositions $Σ⊆P$ has a meet, which amounts to:

$Display mathematics$

Fact 3.2 (Join)

Any set of propositions $Σ⊆P$ has a join, which amounts to:

$Display mathematics$

if Σ≠, and to {} otherwise.

The existence of meets and joins for arbitrary sets of propositions implies that $〈P,⊆〉$ forms a complete lattice. This lattice has a unique strongest element, ⊥:={}, and a unique weakest element, $⊤:=℘(W)$.

Furthermore, just as in the classical setting, for every two propositions P and Q, there is a unique weakest proposition R such that $P∩R$ (p.50) (p.51) entails Q. Recall that this proposition, the pseudo-complement of P relative to Q, can be thought of intuitively as the difference between P and Q.

Fact 3.3 (Relative pseudo-complement)

For any $P,Q∈P$, the pseudo-complement of P relative to Q amounts to:

$Display mathematics$

The existence of relative pseudo-complements implies that $〈P,⊆〉$ forms a Heyting algebra. Finally, recall that the absolute pseudo-complement of a proposition P, denoted P*, is defined as the pseudo-complement of P relative to ⊥. We saw that in the classical setting, P* amounts to the set of worlds that are not in P. In the inquisitive setting, P* amounts to the set of states that are incompatible with any state in P.

Fact 3.4 (Absolute pseudo-complement)

For any proposition $P∈P$:

$Display mathematics$

A state s is incompatible with all states in P just in case it is incompatible with the union $⋃P$ of all these states, which amounts to info(P). In turn, s is incompatible with info(P) just in case $s⊆info(P)¯$. This leads to the following alternative characterization of P*.

Fact 3.5 (Absolute pseudo-complements, alternative characterization)

For any proposition $P∈P$:

$Display mathematics$

This characterization shows in particular that the absolute pseudo-complement of any given proposition P always contains a single alternative, $info(P)¯$, and is therefore never inquisitive.

The algebraic operations that we have identified are exactly the ones that are present in the classical setting. One notable difference, however, is that the absolute pseudo-complement of an inquisitive proposition is not always its Boolean complement. In fact, most inquisitive propositions do not have a Boolean complement at all. To see this, suppose that P and Q are Boolean complements. This means that:

1. (i) $P∩Q=⊥$

2. (ii) $P∪Q=⊤$

(p.52) Since $⊤=℘(W)$, condition (ii) can only be fulfilled if either P or Q contains $W$. Suppose $W∈P$. Then, since P is downward closed, $P=℘(W)=⊤$. But then, in order to satisfy condition (i), we must have that Q = {} = ⊥. So the only two elements of our algebra that have a Boolean complement are ⊤ and ⊥. Hence, the space $〈P,⊧〉$ of inquisitive propositions does not form a Boolean algebra, unlike the space $〈Pcl,⊧〉$ of classical propositions.

This difference has repercussions for the behavior of the logical system that we will specify, in particular for negation (for instance, the law of double negation will no longer hold). However, the similarity between $〈P,⊧〉$ and $〈Pcl,⊧〉$ that we identified, i.e., the fact that both form a Heyting algebra, is much more important for our current purposes. In particular, the existence of meets, joins, and relative and absolute pseudo-complements in $〈P,⊧〉$ will allow us to specify an inquisitive semantics for the language of first-order logic which is, from an algebraic perspective, the exact counterpart of classical first-order logic in the inquisitive setting. We will turn to this in Chapter 4. Before that, however, we will consider two additional operations that are particularly natural to perform on propositions in inquisitive semantics.

# 3.2 Projection operators

We noted in Section 2.4.2 that propositions in inquisitive semantics can be seen as inhabiting a two dimensional space, with non-inquisitive propositions living on one axis and non-informative propositions on the other. Given this picture, it is natural to consider whether it is possible to define general projection operators on this space, i.e., operators that map any given proposition to a corresponding proposition on one of the axes, trivializing either its informative or its inquisitive content. We will refer to such operators as info-cancelling and issue-cancelling projection operators.

Let us first consider more precisely what would be required for an operator π to qualify as an issue-cancelling projection operator. Such an operator should project a proposition onto the axis inhabited by non-inquisitive propositions. This means that, when applied to a proposition P, π should (i) trivialize the inquisitive content of P, i.e., turn P into a non-inquisitive proposition, and (ii) preserve the informative content (p.53) of P, i.e., yield a proposition that has the same informative content as P. This leads us to the following requirements.

