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Solving differential algebraic equations with help from autograd

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This problem is adapted from one in "Problem Solving in Chemical Engineering with Numerical Methods, Michael B. Cutlip, Mordechai Shacham".

In the binary, batch distillation of benzene (1) and toluene (2), the moles of liquid $$L$$ remaining as a function of the mole fraction of toluene ($$x_2$$) is expressed by:

$$\frac{dL}{dx_2} = \frac{L}{x_2 (k_2 - 1)}$$

where $$k_2$$ is the vapor liquid equilibrium ratio for toluene. This can be computed as:

$$k_i = P_i / P$$ where $$P_i = 10^{A_i + \frac{B_i}{T +C_i}}$$ and that pressure is in mmHg, and the temperature is in degrees Celsius.

One difficulty in solving this problem is that the temperature is not constant; it changes with the composition. We know that the temperature changes to satisfy this constraint $$k_1(T) x_1 + k_2(T) x_2 = 1$$.

Sometimes, one can solve for T directly, and substitute it into the first ODE, but this is not a possibility here. One way you might solve this is to use the constraint to find $$T$$ inside an ODE function, but that is tricky; nonlinear algebra solvers need a guess and don't always converge, or may converge to non-physical solutions. They also require iterative solutions, so they will be slower than an approach where we just have to integrate the solution. A better way is to derive a second ODE $$dT/dx_2$$ from the constraint. The constraint is implicit in $$T$$, so We compute it as $$dT/dx_2 = -df/dx_2 / df/dT$$ where $$f(x_2, T) = k_1(T) x_1 + k_2(T) x_2 - 1 = 0$$. This equation is used to compute the bubble point temperature. Note, it is possible to derive these analytically, but who wants to? We can use autograd to get those derivatives for us instead.

The following information is given:

The total pressure is fixed at 1.2 atm, and the distillation starts at $$x_2=0.4$$. There are initially 100 moles in the distillation.

species A B C
benzene 6.90565 -1211.033 220.79
toluene 6.95464 -1344.8 219.482

We have to start by finding the initial temperature from the constraint.

import autograd.numpy as np
from scipy.integrate import solve_ivp
from scipy.optimize import fsolve
%matplotlib inline
import matplotlib.pyplot as plt

P = 760 * 1.2 # mmHg
A1, B1, C1 = 6.90565, -1211.033,  220.79
A2, B2, C2 = 6.95464, -1344.8, 219.482

def k1(T):
return 10**(A1 + B1 / (C1 + T)) / P

def k2(T):
return 10**(A2 + B2 / (C2 + T)) / P

def f(x2, T):
x1 = 1 - x2
return k1(T) * x1 + k2(T) * x2 - 1

T0, = fsolve(lambda T: f(0.4, T), 96)
print(f'The initial temperature is {T0:1.2f} degC.')

The initial temperature is 95.59 degC.

Next, we compute the derivative we need. This derivative is derived from the constraint, which should ensure that the temperature changes as required to maintain the constraint.

dfdx2 = grad(f, 0)

def dTdx2(x2, T):
return -dfdx2(x2, T) / dfdT(x2, T)

def ode(x2, X):
L, T = X
dLdx2 = L / (x2 * (k2(T) - 1))
return [dLdx2, dTdx2(x2, T)]

Next we solve and plot the ODE.

x2span = (0.4, 0.8)
X0 = (100, T0)
sol = solve_ivp(ode, x2span, X0, max_step=0.01)

plt.plot(sol.t, sol.y.T)
plt.legend(['L', 'T']);
plt.xlabel('$x_2$')
plt.ylabel('L, T')
x2 = sol.t
L, T = sol.y
print(f'At x2={x2[-1]:1.2f} there are {L[-1]:1.2f} moles of liquid left at {T[-1]:1.2f} degC')

At x2=0.80 there are 14.04 moles of liquid left at 108.57 degC

<Figure size 432x288 with 1 Axes>

You can see that the liquid level drops, and the temperature rises.

Let's double check that the constraint is actually met. We do that qualitatively here by plotting it, and quantitatively by showing all values are close to 0.

constraint = k1(T) * (1 - x2) + k2(T) * x2 - 1
plt.plot(x2, constraint)
plt.ylim([-1, 1])
plt.xlabel('$x_2$')
plt.ylabel('constraint value')
print(np.allclose(constraint, np.zeros_like(constraint)))
constraint

True

array([ 2.22044605e-16,  4.44089210e-16,  2.22044605e-16,  0.00000000e+00,
1.11022302e-15,  0.00000000e+00,  6.66133815e-16,  0.00000000e+00,
-2.22044605e-16,  1.33226763e-15,  8.88178420e-16, -4.44089210e-16,
4.44089210e-16,  1.11022302e-15, -2.22044605e-16,  0.00000000e+00,
-2.22044605e-16, -1.11022302e-15,  4.44089210e-16,  0.00000000e+00,
-4.44089210e-16,  4.44089210e-16, -6.66133815e-16, -4.44089210e-16,
4.44089210e-16, -1.11022302e-16, -8.88178420e-16, -8.88178420e-16,
-9.99200722e-16, -3.33066907e-16, -7.77156117e-16, -2.22044605e-16,
-9.99200722e-16, -1.11022302e-15, -3.33066907e-16, -1.99840144e-15,
-1.33226763e-15, -2.44249065e-15, -1.55431223e-15, -6.66133815e-16,
-2.22044605e-16])

<Figure size 432x288 with 1 Axes>

So indeed, the constraint is met! Once again, autograd comes to the rescue in making a computable derivative from an algebraic constraint so that we can solve a DAE as a set of ODEs using our regular machinery. Nice work autograd!


text:
Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 6 Quiz Answer Key.

6.1 Multiplying Integers

Find each product.

Question 1.
(- 2) (3) ______________
First, determine if the product wilt be positive or negative
Since – 2 is negative and 3 is positive (they have opposite signs), the product will be negative.
Then, multiply absolute values of given integers:
|- 2| = 2 and |3| = 3
2 × 3 = 6
The result is 6 or – 6, depending on signs of given integers.
Product is negative so final result is – 6.

Result is – 6.
Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 2.
(- 5) (- 7) _____________
First, determine if the product wilt be positive or negative
Since – 5 is negative and – 7 is negative (they have same signs), the product will be positive.
Then, multiply absolute values of given integers:
|- 5| = 5 and |- 7| = 7
5 × 7 = 35
The result is 35 or – 35, depending on signs of given integers.
Product is negative so final result is 35.

Result is 35.
Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 3.
(8) (- 11) ______________
First, determine if the product will be positive or negative
Since 8 is positive and – 11 is negative (they have opposite signs), the product will be negative.
Then, multiply absolute values of given integers:
|8| = 8 and |- 11| = 11
8 × 11 = 88
The result is 88 or – 88, depending on signs of given integers.
Product is negative so final result is – 88

Result is – 88.
Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 4.
(- 3) (2) (- 2) ______________
First, calculate the product:
(-3) (2)
Determine if the product will be positive or negative
Since – 3 is negative and 2 is positive (they have opposite signs, the product will be negative.
Then, multiply absolute values of given integers:
|- 3| = 3 and |2| = 2
3 × 2 = 6
The result is 6 or – 6, depending on signs of given integers.
Product is negative so result is – 6.
Now, find the product:
(- 6) (- 2)
Determine if the product will be positive or negative.

Since – 6 is negative and – 2 is negative (they have the same sign), the product will be positive.
Then, multiply absolute values of given integers:
|- 6| = 6 and |- 2| = 2
6 × 2 = 12
The result is 12 or – 12, depending on signs of given integers.
Product is positive so final result is 12.
Therefore:
(- 3) (2) (- 2) = 12

First, calculate the product (- 3)(2) and then multiply it with – 2.
Find the sign of the product, multiply absolute values of integers and add minus if the sign of product is minus (-).

Question 5.
The temperature dropped 3 °C every hour for 5 hours.
Write an integer that represents the change ¡n temperature. ______________
Use negative integer to represent how many degrees the temperature dropped each hour.
The expression you get is:
5 (- 3)
First, determine if the product will be positive or negative
Since 5 is positive and – 3 is negative (they have opposite signs), the product will be negative.
Then, multiply absolute values of given integers:
|5| = 5 and |- 3| = 3
5 × 3 = 15
The result is 15 or – 15, depending on signs of given integers.
Product is negative so final result is – 15.
The change in temperature is – 15 degrees.
Write the expression that represents the change in temperature and find the product using rules for multiplying integers.

6.2 Dividing Integers

Find each quotient.

Question 6.
$$\frac{-63}{7}$$
First determine if the quotient will be positive or negative
Since – 63 is negative and 7 is positive (they have opposite signs), the quotient will be negative.
Divide given integers:
$$\frac{-63}{7}$$ = – 9

Result is – 9
Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 7.
$$\frac{-15}{-3}$$
First determine if the quotient will be positive or negative
Since – 15 is negative and – 3 is negative (they have same signs), the quotient will be positive.
Divide given integers:
$$\frac{-15}{-3}$$ = 5

Result is 5
Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 8.
$$\frac{0}{-15}$$
First, determine if the quotient wilt be positive or negative
Since dividend is 0 and divisor – 15 is not 0, the quotient is 0.
The result of division is:
$$\frac{0}{-15}$$ = 0

Result is 0.
Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 9.
$$\frac{96}{-12}$$
First determine if the quotient will be positive or negative
Since 96 is positive and – 12 is negative (they have opposite signs), the quotient will be negative.
Divide given integers:
$$\frac{96}{-12}$$ = – 8

Result is – 8.
Find the sign of the quotient (opposite signs of integers then -, the same sign of integers then +) and divide given integers.

Question 10.
An elephant at the zoo lost 24 pounds over 6 months. The elephant lost the same amount of weight each month. Write an integer that represents the change in the elephant’s weight each month.
Use negative integer to represent the change in elephant’s weight
The expression you get is:
$$\frac{- 24}{6}$$
First determine if the quotient will be positive or negative
Since – 24 is negative and 6 is positive (they have opposite signs), the quotient wilt be negative.
Divide given integers:
$$\frac{- 24}{6}$$ = – 4
The change in weight each month is – 4 pounds.
Write the expression that represents the change in elephant’s weight and find the quotient using rules for dividing integers.

6.3 Applying Integer Operations

Evaluate each expression.

Question 11.
(- 4) (5) + 8 _____________
The expression you need to evaluate is:
(- 4)(5) + 8
First find the product using rules for multiplying integers:
– 20 + 8
Then, use rules for adding integers to find the sum:
– 20 + 8 = – 12
Therefore:
(- 4)(5) + 8 = – 12

The value of the expression is – 12
Use rules for multiplying integers and rules for adding integers to find that value.

Question 12.
(- 3) (-6) – 7 _____________
The expression you need to evaluate is:
(- 3) (- 6) – 7
First find the product using rules for multiplying integers:
18 – 7
Then, use rules for subtracting integers to find the difference:
18 – 7 = 11
Therefore:
(- 3) (- 6) – 7 = 11

The value of the expression is 11.
Use rules for multiplying integers and rules for subtracting integers to find the value.

Question 13.
$$\frac{- 27}{9}$$ – 11 _____________
The expression you need to evaluate is:
$$\frac{-27}{9}$$ – 11
First find the quotient using rules for dividing integers:
– 3 – 11
Then, use rules for subtracting integers to find the difference:
– 3 – 11 = – 14
Therefore:
$$\frac{-27}{9}$$ – 11 = – 14

The value of the expression is – 14.
Use rules for dividing integers and rules for subtracting integers to find the value.

Question 14.
$$\frac{-24}{-3}$$ – (- 2) _____________
The expression you need to evaluate is:
$$\frac{-24}{-3}$$ – (- 2)
First find the quotient using rules for dividing integers:
8 – (- 2)
Then, use rules for subtracting integers to find the difference:
8 – (- 2) = 10
Therefore:
$$\frac{-24}{-3}$$ – (- 2) = 10

The value of the expression is 10.
Use rules for dividing integers and rules for subtracting integers to find the value.

Essential Question

Question 15.
Write and solve a real-world problem that can be represented by the expression (- 3) (5) + 10.
Jim is at the entrance to the cave. He descends 5 times, each time 3 feet Then, he ascends 10 feet What is his new position relative to the cave entrance
Use negative integer if he descends and positive integer if he ascends.
The expression you get is:
(- 3) (5) + 10
First, find the product using rules for multiplying integers:
– 15 + 10
Then, use rules for adding integers to find the sum
– 15 + 10 = – 5
His new position is 5 feet below the cave entrance.

Jim is at the entrance to the cave. He descends 5 times, each time 3 feet Then, he ascends 10 feet What is his new position relative to the cave entrance?
Jim’s new position is 5 feet below the cave entrance.
Use rules for multiplying integers and rules for adding integers to find his new position.

Texas Go Math Grade 6 Module 6 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
A diver is at an elevation of – 18 feet relative to sea level. The diver descends to an undersea cave that is 4 times as far from the surface. What is the elevation of the cave?
(A) – 72 feet
(B) – 22 feet
(C) – 18 feet
(D) – 14 feet
(A) – 72 feet

Explaination:
The expression that represents elevation of the cave is:
4(- 18)
Calculate that product to find the elevation.
First, determine if the product will be positive or negative
Since 4 is positive and – 18 is negative (they have opposite signs), the product will be negative.
Then, multiply absolute values of given integers:
|4| = 4 and |- 18| = 18
4 × 18 = 72
The result is 72 or – 72, depending on signs of given integers.
Product is negative so final result is – 72.
The elevation of the cave is – 72 feet.

Question 2.
The football team lost 4 yards on 2 plays in a row. Which of the following could represent the change in field position?
(A) – 12 yards
(B) – 8 yards
(C) – 6 yards
(D) – 2 yards
(B) – 8 yards

Explaination:
Use negative integer because the team lost yards. The expression you get is:
2(- 4)
First, determine if the product will be positive or negative
Since 2 is positive and – 4 is negative (they have opposite signs), the product will be negative.
Then, multiply absolute values of given integers:
|2| = 2 and |- 4| = 4
2 × 4 = 8
The result is 8 or – 8, depending on signs of given integers.
Product is negative so final result is – 8.
The change in field position is – 8 yards.

Question 3.
Clayton climbed down 50 meters. He climbed down in 10-meter intervals. In how many intervals did Clayton make his climb?
(A) 5
(B) 10
(C) 40
(D) 500
(A) 5

Explaination:
Divide total distance by distance of one interval to get number of intervals. The expression you get is:
50 ÷ 10
First, determine if the quotient will be positive or negative
Since 50 is positive and 10 is positive (they have the same sign), the quotient will be positive.
Divide given integers:
50 ÷ 10 = 5
He made his climb in 5 intervals.

Question 4.
Which expression results in a negative answer?
(A) a negative number divided by a negative number
(B) a positive number divided by a negative number
(C) a negative number multiplied by a negative number
(D) a positive number multiplied by a positive number
(B) a positive number divided by a negative number

Explaination:
The quotient of two integers with opposite signs is negative so the correct answer is B.

The quotient of two integers with the same sign is positive as well as the product of two integers with the same sign so results of expressions A), C) and D) are positive

Question 5.
Clara played a video game before she left the house to go on a walk. She started with 0 points, lost 6 points 3 times, won 4 points, and then lost 2 points. How many points did she have when she left the house to go on the walk?
(A) – 20
(B) 12
(C) – 16
(D) 20
(C) – 16

Explaination:
Subtract integer if she lost points and add integer if she won points
The expression you get is:
3(- 6) + 4 – 2
First, find the product using rules for multiplying integers:
– 18 + 4 – 2
– 14 – 2
Use rules for subtracting integers to find the final result:
– 14 – 2 = – 16
Clara had – 16 points when she left the house.

Question 6.
Which expression is equal to
(A) $$\frac{-24}{6}$$ – 4
(B) $$\frac{-24}{-6}$$ + 4
(C) $$\frac{24}{6}$$ + 4
(D) $$\frac{-24}{-6}$$ – 4
(D) $$\frac{-24}{-6}$$ – 4

Explaination:
Calculate value of each expression to see which one is equal to 0.

(A) $$\frac{-24}{6}$$ – 4
Find the quotient using rules for dividing integers:
– 4 – 4
Use rules for subtracting integers to find the difference:
– 4 – 4 = – 8

(B) $$\frac{-24}{-6}$$ + 4
Find the quotient using rules for dividing integers:
4 + 4
Use rules for adding integers to find the sum:
4 + 4 = 8

(C) $$\frac{24}{6}$$ + 4
Find the quotient using rules for dividing integers:
4 + 4
Use rules for adding integers to find the sum:
4 + 4 = 8

(D) $$\frac{-24}{-6}$$ – 4
Find the quotient using rules for dividing integers:
4 – 4
Use rules for subtracting integers to find the difference:
4 – 4 = 0
Therefore, the correct answer is D.

Gridded Response

Question 7.
Rochelle made three $25 withdrawals and then wrote a check for$100. If she started with $200 in her account, find the total amount she has left in her account in dollars. Answer: Use positive integer to represent the amount she had at the beginning and use negative integers to represent the amount she withdrew and the value of the check. The expression you get is: 200 + 3(- 25) + (- 100) First find the product using rules for multiplying integers: 200 – 75 + (- 100) Use rules for subtracting integers: 125 – (- 100) Use rules for adding integers to find the final result: 125 + (- 100) = 25 Rochelle has left$25 in her account

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# Normal dense subspaces of a product of “continuum” many separable metric factors

Is every normal dense subspace of a product of separable metric spaces collectionwise normal? This question was posed by Arkhangelskii in [1] (see Problem I.5.25). A partial positive answer is provided by a theorem that is usually attributed to Corson: If $Y$ is a normal dense subspace of a product of separable metric spaces and if $Y \times Y$ is also normal, then $Y$ is collectionwise normal. In this post, using a simple combinatorial argument, we show that any normal dense subspace of a product of continuum many separable metric space is collectionwise normal (see Corollary 4 below), which is a corollary of the following theorem.

Theorem 1
Let $X$ be a normal space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following holds:

• If $Y$ is a closed and discrete subspace of $X$ with $\lvert Y \lvert=\omega_1$, then $Y$ contains a separated subset of cardinality $\omega_1$.