Definition 3.6 (Requirements on issue-cancelling projection operators)

An operator π qualifies as an issue-cancelling projection operator just in case for any $P∈P$:

• πP is non-inquisitive

• info(πP) = info(P)

Now, in Section 2.4.2 we saw that if P is non-inquisitive, then we always have that P = (info(P)). This means that in order to satisfy the above requirements, πP must amount to (info(P)) for any proposition P. Thus, the semantic behavior of π is uniquely determined by the given requirements.

Fact 3.7 (Unique characterization)

An operator π qualifies as an issue-cancelling projection operator just in case for any $P∈P$:

• πP = (info(P))

Now let us consider which requirements π should fulfill in order to qualify as an info-cancelling projection operator. Such an operator should project a proposition onto the axis inhabited by non-informative propositions. For this, we should require that π trivializes the informative content of the proposition to which it applies, i.e., πP should always be non-informative. But, given this basic requirement, we cannot further demand that π always preserve the inquisitive content of P. For, if P and πP do not have the same informative content, then their inquisitive content will differ as well.

Fortunately, there is a natural way to overcome this obstacle. Namely, what we can require is that π preserve the decision set of P, i.e., the set of states that either settle the issue embodied by P, or contradict the informative content of P and thereby establish that it is impossible to settle the issue altogether.

Definition 3.8 (Contradicting and deciding on a proposition)

Let s be an information state and P a proposition. Then we say that:

• s contradicts P just in case $s∩info(P)=∅$;

• s decides on P just in case s either supports or contradicts P.

(p.54)

Definition 3.9 (Decision set)

The decision set D(P) of a proposition P is the set of states that decide on P.

The decision set of a proposition can be characterized explicitly as follows.

Fact 3.10 (Decision set explicated)

For any proposition P:

• $D(P)=P∪P*$

Now, what we require of an info-cancelling projection operator π is that, besides trivializing the informative content of the proposition it applies to, it preserves the proposition’s decision set. This is a requirement that can in principle be met, since P and πP can very well have the same decision set even if they differ in informative content.

Definition 3.11 (Requirements on info-cancelling projection operators)

An operator π qualifies as an info-cancelling projection operator just in case for any $P∈P$:

• πP is non-informative;

• D(πP) = D(P).

Now suppose that π fulfils these requirements. Then for any P, πP is non-informative, which means that $info(πP)=W$. But then $(πP)*=℘(info(P)¯)=℘(W¯)=℘(∅)={∅}$, and therefore $D(πP)=(πP)∪(πP)*=πP$. But since π should preserve the decision set of P, we also have that $D(πP)=D(P)=P∪P*$. Putting these facts together, we obtain that $πP=P∪P*$. Thus, the requirements we placed on π again uniquely determine its behavior.

Fact 3.12 (Unique characterization)

An operator π qualifies as an info-cancelling projection operator just in case for any $P∈P$:

• $πP=P∪P*$

Thus, by spelling out the natural requirements on issue-cancelling and info-cancelling projection operators we have arrived at a unique characterization of these operators, which we will denote as ! and ?, respectively. (p.55)

Definition 3.13 (Projection operators)

For any proposition P:

• !P:= (info(P))

• $?P:=P∪P*$

As depicted in Figure 3.5, the projection operators ! and ? turn any proposition P into an non-inquisitive proposition !P which has the same informative content as P, and a non-informative proposition ?P which has the same decision set as P. P itself can always be reconstructed as the meet of these two ‘pure components’.

Figure 3.5 Projection operators.

Fact 3.14 (Division)

For any proposition P:

• $P=!P∩?P$

Finally, let us consider how ? and ! are related to the algebraic operations identified in Section 3.1. Notice that ?P is already explicitly characterized in terms of the algebraic operations: it amounts to the join of P and its absolute pseudo-complement P*. It turns out that !P can also be characterized in terms of pseudo-complementation. Namely, for any proposition P, !P amounts to P**, i.e., to the proposition that results from two successive applications of the absolute pseudo-complementation operator to P.

Fact 3.15 (Projection operators and algebraic operators)

For any proposition P:

• !P = P**

• $?P=P∪P*$

(p.56) This concludes our discussion of the basic semantic operations that can be performed on propositions in inquisitive semantics. We end this chapter with a brief remark on the linguistic relevance of these operations, which will be further substantiated in later chapters.