Theorem 1 gives the corollary indicated at the beginning and several other interesting results. The statement $2^\omega<2^{\omega_1}$ means that the cardinality of the power set (the set of all subsets) of $\omega$ is strictly less than the cardinality of the power set of $\omega_1$. Note that the statement $2^\omega<2^{\omega_1}$ follows from the continuum hypothesis (CH), the statement that $2^\omega=\omega_1$. With the assumption $2^\omega<2^{\omega_1}$, Theorem 1 is a theorem that goes beyond ZFC. We also present an alternative to Theorem 1 that removes the assumption $2^\omega<2^{\omega_1}$ (see Theorem 6 below).

A subset $T$ of a space $S$ is a separated set (in $S$) if for each $t \in T$, there is an open subset $O_t$ of $S$ with $t \in O_t$ such that $\left\{O_t: t \in T \right\}$ is a pairwise disjoint collection. First we prove Theorem 1 and then discuss the corollaries.

____________________________________________________________________

Proof of Theorem 1

Suppose $Y$ is a closed and discrete subset of $X$ with $\lvert Y \lvert=\omega_1$ such that no subset of $Y$ of cardinality $\omega_1$ can be separated. We then show that $2^{\omega_1} \le 2^{\omega}$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert \le 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose a maximal disjoint collection $\mathcal{M}_W$ of open sets in $\mathcal{G}_W$. Because no subset of $Y$ of cardinality $\omega_1$ can be separated, each $\mathcal{M}_W$ is countable. If $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that $\lvert \mathcal{P}(Y) \lvert=2^{\omega_1}$. Also note that since $\lvert \mathcal{B} \lvert \le 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert \le 2^\omega$. Thus $2^{\omega_1} \le 2^{\omega}$. $\blacksquare$

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Some Corollaries of Theorem 1

Here’s some corollaries that follow easily from Theorem 1. A space $X$ has the countable chain condition (CCC) if every pairwise disjoint collection of non-empty open subset of $X$ is countable. For convenience, if $X$ has the CCC, we say $X$ is CCC. The following corollaries make use of the fact that any normal space with countable extent is collectionwise normal (see Theorem 2 in this previous post).

Corollary 2
Let $X$ be a CCC space with character $\le 2^\omega$. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 3
Let $X$ be a CCC space with character $\le 2^\omega$. If CH holds, then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ is countable, i.e., $X$ has countable extent.
• If $X$ is normal, then $X$ is collectionwise normal.

Corollary 4
Let $X=\prod_{\alpha<2^\omega} X_\alpha$ be a product where each factor $X_\alpha$ is a separable metric space. If $2^\omega<2^{\omega_1}$, then the following conditions hold:

• If $Y$ is a normal dense subspace of $X$, then $Y$ has countable extent.
• If $Y$ is a normal dense subspace of $X$, then $Y$ is collectionwise normal.

Corollary 4 is the result indicated in the title of the post. The product of separable spaces has the CCC. Thus the product space $X$ and any dense subspace of $X$ have the CCC. Because $X$ is a product of continuum many separable metric spaces, $X$ and any subspace of $X$ have characters $\le 2^\omega$. Then Corollary 4 follows from Corollary 2.

When dealing with the topic of normal versus collectionwise normal, it is hard to avoid the connection with the normal Moore space conjecture. Theorem 1 gives the result of F. B. Jones from 1937 (see [3]). We have the following theorem.

Theorem 5
If $2^\omega<2^{\omega_1}$, then every separable normal Moore space is metrizable.

Though this was not how Jones proved it in [3], Theorem 5 is a corollary of Corollary 2. By Corollary 2, any separable normal Moore space is collectionwise normal. It is well known that collectionwise normal Moore space is metrizable (Bing’s metrization theorem, see Theorem 5.4.1 in [2]).

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A ZFC Theorem

We now prove a result that is similar to Corollary 2 but uses no set-theory beyond the Zermelo–Fraenkel set theory plus axiom of choice (abbreviated by ZFC). Of course the conclusion is not as strong. Even though the assumption $2^\omega<2^{\omega_1}$ is removed in Theorem 6, note the similarity between the proof of Theorem 1 and the proof of Theorem 6.

Theorem 6
Let $X$ be a CCC space with character $\le 2^\omega$. Then the following conditions hold:

• If $X$ is normal, then every closed and discrete subset of $X$ has cardinality less than continuum.

Proof of Theorem 6
Let $X$ be a normal CCC space with character $\le 2^\omega$. Let $Y$ be a closed and discrete subset of $X$. We show that $\lvert Y \lvert < 2^\omega$. Suppose that $\lvert Y \lvert = 2^\omega$.

For each $y \in Y$, let $\mathcal{B}_y$ be a local base at the point $y$ such that $\lvert \mathcal{B}_y \lvert \le 2^\omega$. Let $\mathcal{B}=\bigcup_{y \in Y} \mathcal{B}_y$. Thus $\lvert \mathcal{B} \lvert = 2^\omega$. By normality, for each $W \subset Y$, let $U_W$ be an open subset of $X$ such that $W \subset U_W$ and $\overline{U_W} \cap (Y-W)=\varnothing$. For each $W \subset Y$, consider the following collection of open sets:

$\mathcal{G}_W=\left\{V \in \mathcal{B}_y: y \in W \text{ and } V \subset U_W \right\}$

For each $W \subset Y$, choose $\mathcal{M}_W \subset \mathcal{G}_W$ such that $\mathcal{M}_W$ is a maximal disjoint collection. Since $X$ is CCC, $\mathcal{M}_W$ is countable. It is clear that if $W_1 \ne W_2$, then $\mathcal{M}_{W_1} \ne \mathcal{M}_{W_2}$.

Let $\mathcal{P}(Y)$ be the power set (i.e. the set of all subsets) of $Y$. Let $\mathcal{P}_\omega(\mathcal{B})$ be the set of all countable subsets of $\mathcal{B}$. Then the mapping $W \mapsto \mathcal{M}_W$ is a one-to-one map from $\mathcal{P}(Y)$ into $\mathcal{P}_\omega(\mathcal{B})$. Note that since $\lvert \mathcal{B} \lvert = 2^\omega$, $\lvert \mathcal{P}_\omega(\mathcal{B}) \lvert = 2^\omega$. Thus $\lvert \mathcal{P}(Y) \lvert \le 2^{\omega}$. However, $Y$ is assumed to be of cardinality continuum. Then $\lvert \mathcal{P}(Y) \lvert>2^{\omega_1}$, leading to a contradiction. Thus it must be the case that $\lvert Y \lvert < 2^\omega$. $\blacksquare$

With Theorem 6, Corollary 3 still holds. Theorem 6 removes the set-theoretic assumption of $2^\omega<2^{\omega_1}$. As a result, the upper bound for cardinalities of closed and discrete sets is (at least potentially) higher.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Jones, F. B., Concerning normal and completely normal spaces, Bull. Amer. Math. Soc., 43, 671-677, 1937.

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$\copyright \ 2014 \text{ by Dan Ma}$


text:
MathSciNet bibliographic data MR1111326 (92c:17017) 17B35 (14M17 17B37) Ginzburg, Victor Lagrangian construction of the enveloping algebra $U({\rm sl}\sb n)$$U({\rm sl}\sb n)$. C. R. Acad. Sci. Paris Sér. I Math. 312 (1991), no. 12, 907–912. Links to the journal or article are not yet available

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text:
the distance between two parallel lines is always