# 3.3 Linguistic relevance

Since the algebraic operations on propositions that are associated with the connectives and quantifiers in classical logic are so fundamental, it is to be expected that natural languages will generally have ways to express them as well; just like one would expect, for instance, that basic arithmetic operations like addition and substraction are generally expressible in natural languages. This makes the algebraic operations discussed here of special interest from a linguistic point of view.

Similar considerations apply to the projection operators. Again, since these semantic operators are so fundamental, it is to be expected that they too are expressible in many natural languages. More specifically, it seems plausible to hypothesize that they are expressed in English and many other languages by declarative and interrogative clause type markers. For instance, on a first approximation, we may hypothesize that declarative clause type marking in English invokes the issue-cancelling projection operator ‘!’, and interrogative clause type marking the info-cancelling projection operator ‘?’. A more detailed account of clause type marking in English in terms of the projection operators will be presented in Chapter 6.

# 3.4 Exercises

## Exercise 3.1 Working through some examples

Consider the four propositions depicted in Figure 2.4.

1. 1. Determine the absolute pseudo-complement of each of these propositions.

2. 2. Determine the meet and the join of every pair among these propositions.

3. 3. Determine the outcome of applying the projection operators to each of these propositions.

## (p.57) Exercise 3.2 Meets and joins

Prove Facts 3.1 and 3.2. That is, show that every set of propositions in inquisitive semantics has a meet and a join with respect to entailment.

## Exercise 3.3 Relative pseudo-complementation

Prove Fact 3.3. That is, show that in inquisitive semantics every proposition P has a pseudo-complement relative to any other proposition Q, which amounts to $P⇒Q={s∣for everyt⊆s,ift∈Pthent∈Q}$.

## Exercise 3.4 Projection operators

Suppose we apply both projection operators to a given sentence, one after the other. Does it matter in which order we do this? That is, does the following hold for every proposition P:

$Display mathematics$

## Exercise 3.5 Projection operators

Show that the projection operators are idempotent, meaning that for every proposition P we have !!P = !P and ??P = ?P.

## Exercise 3.6 Division

Prove Fact 3.14. I.e., show that for every proposition P we have $P=!P∩?P$. (p.58)

## Notes:

(1) In fact, as we will discuss, the space of classical propositions ordered by entailment is a special kind of Heyting algebra, namely a Boolean algebra. For us, however, the more general fact that it forms a Heyting algebra will be crucial, because we will find that in inquisitive semantics the space of propositions no longer forms a Boolean algebra, but does still form a Heyting algebra. This will allow us to construe the basic logical operators in inquisitive semantics as the exact counterparts of those in classical logic.


text:
# Ochem energy calculation - correct answer?

1. May 21, 2006

### tandoorichicken

I think the solution manual is wrong for this problem.

The problem:
The naturally occurring sugar glucose exists in two isomeric cyclic forms. These are called [alpha] and [beta], and at equilibrium they are present in a ratio of approximately 64:36. Calculate the free energy difference that corresponds to this equilibrium ratio.

I basically used the free energy equation
$$\Delta G° = -RT \ln{K_{eq}}$$.

I used a value of 0.001986 for R and 298K for T (I assumed STP) and got a value of -0.34 kcal/mol for DG. The solution manual lists a value of -0.81 in kcal/mol.

2. May 21, 2006

### Bystander

Looks about right --- the idiot doing the solution may have taken a natural log and multiplied it by 2.303 rather than realizing that's a hangover from the old days of decimal logs from slide rules.


text:
# Chapter 1-8 - Cumulative Review - Page 679: 64

$\frac{(x+2)(x+5)}{(x+3)(x-8)}$

#### Work Step by Step

Multiplying the two fractions together, $\frac{x+2}{x-8}\times\frac{x+5}{x+3}=\frac{(x+2)(x+5)}{(x+3)(x-8)}$.

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:
# GATE Mechanical 2014 Set 4 | Question: 6

In a statically determinate plane truss, the number of joints $(j)$ and the number of members $(m)$ are related by

1. $j = 2m – 3$
2. $m = 2j + 1$
3. $m = 2j – 3$
4. $m = 2j − 1$

recategorized

## Related questions

For the truss shown in the figure, the forces $F_1$ and $F_2$ are $9$ $kN$ and $3$ $kN$, respectively. The force (in $kN$) in the member QS is $11.25$ tension $11.25$ compression $13.5$ tension $13.5$ compression
A two member truss ABC is shown in the figure. The force (in $kN$) transmitted in member AB is _______
A ladder AB of length $5$ $m$ and weight $(W)$ $600$ $N$ is resting against a wall. Assuming frictionless contact at the floor $(B)$ and the wall $(A)$, the magnitude of the force $P$ (in newton) required to maintain equilibrium of the ladder is _______
A frame is subjected to a load $P$ as shown in the figure. The frame has a constant flexural rigidity $EI$. The effect of axial load is neglected. The deflection at point $A$ due to the applied load $P$ is $\dfrac{1}{3}\dfrac{PL^3}{EI} \\$ $\dfrac{2}{3}\dfrac{PL^3}{EI} \\$ $\dfrac{PL^3}{EI} \\$ $\dfrac{4}{3}\dfrac{PL^3}{EI}$
A uniform slender rod ($8$ $m$ length and $3$ $kg$ mass) rotates in a vertical plane about a horizontal axis $1$ $m$ from its end as shown in the figure. The magnitude of the angular acceleration (in $rad/s^2$) of the rod at the position shown is _______


text:
# Documentation

### This is machine translation

Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.

# Constellation Diagram

Display constellation diagram for input signals

Comm Sinks

## Description

The Constellation Diagram block plots constellation diagrams, signal trajectory, and provides the ability to perform EVM and MER measurements.

The symbols that the Constellation Diagram scope displays are always the most recently available symbols from the time buffer.

## Dialog Box

To change the signal display settings, select View > Configuration Properties to bring up the Visuals—Constellation Properties dialog box. Then, modify the values for the Samples per symbol, Offset and Symbols to display parameters on the Main tab. You can modify the reference constellation parameters on the Reference constellation tab.

## Visuals — Constellation Properties

### Main Pane

#### Samples per symbol

Number of samples used to represent a symbol. This value must be a positive number.

#### Offset (samples)

Number of samples to skip before plotting points. The offset must be a nonnegative integer value less than the value of the samples per symbol.

#### Symbols to display

The maximum number of symbols that can be displayed. Must be a positive integer value.

### Display Pane

#### Show grid

Select this check box to turn on the grid.

#### Show legend

Select this check box to display a legend for the graph.

#### Show signal trajectory

Select this check box to display the trajectory of a modulated signal by plotting its in-phase component versus its quadrature component.

When you set select this check box, the points in the display fade as the interval of time after they are first plotted increases. The default value of this property is `false`. This property is tunable.

#### X-limits (Minimum)

Specify the minimum value of the x-axis.

#### X-limits (Maximum)

Specify the maximum value of the x-axis.

#### Y-limits (Minimum)

Specify the minimum value of the y-axis.

#### Y-limits (Maximum)

Specify the maximum value of the y-axis.

#### Title

Specify a label that appears above the constellation diagram plot. By default, there is no title.

#### X-axis label

Specify the text the scope displays along the x-axis

#### Y-axis label

Specify the text the scope displays along the y-axis

### Reference Constellation Pane

#### Show reference constellation

Select the check box to display the reference constellation.

#### Reference constellation

Select the reference constellation from `BPSK` | `QPSK` | `8-PSK` | `16-QAM` | `64-QAM` | `256-QAM` | `<user-defined>`. If not selected, the reference constellation is specified in the variable `refConst`.

#### Constellation normalization

Select the type of constellation normalization as ```Minimum distance```, `Average power`, or `Peak power`.

#### Reference minimum distance

Specify the minimum distance between symbols in the reference constellation as a positive real scalar. This parameter is available when Constellation normalization is set to ```Minimum distance```.

#### Average reference power

Specify the average power of the reference constellation as a positive real scalar. This parameter is available when Constellation normalization is set to `Average power`.

#### Peak reference power

Specify the peak power of the reference constellation as a positive real scalar. This parameter is available when Constellation normalization is set to `Peak power`.

Specify the phase offset of the reference constellation in radians as a real scalar.

## Measurement Panels

### Measurements Panel Buttons

Each of the Measurements panels contains the following buttons that enable you to modify the appearance of the current panel.

ButtonDescription

Move the current panel to the top. When you are displaying more than one panel, this action moves the current panel above all the other panels.

Collapse the current panel. When you first enable a panel, by default, it displays one or more of its panes. Click this button to hide all of its panes to conserve space. After you click this button, it becomes the expand button .