Nov 2010 57 0. Which can be written as mx â y + c1 = 0. Then, the point O divides the segment PQ in the ratio. The distance between two straight lines in the plane is the minimum distance between any two points lying on the lines. d1= distance of perpendicular from (0,0) to 3x+4y+7 = 0, d2 = (3 × 0 + 4 × 0 + (−5) )/ √(32+42) = −5/5. Example 1: Find the distance of the point (4, –6) from the line 2x – 7y – 24 = 0. Distance between Two Intersecting Lines: The shortest distance between such lines is eventually zero. Example 10: A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. 4x + 6y = -7. Find the coordinates of the vertices of the image after the translation (x, y)â(x, yâ4). Otherwise, draw a diagram and consider Pythagoras' Theorem. Thread starter thamathkid1729; Start date Mar 14, 2011; Tags distance lines parallel; Home. At the equator, the distance is 68.703 miles (110.567 kilometers). The lines can be extended till infinity. The graph of the function, a pair of railroad tracks or the opposite sides of a parallelogram or keys of a piano can be a few examples of parallel lines. a = 4, b = 6, c 1 = 5 and c 2 = 7. There are 14 passengers on the bus. For the two lines to be parallel, the most important thing is that they are drawn in the same plane. At each of the poles, the distance is 69.407 â¦ units.\Delta =\frac{1}{2}\left| \begin{matrix} 0 & 12 & 1 \\ 8 & 0 & 1 \\ -\frac{13}{2} & -1 & 1 \\ \end{matrix}\, \right|=91 \text \ sq. Here, we use a more geometric approach, and end up with the same result. Each degree of latitude is approximately 69 miles (111 kilometers) apart. A common property which can be found in the above examples is that the two railroad tracks never meet, the opposite sides of a parallelogram will never intersect or the piano keys are parallel to each other. Therefore, the distance between the lines (i) and (ii) is. The required equation is y+1=−29(x−1)y+1=\frac{-2}{9}(x-1)y+1=9−2(x−1) that is 2x+9y+7=0.2x+9y+7=0.2x+9y+7=0.. In a Cartesian plane, the relationship between two straight lines varies because they can merely intersect each other, be perpendicular to each other, or can be the parallel lines. Ridhi â¦ By THE EDITORS on November 8, 2010; ... parallel lines may actually diverge or converge. They are always equidistant from each other. x1 and y1 are the two intersection points of the lines with the axis in a cartesian plane, while a1 and b1 are the coefficients of variable x and y of the line. Finally, put all the above values in the distance formula to find the distance between two parallel lines. Or a = 3, b = 4 and c2 = -1/2. Equation of a line passing through (0, 1) and parallel to the x-axis is y=−1. Distance Between 2 Parallel Lines. Parallel lines are those lines which never meet each other. Given lines are  4x+2y = 9 and 2x+y+6 = 0. 6) Parallel lines : Lines are parallel if they are always the same distance apart (called "equidistant"), and will never meet or never intersect â¦ Distance Between Two Parallel Lines In order to find the distance between two parallel lines, first we find a point on one of the lines and then we find its distance from the other line. They are always equidistant from each other. The second equation is 6x + 8y -1 =0, dividing throughout by 2; 3x + 4y -(1/2) = 0. Formula for distance between parallel lines is We want to find the w(s,t) that has a minimum length for all s and t. This can be computed using calculus [Eberly, 2001]. Answer!! Example 6: The graph of the function cosx cos(x+2)−cos2(x+1)\cos x\ \cos (x+2)-{{\cos }^{2}}(x+1)cosx cos(x+2)−cos2(x+1) is, A) A straight line passing through (0,−sin^2 [1]) with slope 2, B) A straight line passing through (0, 0), D) A straight line passing through the point (π/ 2, −sin^2 [1]) and parallel to the x-axis, Example 7: The line 3x + 2y = 24 meets y-axis at A and x-axis at B. The symbol for âparallel toâ is â// â. For example, the equations of two â¦ .I want gf girl who want bf can join in this Google meet âmkr-rphe-qosâ. When $$x^{3}$$ + p$$x^{2}$$ - 7x - 15 by x-3, the remainder is -27. Parallel lines have similarity to railroad tracks. This is what Iâm talking about.. Let the equations of the lines be ax+by+c 1 =0 and ax+by+c 2 =0. When the distance between a pair of lines is the same throughout, it can be called parallel lines. Comparing (1) with general equation of line Ax + By + C = 0, we get, The distance of the given point from given line is d = |Ax1 + By1 + C|/√A2+B2 = 26/7.2 = 3.571. Then PQRS must be a. (-3) * ................................................. the sum of two no is 85 if the no are in ratio 8:9 find the noâ. So, the required ratio is (9/√20) / (6/√5) = 3/4. Considering the following equations of 2 parallel lines, we can calculate the distance between those lines using the, A straight line through the origin O meets the parallel lines 4x + 2y = 9 and 2x + y + 6 = 0 at points P and Q respectively. give the inequality and thats all. Therefore, distance between tâ¦ The hypotenuse of a right triangle is not always the shortest distance between the two points that define it. Because parallel lines in a Euclidean plane are equidistant there is a unique distance between the two parallel lines. Thinking more intuitively, this has to be true since if the lines were getting farther apart from each other, then on the opposite side of the lines would be getting closer (and â¦ Ex 10.3, 6 Find the distance between parallel lines 15x + 8y â 34 = 0 and 15x + 8y + 31 = 0 We know that , distance between two parallel lines Ax + By + C1 = 0 & Ax + By + C2 = 0 is d = |ð¶_1â ð¶_2 |/â(ð´^2 + ðµ^2 ) Equation of first line is 15x + 8y â 34 = 0 Above equation is of the The red line is parallel to the blue line in each of these examples: Example 1. Distance Between Parallel Lines. The equation of line through which equation of the distance formula is written as, Considering the following equations of 2 parallel lines, we can calculate the distance between those lines using the distance formula, Using above 2 equations we can conclude that. Consider two parallel lines and .Pick some point on .Now pick a point to vary along .Say is a point on such that is perpendicular to both lines. Example 8: The diagonals of a parallelogram PQRSare along the lines x + 3y = 4 and 6x − 2y = 7. The two parameters are: Log ratio of distances between two circles whose shifting centers are foci of Apollonius circles, whereas the common orthogonal trajectories to either of the set are circles that includes a second parameter Angle between rays inside a circle segment. It is denoted by â||â. The slopes of two parallel lines are equal. Distance between two lines is equal to the length of the perpendicular from point A to line (2). Say the perpendicular distance between the two lines is , and the distance varies since our point B varies, call this distance . Parallel lines do not result in the formation of any angles. After obtaining the above values, substitute them in the slope-intercept equation to find y. Example 2. The distance of the origin from 4x+2y-9 = 0 is |-9|/√42+22 = 9/√20. \ units.Δ=21∣∣∣∣∣∣∣08−213120−1111∣∣∣∣∣∣∣=91 sq. Pre-University Math Help. Since the slope of the two lines are equivalent, we know that the lines are parallel. Parallel sides, lines, line segments, and rays are two lines that are always the same distance apart and never meet. Example 3: Estimate the distance between the two parallel lines y=2x+7 and y=2x+5. Lines are parallel if they are always the same distance apart (called "equidistant"), and will never meet. The distance between two Parallel Lines always Ask for details ; Follow Report by Rusheenddra 23.06.2019 Log in to add a comment Parallel lines are two lines that are always the same distance apart and never touch. In this article you will learn parallel lines definition , how to find distance between them and solved examples. 4x + 6y + 7 = 0. The required distance d will be PA â PB. The vertices of â³ABC are A(2,â1), B(0, 4), and C(â3, 5). Distance between Two Skew Lines: The distance is equal to the length of the â¦ Given the equations of two non-vertical, non-horizontal parallel lines, = + = +, the distance between the two lines can be found by locating two points (one on each line) that lie on a common perpendicular to the parallel lines and calculating the distance between â¦ We have step-by-step solutions for your textbooks written by Bartleby experts! Parallel Lines. Hi, The first equation is 3x + 4y - 3 = 0 Or a = 3, b = 4 and c1 = -3. The length of the common perpendiculars at different points on these parallel lines is same. Distance between Two Parallel Lines: The distance is the perpendicular distance from any point on one line to the other line. Hence lines x + 3y = 4 and 6x − 2y = 7 are perpendicular to each other. To find distance between two parallel lines find the equation for a line that is perpendicular to both lines and find the points of intersection of that line with the parallel lines. From the above equations of parallel lines, we have. Shortest Distance Between Parallel LinesWatch more videos at https://www.tutorialspoint.com/videotutorials/index.htmLecture By: Er. The equation of the line passing through (1,1) and making an angle of 135∘ is, Co-ordinates of any point on this line are. This video shows the proof of distance between two parallel lines. The perpendicular bisector of AB meets the line through (0,−1) parallel to the x-axis at C. The area of the triangle ABC is, The coordinates of A and B are (0, 12) and (8, 0) respectively. The length of the perpendicular from point A to the line (ii) is of the same length as the distance between two lines. Distance between 2 parallel lines is the perpendicular distance from any point to one of the lines. Forums. The distance from a line to a point not on the line is the length of the segment perpendicular to the line from the point. The distance between two parallel lines with the same slope ( i.e. Required distance is =|d1 − d2| = ∣7/5 − (−5/5)∣  = 12/5. s. Write an inequality to represent the number of passengers, p, that can get on the bus at each of the stops. The distance between the points â¦ The slope of line x + y = 1 is -1. We know that slopes of two parallel lines are equal. Find the distance between the lines  4x +3y+6= 0 and  4x+3y-3= 0. Let P(x 1, y 1) be any point. Itâs quite straightforward â the distance between two parallel lines is the difference between the distances of the lines from a point. Example 4: Calculate the distance between the parallel lines 3x+4y+7=0 and 3x+4y−5=0 from origin. Distance formula d = |a1x1+b1y1+c1| / √a12+b12 , where d is the distance between two parallel lines. The main criteria for any two lines to be parallel are that they have to be drawn on the same plane. Here, lines p and q are skew lines as they both lie in two different planes. Parallel lines are the lines which never meet each other. The distance is then found by taking the dot product of the two vectors since the dot product gives you the length of the projection of the vector V on the direction N. Form a vector that is perpendicular to the lines and lies in the same plane, Then form a vector from a point on one of the lines to a point on the other â¦ |(–m)(–c1/m) + (–c2)|/√(1 + m2) or d = |c1–c2|/√(1+m2). It should be pretty simple to see why intuitively. One of the important elements in three-dimensional geometry is a straight line. Find the distance that separates these linesâ¦ To calculate the distance between the two parallel lines having y = mx + c1. definition of parallel lines for class 9th, Parallel lines:If two lines do not meet at a point if extended to both directions, such lines are called parallel lines. In the case of intersecting lines, the distance between them is zero, whereas in the case of two parallel lines, the distance is the perpendicular distance from any point on one line to the other line. Parallel lines are those lines which never meet each other. And we get point A = (-c2/ m , 0) So through the above explanation of point A, we can further classify equation (4) to find the distance between the two parallel lines, to get. PS=2−12−132=−29PS=\frac{2-1}{2-\frac{13}{2}}=-\frac{2}{9}\\PS=2−2132−1=−92. Let be a vector between points on the two lines. Let PQ and RS be the parallel lines, with equations y = mx + b1 y = mx + b2 The distance between these two lines is the distance between the two intersection points of these lines with the perpendicular line.Let that distance be d. The method for calculating the distance between two parallel lines is as follows: The two parallel lines can be taken in the form. Distance between two parallel lines formula y=2x+7 and y=2x+5 is 2. Answer: Lines that never intersect are Parallel lines are those that never intersect each other. T. thamathkid1729. The value of p is : A bus can hold a maximum of 68 passengers. When the distance between a pair of lines is the same throughout, it can be called parallel lines. At the Tropic of Cancer and Tropic of Capricorn (23.5 degrees north and south), the distance is 68.94 miles (110.948 kilometers). Example 19 Find the distance between the parallel lines 3x â 4y + 7 = 0 and 3x â 4y + 5 = 0 We know that , distance between two parallel lines Ax + By + C1 = 0 & Ax + By + C2 = 0 is d = |ð¶_1 â ð¶_2 |/â(ð´^2 + ðµ^2 ) Distance between the parallel lines 3x â 4y + 7 = The equation of the line passing through (1, 1) and parallel to PS is, S = midpoint of QR=(6+72, −1+32)=(132, 1)QR=\left( \frac{6+7}{2},\,\frac{-1+3}{2} \right)=\left( \frac{13}{2},\,1 \right)\\QR=(26+7,2−1+3)=(213,1) So, the distance between two parallel lines is given by. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line: First, we need to take one of the line and convert it to â¦ This means â¦ This is for Grade 11 (NCERT) Coordinate Geometry. Find the distance between the following two parallel lines. Therefore the parallelogram is a rhombus. Therefore, they are separated by a constant distance. This equal length is called the distance between two parallel lines. Textbook solution for Geometry, Student Edition 1st Edition McGraw-Hill Chapter 4.1 Problem 69SPR. It is denoted by “||”. This implies that two parallel lines are always a constant distance apart from each other, which is another important characteristic of parallel lines. Also the distance of origin from 2x+ y + 6 = 0 is  |6|/√(22 +12 ) = 6/√5. 4x + 6y = -5. When the distance between a pair of lines is always the same, then we call such lines as parallel lines. 4x + 6y + 5 = 0. It makes an angle of 135∘ with the x-axis. 5) Skew lines : The two or more non-coplanar lines which do not intersect each others are called skew lines. Solution : Write the equations of the parallel line in general form. Here, the equations of parallel lines are y = 2x + 7 and y = 2x + 5. x / a = sinh Using the above formula, the distance = |(-3 - â¦ Lines PQ and RS are parallel lines. â¦. This meets (i) at C, Therefore the coordinates of C are (−132,−1)\left( -\frac{13}{2},-1 \right)(−213,−1). Just remember: Always the same distance apart and never touching. m1=m2) is given by |c1-c2|. How to Find the Distance Between Two Parallel Lines, , where d is the distance between two parallel lines. Writing the given equation in the normal form, we get, Example 5: Let PS be the median of the triangle with vertices P(2, 2), Q(6, − 1)P(2,\ 2),\ Q(6,\ -\ 1)P(2, 2), Q(6, − 1) and R(7, 3)R(7,\ 3)R(7, 3). Distance between parallel lines The distance between two parallel lines is the distance between one â¦ Therefore, two parallel lines can be taken in the form y = mx + c1â¦ (1) and y = mx + c2â¦ (2) Line (1) will intersect x-axis at the point A (âc1/m, 0) as shown in figure. The same number of passengers get on the bus at each of the next 6 stop A similar geometric approach was used by [Teller, 2000], but he used a cross product which restricts his method to 3D space whereaâ¦ units. Hence the area of the triangle ABC is Δ=12∣0121801−132−11 ∣=91 sq. This article explains the method of finding the distance between two parallel lines with appropriate examples. For a given line, only one line passing through a point not on that line will be parallel to it, like this: Slopes are same m1 = m2 = 2 and c1 = 7 ,c2 = 5. The main criteria for any two lines to be parallel are that they have to be drawn on the same plane. Mar 14, 2011 #1 The lines 3 x + 4y = 12 and 3x + 4y = 72 are parallel. Then, the point O divides the segment PQ in the ratio, The distance of the origin from 4x+2y-9 = 0 is |-9|/√4, Also the distance of origin from 2x+ y + 6 = 0 is  |6|/√(2, CBSE Previous Year Question Papers Class 10, CBSE Previous Year Question Papers Class 12, NCERT Solutions Class 11 Business Studies, NCERT Solutions Class 12 Business Studies, NCERT Solutions Class 12 Accountancy Part 1, NCERT Solutions Class 12 Accountancy Part 2, NCERT Solutions For Class 6 Social Science, NCERT Solutions for Class 7 Social Science, NCERT Solutions for Class 8 Social Science, NCERT Solutions For Class 9 Social Science, NCERT Solutions For Class 9 Maths Chapter 1, NCERT Solutions For Class 9 Maths Chapter 2, NCERT Solutions For Class 9 Maths Chapter 3, NCERT Solutions For Class 9 Maths Chapter 4, NCERT Solutions For Class 9 Maths Chapter 5, NCERT Solutions For Class 9 Maths Chapter 6, NCERT Solutions For Class 9 Maths Chapter 7, NCERT Solutions For Class 9 Maths Chapter 8, NCERT Solutions For Class 9 Maths Chapter 9, NCERT Solutions For Class 9 Maths Chapter 10, NCERT Solutions For Class 9 Maths Chapter 11, NCERT Solutions For Class 9 Maths Chapter 12, NCERT Solutions For Class 9 Maths Chapter 13, NCERT Solutions For Class 9 Maths Chapter 14, NCERT Solutions For Class 9 Maths Chapter 15, NCERT Solutions for Class 9 Science Chapter 1, NCERT Solutions for Class 9 Science Chapter 2, NCERT Solutions for Class 9 Science Chapter 3, NCERT Solutions for Class 9 Science Chapter 4, NCERT Solutions for Class 9 Science Chapter 5, NCERT Solutions for Class 9 Science Chapter 6, NCERT Solutions for Class 9 Science Chapter 7, NCERT Solutions for Class 9 Science Chapter 8, NCERT Solutions for Class 9 Science Chapter 9, NCERT Solutions for Class 9 Science Chapter 10, NCERT Solutions for Class 9 Science Chapter 12, NCERT Solutions for Class 9 Science Chapter 11, NCERT Solutions for Class 9 Science Chapter 13, NCERT Solutions for Class 9 Science Chapter 14, NCERT Solutions for Class 9 Science Chapter 15, NCERT Solutions for Class 10 Social Science, NCERT Solutions for Class 10 Maths Chapter 1, NCERT Solutions for Class 10 Maths Chapter 2, NCERT Solutions for Class 10 Maths Chapter 3, NCERT Solutions for Class 10 Maths Chapter 4, NCERT Solutions for Class 10 Maths Chapter 5, NCERT Solutions for Class 10 Maths Chapter 6, NCERT Solutions for Class 10 Maths Chapter 7, NCERT Solutions for Class 10 Maths Chapter 8, NCERT Solutions for Class 10 Maths Chapter 9, NCERT Solutions for Class 10 Maths Chapter 10, NCERT Solutions for Class 10 Maths Chapter 11, NCERT Solutions for Class 10 Maths Chapter 12, NCERT Solutions for Class 10 Maths Chapter 13, NCERT Solutions for Class 10 Maths Chapter 14, NCERT Solutions for Class 10 Maths Chapter 15, NCERT Solutions for Class 10 Science Chapter 1, NCERT Solutions for Class 10 Science Chapter 2, NCERT Solutions for Class 10 Science Chapter 3, NCERT Solutions for Class 10 Science Chapter 4, NCERT Solutions for Class 10 Science Chapter 5, NCERT Solutions for Class 10 Science Chapter 6, NCERT Solutions for Class 10 Science Chapter 7, NCERT Solutions for Class 10 Science Chapter 8, NCERT Solutions for Class 10 Science Chapter 9, NCERT Solutions for Class 10 Science Chapter 10, NCERT Solutions for Class 10 Science Chapter 11, NCERT Solutions for Class 10 Science Chapter 12, NCERT Solutions for Class 10 Science Chapter 13, NCERT Solutions for Class 10 Science Chapter 14, NCERT Solutions for Class 10 Science Chapter 15, NCERT Solutions for Class 10 Science Chapter 16, JEE Main Chapter Wise Questions And Solutions, Distance d between two parallel lines y = mx + c. Ensure whether the equations of the given parallel lines are in slope-intercept form (y=mx+c). Pre-Calculus. Example 2: What is the distance of the lines 2x − 3y = 4 from the point (1, 1) measured parallel to the line x+y=1? Consider two lines L1: and L2: . The equation of the perpendicular bisector of AB is  y−6=23(x−4) or 2x−3y+10=0…..(i)y-6=\frac{2}{3}(x-4) \text \ or \ 2x-3y+10=0 …..(i)\\y−6=32(x−4) or 2x−3y+10=0…..(i). x1 and y1 are the two intersection points of the lines with the axis in a cartesian plane, while a. and b1 are the coefficients of variable x and y of the line. Example 5: Find the distance from the line 6x – 4y + 36 = 0 to point (0, 0). The line (i) will intersect the x-axis at point A (–c1/m, 0) as shown in the figure. Lines be ax+by+c 1 =0 and ax+by+c 2 =0 equivalent, we know that slopes of no... Parallel if they are always the same plane finally, put all the above values substitute. Values in the slope-intercept equation to find distance between the points â¦ since the of... D will be PA â PB perpendicular to each other can get on bus! As parallel lines that slopes of two parallel lines definition, how to find the distance between the lines... In each of these examples: example 1 points lying on the lines from a point examples: example:... Lines formula y=2x+7 and y=2x+5 the number of passengers, p, that can get on the slope... ( 1+m2 ) 4x +3y+6= 0 and 4x+3y-3= 0: a bus can hold a of! Is 6x + 8y -1 =0, dividing throughout by 2 ; 3x + 4y - ( 1/2 ) 6/√5... And 3x + 4y = 72 are parallel p is: a bus can hold a maximum of passengers... Passengers get on the bus at each of these examples: example 1 find. Degree of latitude is approximately 69 miles ( 110.567 kilometers ) the distance between the of! Lines are parallel if they are separated by a constant distance passing through ( 0, )! Separated by a constant distance, 1 ) and parallel to the.... 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Is always the same distance apart and never touching will learn parallel lines definition, how to find y p... Gf girl who want bf can join in this Google meet âmkr-rphe-qosâ................................................. the sum of parallel! = 9/√20 of 68 passengers bf can join in this article you will parallel! *................................................. the sum of two no is 85 if the no are in ratio 8:9 find the between. Between the two lines L1: and L2: will never meet each.... 1 the lines 4x +3y+6= 0 and 4x+3y-3= 0 = 1 is -1 6/√5... Pqrsare along the lines which never meet each other s. Write an inequality to the... Perpendicular from point a to line ( 2 ) distance is =|d1 d2|! − d2| = ∣7/5 − ( −5/5 ) ∣ = 12/5 to each other, substitute them the! Called the distance from the above values in the form diagonals of a PQRSare... Plane is the distance between two parallel lines between parallel lines -3 â¦... 3, b = 6, c 1 = 5 and c =...