Expand the current panel. This button appears after you click the collapse button to hide the panes in the current panel. Click this button to display the panes in the current panel and show measurements again. After you click this button, it becomes the collapse button again.

Undock the current panel. This button lets you move the current panel into a separate window that can be relocated anywhere on your screen. After you click this button, it becomes the dock button in the new window.

Dock the current panel. This button appears only after you click the undock button. Click this button to put the current panel back into the right side of the Scope window. After you click this button, it becomes the undock button again.

Close the current panel. This button lets you remove the current panel from the right side of the Scope window.

Some panels have their measurements separated by category into a number of panes. Click the pane expand button to show each pane that is hidden in the current panel. Click the pane collapse button to hide each pane that is shown in the current panel.

### Signal Quality Panel

The Signal Quality panel controls the Settings and Signal Quality panes. Both panels can be independently expanded or collapsed.

You can choose to hide or display the Signal Quality panel. In the Scope menu, select Tools > Measurements > Signal Quality.

### Settings Pane

The Settings pane enables you to define the measurement interval and normalization method the scope uses when obtaining signal measurements.

• Measurement interval — The duration of the EVM or MER measurement, specified as ```'Current Display'```, `'All displays'`, or as a postive integer. The value of this property must be greater than one and less than or equal to the setting for the number of symbols to display. The measurement is computed after number of input data samples is greater than the measurement interval.

• EVM normalization — For the EVM calculations, you may use one of two normalization methods: ```Average constellation power``` or `Peak constellation power`. The scope performs EVM calculations using the `comm.EVM` System object™. For more information, see `comm.EVM`.

• Reference constellation — Select the reference constellation as `BPSK` | `QPSK` | `8-PSK` | `16-QAM` | `64-QAM` | `256-QAM` | `<user-defined>`.

• Constellation normalization — Select the type of constellation normalization as `Minimum distance`, ```Average power```, or `Peak power`.

• Minimum distance — Specify the minimum distance between symbols in the reference constellation as a positive real scalar. This parameter is available when Constellation normalization is set to `Minimum distance`.

• Average reference power — Specify the average power of the reference constellation as a positive real scalar. This parameter is available when Constellation normalization is set to `Average power`.

• Peak reference power — Specify the peak power of the reference constellation as a positive real scalar. This parameter is available when Constellation normalization is set to `Peak power`.

• Reference phase offset — Specify the phase offset of the reference constellation in radians as a real scalar.

### Signal Quality Pane

The Signal Quality pane displays the calculation results.

• EVM — An error vector is a vector in the I-Q plane between the ideal constellation point and the actual point at the receiver. The root mean square error vector magnitude, EVMRMS, is measured for the average and peak constellation power.

On the constellation diagram, you can display the EVMRMS measurements normalized by the `Average constellation power` or ```Peak constellation power``` as computed using these algorithms.

EVM Normalization MethodAlgorithm
`Average constellation power`

EVMRMS in percent for average constellation power normalization

`$EV{M}_{RMS}\left(percent\right)=\sqrt{\frac{\frac{1}{N}\sum _{k=1}^{N}\left({e}_{k}\right)}{{P}_{avg}}}*100$`

`Peak constellation power`

EVMRMS in percent for peak constellation power normalization

`$EV{M}_{RMS}\left(percent\right)=\sqrt{\frac{\frac{1}{N}\sum _{k=1}^{N}\left({e}_{k}\right)}{{P}_{\mathrm{max}}}}*\text{}100$`

The display shows the average and peak EVMRMS in percent and in decibels. The EVM reported in decibels is computed as EVM (dB) = 10*log10(EVMMS) = 20*log10(EVMRMS)

Where:

• ek = ${\left({I}_{k}-\stackrel{~}{{I}_{k}}\right)}^{2}+{\left({Q}_{k}-{\stackrel{~}{Q}}_{k}\right)}^{2}$

• Ik = In-phase measurement of the kth symbol in the burst

• Qk = Quadrature phase measurement of the kth symbol in the burst

• N = Input vector length

• Pavg = The value for Average constellation power

• Pmax = The value for Peak constellation power

• Ik and Qk represent ideal (reference) values. ${\stackrel{~}{I}}_{k}$ and ${\stackrel{~}{Q}}_{k}$ represent measured (received) symbols.