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text:
# Is there a unified formula for the lengths of the median, angle bisector, and altitude of a triangle?

The (squares of the) lengths of the median, angle bisector, and altitude of an $$a$$-$$b$$-$$c$$ triangle with base $$a$$ are given by the following:

\begin{align} d_{\text{med}}^2 &= \frac14\left( - a^2 + 2 b^2 + 2 c^2 \right) \tag{1}\\[4pt] d_{\text{bis}}^2 &= \frac{b c}{(b+c)^2}(a+b+c)(-a+b+c) \tag{2}\\[4pt] d_{\text{alt}}^2 &= \frac{1}{4a^2}(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{3}\\[4pt] \end{align}

(These can be derived using, for instance, Stewart's formula.)

I'm wondering if there's a unified formula for $$(1)$$, $$(2)$$, $$(3)$$, with the instances distinguished by some auxiliary parameter. In other words,

• Is there a function $$f(a,b,c,n)$$ such that $$f(a,b,c,n_\text{med})=d_\text{med}\qquad f(a,b,c,n_\text{bis})=d_\text{bis}\qquad f(a,b,c,n_\text{alt})=d_\text{alt}$$ for some $$n_\text{med}$$, $$n_\text{bis}$$, $$n_\text{alt}$$ that are independent of $$a$$, $$b$$, $$c$$?

• And, is it possible that these particular values of $$n$$ can be taken as (consecutive) integers?

• And, further, might other special cevians be associated with other specific (possibly integer) values of $$n$$?

• How are you associating $n=0, 1, 2$ to the altitude, angle bisector, and median? And what are the "interesting results" associated with $n=-\infty, -1, \infty$, and what justifies the association of those value of $n$ to those results? (Hmmm ... After an edit, I read that the associations with $n=-1,0,1,2$ aren't necessarily those you give.)... In any case, as you note, the formula for the length of a cevian is already known; it either is or isn't an analogue of the "generalized mean" in the way you want. You seem to be asking us to read your mind about how this is supposed to work. – Blue Jun 3 at 3:33
• BTW: Instead of posting a drafty question and then making a zillion edits, you should use the sandbox in Meta to get your question to a reasonably-stable form. Rampant "live" editing makes it impossible to know when a question is ready for proper consideration. – Blue Jun 3 at 3:38
• The description is quite confusing in this regard. Here's my understanding of what you're asking ... "Is there a function $f(a,b,c,n)$ that, for a specific $n=n_{\text{alt}}$, yields the length of the altitude of the $a$-$b$-$c$ triangle (with base $a$)? and likewise, for $n=n_{\text{bis}}$ and $n=n_{\text{med}}$, yields the lengths of the angle bisector and median? And is it possible that these particular values of $n$ can be taken as (consecutive) integers? And, further, might other special cevians be associated with other specific (possibly integer) values of $n$?" – Blue Jun 3 at 3:57
• I suggest editing. While our comment-discussion may help clarify your intent, people shouldn't have to consult the comments (and very often don't) to understand a question. ... Speaking of which: I'm getting the "avoid extended discussions" warning. I will likely not be responding further. Good luck! – Blue Jun 3 at 4:21
• I went to fix some formatting in the question, but couldn't seem to stop myself from significantly revising the question. :) Since you used my commented suggestion for your own revision, I didn't think you'd mind. In any case ... Please double-check that I've properly captured your intention. – Blue Jun 3 at 11:36


text:
# Proving induction from basic recursion lemma

Induction Principle: Let $A$ be a set such that $0 \in A$ and $n \in A \implies n + 1 \in A$. Then for all $n \in \mathbb{N}$, $n \in A$.

Basic Recursion Lemma: For all sets $X, W$ and given functions $g : X \to W$, $h : W \times \mathbb{N} \times X \to W$, there exists a unique function $f : \mathbb{N} \times X \to W$ such that $$f(0, x) = g(x)$$ $$f(n + 1, x) = h(f(n, x), n, x).$$

My book tells me that these two statements are equivalent. I know how to prove the basic recursion lemma from the induction principles, but I cannot figure out how to do the reverse. Everything I have tried ends up requiring me to use induction. Can anyone point me in the right direction?

Thanks.

-

Assume $0\in A$ and $n\in A\Rightarrow n+1\in A$.

Let $X=\{\bullet\}$, $W=A$. Then we can define $g\colon X\to W$ by $g(\bullet)=0$ because $0\in A$. And we can define $h\colon W\times \mathbb N\times X\to W$ by $$h(x,y,z)=\begin{cases}y+1&\text{if }y+1\in A\\x&\text{otherwise}\end{cases}$$ Then there exists a unique function $f\colon \mathbb N\times X\to W$ such that $$f(0,x)=g(x)$$ $$f(n+1,x)=h(f(n,x),n,x)$$ Note that for any element $a\in A$, the function $f_a\colon \mathbb N\times X\to W$, $f_a(n,x)=\begin{cases}n&\text{if }n\in A\\a&\text{otherwise}\end{cases}$ obeys this recursion:

• $f_a(0,\bullet)=0=g(\bullet)$ because $0\in A$.
• If $n+1\in A$ then $f_a(n+1,\bullet)=n+1=h(f_a(n,\bullet),n,\bullet)$
• If $n+1\notin A$ then also $n\notin A$, hence $f(n+1,\bullet)=a =h(a,n,\bullet)=h(f_a(n,\bullet),n,\bullet)$

By uniqueness, any two such functions are equal. As $0\in A$ immediately and $1=0+1\in A$, we have $f_0=f_1$. Because $0\ne 1$, this is only possible if the "otherwise" in the definition of $f_a$ never occurs, that is: for all $n\in \mathbb N$, we also have $n\in A$.

-
I see how $f_a$ satisfies $f(0, x) = g(x)$. But if we have $n + 1 \not \in A$, then $n \not \in A$. Therefore $f_a(n, x) = a$ and $f_a(n + 1, x) = a$. So $f_a(n + 1, x) \neq f_a(n, x) + 1 = h(f(n, x), n, x)$. Right? Is there another way to say $f_a$ obeys the recursion without assuming all of $n$ is already in $A$? –  Danikar Feb 28 '13 at 23:40
@Danikar Sorry, it was necessary to adjust $h$ a bit. I hope now you see how it works. –  Hagen von Eitzen Feb 28 '13 at 23:48
thanks, I got it. –  Danikar Feb 28 '13 at 23:57

Okay, I fixed my answer. It's similar to Hagen's now, but I think it's a bit simpler.

We may assume without loss of generality that $A$ is an initial segment of $\mathbb{N}$.

Let $h:A \cup \{\infty_0,\infty_1\} \to A \cup \{\infty_0,\infty_1\}$ by given by $h(n) = n+1$ for $n \in A$ and $h(\infty_i) = \infty_i$ for $i<2$.

For $i<2$, define the function $f_i: \mathbb{N} \to A \cup \{\infty_0,\infty_1\}$ by $$f_i(n) = \begin{cases} n & \text{if n \in A}\\ \infty_i & \text{otherwise} \end{cases}$$ These functions both satisfy $f_i(0) = 0$ and $f_i(n+1) = h(f_i(n))$ so by the uniqueness assumption for recursive definitions, they are the same (we do not need the existence assumption.) Assuming $\infty_0 \ne \infty_1$ this means that the second case in the definition cannot occur, so $A$ is all of $\mathbb{N}$.

-
If I could show that $f(n) = n$, I would be done. But don't I need induction to show this? $f(n) - n \neq 0 \implies f(n - 1) - (n - 1) \neq 0$ is the only kind of thing I could come up with. –  Danikar Feb 28 '13 at 23:29
@Danikar Good point. I was thinking that $f = \text{id}$ followed from uniqueness, but one would really need uniqueness of $f \restriction n$ for all $n$. –  Trevor Wilson Feb 28 '13 at 23:34


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Solved

# Powershell - "open with" equal

Posted on 2010-08-28
Medium Priority
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I have several XML files that are encrypted and require opening with a specific application.  I would like to use powershell to:

1. open the file in the specific application
2. save the file as txt or xml
3. parse the file using XML controls.

I will address step 3 later.  1 and 2 is where I could use some help. Is there a switch or bar I could add to a get cmdlet to specify an application to access a file with?

The application I need to use is not an installed app.  It is a single exe file.
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Question by:kabaam
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Accepted Solution

soostibi earned 2000 total points
ID: 33550664
Let's assume, that your specific application is notepad.exe. You can open a file from PowerShell:
$sh = new-object -com wscript.shell$sh.Run("notepad c:\ee\10.txt")

(it might not work for you app, but you have to test it. You could also open it by:
but we have to further control the app, see below.)

To save it, it's a bit harder to do. But it might work like this with the help of the wscript.shell com object and sending keys to the app:

$sh.AppActivate("notepad")$sh.SendKeys("%(F)A")                   # send Alt+F and A to notepad to open the Send As dialog
start-sleep 1                              # give time for the dialog to open
$sh.SendKeys("abcdefghijk.txt") # send a filename to Save As dialog start-sleep 1 # wait till the characters are entered$sh.SendKeys("~")                        # send an Enter
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LVL 11

Author Comment

ID: 33550839
wow, that is insanely closer to what I need than I thought I would get.  One additional request if you can?

in the save as dialog box I would like to change the default file type to txt.

the above works great...thanks
0

LVL 11

Author Comment

ID: 33550872
Okay, I am almost there.  I added these two lines and it almost works but 1 flaw.
the Alt-f is applied to the window with focus.  Once the app opens it loses focus to the script.

$sh.AppActivate($applicationName)
$sh.SendKeys("%(F)A") # send Alt+F and A to notepad to open the Send As dialog start-sleep 1 # give time for the dialog to open$sh.SendKeys("$newFile") # send a filename to Save As dialog$sh.SendKeys("{tab}")
$sh.SendKeys("t") start-sleep 1 # wait till the characters are entered$sh.SendKeys("~")                        # send an Enter

thanks a bunch
0

LVL 11

Author Comment

ID: 33550936
ahh.. figured it out.
First up was the fact that my app doesn't like the use of atl-F shortcut.  So what I did was :
send Alt
send down
send down
send down
send down
send enter

this still did not allow focus on the correct app and I found that was due to the application was not fully awake yet so I put in a two second sleep.

Thank you for the code and especially for adding the comments so it was easy for me to figure out and to experiment on.

$sh = new-object -com wscript.shell$sh.Run("$applicationName$filename")                           #open application with specific file
start-sleep 1                                                   #give time for application to open
$sh.AppActivate($strApplicationName);$sh.SendKeys("%") #sends an ALT key press$sh.AppActivate($strApplicationName);$sh.SendKeys("{DOWN}")     #sends a Down arrow key press
$sh.AppActivate($strApplicationName);$sh.SendKeys("{DOWN}") #sends a Down arrow key press$sh.AppActivate($strApplicationName);$sh.SendKeys("{DOWN}")     #sends a Down arrow key press
$sh.AppActivate($strApplicationName);$sh.SendKeys("{DOWN}") #sends a Down arrow key press$sh.AppActivate($strApplicationName);$sh.SendKeys("~")          #sends an Enter key press
start-sleep 1                                                   #give time for the dialog to open
$sh.AppActivate($strApplicationName);$sh.SendKeys("$newFile")   #send a filename to Save As dialog
$sh.AppActivate($strApplicationName);$sh.SendKeys("{tab}") #sends a Tab key press$sh.AppActivate($strApplicationName);$sh.SendKeys("t")          #sends an "T" key press
start-sleep 1                                                   #wait till the characters are entered
\$sh.SendKeys("~")                                               #send an Enter

0

LVL 11

Author Comment

ID: 33550938
for those that found this question on a search check out this link for many more sendkeys options
http://technet.microsoft.com/en-us/library/ff731008.aspx
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# Calculating the congruence between languagesClosed, ResolvedPublicActions

Assigned To
 Livnetata
Authored By
 Livnetata Dec 28 2014, 2:54 PM2014-12-28 14:54:43 (UTC+0)
Referenced Files
None
Subscribers

# Description

Calculating how many linked articles exist in both languages (for example Paris exists in both HE and EN but Elephant only exists in EN).
It should only include pages on the main namespaces not including categories.

Output:

• A matrix of (287X287 representing the different wikipedias) that counts the number of pages that exist for both languages. This matrix is symmetric.
• A matrix that presents the percentages of the overlapped pages compared to a specific language (that is: $\frac{EN\cap HE}{EN}$). This matrix is not symmetric.
NOTE: Right now the tables are not planned to be updated in real time. Maybe in a later task.


text:
# Plugin PivotTable

Introduced in Tiki 16.2

Use this wiki plugin to create dashboards with summaries of data in Tiki objects through the unified search index to produce a pivot table report of your choice. Initially this plugin works with tracker data, but other Tiki objects can be connected later with this plugin. Results for the variables of interest (tracker fields, as well as creation_date, modification_date and tracker_status of the items) are aggregated by criteria selected by the user.

It produces the JavaScript Pivot Table (aka Pivot Grid, Pivot Chart, Cross-Tab) implementation from Nicolas Kruchten with drag'n'drop (see the list of changes in each version).

## Parameters

Introduced in Tiki 16.1. Required parameters are in bold.
Preferences required: wikiplugin_pivottable
Parameters Accepted Values Description Default Since
(body of plugin) Leave one space in the box below to allow easier editing of current values with the plugin popup helper later on
data text
separator: :
For example 'tracker:1' or 'activitystream' 0
chartTitle text Override title when using Chart renderers. 16.3
menuLimit digits Pivottable menuLimit option override - number of entries to consider the menu list too big when filtering on a particular column or row. 16.2
inclusions text Filter values for fields in rows or columns. Contains JSON encoded object of arrays of strings.
xAxisLabel text Override label of horizontal axis when using Chart renderers. 16.3
yAxisLabel text Override label of vertical axis when using Chart renderers. 16.3
height word Height of pivot table. Units: px 400px
width word Width of pivot table. Units: % or px. 100%
aggregateDetails text
separator: :
When enabled, clicking a table cell will popup all items that were aggregated into that cell. Specify the name of the field or fields to use to display the details separated by colon. Enabled by default. To disable, set contents to an empty string. 16.2
aggregatorName Count| Count Unique Values| List Unique Values| Sum| Integer Sum| Average| Minimum| Maximum| Sum over Sum| 80% Upper Bound| 80% Lower Bound| Sum as Fraction of Total| Sum as Fraction of Rows| Sum as Fraction of Columns| Count as Fraction of Total| Count as Fraction of Rows| Count as Fraction of Columns Function to apply on the numeric values from the variables selected. Count
chartHoverBar y
n
Display the Chart hover bar or not. y 16.3
heatmapColors text
separator: :
17
highlightGroupColors text
separator: :
18.1
highlightGroup (blank)
y
n
Highlight items' values belonging to one of my groups in Charts. n 16.3
highlightMine (blank)
y
n
Highlight owned items' values in Charts. n 16.3
overridePermissions (blank)
y
n
Return all tracker items ignoring permissions to view the corresponding items. n 18.1
rendererName Table| Table Barchart| Heatmap| Row Heatmap| Col Heatmap| Line Chart| Bar Chart| Stacked Bar Chart| Area Chart| Scatter Chart| Treemap Display format of data Table
translate (blank)
n
y
Use translated data values for calculations and display. Default value: No n 18.3
vals text
separator: :
Variable with numeric values or tracker field permNames, on which the formula from the aggregator is applied. It can be left empty if aggregator is related to Counts. Use permanentNames in case of tracker fields, separated by : in case of multiple fields function.
heatmapDomain text
separator: :
17
cols text
separator: :
Which field or fields to use as table columns. Leaving blank will use the first available field. Use permanentNames in case of tracker fields. Separated by colon (:) if more than one.
rows text
separator: :
Which field or fields to use as table rows. Leaving blank will remove grouping by table rows. Use permanentNames in case of tracker fields. Separated by colon (:) if more than one.