• EVMRMS = sqrt(EVMMS)

The max EVM is the maximum EVM value in a frame or $EV{M}_{\mathrm{max}}=\underset{k\in \left[1,...,N\right]}{\mathrm{max}}\left\{EV{M}_{k}\right\}\text{\hspace{0.17em}},$ where k is the kth symbol in a burst of length N.

EVM NormalizationAlgorithm
`Average constellation power`

Average constellation power normalization

`$EV{M}_{k}=\sqrt{\frac{{e}_{k}}{{P}_{avg}}}*100$`

`Peak constellation power`

Peak constellation power normalization

`$EV{M}_{k}=\sqrt{\frac{{e}_{k}}{{P}_{\mathrm{max}}}}*\text{}100$`

For more information, see `comm.EVM`.

• MER — MER is the ratio of the average power of the transmitted signal to the average power of the error vector. The scope indicates the measurement result in decibels.

MER is a measure of the SNR in a modulated signal calculated in dB. The MER over N symbols is

`$MER=10*\text{}{\mathrm{log}}_{10}\left(\frac{\sum _{n=1}^{N}\left({I}_{k}^{2}+{Q}_{k}^{2}\right)}{\sum _{n=1}^{N}\left({e}_{k}\right)}\right)\text{\hspace{0.17em}}\text{dB}\text{.}$`

where:

• ek = ${\left({I}_{k}-\stackrel{~}{{I}_{k}}\right)}^{2}+{\left({Q}_{k}-{\stackrel{~}{Q}}_{k}\right)}^{2}$

• Ik = In-phase measurement of the kth symbol in the burst

• Qk = Quadrature phase measurement of the kth symbol in the burst

• Ik and Qk represent ideal (reference) values. ${\stackrel{~}{I}}_{k}$ and ${\stackrel{~}{Q}}_{k}$ represent measured (received) symbols.

For more information, see `comm.MER`.

## Style Dialog Box

In the Style dialog box, you can customize the style of displays. You are able to change the color of the figure containing the displays, the background and foreground colors of display axes, and properties of lines in a display. From the scope menu, select View > Style to open this dialog box.

### Properties

The Style dialog box allows you to modify the following elements of the scope figure:

#### Figure color

Specify the color that you want to apply to the background of the scope figure. By default, the figure color is gray.

#### Axes colors

Specify the color that you want to apply to the background of the axes for the active display. Using a second drop down, you can also specify the color of the ticks, labels, and grid lines.

Specify the marker shape, marker size, marker line width, and color for the signal on the active display. The marker shape cannot be set to `none` unless the `ShowTrajectory` property is `true`.

#### Signal trajectory

Specify the line type, width, and color for the signal trajectory plot. The line type can only be set to something other than `no line` when the `ShowTrajectory` property is `true`. Conversely, the line type must be `no line` when the `ShowTrajectory` property is `false`.

#### Reference constellation

Specify the marker shape, marker size, marker line width, and color for the reference constellation shown on the active display. These settings are only applicable when the `ShowReferenceConstellation` property is `true`.

#### Marker symbols

Specify the markers for the selected signal and the reference constellation on the active display. This parameter is similar to the `Marker` property for the MATLAB® Handle Graphics® plot objects.

SpecifierMarker Type
noneNo marker
Circle
Square
Cross
Point (default)
Plus sign
Asterisk
Diamond
Downward-pointing triangle
Upward-pointing triangle
Left-pointing triangle
Right-pointing triangle
Five-pointed star (pentagram)
Six-pointed star (hexagram)

### Properties

The Tools—Axes Scaling Properties dialog box appears as follows.

#### Axes scaling

Specify when the scope automatically scales the axes. You can select one of the following options:

• `Manual` — When you select this option, the scope does not automatically scale the axes. You can manually scale the axes in any of the following ways:

• Select Tools > Axes Scaling Properties.

• Press one of the Scale Axis Limits toolbar buttons.

• When the scope figure is the active window, press and simultaneously.

• `Auto` — When you select this option, the scope scales the axes as needed, both during and after simulation. Selecting this option shows the Do not allow Y-axis limits to shrink check box.

• `After N Updates` — Selecting this option causes the scope to scale the axes after a specified number of updates. This option is useful and more efficient when your scope display starts with one axis scale, but quickly reaches a different steady state axis scale. Selecting this option shows the Number of updates edit box.

By default, this property is set to `Auto`. This property is Tunable (Simulink).

#### Do not allow Y-axis limits to shrink

When you select this property, the y-axis is allowed only to grow during axes scaling operations. If you clear this check box, the y-axis or color limits may shrink during axes scaling operations.

This property appears only when you select `Auto` for the Axis scaling property. When you set the Axes scaling property to `Manual` or ```After N Updates```, the y-axis or color limits are allowed to shrink. Tunable (Simulink).

Specify as a positive integer the number of updates after which to scale the axes. This property appears only when you select ```After N Updates``` for the Axes scaling property. Tunable (Simulink).

#### Scale axes limits at stop

Select this check box to scale the axes when the simulation stops. The y-axis is always scaled. The x-axis limits are only scaled if you also select the Scale X-axis limits check box.

#### Y-axis Data range (%)

Set the percentage of the y-axis that the scope uses to display the data when scaling the axes. Valid values are from 1 through 100. For example, if you set this property to `100`, the Scope scales the y-axis limits such that your data uses the entire y-axis range. If you then set this property to `30`, the scope increases the y-axis range such that your data uses only 30% of the y-axis range. Tunable (Simulink).

#### Y-axis Align

Specify where the scope aligns your data along the y-axis when it scales the axes. You can select `Top`, `Center`, or `Bottom`. Tunable (Simulink).

#### Autoscale X-axis limits

Check this box to allow the scope to scale the x-axis limits when it scales the axes. If Axes scaling is set to `Auto`, checking Autoscale X-axis limits only scales the data currently within the axes, not the entire signal in the data buffer. If Autoscale X-axis limits is on and the resulting axis is greater than the span of the scope, trigger position markers will not be displayed. Triggers are controlled using the Trigger Measurements panel. Tunable (Simulink).

#### X-axis Data range (%)

Set the percentage of the x-axis that the scope uses to display the data when scaling the axes. Valid values are from 1 through 100. For example, if you set this property to `100`, the scope scales the x-axis limits such that your data uses the entirex-axis range. If you then set this property to `30`, the scope increases the x-axis range such that your data uses only 30% of the x-axis range. Use the x-axis Align property to specify data placement along the x-axis.

This property appears only when you select the Scale X-axis limits check box. Tunable (Simulink).

#### X-axis Align

Specify how the scope aligns your data along the x-axis: `Left`, `Center`, or `Right`. This property appears only when you select the Scale X-axis limits check box. Tunable (Simulink).

## Examples

### View Constellation Diagram

This example shows how to use the Constellation Diagram block to visualize the constellation or scatter plot of a modulated signal.

Open the model, doc_constellation_diagram_example, from the MATLAB command prompt.

`doc_constellation_diagram_example`

The model includes:

Run the model and observe the 8-PSK constellation. The received data points are shown in yellow while the red ‘+’ symbols represent the ideal constellation locations.

Click on the button to display the signal trajectory of the modulated signal. Observe that the constellation diagram icon in the model has changed to reflect that the diagram is now displaying a trajectory and the constellation diagram now shows the signal trajectory.


text:
# General solution of the Schrodinger equation for a free particle?

## Homework Statement

I'm trying to figure out how the general solution of the Schrodinger equation for a free particle when v=0 relates to anything I have learned in class...

## Homework Equations

For Eψ=(hbar2/2m)d2ψ/dx2

## The Attempt at a Solution

I really have no idea- what is confusing me is that ψ is a function rather than a constant so I don't know how to treat it...

Thank you all for any help!

And I really apologise for the equation... I don't really understand how to use LaTeX

ehild
Homework Helper

## Homework Statement

I'm trying to figure out how the general solution of the Schrodinger equation for a free particle when v=0 relates to anything I have learned in class...

## Homework Equations

For [STRIKE]Eψ=(hbar2/2m)d2ψ/dx2
[/STRIKE]

You miss the negative the sign: ##-\hbar^2/(2m)d^2ψ/dx^2=Eψ##

That is a differential equation of form y"+kx=0. What is the solution?

The former equation is the time independent equation for one dimension. The time dependent equation is

##i\hbar \frac{\partial ψ}{\partial t}=-\hbar^2/(2m)\frac{\partial^2}{\partial x^2}##

ehild

Last edited:
haruspex
Homework Helper
Gold Member
2020 Award
That is a differential equation of form y"+kx=0.
y"+ky=0?

ehild
Homework Helper
I meant it so. Thanks.

ehild