Notes on aggregateDetails:

• The aggregateDetails accepts multiple field names or permNames separated by colon.
• The aggregateDetails parameter is also enabled by default and can be disabled setting aggregateDetails to an empty string.
• Each item has the associated object_link available by default and clickable in the popup where the aggregateDetails field data is shown.
• It will work with other unified search index content entries (not only tracker items) but might be slow for large result sets.
• It is only activated if aggregateDetails is not disabled. Therefore, there is a workaround to disable this feature for large sets of data (e.g. containing several or hundreds of thousands of items).

## Basic Usage

Basic usage requires just to provide the data source (e.g. a tracker with id 1: "tracker:1" since Tiki16, or activitystream also since Tiki19 ), and the rest will be taken as default values by the pivot table plugin, and you will be able to edit it through the PivotTable UI itself. That will allow you to display all field names of the tracker, and will let you drag and drop them in rows or columns of the pivot table editor.

That will cover most use cases. However, if your dataset is huge, or the tracker has many fields, and some of them carrying heavy data (long text fields, or big files/images attached to the tracker items in files tracker fields), you can use an advanced syntax to filter the number of items or reduce the amount of tracker fields exposed to the pivot table to work with, so that performance of the pivot table plugin is fast again. See below for "Advanced Usage"

### Example 1

After installing the Bug_Tracker_16 profile on a brand new Tiki 16, you will get a new tracker with id 1 to hold the data of the bug reports/issue tickets. When you add a few dozen items, you can use some syntax like the one indicated below to produce some demo pivot tables table with default values as a starting point, to let you start reviewing the data as wiki-wiki (quick) as possible.

This code:

{pivottable data="tracker:1"}

Would produce with the data from that profile (at the time of this writing):

Click to expand

Once saved, you can click on any cell of the pivottable report, and you will be shown a popup with the information tracker items that produced the count for that cel, with a link to view the full record of each of the tracker items.

Click to expand

From there we can edit the Pivottable again through the PivotTable UI itself, and modify the variables to be used as row or column data, or add new variables in columns, change the type of table or chart produced, etc.

A table can even consider more than one value in a single dimension. The following example therefore uses both Status and Priority on the horizontal axis (meaning a column can have subcolumns):

Click to expand

### Example 2

A default configuration for each parameter of the plugin can also be specified. For instance, the values considered in both dimensions can be specified, using the rows and cols parameters, as in the following example (which considers 2 values on the horizontal axis, as in the previous screenshot).

This code:

{PIVOTTABLE(data="tracker:1" width="100%" height="500px" rows="bug_tracker_severity" cols="bug_tracker_bug_status:bug_tracker_priority" rendererName="Heatmap" aggregatorName="Count as Fraction of Columns" vals="bug_tracker_priority")} {PIVOTTABLE}

Would produce with the data from that profile (at the time of this writing):

Click to expand

### Example 3

You can also make some charts:

• Line Chart
• Bar Chart
• Stacked Bar Chart
• Area Chart
• Scatter Chart

For instance...

This code:

{PIVOTTABLE(data="tracker:1"  width="400px" height="300px" rows="bug_tracker_severity" cols="bug_tracker_bug_status" rendererName="Stacked Bar Chart" aggregatorName="Count")}
{PIVOTTABLE}

Would produce:

Click to expand

### Example 4 (subtotals since Tiki 18)

Since Tiki18 new renderers were added to allow displaying subtotal sums for rows in the table, through the addition of subtotal.js to the plugin:

• Table With Subtotal
• Table With Subtotal Bar Chart
• Table With Subtotal Heatmap
• Table With Subtotal Row Heatmap
• Table With Subtotal Col Heatmap

Click to expand

If you click on the triangle at the left of each row name ("Severity" values, in this example), you will get the options of the next column ("Bug Status", in this example) contracted, hiding the different values of this other column, and showing only the subtotals for the field where you first clicked at (a "severity" value, or the whole column "Severity").

Click to expand

### Example 5 (activity stream since Tiki19)

Since Tiki19, you can display data from the Activity Stream into the Plugin PivotTable.

Minimum syntax to let the user choose options throught the PivotTable UI:

{pivottable data="activitystream"}

Example:

{pivottable data="activitystream" rows="object:type" cols="modification_date" width="100%" height="1000px" rendererName="Bar Chart" aggregatorName="Count" inclusions="{}" menuLimit="500" aggregateDetails="object_type"}

If your dataset is huge (many thousands), or the tracker has many fields (many hundreds), and some of them carrying heavy data (long text fields, or big files/images attached to the tracker items in files tracker fields), you can use an advanced syntax to filter the number of items or reduce the amount of tracker fields exposed to the pivot table to work with, so that the good performance of the pivot table plugin is preserved.

You can use the filter or display commands (both from PluginList ) to indicate which items (filter) or tracker fields (display) you want to use, respectively, in the pivot table plugin.

See:

### Add creation_date, modification_date and status

You can also indicate if you want the creation_date, modification_date and status if the tracker items to be displayed as optional variables to be used in the report.

{display name="creation_date" format="datetime"}
{display name="modification_date" format="datetime"}
{display name="tracker_status"}

### Customize aggregation date values

You can also customize the date format of these dates to show only the date, not the time, for instance, And also:

• only the day number of the month with %d , or
• only the day number of the year with %z , or
• only or the week number of the year with %W , or
• only the month number with %m , or
• only the year with %Y , etc

This way, you can get the aggregated values of interest for your report.

Example of code to display only the month number from these dates so that you can draw charts with counts per month, etc:

{display name="creation_date" format="date" dateFormat="%m"}
{display name="modification_date" format="date" dateFormat="%m"}

Full list of codes available here:
http://php.net/manual/en/function.date.php

This code:

{PIVOTTABLE(data="tracker:4" rows="bug_tracker_submitted_by:bug_tracker_severity:" cols="bug_tracker_bug_status:bug_tracker_priority:" rendererName="Heatmap" aggregatorName="Count as Fraction of Total")}
{display name="tracker_field_bug_tracker_submitted_by" default=""}
{display name="tracker_field_bug_tracker_severity" default=""}
{display name="tracker_field_bug_tracker_bug_status" default=""}
{display name="tracker_field_bug_tracker_priority" default=""}
{display name="tracker_field_bug_tracker_version" default=""}
{PIVOTTABLE}

Would produce with the data from that profile (at the time of this writing):

Click to expand

And once you click at the Edit Pivot Table button, you would see the controls to edit variable selection, but notice that you have less amount of variables to choose from than before; only the ones you have selected in the display commands of the plugin body above:

Click to expand

This code:

{PIVOTTABLE(data="tracker:4" rows="bug_tracker_submitted_by:bug_tracker_severity:" cols="bug_tracker_bug_status:bug_tracker_priority:" rendererName="Heatmap" aggregatorName="Count as Fraction of Total")}

{filter field="tracker_field_bug_tracker_bug_status" content="new"}

{display name="tracker_field_bug_tracker_submitted_by" default=""}
{display name="tracker_field_bug_tracker_severity" default=""}
{display name="tracker_field_bug_tracker_bug_status" default=""}
{display name="tracker_field_bug_tracker_priority" default=""}
{display name="tracker_field_bug_tracker_version" default=""}
{PIVOTTABLE}

Would produce the same as before, but restricting the data set to only those items tagged as new bugs (bug status is "new"):

Click to expand

Again, if you edit the pivot table, you will see that also have the restricted the number of fields, as well as the data points, that comply with your filtering criteria:

Click to expand

Since Tiki 16.2, any plugin using unified index search formatter and wikibuilder (aka filter, output, display, format, etc. wiki syntax, such as PluginPivottable ) now accepts {filter field=... editable=...} syntax to allow user enter a search value instead of hard-coding it. This means a trackerfilter-like functionality for unified index-based plugins.

You can see this feature in action if you apply profile Bug_Tracker_16

Therefore, this code:

{PIVOTTABLE(data="tracker:4" rows="bug_tracker_severity" cols="bug_tracker_bug_status" rendererName="Heatmap" aggregatorName="Count")}
{filter field="tracker_field_bug_tracker_priority" editable="content"}
{filter field="tracker_field_bug_tracker_assignee" editable="content"}
{filter field="tracker_field_bug_tracker_summary" editable="content"}
{PIVOTTABLE}

Would produce the expected pivottable report, with some fields on top to allow the user to filter results before re-drawing the table or chart:

Click to expand

## Aliases

### Keywords

Keywords serve as "hubs" for navigation within the Tiki documentation. They correspond to development keywords (bug reports and feature requests):

Accessibility (WAI and 508)
Accounting
Articles and Submissions
Banners
Batch
BigBlueButton audio/video/chat/screensharing
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Bookmark
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Category
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Watch
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XMLRPC